15fgbfgbfg by dfvdfvdfv345345ghnyuny

19 43

70 35 60

Vol. XXXV

No. 9

September 2017

Corporate Office: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 e-mail : info@mtg.in website : www.mtg.in Regd. Office: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029. Managing Editor : Mahabir Singh Editor : Anil Ahlawat

8 72

Competition Edge

Class XII

Class XI

CONTENTS 8

Brain @ Work

Concept Boosters

Ace Your Way (Series 5)

MPP-5

Concept Map

Concept Boosters

Ace Your Way (Series 5)

MPP-5

Maths Musing Problem Set - 177

JEE Work Outs

Mock Test Paper - JEE Main 2018 (Series 3)

Challenging Problems

Math Archives

Maths Musing Solutions

83 46

Subscribe online at

www.mtg.in

Individual Subscription Rates Mathematics Today Chemistry Today Physics For You Biology Today

1 yr.

2 yrs.

3 yrs.

330 330 330 330

600 600 600 600

775 775 775 775

Combined Subscription Rates 1 yr. PCM PCB PCMB

900 900 1000

2 yrs.

3 yrs.

1500 1500 1800

1900 1900 2300

Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot 99, Sector 44 Institutional Area, Gurgaon - 122 003, Haryana. We have not appointed any subscription agent. Owned, Printed and Published by MTG Learning Media Pvt. Ltd. 406, Taj Apartment, New Delhi - 29 and printed by HT Media Ltd., B-2, Sector-63, Noida, UP-201307. Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.

MATHEMATICS TODAY | SEPTEMBER‘17

TWO DIMENSIONAL GEOMETRY 1. The number of points on the line 3x + 4y = 5, which are at a distance of sec2q + 2 cosec2q, q ∈ R, from the point (1, 3), is (a) 1 (b) 4 (c) 3 (d) none of these

7. Equation of straight line ax + by + c = 0, where 3a + 4b + c = 0, which is at least distance from (1, –2) is (a) 3x + y – 17 = 0 (b) 4x + 3y – 24 = 0 (c) 3x + 4y – 25 = 0 (d) x + 3y – 15 = 0

2. Let ax + by + c = 0 be a variable straight line, where a, b and c are 1st, 3rd and 7th terms of an increasing A.P. respectively. Then the variable straight line always passes through a fixed point which lies on (b) x2 + y2 = 5 (a) y2 = 4x (c) 3x + 4y = 9 (d) x2 + y2 = 13

1  8. The curve y =  x 2 − 2x + 27  where [x] is the 6   greatest integer less than or equal to x, 6 < x < 9 represents (a) a parabola (b) part of the parabola (c) a straight line (d) two straight line segments

3. The vertices of a triangle are A(–1, –7), B(5, 1) and C(1, 4). The equation of the bisector BD of ∠ABC is (a) x + 7y + 2 = 0 (b) x – 7y + 2 = 0 (c) x – 7y – 2 = 0 (d) x + 7y – 2 = 0

9. Equation of circle through intersection of x2 + y2 + 2x = 0 and x – y = 0, having minimum radius is (a) x2 + y2 – 1 = 0 (b) x2 + y2 – x – y = 0 (c) x2 + y2 – 2x – 2y = 0 (d) none of these

4. The equation of the base of an equilateral triangle ABC is x + y = 2 and the vertex is (2, –1). The area of the triangle ABC is (sq. units) 2 3 3 (b) (c) (d) none of these 6 6 8 5. The line 3x – 4y + 7 = 0 is rotated through an angle p/4 in the clockwise direction about the point (–1, 1). The equation of the line in its new position is (a) 7y + x – 6 = 0 (b) 7y – x – 6 = 0 (c) 7y + x + 6 = 0 (d) 7y – x + 6 = 0 (a)

6. The equation of a line through the point (1, 2) whose distance from the point (3, 1) has the least possible value is (a) x + 2y = 3 (b) y = 2x (c) y = x + 1 (d) x + 2y = 5

