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Vol. XXXVI

No. 1

23 58

January 2018

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MATHEMATICS TODAY | JANUARY ‘18

This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging. *ALOK KUMAR, B.Tech, IIT Kanpur

ELLIPSE z

The Ellipse is a conic whose eccentricity is less than unity i.e., e < 1. Standard equation of an ellipse referred to its principal x2 y2 axes along the co-ordinate axes is 2 + 2 = 1 , a b where a > b 2 b Eccentricity : e = 1 − 2 ⇒ a2e2 = a2 – b2(0 < e < 1) a

Foci : S ≡ (ae, 0) and S′ ≡ (–ae, 0)

a a and x = − e e Vertices : A′ ≡ (–a, 0) and A ≡ (a, 0) The equation of directrices are x =

The line segment A′A of length 2a(a > b) is called the major axis of the ellipse. The y-axis intersects the ellipse at the points B′ ≡ (0, –b) and B ≡ (0, b). The line segment B′B of length 2b(b < a) is called the minor axis of the ellipse.

2b2 = 2a(1 − e 2 ) a The sum of the focal distances of any point on the ellipse is equal to the major axis. Hence, distance of focus from the extremity of a minor axis is equal to semi major axis. i.e., BS = CA. The point P(x 1 , y 1 ) lies outside, inside or on x2 y2 the ellipse according as, 1 + 1 − 1 >, < or = 0 a 2 b2 (respectively) A circle described on major axis as diameter is called the auxiliary circle. Let Q be a point on the auxiliary circle x2 + y2 = a2 such that QP produced is perpendicular to the x-axis then P and Q are called as the corresponding points on the ellipse and the auxiliary circle respectively ‘q’ is called the eccentric angle of the point P on the ellipse (0 ≤ q < 2p). =

The major and minor axis together are called principal axis of the ellipse. A chord which passes through focus is called a focal chord and which passes through the centre is called a diameter of the conic. A chord perpendicular to the major axis is called a double ordinate. The focal chord perpendicular to the major axis is called the latus rectum. Length of latus rectum (LL′)

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.

MATHEMATICS TODAY | JANUARY ‘18

circle. The equation to this locus is x2 + y2 = a2 + b2 i.e. a circle whose centre is the centre of the ellipse and whose radius is the length of the line joining the ends of the major and minor axis. HYPERBOLA z

The equations x = a cosq and y = b sinq together x2 y2 represent the ellipse + = 1, where q is a a 2 b2 parameter. Note : If P(q) ≡ (a cos q, b sin q) is on the ellipse then, Q(q) ≡ (a cos q, a sinq) is on the auxiliary circle. z

The line y = mx + c meets the ellipse

z z

−

= 1.

⇒ a2 e2 = a2 + b2 a2 Foci : S ≡ (ae , 0) and S′ ≡ (–ae , 0).

Eccentricity : e =

= 1 if c2 = a2m2 + b2. a 2 b2 The equation to the chord of the ellipse joining two points with eccentric angles a and b is given by x α +β y α +β α −β cos   + sin   = cos  .  a 2 b 2 2  xx1 yy1 + = 1 is tangent to the ellipse at (x1, y1). a2 b2 For general ellipse replace x2 by xx1, y2 by yy1, 2x by x + x1, 2y by y + y1, 2xy by xy1 + yx1. +

y = mx ± a2m2 + b2 is tangent to the ellipse for all values of m. x cos θ y sin θ + = 1 is tangent to the ellipse at the a b point (a cosq, b sinq). The eccentric angles of point of contact of two parallel tangents differ by an angle p. Equation of the normal at (x1, y1) is

a 2 x b2 y = a2 – b2 = a2 e2 . − y1 x1 Equation of the normal at the point (a cosq, b sinq) is ax ⋅ secq –by ⋅ cosecq = a2 – b2. Equation of a normal in terms of its slope ‘m’ is y y = mx −

Standard equation of the hyperbola is

= 1 in a 2 b2 two points real, coincident or imaginary according as c2 is <, = or > a2m2 + b2 . Hence, y = mx + c is tangent to the ellipse x2

The hyperbola is a conic whose eccentricity is greater than unity i.e., e > 1.

(a2 − b2 ) m

. a2 + b2m2 Locus of the point of intersection of the tangents which meet at right angles is called the director MATHEMATICS TODAY | JANUARY ‘18

a a and x = − e e Vertices : A ≡ (a, 0) and A′ ≡ (–a, 0). 2b2 Latus rectum : l = = 2a(e 2 − 1) . a The line segment A′A of length 2a in which the foci S′ and S both lie is called the transverse axis of the hyperbola. The line segment B′B between the two points B′ ≡ (0, – b) and B ≡ (0, b) is called as the conjugate axis of the hyperbola. The transverse axis and the conjugate axis of the hyperbola are together called the principal axes of the hyperbola. The difference of the focal distances of any point P on the hyperbola is constant and equal to transverse axis i.e. ||PS| – |PS′|| = 2a. The distance SS′ = focal length. Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & the transverse axes of the other are called conjugate hyperbolas of each other . Equation of directrices : x =

x2 y2 x2 i.e. 2 − 2 = 1 and − 2 + a b a hyperbola of each other.

= 1 are conjugate

MATHEMATICS TODAY | JANUARY â&#x20AC;&#x2DC;18

If e1 and e2 are the eccentricities of the hyperbola and its conjugate then e1–2+ e2–2 = 1. The particular kind of hyperbola in which the lengths of the transverse and conjugate axis are equal is called an equilateral hyperbola. Note : Eccentricity of the rectangular hyperbola is 2 and the length of its latus rectum is equal to its transverse or conjugate axis.

2 2 2 Also, y = m x ± a m − b can be taken as the

tangents to the hyperbola z

= 1. a 2 b2 E qu at i on of a c h ord j oi n i ng a an d b i s α −β y α +β α +β x cos − sin = cos . a 2 b 2 2 The equation of the normal to the hyperbola x2

−

= 1 at the point p(x1, y1) on it is

a 2 x b2 y + = a 2 + b2 = a 2e 2 . x1 y1 z

A circle drawn with centre C & transverse axis as a diameter is called the auxiliary circle of the hyperbola. Equation of the auxiliary circle is x2 + y2 = a2. Note from the figure that P and Q are called the ‘‘corresponding points” on the hyperbola and the auxiliary circle. ‘q’ is called the eccentric angle of the point ‘P ’ on the hyperbola (0 ≤ q < 2p). The point (x1, y1) will be outside, on or inside the x2 y2 hyperbola according as 1 − 1 − 1 >, = or < 0 a 2 b2 (respectively) The straight line y = mx + c is a secant, a tangent or x2 y2 passes outside the hyperbola − = 1 according 2 2 a b 2 2 2 2 as c >, =, < a m – b . Hence, y = mx + c is tangent to the hyperbola x2 y2 − = 1 if c2 = a2m2 – b2. a 2 b2 2 2 Equation of the tangent to the hyperbola x − y = 1 a 2 b2 xx1 yy 1 at the point (x1, y1) is − 2 = 1. a2 b In general two tangents can be drawn from an external point (x1, y1) to the hyperbola and they are y – y1 = m1 (x – x1) and y – y2 = m2 (x – x2), where m 1 and m 2 are roots of the equation (x12 – a2)m2 – 2x1y1m + y12 + b2 = 0. 2 2 Equation of the tangent to the hyperbola x − y = 1 a 2 b2 x sec θ y tan θ at the point (a secq, b tanq) is − = 1. a b MATHEMATICS TODAY | JANUARY ‘18

The equat ion of t he nor mal at t he p oint 2 2 P(a secq, b tanq) on the hyperbola x − y = 1 is a 2 b2 by ax + = a 2 + b2 = a 2 e 2 . sec θ tan θ

The locus of the intersection of tangents which are at right angles is known as the director circle of the hyperbola. The equation to the director circle is x2 + y2 = a2 – b2 . If b2 < a2 this circle is real; if b2 = a2 the radius of the circle is zero and it reduces to a point circle at the origin. In this case, the centre is the only point from which the tangents at right angles can be drawn to the curve. If b2 > a2 , the radius of the circle is imaginary, so that there is no such circle and so no tangents at right angle can be drawn to the curve. Let y = mx + c is the asymptote of the hyperbola x2

−

= 1. a 2 b2 Solving these two we get the quadratic as (b2 – a2m2)x2 – 2a2mcx – a2(b2 + c2) = 0 x– y a b=0 B (0, b)

(–a, 0) A

B (0, –b)

A (a, 0)

x+y a b =0

...(i)

MATHEMATICS TODAY | JANUARY â&#x20AC;&#x2DC;18

In order that y = mx + c be an asymptote , both roots of equation (i) must approach infinity, the conditions for which are : coeff. of x2 = 0 and coeff. of x = 0. b ⇒ b2 – a2m2 = 0 or m = ± a and a2mc = 0 ⇒ c = 0. \ Equations of asymptote are x y x y + = 0 and − = 0 a b a b On combining equation to the asymptotes, we get x2 y2 − =0 a 2 b2 z When b = a, then, the asymptotes of the rectangular hyperbola x2 – y2 = a2 or y = ± x which are at right angles. z If a hyperbola is equilateral then the conjugate hyperbola is also equilateral. z A hyperbola and its conjugate have the same asymptote. z The equation of the pair of asymptotes differ the hyperbola and the conjugate hyperbola by the same constant only. ∂f ∂f z Find and .Then the point of intersection of ∂x ∂y ∂f ∂f = 0 and = 0 gives the centre of the hyperbola. ∂x ∂y Important Points on Rectangular Hyperbola z Equation is xy = c2 with parametric representation x = ct, y = c/t, t ∈ R. z Equation of a chord joining the points P(t1) and Q(t2) is x + t1t2 y = c(t1 + t2). z Equation of the tangent at P(x1, y1) is xy1 + x1y = 2c2 x and at P(ct, c/t) is + ty = 2c. t z Chord with a given middle point as (h, k) is kx + hy = 2hk . z z

The pole of the straight line x + 4y = 4 with respect to ellipse x2 + 4y2 = 4 is (a) (1, 4) (b) (1, 1) (c) (4, 1) (d) (4, 4)

= 1, the equation of diameter a 2 b2 b conjugate to the diameter y = x , is a b a (a) y = − x (b) y = − x a b b (d) none of these (c) x = − y a Minimum area of the triangle by any tangent to the

In the ellipse

ellipse

a 2 + b2 2

= 1 with the coordinate axes is

(a + b)2 2 (a − b)2 (c) ab (d) 2 The eccentricity of the ellipse 25x2 + 16y2 – 150x – 175 = 0 is (a) 2/5 (b) 2/3 (c) 4/5 (d) 3/5 (a)

(b)

If 5x2 + ly2 = 20 represents a rectangular hyperbola, then l equals (a) 5 (b) 4 (c) –5 (d) none of these

The locus of the point of intersection of the lines bxt – ayt = ab and bx + ay = abt is (a) a parabola (b) an ellipse (c) a hyperbola (d) none of these

The eccentricity of the hyperbola can never be equal to 1 1 9 (a) (b) 2 (c) 3 (d) 2 9 8 5 The locus of the centre of a circle, which touches externally the given two circles, is (a) circle (b) parabola (c) hyperbola (d) ellipse

Equation of the normal at P(ct, c/t) is xt3 – yt = c(t4 –1). Vertex of this hyperbola is (c, c) and (–c, –c), focus is ( 2c , 2c) and (− 2c, − 2c) , the directrices are x + y = ± 2 c and latus rectum = 2 2c.

PROBLEMS Single Correct Answer Type The eccentricity of the hyperbola conjugate to x2 – 3y2 = 2x + 8 is 2 (b) (a) 3 3 (c) 2 (d) none of these

10. The foci of the hyperbola 9x2 – 16y2 = 144 are (a) (±4, 0) (b) (0, ±4) (c) (±5, 0) (d) (0, ±5)

MATHEMATICS TODAY | JANUARY ‘18

11. The equation x2 + 4xy + y2 + 2x + 4y + 2 = 0 represents (a) an ellipse (b) a pair of straight lines (c) a hyperbola (d) none of these

12. The auxiliary equation of circle of hyperbola x

a2 (a) (c)

−

= 1, is b2 x2 + y2 = a2 x2 + y2 = a2 + b2

to the hyperbola (b) x2 + y2 = b2 (d) x2 + y2 = a2 – b2

13. If the line y = 2x + l be a tangent to the hyperbola 36x2 – 25y2 = 3600, then l = (a) 16 (b) –16 (c) ±16 (d) none of these 14. The equation of the tangents to the conic 3x2 – y2 = 3 perpendicular to the line x + 3y = 2 is (a) y = 3x ± 6 (b) y = 6 x ± 3 (c) y = x ± 6

(d) y = 3x ± 6

15. The locus of the point of intersection of any two perpendicular tangents to the hyperbola is a circle which is called the director circle of the hyperbola, then the equation of this circle is (a) x2 + y2 = a2 + b2 (b) x2 + y2 = a2 – b2 2 2 (c) x + y = 2ab (d) none of these 16. The equation of the tangents to the hyperbola 3x2 – 4y2 = 12 which cuts equal intercepts from the axes, are (a) y + x = ±1 (b) y – x = ±1 (c) 3x + 4y = ±1 (d) 3x – 4y = ±1 17. If m1 and m2 are the slopes of the tangents to the

2 2 hyperbola x − y = 1 which pass through the 25 16 point (6, 2), then

24 2 (a) m1 + m2 = (b) m1m2 = 11 11 11 48 (d) m1m2 = (c) m1 + m2 = 20 11 18. The value of m for which y = mx + 6 is a tangent to 2 2 the hyperbola x − y = 1, is 100 49

20 17 (b) 17 20 3 20 (d) (c) 20 3 19. The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is (a) x + 2 = 0 (b) 2x + 1 = 0 (c) x – 2 = 0 (d) x + y + 1 = 0 (a)

20. If the straight line x cosa + y sina = p be a tangent

(a) (b) (c) (d)

= 1, then a 2 b2 a2 cos2a + b2 sin2a = p2 a2 cos2a – b2 sin2a = p2 a2 sin2a + b2 cos2a = p2 a2 sin2a – b2 cos2a = p2 −

21. If the tangent on the point (2 secf, 3tanf) of the

x2 y2 − = 1 is parallel to 3x – y + 4 = 0, 4 9 then the value of f is (a) 45° (b) 60° (c) 30° (d) 75° hyperbola

22. What is the slope of the tangent line drawn to the hyperbola xy = a(a ≠ 0) at the point (a, 1)? (a) 1/a (b) –1/a (c) a (d) – a 23. The straight line x + y = 2 p will touch the hyperbola 4x2 – 9y2 = 36, if (a) p2 = 2 (b) p2 = 5 2 (c) 5p = 2 (d) 2p2 = 5 x2 y2 + = 1 and C be the circle 9 4 x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E

24. Let E be the ellipse

25. The distance between the directrices of the ellipse x2 y2 + = 1 is 36 20 (a) 8 (b) 12

(d) 24

26. The equation of the ellipse whose vertices are (±5, 0) and foci are (±4, 0) is (b) 25x2 + 9y2 = 225 (a) 9x2 + 25y2 = 225 (c) 3x2 + 4y2 = 192 (d) none of these 27. The equation of the ellipse whose one of the vertices is (0, 7) and the corresponding directrix is y = 12, is (a) 95x2 + 144y2 = 4655 (b) 144x2 + 95y2 = 4655 (c) 95x2 + 144y2 = 13680 (d) none of these 28. For the ellipse 3x2 + 4y2 = 12, the length of latus rectum is 8 3 3 (a) (b) 3 (c) (d) 3 2 2 MATHEMATICS TODAY | JANUARY ‘18

29. Eccentricity of the ellipse 9x2 + 25y2 = 225 is 34 4 9 (b) (c) (d) (a) 3 5 5 25 5 30. What is the equation of the ellipse with foci (±2, 0) 1 and eccentricity = ? 2 (b) 4x2 + 3y2 = 48 (a) 3x2 + 4y2 = 48 2 2 (c) 3x + 4y = 0 (d) 4x2 + 3y2 = 0 31. The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrices is 18, is (b) 9x2 + 5y2 = 180 (a) 5x2 – 9y2 = 180 2 2 (c) x + 9y = 180 (d) 5x2 + 9y2 = 180 32. The centre of the ellipse 4x2 + 9y2 – 16x – 54y + 61 = 0 is (a) (1,3) (b) (2, 3) (c) (3, 2) (d) (3, 1) 33. The equation of an ellipse whose eccentricity is 1/2 and the vertices are (4, 0) and (10, 0) is (a) 3x2 + 4y2 – 42x + 120 = 0 (b) 3x2 + 4y2 + 42x + 120 = 0 (c) 3x2 + 4y2 + 42x – 120 = 0 (d) 3x2 + 4y2 – 42x – 120 = 0 34. The eccentricity of the ellipse 9x2 + 5y2 – 18x – 20y – 16 = 0 is (a) 1/2 (b) 2/3 (c) 1/3

(d) 3/4

35. The position of the point (1, 3) with respect to the ellipse 4x2 + 9y2 – 16x – 54y + 61 = 0 (a) outside the ellipse (b) on the ellipse (c) on the major axis (d) on the minor axis 36. The locus of the point of intersection of mutually x2 y2 perpendicular tangent to the ellipse + = 1, is a 2 b2 (a) a straight line (b) a parabola (c) a director circle (d) none of these 37. The equation of the tangents drawn at the ends of the major axis of the ellipse 9x2 + 5y2 – 30y = 0, are (a) y = ±3 (b) x = ± 5 (c) y = 0, y = 6 (d) none of these 38. The line y = mx + c is a normal to the ellipse x2

+ = 1, if c = a 2 b2 (a) –(2am + bm2) (c) − 16

(a2 − b2 )m a2 + b2m2

(b) (d)

(a2 + b2 )m a2 + b2m2 (a2 − b2 )m a 2 + b2

MATHEMATICS TODAY | JANUARY ‘18

39. The equation of the normal at the point (2, 3) on the ellipse 9x2 + 16y2 = 180, is (a) 3y = 8x – 10 (b) 3y – 8x + 7 = 0 (c) 8y + 3x + 7 = 0 (d) 3x + 2y + 7 = 0 40. If (4, 0) and (–4, 0) be the vertices and (6, 0) and (–6, 0) be the foci of a hyperbola, then its eccentricity is (a) 5/2 (b) 2 (c) 3/2 (d) 2 41. If (0, ±4) and (0, ±2) be the foci and vertices of a hyperbola, then its equation is

2 2 2 2 (a) x − y = 1 (b) x − y = 1 4 12 12 4 y2 x2 y2 x2 (c) (d) − =1 − =1 4 12 12 4 42. The eccentricity of the hyperbola 4x2 – 9y2 = 16, is 8 4 5 13 (b) (c) (d) (a) 3 3 4 3 43. The eccentricity of the hyperbola 2x2 – y2 = 6 is (b) 2 (c) 3 (d) (a) 2 3 2 2 44. The equation x – 16xy – 11y – 12x + 6y + 21 = 0 represents (a) parabola (b) ellipse (c) hyperbola (d) two straight lines

45. The equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is 16 and eccentricity is 2, is (a) x2 – y2 = 16 (b) x2 – y2 = 32 2 2 (c) x – 2y = 16 (d) y2 – x2 = 16 x2 y2 + = 1 and the 16 b2 2 2 y x 1 − = coincide, then the value hyperbola 144 81 25 2 of b is (a) 1 (b) 5 (c) 7 (d) 9

46. If the foci of the ellipse

47. The eccentricity of the hyperbola 1999 2 (x − y 2 ) = 1 is 3 2 (b) (c) 2 (a) 3

(d) 2 2

Assertion & Reason Type

48. Statement-1 : The foci of the hyperbola xy = 36 are (6 2 , 6 2 ) and (−6 2 , − 6 2 ). Statement-2 : The foci of the hyperbola x2

−

54. The maximum possible length of semi latus rectum is (a) 2 + 3 (b) 3 + 3 (c) 4 + 3 (d) 1 + 3 Paragraph for Q.No. 55 - 57

= 1 are  ± a2 − b2 , 0  .   b 2

49. Statement-1 : The angle of intersection between the x2 y2 ellipse + = 1 and the circle x2 + y2 = ab is 2 2 a b b − a   tan −1  .  ab 

Let P, Q, R be three points on the ellipse

 ab y2 + = 1 and x 2 + y 2 = ab is  , 2 2 a b  a+ b

55. The maximum area of the triangle PQR is (a)

ab  . a + b 

50. Statement-1 : The condition on a and b for which two distinct chords of the ellipse

xx yy x2 y2 i.e. 21 + 21 − 1 = 12 + 12 − 1. a b a b

51. Statement-1 : If the point (x, y) lies on the curve 2x2 + y2 – 24y + 80 = 0 then the maximum value of x2 + y2 is 400. Statement-2 : The point (x, y) is at a distance of x 2 + y 2 from origin. Comprehension Type

Paragraph for Q.No. 52 - 54 A parabola is drawn through two given points A(1, 0) and B(–1, 0) such that its directrix always touches the circle x2 + y2 = 4. Then 52. The equation of directrix is of the form (a) x cosa + y sina = 1 (b) x cosa + y sina = 2 (c) x cosa + y sina = 3 (d) x tana + y seca = 2 53. The locus of focus of the parabola is (a)

x2 y2 + =1 4 3

x2 y2 + =1 (c) 3 4

(b)

x2 y2 + =1 4 5

(d)

x2 y2 + =1 5 4

3 3 ab 4

(b)

3 ab 4 Area of ∆PQR

56.

3 3 ab 2

(d) pab

(c)

2a2 2b2 passing through (a, –b) are bisected by x + y = b is a2 + 6ab – 7b2 > 0. x2 y2 Statement 2 : Equation of the chord of + =1 a 2 b2 whose mid point is (x1, y1) of the form T = S1

=1 a 2 b2 and let P′, Q′, R′ be their corresponding points on it’s auxiliary circle, then

Statement-2 : The point of intersection of the ellipse x2

Area of ∆P ′Q ′R ′ a (a) b 1 (c) 2

= (b)

b a

(d) none of these

57. When the area of triangle PQR is maximum, the centroid of triangle P′Q′R′ lies at (a) one focus (b) one vertex (c) centre (d) on one directrix Paragraph for Q.No. 58 - 60 Suppose an ellipse and a circle are respectively given by the equation

= 1 ...(i) and a 2 b2 ...(ii) x2 + y2 + 2gx + 2fy + c = 0 The equation,  x2 y2  2 2 ...(iii)  2 + 2 − 1 + λ(x + y + 2 gx + 2 fy + c) = 0 b a  represents a curve which passes through the common points of the ellipse (i) and the circle (ii). We can choose l so that the equation (iii) represents a pair of straight lines. In general we get three values of l, indicating three pair of straight lines can be drawn through the points. Also when (iii) represents a pair of straight lines they are parallel to the lines x2 2

y2 2

+ λ(x 2 + y 2 ) = 0, which represents a pair of

a b lines equally inclined to axes (the term containing y is absent). Hence two straight lines through the points MATHEMATICS TODAY | JANUARY ‘18

of intersection of an ellipse and any circle make equal angles with the axes.

63. The minimum distance of 4x2 + y2 + 4x – 4y + 5 = 0 from the line –4x + 3y = 3 is

58. The radius of the circle passing through the point

64. The distance between the directrices of the ellipse K (4 x − 8)2 + 16 y 2 = (x + 3 y + 10)2 is K then is 2 x2 y2 + = 1 from 65. Number of points on the ellipse 50 20 which pair of perpendicular tangents are drawn to

of intersection of (a) (c)

y b

a 2 + b2 a 2 − b2

= 1 and x 2 − y 2 = 0 is 2 ab

(b)

a 2 + b2 a 2 + b2

(d)

a 2 + b2 a 2 + b2 59. Suppose two lines are drawn through the common points of intersection of

−

= 1 and x2 + y2 +

a 2 b2 2gx + 2fy + c = 0. If these lines are inclined at an angle a, b to x-axis then π (a) a = b (b) α + β = 2 b (c) a + b = p (d) α + β = 2 tan −1   a

60. The no. of pair of straight lines through the points of intersection of x2 – y2 = 1 and x2 + y2 – 4x – 5 = 0 is (a) 0 (b) 1 (c) 2 (d) 3 Matrix–Match Type 61. Match the following. (A)

(B)

(C)

(D)

Column-I The locus of mid-points of chords of an ellipse which are drawn through an end of minor axis, is The locus of an end of latus rectum of all ellipses having a given major axis is The locus of the foot of perpendicular from a focus of the ellipse on any tangent is A variable line is drawn through a fixed point cuts axes at A and B. The locus of the mid point of AB is

Column-II (p) hyperbola

(x − 1)2 y 2 − =1 9 3 (x − 1)2 y 2 + =1 Conjugate of this hyperbola is − 9 3 ⇒ (x − 1)2 − 3 y 2 = 9 ⇒

 a 2 + b2  and its eccentricity (e) =    b2  9+3 Here, a2 = 9, b2 = 3 ∴ e = =2 3 2. (b) : We know that equation of polar at point (h, k) is

(q) circle

(r) parabola

(s) ellipse

=1 ⇒

hx ky + = 1 ⇒ hx + 4ky = 4 4 1

...(i) a b which is similar to given straight line x + 4y = 4 ...(ii) Comparing (i) and (ii), we get h = 1, k = 1. Hence, the point is (1, 1). 2

62. If e is the eccentricity of the hyperbola (5x – 10)2 + 25e (5y + 15)2 = (12x – 5y + 1)2 then is equal to 13 MATHEMATICS TODAY | JANUARY ‘18

x2 y2 + = 1 and the circle x2 + y2 = 16 intercepted 25 4 3L is by the coordinate axis then 2 SOLUTIONS 1. (c): Given, equation of hyperbola is x2 – 3y2 = 2x + 8 ⇒ x2 – 2x – 3y2 = 8

Integer Answer Type

x2 y2 + = 1 is 16 9 66. If L be the length of common tangent to the ellipse the ellipse

−b x (Two diameters y = m1x and y = m2x a b2 will be conjugate diameters, if m1m2 = − 2 ). a 4. (c) : Equation of tangent at (a cosq, b sinq) is y x Y cos θ + sin θ = 1 Q a b (a cos , b sin )  a  P= , 0  cos θ  P X O 3. (a) : y =

b   Q =  0,  sin θ 

Area of ∆OPQ = \

1  a  b  ab =     2  cos θ   sin θ  |sin 2θ |

(Area)min = ab.

5. (d) : 25(x – 3)2 + 16y2 = 400 (x − 3)2 y 2 + =1 16 25

e = 1−

16 3 = 25 5

6. (c) : Since the general equation of second degree represents a rectangular hyperbola, if D ≠ 0, h2 > ab and coefficient of x2 + coefficient of y2 = 0. Therefore the given equation represents a rectangular hyperbola, if l + 5 = 0 i.e., l = – 5. 7. (c) : Multiplying both, we get (bx )2 − (ay )2 = (ab)2 ⇒

−

=1 a 2 b2 which is the standard equation of hyperbola. 2 8. (b) : Since e > 1 always for hyperbola and < 1. 3 9. (c) : We know that when a circle touches externally to the two given circles, then the locus of the circle will be hyperbola. x2 y2 10. (c) : The equation of hyperbola is − =1 16 9 5 Now b2 = a2 (e 2 − 1) ⇒ e = 4 5   Hence foci are (±ae, 0) ⇒  ± 4 ⋅ , 0  i.e., (±5, 0).  4  11. (c) : Since h2 > ab Hence it is a hyperbola. 12. (a) : The equation is (x – 0)2 + (y – 0)2 = a2. 13. (c) : If y = 2x + l is tangent to given hyperbola, then λ = ± a2m2 − b2 = ± (100)(4) − 144 = ± 256 = ±16 x2 y2 − = 1 and perpendicular to 1 3 x + 3y – 2 = 0 are given by y = 3x ± 9 − 3 = 3x ± 6. 14. (a) : Tangents to

15. (b) : Equation of hyperbola is

x2 a2

−

y2 b2

Tangents to hyperbola are y = mx ± a2m2 − b2 ...(i) −1 a2 x± − b2 2 m m Eliminating m, we get x2 + y2 = a2 – b2. y x 16. (b) : The tangent at (h, k) is − =1 4/h 3/k

Tangents perpendicular to (i) are y =

4 3 h 4 ...(i) = ⇒ = h k k 3 ...(ii) 3h2 – 4k2 = 12 Using (i) and (ii), we get the tangents as y – x = ± 1. 17. (a) : The line through (6, 2) is y – 2 = m(x – 6) ⇒ y = mx + 2 – 6m Now from condition of tangency, (2 – 6m)2 = 25m2 – 16 ⇒ 36m2 + 4 – 24m – 25m2 + 16 = 0 ⇒ 11m2 – 24m + 20 = 0 Obviously its roots are m1 and m2, therefore 24 20 m1 + m2 = and m1m2 = 11 11 x2 y2 18. (a) : If y = mx + c touches − 2 = 1, 2 a b 2 2 2 2 then c = a m – b . 2 2 Here, c = 6, a = 100, b = 49 17 ∴ 36 = 100m2 − 49 ⇒ 100m2 = 85 ⇒ m = . 20 19. (c) : Equation of the tangent to x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is 2x – y – 4(x + 2) + (y + 1) + 11 = 0 or x = 2. 20. (b) : x cosa + y sina = p ⇒ y = – cota x + p coseca ∴

y2 − = 1. a 2 b2 Therefore, p2 cosec2a = a2 cot2 a – b2 ⇒ a2 cos2a – b2 sin2a = p2. 21. (c) : We have, x = 2 secf and y = 3 tanf dy dx = 3 sec2 φ = 2 secftanf and ⇒ dφ dφ dy dy / dφ 3 sec2 φ \ = = dx dx / dφ 2 sec φ tan φ dy 3 ...(i) = cosecφ dx 2 But, tangent is parallel to 3x – y + 4 = 0 ...(ii) 3 By (i) and (ii), cosecφ = 3 ⇒ cosecφ = 2 ∴ φ = 30°. 2 22. (b) : Given equation of hyperbola is xy = a Slope of tangent at point (x1, y1) is It is tangent to the hyperbola

xdy dy − y  dy  m=  ∴ + y =0 ⇒ =  dx ( x , y ) dx dx x 1 1 1  dy  At point (a, 1), we have m =   =−  dx (a,1) a 23. (d) : The condition for the line y = mx + c will touch the hyperbola

−

= 1 if c2 = a2m2 – b2

MATHEMATICS TODAY | JANUARY ‘18

Here, m = − 1, c = 2 p, a2 = 9, b2 = 4 \

We get 2p2 = 5

x2 y2 24. (d) : The given ellipse is + = 1. The value of 9 4 x2 y2 + − 1 is positive for x = 1, y = 2 the expression 9 4 and negative for x = 2, y = 1. Therefore P lies outside E and Q lies inside E. The value of the expression x2 + y2 – 9 is negative for both the points P and Q. Therefore P and Q both lie inside C. Hence, P lies inside C but outside E. 25. (c) : We have, a = 6, b = 2 5 20 16 2 = (1 − e 2 ) ⇒ e = = 36 36 3 a But directrices are x = ± e 6 Hence, distance between them is 2⋅ = 18. 2/3 \

b2 = a2 (1 − e 2 ) ⇒

26. (a) : Vertices ≡ (±5, 0) ≡ (±a, 0) ⇒ a = 5 4 Foci (±4, 0) ≡ (±ae, 0) ⇒ e = 5  16  ∴ b2 = (25) 1 −  = 9 ⇒ b = 9  25  x2 y2 + = 1 i.e., 9x2 + 25y2 = 225 25 9

Hence equation is

27. (b) : We have, vertex = (0,7) ⇒ b = 7 7 b Directrix y = 12 ⇒ = 12 ⇒ e = e 12 95 7 95 = 144 12 Hence equation of ellipse is 144x2 + 95y2 = 4655. Also, a = 7

28. (b) :

x2 y2 2b2 + = 1. Latus rectum = = 3. 4 3 a

29. (b) : The ellipse is

x2 y2 + =1 25 9

b 9 4 ∴ e = 1− 2 = 1− = 25 5 a 30. (a) : Since, ae = ± 2 ⇒ a = ± 4 ( e = 1 / 2) Now b2 = a2(1 – e2) ⇒ b2 = 16(1 – 1/4) ⇒ b2 = 12 Hence ellipse is 20

y x + = 1 ⇒ 3x 2 + 4 y 2 = 48 16 12

MATHEMATICS TODAY | JANUARY ‘18

31. (d) : We have, 2ae = 8,

2a = 18 ⇒ a = 4 × 9 = 6 e

2 4 6 e = , b = 6 1− = 5 =2 5 3 9 3 Hence the required equation is

x2 y2 + =1 36 20

i.e., 5x2 + 9y2 = 180. 32. (b) : 4(x – 2)2 + 9(y – 3)2 = 36 Hence, the centre is (2, 3). 33. (a) : Major axis = 6 = 2a ⇒ a = 3 1 1 3 3 ⇒ b = 3 1− = 2 4 2 Also centre is (7, 0) Now, e =

y2 (x − 7)2 + =1 9 (27 / 4) ⇒ 3x2 + 4y2 – 42x + 120 = 0 34. (b) : Given equation can be written as Equation is

(x − 1)2 ( y − 2)2 b2 − a 2 9−5 2 + =1 ∴ e = = = 2 5 9 9 3 b 35. (c) : E ≡ 4 + 9(3)2 – 16(1) – 54(3) + 61 < 0 Therefore, the point is inside the ellipse.