10. The common chord of the circle x2 + y2 + 6x + 8y – 7 = 0 and a circle passing through the origin, and touching the line y = x, always passes through the point (a) (– 1/2, 1/2) (b) (1, 1) (c) (1/2, 1/2) (d) none of these 11. Three equal circles each of radius r touch one another. The radius of the circle touching all the three given circles internally is (a) (2 + 3) r (c)

(2 − 3) 3

(b)

MATHEMATICS TODAY | SEPTEMBER ‘17

(d) (2 − 3) r

Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 8

(2 + 3)

12. The set of points P(x, y) such that their distance from (3, 0) is 2 times their distance from (0, 2) form a circle (a) whose radius is greater than 5 units (b) whose radius is equal to 5 units (c) whose radius is less than 5 units (d) none of these 13. If the circles ax2 + ay2 + 2bx + 2cy = 0 and Ax2 + Ay2 + 2Bx + 2Cy = 0 touch each other, then (a) bC = cB (b) aC = cA a b c (c) aB = bA (d) = = A B C 14. Let f(x , y) = 0 be the equation of a circle. If f(0, l) = 0 has equal roots l = 2, 2 and f(l, 0) = 0 4 has roots λ = , 5 then the centre of the circle is 5 (a) (2, 29/10) (b) (29/10, 2) (c) (– 2, 29/10) (d) none of these 15. The minimum distance between the circle x2 + y2 = 9 and the curve 2x2 + 10y2 + 6xy =1 is (b) 2 (a) 2 2 1 (c) 3 − 2 (d) 3 − 11 16. If one end of the diameter of a circle which touches x-axis is (3, 4) then the locus of other end of the diameter of the circle is/an (a) parabola (b) hyperbola (c) ellipse (d) circle 17. If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2 = 4ax and x2 = 4ay, then (a) d2 + (3b – 2c)2 = 0 (b) d2 + (3b + 2c)2 = 0 (c) d2 + (2b – 3c)2 = 0 (d) d2 + (2b + 3c)2 = 0 18. The equation of the tangent to the parabola y = (x – 3)2 parallel to the chord joining the points (3, 0) and (4, 1) is (a) 2x – 2y + 6 = 0 (b) 2y – 2x + 6 = 0 (c) 4y – 4x + 13 = 0 (d) 4x + 4y = 13 19. The abscissa and ordinate of the end points A and B of a focal chord of the parabola y2 = 4x are respectively the roots of x2 – 3x + a = 0 and y2 + 6y + b = 0. The equation of the circle with AB as diameter is 10

MATHEMATICS TODAY | SEPTEMBER ‘17

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

x2 + x2 + x2 + x2 +

y2 – 3x + 6y + 3 = 0 y2 – 3x + 6y – 3 = 0 y2 + 3x + 6y – 3 = 0 y2 – 3x – 6y – 3 = 0

20. AB is a chord of the parabola y2 = 4ax with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the x-axis is (a) a (b) 2a (c) 4a (d) 8a 21. If normal at point P on parabola y2 = 4ax, (a > 0) meet it again at Q in such a way that OQ is of minimum length where O is vertex of parabola, then DOPQ is (a) a right angled triangle (b) obtuse angled triangle (c) acute angled triangle (d) none of these 22. If two distinct chords drawn from the point (4, 4) on the parabola y2 = 4ax are bisected on the line y = mx, then the set of value of m is given by 1− 2 1+ 2  (a)  (b) R , 2   2 (c) (0, ∞)

(d) (–2, 2)

23. If t1 and t2 be the ends of a focal chord of the parabola y2 = 4ax, then the equation t1x2 + ax + t2 = 0 has (a) imaginary roots (b) both roots positive (c) one positive and one negative root (d) both roots negative 24. If the normal at three points (ap2, 2ap), (aq2, 2aq) and (ar2, 2ar) are concurrent then the common root of equations px2 + qx + r = 0 and a(b – c)x2 + b(c – a)x + c(a – b) = 0 is (a) p (b) q (c) r (d) 1 25. The second degree equation x2 + 4y2 + 2x + 16y +13 = 0 represent itself as (a) a parabola (b) a pair of straight line (c) an ellipse (d) a hyperbola 4 to the ellipse 3 2 2 x y + = 1 intersects the major and minor axes at 18 32 A and B. If O is the origin, then the area of DOAB is (a) 48 sq. units (b) 9 sq. units (c) 24 sq. units (d) 16 sq. units