4(x − 2)2 9( y − 3)2 + =1 36 36 Equation of major axis is y – 3 = 0 and point (1, 3) lies on it. 36. (c) : It is the director circle. 37. (c) : Change the equation 9x2 + 5y2 – 30y = 0 in standard form 9x2 + 5(y2 – 6y) = 0 x 2 ( y − 3)2 ⇒ 9 x 2 + 5( y 2 − 6 y + 9) = 45 ⇒ + =1 5 9  a2 < b2 , so axis of ellipse is on y-axis. At y axis, put x = 0, so we get 0 + 5y2 – 30y = 0 ⇒ y = 0, y = 6 i.e., tangents at vertex is y = 0, y = 6. 38. (c) : As we know that the line lx + my + n = 0 is normal to

(a2 − b2 )2

. But in a 2 b2 l 2 m2 n2 this condition, we have to replace l by m, m by –1 and n +

= 1, if

by c, then the required condition is c = ± 39. (b) :

x − x1

y − y1

x1 / a y1 / b2 normal at point (x1, y1).

(a2 − b2 )m a2 + b2m2

, which is the equation of

180 In the given ellipse, a2 = 20, b2 = 16 Hence the equation of normal at the point (2, 3) is y −3 x −2 = ⇒ 40(x − 2) = 15( y − 3) 2 / 20 48 / 180 ⇒ 8x – 3y = 7 ⇒ 3y – 8x + 7 = 0. 40. (c) : Vertices (±4, 0) ≡ (±a, 0) ⇒ a = 4 Foci (±6, 0) ≡ (±ae, 0) ⇒ e =

6 3 = 4 2

41. (c) : Foci = (0, ±4) ≡ (0, ±be) ⇒ be = 4 Vertices (= 0, ±2) ≡ (0, ±b) ⇒ b = 2 ⇒ a = 2 3 y2

y2 x2 − =1 + = 1 or Hence equation is 4 12 (2 3 )2 (2)2 −x2

2 2 42. (c) : Given equation of hyperbola, x − y = 1, 4 (16 / 9) 4 ∴ a = 2, b = . As we know, b2 = a2(e2 – 1) 3 16 13 13 ⇒ = 4(e 2 − 1) ⇒ e 2 = ∴e= 9 9 3

43. (d) :

y2 x2 − = 1 ⇒ a2 = 3 and b2 = 6 (6 / 2) 6

Therefore e =

b2 a2

+1 ⇒ e = 3

44. (c) : D ≠ 0, h2 > ab 45. (b) : 2ae = 16, e = 2 ⇒ a = 4 2 and b = 4 2 ∴ Equation is

x2 2

(4 2 )

46. (c) : Hyperbola is a=

−

y2 2

(4 2 ) 2

= 1 ⇒ x 2 − y 2 = 32

y x 1 − = 144 81 25

144 81 81 225 15 5 ,b= , e1 = 1 + = = = 25 25 144 144 12 4

 12 5  Therefore, foci = (±ae1 , 0) =  ± . , 0  = (±3, 0)  5 4  Also, focus of ellipse = (4e, 0). 3 9  ⇒ e = . Hence, b2 = 16 1 −  = 7  4 16  47. (b) : Here a = b, so it is a rectangular hyperbola. Hence, eccentricity, e = 2.

48. (c) : Foci of xy = 36 are (c 2 , c 2 ) and (−c 2 , − c 2 ) = (6 2 , 6 2 ), (−6 2 , − 6 2 ) \ Statement-1 is true. x2

= 1 are  ± a2 + b2 , 0    a b \ Statement-2 is false.  a 2b ab2  , , 49. (c) : The point of intersection is,   a + b a + b  −b2 a a , m2 = − . with m1 = 2 b b a The foci of

−

 m − m2  b −a  ⇒ θ = tan −1  1 = tan −1  .   ab  1 m m +  1 2 50. (a) : Let the mid point (t, b – t). 2

(b − t ) y = t 2 + (b − t ) + 2a2 2b2 2a2 2b2 It passes through (a, −b ) . tx

b (b − t )

2 b −t) ( +

− = 2 2a2 2b2 2a 2b2 ⇒ t2(a2 + b2) – ab(3a + b)t + 2a2b2 = 0 For real t, a2b2(3a + b)2 – 4(a2 + b2)2a2b2 > 0 9a2 + 6ab + b2 – 8a2 – 8b2 > 0 a2 + 6ab – 7b2 > 0 51. (a) : Given equation of ellipse is 2

x 2 ( y − 12) + = 1. The maximum value of 32 64 is the distance between (0, 0) & (0, 20). 52. (b) 53. (a) 54. (a)

x2 + y2

52 - 54 : Any point on circle x2 + y2 = 4 is (2cosa, 2sina) \ Equation of directrix is x(cosa) + y(sina) – 2 = 0 Let focus be (x1, y1). Then as A(1, 0), B(–1, 0) lie on parabola we must have (x1 − 1)2 + y12 = (cos α − 2)2   ⇒ x1 = 2 cos α, (x1 + 1)2 + y12 = (cos α + 2)2  y1 = ± 3 sinα 2 2 y x \ Locus of focus is + = 1 and focus is of the 4 3 form (2 cos α, ± 3 sin α). \ Length of semi latus rectum of parabola = ⊥r distance from focus to directrix 2 ± 3 sin2 α.

Hence maximum possible length = 2 + 3 MATHEMATICS TODAY | JANUARY ‘18

55. (a) : Let P = (a cosa, b sina), P ′ = (a cosa, a sin a) Q = (a cos b, b sinb), Q′ = (a cosb, a sinb) R = (a cosg, b sing), R′ = (a cosg, a sing). α −β β − γ γ −α Area of DPQR is 2ab sin sin sin 2 2 2  3 Its max value is 2ab .    2  56. (b) :

(

3 /2

)(

)

3 /2 =

1 2 cos α − cos γ sin α − sin γ a 2 cos α − cos β sin α − sin β

b a

57. (c) : Area of DPQR is max when a – b = b – g = g – a = 120° and DP ′Q′R′ is equilateral hence its centroid is (0, 0) centre of the ellipse.

Radius of the circle =

2a2b2 2

2ab

a + b2 59. (c) : As the lines joining common point of intersection must be equally inclined to the axis tana = – tanb ⇒ a + b = np a +b

60. (c) : Any curve through their point of intersection is x2 + y2 – 4x – 5 + l(x2 – y2 – 1) = 0 ⇒ (1 + l)x2 + (1 – l)y2 – 4x – 5 – l = 0 Curve will be pair of straight lines, if abc + 2fgh – af2 – by2 – ch2 = 0 ⇒ (1 + l)(1 – l)(–5 –l) + 0 – (1 + l)⋅0 – (1 – l)⋅4 + (5 + l)⋅0 = 0 ⇒ (l – 1)(l + 3)2 = 0 ⇒ l = 1, –3 \ Two pair of straight lines can be drawn. 61. (A) → (s), (B) → (r), (C) → (q), (D) → (p) (A) The chord with mid point (h, k) x2 y2 y ky h2 k 2 + = + ∴ Locus is + = a 2 b2 a 2 b2 a 2 b2 b

(B) (h, k) be the end of the latus rectum. h = ae, k = a(1 – e2) h2 = –a(k – a) ⇒ x2 = –a(y – a), parabola. (C) y − mx = ± a2m2 + b2 22

α β + =1 a b a b h = ,k = 2 2 α β α β  αβ  + = 2 ⇒ x −   y −  =    x y 2 2 4

62. (5) : Equation can be rewritten as

1 cos α − cos γ sin α − sin γ ab 2 cos α − cos β sin α − sin β

2a2b2 58. (b) : x 2 + y 2 = a 2 + b2

∴

3 3ab 4

Area of ∆PQR Area of ∆P ′Q ′ R′

x y + = 1 passes through fixed point (a, b) a b

(D) Line

x2 + y2 = a2

MATHEMATICS TODAY | JANUARY ‘18

(x − 2)2 + ( y + 3)2 =

13 13 12 x − 5 y + 1 So, e = . 5 13 5

 1  63. (1) : The given curve represents the point  − , 2   2  \ Minimum distance = 1. 2 2  1  (x + 3 y + 10) 64. (8) : (x − 2)2 + y 2 =   ⇒ e = 1/2 2 4 Perpendicular distance from (2, 0) to x + 3 y + 10 = 0 a a is − ae ⇒ 2a − = 6 ⇒ a = 4 e 2 2a Distance between directrices = = 16 = K e

x2 y2 + = 1 is x2 + y2 = 25 16 9 x2 y2 The director circle will cut the ellipse + = 1 at 50 20 4 points. 66. (7) : The equation of the tangent at (5 cosq, 2sinq) y x is cos θ + sin θ = 1 5 2 1 If it is a tangent to the circle then =4 cos2 θ sin2 θ + 10 3 25 4 ⇒ cos θ = ,sin θ = 4 7 2 7 Let A and B be the points where the tangent meets the coordinate axis then 2   5   A , 0 , B  0,  cos θ   sin θ  65. (4) : Director circle of

25 2

cos θ

4 2

sin θ

14 3

Solution Sender of Maths Musing SET-180 • N. Jayanthi (Hyderabad)



CLASS XI

Series 9

CBSE Statistics | Probability STATISTICS X

Range = Maximum value – Minimum value n   ∑ fidi   i =1  ×h Mean, x = A + N Where A = assumed Mean, h = width of class x −A intervals and di = i h Mean deviation for ungrouped data M .D.( x ) =

Σ xi − x

, M .D.( M ) = n Mean deviation for grouped data M .D.(x ) =

Σfi xi − x

where N = fi X

σ2 = X

Σfi xi − M N

1 Σ( xi − x )2 N

Variance and standard deviation of a continuous frequency distribution 1 1 σ 2 = Σfi ( xi − x )2 , σ = N Σfi xi2 − (Σfi xi )2 N N Shortcut method to find variance and standard deviation.

h2  N Σfi yi2 − (Σfi yi )2  , 2 N x −A h σ= N Σfi yi2 − (Σfi yi )2 , where yi = i , h N A = assumed mean, h = width of class intervals σ Coefficient of Variation (C.V.) = × 100, x– ≠ 0. x For series with equal means, the series with lesser standard deviation is more consistent or less scattered.

σ2 =

Σ xi − M

, M .D.(M ) =

1 Σ( xi − x )2 , σ = N

, X

Variance and standard deviation of a discrete frequency distribution

PROBABILITY X

Event A or B = A

B = {x : x ∈ A or x ∈ B}

(i)

0 ≤ P( i) ≤ 1

Event A and B = A

B = {x : x ∈ A and x ∈ B}

(ii)

P( i) = 1

Set A′ = S – A where S = sample space

A – B = Event ‘A but not B’ = A

For disjoint sets, A

(iii) P(A) = P( i)

B′

B=f

Number P( i ) associated with sample point such that

i X

∈S i

∈A

P(A

B) = P(A) + P(B) – P(A

P(A P (B

B C) = P(A) + P(B) + P(C) – P(A C) – P(C A) + P(A B C)

B) –

P(A′) = 1 – P(A) MATHEMATICS TODAY | JANUARY ‘18

13. The following is the record of goals scored by team A in football session.

WORK IT OUT VERY SHORT ANSWER TYPE

1. Calculate the mean deviation about median from the following data: 340, 150, 210, 240, 300, 310, 320. 2. There are 4 red and 3 black balls in a bag. If one ball is taken out of this bag at random, then represent the sample space and the event of this ball being black. 3. Two dice are thrown simultaneously. What is the probability of obtaining a total score of 7? 4. For a sample of size 60, we have the information

∑ xi2 = 18000 and ∑ xi = 960.

Find the variance.

5. What is the probability of getting a total of less than 12 in the throw of two dice? SHORT ANSWER TYPE

6. The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14, find the remaining observations. 7. If a number of two digits is formed with the digits 2, 3, 5, 7, 9, without repetition of digits, what is the probability that the number formed is 35? 8. Three coins are tossed together. Find the probability of getting (i) at least two heads. (ii) at least one head and one tail. 9. The probability that a student will receive A, B, C or D grade are 0.35, 0.45, 0.09 and 0.11 respectively. Find the probability that the student will receive (i) A or B grade (ii) at most a C grade.

No. of goals scored by team A No. of matches

For team B, the number of goals scored per match was 2.5, with a S.D. of 1.25. Find which team may be considered more consistent. 14. What is the probability that in a group of 3 people at least two will have the same birthday? Assume that there are 365 days in a year and no one has his/her birthday on the 29th of February. 15. A box contains 4 red, 5 white and 6 black balls. A person draws 4 balls from the box at random. Find the probability of selecting at least one ball of each colour. LONG ANSWER TYPE - II

16. Let x1, x2, x3, ..., xn be n values of a variable X, and let xi = a + hui, i = 1, 2, ..., n, where u1, u2, ..., un are the values of variable U. Then, prove that Var (X) = h2 Var (U), h ≠0. 17. There are n articles to be distributed among N people. Find the probability of a particular person getting r(< n) articles. 18. Find the mean and standard deviation of the first n terms of an A.P. whose first term is a and common difference is d.

10. In a race, the odds in favour of horses P, Q, R, S are 1 : 2, 1 : 3, 1 : 4 and 1 : 5 respectively. Find the probability that one of them wins the race.

19. Find the mean and variance of the data 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...... n, n, n,.........., n where the number r occurs r times, r = 1, 2, 3, ..., n.

LONG ANSWER TYPE - I

20. The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases : (i) If the wrong item is omitted. (ii) If it is replaced by 12.

11. Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting (i) all the four cards of the same suit (ii) all the four cards of the same number. 12. Calculate the mean deviation from the median of the following data: Wages per week 10-20 (in `)

20-30

30-40 40-50 50-60 60-70 70-80

No. of workers

MATHEMATICS TODAY | JANUARY ‘18

SOLUTIONS

1. Arranging the observations in ascending order of magnitude, we obtain 150, 210, 240, 300, 310, 320, 340. Clearly, the middle observation is 300. So, median = 300.

|di| = |xi – 300|

340 150 210 240 300 310 320

40 150 90 60 0 10 20 |di| = Σ|xi – 300| = 370

∴ M.D.(M ) =

370 1 1 = 52.8 | di | = ∑| xi − 300 | = ∑ n 7 7

2. Let Ri denotes red balls and Bi denotes black balls. S = Set of all possible outcomes when one ball is drawn from the bag = {R1, R2, R3, R4, B1, B2, B3} and E = Event of ball being black = {B1, B2, B3}

3. Let S be the sample space and E be the event of "obtaining a total of 7." Then, n(S) = 6 × 6 = 36 Also, E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} \ n(E) = 6 n (E) 6 1 Now, required probability, P( E ) = = = . n (S) 36 6 4. Since, σ 2 = =

∑ xi2

 ∑ xi  −  60  60

18000  960  − = 300 − 256 = 44.  60  60

5. Let S be the sample space and E be the event that the sum of numbers coming up is 12 Then, n(S) = 6 × 6 = 36 and E = {(6, 6)} \ n(E) = 1 n (E) 1 ∴ P( E ) = = n (S) 36 35 \ Required probability, P( E ′) = 1 − P( E ) = 36 6. Let a and b be the two observations. a + b + 2 + 4 + 10 + 12 + 14 x =8⇒ =8 7 \ a + b = 56 – 42 = 14 a2 + b2 + 22 + 42 + 102 + 122 + 142 2 − 8 = 16 7 \ a2 + b2 = 7(64 + 16) – 460 = 100 a + b = 14, a2 + b2 = 100 ⇒ a = 8, b = 6 or a = 6, b = 8 σ2 = 16 ⇒

7. Let S = the sample space and E = the event that the number formed is 35 Now n(E) = 1 and n(S) = total number of numbers of two digits formed with the digits 2, 3, 5, 7, 9 without repetition = 5P2 = 5 × 4 = 20 n (E) 1 \ Required probability, P( E ) = = n (S) 20 8. Let S be the sample space for the given experiment. A = the event of getting "at least two heads" B = the event of getting "at least one head and one tail" Now S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, H, T), (T, T, H), (H, T, T), (T, T, T)} \ n(S) = 8 (i) A = {(H, H, H), (H, H, T), (H, T, H), (T, H, H)} n ( A) 4 1 \ Required probability , P ( A) = = = . n (S) 8 2 (ii) B = {(H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H)} n ( B) 6 3 \ Required probability, P (B) = = = . n (S) 8 4 9. Let E1, E2, E3 and E4 denote the events of the student receiving A, B, C and D grades respectively. Then, P(E 1) = 0.35, P(E2) = 0.45, P(E3) = 0.09 and P(E4) = 0.11 (i) Required probability = P(A or B grade) = P(E1 E2) = P(E1) + P(E2) = 0.35 + 0.45 = 0.80 (ii) Required probability = P(at most a C grade) = P(C or D grade) = P(E3 E4) = P(E3) + P(E4) = 0.09 + 0.11 = 0.20

10. Let A, B, C and D be the events that the horses P, Q, R and S respectively win the race. 1 1 1 1 Then, P( A) = , P( B) = , P(C ) = and P(D ) = 3 4 5 6 Since there can be only one winner, \ A, B, C and D are mutually exclusive events

MPP-9 CLASS XI 1. (c) 6. (b) 11. (a,b,c,d) 16. (c)

2. 7. 12. 17.

(b) (a,c) (c,d) (2)

3. 8. 13. 18.

ANSWER KEY (c) (c,d) (a,d) (3)

4. 9. 14. 19.

MATHEMATICS TODAY | JANUARY ‘18

5. (d) 10. (a,b,c) 15. (d) 20. (2) 25

Now, required probability = P(A = P(A) + P(B) + P(C) + P(D) 1 1 1 1 19 = + + + = 3 4 5 6 20

\ l = 40, f = 20, h = 10, C = 20.

So, Median = l +

11. Let S = the sample space A = event that all the four cards are of the same suit. and B = event that all the four cards have the same number n(S) = total number of ways of drawing 4 cards from a pack of 52 cards = 52C4 (i) The four suits are club, space, heart and diamond, each having 13 cards. Now, n(A) = number of ways of getting all the four cards of the same suit = 13C4 + 13C4 + 13C4 + 13C4 = 4(13C4) \ Required probability, n ( A) 4 (13 C4 ) 44 P ( A) = = = 52 n (S) 4165 C4 (ii) Four cards bearing the same number can be drawn in the following ways : (2, 2, 2, 2), ...., (10, 10, 10, 10), (A, A, A, A), (J, J, J, J), (Q, Q, Q, Q), (K, K, K, K) \ n(B) = number of favourable cases = 13 n ( B) 13 \ Required probability, P( B) = = n (S) 52 C4 1 = 20825 12. Wages Mid- Frequency Cummulative per Values week ( x i) (in `)

frequency

|di|= fi|di| |xi – 45|

10-20

120

20-30

120

30-40

100

40-50

50-60

100

60-70

120

70-80

120

N = Σ fi = 60

Σ fi |di| = 680

N = 30. The cumulative frequency 2 N just greater than = 30 is 40 and the corresponding 2 class is 40-50. So, 40-50 is the median class. Here, N = 60. So,

MATHEMATICS TODAY | JANUARY ‘18

N /2 − C 30 − 20 × h = 40 + × 10 = 45. f 20

Thus, we have fi|xi – 45| = fi|di| = 680 and N = 60. \ Mean deviation from median =

∑ fi | di | 680 = = 11.33 N 60

13. Let the assumed mean for team A = 2. No. of goals No. of di = xi – A by team A matches = xi – 2 (fi)

fi di

fi di2

–2

–1

–9

fidi = 0

fidi2 = 30

30 = 25

6 = 1. 1 5

fi = 25 A . M. = x = A +

S. D. = σ =

Σ fi di 0 = 2+ =2 Σ fi 25 2

Σ fi di2  Σ fi di  − = Σ fi  Σ fi 

For team A : Coefficient of variation =

σ 1. 1 × 100 = × 100 = 55 x 2

1.25 × 100 = 50 2. 5 Clearly 55 > 50. Hence team 'B' is more consistent.

For team B : Coefficient of variation =

14. Let S = the sample space and E = the event that "at least two people have the same birthday" \ E ′ = all the three people have distinct birthdays Then, n(S) = number of ways in which three persons may have their birthdays = 365 × 365 × 365 = 3653 and n(E ′) = 365 × 364 × 363 n ( E ′) 365 × 364 × 363 364 × 363 Now, P( E ′) = = = n (S) 3653 3652 Hence, required probability, 364 × 363 P ( E ) = 1 − P ( E ′) = 1 − 3652

15. Let S = the sample space Selecting the balls such that there is at least one ball of each colour can be drawn in the following mutually exclusive ways : (i) 1 red, 1 white and 2 black balls (ii) 1 red, 2 white and 1 black balls (iii) 2 red, 1 white and 1 black balls Let A = event that 1 red, 1 white and 2 black balls are drawn B = event that 1 red, 2 white and 1 black balls are drawn C = event that 2 red, 1 white and 1 black balls are drawn Here A, B and C are mutually exclusive events. Hence required probability = P(A B C) = P(A) + P(B) + P(C) 4 C1 ⋅5 C1 ⋅6 C2 4 C1 ⋅5 C2 ⋅6 C1 4 C2 ⋅5 C1 ⋅6 C1 48 = + + = 15 15 15 91 C4 C4 C4 16. We have, xi = a + h ui, i = 1, 2, ..., n ⇒ ⇒

i =1 n

i =1

∑ xi = ∑ (a + h ui ) n

∑ xi = na + h ∑ ui i =1

i =1

1 n  1 ⇒ ∑ xi = a + h  ∑ ui  n i =1  n i =1   1 n 1 n   X = ∑ xi and U = ∑ ui  n i =1 n i =1   \ xi – X = (a + h ui) – (a + h U), i = 1, 2, ..., n ⇒ xi – X = h (ui – U), i = 1, 2, ..., n ⇒ (xi – X)2 = h2(ui – U)2, i = 1, 2, ..., n  1 n 1  ⇒ ∑(xi − X )2 = h2  ∑ (ui − U )2  n  n i =1  2 ⇒ Var (X) = h Var (U). 17. Let S = the sample space E = the event that a particular person gets r(< n) articles. Now, each article can be given to any one of the N people. \ Each article can be distributed in N ways. \ n(S) = number of ways of distributing n articles among N people = N × N × ... × N = Nn. r articles out of n articles can be given to a particular person in nCr ways and the remaining (n – r) articles can be distributed to remaining (N – 1) people in (N – 1)n – r ways. ⇒ X=a+hU

\ n(E) = number of ways in which a particular person gets r articles = nCr × (N – 1)n – r Hence, required probability, n (E )

P (E ) =

n−r

Cr × (N − 1)

n (S) Nn 18. Here the first n terms of the A.P. would be a, a + d, a + 2d + ... + a + (n – 1)d respectively. Let the assumed mean be a. xi a a+d a + 2d ... ... a + (n – 1)d

di2 0 d2 4d2 ... ... (n – 1)2 d2

di = xi – a 0 d 2d ... ... (n – 1)d

Now, di = 0 + d + 2d + .... + (n – 1)d n (n − 1) d = [1 + 2 + ... + (n − 1)]d = 2 and di2 = 0 + d2 + 4d2 + .... + (n – 1)2d2 = [12 + 22 + ... + (n – 1)2]d2 =

n(n − 1)(2n − 1) d 2

6 Σ di (n − 1) d The actual mean, x = a + =a+ n 2 Standard deviation, σ =

Σ di2

 Σd  − i  n  n

EXAM CORNER 2018 Exam

Date

VITEEE

4th April to 15th April

JEE Main

8th April (Offline), 15th & 16th April (Online)

SRMJEEE

16th April to 30th April

Karnataka CET

18th April & 19th April

WBJEE

22nd April

Kerala PET

23rd April & 24th April

AMU (Engg.)

29th April

COMEDK (Engg.) 13th May BITSAT

16th May to 31st May

JEE Advanced

20th May

AIIMS

27th May

MATHEMATICS TODAY | JANUARY ‘18

(n − 1)(2n − 1) d 2

(n − 1)(n + 1) (n − 1)  2n − 1 n − 1  − =d   2  3 2  12

n2 − 1 12

19. x = =

= =

1 + 2 + 2 + ..... + n + n 1 + 2 + 3 + .... + n

12 + 22 + 32 + ... + n2 n(n + 1) / 2

σ2 =

n(n + 1)(2n + 1) 6n(n + 1) / 2

1 ⋅ 1 + 2 ⋅ 2 + .... + n ⋅ n n(n + 1) / 2

13 + 23 + 33 + .... + n3 n(n + 1) / 2

 2n + 1  −  3 

= 2

n(n + 1) (2n + 1)2 n2 + n − 2 (n − 1)(n + 2) − = = 2 9 18 18

20. We have, n = 20, x = 10 and = 2 1 ∴ x = Σ xi ⇒ Σxi = nx = 20 × 10 = 200 n \ Incorrect xi = 200 and = 2 ⇒ 2 = 4 1 1 ⇒ Σ x2 − x 2 = 4 ⇒ Σ x 2 − 100 = 4 n i 20 i ⇒ xi2 = 2080 \ Incorrect xi2 = 2080

(i) If we omit the wrong item, 8 from the observations, then 19 observations are left Correct xi + 8 = incorrect xi \ Correct xi = 200 – 8 = 192 192 ∴ Correct mean = = 10.10 19 and correct xi2 + 82 = incorrect xi2 \ Correct xi2 = 2080 – 64 = 2016 1 Correct variance = (correct xi2) 19 – (correct mean)2

 n − 1 2 − d  2 

MATHEMATICS TODAY | JANUARY ‘18

2n + 1 3

2016  192  1440 −  =  19 19 361 1440 \ Correct standard deviation = = 1.997 361 (ii) If we replace the wrong item by 12, then Incorrect xi – 8 + 12 = correct xi \ Correct xi = 200 + 4 = 204 204 and correct mean = = 10.2 20 and incorrect xi2 – 82 + 122 = correct x2i \ Correct xi2 = 2080 – 82 + 122 = 2160 1 Correct variance = (correct xi2) 20 – (correct mean)2 =

2160  204  1584 − =  20  20 400 \ Correct standard deviation 1584 = = 1.9899 400  =

MPP-9

Class XI

his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Conic Sections Total Marks : 80

Time Taken : 60 Min.

Only One Option Correct Type 1. Equation of the smallest circle passing through the centre of the circle x2 + y2 + 2x + 4y + 4 = 0 and touching the circle x2 + y2 – 4x – 2y – 3 = 0 externally is (a) x2 + y2 – x + y – 4 = 0 (b) x2 + y2 + x – y – 6 = 0 (c) x2 + y2 + x + 3y + 2 = 0 (d) none of these 2. The equation of the parabola whose vertex and focus lie on the axis of x at distances a and a1 from the origin respectively is (b) y2 = 4(a1 – a)(X – a) (a) y2 = 4(a1 – a)X 2 (c) y = 4(a1 – a)(X – a1) (d) none of these 3. The curve described parametrically by x = t2 + t + 1, y = t2 – t + 1 represents (a) a pair of straight line (b) an ellipse (c) a parabola (d) a hyperbola 4. If angle between the tangents at the end points of x2 y2 − = 1 is 90°, then the chords of the hyperbola 9 4 locus of the mid points of the chords is  x2 y2  2 2 (a) x + y = 9  −  4   9

 x2 y2  (b) x 2 + y 2 = 4  −  4   9

 x2 y2  (c) x 2 + y 2 = 5  −  (d) none of these 4   9 5. If number of common tangents of the circles x2 + y2 + 2x + 6y + 1 = 0 and x2 + y2 – 6x + k = 0 is three, then value of k is.

(a) 2 (b) 4 (c) 6 (d) 5 6. Tangent is the drawn to the ellipse x2   π  + y = 1at (3 3 cos θ,sin θ)  where θ ∈  0,   .  2   27 Then the value of q such that sum of intercepts on axes made by this tangent is minimum is π π π π (a) (b) (c) (d) . 3 4 8 6 One or More Than One Option(s) Correct Type 7. Parabola y2 = 4x and the cirlce having it's centre at (6, 5) intersect at right angle. Possible point of intersection of these curves can be (a) (9, 6) (b) (2, 8 ) (c) (4, 4) (d) (3, 2 3) 8. If the tangent at the point P(q) to the ellipse 16x2 + 11y2 = 256 is also a tangent to the circle x2 + y2 – 2x = 15, then q is equal to 4π π 2π 5π (b) (d) (a) (c) 3 3 3 3 2 2 x y 9. If foci of 2 − 2 = 1 coincide with the foci of a b x2 y2 − = 1 and eccentricity of the hyperbola is 2, 25 9 then (a) a2 + b2 = 16 (b) there is no director circle of the hyperbola (c) centre of the director circle is (0, 0) (d) length of latusrectum of the hyperbola = 12 10. At a point P on the parabola y2 = 4ax, tangent and normal are drawn. Tangent intersects the x-axis at Q and normal intersects the curve at R such that chord PR subtends an angle of 90° at its vertex. Then MATHEMATICS TODAY | JANUARY ‘18

(a) PQ = 2a 6

Matrix Match Type 16. Match the following.

(b) PR = 6a 3

=1 a 2 b2 whose eccentric angles are differ by 90°, then (a) locus of point of intersection of tangents at x2 y2 P and Q is + =2 a 2 b2 x2 y2 1 (b) locus of mid-point (P, Q) is 2 + 2 = 2 a b (c) product of slopes of OP and OQ where O is the −b2 centre is 2 a 1 (d) max. area of DOPQ is ab 2 12. The director circle of a hyperbola is x2 + y2 – 4y = 0. One end of the major axis is (2, 0) then the focus is (a) (c)

( (

) 6)

( (−

Q. R.

(a) (b) (c) (d)

) 6)

3, 2 − 3

(b) − 3, 2 + 3

6, 2 −

(d)

6, 2 +

13. If the ellipse x2 + 2y2 = 4 and the hyperbola S = 0 have same end points of the latus rectum, then the eccentricity of the hyperbola can be π π (b) cosec (a) cosec 4 3 π π π π (c) 2 sin + sin (d) 2 sin + sin 3 4 3 4 Comprehension Type Tangent to the parabola y = x2 + ax + 1, at the point of intersection of y-axis also touches the circle x2 + y2 = r2. Also, no point of the parabola is below the x-axis. 14. The radius of circle when a attains its maximum value 1 1 (a) (b) (c) 1 (d) 5 5 10 15. The minimum area bounded by the tangent and the coordinates axes is 1 1 1 (c) (d) (a) 1 (b) 2 4 3

Column I The foci of the hyperbola 8x2 – y2 – 64x + 10y + 71 = 0 are Director circles of 3x2 + 2y2 = 6 and 3x2 – 2y2 = 6 are x2 + y2 = k, where k is The foci of the hyperbola 9x2 – 16y2 – 36x + 96y + 36 = 0 are Two circles x2 + y2 + px + py – 7 = 0 and x2 + y2 – 10x + 2py + 1 = 0 will orthogonally, then the value p is P Q R S 2 1 3 4 1 3 4 2 3 1 4 2 3 4 1 2 Integer Answer Type

Column II 1.

2, 3

3. (10, 5)

(2, 8)

17. Number of distinct normal lines that can be drawn x2 y2 to ellipse + = 1 from the point P (0, 6) is 169 25 18. From a point on y = x + 1 tangents are drawn to x2 − y 2 = 1 such that the chord of contact passes 2 x +y through fixed point (x1, y1), then 1 1 is y1 19. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)2 + (y + 2)2 = r2, then the value of r2 is 20. The straight line 2x – 3y = 1 divides the circular region x2 + y2 ≤ 6 into two parts.  3   5 3   1 1   1 1  If S =   2,  ,  ,  ,  ,  ,  ,    4   2 4   4 4   8 4  then the number of points(s) in S lying inside the smaller part is  Keys are published in this issue. Search now! J

Check your score! If your score is No. of questions attempted …… …… No. of questions correct Marks scored in percentage ……

> 90%

EXCELLENT WORK ! You are well prepared to take the challenge of final exam.

90-75%

GOOD WORK !

You can score good in the final exam.

74-60%

SATISFACTORY !

You need to score more next time.

< 60%

MATHEMATICS TODAY | JANUARY ‘18

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

MATHEMATICS TODAY | JANUARY â&#x20AC;&#x2DC;18

Trigonometry This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

*ALOK KUMAR, B.Tech, IIT Kanpur

TRIGONOMETRIC FUNCTIONS An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric equation. The trigonometric equation may have infinite number of solutions (because of their periodic nature) and can be classified as : • Principal solution • General solution If sinq = sina ⇒ q = np + (–1)n a, n ∈ Z If cosq = cosa ⇒ q = 2np ± a n ∈ Z If tanq = tana ⇒ q = np + a, n ∈ Z If sin2q = sin2a ⇒ q = np ± a, n ∈ Z If cos2q = cos2a ⇒ q = np ± a, n ∈ Z If tan2q = tan2a ⇒ q = np ± a, n ∈ Z sinq = 0 ⇒ q = np, n ∈ Z π cosq = 0 ⇒ θ = (2n + 1) , n ∈ Z 2 tanq = 0 ⇒ q = np, n ∈ Z Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ± cosx = t Trigonometric equations of the form a sinx + b cosx = c, where a, b, c ∈ R, can be solved by dividing both sides of the equation by a2 + b2 Trigonometric equations can also be solved by transforming a product of trigonometric ratios into their sum or difference and viceversa.

Application of Trigonometric Functions Sine Formula In any triangle ABC, the sines of the angles are proportional to the opposite sides a b c = = = 2R sin A sin B sin C where R is circumradius of the DABC. Cosine Formula cos A = cosC =

b2 + c2 − a2 2b c

2c a

2 ab Projection Formula a = b cosC + c cosB b = c cosA + a cosC c = a cosB + b cosA Napolian or Napier's Analogy C A− B a−b tan  cot =  2  a + b 2 A  B −C  b−c tan  = cot   2  b+c 2 B C − A c − a tan  cot =  2  c + a 2 Half Angle Formula sin

( s − b) (s − c ) A = 2 bc

He trains IIT and Olympiad aspirants.

MATHEMATICS TODAY | JANUARY ‘18

c2 + a2 − b2

a2 + b2 − c2

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). 34

cos B =

INVERSE TRIGONOMETRIC FUNCTIONS

cos

s ( s − a) A = 2 bc

tan

( s − b) (s − c ) A = 2 s ( s − a)

where s =

a+b+c 2

is semi perimetre of triangle.

Area of the DABC 1 (∆) = ab sin C 2 1 1 = bc sin A = ca sin B 2 2 (m + n)cotq = mcota – ncotb a b c abc R= = = = 2 sin A 2 sin B 2 sin C 4 ∆ ∆ , where r is in radius of DABC. s A B C r = (s − a)tan = (s − b)tan = (s − c)tan 2 2 2 r=

B C a sin sin 2 2 r= A cos 2 A B C r = 4 R sin sin sin 2 2 2 If r1, r2 and r3 are ex radius of DABC respectively then, A B ∆ ∆ r1 = = s tan ; r2 = = s tan 2 2 s−a s−b r3 =

∆ C = s tan 2 s−c

Length of an angle bisector from the angle A, A 2 bc cos 2 βa = b+c Length of median from the angle A 1 2 b2 + 2 c2 − a2 2 3 Also, ma2 + mb2 + mc2 = (a2 + b2 + c2 ) 4 2∆ Length of altitude from the angle A = a ma =

If f : A → B such that f (x) = y is one-one and onto, then there exists a unique function f –1, where f –1: B → A and f –1(y) = x, for all x ∈ X, y ∈ Y. –1

i.e., domain of f

= range of f and range of f

–1

= domain of f D o m a in a n d R a n g e o f I n v e r s e T r ig o n o m e tr ic a l F u n c t i no s F u n c t i on

D om

ai n

R an

sin–1x

[–1, 1]

[–p/2, p/2]

cos–1x

[–1, 1]

[0, p]

tan–1x

(–

, )

(–p/2, p/2)

cot–1x

(–

, )

(0, p)

sec–1x

(– , –1]

cosec–1x (–

[0, p/2)

(p/2, p]

[1, ) [–p/2, 0)

(0, p/2]

[1, )

, –1]

(

)

sin–1x + sin–1y = sin–1 x 1 − y 2 + y 1 − x 2 , –1 ≤ x, y ≤ 1,

≤1

{

}

sin –1 x – sin –1 y = sin –1 x 1 − y 2 − y 1 − x 2 , x2

–1 ≤ x, y ≤ 1 and + ≤ 1 cos–1x + cos–1y = cos–1 xy − 1 − x 2 1 − y 2 , –1 ≤ x, y ≤ 1, x + y

(

)

{

}

cos–1x – cos–1y = cos–1 xy + 1 − x 2 1 − y 2 , –1 ≤ x, y ≤ 1 and x – y ≤ 0 −1  x + y  tan–1x + tan–1y = tan   , xy < 1  1 − xy 

 x− y  tan–1x – tan–1y = tan–1   , xy > –1  1 + xy   2x  2 tan–1 x = sin–1   , |x| ≤ 1  1 + x2   1 − x2   2x  –1 –1  = cos   ,–1< x<1 2  , x 0 = tan   1 − x2  1+ x  1 2 sin–1x = sin–1 2 x 1 − x 2 , – 1 ≤ x ≤ 2 2

(

)

MATHEMATICS TODAY | JANUARY ‘18

(

)

2 cos–1x = sin–1 2 x 1 − x 2 , 3 sin–1 x = sin–1 (3x – 4x3), –

1 2

≤x≤1

1 1 ≤x≤ 2 2

1 ≤x≤1 2  3 x − x3  1 1 , − <x<   2  3 3  1 − 3x 

3 cos–1 x = cos–1 (4x3 – 3x), 3 tan–1 x = tan–1

PROBLEMS Single Correct Answer Type

1. (a) (c) 2. (a) (c)

General solution of the equation cotq – tanq = 2 is π nπ π nπ + (b) + 4 2 8 nπ π ± (d) none of these 4 8 5 If cos θ + sec θ = , then the general value of q is 2 π π nπ ± , n ∈ Z (b) 2nπ ± , n ∈ Z 3 6 π π nπ ± , n ∈ Z (d) 2nπ ± , n ∈ Z 3 6

3. If tan2 θ − (1 + 3 )tan θ + 3 = 0 , then the general value of q (where n ∈Z) is π π π π (a) nπ + , nπ + (b) nπ − , nπ + 4 3 4 3 π π π π (c) nπ + , nπ − (d) nπ − , nπ − 4 3 4 3 4. If cos2q + 3cosq = 0, then the general value of q is  −3 + 17  (a) 2nπ ± cos−1   , n ∈ Z  4  −3 − 17  (b) 2nπ ± cos−1   , n ∈ Z  4  −3 + 17  (c) nπ ± cos−1   , n ∈ Z  4  −3 − 17  (d) nπ ± cos−1   , n ∈ Z  4 5. If tan θ − 2 sec θ = 3 , then the general value of q (where n∈Z) is nπ π nπ π (a) nπ + (−1) − (b) nπ + (−1) − 4 3 3 4 π π nπ π n (d) nπ + (−1) + (c) nπ + (−1) + 3 4 4 3 36

MATHEMATICS TODAY | JANUARY ‘18

6. If tanq + tan2q + tan3q = tanq tan2q tan3q, then the general value of q is nπ (a) np, n ∈Z (b) , n ∈Z 6 π nπ (c) nπ ± , n ∈ Z , n ∈Z (d) 3 2 7. If cospq = cosqq, p ≠ q, n ∈Z then 2nπ (a) q = 2np (b) θ = p±q nπ (c) θ = (d) none of these p+q 8. If sin3a = 4sina sin(x + a) sin(x – a), then x = π π (a) nπ ± , n ∈ Z (b) nπ ± , n ∈ Z 3 6 π π (c) nπ ± , n ∈ Z (d) nπ ± , n ∈ Z 4 2 9. The general solution of the equation ( 3 − 1)sin θ + ( 3 + 1)cos θ = 2 (where n ∈Z) is π π nπ π (a) 2nπ ± + (b) nπ + (−1) + 4 12 4 12 π π π π n (c) 2nπ ± − (d) nπ + (−1) − 4 12 4 12 10. The general value q is obtained from the equation cos2q = sina is π (a) 2θ = − α 2 π  (b) θ = 2nπ ±  − α  , n ∈ Z 2  (c) θ =

nπ + (−1)n α , n ∈Z 2

π α (d) θ = nπ ±  −  , n ∈ Z 4 2 11. The equation 3sin2x + 10cosx – 6 = 0 is satisfied, if −1 −1 (a) x = nπ ± cos (1 / 3) (b) x = 2nπ ± cos (1 / 3) −1 −1 (c) x = nπ ± cos (1 / 6) (d) x = 2nπ ± cos (1 / 6)

12. The general value of q in the equation 2 3 cos θ = tan θ , is π π (a) 2nπ ± , n ∈ Z (b) 2nπ ± , n ∈ Z 6 4 π nπ (c) nπ + (−1) , n ∈ Z (d) nπ + (−1)n , n ∈ Z 3 4 13. The set of values of x for which the expression tan 3x − tan 2 x = 1 , is 1 + tan 3x tan 2 x

(a) f (b) p/4 π   (c) nπ + : n = 1, 2, 3..... 4   π   (d) 2nπ + : n = 1, 2, 3..... 4   14. The equation sinx cosx = 2 has (a) one solution (b) two solutions (c) infinite solutions (d) no solutions 15. If sin5x + sin3x + sinx = 0, then the value of x other π than 0 lying between 0 ≤ x ≤ is 2 (a) p/6 (b) p/12 (c) p/3 (d) p/4 16. The number of values of x in the interval [0, 5p] satisfying the equation 3sin2x – 7sinx + 2 = 0 is (a) 0 (b) 5 (c) 6 (d) 10 17. The equation sinx + siny + sinz = –3 for 0 ≤ x ≤ 2p, 0 ≤ y ≤ 2p, 0 ≤ z ≤ 2p has (a) one solution (b) two sets of solutions (c) four sets of solutions (d) no solution 18. If 2sin2q = 3cosq where 0 ≤ q ≤ 2p, then q = π 7π π 5π (b) , , 6 6 3 3 π 7π (d) none of these (c) , 3 3 19. The domain of definition of the function f (x) = sin–1(|x – 1| – 2) is (a)

(a) [–3, 0]

[1, 3]

(b) [–2, 0]

[1, 4]

[2, 4]

(d) [–2, 0]

[1, 2]

20. If (2cosx – 1)(3 + 2cosx) = 0, 0 ≤ x ≤ 2p, then x = π 5π π , (b) (a) 3 3 3 π 5π 5π  3 (c) , , cos−1  −  (d)  2 2 3 3 21. If sin θ = 3 cos θ, − π < θ < 0 , then q = 5π 4π 4π 5π (b) − (d) (a) − (c) 6 6 6 6 22. The most general value of q which will satisfy both 1 1 the equations sin θ = − and tan θ = is 2 3 π π (b) nπ + (a) nπ + (−1)n 6 6 π (c) 2nπ ± (d) none of these 3

23. The smallest positive angle which satisfies the equation 2 sin2 θ + 3 cos θ + 1 = 0 , is 2π π 5π π (a) (c) (d) (b) 3 3 6 6 24. If cos40° = x and cosq = 1 – 2x2, then the possible values of q lying between 0° and 360° is (a) 100° and 260° (b) 80° and 280° (c) 280° and 100° (d) 110° and 260° π  25. If tan(pcosq) = cot(psinq), then sin  θ +  equals  4 (a)

(b)

1 2

(c)

1 2 2

(d)

3 2

26. The general solution of sin x − cos x = 2 , for any integer n is 3π (a) np (b) 2nπ + 4 (c) 2np (d) (2n + 1)p 27. If 12cot2q – 31cosecq + 32 = 0, then the value of sinq is −2 2 3 or (a) or 1 (b) 3 3 5 4 3 1 or (d) ± (c) 5 4 2 sin θ + sin 2θ is 28. Period of cos θ + cos 2θ (a) 2p (b) p (c) 2p/3 (d) p/3 29. Period of sin θ − 3 cos θ is (a) p/4 (b) p/2 (c) p

(d) 2p

 2x   3x  30. The period of the function sin   + sin   is  3   2  (a) 2p (b) 10p (c) 6p (d) 12p 31. Which of the following functions has period 2p? π π   (a) y = sin  2 πt +  + 2 sin  3πt +  + 3 sin 5πt  3 4 π π (b) y = sin t + sin t 3 4 (c) y = sint + cos2t (d) none of these 32. The possible values of x, which satisfy the trigonometric  x − 1  x + 1 π + tan −1  = are equation tan −1    x − 2  x + 2  4 (a) ±

1 2

(b) ± 2

1 2

(d) ± 2

MATHEMATICS TODAY | JANUARY ‘18

π 2π −1 −1 33. If sin −1 x + sin −1 y = and cos x − cos y = 3 3 Then, (x, y) is equal to (a) (0, 1) (b) (1/2, 1) (c) (1, 1/2) (d) ( 3 / 2, 1)

34. In DABC, asin(B – C) + bsin(C – A) + csin(A – B) = (a) 0 (b) a + b + c (c) a2 + b2 + c2 (d) 2(a2 + b2 + c2) 35. In triangle ABC, if a, b, c are in A.P., then the value A C sin sin 2 2 = of B sin 2 (a) 1 (b) 1/2 (c) 2 (d) –1 36. In a DABC, if C = 30°, a = 47 cm and b = 94 cm, then the triangle is (a) Right angled (b) Right angled isosceles (c) Isosceles (d) none of these 37. If a = 9, b = 8 and c = x satisfies 3cosC = 2 then (a) x = 5 (b) x = 6 (c) x = 4 (d) x = 7 C A 3b + c cos2 = , then its 38. If in a triangle, a cos 2 2 2 sides will be in (a) A.P. (b) G.P. (c) H.P. (d) A.G. 2

39. In a triangle ABC if 2a2b2 + 2b2c2 = a4 + b4 + c4 , then angle B is equal to (a) 45° or 135° (b) 135° or 120° (c) 30° or 60° (d) none of these 40. If a = 2, b = 3, c = 5 in DABC, then C = (a) p/6 (b) p/3 (c) p/2 (d) none of these 41. Point D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If BAD = x, DAE = y, sin(x + y )sin( y + z ) EAC = z, then the value of = sin x sin z (a) 1 (b) 2 (c) 4 (d) none of these 42. If a1, a2, a3 are in A.P. and if d is the common difference,  d   d  + tan −1  then tan −1  =  + 1 a a 1 + a a 2 3 1 2   (a) tan

−1 

2d  1+ a a  1 3 

 2d  (c) tan −1    1 + a2 a3  38

(b) tan

−1 

d  1+ a a  1 3 

 2d  (d) tan −1    1 − a1a3 

MATHEMATICS TODAY | JANUARY ‘18

1 then the value of 2   2   tan sin −1  x + 1 − x  − sin −1 x  is   2 2  

43. Given 0 ≤ x ≤

(a) 1

(b)

(d)

3 31 44. In a DABC, a = 5, b = 4 and cos( A − B) = , then 32 side c is equal to

(a) 6 (c) 9 45. 6+ (a) (c)

(b) 7 (d) none of these

The smallest angle of the triangle whose sides are 12 , 48 , 24 is p/3 (b) p/4 p/6 (d) none of these

46. If angles of a triangle are in the ratio of 2 : 3 : 7, then the sides are in the ratio of 2 : 2 : ( 3 + 1) (a) (b) 2 : 2 : ( 3 + 1) (c)

2 : ( 3 + 1) : 2

(d) 2 : ( 3 + 1) : 2

47. If sides of a triangle are 2 cm, 6 cm and ( 3 + 1) cm . Then, the smallest angle of the triangle is (a) 30° (b) 45° (c) 60° (d) 75° A B − tan 2 2 = 48. In any triangle ABC , A B tan + tan 2 2 a − b a −b (a) (b) c a +b c a −b (d) (c) a +b a +b+c tan

Assertion & Reason Type

Directions : In the following questions, Statement-1 is followed by Statement-2. Mark the correct choice as : (a) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 49. Statement-1: The number of integral values of l, for which the equation 7cosx + 5 sinx = 2l + 1 has a solution, is 8. Statement-2 : acosq + bsinq = c has atleast one solution 2 2 if c > a + b .

50. Statement-1: sin 2 > sin 3.

Comprehension Type

π  Statement-2 : If x , y ∈  , π  , x < y , then sinx > siny 2  51. Let a, b, g > 0 and α + β + γ =

π . 2

a! b! Statement-1: If tan α tan β − + tan β tan γ − 6 2 c! + tan γ tan α − ≤ 0 , where n! = 1·2 …. n) 3 then tan α tan β, tan β tan γ , tan γ tan α are in A.P. Statement-2 : tan α tan β + tan β tan γ + tan γ tan α = 1 52. ABC is an isosceles triangle. ha is the length of the altitude from A. hb has a similar meaning. Given a ≤ ha and b ≤ hb. Statement-1 : a : b : c = 1 : 1 : 2 Statement-2 : In any triangle the length of an altitude does not exceed that of the corresponding median. 53. Statement-1 : In a DABC, if a < b < c and r is inradius and r1, r2, r3 are the exradii opposite to angle A, B, C respectively then r < r1 < r2 < r3. r1 r2 r3 Statement-2: For, DABC, r1r2 + r2r3 + r3r1 = r 54. Let a, b, c be 3 positive real numbers, such that a3 + b3 + c3 + 3abc > max(a, b, c ), then 2 Statement-1 : It is impossible to form a triangle whose sides have length a, b and c. Statement-2 : p > max(a, b, c), p > a, p > b and p > c. 3

55. Statement-1 : In a DABC, if cosA + 2 cosB + cosC = 2, then a, b, c are in A.P. Statement-2 : In a DABC, we have cosA + cosB + cosC = 1 + 4 sin

A B C A (s − b)(s − c ) sin sin and sin = , 2 2 2 2 bc

(s − a)(s − b) C (s − c )(s − a) B sin = = 2 ab 2 ca where 2s = a + b + c

Paragraph for Q. No. 57 – 59 If curve of y = f(x) and y = g(x) intersects at n different points x = x1, x2, x3 ...... . Then equation f(x) = g(x) is said to have n solutions. 57. Number of solutions of |cosx| = 2[x] is (where [x] is integral part of x) (a) 0 (b) 1 (c) 2 (d) infinite 58. The number of solutions of sinpx = |loge|x|| is (a) 0 (b) 6 (c) 4 (d) 8 59. Number

solutions of the equation 1 1 (sinx ≠ cosx) is sin5 x − cos5 x = − cos x sin x (a) 0 (b) 1 (c) 2 (d) infinite Paragraph for Q. No. 60 – 62 The sides of a triangle ABC are 7, 8, 6 the smallest angle being ‘C’.

60. The length of the altitude from vertex ‘C’ is 35 (b) (a) 5 3 4 7 7 15 (c) (d) 15 3 4 61. The length of the median from vertex C is 95 95 95 95 (b) (c) (d) 2 2 4 3 62. The length of the internal bisector of angle C is 14 14 (d) 2 6 (a) 6 (c) 30 (b) 5 5 Paragraph for Q. No. 63 – 65 Let ABC be any triangle and P be a point inside it such π π π 2π that ∠PAB = , ∠PBA = , ∠PCA = , ∠PAC = . 9 18 9 6 Let PCB = x. (a)

sin

63. x = (a) p/9 (c) p/3

(b) 2p/9 (d) none of these

56. The angles of a right angled triangle ABC are in A.P.

64. DABC is (a) equilateral (c) scalene

(b) isosceles (d) right angled

r 3 −1 Statement-1 : = R 2 r 2− 3 Statement-2 : = s 3

65. Which of the following is true? (a) BC > AC (b) AC < AB (c) AC > AB (d) BC = AC MATHEMATICS TODAY | JANUARY ‘18

Paragraph for Q. No. 66 – 68 n  xr − xr − 1  −1 −1  = ∑ (tan xr − tan xr − 1 ) r − 1xr  r = 1 

∑ tan−1  1 + x

r =1

= tan–1xn – tan–1x0,

n∈N

66. The value of cosec −1 5 + cosec −1 65 + cosec −1 (325) + ... to ∞ is π π 3π (a) p (b) (c) (d) 2 4 4 67. The sum to infinite terms of the series 1 1 1    cot −1  22 +  + cot −1  23 + 2  + cot −1  24 + 3  + ... is     2 2 2  π π (a) (b) (c) cot–12 (d) – cot–12 2 4

68. The sum to infinite terms of the series cot–13 + cot–17 + cot–113+... is π (b) cot–12 (a) 2 π (d) (c) tan–12 4 Matrix–Match Type

69. Match the following. If in DABC, r = 2, r1 = 4, s = 12 and a < b < c, then Column-I (A) Area of DABC is (B) 4 + 4R (C) A of triangle is

(q) 24 (r) sin–1(4/5)

(D)

(s) sin–1(3/5)

B of triangle is

Column-II (p) 22

70. Match the following. Column-I Column-II (A) If principal values of (p) λ + µ = π −1  1  −1 2 sin  −  + tan ( 3 ) and  2 cos

1 are l and −  2 

−1 

respectively, then

(B) If principal values of  7π  sin −1  sin  and  6   5π   cos −1  − sin    are l  6  

and 40

respectively, then

MATHEMATICS TODAY | JANUARY ‘18

(q) λ + µ = 5π 6

π (C) If principal values of (r) λ + µ = − 6  3 sin −1  −  and  2    3 sin −1 cos  sin −1   are  2   l and respectively, then 5π (s) µ − λ = 6 Integer Answer Type

cot–1

71. If q = cotq must be

7 + cot–1 8 + cot–118, then the value of

72. If q ∈ [0, 5p] and r ∈ R such that 2sinq = r4 – 2r2 + 3 then the maximum no. of values of the pair (r, q) is. 73. If l = cos4 [tan–1 {sin (cot–15)}], then the value of 729 l must be 169 74. If sin–1 x + sin–1y = p and, if x = ly, then the value of l must be 75. The number of solutions of x, which satisfy the equation log|sinx| (1 + cosx) = 2 for x ∈ [0, 2p] is 50 2r   76. If S = ∑ tan −1  , then the value of 2 4   2+r +r r =1 2550 cot S must be 638 C 7 77. In a DABC, a = 5, b = 4 and tan = , then 2 9 measure of side ‘c’ is

π and c1 and 6 c2 are the two possible values of third side then |c1 – c2| is

78. In triangle ABC, a = 5 ; b = 2; ∠A =

79. In DABC, 3a = b + c then cot B/2 cot C/2 is 80. In DABC, if r1 = 6, R = 5, r = 2, then the value of 6tanA is SOLUTIONS 1. (b) : On simplification, given equation reduces to cos2q = sin2q π π nπ π ⇒ tan 2θ = tan ⇒ 2θ = nπ + ⇒ θ = + , n ∈Z 4 4 2 8 5 2. (d) : cos2 θ − cos θ + 1 = 0 2 (5 / 2) ± (25 / 4) − 4 5 ± 3 ⇒ cos θ = = 2 4

⇒ cos θ =

1 π π = cos   ⇒ θ = 2nπ ± , n ∈ Z   2 3 3

3. (a) : tan2 θ − tan θ − 3 tan θ + 3 = 0 ⇒ tan θ(tan θ − 1) − 3 (tan θ − 1) = 0

π π ⇒ (tan θ − 3 ) (tan θ − 1) = 0 ⇒ θ = nπ + , nπ + 3 4

4. (a)

5. (d)

6. (b) : Given, tanq + tan2q + tan3q = tanqtan2qtan3q tan 6θ =

tan θ + tan 2θ + tan 3θ − tan θ tan 2θ tan 3θ 1 − ∑ tan θ tan 2θ

= 0, (from the given condition) ⇒ 6q = np ⇒ q = np/6, n ∈ Z 7. (b) : cos pq = cos qq ⇒ pq = 2np ± qq ⇒ θ=

2nπ , n ∈Z p±q

8. (b) : 3 sin α − 4 sin3α = 4 sin α(sin2 x − sin2 α)  3 ∴ sin2 x =    2 

⇒ sin2 x = sin2 π / 3

⇒ x = nπ ± π / 3, n ∈ Z 9. (a) π  π  10. (d) : cos 2θ = cos  − α  ⇒ 2θ = 2nπ ±  − α  2  2  π α ⇒ θ = nπ ±  −  , n ∈ Z 4 2 11. (b) : We have, 3sin2x + 10cosx – 6 = 0 ⇒ 3(1 – cos2x) + 10 cosx – 6 = 0 On solving, (cosx – 3)(3cosx – 1) = 0 Either cosx = 3 (not possible) 1 or cos x = ⇒ x = 2nπ ± cos−1(1 / 3), n ∈ Z 3 12. (c) π 13. (a) : tan(3x – 2x) = tanx = 1 ⇒ x = nπ + , n ∈ Z 4 But this value does not satisfy the given equation. Hence option (a) is the correct answer. 14. (d) : sinxcosx = 2 or sin2x = 4, which is impossible. 15. (c) : Given, sin5x + sin3x + sinx = 0 ⇒ –sin3x = sin5x + sinx = 2sin3xcos2x ⇒ sin3x = 0 ⇒ x = 0

π 1  π or cos 2 x = − = − cos   = cos  π −     2 3 3 2π π ⇒ x = nπ ±    3 3 π π For x lying between 0 and , we get x = 3 2 16. (c) 17. (a) : Given, sinx + siny + sinz = –3 which is satisfied 3π only when x = y = z = , for x , y , z ∈[0, 2 π] 2 18. (b) : 2 – 2cos2q = 3cosq ⇒ 2cos2q + 3cosq – 2 = 0 −3 ± 9 + 16 −3 ± 5 ⇒ cos θ = = 4 4 Neglecting (–) sign, we get 1 π π cos θ = = cos   ⇒ θ = 2nπ ± . 3 2 3 π 5π The values of q between 0 and 2p are , . 3 3 19. (c) 20. (b) : Given, (2cosx – 1)(3 + 2cosx) = 0 −3 1 Then, cos x = as cos x ≠ 2 2 π   for n = 0, x =  π  3  ⇒ x = 2nπ ± ;   5π 3  for n = 1, x =   3  π π 21. (b) : tan θ = 3 = tan ⇒ θ = nπ + 3 3 for − π < θ < 0 ⇒ 2 x = 2nπ ±

Put n = −1, we get θ = − π +

π −2 π −4 π = or 3 3 6

1 π   π 22. (d) : sin θ = − = sin  −  = sin  π +    6 2 6 tan θ =

π π  π  = tan   = tan  π +  ⇒ θ =  π +       6 6 6 3

7π . 6 23. (a) : 2 − 2 cos2 θ + 3 cos θ + 1 = 0

Hence general value of q is 2nπ + ⇒ 2 cos2 θ − 3 cos θ − 3 = 0 ⇒

cos θ =

⇒

θ=

5π 6

3 ± 3 + 24 3(1 ± 3)  1 = = 3 −   2 4 4

MATHEMATICS TODAY | JANUARY ‘18

24. (a) : Here, cos θ = 1 − 2 cos2 40° = −(2 cos2 40° − 1) = − cos(2 × 40°) = − cos 80° = cos(180° + 80°) = cos(180° − 80°) Hence, cos 260°and cos100° i.e., θ = 100° and 260° π  25. (c) : tan(π cos θ) = tan  − π sin θ  2  1 π 1  ∴ sin θ + cos θ = ⇒ sin  θ +  =  2 4 2 2 π  = 1 ⇒ cos  x +  = −1  4 2 2 π 3π 5π x + = 2nπ ± π ⇒ 2nπ + or 2nπ − 4 4 4 2 (c) : Given, 12 cot θ − 31 cosecθ + 32 = 0 12(cosec2θ − 1) − 31cosecθ + 32 = 0 12cosec2θ − 31cosecθ + 20 = 0 12 cosec2θ − 16 cosecθ − 15cosecθ + 20 = 0 (4 cosecθ − 5)(3 cosecθ − 4) = 0

26. (b) : sin x ⇒ 27. ⇒ ⇒ ⇒ ⇒

− cos x

5 4 ⇒ cosecθ = , 4 3

4 3 ∴ sin θ = , 5 4

 3θ  θ 2 sin   cos     2 sin θ + sin 2θ 2  3θ  28. (c) : = = tan   2 cos θ + cos 2θ  3θ  θ 2 cos   cos   2 2 2π Hence period = 3 π  29. (d) : sin θ − 3 cos θ = 2 sin  θ −   3 Hence period = 2p

 2x  2π 30. (d) : Period of sin   = = 3π  3  2/3  3x  2 π 4 π Period of sin   = =  2  3/2 3 L.C.M. of 3p and 31. (c)

4π = 12p . Hence period is 12p. 3

−1  x + 1  −1 −1  x − 1  32. (a) : tan   = tan 1 − tan   x+2 x − 2

⇒

x +1 = x+2

⇒ 2x2 = 1 42

x −1 x −2 x −1 1+ x −2 1−

⇒

x +1 −1 = x + 2 2x − 3

⇒

x=±

1 2

MATHEMATICS TODAY | JANUARY ‘18

33. (b) 34. (a) : a sin(B − C ) + b sin(C − A) + c sin( A − B) = k (Σ sin A sin(B − C ) = k {Σ sin(B + C )sin(B − C )}  1  = k Σ (cos 2C − cos 2 B) = 0  2  A C sin 2 2 35. (b) : B sin 2 ac(s − b)(s − c)(s − b)(s − a) s − b = = (s − a)(s − c)bc × ab b sin

But a, b and c are in A. P. ⇒ 2b = a + c 3b s−b 2 −b 1 Hence, = = b b 2 36. (d) : cosC = ⇒

3 (47)2 + (94)2 − c2 = ⇒ c = 58.24 2 2 × 47 × 94

37. (d) : cosC = ⇒

a2 + b2 − c2 2ab

2 81 + 64 − x 2 = 3 2⋅9⋅8

x 2 = 49 ⇒ x = 7

38. (a) : a

s( s − c ) s(s − a) 3b +c = ab bc 2

⇒ 2s(s − c + s − a) = 3b2 ⇒ 2s(b) = 3b2 ⇒ 2s = 3b ⇒

a + b + c = 3b ⇒ a + c = 2b ⇒ a, b, c are in A.P.