26. A tangent having a slope −

MATHEMATICS TODAY | SEPTEMBERâ&#x20AC;&#x2DC;17

y2 + =1 a 2 b2 with foci S1 and S2. If A be the area of DPS1S2, then the maximum value of A (a) ab sinq (b) abe (c) a sinq (d) b sinq

27. Let P be a variable point on the ellipse

28. Maximum distance of any point on the circle (x − 7)2 + ( y − 2 30 )2 = 16 from the centre of the ellipse 25x2 + 16y2 = 400 is 1− 3 2 (c) 3

3 2 (d) none of these

(a)

(b)

29. If the eccentricity of the ellipse be

1 3

(b)

18 6

(c)

10 6

y + 2 =1 a +2 a +5 2

(d) 8

30. The number of rational points on the ellipse x2 y2 + = 1 is 9 4 (a) infinite (b) 4 (c) 0 (d) 2 31. The radius of the circle passing through the points of intersection of ellipse (a) (c)

ab a 2 + b2

a 2 − b2

y2 2 2 + = 1 and x – y = 0 is 2 2 a b 2 ab (b) a 2 + b2 (d)

a 2 + b2

a 2 + b2 a 2 − b2 32. The pair of lines joining origin to the intersection of

y2 + = 1 by the line lx + my + n = 0 a 2 b2 are coincident if the curve

(a) a2l2 + b2m2 = n2 (b) (c)

b2 1 + = 2 2 2 l m n

(d) none of these + =n a 2 b2 33. The auxiliary circle of a family of ellipse passes through origin and makes intercept of 8 and 6 units on x-axis and y-axis respectively. If eccentricity of all such family is 1/2 then locus of focus will be 12

MATHEMATICS TODAY | SEPTEMBER ‘17

x2 y2 − = 25 16 9 (d) none of these

(c)

34. Equation of a common tangents to the curves y2 = 8x and xy = –1 is (a) 3y = 9x + 2 (b) y = 2x + 1 (c) 2y = x + 8 (d) y = x + 2 35. If the foci of the ellipse 16x2 + 7y2 = 112 and of the

, then length of latus rectum of ellipse is

(a) 4

x2 y2 + = 25 16 9 (b) 4x2 + 4y2 – 32x –24y + 75 = 0 (a)

hyperbola (a) 3

y2 x2 1 coincide, then a = − 2 = 144 a 25 (b) 9 (c) 81 (d) 8

36. The point of the curve 3x2 – 4y2 = 72 which is nearest to the line 3x + 2y – 1 = 0 is (a) (6, 3) (b) (6, –3) (c) (6, 6) (d) (6, 5) 37. Let PQ be a double ordinate of hyperbola x2

y2 − = 1 . If O be the centre of the hyperbola and a 2 b2 OPQ is an equilateral triangle then the eccentricity e is (b) > 2 (a) > 3 2 (c) > (d) none of these 3

38. If x = 9 is the chord of contact of the hyperbola x2 – y2 = 9, then the equation of the corresponding pair of tangents is (a) 9x2 – 8y2 + 18x – 9 = 0 (b) 9x2 – 8y2 – 18x + 9 = 0 (c) 9x2 – 8y2 – 18x – 9 = 0 (d) 9x2 – 8y2 + 18x + 9 = 0 2 2 2 2 39. The equation x + ( y − 1) − x + ( y + 1) = K will represent a hyperbola for (a) K ∈ (0, 2) (b) K ∈ (0, ∞) (c) K ∈ (1, ∞) (d) none of these

40. A(3, 4), B(0, 0) and C(3, 0) are vertices of DABC. If 'P' is a point inside DABC, such that d(P, BC) < min. {d(P, AB), d(P, AC)}. Then maximum of d(P, BC) is [d(P, BC) represent distance between P and BC] (a) 1 (b) 1/2 (c) 2 (d) none of these