39. (a) : 2a2b2 + 2b2c2 = a4 + b4 + c4 Also, (a2 − b2 + c2 )2 = a4 + b4 + c4 − 2(a2b2 + b2c2 − c2 a2 ) ⇒ (a2 − b2 + c2 )2 = 2c2a2 ⇒

1 a2 − b2 + c2 =± = cos B 2ca 2

⇒

∠ B = 45° or 135°

a2 + b2 − c2 = −1 2ab ⇒ C = 180° (not possible) 41. (c) 42. (a) : Since, a1, a2, a3 are in A.P. 40. (d) : cosC =

⇒ a2 – a1 = d = a3 – a2  d   d  Now, tan −1  + tan −1     1 + a1a2   1 + a2 a3 

⇒ − 9.6 ≤ 2 λ ≤ 7.6 ⇒ − 4.8 ≤ λ ≤ 3.8 ∴ λ = −4, − 3, − 2, − 1, 0, 1, 2, 3 acosq + bsinq = c has no solution if | c |> a2 + b2 50. (a) :

 a −a   a −a  = tan −1  2 1  + tan −1  3 2   1 + a1a2   1 + a2 a3  = tan–1a2 – tan–1a1 + tan–1a3 – tan–1a2 = tan–1a3 – tan–1a1  a −a   (a − a ) + (a2 − a1 )   2d  = tan −1  3 1 = tan −1  3 2 = tan −1    1 + a1a3  1 + a1a3     1 + a1a3 

43. (a) 1 − cos( A − B) A− B 44. (a) : We have, tan  =  2  1 + cos( A − B) = ⇒

1 1 − (31 / 32) 1 a −b C = ⇒ cot = 2 1 + (31 / 32) a +b 63 63 7 1 C 1 C cot = ⇒ tan = 2 3 9 2 63

Now, cos C = ∴ ⇒

1 − tan2 (C / 2) 2

1 + tan (C / 2)

⇒ cos C =

1 − (7 / 9) 1 = 1 + (7 / 9) 8

c2 = a2 + b2 − 2ab cos C 1 c = 25 + 16 − 40 × = 36 ⇒ c = 6 8 2

45. (c) : Let a = 6 + 12 , b = 48 , c = 24 Clearly, c is the smallest side. Therefore, the smallest angle C is given by cosC =

a2 + b2 − c2 3 π = ⇒ ∠C = 2ab 2 6

46. (a) : Obviously, the angles are 30°, 45°, 105° . \ a : b : c = sin30° : sin45° : sin105° 1 1 3 +1 = : : = 2 : 2 : ( 3 + 1) 2 2 2 2 47. (b) : Let a = 2, b = 6 , c = 3 + 1 \ Smallest angle A is given by ∴ cos A =

6 + 3 +1+ 2 3 − 4 2 6 ( 3 + 1)

⇒ A = 45°

48. (b) 49. (c) : 7cosx + 5sinx = 2l + 1 2 λ + 1 ≤ 49 + 25 ⇒ 2 λ + 1 ≤ 74 − 74 ≤ 2 λ + 1 ≤ 74

51. (d) 52. (b) : a ≤ ha = b sinC and b ≤ hb = asinC ⇒ ab ≤ ab sin2C ⇒ 1 ≤ sin2C π Hence, sinC = 1 ⇒ ∠C = 2 Hence, Statement-1 is true. Clearly, Statement-2 is true but has no relevance with Statement-1. 53. (b) 3 3 3 54. (d) : a + b + c + 3abc > a3 2

⇒ b3 + c3 + (−a)3 − 3(−a)(b)(c) > 0

1 (b + c − a)[(b − c)2 + (c + a)2 + (b + a)2 ] > 0 2 ⇒ b+c–a>0 ⇒ b+c>a Similarly, c + a > b and a + b > c ⇒ a, b, c will form the sides of a triangle. 55. (a) : We have, cosA + cosC = 2(1 – cosB) A+C A−C B ⇒ 2 cos cos = 4 sin2 2 2 2 A − C   cos   2  ⇒ = 2 ⇒ 2s – b = 2b ⇒ a + c = 2b B sin 2 Thus, a, b, c are in A.P. Clearly, statement-2 is true. 56. (b) : (a – b) + a + (a + b) = p and a + b = p/2 ⇒ angles are p/6, p/3, p/2 ⇒ a = R, b = 3 R, c = 2R 3 3 (3 − 3 ) 3 −1 ∆ 3 r ∴ r= = R ⇒ = = = 6 2 s 3+ 3 R 3+ 3 ⇒

Also,

3R r 1 1 = × = 2s 3 + 3 (3 + 3 ) R (3 + 3 )( 3 + 1) MATHEMATICS TODAY | JANUARY ‘18

⇒

2− 3 r 1 r 2− 3 = = ⇒ = 2s 6 + 4 3 2 3 s 3

PB sin x PC sin 40° = , = PC sin(80° − x ) PA sin 30°

Similarly,

57. (a) : Use the properties of modulus and greatest integer function. 58. (b) : PA PB PC × × =1 PB PC PA

Now, ⇒

The graph shows 6 solutions. 59. (a) : sin5 x − cos5 x =

sin x − cos x sin x cos x

 sin5 x − cos5 x  ⇒ sin x cos x   =1  sin x − cos x  ⇒ sin2x[sin4 x + sin3xcosx + sin2xcos2x + sinxcos3x + cos4 x] = 2 2 ⇒ (sin2x – 2) (sin2x + 2) = 0 ⇒ sin2x = ±2, which is not possible. 2∆ 2 21 15 7 15 = × = c 6 4 4 2 2 2 61. (c) : 2(9 + m ) = 7 + 8 where ‘m’ is the length of the median from vertex ‘C’ 60. (d) : Altitude from 'C ' =

⇒ ⇒ ⇒ 64. 65. 66.

sin 20° sin x sin 40° × × =1 sin 10° sin(80° − x ) sin 30° sin (80° – x) – sinx = 2 sin 50° sin x 2 sin (40° – x) cos 40° = 2 sin 50° sin x sin(40° – x) = sinx ⇒ x = 20° (b) : A = C = 50° (c) : ABC = 80° ⇒ AC is longest side (d)

67. (c) :



r =1

= lim

 2r + 1 − 2r  r + 1  1 + 2r ⋅ 2

r =1 n

r +1

∑ (tan−1 2

n → ∞ r =1 n→∞

= tan–1 68. (d)

21  21   − 6  10584 14 6 2 2 = = 7+8 15 5

63. (a) :

PA PB PA sin 20° = ⇒ = sin 20° sin10° PB sin10° MATHEMATICS TODAY | JANUARY ‘18

 r + 1  1 + 2r ⋅ 2

n +1

2 7⋅8⋅

1  2r 

∑ tan−1  n→∞

= lim

= lim [tan −1 2

62. (b) : Length of the internal bisector of angle C ' 2 abs(s − c ) = a+b

r =1

tan −1  ∑ n→∞ 

= lim

 r +1

∑ cot−1  2 n→∞ lim

− tan −1 2r )

− tan −1 2]

–tan–12 =

π − tan −1 2 = cot–12 2

69. (A) → (q), (B) → (q), (C) → (s), (D) → (r) rr r r1r2 + r2r3 + r3r1 = 1 2 3 = s2 = 144 r ⇒ 4(r2 + r3) + r2r3 = 2r2r3 = 144 ⇒ r2r3= 72, ⇒ r2 + r3 = 18 ⇒ |r2 – r3| = 6 ⇒ r2 = 6, r3 = 12 r s−a 1 Now, = = ⇒ a=6 r1 s 2 Similarly, b = 8, c = 10 ⇒ ABC is right angle triangle. 3 1 Smallest angle is sin−1 , ∆ = × 6 × 8 = 24 sq. units 5 2 R=5

70. (A) → (q), (B) → (p, s), (C) → (r) π π π 2π (A) λ = − + = and µ = 3 6 3 6 π 2 π 5π 2π π π ∴ λ+µ = + = and µ − λ = − = 6 3 6 3 6 2 (B) λ = π −

7π π 5π   =− and µ = cos−1  − sin   6 6 6 

2π  2π  1  = cos−1  −  = cos−1  cos  =  2  3  3 π 2π π 2 π π 5π ∴ λ+µ = − + = ,µ−λ= + = 2 3 6 6 6 3    π  3 π = sin −1  sin  −   = − (C) λ = sin −1  −  2    3  3   π 3  −1  µ = sin −1  cos  sin −1   = sin  cos    2  3 1 π = sin −1   = 2 6 ∴ λ+µ = −

π π and µ − λ = 2 6

71. (3) : q = cot–17 + cot–18 + cot–118 1 1 1 = tan −1   + tan −1   + tan −1   7 8  18   1 1 1 1 1 1  + + − ⋅ ⋅   = tan −1  7 8 18 7 8 18  1 1 1 1 1 1  1 − ⋅ − ⋅ − ⋅  7 8 8 18 18 7 1 = tan −1   = cot −1 3 3 \ cotq = 3 72. (6) : 2sinq = (r2 – 1)2 + 2 This is possible when sinq = 1, r2 = 1, r = ±1 π 5π 9 π ∴ θ= , , 2 2 2 \ Number of values of the pair(r, q) is 6. 73. (4) 74. (1) :  π − cos−1 x + π − cos−1 y = π 2 2 ⇒ cos–1x + cos–1y = 0 ⇒ cos−1{ xy − (1 − x 2 ) (1 − y 2 } = 0 ⇒ xy − (1 − x 2 )(1 − y 2 ) = 1

⇒ ⇒ ⇒ \

(xy – 1)2 = (1 – x2) (1 – y2) x2y2 + 1 – 2xy = 1 – x2 – y2 + x2y2 x2 + y2 – 2xy = 0 ⇒ (x – y)2 = 0 ⇒ x = y l=1

75. (0) : log|sinx| (1 + cosx) = 2 ⇒ 1 + cosx = |sin x|2 ⇒ 1 + cosx = 1 – cos2x ⇒ cosx(1 + cosx) = 0 But (1 + cosx) ≠ 0 ⇒ cos x = 0 ⇒ sinx = 1. But sinx = 1 is not possible because the base of log can not be 1. Hence no solution. 76. (4) : S = =

 (1 + r + r 2 ) − (1 − r + r 2 )  2 2   1 + (1 + r + r )(1 − r + r ) 

∑ tan−1 

r =1

∑ [tan−1(1 + r + r 2 ) − tan−1(1 − r + r 2 )]

r =1

= {at r = 50 value of tan–1 (1 + r + r2)} – {at r = 1 value of tan–1 (1 – r + r2)} –1 = tan (1 + 50 + 502) – tan–11  1 + 50 + 502 − 1   2550  = tan −1  = tan −1   1 + 1 + 50 + 502   2552  ∴ tan S =

2550 2550 ∴ cot S = 4 2552 638

7 9 =1 77. (6) : cosC = 7 8 1+ 9 1 ∴ c2 = 25 + 16 − 2 ⋅ 5 ⋅ 4 ⋅ = 36 ⇒ c = 6 8 1−

78. (4) : a2 = b2 + c2 – 2bc cosA 3 ⋅ 4c − 5 = 0 ⇒ c 2 − 2 3 ⋅ c − 1 = 0 2 Now, c1 + c2 = 2 3 ; c1c2 = −1 4 + c2 −

2 2 (c1 – c2)2 = (c1 + c2 ) − 4c1c2 = (2 3 ) − 4(−1) = 16 79. (2) : 3a = b + c ⇒ 4a = 2s ⇒ s = 2a B C s 2a ∴ cot cot = = =2 2 2 s − a 2a − a

A A ⇒ 4 = 20 sin2 2 2 1 2× A 1 2 = 4 ⇒ 6 tan A = 8 ⇒ sin = tan A = 1 2 3 5 1− 4 

2 80. (8) : r1 − r = 4 R sin

MATHEMATICS TODAY | JANUARY ‘18

PROBABILITY 1. Three ordinary and fair dice are rolled simultaneously. The probability of the sum of outcomes being atleast equal to 8, is equal to (a) 81/108 (b) 27/216 (c) 81/216 (d) 181/216 2. A person takes a step forward with probability p and takes a step backward with probability q, where p + q = 1. The probability that after (2n + 1) steps the person is only one step away from his initial position, is equal to (a) 2n + 1Cn . pn + 1 . qn (b) 2n + 1Cn . pn . qn+1 (c) 2n + 1Cn . pn . qn (d) 2 . 2n + 1Cn . pn qn+1 3. Three numbers are selected simultaneously from the set {1, 2, 3, ...., 25}. The probability that the product of selected numbers is divisible by 4, is equal to (a) 1/115 (b) 98/115 (c) 773/1150 (d) 963/1150 4. Two numbers n1 and n2 are chosen at random (without replacement) from the set {1, 2, 3, ..........., 5n},

the probability that n14 − n24 is divisible by 5, is equal to 4(4n − 1) n −1 (a) (b) 5(5n − 1) 5n − 1 17n − 5 8n (c) (d) 5(5n − 1) 5(5n − 1)

5. ‘X’ follows a binomial distribution with parameters ‘n’ and ‘p’. ‘Y’ follows a binomial distribution with parameters ‘m’ and ‘p’. If X and Y are independent then P(X = r|X + Y = r + s) is equal to (a) (c)

Cr . mCs

m +n m

Cr n

Cs . Cr

m +n

Cs −1

(b) (d)

Cr . mCs

m +n m

Cr + s

Cs . nCr

m +n

Cr + s −1

3 If two events A and B are such that P ( A) = , 10 2 1 P (B) = and P ( A ∩ B ) = , then P (B | A ∪ B ) is equal 5 2 to

(a) 1/2

(b) 1/3

7. 3m fair dice are each rolled 2n times. The probability that the scores 1, 2, 3, 4, 5, 6 each appear mn times, is equal to (6mn)!

(a)

1 . 6  6  6 !((mn)!)

(c)

(6mn)!  1  .  ((mn)!)6  6 

MATHEMATICS TODAY | JANUARY ‘18

(6mn)!

(b)

1 . 6  6  (6 !(mn)!)

(d)

(6mn)!  1  .  ((mn)!)6  6 

6 mn

8. Two persons are selected randomly from n persons seated in a row (n 3). The probability that the selected persons are not seated consecutively, is equal to n−2 n −1 n+2 n−2 (a) (d) (b) (c) n n n+3 n −1 9. A real estate man has eight master keys to open several new homes. Only one master key will open any given home. If 40% of these homes are usually left unlocked, the probability that the real estate man can get into a specific home, if it is given that he selected 3 keys randomly before leaving his office, is equal to (a) 5/8 (b) 3/8 (c) 3/4 (d) 1/4 10. A man alternatively tosses a fair coin and rolls a fair ordinary dice. He starts with the coin. The probability that he gets a tail on the coin before getting 5 or 6 on the dice, is equal to (a) 3/4 (b) 1/2 (c) 1/3 (d) 2/3 11. For three events E1, E2 and E3. P(exactly one of the events E1 or E2 occur) = P(exactly one of the event E2 or E3 occur) = P(exactly one of the events E3 or E1 occur)

Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 46

(d) 2/3

= p and P(all the three events occur simultaneously)  1 = p2 , where p ∈  0,  . The probability that atleast one  2 of these events occur is

3 p + 2 p2 p + 3 p2 (b) 2 2 3 p + p2 3 p + 2 p2 (d) (c) 2 4 12. A bag contains ‘W’ white balls and ‘R’ red balls. Two players P1 and P2 alternatively draw a ball from the bag, replacing the ball each time after the draw, till one of them draws a white ball and wins the game. ‘P1’ begins the game. The probability of P2 being the winner, is equal to R W2 (b) (a) (2W + R) (W + 2R)R

(a)

(c)

R2 (2W + R)W

(d)

R (W + 2R)

13. A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4, is equal to (a) 5/12 (b) 5/24 (c) 7/12 (d) 19/24 n n 14. A number of the form 7 1 + 7 2 is formed, by selecting the numbers n1 and n2 from the set {1, 2, 3, ....., 99, 100} with replacement. The probability of the formed number being divisible by 5, is equal to (a) 1/8 (b) 1/4 (c) 1/16 (d) 1/2

15. ‘P1’ is the probability that a statement by A is true and ‘P2’ has the similar meaning for B. A and B agree in making a statement S. The probability that statement is correct is P1P2 (a) P1P2 + P1 (1 − P2 )(1 − P1 ) P1P2 (b) P1P2 + P2 (1 − P1 )(1 − P2 ) P1P2 P1P2 (d) P1 + P2 + 1 1 − P1 − P2 + 2P1P2 16. A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let Ei(i = 1, 2, 3) denote the event that ith digit on the selected ticket is 2, then which of the following is not correct? (a) E1 and E2 are independent (b) E2 and E3 are independent (c)

(c) E3 and E1 are independent (d) E1, E2, E3 are independent 17. A bag B1 has 3 white balls and 2 red balls. Another bag B2 has 4 white and 6 red balls. A ball is drawn randomly from bag B1 and without seeing it’s colour, is being put in bag B2. Now a ball is drawn from bag B2. The probability of both the drawn balls, of being same colour, is (a) 41/55 (b) 31/55 (c) 29/55 (d) none of these 18. A natural number ‘n’ is selected at random from the set of first 100 natural numbers. The probability that 100 ≤ 50 is equal to n+ n (a) 9/10 (b) 39/50 (c) 9/20 (d) none of these 1 and 19. For two events A and B, P(A) = P(A | B) = 4 1 P(B | A) = , then which of the following is not correct? 2 1 3 (a) P ( A′ | B) = (b) P (B ′ | A′) = 2 4 (c) P ( A ∪ B) =

3 4

(d) P ( A ∩ B) =

1 8

20. If the probability of choosing an integer ‘n’ out of 2m integers {1, 2, 3, .........., 2m – 1, 2m} is inversely proportional to n4( 1 ≤ n ≤ 2m), then the probability of the chosen number being odd, is (a)

1 2

(b) <

1 2

1 (d) none of these 2 21. In a bag there are 15 red and 5 white balls. Two balls are drawn in succession, without replacement. The first drawn ball is found to be red. The probability that second ball is also red, is equal to (a) 3/10 (b) 7/10 (c) 5/19 (d) 14/19

22. Two squares are chosen from squares of a ordinary chess board. It is given that the selected squares do not belong to the same row or column. The probability that the chosen squares are of same colour, is equal to (a) 25/49 (b) 32/49 (c) 25/64 (d) 1/2 23. Two fair dice are rolled simultaneously. One of the dice shows four. The probability of other dice showing six, is equal to (a) 2/11 (b) 1/18 (c) 1/6 (d) 1/36 MATHEMATICS TODAY | JANUARY ‘18

24. Four integers are selected randomly and are multiplied. The probability of this product being divisible by 5 but not by 10, is equal to 175 369 3471 1 (b) (c) (d) (a) 4 4 4 32 10 10 10 25. Two events A and B are such that P(A) > 0 and P(B) ≠ 1. The expression P ( A | B ) is also equal to (a)

1 – P (A | B)

(b) 1− P ( A | B)

1− P ( A ∪ B) 1− P ( A ∩ B) (d) P (B ) P (B ) 26. Consider the set of integers {10, 11, 12, ....., 98, 99}. By seeing the number a person A will laugh if the product of the digits of the number is 12. He chooses three numbers one at a time from this set of integers, randomly and with replacement. The probability that he will laugh atleast once, is equal to (c)

 43  (a) 1 −    45  4 (c) 1 −    25 

3 (b) 1 −   5

(d) none of these

27. The sum of two natural numbers n1 and n2 is known to be equal to 100. The probability that their product being greater than 1600, is equal to (a) 20/33 (b) 58/99 (c) 13/33 (d) 59/99 28. Let ‘head’ means one and ‘tail’ means two and the coefficients of the equation ax2 + bx + c = 0 are chosen by tossing three fair coins. The probability that the roots of the equation are non-real, is equal to (a) 5/8 (b) 7/8 (c) 3/8 (d) 1/8 29. A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign from the committee, then the probability of exactly two of the resigned experts being from the same institution, is equal to (a) 4/7 (b) 25/84 (c) 55/84 (d) 37/84

3 (a) 1 −   4 (c)

3 (b)   4 (d)

2 3n 32. Consider all functions that can be defined from the set A = {1, 2, 3} to the set B = {1, 2, 3, 4, 5}. A function f(x) is selected at random from these functions. The probability that, selected function satisfies f(i) ≤ f(j) for i < j, is equal to (a) 6/25 (b) 7/25 (c) 2/5 (d) 12/25 33. A fair coin is tossed ‘n’ number of times. The probability that head will turn up an even number of times, is equal to n −1 1 (a) (b) 2n 2 n−1 (d) 2 − 1 2n 34. Two numbers n1 and n2 are selected from the set {1, 2, 3, ......, n} without replacement. The value of P(n1 ≤ r | n2 ≤ r) , where r ≤ n is equal to

(c)

n +1 2n

r n(n −1)

(b) r n r (r − 1) r −1 (c) (d) n(n − 1) n −1 35. A and B play a game where each is asked to select a number from 1 to 25. If the numbers selected by A and B match, both of them win a prize. The probability that they win their third prize on 5th game is equal to 6.(24)2 6.(21)2 (b) (a) (25)5 (25)5

(a)

(c)

(24)2 (25)2

(d)

(21)2 (25)5

30. A person while dialing a telephone number, forgets the last three digits of the number but remembers that exactly two of them are same. He dials the number randomly. The probability that he dialed the correct number, is equal to (a) 1/135 (b) 1/27 (c) 1/54 (d) 1/270

36. The probability that an archer hits the target when it is windy is equal to 2/5, when it is not windy his probability of hitting the target is 7/10. On any shot the probability of gust of wind is 3 . The probability that 10 there is no gust of wind on the occasion when he missed the target, is equal to (a) 5/13 (b) 18/39 (c) 7/13 (d) 23/39

31. Two subsets A and B of a set S consisting of ‘n’ elements are constructed randomly. The probability that A B = f and A B = S is equal to

37. A and B each throw a fair dice. The probability that A’s throw is not greater than B’s throw, is equal to (a) 1/3 (b) 2/3 (c) 7/12 (d) 5/12

MATHEMATICS TODAY | JANUARY ‘18

38. A fair dice is rolled three times. Let E1 be the event that even number comes up on the first roll, E2 be the event that even number appears on second and third roll and E3 be the event that all the roll result in same number. Which of the following statements is not correct? (a) E1 and E2 are not independent (b) E2 and E3 are independent (c) E1 and E3 are independent (d) None of these 1 39. A and B are two events such that P ( A) = , 2 2 P (B) = . Then which of the following is not correct? 3 1 2 (a) P ( A ∪ B) ≥ (b) P ( A ∩ B ′) ≤ 3 3 1 (c) 0 ≤ P ( A ∩ B) ≤ (d) None of these 2 40. Three numbers are selected from the set {1, 2, 3, ......., 23, 24}, without replacement. The probability that the formed numbers form an A.P. is equal to (a) 11/23 (b) 12/23 (c) 3/46 (d) None of these SOLUTIONS 1. (d) : Let the outcome on dice be n1, n2 and n3, we first try to find the number of ways in which sum of outcomes is at the most equal to seven. For this, n1 + n2 + n3 ≤ 7, where ni ∈ [1, 6] ⇒ (n1 – 1) + (n2 – 1) + (n3 – 1) + y4 = 4 i.e., y1 + y2 + y3 + y4 = 4, where y1, y2, y3, y4 are non-negative integers. This can happen in 4 + 4 – 1C4 i.e., 7C4 ways. 7 C 181 Thus, required probability = 1 − 4 = 3 216 6 2. (c) : Clearly, the person should take (n + 1) steps in one direction and n steps in the another direction. Thus, required probability = 2n + 1Cn pn . qn + 1 + 2n + 1Cnqn . pn + 1 = 2n + 1Cn pn . qn(p + q)= 2n + 1Cnpn . qn 3. (c) : Product can’t be divisible by 4 in the following mutually exclusive cases: (i) All the selected numbers are odd : Corresponding number of ways = 13C3 = 286 (ii) Two of the selected numbers are odd and another is a multiple of 2 only. Corresponding number of ways = 13C2 . 6C1 = 468 (286 + 468) 773 = Thus, required probability = 1 − 25 1150 C3

4. (c) : n14 − n24 = (n12 + n22 )(n1 + n2 )(n1 − n2 ) We can segregate the numbers as multiple of five 5l, 5l + 1, 5l + 2, 5l + 3, 5l + 4, as follows : 6 .............5n – 4 → 5l + 1 R1 → 1 R2 → 2 7 .............5n – 3 → 5l + 2 R3 → 3 8 .............5n – 2 → 5l + 3 R4 → 4 9 ............5n – 1 → 5l + 4 R5 → 5 10 ............5n → 5l We can select either both the number from R5, or any two numbers from the first four rows. (As square of any number that is not a multiple of three will be in the form of 5l' + 1 or 5l' – 1). Thus, required probability =

C2 + 4nC2 5n

n(n − 1) + 4n(4n − 1) 17n − 5 = 5n(5n − 1) 5(5n − 1)

5. (b) : P(X = r|X + Y = r + s) P ( X = r ∩ X + Y = r + s) P ( X = r ) . P (Y = s) = = P ( X + Y = r + s) P ( X + Y = r + s) Now, P(X = r) = nCr pr . (1 – p)n – r, P(Y = s) = mCs ps . (1 – p)m – s, n

and, P ( X + Y = r + s) = ∑ P ( X = i) i =0

P(Y = r + s – i) = ∑ n Ci pi (1 − p)n−i . mCr + s −i i =0

. pr + s – i. (1 – p)m – r – s + i

= pr + s ⋅ (1 − p)m+n−r − s ∑ n Ci . mCr + s −i =

pr + s

i =0

. (1 –

p)m + n – r – s

m + nC r+s

n .m C Cs Thus, required probability = m+rn Cr + s

6. (c) : P (B | A ∪ B ) = P (B ∩ ( A ∪ B )) = P ( A ∩ B) P(A ∪ B) P(A ∪ B) Now, P ( A ∩ B ) = P ( A) − P ( A ∩ B)

7 1 1 − = 10 2 5 Also, P ( A ∪ B ) = P ( A) + P (B ) − P ( A ∩ B ) 7 3 1 4 1 = + − = ⇒ P (B | A ∪ B ) = 10 5 2 5 4

⇒ P ( A ∩ B) = P ( A) − P ( A ∩ B ) =

7. (d) : Total number of ways of arranging 6mn objects to equal groups, consisting of mn objects in each groups (6mn)! are numbered as 1, 2, ......., 6 = (mn !)6 MATHEMATICS TODAY | JANUARY ‘18

Corresponding probability for each of these ways 1 =  6

.  1  6

1 ........   6

Thus, required probability =

1 =  6

6mn

(6mn)! .  1    (mn !)6  6 

6 mn

8. (a) : Total ways of selecting two persons = nC2 n(n − 1) = 2 Total ways of selecting two consecutively seated persons = n – 1C1 = (n – 1) (n − 1) 2 n − 2 = Thus, required probability = 1 − n(n − 1) n 9. (a) : E1 : Specific home is locked. E2 : Specific home is unlocked. A : Real estate man get into the home. 40 2 3 = , P (E1 ) = 100 5 5 . P(A) = P(E1) P(A/E1) + P(E2) . P(A/E2) 25 5 3 3 2 = . + .1 = = 5 8 5 40 8 10. (a) : Probability of getting 5 or 6 on a specific roll 2 1 of dice = = 6 3 1 and, probability of getting a tail = 2 The desired outcome can happen, in general, on (2r + 1)th trial. That means first 2r trials should result neither in tail nor in 5 or 6 and (2r + 1)th trial must result in tail. If the corresponding probability is p, then P (E2 ) =

1 2 1 1 1 pr =   .   . = .   2 3 2 2 3 Thus, required probability ∞

= ∑ pr = r =0

1 ∞ 1 3 ∑  = 2 r =0  3  4

11. (a) : Required probability = P(Ei) – P(Ei Ej) + P(E1 E2 E3) We have, P(E1) + P(E2) – 2P(E1 E2) = p P(E2) + P(E3) – 2P(E2 E3) = p P(E1) + P(E3) – 2P(E1 E3) = p 3p ⇒ ΣP (Ei ) − ΣP (Ei ∩ E j ) = 2 3 p + 2 p2 3p Thus, required probability = + p2 = 2 2 54

MATHEMATICS TODAY | JANUARY ‘18

12. (d) : Probability of drawing a white ball at any draw W = W +R Now, P2 can be the winner in general on 2rth, r 1, draw. That means in the first (2r – 1) draws no player draws a white ball and 2rth draw results in a white ball for P2. If the corresponding probability is pr, then  R  p = r  W + R 

 R  .  W + R 

r −1

 W   R  . =  W + R   W + R  ∞

Hence, required probability = ∑ pr = ∞

= ∑ pr = r =1

∞

W  R  ∑  R r =1  W + R 

r =1

W R

W ∞ R  ∑  R r =1  W + R 

R (W + 2R)

13. (b) : E1 : First bag is chosen, P(E1) =

1 2

1 E2 : Second bag is chosen, P(E2) = 2 A : Drawn number is 4. Now, P(A) = P(E1) . P(A/E1) + P(E2) . P(A/E2) 1 1 1 1 5 = . + . = 2 4 2 6 24 14. (a) : We know that 74l ends with 1, 74l + 1 ends with 7, 74l + 2 ends with 9, and 74l + 3 ends with 3. Now, 7l1 + 7l2 will be divisible by 5, if it ends with either 0 or 5. But it can never end with 5 and can end with zero in the two cases : (i) n1 = 4l, n2 = 4l + 2 (ii) n1 = 4l + 1, n2 = 4l + 3 Thus, total favourable ways of selecting n1 and n2. = 25C1 . 25C1 + 25C1 . 25C1 = 2 . 25 . 25 2 . 25 . 25 1 Thus, required probability = = 100 .100 8 15. (d) : Let E1 : Statement is true E2 : Statement is false M : A and B agree 1 P(E1) = P(E2) = 2 P(M) = P(E1) . P(M/E1) + P(E2) . P(M/E2) 1 . 1 . = P1P 2 + (1 – P1) (1 – P2) 2 2 P P + (1 − P1 )(1 − P2 ) = 1 2 2

R (W + 2R)

Now, P (E1 / M ) =

P ( A ∩ E1 )

P (E1 ) . P ( A / E1 )