MATHEMATICS TODAY | SEPTEMBERâ&#x20AC;&#x2DC;17

41. If coordinates of the vertices of a triangle are (2, 0), (6, 0) and (1, 5) then distance between its orthocentre and circumcentre is (a) 4 (b) 6 (c) 5 (d) none of these 42. If the vertices of a triangle are A(1, 4), B(3, 0) and C(2, 1) then the length of the median passing through C is (a) 1 (b) 2 (c) 2 (d) 3 43. The equation to the locus of a point which moves so that its distance from x-axis is always one half its distance from the origin is (b) x2 – 3y2 = 0 (a) x2 + 3y2 = 0 (c) 3x2 + y2 = 0 (d) 3x2 – y2 = 0 44. A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The coordinates of the point A are  13  (a)  , 0  5 

5  (b)  , 0   13 

(d) none of these

45. The locus of the point of intersection of the lines x + 4y = 2a sinq, x – y = a cosq where q is a variable parameter is (a) 5x2 + 20y2 = a2 (b) 5x2 + 20y2 = 2a2 (c) 5x2 + 20y2 = 3a2 (d) 5x2 + 20y2 = 4a2 1. (d) : The perpendicular distance of (1, 3) from the line 3x + 4y = 5 is 2 units while, sec2q + 2cosec2q ≥ 3 {as sec2q, cosec2q ≥ 1} 2. (d) : Let the common difference of A.P. is d then b = a + 2d and c = a + 6d, so variable straight line will be ax + (a + 2d)y + a + 6d = 0 ⇒ a(x + y + 1) + d(2y + 6) = 0 which always passes through (2, –3). 3. (b) : We have AB = 10, BC = 5. By bisector property

AD 10 2 = = DC 5 1

1 1 ⇒ Co-ordinates of D are  ,  . 3 3 Hence equation of BD is  (1 / 3) − 1  y −1 =  (x − 5) or x − 7 y + 2 = 0  (1 / 3) − 5  14

MATHEMATICS TODAY | SEPTEMBER ‘17

4. (b) : Let side AB is x |2 −1− 2 | 1 = \ Length AD = 2 2 1 In DABD, sin60º = 2x

60°

3 1 2 = ⇒ x= 2 3 2x \ Area of an equilateral triangle 32 3 sq. units = = 4 3 6 5. (a) : As (–1, 1) is a point on 3x – 4y + 7 = 0, the rotation is possible. 3 Slope of the given line = 4 3 −1 1 Slope of the line in its new position = 4 =− 3 7 1+ 4 1 The required equation is y − 1 = − (x + 1) 7 or 7y + x – 6 = 0. 6. (b) : Line at least distance from (3, 1) points will be perpendicular to line joining given two points (1, 2) and (3, 1). Slope of the line joining the points (1, 2) and (3, 1) is 2 −1 1 m= =− 1− 3 2 Hence the slope of the required line is 2 and its equation is y – 2 = 2(x – 1). 7. (d) : It passes through a fixed point (3, 4) −6 =3 Slope of line joining (3, 4) and (1, – 2) is −2 \ Slope of required line = –1/3 1 Equation is y − 4 = − (x − 3) 3 ⇒ x + 3y – 15 = 0 1  8. (d) : Given, y =  x 2 − 2x + 27  6  1 2 Let f (x) = x − 2x + 27 and f ′(x) ≥ 0 in given interval 6 6 < x < 9. Hence range of f (x) = [f (6), f(9)]. For x ∈[6, 9], y take only two integral values. 9. (d) : Equation of required circle is x2 + y2 + 2x + l(x – y) = 0  2+λ λ whose centre of circle is  −  ,   2  2 