P (M ) P (M ) P1P2 P1P2 = = P1P2 + (1 − P1 )(1 − P2 ) 1 − P1 − P2 + 2P1P2

2 1 2 1 2 1 16. (d) : P (E1 ) = = , P (E2 ) = = , P (E3 ) = = 4 2 4 2 4 2 1 P (E1 ∩ E2 ) = = P (E1 ) ⋅ P (E2 ) 4 ⇒ ‘E1’ and ‘E2’ are independent. 1 P (E1 ∩ E3 ) = = P (E1 ) ⋅ P (E3 ) 4 ⇒ ‘E1’ and ‘E3’ are independent. 1 P (E2 ∩ E3 ) = = P (E2 ) ⋅ P (E3 ) 4 ⇒ ‘E2’ and ‘E3’ are independent. 1 P (E1 ∩ E2 ∩ E3 ) = ≠ P (E1 ) ⋅ P (E2 ) ⋅ P (E3 ) 4 ⇒ E1, E2 and E3 are not independent. 17. (c) : E1 : Ball drawn from B1 is white. E2 : Ball drawn from B1 is red. A : Both the drawn balls are of same colour. P(A) = P(E1) . P(A/E1) + P(E2) . P(A/E2) 3 5 2 7 29 = . + . = 5 11 5 11 55 100 18. (c) : n + ≤ 50 ⇒ n2 – 50n + 100 ≤ 0 n

20. (c) : Let’s denote the probability of choosing the integer n by pn, then λ pn = 4 , n ∈[1, 2m] n 2m 2m 1 1 Now, ∑ pn = λ ∑ = 1 ⇒ λ = 2m 4 1 n=1 n=1 n

∑

4 n=1 n

Now, required probability m

=∑p n=1

Now,

2n−1

= λ. ∑

4 n=1 (2n − 1)

∑ (2n − 1)4

= n=12m

∑ n4

n=1

∑ n4 = ∑ (2n)4 + ∑ (2n − 1)4 < 2 ∑ (2n − 1)4

n=1

1 Thus, required probability > . 2 21. (d) : Total number of ways of selecting the second ball = 19C1 = 19. Total number of ways of selecting the second ball red = 14C1 = 14 14 Thus, required probability = 19

⇒ 25 − 5 21 ≤ n ≤ 25 + 5 21 ⇒ n = 3, 4, 5, .........., 47 Thus, favourable number of ways = 45. 45 9 Thus, required probability = = 100 20 1 19. (c) : P ( A) = P ( A / B) = 4 ⇒ A and B are independent. 1 Hence, P (B / A) = P (B) = 2 Now, A′, B; A′, B′ will also independent. 3 Thus, P ( A ′ / B) = P ( A ′) = 1 − P ( A) = 4 1 P ( B ′ / A ′) = P ( B ′) = 1 − P ( B) = 2 P(A B) = P(A) + P(B) – P(A B) 1 1 1 1 5 = + − . = 4 2 4 2 8 1 P ( A ∩ B) = P ( A) ⋅ P (B) = 8 MATHEMATICS TODAY | JANUARY ‘18

22. (a) : Let the first chosen square is white. In this case total number of squares that don’t belong to the row or column associated with the selected square is 49. Out of these 25 are white and 24 are black. 1 25 Thus, corresponding probability = . 2 49 Similarly, if the first square is black, then the 1 25 corresponding probability = . 2 49 1  25 25  25 Thus, required probability =  +  = 2  49 49  49 23. (a) : Following equally likely outcomes may occur when one of the dice show four {(4, 1), (1, 4), (4, 2), (2, 4), (4, 3), (3, 4), (4, 4), (4, 5), (5, 4), (4, 6), (6, 4)} Out of these eleven outcomes exactly 2 outcomes favor the cause of second dice showing a six. 2 Thus, required probability = 11 24. (b) : In the case end digit of the product should be 5. Now, if end digit of each number is either 1, 3, 5, 7 or 9 then product of these numbers will end with either 1, 3, 5, 7 or 9. On the other hand, end digit of the product will be 1, 3, 7, or 9, if end digit of each integer is either 1, 3, 7 or 9. Thus, total ways in which the product ends with 5 is 54 – 44, also total ways of selecting the end digit of the number are 10. 369 Thus, required probability = 104 P ( A ∩ B ) 1 − P ( A ∪ B) 25. (c) : P ( A / B ) = = P (B ) P (B ) 26. (a) : Person will laugh if the selected number is (43, 34, 62, 26). Thus, probability that person will not laugh on seeing 86 43 the number = = 90 45 3  43  = 1 − Thus, required probability  45  27. (d) : Total number of ways in which n1 + n2 = 100 is equal to 99. Now, n1 . n2 > 1600 ⇒ n1(100 – n1) > 1600 ⇒ n12 – 100n1 + 1600 < 0 ⇒ (n1 – 80) (n1 – 20) < 0 ⇒ 20 < n1 < 80 ⇒ 21 ≤ n1 ≤ 79 Thus, number of favourable ways = 79 – 21 + 1 = 59 59 Hence, required probability = 99 56

MATHEMATICS TODAY | JANUARY ‘18

28. (b) : Roots will be non real, if b2 < 4ac. Now, if ‘b’ is one then 4ac is definitely greater than one (that is ‘a’ and ‘c’ can be chosen in 4 ways). If b = 2 and b2 < 4ac, then ‘a’ and ‘c’ both at the same times should not be equal to one. That is for b = 2, we must have a = 1, c = 2 or a = 2, c = 1 or a = 2, c = 2 (i.e., ‘a’ and ‘c’ can be chosen in three ways) 7 Thus, required probability = . 8 29. (c) : There are three mutually exclusive cases : (i) 2 are from A and another is either from B or C. 2 . 3 C ( C + 4C1 ) 7 = Corresponding probability = 2 9 1 84 C3 (ii) 2 are from B and another is either from A or C. 3 . 2 C ( C + 4C1 ) 18 = Corresponding probability = 2 9 1 84 C3 (iii) 2 are from C and another is either from A or B. Corresponding probability =

C2 .(2 C1 + 3C1 ) 9

30 84

Thus, required probability =

7 + 18 + 30 55 = 84 84

30. (d) : Total number of ways of dialing the last three digits such that exactly 2 of them are same 3! = 10C2 ⋅ 2 ⋅ = 270 2! 1 Thus, required probability = 270 31. (c) : Let ‘A’ has ‘r’ elements, then ‘B’ have all the remaining (n – r) elements and none of the elements that are already present in A. n

Thus, total number of favourable ways = ∑ n Cr = 2n 2n

r =0

Hence, required probability = n = n 4 2

MPP-9 CLASS XII 1. (b) 6. (c) 11. (b,d) 16. (a)

2. 7. 12. 17.

(a) (c) (a, b) (7)

3. 8. 13. 18.

ANSWER KEY

(a) (a,c) (a,c,d) (3)

4. 9. 14. 19.

5. 10. 15. 20.

32. (b) : Total number of functions = 53 Now let us count the favourable number of functions. Let ‘1’ is assigned to r, where r = 1, 2, 3, 4, 5. Now ‘2’ can be assigned to r, r + 1, ...... 5. Let ‘2’ is assigned to j, where j = r, r + 1, .... 5. Finally, ‘3’ can be assigned in (6 – j) ways. Thus, total number of favourable ways 5  5  5 (6 − r )(7 − r ) = ∑  ∑ (6 − j)  = ∑   r =1 2 r =1  j =r 1 5 ∑ (42 − 13r + r2 ) = 35 2 r =1 35 7 Thus, required probability = 3 = 25 5 33. (b) : Total outcomes = 2n Total number of favourable outcomes = nC0 + nC2 + nC4 + ....... nCn/2 = 2n – 1 =

Thus, required probability = 34. (c) : P (n1 ≤ r / n2 ≤ r ) = =

C2 / C2 r − 1 = r /n n −1

2n−1 n

1 2

2 P (n1 ≤ r ∩ n2 ≤ r ) P (n2 ≤ r )

35. (a) : Probability of winning the prize in single game 25 1 = = 2 25 25 In this case first 4 games, must result in exactly two prizes and 5th game must result in prize. Thus, required probability 2 2  1   24  1 6.(24)2 = 4C2   .   . =  25   25  25 (25)5 36. (c) : E1 : There is a gust of wind. E2 : There is no gust of wind. A : Archer misses the target. P(A) = P(E1) . P(A/E1) + P(E2) . P(A/E2) 3 3 7 3 39 = . + . = 10 5 10 10 100 P (E2 ∩ A) Now, required probability = P (E2 / A) = P ( A) 7.3 P (E2 ) . P ( A / E2 ) 10 10 7 = = = 39 P ( A) 13 100 37. (c) : Let A’s throw is r, (r = 1, 2, 3, ......, 6), then B’s outcome can be (r, r + 1, ......, 6)

7−r 6 r 7 =∑ = 36 36 12 r =1 r =1 3 1 3 3 38. (a) : P (E1 ) = = ; P (E2 ) = . = 1 ; 6 2 6 6 4 6 1 P (E3 ) = = 63 36 Now, P(E1 E2) = Probability of getting even number 1 1 1 1 on each throw = . . = = P(E1) . P(E2) 2 2 2 8 P(E1 E3) = Probability of getting even number on 1 1 1 1 each throw = . . = ≠ P (E1 ) ⋅ P (E3 ) 2 2 2 8 P(E2 E3) = Probability of getting even number on 1 1 1 1 each throw = . . = ≠ P (E2 ) ⋅ P (E3 ) 2 2 2 8 2 39. (d) : P(A B) maximum {P(A), P(B)} = 3 1 P(A B′) ≤ minimum {P(A), P(B′)} = 3 P(A B) 0 1 and P(A B) ≤ minimum {P(A), P(B)} = 2 Thus, all the statements are correct. 40. (c) : Let the selected numbers be n1, n2 and n3. We must have 2n2 = n1 + n3. Thus, n1 + n3 must be even. That means n1 and n3 both must have same nature (either odd or even). 2 . 12C 3 Thus, required probability = 24 2 = 46 C3  Thus, required probability = ∑

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CLASS XII

Series 9

CBSE Linear Programming and Probability ImportantFFormulae IMPORTANT ORMULAE LINEAR PROGRAMMING

TERMINOLOGIES RELATED TO L.P.P. Term

Definition

The restrictions or linear inequalities or equations in the variables of an Constraints L.P.P., which describe the conditions under which the optimisation is to be accomplished. The assumption that negative values Nonof variables are not possible in the negativity solution. They are descr ibed as Constraints x ≥ 0, y ≥ 0 Objective Function

The linear function which is to be maximised or minimised under given constraints.

Feasible Region and Feasible Solution

The common region determined by all the constraints including non-negative constraints of a L.P.P. is called the feasible region or solution region. Feasible region is always a convex set. Feasible region may be bounded or unbounded.

Infeasible Region and Infeasible Solution

The region other than the feasible region is called infeasible region of the L.P.P. and the points which come under infeasible region are called infeasible solution.

FORMULATION OF A L.P.P. X The three steps in the mathematical formulation of an L.P.P. are as follows: (i) Identify the objective function as a linear combination of variables (x and y) and construct all constraints i.e.linear equations and inequations involving these variables. Thus, an L.P.P. can be stated mathematically as Maximise (or minimise) z = ax + by Subject to the constraints: aix + biy ≤ (or or = or > or <) ci, where i = 1 n, x 0, y 0 (non-negative constraints). (ii) Find the solutions (feasible region) of these equations and inequations by some mathematical method. (iii) Find an optimal solution i.e. select particular values of the variables x and y that give the desired value (maximum/minimum) of the objective function.

PROBABILITY

CONDITIONAL PROBABILITY Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A under the condition that B has already occurred

MATHEMATICS TODAY | JANUARY ‘18

and P(B) ≠ 0, is called the conditional probability. P ( A ∩ B) i.e., P(A/B) = P ( B)

If A and B are two events associated with a random experiment, then P(A B) = P(A) ⋅ P(B/A), if P(A) ≠ 0 P(A B) = P(B) ⋅ P(A/B), if P(B) ≠ 0 Events are said to be independent, if the occurrence or non-occurrence of one does not affect the probability of the occurrence of the other. i.e., P(A/B) = P(A), P(B) ≠ 0; P(B/A) = P(B), P(A) ≠ 0 Two events A and B are independent events associated with a random experiment, if P(A B) = P(A) ⋅ P(B) If A1, A2, A3, .... An, are n independent events, then P(A1 A2 ...... An) = 1 – P( A1) ⋅ P( A2 )......P( An ) If A is any event which occurs with E1 or E2 ....or En, then P(A) = P(E1) P(A/E1) + P(E2) P(A/E2) + .... + P(En) P(A/En) Baye's theorem : If A is any event which occurs with E1 or E2 .... or En, then WORK IT OUT

VERY SHORT ANSWER TYPE 1. If a die is tossed, then the sample space S = {1, 2, 3, 4, 5, 6}. Let A = {1, 3, 5}, B = {2, 4, 6} and C = {2, 3, 5}. Find : (i) P(A/C) (ii) P(C/B). 2. Find the maximum value of z = 4x + 3y subject to x + y ≤ 4, x, y 0. 3. If P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6. Find P(A/B) and P(A B). 4. If A and B are mutually exclusive events, find P(A/B). 5. X is taking up subjects-Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets grade A in all subjects. SHORT ANSWER TYPE 6. Consider z(x, y) = px + qy subject to 2x + y ≤ 0, x + 3y ≤ 15, x, y 0. If z is maximum at both the points (3, 4) and (0, 5), then find q in terms of p. 7. A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. If the second selected ball happens to be red, what is the probability that the first selected ball is also red?

P(Ei/A) =

P (Ei ) ⋅ P ( A / Ei )

∑ P(Ei ) ⋅ P(A / Ei )

, i = 1, 2, ...., n

i =1

A real valued function X, defined on a sample space S, of a random experiment i.e., X : S → R is called a random variable. Binomial Distribution : The probability of 'r' success in n trials is nCr ⋅ pr ⋅ qn–r, where p is the probability of success and 'q' is the probability of failure in a single trial. Mean of Binomial Distribution = np Variance = npq, standard deviation = npq Let X be a random variable whose possible values x1, x2, x3, ..., xn occur with probabilities p1, p2, p3, ..., pn respectively. Then we define expectation of n

X, denoted by E(X) and is given by E(X) = ∑ xi pi . i =1 Var (X) = E(X2) – [E(X)]2

8. If A and B are independent events, prove that P(A B) · P(A′ B′) ≤ P(C), where C is an event defined that exactly one of A and B occurs. 9. Find the minimum value of z = 2x +y subject to : x + 2y 3, 2x + y 6, x, y 0. 10. The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P(X = 1) = p and E(X 2) = E(X), find the value of p. LONG ANSWER TYPE-I 11. Find the maximum value of z = x + y subject to x + y ≤ 3, –2x + y ≤ 1, x ≤ 2, x, y 0. 12. The ratio of the number of boys to the number of girls in a class is 1 : 2. It is known that the probabilities of a girl and a boy getting a first division are 0.25 and 0.28 respectively. Find the probability that student chosen at random will get first division. 13. A brick manufacturer has two depots A and B, with stocks of 30000 and 20000 bricks respectively. He receives orders from three builders P, Q and R for 15000, 20000 and 15000 bricks respectively. The cost (in `) of transporting 1000 bricks to the builders from the depots are as given in the table : To Transportation cost per 1000 bricks (in `) P Q R From 40 20 20 A 20 60 40 B MATHEMATICS TODAY | JANUARY ‘18

The manufacturer wishes to fulfil the order so that transportation cost is minimum. Formulate the LPP. 14. There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3, then a ball is drawn from the first box but if it shows a number more than 3 then a ball is drawn from the second box. If the ball drawn is black, then find the probability that the ball was drawn from the first box. 15. A bag X contains 3 white and 2 black balls; another bag Y contains 2 white and 4 black balls. A bag and a ball out of it is picked at random. What is the probability that the ball is white? LONG ANSWER TYPE-II 16. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of proteins and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs ` 5 per kilogram and rice costs ` 4 per kilogram. 17. The sum of mean and variance of a binomial distribution is 15 and the sum of their squares is 117. Determine the distribution. 18. A can hit a target 4 times in 5 shots, B 3 times in 4 shots, and C 2 times in 3 shots. Calculate the probability that (i) B, C may hit and A may not. (ii) Any two of A, B and C will hit the target (iii) none of them will hit the target 19. A manufacturer makes two products, A and B. Product A sells at ` 200 per unit and takes 30 minutes to make. Product B sells at ` 300 per unit and takes 1 hour to make. There is a permanent order of 14 units of product A and 16 units of product B. A working week consists of 40 hours of production and the weekly turnover must not be less than ` 10000. If the profit on each of product A is ` 20 and on product B is ` 30, then how many product A and B respectively should be produced so that the profit is maximum? Also, find the maximum profit. 60

MATHEMATICS TODAY | JANUARY ‘18

20. Sixteen players P1, P2, ..., P16 play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pairs. Assuming that all the players are of equal strength, find the probability that exactly one of the two players P1 and P2 is among the eight winners. SOLUTIONS

P ( A ∩ C ) P {3, 5} 2 2 2 = ⋅ = = 6 1 3 P (C ) 1/ 2 P (C ∩ B) P {2} 1 2 1 (ii) P(C/B) = = = ⋅ = P ( B) 1/ 2 6 1 3

1. (i) P(A/C) =

2. The corners of the feasible region are O(0, 0), A(4, 0), B(0, 4).  z = 4x + 3y ⇒ z(O) = 0, z(A) = 16 z(B) = 12 \ Maximum value is z = 16 at A. 3. P(A

B) = P(A) · P(B/A) = (0.4)(0.6) = 0.24

Now, P ( A / B) = P(A

P ( A ∩ B) 0.24 = = 0. 3 P ( B) 0. 8

B) = P(A) + P(B) – P(A B) = 0.4 + 0.8 – 0.24 = 0.96

4. Since A and B are mutually exclusive events, \ A B = f. P ( A ∩ B) P (φ) = =0 Hence, P ( A / B) = P ( B) P ( B) 5. Let M, P ′ and C denote the events that X gets grade A in Mathematics, Physics and Chemistry respectively. Given, P(M) = 0.2, P(P ′) = 0.3, P(C) = 0.5 \ P(grade A in all subjects) = P(M P ′ C) = P(M) · P(P ′) · P(C) = 0.2 × 0.3 × 0.5 = 0.03 6. The feasible region is shown in the graph.

Since z = px + qy

\ z(O) = 0, z(A) = 5p, z(E) = 3p + 4q, z(D) = 5q Also, z(E) = z(D) ⇒ 3p + 4q = 5q \ q = 3p. 7. Let A be the event of drawing a red ball in first draw and B be the event of drawing a red ball in second draw. 3 C 3 \ P ( A) = 10 1 = C1 10 Now, P(B/A) = Probability of drawing a red ball in the second draw, when a red ball already has been 2 C 2 drawn in the first draw = 9 1 = C1 9 \ Required probability = P(A B) = P(A)⋅P(B/A) 3 2 1 = × = 10 9 15 8. L.H.S. = P(A B) ⋅ P(A′ B′) = P(A B) P(A ) P(B ) ≤ (P(A) + P(B)) P(A ) P(B ) = P(A) P(A ) P(B ) + P(B) P(A ) P(B ) ≤ P(A)P(B ) + P(B) P(A ) [since P(A ) ≤ 1, P(B ) ≤ 1] = P(A B ) + P(A B) = P(C) = R.H.S. 9. The feasible region is unbounded. \ z(A) = 2 × 3 = 6 z(C) = 6 \ Minimum value of z is 6 on the line 2x + y = 6. Maximum value of z does not exist. 10. Clearly, P(X = 0) + P(X = 1) + P(X = 2) = 1 ⇒ p + p + P(X = 2) = 1 ⇒ P(X = 2) = 1 – 2p. So, the probability distribution of X is as given below : xi pi

2 1 – 2p

\ E(X) = (0 × p) + (1 × p) + 2 (1 – 2p) = 2 – 3p and E(X2) = 02 × p + 12 × p + 22(1 – 2p) = 4 – 7p 1 Now, E(X2) = E(X) ⇒ 4 – 7p = 2 – 3p ⇒ p = . 2 11. The corners points of feasible region are O(0, 0), A(2, 0), 2 7 B(2, 1), C  ,  , 3 3 D(0, 1). Now, z = x + y ⇒ z(A) = 2, z(B) = 3, z(C) = 3, z(D) = 1, z(O) = 0 \ Maximum z = 3 along line BC.

12. Let E1 = event that a student selected at random is a girl and E2 = event that a student selected at random is a boy. E = event that a student selected at random will get first division Given, Number of boys : Number of girls = 1 : 2 2 1 \ P (E1 ) = and P (E2 ) = 3 3 By rule of total probability P(E) = P(E1) · P(E/E1) + P(E2) · P(E/E2) 2 1 0.78 = (0.25) + (0.28) = = 0.26 3 3 3 13. The given information can be expressed as given in the diagram. We assume that 1 unit = 1000 bricks

Suppose that depot A supplies x units to P and y units to Q, so that depot A supplies (30 – x – y) units of bricks to builder R. Now as P requires a total of 15000 bricks, it requires (15 – x) units from depot B. Similarly Q requires (20 – y) units from B and R requires 15 – (30 – x – y) = x + y – 15 units from B. Total transportation cost, Z = 40x + 20y + 20(30 – x – y) + 20(15 – x) + 60(20 – y) + 40(x + y – 15) = 40x – 20y + 1500 Obviously the constraints are that all quantities of bricks supplied from A and B to P, Q, R are non-negative. \ x ≥ 0, y ≥ 0, 30 – x – y ≥ 0, 15 – x ≥ 0, 20 – y ≥ 0, x + y – 15 ≥ 0 Since, 1500 is a constant, hence instead of minimizing Z = 40x – 20y + 1500, we minimize Z = 40x – 20y. Hence, mathematical formulation of the given LPP is Minimize Z = 40x – 20y, subject to the constraints : x + y ≥ 15, x + y ≤ 30, x ≤ 15, y ≤ 20, x ≥ 0, y ≥ 0 MATHEMATICS TODAY | JANUARY ‘18

14. Let E1 = Event of drawing a ball from the first box. E2 = Event of drawing a ball from the second box. B = Event of drawing a black ball. 3 1 3 1 \ P(E1) = = and P (E2 ) = = 6 2 6 2 Clearly, E1 and E2 are mutually exclusive and exhaustive.  B  4 B 3 Now, P   = and P  = .  E1  7 E2  7 By Bayes' theorem, required probability B P (E1 ) ⋅ P    E1  E  =P 1= B B B P (E1 ) ⋅ P   + P (E2 ) ⋅ P    E1   E2  1 3 ⋅ 3 2 7 = = 1 3 1 4 7 ⋅ + . 2 7 2 7 15. Let E1 = the event of selecting bag X E2 = the event of selecting bag Y E = the event of drawing a white ball 1 1 \ P(E1) = , P(E2) = 2 2 Let A = E1 E and B = E2 E \ Required probability, P(E) = P(A B) = P(A) + P(B) [Since events A and B are mutually exclusive] = P(E1 E) + P(E2 E) = P(E1) · P(E/E1) + P(E2) · P(E/E2) 1 3 1 2 7 = ⋅ + ⋅ = 2 5 2 6 15 16. Let the cereal contain x kg of bran and y kg of rice. Let z be the cost of the new cereal. According to question, 80 100 88 or 20x + 25y ≥ 22 x× +y× ≥ 1000 1000 1000 40 30 36 x× + y× ≥ or 20x + 15y ≥ 18 1000000 1000000 1000000 x ≥ 0, y ≥ 0

MATHEMATICS TODAY | JANUARY ‘18

Mathematical formulation of the LPP is : Minimize z = 5x + 4y subject to constraints : 20x + 25y 22, 20x + 15y 18, x 0, y 0 The corner points of unbounded feasible region as A(1.1, 0), E(0.6, 0.4) and D(0, 1.2) At A(1.1, 0) : z = 5 × 1.1 + 4 × 0 = 5.5 At E(0.6, 0.4) : z = 5 × 0.6 + 4 × 0.4 = 4.6 At D(0, 1.2) : z = 5 × 0 + 4 × 1.2 = 4.8 \ The minimum cost of producing this cereal is ` 4.6 17. Let n and p be the parameters of the distribution. \ Mean = np and Variance = npq. Given, np + npq = 15 and n2p2 + n2p2q2 = 117 ⇒ np(1 + q) = 15 and n2p2(1 + q2) = 117 ⇒ n2p2(1 + q)2 = 225 and n2p2(1 + q2) = 117 ⇒

n2 p2 (1 + q)2 2 2

225 117

n p (1 + q ) 1 + q2 + 2q 225 = ⇒ 117 1 + q2 ⇒

1 + q2 13 = 2q 12 2

1+ q ⇒  = 25  1 − q 

⇒

(1 + q)2 2

225 117

(1 + q ) 2q 225 = ⇒ 1+ 2 117 1+ q 1 + q2 + 2q 2

1 + q − 2q

⇒

13 + 12 13 − 12

1+ q = 5 ⇒ 6q = 4 1− q

2 3

2 1 \ p = 1 – q = 1− = 3 3 1 2 Putting p = , q = in np + npq = 15, we get 3 3 n 2n 5n + = 15 ⇒ = 15 ⇒ n = 27 3 9 9 1 2 Thus, n = 27, p = and q = 3 3 Hence, the distribution is given by ⇒ q=

P(X = r) =

 1  2 Cr      3  3

27 −r

, r = 0, 1, 2, ..., 27.

18. Consider the following events : E = A hits the target, F = B hits the target and G = C hits the target 4 3 2 We have, P(E) = , P(F) = and P(G) = 5 4 3 (i) P(B, C may hit and A may not) = P(E F G) = P(E) P(F) P (G) [ E, F, G are independent events]  4 3 2 1 = 1 −  × × =  5  4 3 10

MATHEMATICS TODAY | JANUARY â&#x20AC;&#x2DC;18

(ii) P (Any two of A, B and C will hit the target) = P(E F G) (E F G) (E F G)) = P(E F G) + P(E F G) + P (E F G)) = P(E) P(F) P(G) + P(E) P(F) P(G) + P(E) P(F) P(G)  4 3 1   1 3 2   4 1 2  13 =  × ×  + × ×  + × ×  =  5 4 3   5 4 3   5 4 3  30 (iii) P(None of A, B and C will hit the target) 1 1 1 1 = P(E F G) = × × = 5 4 3 60 19. Let x be the number of units of the product A and y be the number of units of the product B produced. Let z be the total profit. \ z = 20x + 30y x and y must satisfy the following conditions 200x + 300y ≥ 10000 or 2x + 3y ≥ 100 1 x + 1⋅y ≤ 40 or x + 2y ≤ 80 2 x ≥ 14, y ≥ 16, x ≥ 0, y ≥ 0 The mathematical formulation of the LPP will be Maximize z = 20x + 30y subject to constraints : 2x + 3y ≥ 100, x + 2y ≤ 80, x ≥ 14, y ≥ 16, x ≥ 0, y ≥ 0 On plotting, we get S(26, 16), P(48, 16), Q(14, 33) and R(14, 24) as corners of feasible region.

 Maximize z = 20x + 30y \ At S(26, 16), z = 1000 At P(48, 16), z = 1440 At Q(14, 33), z = 1270 At R(14, 24), z = 1000 Hence, the maximum profit is `1440 and it is attained when 48 units of product A and 16 units of product B are produced. 20. Let E1 and E2 denote the event that P1 and P2 are paired and not paired together and let A denote the event that one of two players P1 and P2 is among the winners. 64

MATHEMATICS TODAY | JANUARY ‘18

Since P1 can be paired with any of the remaining 15 players. We have, P(E1) =

1 15

and P(E2) = 1 – P(E1) = 1 −

1 14 = . 15 15

In case E1 occurs, it is certain that one of P1 and P2 will be among the winners. Let C and D be the events that P1 and P2 wins respectively. In case E2 occurs, the probability that exactly one of P1 and P2 is among the winners is P{(C

D)} = P(C) P(D) + P(C) P(D)

1 1  11 1 1 1 =   1 −  + 1 −    = + = 2 2  22 4 4 2 i.e., P (A/E1) = 1 and P (A/E2) = 1 2

By the total probability rule, 







P(A) = P (E1) · P  A  + P( E2 )·P  A   E1   E2  =

14  1  8 1 = (1) + 15 15  2  15



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Class XII

Vector Algebra and Three Dimensional Geometry Total Marks : 80

Time Taken : 60 Min.

Only One Option Correct Type

One or More Than One Option(s) Correct Type

1. A unit tangent vector at t = 2 on the curve x = t2 + 2, y = 4t – 5, z = 2t2 – 6t is 1 ˆ ˆ ˆ (a) (b) 1 (2iˆ + 2 jˆ + kˆ ) (i + j + k ) 3 3 1 ˆ ˆ ˆ (c) (d) none of these (2i + j + k ) 6 2. The projection of a line segment on the coordinates axes are 2, 3, 6. Then, the length of the line segment is (a) 7 (b) 5 (c) 1 (d) 11

7. Let A1, A2,..., An (n > 2) be the vertices of a regular polygon of n sides with its centre at the origin.  Let ak be the position vector of the point Ak,

3. The angle between a line with direction ratios proportional to 2, 2, 1 and a line joining (3, 1, 4) to (7, 2, 12), is −1  2  −1  2  (b) cos  −  (a) cos   3 3 2 (c) tan −1   3

(d) none of these

4. Let a , b , c be three non-coplanar vectors and d be a non-zero vector which is perpendicular to a + b + c . If d = x b × c + y c × a + z a × b , then (b) x = y = z (a) xy + yz + zx = 0 (c) x3 + y3 + z3 = 3xyz (d) x + y + z = 1 5. The line through the points (5, 1, a) and (3, b, 1) 

17 −13 

crosses yz plane at the point  0, , . Then  2 2  (a) a = 2, b = 8 (b) a = 4, b = 6 (c) a = 6, b = 4 (d) a = 8, b = 2    2  2 6. If (a × b ) + (a ⋅ b ) = 144 and | a | = 4, then | b | is equal to (a) 16 (b) 8 (c) 3 (d) 12

k = 1, 2,..., n. If

n−1



n−1

 

∑ (ak × ak+1 ) = ∑ (ak ⋅ ak+1 ) , then k =1

the minimum value of n is (a) 1 (b) 2 (c) 8

k =1

(d) 9

8. Let P1 denote the equation of the plane to which the vector (jˆ + iˆ ) is normal and which contains the line L whose equation is r = i + j + k + λ( i − j − k ). Let P2 denote the equation of the plane containing the

line L and a point with position vectors j . Which of the following holds goods? (a) Equation of P1 is x + y = 2 (b) Equation of P2 is r ⋅ ( i − 2 j + k ) = 2.

tan −1 3 . 9. If a = iˆ + jˆ + kˆ and b = iˆ − jˆ , then the vectors (a ⋅ iˆ )iˆ + (a ⋅ jˆ )jˆ + (a ⋅ kˆ )kˆ , (b ⋅ iˆ )iˆ + (b ⋅ jˆ )jˆ + (b ⋅ kˆ )kˆ , and iˆ + jˆ − 2kˆ (a) (b) (c) (d)

are mutually perpendicular are coplanar form a parallelopiped of volume 6 units form a parallelopiped of volume 3 units x −1 y +1 z − 3 = = on 10. The projection of the line 2 −1 4 the plane x+ 2y + z = 9 is the line of intersection of which of the following planes? MATHEMATICS TODAY | JANUARY ‘18

(a) x + 2y + z – 9 = 0 (b) 9x + 2y + 3z – 4 = 0 (c) x – 2y + z + 3 = 0 (d) 9x – 2y – 5z + 4 = 0 A1 11. The volume of a right triangular prism ABCA1B1C1 is equal to 3. B1 C1 Find the co-ordinates of the vertex A1, if the co-ordinates of the base A vertices of the prism are A(1, 0, 1), B C B(2, 0, 0) and C(0, 1, 0). (a) (–2, 0, 2) (b) (0, –2, 0) (c) (0, 2, 0) (d) (2, 2, 2)

Matrix Match Type 16. Match the following. Column I

The number of independent P. constants in the equation of a 1. 2 straight line in 3 dimensions is A plane passes through (1, –2, 1) and is perpendicular to two planes 2x – 2y + z = 0, x – y + 2z = 4. 2. 9 Q. If the distance of (1, 2, 2) from it, is k 2 units. Then k equals

12. The equations of two lines through the origin which x −3 y −3 z = = at an angle 60° is intersect the line 2 1 1 x y z x y z = (a) (b) = = = 1 −1 2 1 2 −1 x y z x y z = = (c) (d) = = 2 1 −1 2 −1 1 13. The line of intersection of the planes r ⋅ ( i + 2 j + 3k ) = 0 and r ⋅ (3 i + 2 j + k ) = 0 is (a) equally inclined to i and kˆ (b) equally inclined to i and jˆ

x − 1 y +1 z − 1 = = If the lines 2 3 4 x−3 y−k z R. = = and 1 2 1 are coplanar, then 2k is (a) (b) (c) (d)

2 with jˆ 3  −1  with iˆ (d) inclined at an angle cos −1   6  (c) inclined at an angle cos −1

Comprehension Type If a , b and c be any three non-coplanar vectors. T h e n t h e s y s t e m a ′, b ′ and c ′ w h i c h s a t i s f i e s a ⋅ a′ = b ⋅ b ′ = c ⋅ c ′ = 1 and a ⋅ b ′ = a ⋅ c ′ = b ⋅ a ′ = b ⋅ c ′ = c ⋅ a ′ = c ⋅ b ′ = 0 is called the reciprocal system to the vectors a , b and c . 14. The value of (a × a ′) + (b × b ′) + (c × c ′) (b) = a ′ + b ′ + c ′ (a) = a + b + c (c) ≠ 0 (d) = 0

Column II

P 3 3 1 2

Q 1 2 2 1

3. 4

R 2 1 3 3

Integer Answer Type 17. The distance of the point iˆ + 2 jˆ + 3kˆ from the plane r ⋅ (iˆ + jˆ + kˆ ) = 5 measured parallel to the vectors 2iˆ + 3 jˆ − 6kˆ must be 18. The distance of the point (3, 0, 5) from the line x – 2y + 2z – 4 = 0 = x + 3z – 11 is 19. If the lines x – y + z = –1 = 3x – y + az + 1 and x + 4y – z – 1 = 0 = x + 2y + bz + 1 are perpendicular, then 3ab + a – b + c = 0, where c is 20. Let p = 2 i − j + k , q = i + 2 j − k and r = i + j − 2k .

15. The value of [a ′ b ′ c ′]−1 (b) = [a b c ] (a) < [a b c ] (d) = 0 (c) > [a b c ]

If V = p + λq and projection of V on r is value of l is equal to

, then 

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MATHEMATICS TODAY | JANUARY ‘18

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

Vectors and 3D Geometry 6.

PROBLEMS Single Correct Answer Type

divides the line joining the points −2 i + 4 j + 7 k ^

and 3 i − 5 j + 8 k is (a) 3 : 5 (b) 1 : 5 2.

[a b c ]

(c)

(a)

(d) 3 : 10

The length of the perpendicular from origin to the plane passing through a point c and containing the line r = b + µa is [c b a] [a b c ] (a) (b) |b × a + a × c | |b × c + c × a |

[a b c ]

(d)

r ⋅(α k + β i ) = 0 and r ⋅ (β i + γ j + α k) = λ, then value of tetrahedron (in cubic units) (if b, y, a, l each > 0) is

The ratio in which the plane r ⋅ (i^ − 2 j^+ 3 k^) = 17 ^

If the four faces of a tetrahedron are represented ^ ^ ^ ^ by the equations r ⋅ (β i + γ j ) = 0, r ⋅ ( γ j + α k ) = 0,

1 λ3 6 (αβy )3 λ3 3αβy

(b)

λ3 αβy

(d)

λ3 6αβγ

x − 2 y −1 z + 2 lie in the plane = = 3 −5 2 x + 3y – az + b = 0, then (a, b) equals (a) (–6, 7) (b) (–5, 5) (c) (6, –7) (d) (–6, –7)

Let the line

The distance of the point (1, –2, 3) from the plane x – y + z – 5 = 0 measured parallel to the line 3. ^ ^ ^ ^ x^= y ^= z is of the point P(1, 2, 3) on the line (6 i + 7 j + 7 k ) + λ(3 i + 2 j − 2 k) 2 3 −6 ^ ^ ^ ^ ^ ^ (6 i + 7 j + 7 k ) + λ(3 i + 2 j − 2 k) is (a) 4 (b) 3 (c) 2 (d) 1

|a × b + b × c | |a ×b +b ×c + c ×a | The position vector of foot of the perpendicular

(b) 3 i + 5 j − 9 k

(d) none of these

(a) 3 i − 5 j + 9 k

A parallelopiped is formed by planes drawn parallel to the co-ordinate axes through the points A = (2, 3, 4) and B = (11, 9, 8). Then, the volume of the parallelepiped is equal to (in cubic units) (a) 108 (b) 36 (c) 72 (d) 216 A line passes through P(2, 3, 4) and having direction ratios (4, 5, 6) meet the plane x + 2y – 3z + 16 = 0 at Q then length of PQ is equal to (b) 693 units (a) 936 units (c) 963 units (d) 369 units

The equation of the plane passing through the point (1, –1, 2) and (2, –2, 2) and is perpendicular to the plane 6x – 2y + 2z = 9 is (a) x + y + 2z + 4 = 0 (b) x + y + 2z – 4 = 0 (c) x + y – 2z + 4 = 0 (d) none of these More than One Correct Answer Type

10. If a , b , c are three non-zero vectors, then which of the following statement(s) is/are true? (a) a × (b × c ) + b × (c × a ) + c (a × b ) = 0 (b) If a + b + c = 0 then a ⋅ b + b ⋅ c + c ⋅ a < 0 (c × a ) ⋅ (b × c ) = 1 if a + b + c = 0 (c) (b × c ) ⋅ (a × b ) (d) none of these

By : R. K. Tyagi, Retd. Principal, HOD Maths, Samarth Shiksha Samiti, New Delhi

MATHEMATICS TODAY | JANUARY ‘18

11. If a and b are any two unit vectors then which of the following statement(s) in the range of 5 | a + b | + 6 | a − b | is/ are true? 2 (a) The maximum integer of the range is 13 (b) The minimum integer of the range is 5 (c) The most middle integer of the range is 9 (d) The average value of the maximum and minimum integer of the range is 9

(b) r ⋅ (4 ^j − 3 k^) + 1 = 0 is plane through the line (i) ^ ^ and r ⋅ (5 j − 4 k ) + 1 = 0 is plane through the line (ii) (c) Lines (i) and (ii) are coplanar (d) Lines (i) and (ii) intersecting Comprehension Type

Paragraph for Q. No. 17-19 Let a plane P1 passes through the point (2, 3, 4) and is parallel to the plane P2 whose equation is 12. A vector of magnitude 6 along the bisector of the ^ ^ ^ ^4x +^ 4y –^ 2z – 6 = 0. angle between the two vectors 4 i − 4 j + 2 k and 2 i + 4 j − 4 k is ^ ^ ^ ^ ^ ^ 17. The distance of the point (1, –1, 3) from the plane 4 i − 4 j + 2 k and 2 i + 4 j − 4 k is P1 in units is ^ ^ 3 ^ ^ ^ 6 (3 i − k ) (i − 4 j − 3 k) (a) (b) (a) 5 (b) 4 (c) 3 (d) 2 10 26 18. The coordinates of the foot of the perpendicular 6 ^ ^ ^ 6 ^ ^ (i − 4 j + 3 k) (c) (d) ( i − 3 j) drawn from the point (4, 5, 6) to the plane P2 are 26 5 (a) (–2, 3 7) (b) (2, 3, 7) 13. Let A(1, –2, 3), B (1, 1, 1), C(1, 2, –3), D(–1, –1, 1) (c) (2, –3, 7) (d) (2, 3, –7) be four points and 3x – y + 4z – 5 = 0 be the equation of plane then which of the following line 19. The distance between the planes P1 and P2 is segments are intersected by the plane? (a) 1 units (b) 2 units (a) AC (b) BC (c) BD (d) AD (c) 3 units (d) 4 units ^

^ ^

14. Let r = 2 i − 2 j + 3 k + λ( i − j + 4 k) be a line and ^

^ ^

Three vectors are a , b , c forming a right handed system and a × b = c , b × c = d , c × a = b .

equation of plane is r .( i + 5 j + k ) = 5. Then, (a) line is parallel to the plane 10 (b) the distance between line and plane 3 3

10 (c) the ⊥ distance between line and plane is 3 3 units (d) the point (2, –2, 3) lies on the plane 15. Let A(2, 3, 2), B(1, 1, 1), C(3, –2, 1), D(7, 1, 4). Then, (a) Points A, B, C, D are non coplanar 21 sq. units (b) Area of DBCD is 2 (c) Equation of the plane BCD is 3x + 2y – 6z + 1 = 0 1 (d) Volume of tetrahedron ABCD is cubic units 2 16. Consider the equations ^

r = ( i + 2 j + 3 k ) + λ(2 i + 3 j + 4 k) ^

...(i)

...(ii) r = (2 i + 3 j + 4 k) + µ(3 i + 4 j + 5 k ) then which of the following statement(s) is/are true? ^ ^ (a) r ⋅ (4 i − 3 j) + 1 = 0 is the plane through the ^ ^

line (ii) and r ⋅ (2 i − k ) + 1 = 0 and is plane through the line (i) 68

MATHEMATICS TODAY | JANUARY ‘18

Paragraph for Q.No. 20-23

20. The value of 3a ⋅ (b × c ) + 4b ⋅ (c × a ) + 5c ⋅ (a × b ) equals (a) 12 (b) 50 (c) 50 (d) none of these 21. If the vectors 3a − 4b + 3c , a + 2b − c , 2c − yb + 2c are coplanar then value of y equals. 3 8 (b) (a) 8 3 (c) 0 (d) none of these 22. If x = a − 2b + c , y = −a + 3b + c , z = 2a − b + 3c the unit vector normal to the vector xx++ yy and and xx++zz isisis 1 (a) (10a − 6b − 3c ) 145 1 (b) (−10a + 6b − 3c ) 145 1 (c) (10a + 6b − 3c ) 145 (d) none of these

^ ^ ^ 23. Let x = 2a + b and y = a − 2b then the point of Let a = 3 i + 2 j + 2 k and andr × y = x ×C.y is ^ ^ ^ intersection of the straight lines r × x = y × x and b = i + 2 j + 2 k then d.r.’s of r × x = y × x and r × y = x × y is (a + b ) × (a − b ) are (b) a − b (a) a + b A line passes through A(–3, –2, 4) (c) 3a + 2b (d) 3a − b D. and B(5, 6, 4) then d.r. of the Matrix Match Type line AB are 24. Match the following. Integer Type

Column-I

Column-II

A(0, 3, 5) B(3, 6, 5) C(a, 4, b) are vertices of triangle ABC such that the median through A. p. 5 B is equally inclined to the co-ordinate axes then the value of (9a + b) is ^

^ ^

Let a = i + 2 j − k ^ ^

and b = i − j + 2 k ^

and c = i + (x − 3) j − 2 x k . If the vector c lies in the plane of a and b then x2 equals

q. 21

Vectors a , b , c satisfies the condition a + b + c = 0 such C. that | a |= 1, | b |= 4, | c |= 2 and

r. 16

vectors then the equals

s. 2, 5, 3

26. If a , b and c are unit vectors perpendicular to each other and | b | = 3, | c |= 4 and a × b = c then, the least value of 3| a − b | is 27. Let a is vector whose magnitude is unity and coplanar with the vectors b and c , where ^ ^ ^ ^ ^ ^ b = i − j + 3 k and c = 3 i − j + k such that a is perpendicular to b . If l is the projection of a 11 2 λ is along c then the value of 4 28. Let A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) are three vertices of a parallelogram ABCD if coordinates of x+y+z D(x, y, z) then is 17 29. The value of the tetrahedron which is included between the 4x – 5y + 6z = 120 and the co-ordinate planes is 600 l where l is 30. Let A(1, 2, –3) and B(3, –4, 2) then the moment about the point C(–2, 4, –6) of the force represented 341 . Then in magnitude and position by AB is λ value of l is

= a ⋅ b + b ⋅ c + c ⋅ a then |2 | equals If a , b , c be three vectors such that | a |= 3, | b |= 4 ,| c |= 5 and each vector is perpendicular D. to the sum of the other two

r. 8, 8, 0

s. 9

a +b +c 2

25. Match the following. Column-I Column-II If the co-ordinates of the points A (3, 4, 6) and B(8, 6, 9) A. p. 0, 8, –8 then projections of the line AB to the co-ordinate axis are The equation of a line is ^ ^ ^ B. r = 3 ^i + 4 ^j + 5 k^ + λ(2 i + 5 j + k) q. 5, 2, 3 then d.r.’s of the line are

SOLUTIONS ^

1. (d) : Let the plane r ⋅ ( i − 2 j + 3 k ) = 17 divides the ^

line joining the points −2 i^ + 4 ^j + 7 k^ and 3 i − 5 j + 8 k in the ratio l : 1 R(x, y, z) (–2, 4, 7) (3, –5, 8) l:1  3λ − 2 −5λ + 4 8 λ + 7  ∴ R(x , y , z ) =  , ,  λ +1 λ +1 λ + 1  Using R(x, y, z) in the equation of the plane, we get  3λ − 2 ^ 4 − 5λ ^ 8 λ + 7 ^  ^ ^ ^  λ + 1 i + λ + 1 j + λ + 1 k  ⋅ ( i − 2 j + 3 k) = 17   ⇒ (3l – 2)(1) + (4 – 5l)(–2) + (8l + 7)(3) – 17(l + 1) = 0 3 ⇒ 20l = 6 ⇒ λ = 10 \ Required ratio is 3 : 10 MATHEMATICS TODAY | JANUARY ‘18

 2. (b) : As the given plane passes through c and is    parallel to the vector b − c and a so it is normal to    (b − c ) × a . Hence its equation is given as      ...(i) (r − c ) ⋅[(b − c ) × a ] = 0 \ Length of perpendicular from origin O(0, 0, 0) to (i) is given by        | 0 ⋅ (b × a + a × c ) − [c b a ]|  1 =     [c b a ]     |b × a + a × c | |b × a + a × c | 3. (c) : Equation of line is given by  ^ ^ ^ ^ ^ ^ r = 6 i + 7 j + 7 k + λ(3 i + 2 j − 2 k ) x −6 y −7 z −7 ⇒ = = = λ (say ) 3 2 −2 \ Coordinates of M are (3l + 6, 2l + 7, –2l + 7) \ d.r.’s of PM are (3l + 5, 2l + 5, –2l + 4) and d.r.’s of AB are (3, 2, –2) Now, PM ⊥ AB ⇒ 3(3l + 5) + 2(2l + 5) – 2(–2l + 4) = 0 ⇒ 17l = –17 ⇒ l = –1 \ Foot of the perpendiculars, M (3, 5, 9) ^

r =of a+ λb = (2 i + 3 j + 4 k) + λ(4 i + 5 j + 6 k) Equation line Any point on the line is (2 + 4l, 3 + 5l, 4 + 6l), l ∈ R As the line meet the plane therefore any point that lie on the line satisfies the equation of the plane \ (2 + 4l) + 2(3 + 5l) – 3(4 + 6l) + 16 = 0 ⇒ –4l = –12 \ l = 3 \ Coordinates of Q = (14, 18, 22) \

PQ = (14 − 2)2 + (18 − 3)2 + (22 − 4)2 = 144 + 225 + 324 = 693 units

6. (d) : Cartesian equations of the planes are ⇒ bx + gy = 0 ...(i); gy + az = 0 az + bx = 0 ...(iii) and bx + gy + az = l λ  Now, plane (iv) cuts the plane (i) at  0, 0,   α 70

MATHEMATICS TODAY | JANUARY ‘18

x − 2 y −1 z + 2 and = = 3 −5 2

plane is x + 3y – az + b = 0. The d.r.’s of the line are (3, –5, 2) As the line lies in the plane, so every point of line satisfies the equation on plane and normal to plane must be ⊥ to the line \ 3(1) + (–5)(3) + 2(–a) = 0 ⇒ a = –6 and 2 + 3(1) + 2a + b = 0 ⇒ 2 + 3 + 2(–6) + b = 0 \ b = 7 Thus (a, b) = (–6, 7)

1λ λ λ λ3 Volume of tetrahedron =  ⋅ ⋅  = 6  β γ α  6αβγ

7. (a) : Equation of line is

5. (b) : As the line passes through P(2, 3, 4) and has d.r.’s (4, 5, 6) ^

 λ  the plane (iv) cuts the plane (iii) at  0, , 0   γ  thus the planes (i) , (ii) and (iii) are coordinate planes intersecting at origin (0, 0, 0). Therefore the lengths of the λ λ λ coterminous edges of the tetrahedron are  , ,  β γ α \

Position vector of M is 3 i + 5 j + 9 k 4. (d) : Translating the axes through A(2, 3, 4) then, new position of A is (0, 0, 0) and B is (9, 6, 4) \ Coterminous edges are of lengths 9, 6, 4 respectively \ Volume of parallelepiped = 9 × 6 × 4 = 216 cu. units

\ \

λ  the plane (iv) cuts the plane (ii) at  , 0, 0  and β 

...(ii); ...(iv)

8. (d) 9. (c) : Equation of the plane through (1, –1, 2) will be a(x – 1) + b(y + 1) + c(z – 2) = 0 which passes through (2, –2, 2) \ a(2 – 1) + b(–2 + 1) + c(2 – 2) = 0 a–b=0 ⇒ a=b Also, the plane a(x – 1) + b(y + 1) + c(z – 2) = 0 is ⊥ to 6x – 2y + 2z = 9 \ 6a – 2b + 2c = 0 ⇒ 3a – b + c = 0 ⇒ c = –2a ( a = b) \ a(x – 1) + a(y + 1) –2a(z – 2) = 0 ⇒ x – 1 + y + 1 – 2(z – 2) = 0 ⇒ x + y – 2z + 4 = 0 10. (a,b,c) : (a) We know that a × (b × c ) = (a ⋅ c )b − (a ⋅ b )c b × (c × a ) = (b ⋅ a )c − (b ⋅ c )a c × (a × b ) = (c ⋅ b )a − (c ⋅ a )b Adding (i), (ii) and (iii) we get a × (b × c ) + b × (c × a ) + c × (a × b ) = 0 2 (b) ∵ a + b + c = 0 ⇒ (a + b + c ) = 0 2

⇒

a + b + c + 2(a ⋅ b + b ⋅ c + c ⋅ a ) =

⇒

a + b + c = −2(a ⋅ b + b ⋅ c + c ⋅ a )

...(i) ...(ii) ...(iii)

⇒ a ⋅ b + b ⋅ c + c ⋅ a < 0 ( L.H.S. is a positive quantity)

(c) ∵ a + b + c = 0 ⇒ (a + b ) = −c ⇒ c × (a + b ) = 0 ⇒ c × a = b × c ...(i) ⇒ b × (a + b ) = −b × c Again , (a + b ) = −c ⇒ a ×b =b ×c

By (i) and (ii) we get

f(0)=5(1) + 12(0) = 5

  12   12   5(5) + 12   = 13 f  θ = 2 tan−1    =  13    5  13  and f(q = p) = 5(0) + 12(1) = 12 \ Range = [5, 13] \ Maximum integral value of range = 13 Minimum integral value of range = 5 Most middle integral value of range 5, 6, 7, 8, 9, 10, 11, 12, 13 is 9. Average of minimum integral value and maxima 5 + 13 integral value of the range is =9 2 ^ ^ ^ ^ ^ ^ 12. (b,d) : Let a = 4 i − 4 j + 2 k , b = 2 i + 4 j − 4 k ∴ | a |=| b |= 6 \ Vectors along bisectors is ^ ^ ^ a b 1 ^ ^ ^ ± = [(4 i − 4 j + 2 k) ± (2 i + 4 j − 4 k )] |a | |b | 6 \ Vectors along bisectors are ^ ^ ^ 1 ^ ^ ^ ^ ^ ^ = [(4 i − 4 j + 2 k + 2 i + 4 j − 4 k), (4 i − 4 j + 2 k 6 ^ ^ ^ −2 i − 4 j + 4 k)]   ^ ^ ^ ^ ^ 1^ i 4^ ^ 1 ^ ^ = [(6 i − 2 k ),(2 i − 8 j + 6 k)] =  i − k, − j + k   3 3 3  6

^ 1 ^  i− 3 k   10 9

i 4^ ^ − j+ k 3 3

 1   −4  2   +   + 1 3 3  ^ ^  ^ ^ ^  3 i − k   i − 4 j+ 3 k  = 6  , 6⋅   10   26 

...(ii)

(c × a ) ⋅ (b × c ) =1 (b × c ) ⋅ (a × b) 11. (a,b,c,d) : ∵ a , b are unit vectors ∴ | a |=| b |= 1, let q be the angle between vectors a and b . 2 2 θ 2 Now a + b = a + b + 2a ⋅ b ⇒ | a + b |= 2 cos 2 2 2 θ 2 and a − b = a + b − 2a ⋅ b ⇒ | a − b |= 2 sin 2 5 θ θ  Let f (θ) =  2 cos  + 6  2 sin  , θ ∈[0, π]  2 2 2 −5 θ θ ∴ f ′(θ) = sin + 6 cos 2 2 2 θ 12 f ′(θ) = 0 ⇒ tan = 2 5 θ 12 θ 5 ∴ sin = , cos = 2 13 2 13

Required vectors are

6 10

^ ^

(3 i − k ),

6 26

(i − 4 j + 3 k)

13. (a, b, c, d) : For point A(1, –2, 3) and the plane 3x – y + 4z – 5 = 0 we have 3(1) + 2 + 4(3) – 5 > 0 For point B(1, 1, 1) 3x – y + 4z – 5 = 3 – 1 + 4 – 5 > 0 For point C(1, 2, –3), 3x – y + 4z – 5 = 3 – 2 – 12 – 5 < 0 For point D(–1, –1, 1), 3x – y + 4z – 5 = –3 + 1 + 4 – 5 < 0 \ Points A, B are on one side of the plane and C, D on the other side of the plane.

The line segments AC, AD, BC, BD intersect the plane.

14. (a,c) 15. (a, b, c, d) : Equation of the plane through the points A, B, C is x −2 y −3 z −2 −1 −2 −1 = 0 (a) 1 −5 −1 ⇒ (x – 2) (–3) – (y – 3)(2) + (z – 2)(7) = 0 ...(i) Using D(7, 1, 4) in (i), we have (7 – 2) (–3) – (1 – 3)(2) + (4 – 2)(7) ≠ 0 \ Points A, B, C, D are non-coplanar . (b) Area of DBCD : B(1, 1, 1), C(3, –2, 1), D(7, 1, 4) = B(x1, y1, z1), C(x2, y2, z2), D(x3, y3, z3) \

Area of ∆BCD = ∆ = ∆2x + ∆2y + ∆2z y1 y2 y3 1 where ∆ x = z1 z2 z3 and so on 2 1 1 1 1 −2 1 1 3 7 1 3 7 1 1 1 ∆ x = 1 1 4 , ∆ y = 1 1 4 , ∆ z = 1 −2 1 2 2 2 1 1 1 1 1 1 1 1 1 MATHEMATICS TODAY | JANUARY ‘18

9 ∆ x = , ∆ y = 3, ∆ z = 9 2

81 21 + 9 + 81 = sq. units 4 2 (c) Equation of the plane B(1, 1, 1), C(3, –2, 1), D(7, 1, 4) is x −1 y −1 z −1 −3 2 0 =0 6 0 3

Area of ∆BCD = ∆ =

⇒ (x – 1)(–9) – (y – 1)(6) + (z – 1)(18) = 0 ⇒ 3x + 2y – 6z + 1 = 0 (d) Volume of tetrahedron ABCD 1 = × area of DBCD 3 × ⊥ distance from A to the plane of BCD 1 21 1 1 = × × = cubic units 3 2 7 2 ^

16. (a,b,c,d) : As r = ( i + 2 j + 3 k ) + λ(2 i + 3 j + 4 k) x −1 y − 2 z − 3 ...(i) ⇒ = = =λ 2 3 4 ^

and r = (2 i + 3 j + 4 k ) + µ(3 i + 4 j + 5 k) x −2 y −3 z −4 ...(ii) ⇒ = = 3 4 5 \ Equation of the planes through (i) are x −1 z − 3 x −1 y − 2 y − 2 z − 3 and = = , = 2 4 2 3 3 4 ⇒ 3x – 2y + 1 = 0, 4y – 3z + 1 = 0 and 2x – z + 1 = 0 ^

⇒ r ⋅ (3 i − 2 j) + 1 = 0, r ⋅ (4 j − 3 k) + 1 = 0 ^ ^

and r ⋅ (2 i − k ) + 1 = 0 Again, equation of planes through line (ii) are x −2 y −3 y −3 z −4 x −2 z −4 , and = = = 3 4 4 5 3 5 ⇒ 4x – 3y + 1 = 0, 5y – 4z + 1 = 0 and 5x – 3z + 2 = 0 ^

⇒ r ⋅ (4 i − 3 j) + 1 = 0, r ⋅ (5 j − 4 k) + 1 = 0 ^

and r ⋅ (5 i − 3 k) + 2 = 0 Again any point on the line (1) is P(2l + 1, 3l + 2, 4l + 3) and 4x – 3y + 1 = 0, 5x – 3z + 2 = 0, 5y – 4z + 1 = 0 or 5(2l + 1) – 3(4l + 3) + 2 = 0 each gives the same value of l = –1. So the lines (i) and (ii)are coplanar. Also, the lines are coplanar therefor they will intersect and it can be verified by using the formula, d = Shortest distance = ^ ^ ^

where a2 − a1 = i + j + k 72

(a2 − a1 ) ⋅ (b1 × b2 ) b1 × b2

MATHEMATICS TODAY | JANUARY ‘18

i j k ^ ^ ^ Now, b1 × b2 = 2 3 4 = − i + 2 j − k 3 4 5 \ \

^ ^ ^

^ ^

(a2 − a1 ) ⋅ (b1 × b2 ) = ( i + j + k ) ⋅ (− i + 2 j − k) = 0 d = 0.

17. (c) : Equation of the plane P1 passing through the point (2, 3, 4) and parallel to plane 4x + 4y – 2z – 6 = 0 is given by 4(x – 2) + 4(y – 3) – 2(z – 4) = 0 ⇒ 4x + 4y – 2z – 12 = 0 ...(i) \ Distance from (1, –1, 3) to the plane (i) 4(1) + 4(−1) − 2(3) − 12 18 = = = 3 units 6 42 + 42 + (−2)2 18. (b) : Equation of the line AB having d.r.’s (4, 4, –2) through the point (4, 5, 6) is x −4 y −5 z −6 = = = λ (say ) 4 4 −2 ⇒ x – 4 = 4l, y – 5 = 4l, z – 6 = –2l ⇒ x = 4l + 4, y = 4l + 5, z = –2l + 6 Using this point on the plane P2, we have 4(4l + 4) + 4(4l + 5) – 2(–2l + 6) – 6 = 0 ⇒ 16l + 16l + 4l = –16 – 20 + 12 + 6 −1 ⇒ 36 λ = −18 ⇒ λ = 2 and foot of the ⊥ are x = 4(–1/2) + 4, y = 4(–1/2) + 5, z = –2(–1/2) + 6 x = 2, y = 3, z = 7 \ (x, y, z) = (2, 3, 7) 19. (a) : Given P1 : 4x + 4y – 2z – 12 = 0 P2 : 4x + 4y – 2z – 6 = 0 Let (x1, y1, z1) be any point on P2 \ 4x1 + 4y1 – 2z1 – 6 = 0 ...(*) Now distance from, (x1, y1, z1) to plane P1 \ Required distance | 4 x1 + 4 y1 − 2z − 6 − 6 | 6 = = = 1 unit 6 42 + 42 + (−2)2 20. (a) : ∵ a , b , c forming a right handed system ∴ a ⋅ (b × c ) = b ⋅ (c × a ) = c ⋅ (a × b ) = [a b c ] and a × b = c ⇒ c ⋅ (a × b ) = c ⋅ c ⇒ [a b c ] = 1 = (3 + 4 + 5) a ⋅ (b × c ) = 12[a b c ] = 12 21. (b) : ∵ 3a − 4b + 3c , a + 2b − c , 2a − yb + 2c are coplanar

∴

3 −4 3 1 2 −1 = 0 2 −y 2

⇒ 3(4 – y) + 4(4) + 3(–y – 4) = 0 8 ⇒ −6 y = −16 ⇒ y = 3 22. (c) : x = a − 2b + c , y = −a + 3b + c , z = 2a − b + 3c z = 2a − b + 3c So, x + y = b + 2c , x + z = 3a − 3b + 4c a b c ∴ (x + y ) × x + z ) = 0 1 2 3 −3 4 ∴ a(10) + 6b + c (−3) = 10a + 6b − 3c \ Unit vector in the direction of 10a + 6b − 3c (x + y ) × (x + z ) = 145

\ B. ∴ ⇒ ⇒ C. ∴ ⇒ D. ∴

Now, | a + b + c |2 = (a + b + c )(a + b + c ) = a ⋅ a + b ⋅ b + c ⋅ c = a2 + b2 + c2 = 50 a +b +c ∴ | a + b + c |= 5 2 ⇒ =5 2

23. (d) : Given x = 2a + b , y = a − 2b Now, r × x = y × x = (r − y ) × x = 0 ⇒ r − y || x ⇒ r − y = λx ⇒ r = y + λx

...(i)

and r × y = x × y ⇒ (r − x ) × y = 0 ⇒ r − x || y ⇒ r − x = µ y ⇒ r = x + µy From (i) and (ii) we have y + λx = x + µy

β+5 −5 −5 = ⇒ β=0 2 2 9a + b = 9(1) + 0 = 9 As vectors a , b , c are coplanar 1 2 −1 1 −1 2 =0 1 x − 3 −2 x 1(2x – 2x + 6) – 2(–2x – 2) –1(x – 2) = 0 x = –4 \ x2 = 16 Given a + b + c = 0 and µ = a ⋅ b + b ⋅ c + c ⋅ a (a + b + c ) ⋅ (a + b + c ) = 0 a2 + b2 + c2 + 2 = 0 ⇒ 2 = –21 \ |2 | = 21  Each vector is ⊥ to the sum of other two vector ...(*) a ⋅ (b + c ) = 0, b ⋅ (a + c ) = 0, c ⋅ (a + b ) = 0

and

...(ii)

⇒ (a − 2b ) + λ(2a + b ) = (2a + b ) + µ(a − 2b ) ⇒ a(1 + 2 λ) + b (−2 + λ) = a(2 + µ) + b (1 − 2µ) ⇒ 1 + 2l = 2 + and –2 + l = 1 – 2 ⇒ l = = 1 \ Point of intersection from (i) and (ii) is 3a − b . 24. A → s, B → r, C → q, D → p α 7 β +5 A. Let M is mid point of AC = , ,  2 2  2  7 α β+5  d.r.’s of the median BM are  − 3, − 6, − 5 2  2 2

25. A → q, B → s, C → p, D → r A. Given A(3, 4, 6) and B (8, 6, 9) \ x2 – x1 = 5, y2 – y1 = 2, z2 – z1 = 3 and d.c.’s of x, y, z axes are (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively. Also the projection of a line segment having d.r.’s x2 – x1, y2 – y1, z2 – z1 on a line whose d.c.’s are l, m, n is given by (x2 – x1) l + (y2 – y1) m + (z2 – z1)n \ The projections of the line AB on the co-ordinate x axis are (x2 – x1)l + (y2 – y1)m + (z2 – z1)n = 5(1) + 2(0) + 3(0) = 5 Similarly projection of y and z axes are 2, 3 respectively \ Projection of line AB on axis are 5, 2, 3 ^

⇒ b = 2 i + 5 j + 3 k ∴ d.r. ’s = 2, 5, 3 ^

As a = 3 i + 2 j + 2 k , b = i + 2 j + 2 k

∴

a + b = 4 i + 4 j + 4 k and a − b = 2 i + 0 j + 0 k ^

∴

i j k (a + b ) × (a − b ) = 4 4 4 2 0 0 ^

But the median is equally inclined to the axes \ Its d.c.’s are equal 7 α 7 α ∴ −3= −6 ⇒ = −3 ⇒ α =1 2 2 2 2

B. We have, r = (3 i + 4 j + 5 k ) + λ(2 i + 5 j + 3 k )

= i (0) − j(−8) + k(−8) = 0 i + 8 j − 8 k ∴ d.r.’s of (a + b ) × (a − b ) are (0, 8, −8) D. We have, A(–3, –2, 4) , B(5, 6, 4) \ d.r.’s of AB = x2 – x1, y2 – y1, z2 – z1 = 5 – (–3), 6 – (–2), 4 –4 = 8, 8, 0 MATHEMATICS TODAY | JANUARY ‘18

26. (4) : Given | b |= 3, | c |= 4 ∵ a × b = c ⇒ | a × b |=| c | ⇒ | a || b |sin α =| c | 4 ⇒ | a |= cos ecα 3 Now | a − b |2 =| a |2 + | b |2 −2a ⋅ b

...(*)

16 cos ec2 α + 9 − 2 ⋅ 4 cos ecα ⋅ cos α (using (*)) 9 16 = (1 + cot2 α) + 9 − 8 cot α 9 16  16 2  = +  cot α − 8 cot α + 9   9 9 =

16 16  4  +  cot α − 3  ≥ 9  9 3 44 16 ⇒ ∴ | a − b |2 ≥ ⇒ | |aa−−bb| |≥≥ ⇒ 3 | a − b | ≥ 4 33 9 ⇒ least value of 3 | a − b |= 4 ^ ^

^ ^ ^

27. (3) : Given b = i − j + 3 k , c = 3 i − j + k ⇒ | b |=| c |= 11

Let a = xb + yc where x and y ∈ R and| a | = 1 ∴

^ ^

...(*)

^ ^ ^

30 0 0 1 1  20 × 24 × 30  = 0 −24 0 =   6 6 −1 0 0 20

^ ^

⇒ [ i (x + 3 y ) + j(− x − y ) + k(3x + y )] ⋅[ i − j + 3 k] = 0 ⇒ x + 3y + x + y + 3(3x + y) = 0 −11 ⇒ 11x + 7 y = 0 ⇒ y = x 7 ^ 3(−11x )  ^  11  ^  11  ∴ a = i x + + j  − x + x  + k  3x − x     7  7   7  ^  −26  ^  4 x  ^  10 x  =i x +j +k  7   7   7   (26)2 42 102  2 ⇒ | a |= 1 ⇒  + +  x =1 49 49   49 49 7 7 ⇒ x2 = ∴ x=± or ± 792 792 6 22 7  26 ^ 4 ^ 10 ^  ∴ a=±  − i + 7 j+ 7 k  6 22  7  |a ⋅c | Now projection of a along c = λ = |c | 74

MATHEMATICS TODAY | JANUARY ‘18

6 × 11 2

29. (4) : Given equation of plane is 4x – 5y + 6z = 120 which meet the coordinate axes at P(30, 0, 0), Q(0, –24, 0) and R(0, 0, 20) and the coordinate of the origin are (0, 0, 0) \ Volume of tetrahedron

a = i (x + 3 y ) + j(− x − y ) + k(3x + y ) Now a ⊥ b ∴ a ⋅ b = 0

11 2 λ = 3. 4 28. (1) :  ABCD is a parallelogram diagonals AC and BD bisect each other i.e., the mid point of the segment AC is same as the mid point of the segment BD  2 +1 2 + 3 3 + 2   x −1 y − 2 z −1 , , , , ⇒ M =  2 2 2   2 2 2  ⇒ x – 1 = 3, y – 2 = 5, z – 1 = 5 x = 4, y = 7, z = 6 \ x + y + z = 4 + 7 + 6 = 17

a = x( i − j + 3 k ) + y(3 i − j + k ) ^

∴

=| a |2 +9 − 2 | a || b | cos α

7  −26 ^ 4 ^ 10 ^  ^ ^ ^ i + j + k  ⋅ (3 i − j + k)  7 7  6 22  7

= 2400 cubic units

^ ^ ^ 30. (1) : If 'O' is the origin then OA = i + 2 j − 3 k and ^ ^ ^ ^ ^ ^ OB = 3 i − 4 j + 2 k and OC = −2 i + 4 j − 6 k ^ ^ ^ ^ ^ ^ ∴ AB = OB − OA = (3 i − 4 j + 2 k) − ( i + 2 j − 3 k) ^

= 2 i − 6 j+ 5 k

∴

^ ^ ^ ^ ^ ^ CA = OA − OC = ( i + 2 j − 3 k ) − (−2 i + 4 j − 6 k ) ^

= 3 i − 2 j+ 3 k

∴

i j k Required moment = CA × AB = 3 −2 3 2 −6 5 ^

= i (−10 + 18) − j(15 − 6) + k(−18 + 4) = 8 i − 9 j − 14 k ^ ^ ^ ∴ | CA × AB | = | 8 i − 9 j − 14 k | = 82 + (−9)2 + (−14)2 = 341



aths Musing was started in January 2003 issue of Mathematics Today. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

Set 181 JEE MAIN

1. If 1, , 2, ..., of (9 – ) (9 –

n–1

are, nth roots of unity, the value ) ... (9 – n–1) will be 9n − 1 9n + 1 (a) n (b) 0 (c) (d) 8 8 2 sin 2x – 1 4 – 2 sin 2x 2. The numbers 3 , 14, 3 form first three terms of an A.P., its fifth term is equal to (a) –25 (b) –12 (c) 40 (d) 53 3. The value of (a) 3n – 1 (c) 2n

∑ ∑ nC j jCi , i ≤ j is

i =0 j =1

(b) 0 (d) none of these

t 2 + 3t t − 1 t − 3 4 3 2 4. If At + Bt + Ct + Dt + E = t + 1 2 − t t − 3 , t − 3 t + 4 3t then E equals (a) 33 5. The

(b) –39

rank

the

(where a = –6) is (a) 1 (b) 2 (c) 3 JEE ADVANCED

(d)  −1 2 1  (d)

24 2 5  −4 a − 4  −2 a + 1  4

6. Let f : R → R, g : R → R, be two given functions such that f is injective and g is surjective, then which of the following is injective? (a) gof (b) fog (c) gog (d) fof COMPREHENSION Let f(x) be a continuous function defined on the closed interval [a, b], then n −1 1 r  1 lim ∑ f   = ∫ f (x)dx 0 n→∞ r =0 n  n    n2 n2 1 1 7. The value of lim  + + + ... +  is n→∞ n 8n   (n + 1)3 (n + 2)3  (a) 5/4

(b) 3/4

(d) 3/8

8. The value of 1/n  π  2π   3π   nπ   lim tan   tan   tan   ... tan    is  2n   2n   2n   2n   n→∞  (a) 1 (b) 2 (c) 3 (d) not defined INTEGER TYPE sin(sin x) − sin x 1 = − , then find the 9. If lim 3 5 x →0 ax + bx + c 12 value of a. MATRIX MATCH 10. Match the following. List-I List-II –1 3 P. y = sin (3x – 4x ), dy then is dx –1 3 Q. y = cos (4x – 3x), dy then 2 is dx  3x − x 3  y = tan–1  , R.  1 − 3x 2  dy is then dx If f (x) S. = (logcos x sin x) × (logsin x cos x)–1  2x  + sin −1  ,  1 + x 2   π then f ′   =  4

(a) (b) (c) (d)

P 4 1 1 2

Q 3 3 4 3

 1 1  , x ∈ − ,   3 3 1+ x

−8 32 + ln 2 π2 + 16

3. −

, 1 − x2 1 1   x ∈  −1, −  ∪  , 1  2 2 

R 1 2 2 1

 1 1 , x ∈ − ,   2 2 1− x 2

S 2 4 3 4

See Solution Set of Maths Musing 180 on page no 88

MATHEMATICS TODAY | JANUARY ‘18

Exam Dates OFFLINE : 8th April ONLINE : 15th & 16th April

Series-7 Time: 1 hr 15 min. The entire syllabus of Mathematics of JEE MAIN is being divided into eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given below:

Unit No.7

Topic

Syllabus In Detail

Co-ordinate Direction ratios and direction cosines. Angle between two intersecting lines. Geometry-3D Skew lines, the shortest distance between them and its equation. Equation of a line and a plane in different forms, intersection of a line and a plane, coplanar lines. Differential Rolle’s and Lagrange’s Mean value theorems, Applications of derivatives: Calculus Rate of change of quantities, monotonic-increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. Integral Integral as an anti-derivative. Fundamental integrals involving algebraic, Calculus trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of type:

∫

x ±a

,∫

x ±a

,∫

a −x

 1. If r is equally inclined to the co-ordinate axis and   magnitude of r is 6, then r equals ^ ^ ^ 2( i + j + k) ^ ^ (a) 3 (1 + i + j ) (b) 3 ^

(c)

^ ^

(i + j + k) 3

^ ^

(d) ± 2 3( i + j + k )

2. The equation of the plane bisecting the acute angle between the planes x – 2y + 2z + 3 = 0 and 3x – 6y – 2z + 2 = 0 is (a) 2(8 x − 16 y + 4 z ) + 27 = 0 (b) 8 x − 16 y + 4 z + 27 = 0 (c) 16 x − 32 y + 8z − 27 = 0 (d) 16 x + 32 y + 8z + 27 = 0

,∫

a −x

∫

ax + bx + c

MATHEMATICS TODAY | JANUARY ‘18

ax + bx + c

,∫

(px + q )dx

ax 2 + bx + c

 ^ ^ ^ 3. The distance between the planes r ⋅ (2 i − 3 j + 6 k ) = 5  ^ ^ ^ and r ⋅ (6 i − 9 j + 18 k) + 20 = 0 is (a) 2/3 (b) 4/3 (c) 5/3 (d) 7/3  4. Let n be a vector of magnitude 2 3 such that it makes acute equal angles with the co-ordinate axes, the vector form of equation of plane passing  through (1, –1, 2) and normal to n is  ^ ^ ^ (a) r ⋅ ( i + 2 j + 3 k ) = 2  ^ ^ ^ (b) r ⋅ (2 i − 3 j + k ) = 2  ^ ^ ^ (c) r ⋅ (2 i + j + k) = 2  ^ ^ ^ (d) r ⋅ ( i + j + k ) = 2

By : Sankar Ghosh, S.G.M.C, Mob : 09831244397.

,∫

5. The S.D. between the lines  ^ ^ ^ ^ ^ ^ r = (5 i + 7 j + 3 k) + λ(5 i − 16 j + 7 k) and  ^ ^ ^ ^ ^ ^ r = (9 i + 13 j + 15 k) + λ(3 i − 8 j − 5 k ) is (a) 8 units (b) 5 units (c) 10 units (d) none of these

13. If f(x) and g(x) are differentiable functions in [0, 1] such that f(0) = 2, f(1) = 6, g(0) = 0, g(1) = 2, then there exists 0 < c < 1 such that (a) f ′(c) = g′(c) (b) f ′(c) = –g′(c) (c) f ′(c) = 2g′(c) (d) 2f ′(c) = g′(c)

6. The distance of the point (1, –2, 3) from the plane x y z x – y + z = 5 measured parallel to the line = = 2 3 −6 is (a) 5 (b) 3 (c) 1 (d) none of these

14. If f (x ) = ∫

7. The distance from the point − i + 2 j + 6 k to the straight line passing through the point with position ^ ^ ^ vector 2 i + 3 j − 4 k and parallel to the vectors ^

6i +3j − 4k (a) 10 (b) 7

(d) 3

8. The equation of the plane through the line  ^ ^ ^ of intersection of r ⋅ ( i − 2 j + 2 k ) = 1 and  ^ ^ r ⋅ (2 i − j ) + 2 = 0 and perpendicular to  ^ ^ ^ r ⋅ ( i + j + 2 k ) + 9 = 0 is  ^ ^ ^ (a) r ⋅ (−5 i + j + 2 k ) = 7  ^ ^ ^ (b) r ⋅ (5 i + j + 2 k ) = 7  ^ ^ ^ (c) r ⋅ (5 i − j − 2 k ) = 7 (d) none of these 9. The reflection of the plane 2 x − 3 y + 4 z − 3 = 0 in the plane x – y + z – 3 = 0 is the plane (a) 4x – 3y + 2z – 15 = 0 (b) x – 3y + 2z – 15 = 0 (c) 4x + 3y – 2z + 15 = 0 (d) none of these 10. If P(x, y, z) is a point on the line segment joining A(2, 2, 4) and B(3, 5, 6) such that projection of OP 13 19 26 on axes are respectively, then P divides , , 5 5 5 AB in the ratio (a) 3 : 2 (b) 2 : 3 (c) 1 : 2 (d) 1 : 3 11. If f(x) = xa lnx and f(0) = 0 then the value of a for which Rolle's theorem can be applied in [0, 1] is (a) –2 (b) –1 (c) 0 (d) 1/2 12. For all x in [0, 1], let the second derivative f ′′(x) of a function exist and satisfy |f ′′(x)| ≤ 1. If f(0) = f(1), then in (0, 1) (a) |f(x)| > 1 (b) |f(x)| < 1 (c) |f ′(x)| > 1 (d) |f ′(x)| < 1

(a) (b) (c) (d)

dt , x > 0 , then log t x2

f(x) is maximum at x = 1 f (x) is an increasing function x ∈ R+ only f(x) is minimum at x = 1 f (x) is neither maximum nor minimum at x=1 x

2 15. If y = f (x ) = ∫ 2 − t dt . Then

(a) (b) (c) (d)

f(x) increases if | x | < 2 f(x) decreases if |x| < 2 f(x) increases if |x| > 2 none of these

16. A cone is inscribed in a sphere of radius r. The height of the cone when its volume is maximum is (a) 2r

(b)

3r 2

(c)

(d)

4r 3

3 17. Three normals are drawn from the point (c, 0) to the curve y2 = x. If two of the normals are perpendicular to each other, then c = (a) 1/4 (b) 1/2 (c) 3/4 (d) 1 18. The curve y = ax3 + bx2 + cx + 5 touches the x-axis at P(–2, 0) and cut the y-axis at a point Q where its gradient is 3. Then, a + b + c = (a) 9/4 (b) 7/4 (c) 5/4 (d) 3/4 19. The slope of the straight line which is both tangent and normal to the curve 4x3 = 27y2 is 1 1 (c) ± (d) ± 2 (a) ±1 (b) ± 2 2 20. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is 1 1 cm/min. (a) (b) cm/min. 18π 36π 1 5 (d) (c) cm/min. cm/min. 54π 6π MATHEMATICS TODAY | JANUARY ‘18

21. ∫

dx 2

sin x + tan x

26. The integral ∫

 tan x  1 1 tan−1  +c (a) − cot x −  2  2 2 2 (b)

 x  1 1 cot x − tan−1  +c  2 2 2 2

(4 + 3 x 2 ) 3 − 4 x 2

(a)

(d) 6 3 x2 − 33 x + 6 log 1 + 3 x + c dx = log 1 − f (x ) + f (x ) + c, then f(x) x(1 + xe x )2 1 1 (a) (b) x 1+ xe x x +e 1 1 (c) (d) x 2 (1 + xe ) (x + e x )2

1 −1 2x tan +c 2 5 3 − 4x

  1 5x tan−1  +c (b) 10  2 3 − 4 x2  1 −1 5x tan +c 5 2 3 − 4 x2

(d)

 5x 1 tan−1  10  3 − 4 x2

 +c 

 tan x + 1 + 2 tan x  + b ln   + c, then a + b =  tan x + 1 − 2 tan x  1 1 1 1 (b) (c) (d) (a) 2 2 4 2 2   π sin  x −     π π 6     24. ∫ cosec  x −  cosec  x −  dx = a ln   + b,   6 3  sin  x − π     3   then a is equal to

(a)

3 (b) − (c) 2 2 r

(d) –2 4n−1

25. If Sr = ∫ sin xd(i x ) where (i = −1), then ∑ Sr , is r =1 (n ∈ N) − cos x (b) cos2x + c (a) +c i 4n (c) 0 (d) none of these 78

MATHEMATICS TODAY | JANUARY ‘18

cos3 x + cos5 x

dx is sin2 x + sin4 x (a) sin x − 6 tan−1 (sin x ) + c 2 − 6 tan−1 (sin x ) + c (b) sin x − sin x 2 + 5 tan−1 (sin x ) + c (c) sin x − sin x 2 + 6 tan−1 (sin x ) + c (d) sin x + sin x

28. ∫

−1  tan x − 1  23. If ∫ tan x dx = a tan   2 tan x 

(x + 1)

27. ∫

(c)

 +c  +c

−1  tan x  +c (d) − cot x + tan   2 

represents the function

x + x2

(a) 6  3 x2 − 3 x + log 1 + 3 x  (b) 33 x − 6 6 x + 6 log 1 + 6 x

1 1 tan x + c (c) cot x − 2 2

22. ∫

29. If ∫

cos x − cos3 x 1 − cos3 x

( ( (

dx = f ( x ) + c, then f ( x ) =

) ) )

3 −1 sin cos3/2 x + k 2 2 (b) cos−1 cos3/2 x + k 3 2 (c) sin−1 cos3/2 x + k 3 (d) none of these

(a)

30. If ∫ f (x )sin x cos x dx =

1 2

2(b − a2 )

log( f (x )) + c,

then ∫ f (x ) dx = (a)

1  a tan x  −1  a tan x  tan−1  (b) ab tan    b   b  ab

(c)

 1 1  b tan x   bx   tan−1  tan−1  tan    (d)   a   a  ab ab 

SOLUTIONS

  ^ ^ ^ 1. (d) : Let r = x i + y j + z k and r makes equal angle q to each of the axes. \ Direction cosines are l = cosq, m = cosq and n = cosq Now, cos2q + cos2q + cos2q = 1 1 ⇒ 3cos2q = 1 ⇒ cos θ = ± 3    We know, x = l | r |, y = m | r | and z = n | r | 1 ∴ x=± ×6= ±2 3 [∵ | r | = 6] 3 1 y=± × 6 = ± 2 3 and z = ± 2 3 3 So, r = x i^ + y ^j + z k^ = ± 2 3 ( i^ + ^j + k^)

2. (a) : The given equation of the planes are x – 2y + 2z + 3 = 0 \ (a1, b1, c1) = (1, –2, 2) and 3x – 6y – 2z + 2 = 0 \ (a2, b2, c2) = (3, –6, –2) Now, a1a2 + b1b2 + c1c2 = 3 + 12 – 4 = 11 > 0 \ Equation of acute angle bisector is x − 2 y + 2z + 3 (3x − 6 y − 2z + 2) =− 3 7 ⇒ 16x – 32y + 8z + 27 = 0 3. (c) : The given equation of planes are  ^ ^  ^ ^ ^ ^ r ⋅ (2 i − 3 j + 6 k ) = 5 and r ⋅ (6 i − 9 j + 18 k) + 20 = 0 Now changing to cartesian form of the above plane we get, 2x – 3y + 6z – 5 = 0 and 6x – 9y + 18z + 20 = 0 respectively. Now distance of the plane 6x – 9y + 18z + 20 = 0 from a point (x 1 , y 1 , z 1 ) which lies on the plane 2x – 3y + 6z – 5 = 0 is | 6 x1 − 9 y1 + 18z1 + 20 | 62 + (−9)2 + 182 | 3(2 x1 − 3 y1 + 6z1 − 5) + 15 + 20 | 0 + 15 + 20 = 21 21 35 5 = = units 21 3 4. (d) : Given that a = b = g ⇒ cosa = cosb = cosg 1 i.e. l = m = n ∴ l = ( l2 + m2 + n2 = 1) 3 =

 ^ ^ ^ ^ ^ ^ ^ ^ ^ r ⋅ (2 i + 2 j + 2 k) = ( i − j + 2 k)(2 i + 2 j + 2 k)  ^ ^ ^ ⇒ r ⋅ (i + j + k) = 2 5. (d) : The given lines are  ^ ^ ^ ^ ^ ^ r = (5 i + 7 j + 3 k ) + λ(5 i − 16 j + 7 k )  ^ ^ ^ ^ ^ ^ and r = (9 i + 13 j + 15 k ) + λ(3 i − 8 j − 5 k )  ^ ^ ^  ^ ^ ^ Here a1 = 5 i + 7 j + 3 k , a2 = 9 i + 13 j + 15 k  ^ ^ ^  ^ ^ ^ and b1 = 5 i − 16 j + 7 k, b2 = 3 i − 8 j − 5 k ^

i j k   Now, b1 × b2 = 5 −16 7 = 136 ^i + 46 ^j + 8 k^ 3 −8 −5   ^ ^ ^ and (a2 − a1 ) = 4 i + 6 j + 12 k     (b1 × b2 ) ⋅ (a2 − a1 ) 916 ∴ S.D. = = = 6.37   143.8 | b1 × b2 |

6. (c) : The equation of the line passing through the x y z is point (1, –2, 3) and parallel to the line = = 2 3 −6 x −1 y + 2 z − 3 ...(i) = λ (say) = = 2 3 −6 \ x = 2l + 1, y = 3l – 2 and z = –6l + 3 Now point of intersection of the line (i) and plane x – y + z = 5 ...(ii) can be obtained by putting (2l + 1, 3l – 2, –6l + 3) in (ii), which gives (2l + 1) – (3l – 2) + (–6l + 3) = 5 1 ⇒ − 7 λ = −1 ⇒ λ = 7 and hence the required point of intersection of line and plane is  9 −11 15  (x , y , z ) ≡  , ,  7 7 7 7. (b) : Let C is a point with position vector ^

− i + 2 j + 6 k and D with position vector 2 i + 3 j − 4 k ^

∴ CD = 3 i + j − 10 k

 ^ ^ ^ i + j +k  ^ ^ ^ ∴ n = 2 3   [∵ r = | r |(l i + m j + nk )]  3 Thus the required vector equation of plane is MATHEMATICS TODAY | JANUARY ‘18

Let x be any point in [0, 1] other then c By Lagranges theorem f ′(x) – f ′(c) = (x – c) |f ′′(c1)|, x < c |f ′(x)| = |x – c| |f ′′(c1)| < 1 Since |x – c| < 1, |f ′′(c1)| ≤ 1

CN = Projection of CD on CN = l(x2 – x1) + m(y2 – y1) + n(z2 – z1) where l, m and n are

6 61

−4

61 61 3 3 (−10)(−4) =6× + ×1+ = 61 61 61 61

∴ DN = (CD)2 − (CN )2 = 32 + 12 + (−10)2 − 61 = 7 8. (a) : The equation of plane through the line of  ^ ^ ^  ^ ^ intersection of r ⋅ ( i − 2 j + 2 k) = 1 and r ⋅ (2 i − j ) + 2 = 0 is  ^ ^ ^  ^ ^ [r ⋅ ( i − 2 j + 2 k ) − 1] + λ [r ⋅ (2 i − j ) + 2] = 0 ^ ^ ^  ...(i) ⇒ r ⋅[(1 + 2 λ) i + (−2 − λ) j + 2 k] = 1 − 2 λ  ^ ^ ^ Since (i) is perpendicular to r ⋅ ( i + j + 2 k) + 9 = 0 \ 1(1 + 2l) + (–2 – l)1 + 2 · 2 = 0 ⇒ l = –3 ^ ^ ^  Required equation of plane is r ⋅ (−5 i + j + 2 k ) = 7 9. (a) : As we know that reflection of a′x + b′y + c′z + d′ = 0 in the plane ax + by + cz + d = 0 is given by 2(aa′ + bb′ + cc′)(ax + by + cz + d) = (a2 + b2 + c2)(a′x + b′y + c′z + d′) \ 2(2 + 3 + 4)(x – y + z – 3) = 3(2x – 3y + 4z – 3) ⇒ 6(x – y + z – 3) = 2x – 3y + 4z – 3 ⇒ 4x – 3y + 2z – 15 = 0 is the required equation of the plane. 10. (a) : Projection of OP on coordinate axis are 13 19 26 , , 5 5 5 13 19 26 So OP = i^ + ^j + k^ 5 5 5 Let P divides AB in the ratio l : 1  3λ + 2 5λ + 2 6 λ + 4  ∴ P , ,   λ +1 λ +1 λ +1  3λ + 2 13 3 ∴ = ⇒ 2λ = 3 ⇒ λ = λ +1 5 2 \ Required ratio is l : 1 i.e. 3 : 2

MATHEMATICS TODAY | JANUARY ‘18

1 dt log t x2

14. (d) : Here f (x ) = ∫ ∴ f ′(x ) =

3x2 2x − 3 log x 2 log x

⇒ f ′( x ) =

x2 − x x(x − 1) = log x log x

For extremum, we put f ′(x) = 0. Now, f ′(x) = 0 gives x(x – 1) = 0 \ Critical points are x = 0 and x = 1 Let us consider x = 1 as x > 0 Now for x < 1, f ′(x) = (–ve)(–ve) = (+)ve > 0 and for x > 1, f ′(x) = (+ve)(+ve) = (+)ve > 0 \ f ′(x) does not change its sign in the immediate neighbourhood of x = 1. So x = 1 is neither the point of maxima nor minima. x

15. (a) : f (x ) = ∫ 2 − t 2 dt 1

d (x ) = 2 − x2 dx For an increasing function f ′( x ) = 2 − x 2

⇒

2 − x2 > 0 ⇒ 2 − x2 > 0

⇒ ( 2 + x )( 2 − x ) > 0

c1 x

f ′(x) > 0

11. (d) : Here f(0) = f(1) = 0 and f(x) is differentiable in [0, 1]. f ′(x) = xa – 1(1 + a lnx) f (x) is continuous in [0, 1] Now, f ′(x) = 0, only if a > 0, x → 0+. 12. (d) : f(0) = f(1) ⇒ f ′(c) = 0, 0 < c < 1 0 By Rolle's theorem

13. (c) : Let F(x) = f(x) – 2g(x) F(0) = f(0) – 2g(0) = 2 F(1) = f(1) – 2g(1) = 2 F(x) satisfies the condition for Rolle's theorem. So, there exist c, such that F′(c) = 0 or f ′(c) – 2g′(c) = 0 for 0 < c < 1.

⇒ (x − 2 )(x + 2 ) < 0 ⇒ | x | < 2 16. (d) : Let x be the radius of the base and y the height A of the cone. 2 2 2 2 x = r – (y – r) = 2ry – y r 4 4 V = πx2 y = π(2ry2 − y3 ) 3 3 r y–r r dV 4 C B = 0 ⇒ 4ry = 3 y2 ⇒ y = x dy 3

17. (c) : x = t 2 , y = t ,

dy 1 = dx 2t

Equation of normal is 1 ( y − t ) + x − t2 = 0 2t 1 It passes through (c, 0) ⇒ t 2 + − c = 0 2 1 t1 , t2 are roots ⇒ t1t2 = − c, 2 1 1 Slopes m1 = ,m = 2t1 2 2t2 1 3 = −1 = 2 − 4c ⇒ c = m1m2 = −1 ⇒ 4t1t2 4 18. (b) : y = f(x) \ f(–2) = f ′(–2) = 0, f ′(0) = 3 f(0) = 5 5  f (x ) = (x + 2)2  ax +   4 5  f ′(x ) = 2(x + 2)  ax +  + (x + 2)2 a  4 1 f ′(0) = 3 ⇒ 5 + 4a = 3 ⇒ a = − 2  x 5 f (x ) = (x + 2)2  − +   2 4 27 7 ∴ a + b + c + 5 = f (1) = ⇒ a +b+c = 4 4 19. (d) : The given equation of the curve is 4x3 = 27y2 ⇒ x = 3t2; y = 2t3 dy ∴ =t dx The tangent at t, y – 2t3 = t(x – 3t2) ⇒ tx – y = t3 ....(i) The normal at t1, ...(ii) t y + x = 2t 4 + 3t 2 1

As (i) and (ii) are identical 1 t t3 ∴ =− = 1 t1 2t14 + 3t12 1 ⇒ t1 = − and − t 3 = 2t 31+3t1 t Eliminating t1, we get t6 = 2 + 3t2 ⇒ t 2 = 2, t = ± 2

The lines are y = ± 2 (x − 2). 4 20. (a) : v = π( y + 10)3 where y is thickness of ice 3 \ dv dy 50  dy  = 4 π(10 + y )2 ⇒   dt  t = 5 4 π (15 )2 dt dt

 dv  3  dt = 50 cm / min    =

1 cm / min. 18π

21. (a) : Let I =

10 + y

∫ sin2 x + tan2 x

=∫

Put tanx = t dt 1 1 1  ∫ t 2 (2 + t 2 ) = 2 ∫  t 2 − 2 + t 2  dt t 1 1 =− − tan−1 +c 2t 2 2 2

sec2 xdx tan2 x(2 + tan2 x )

 1  1 1 = − cot x − tan−1  tan x  + c  2  2 2 2 dx 22. (b) : Let I = ∫ (4 + 3 x 2 ) 3 − 4 x 2   3 dθ = 2∫ Putting x = sin θ  2   9 sin2 θ + 16 2 2 sec θdθ sec θdθ = 2∫ = 2∫ 2 2 9 tan θ + 16 sec θ 16 + 25 tan2 θ d ( tan θ) 1 1 5  = 2∫ = 2 × × tan−1  tan θ  + c 2 2 4  5 4 (4) + (5 tan θ) =

 1 1 5x 5  tan−1  tan θ  + c = tan−1  4  10 10  2 3 − 4 x2

 +c 

23. (c) : Let I = ∫ tan x dx

Put tan x = t2 t 2dt t2 + 1 + t2 − 1 =∫ dt I = 2∫ 4 t +1 t4 +1 =∫

1+ 2

t +

dt + ∫

1− 2

t +

dt = ∫

 1 d t −   t 2

 1  t − t  + 2

+∫

 1 d t +   t 2

 1  t + t  − 2

 1  1  t− t+ − 2 1    t  −1 t = tan  ln  + +c 1 2 2 2 2   t + + 2     t  1

 tan x + 1 + 2 tan x   tan x − 1  1 +c tan−1  − ln    2 tan x  2 2  tan x + 1 − 2 tan x  2 1 1 1 ∴ a +b = − = 2 2 2 2 2 =

MATHEMATICS TODAY | JANUARY ‘18

π π   24. (c) : Let I = ∫ cos ec  x −  cos ec  x −  dx    6 3 π 2 sin 6 =∫ dx π  π  sin  x −  sin  x −   6  3  π  π  sin   x −  −  x −    6  3  = 2∫  dx π  π  sin  x −  sin  x −   6  3

  π π   = 2 ∫ cot  x −  − cot  x −   dx  3 6      π  sin  x − 3   = 2 ln  +b  sin  x − π     6   25. (d) 26. (b) : Let I = 5

∫

dx x + 3 x2

[put x = t 6 ]

6t dt t dt 1   = ∫ 3 4 = 6∫ = 6∫ t − 1 +  dt  + + t 1 t 1 t +t = 3t2 – 6t + 6log|t + 1| + c = 33 x − 6 6 x + 6 log | 6 x + 1 | +c

27. (b) : Let I =

(x + 1)

∫ x(1 + xex )2 dx

Put xe x = t ⇒ e x (1 + x )dx = dt dt \ I=∫ 2 t (1 + t )

A B C + + 2 t 1 + t (1 + t )2 t (1 + t ) ⇒ 1 = A(1 + t)2 + Bt(1 + t) + Ct Put t = 0, then, 1 = A Put t = –1, then C = –1 Equating the coefficients of t2, we get 0 = A + B ⇒ B = –A = –1 dt dt 1 −∫ \ I = ∫ dt − ∫ t 1+ t (1 + t )2 1 = log | t | − log | 1 + t | + +c 1+ t t 1 1  1  = log + + c = log 1 − + +c    1+ t 1+ t 1+ t 1+ t Now, let

MATHEMATICS TODAY | JANUARY ‘18

28. (b) : Let I =

cos3 x + cos5 x

∫ sin2 x + sin4 x dx

cos x{(1 − sin2 x ) + (1 − sin2 x )2 } dx sin2 x + sin4 x Put sin x =t =∫

I =∫ =∫

(1 − t 2 ) + (1 − t 2 )2 t2 + t 4

dt = ∫

2 + t 4 − 3t 2 t2 + t 4

(t 2 − 1)(t 2 − 2)

dt t2 + t 4  2 6  = ∫ 1 + − dt (using Partial fractions) 2  t 1 + t 2  2 2 = t − − 6 tan−1 t + c = sin x − − 6 tan−1 (sin x ) + c t sin x 29. (b) : Let I = ∫ =∫

cos1/2 x sin x 1 − (cos3/2 x )2

cos x(1 − cos2 x ) 1 − cos3 x

Put t = cos3/2 x 2 1 2 I=− ∫ dt = − sin−1(cos3/2 x ) + c 3 1 − t2 3

2π 2  = −  − cos−1 cos3/2 x  + c = cos−1 (cos3/2 x ) + k  32 3 30. (a) : Let I = ∫ f ( x ) sin x cos x dx 1 = log f ( x ) + c 2 2(b − a2 )  f ′( x )  1 ∴ f (x )sin x cos x = +c 2 2  f (x )  2(b − a )   ⇒ 2(b2 − a2 )sin x cos x =

f ′( x )

(( f (x ))2 −1 ⇒ 2 ∫ (b2 − a2 ) sin x cos x dx = f (x ) ⇒ b2 ∫ 2 sin x cos xdx − a2 ∫ 2 sin x cos x dx = − ⇒ − b2 cos2 x − a2 sin2 x = − ⇒ f (x ) =

1 f (x ) =

1 f (x )

sec2 x

a2 sin2 x + b2 cos2 x b2 + a2 tan2 x sec2 x 1  a tan x  ∴ ∫ f (x )dx = ∫ dx = tan−1  2 2 2  b  ab b + a tan x 

TWO DIMENSIONAL GEOMETRY PAPER-I ONE OF MORE THAN ONE OPTION(S) CORRECT TYPE This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE may be correct. [Correct ans. 3 marks & wrong ans., no negative mark]

1. A curve that passing through (2, 4) and having subnormal of constant length of 8 units can be (a) y2 = 16x – 16 (b) y2 = –16x + 48 (c) x2 = 16y – 60 (d) x2 = –16y + 68 2. If the conjugate of (x + iy) (1 – 2i) be 1 + i, then 1 1 (a) x = (b) x + iy = (3 + i) 5 5 1 1− i (c) x − iy = (3 + i) (d) x + iy = 5 1 − 2i 3. If a + b + c = 0, then the roots of the equation (b + c – a)x2 + (c + a – b)x + (a + b – c) = 0 can be (a) imaginary (b) real and equal (c) real and unequal (d) none of these 2

dy  dy  4. The solution of   + 2 y cot x = y2 is  dx  dx c c = 0 (b) y = (a) y − 1 − cos x 1 + cos x (c) x = 2 sin 5.

−1 

c   2 y  (d) none of these  

∑ ∑ ∑ 1 is equal to

i =1 j =1 k =1

n(n + 1)(n + 2) 6 (c) nC3 (a)

(b) Sn2 (d)

n+2C

6. If A and B are two matrices such that AB = BA, then "n∈N (a) AnB = BAn (b) (AB)n = AnBn (c) (A + B)n = nC0An + nC1An – 1B + nC2An–2B2 + ... + nCn Bn 2n 2n n n n n (d) A – B = (A – B )(A + B ) 7. If the graph of the function f (x) is symmetrical about two lines x = a and x = b then f (x) must be periodic with period b−a (a) (b) b – a 2 (c) 2(b – a) (d) none of these  | x − 3| , x ≥1  is 8. The function f (x ) =  x2 3x 13  − + , x <1 4 2 4 (a) continuous at x = 1 (b) differentiable at x = 1 (c) continuous at x = 3 (d) differentiable at x = 3 9. Let f(x) = sin(px) – 4x(1 – x), then sin(πx ) > 4 ∀ x ∈(0, 1) (a) x(1 − x ) 5 3 (b) f ′   + f ′   = 0 8 8

By : Vidyalankar Institute, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367 MATHEMATICS TODAY | JANUARY ‘18

(c) f ′′(x) = 0 has no solution in (0, 1) (d) Rolle's theorem cannot be applied to f ′(x) in  1  1 α, 2  for some α ∈  0,    2 10. Let ∫

x 1− x

dx =

2 g ( f (x )) + c. Then 3

−1 (a) g (x ) = cot (x ) and f (x ) =

(b) g (x ) = tan−1 (x ) and f (x ) = (c) g (x ) = cot−1 (x ) and f (x ) = −1 (d) g (x ) = tan (x ) and f (x ) =

1 − x3 x3 1 − x3 x3 x3 1 − x3 x3 1 − x3

INTEGER ANSWER TYPE This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). [Correct ans. 3 marks & wrong ans., no negative mark]

11. If a, b, g are the roots of the equation x3 – 9x2 + 14x + 24 = 0, then find the value of |a + b + g + ab + bg + ga + abg|. 12. The least degree of a polynomial with integer coefficient whose one of the roots may be cos12° is 13. If sin–1x + sin–1y = p and, if x = ly, then the value of 392l + 5l – 1525 must be 14. The number of solutions of the equation

3 x +1  2 28t  2 is + 4  dt = ∫  8t +  3 log x +1 x + 1 −1  x

15. In a DABC, if a is the arithmetic mean and b, c are two geometric means between any two positive sin3 B + sin3 C numbers. Then is equal to sin A sin B sin C 16. Let m denotes the number of ways in which 4 different balls of green colour and 4 different balls of red colour can be distributed equally among 4 persons if each person has balls of the same colour and n be corresponding figure when all the four (m + n) . persons have balls of different colour. Find 132 17. Let f(x) = [3 + 4 sin x] (where [ ] denotes the greatest integer function). If sum of all the values of x in kπ [p, 2p] where f(x) fails to be differentiable, is , 2 k then the value of is 8 18. The polynomial p(x) = 1 – x + x2 – x3 + ... + x16 – x17 can be written as a polynomial in y where y = x + 1,  k  then let coefficient of y2 be k then   (where [.]  400  denote greatest integer function) is ____ 19. Let A be the centre of the circle x2 + y2 – 2x – 4y – 20 = 0. The tangents at the points B(1, 7) and C(4, –2) on the circle meet at the point D. If D denotes the area of the quadrilateral ABDC, then

∆ is equal to 3

20. Find the integer n for which (cos x − 1)(cos x − e x ) is a finite nonzero number. lim x →0 xn

PAPER-II ONLY ONE OPTION CORRECT TYPE This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) for its answer, out of which ONLY ONE is correct. [Correct ans. 3 marks and wrong ans. –1]

1. If x and y are positive real number and m, n are any xn y m positive integers and E = then (1 + x2n )(1 + y2m ) 1 1 1 (a) E > 1 (b) E > (c) E ≤ (d) E < 2 4 8 4 84

MATHEMATICS TODAY | JANUARY ‘18

2. Four persons are selected at random out of 3 men, 2 women and 4 children. What is the chance that exactly 2 of them are children? 9 10 11 10 (a) (b) (c) (d) 21 23 24 21 3. For the equation x2 – (k + 1) x + (k2 + k – 8) = 0 if one root is greater than 2 and other is less than 2, then k can take any value between  11  (a) (–2, 3) (b)  − , 3   3 

11 (d) (0, 5) (c)  − , −2   3  4. If z is a complex number satisfying |z + 1 – i| ≤ 1, then the maximum value of |z| is (a) 2 (b) 2 − 1 (c) 2 + 1 (d) 1 5. The value of (a) 1

e37

∫ 1

π sin(π ln x ) dx is x

(b) –1

(d) 4

6. For a parabola having focus at S, vertex at A such that SA = l1 units and focal chord PQ of length l2 units is given, then ar(D APQ) is l3 (a) l1l2 (b) l1 l1l2 (c) l2 l1l2 (d) 2 2l1 1 and sin4q cosec2a are in A.P., 2 1 then cos8 θ sec6α, and sin8qcosec6a are in 2 (a) A.P. (b) G.P. (c) H.P. (d) A.G.P.

7. If cos4 θ sec2 α,

say, then draw the graphs of y = f(x) and y = g(x). If graphs of y = f(x) and y = g(x) cuts at one, two, three, …, no points, then number of solutions are one, two, three, …, zero respectively. |x| is 10 (d) 10

11. The number of solutions of sin x = (a) 4

(b) 6

12. Total number of solutions of the equation 5π 3x + 2 tan x = in x ∈[0, 2p] is equal to 2 (a) 1 (b) 2 (c) 3 (d) 4 Paragraph for Q. No. 13 and 14 A and B are two matrices of same order 3 × 3, where 1 2 3  3 2 5    A = 2 3 4 and B = 2 3 8       5 6 8   7 2 9  13. The value of adj (adj A) is equal to (a) 2A (b) 4A (c) 8A (d) none of these

8. The point of intersection of the plane ^ ^ ^ r ⋅ 3 i − 5 j + 2 k = 6 with the straight line passing through the origin and perpendicular to the plane 2x – y – z = 4 is (a) (1, –1, –1) (b) (–1, –1, 2)  4 −2 −2  (c) (4, 2, 2) (d)  , ,  3 3 3  2x is increasing 9. The function f (x ) = loge (1 + x ) − 2 +x on (a) (–1, ) (b) (– , 0) (c) ( – , ) (d) none of these

Consider two functions f (x ) = lim  cos x  and   n→∞  n

2 10. If lim f (x ) = a (a finite number), then which of

g(x) = –x4b, where b = lim

x →∞

the following is/are true? 2 2 2 2 (a) lim x f ′(x ) = 0 (b) lim x f ′(x ) = 2a x →∞

x →∞

COMPREHENSION TYPE This section contains 3 paragraph. Based upon each of the paragraphs 2 multiple choice questions have to be answered Each of these questions has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. [Correct ans. 3 marks & wrong ans. –1]

Paragraph for Q. No. 11 and 12 Suppose equation is f(x) – g(x) = 0 or f(x) = g(x) = y

14. The value of

adj(adj B) (24)4

(a) 9 (c) 25

is equal to

(b) 16 (d) 1 Paragraph for Q. No. 15 and 16

x →∞

(

)

x2 + x + 1 − x2 + 1 .

15. The function f(x) is (a) e− x

(b) e

−

x2 2

(d)

x2 e2

16. Number of solutions of f(x) + g(x) = 0 is (a) 0 (b) 1 (c) 2 (d) 4 MATRIX MATCH TYPE This section contains 4 questions, each having two matching columns. Choices for the correct combination of elements from column-I and column-II are given as options (a), (b), (c) and (d), out of which one is correct. [Correct ans. 3 marks & wrong ans. –1] MATHEMATICS TODAY | JANUARY ‘18

17. Match the following. (Q) Exact value of 2. –1/3 Column-I Column-II cos40°(1 – 2 sin10°) is 1. –1 (P) The value of sin(sin–11) is ∞ π (R) The value of ∑ sink   is 3. 1/4 (Q) The points (k, 2 – 2k), 2. 1  6 k =3 (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear if k is equal to (S) The value of l for which the 4. 5/4 lines are concurrent 3. 1/2 (R) The value of x +y+1=0;  π π π 1  4 π 3 5 7       cos   + cos4   + cos4   + cos4     3x + 2ly + 4 = 0; 8  8   8   8  3 x + y – 3l = 0 is/are 1  4 π  3π   5π   7π  cos   + cos4   + cos4   + cos4    is Codes :  8  8   8   8  3 P Q R S P Q R S (S) If a, b, c are all different from 4. (a) 4 3 1 2 (b) 3 4 1 2 2 ±1 zero, and (c) 1 3 2 4 (d) 4 1 3 2 1+ a 1 1 20. Match the following. ∆ = 1 1+ b 1 = 0 Column-I Column-II 1 1 1+ c (P) Let a function f is defined as 1. 9 then the value of f : {1, 2, 3, 4} → {1, 2, 3, 4}. a–1 + b–1 + c–1 is Codes : If f satisfy f(f(x)) = f(x), P Q R S P Q R S x ∈ {1, 2, 3, 4} then number (a) 4 1, 4 2 3 (b) 2 1 2, 4 3 of such functions is (c) 1, 3 3 2 4 (d) 2 1, 3 3 1 (Q) If m and M are the least and 2. 10 18. Match the following. greatest value of Column-I Column-II f(x) = (cos–1x) 2 + (sin–1x) 2, t he n M / m has t he v a lu e (P) L et f(x) = [x – 1] + 1. continuous equal to [1 – x], [x] is the greatest at x = a integer function, a is an (R) L e t x an d y b e t wo re a l 3. 41 integer, then numbers such that lim f (x ) (Q) Let f be as in (P) but a is 2. 2sinx siny + 3 cosy x →0 not an integer, then + 6cosx siny = 7. does not exist The value of tan2x + 2tan2y is (R) Let f(x) = cotx, then 3. f(a) = 0 Codes : P Q R P Q R cot x cos x lim f (x ) −a a x → π / 2 f ( x ) = (a) 3 2 1 (b) 2 1 3 , If 4. (S) cot x − cos x = log a (c) 2 3 1 (d) 3 1 2 then Codes : ANSWER KEYS P Q R S P Q R S Paper-I (a) 4 1 2 3 (b) 2 1 4 3 1. (a,b) 2. (b,d) 3. (b,c) 4. (a,b,c) 5. (a,d) (c) 1 3 2 4 (d) 3 1 2 4 6. (a,b,c,d) 7. (c) 8. (a,b,c) 9. (b) 10. (a,d) 11. (1) 12. (4) 13. (1) 14. (1) 15. (2) 19. Match the following. 16. (6) 17. (3) 18. (2) 19. (5) 20. (3) Column-I Column-II Paper-II (P) The value of the expression 1. 1/2 π 1. (c) 2. (d) 3. (a) 4. (c) 5. (c)  f (x ) = sin2 x + sin2  x +  6. (b) 7. (a) 8. (d) 9. (a) 10. (d)  3 11. (b) 12. (c) 13. (d) 14. (d) 15. (b) π   + cos x ⋅ cos  x +  , x ∈ R 16. (c) 17. (d) 18. (d) 19. (d) 20. (a)  3 is equal to  86

MATHEMATICS TODAY | JANUARY ‘18

51% BTech seats vacant, 153 colleges face closure, 770 shut IT branch NUMBER OF ENGINEERING INSTITUTES (OFFERING B.TECH) THAT HAVE CLOSED AND NEW ONE THAT HAVE OPENED Institutes closed inlast 5 years 128 New Insitutes 276 120 100 110

54 48

39 15 34

30 2016-17

2015-16

2012-13

2013-14

2014-15

TOP FIVE UNPOPULAR ENGINEERING BRANCHES Institutes that have discontinued programmes over last 5 years (2012-13 to 2016-17)*

770

Information Technology

635

Electrical and Electronics

234 Computer Science 185 Mechanical 139 Civil Max number of engineering institutes that discontinued IT programme were from Telangana (157), Tamil Nadu (104) & Andhra Pradesh (128)

*Approximate figures

Of the 15.5 lakh BE/BTech seats in 3,291 engineering colleges across the country, over half — 51 per cent — were vacant in 2016-17, according to data obtained by The Indian Express from the All India Council for Technical Education (AICTE), the apex body for technical education in the country. Last year, roughly eight lakh BE/BTech students graduated, but only about 40 per cent got jobs through campus

placement. According to AICTE data, campus placements has been under 50 per cent for the last five years. This mismatch that underlines the reality of unfilled seats has got AICTE to consider asking technical education institutes which have had 70 per cent or more vacant seats for the last five years to wind up and leave. As part of a three-month-long investigation to find out why engineering seats were going unfilled and what this signifies, The Indian Express analysed AICTE enrollment data for the last five years (from 2012-13 to 2016-17), visited 10 colleges across three states that are among those on the AICTE’s radar for low admissions and spoke to principals, students, academics and experts. The picture that emerged is of glaring gaps in regulation, including alleged corruption; a vicious circle of poor infrastructure, labs and faculty; nonexistent linkages with industry; the absence of a technical ecosystem that can nurture the classroom — all this accounting for low employability of graduates and, therefore, an abysmal record of job placement. The glaring gaps Consider these: • Close to 30 lakh students in the science stream cleared their Class 12 Board exam in 2015-16. Even if all of them were to aim for an engineering seat, at 15.5 lakh undergraduate engineering seats across the country, there is roughly one seat for every two students. A case of too few people chasing too many seats. MBBS and dentistry, on the other hand, has less than a lakh seats nationwide. • From 87,059 BTech and MTech seats in 1990-91, the number has risen to 16.62 lakh in 2017-18, a staggering 18 times in less than three decades. • Ten states — Tamil Nadu, Andhra Pradesh, Maharashtra, UP, Telangana, Karnataka, MP, Gujarat, Kerala, and Haryana — together account for 80 per cent of the total seats in the country. They also account for 80 per cent of the total vacant seats in the country. • Enrollment data of these 10 states show the crisis is at its worst in Haryana. At 74 per cent, the state has the highest proportion of vacant BTech seats in 2016-17. Uttar Pradesh is second with 64 per cent unfilled seats. Tamil Nadu, which has the highest number of engineering seats — 2.79 lakh — has 48 per cent unoccupied seats.

• Of the nearly 370 technical colleges that are on AICTE’s radar for low admissions — 30 per cent or less admissions in the last five years — and which run the risk of being closed next year, 153 are engineering colleges. Most of these are in Maharashtra (26), Andhra Pradesh (19), Haryana (17), Odisha (17), Telangana (16) and Uttar Pradesh (11). Last year, a record 49 engineering colleges went bust and shut down. • At least half the 153 institutes with low admissions were set up in the last decade. • Information Technology (IT) has emerged as the least popular branch, with 770 institutes discontinuing the discipline between 2012-13 and 2016-17. That’s followed by Electricals and Electronics (635 colleges have stopped the branch), Computer Science (234), Mechanical Engineering (185) and Civil Engineering (139). The maximum number of institutes that discontinued IT were in Telangana (157), followed by Andhra Pradesh (128) and Tamil Nadu (104). Those on their last legs are now taking desperate measures — from offering fee concessions to diluting admission criteria; from paying middlemen to bring in students to hiring underqualified faculty; and, as the Bhagwan Parashuram college in Sonepat has done, letting out part of the campus or even converting the colleges into schools. What led to this? Several factors, say experts, but most of them point to what they call the engineering boom that started in 1995 and peaked in the 2000s, fuelled by the IT phenomemon and the Y2K bug. Speaking to The Indian Express, AICTE chairman Anil Sahasrabuddhe says: “A large number of people were required for coding then. Your engineering branch did not matter. There was always a job for an engineer in an IT company. As a result, several private institutes came up to feed the industry’s appetite for engineers. “When there was a demand for engineers, the private sector stepped in. A large number of government colleges did not immediately get into modern branches of engineering such as IT and computer science. Our entire IT industry would have collapsed had it not been for these private institutes,” says retired IISc professor D K Subramaniam, who is on TCS’s Research Advisory Board. The boom, however, ended in a problem of plenty.

Early warning ignored Alarm bells first went off about 15 years ago, in the shape of the U R Rao Committee report of 2003. Rao, former chairman of the Indian Space Research Organisation, had been tasked by the NDA-1 government to review AICTE’s performance. The report had observed that the pace of expansion of technical education was unsustainable and that the explosion in the number of private institutions was fuelled more by speculative rather than real demand. To alleviate this “serious situation”, the committee suggested a fiveyear moratorium on all approvals for undergraduate technical institutions in states where the student intake exceeded the then national average of 150 seats per million population. This figure was 1,047 for the southern states, 486 in the west, 131 in the east and 102 in the north. (Currently, the national average of BE/BTech intake, alone, is 1,286 seats per million population.) However, Rao’s recommendation was never acted upon. Indeed, the reverse happened. According to AICTE data, 2008-09 witnessed an increase of almost 30 per cent in engineering intake over the previous year — the highest in a single year since 2001 — with over 700 new institutes being approved. Many point out that it coincided with a period when AICTE was rocked by allegations of rampant corruption. That year, the CBI caught then AICTE member-secretary K Narayan Rao accepting a bribe from the owner of an engineering college in Andhra Pradesh. The incident eventually led to the suspension of then AICTE chairman R A Yadav. The CBI registered three cases against him, but did not chargesheet him. The effects of this indiscriminate expansion in the sector were probably first felt after the global economic crisis of 2008, when growth slowed in the US and Europe, the main markets for IT companies. “The economy is in a bit of a stationary mode. Industries are not making any investments and so there aren’t enough engineering jobs in the market at present,” says former IIT-Kanpur director Sanjay Dhande. An engineering degree in this climate offers little return on investment.  Courtesy : Indian Express

MATHEMATICS TODAY | JANUARY ‘18

(

SOLUTION SET-180

1. (a) : The sum of all two-digit numbers 90 = 10 + 11 + 12 + ... + 99 = (10 + 99) = 4905. There are 2 5 × 5 = 25 numbers with two odd digits. The sum of these numbers = 5(1 + 3 + 5 + 7 + 9) × 11 = 55 × 25 = 1375 Likewise, the sum of numbers with two even digits = 5(0 + 2 + 4 + 6 + 8) × 11 – (0 + 2 + 4 + 6 + 8) = 54 × 20 = 1080 \ S = 4905 – 1375 – 1080 = 2450 = 2 . 52 . 72    2. (a) : OA = a , OB = b , OC = c , a = 3, b − c = 2 The shortest distance between OA and BC is 2.         ∴ (b − a ) ⋅ a × (b − c ) = 2 a × (b − c )  ⇒ [a b c ] = 2 ⋅ 3 ⋅ 2 ⋅ sin 30° 1  ∴ Volume = [a b c ] = 1 6 3. (d) : f(x) = 2(log 8 3)cos = 2(log 8 3)cos

1/2

+ 3(log 8 2)sin

+ 2(log 8 3)sin

⇒ f(x) ≥ 21 (2(log 8 3)cos = 2 (2log8 3 )

x 2

⋅ 2(log 8 3)sin x ) (using A.M. ≥ G.M.)

= 21+ log8 3  z 0 + 1 z0 + i 4. (a) : z1 = f | z0 | = , z2 = i   , z3 = z0 z0 − i  z 0 − 1 1 z0 + i 5 + i + i = 1 + 10i ∴ z2011 = z1 = = z0 − i 1 +i −i 5 5. (d) : Let ∠ADC = q. By cosine rule, AB2 = m2 = n2 + 92 + 2 . 9 . n cosq AC2 = m2 = n2 + (21)2 – 2 . 21 . n cosq Eliminating cosq, we get m2 – n2 = 189 \ m – n = 1, m + n = 189, m = 95 m – n = 3, m + n = 63, m = 33 m – n = 7, m + n = 27, m = 17 6. (c) : The angles are invariant under translation. We can take the vertices of the triangle as O(0, 0), A(1, 0), B(0, 1). AD and BE are the bisectors of ∠A and ∠B. π 3π ∠OAD = , ∠OEB = 8 8 π m1 = − tan = − ( 2 − 1) 8 88

MATHEMATICS TODAY | JANUARY ‘18

)

3π = − 2 +1 8 2 2 −1 m1 ∴ = = 2 −1 = 3 − 2 2 m2 2 +1 7. (b) : Let r be the radius of the circle and q be the angle subtended by the chord AF at the centre. 5θ θ 4 = 2r sin , 11 = 2r sin 2 2 5θ 11 sin 2 sin 3θ + sin 2θ = = = 3 − 4 sin2 θ + 2 cos θ \ θ sin θ 4 sin 2 15 ⇒ 4 cos2 θ + 2 cos θ = 4 3 7 θ 1 ∴ cos θ = , sin θ = , sin = 4 4 2 2 2 1 4 = 2r ⋅ ⇒ r=4 2 2 2 7 AE = 2r sinθ = 8 2 ⋅ = 2 14 4 3θ 8. (a) : AD = 2r sin = 10 2 9. (5) : (1 + x)5 = C0 + C1x + C2x2 + C3x3 + C4 x4 + C5x5 Differentiating 5(1 + x)4 = C1 + 2C2x + 3C3x2+ 4C4x3 + 5C5x4 m2 = − tan

)

4 3C 5C 4C C 1  5  1 +  = C1 + 2 2 + 23 + 34 + 45  x x x x x Multiplying the above two series and considering the

∴

coefficient of x is (1 + x )8

∑ (r − 1)r ⋅ Cr −1Cr = coefficient of x in

r=2

 8  8 = 25   = 25   = 1400 5  3   x 10. (a) : (P) → (1), (Q) → (2), (R) → (4), (S) → (3) (P) We have, 12x + 12x+1 = 3x + 3x+1 + 3x+2 ⇒ 12x(13) = 3x(13) ⇒ 3x(4x – 1) = 0 ⇒ 3x = 0 (not possible) So, 4x – 1 = 0 ⇒ 4x = 1 ⇒ x =0 x dt π π π π (Q) ∫ = ⇒ sec −1 x − = or sec −1 x = , x = 2 2 12 4 12 3 2 t t −1 25

(R) 5(2cos2q – 1) + 1 + cosq + 1 = 0

1 3 ⇒ 10cos2q + cosq – 3 = 0 ⇒ cos θ = , − 2 5 One values of q in (0, p/2). (S) 32n + 2 – 8n – 9 = (1 + 8)n + 1 – 8n – 9  n + 1 2 =  2  8 + ...,   which is divisible by 82 = 26 ⇒ m = 6.



SRM University AP - Amaravati announces the setting up of School of Liberal Arts and Basic Sciences (SLABS) B.A., B.B.A, B.Com. & B.Sc. courses will be offered from 2018 in 12 subjects

resident of SRM University, Dr. P. Sathyanarayanan, announced the setting up of School of Liberal Arts and Basic Sciences (SLABS) at its University in Amaravati in the presence of the Honorary Pro Chancellor of SRM University, AP - Amaravati, Prof. Nicholas Dirks (Chancellor Emeritus, University of California, Berkley), and Dr. D Narayana Rao, Pro Vice Chancellor of SRM University, AP – Amaravati. SLABS will be SRM Amaravati’s home for fundamental research, where free, open, and critical inquiry is pursued across disciplines, finding answers and solutions to world’s most challenging problems and daunting issues. SLABS will have its first intake of students in 2018 and will offer B.A., B.B.A., B.Com and B.Sc. programs across 12 departments –Economics, English, History, Journalism, Psychology, Business Studies, Commerce, Physics, Chemistry, Mathematics, Biology and Computer Science. “Today, we face increasingly complex issues and challenges, and tackling these, calls for multi-dimensional thought processes and problem solving skills. Education needs to focus on this and much more. We aim to help students develop such skills through the liberal arts and basic sciences education offered at SRM SLABS. For this, we are looking at hands-on guidance from Prof. Nicholas Dirks, given his background as a renowned anthropologist, and his rich experience in Liberal Arts. SRM SLABS also has a strong faculty base of international caliber who will bring a global perspective to liberal arts education.” They will assist in creating a holistic approach to education, which will become, we hope, the calling card for SRM Amarvati’s SLABS”, says Dr. P Sathyanarayanan, President SRM Amaravati. “I am very pleased to be involved with SRM Amaravati as it establishes its School of Liberal Arts and Basic Sciences. In our program, students will learn the skills of critical thinking and knowledge creation in a range of fields in the humanities, social sciences, and sciences. They will have an innovative multi-disciplinary education, in close proximity as well to breaking new teaching and research in areas ranging from machine learning and data science to public policy and social analysis”, says Prof. Nicholas Dirks, Honorary Pro Chancellor, SRM Amaravati. “SRM is committed to offering a distinctive form of learning empowering young students and thinkers with historical and cultural perspectives, as well as language, critical thinking, and communication skills— ideal traits to survive the modern world. The multi-disciplinary focus of SLABS will ensure that the students would have both breadth as well as depth of knowledge about a wide range of subjects”, says Dr. D Narayana Rao, Pro. Vice Chancellor, SRM Amaravati. About SRM University, AP-Amaravati SRM University, AP – Amaravati, is envisaged to be a multi-disciplinary institution starting off with programs in engineering, followed by liberal arts and later on in fields of management, law, medical sciences, and pure sciences. SRM envisions to emerge as a world-class university in creating and disseminating knowledge and providing students a

unique learning experience in their chosen field of scholarship that would best serve the society. The focus is on developing into an inter-disciplinary institution combining academic rigour, excitement of discovery, creativity and entrepreneurship that delivers cutting-edge research based education, creating new knowledge and innovations. The School of Engineering and Applied Sciences is already functional with the first batch of engineering students having commenced their courses in August 2017. For more information, please visit: www.srmap.edu.in FACT SHEET: School of Liberal Arts and Basic Sciences (SLABS), at SRM University, AP-Amaravati Beginning from Courses Offered Program Duration

2018 session B.A., BBA, B. Com., B. Sc. 3 years + 1 Additional year (Optional) resulting in a Diploma / Certificate Disciplines Physical and Natural Sciences, Arts, Humanities, Social Sciences, Business Studies, Commerce Subjects Economics, English, History, Journalism, Psychology, Business Studies, Commerce, Physics, Chemistry, Mathematics, Biology and Computer Science. Admission Criterion Merit, based on Std. XII Exam Results Faculty Profile Top in the category from India and Abroad. 100% PhDs 75% have international Exposure in Research/Teaching 15% Foreign nationals Visiting faculty comprising of local and global experts/ academics SLABS Approach Multi-disciplinary. No restriction on courses across disciplines In class discussions, Field Trips Presentations / Movies / Hands on Assignments

 MATHEMATICS TODAY | JANUARY ‘18

MATHEMATICS TODAY | JANUARY â&#x20AC;&#x2DC;18

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