2 (λ + 1)2 + 1 2+ λ λ + =   2 2  4 Radius of circle is minimum when l = –1. Hence, required equation of circle is x2 + y2 + x + y = 0

and radius of circle is

10. (c): Let circle touching to the line y = x at (0, 0) point is S1 : x2 + y2 + l(x – y) = 0 and S2 : x2 + y2 + 6x +8 y – 7 = 0 \ Common chord is S1 – S2 = 0 ⇒ l(x – y) – 6x – 8y + 7 = 0 ⇒ x = y = 1/2 11. (b) : Q DDEF is an equilateral with side 2r if radius of circumcircle DEF is R1, then 3 (2r)2 = 3r 2 4 2r ⋅ 2r ⋅ 2r 2r 3r 2 = ⇒ R1 = 4R1 3 \ Radius of the circle touching all the three given

Area of DDEF =

circles = r + R1 = r +

2r 3

(2 + 3)r 3

12. (a) : The prescribed condition gives (x – 3)2 + y2 = 2{x2 + (y – 2)2}

i.e., x2 + y2 + 6x – 8y – 1 = 0 which is a circle with centre (–3, 4) and radius = 9 + 16 + 1 = 26 > 5 13. (a) : Obviously both circles, pass through the origin and therefore O must be the point of contact. \ Centres P, O, Q are collinear.  b c  − , −  , (0, 0) and a a

 B C  − , −  are collinear A A

c C = or cB = bC b B 14. (b) : Given that circle touches y-axis at (0, 2) point and intersects the x-axis at (4/5, 0) and (5, 0).  29  Centre:  , 2   10  ∴

15. (b) : Let (r cosq, r sinq) be any point on the curve 2x2 + 10y2 + 6xy =1 then 2r2 cos2q + 10r2 sin2q + 6r2sinq cosq = 1

Also, r 2 =

2 + 8 sin θ + 3 sin 2θ 1 ⇒ rmax = 1 = 6 + 3 sin 2θ − 4 cos 2θ Minimum distance between curve = 2. 16. (a) : Let other end is (h, k) h+3 k + 4 So centre ≡  ,  touching x-axis means  2 2  k+4 r= 2

k+4 h+3  k + 4  =  − 3 +  − 4  2   2  2

Hence locus is x2 – 6x – 16y + 9 = 0 Which is a parabola. 17. (d) : The two parabolas intersect at (0, 0) and (4a, 4a). The equation of their common chord must be y = x which must be same as given line 2bx + 3cy + 4d = 0 2b = –3c, d = 0 ⇒ (2b + 3c)2 + d2 = 0 ⇒ 7 1 18. (c): y′ = 2(x – 3) = 1 gives the point  ,  and the 2 4 1  7 required tangent is y − = 1  x −   2 4 or 4y – 4x + 13 = 0. 19. (b) : t1t2 = –1 as AB is focal chord x2 – 3x + a = 0 ; x1 + x2 = 3 and x1x2 = a y2 + 6y + b = 0 ; y1 + y2 = – 6 and y1y2 = b 1 x1x2 = 2 ⋅ t12 = 1 = a t1

 2 y1 y2 = 2t1  −  = −4 = b  t1  \ a = 1, b = – 4 Equation of circle with AB as diameter is x2 + y2 – 3x + 6y – 3 = 0

20. (c): Let B be (at2, 2at) 2 Slope of AB = . t Equation of BC is t y − 2at = − (x − at 2 ) 2 This meets y = 0 at C whose x-coordinate = 4a + at2 Also, D = (at2, 0) \ DC = 4a + at2 – at2 = 4a. MATHEMATICS TODAY | SEPTEMBER‘17

21. (a) : \ Normal at P(t1) meets at Q(t2) 2 t2 = − − t1 t1 | t2 | ≥ 2 2

Hence A = (6, 0), B = (0, 8)

For minimum length of OQ, |t2| should be minimum i.e. | t2 | = 2 2

1 × 6 × 8 = 24 sq. units 2 27. (b) : Let P(a cosq, b sinq) be the variable point on Area of DOAB =

If t2 = −2 2 ⇒ t1 = 2 Slope of OQ =

2 2 = m1 and of OP = = m2 t2 t1

\ m1m2 = –1 ⇒ DOPQ is right angled triangle. 22. (a) : Any point on the line y = mx can be taken as (t, mt). Equation of the chord of parabola with this as mid point ymt – 2(x + t) = m2t2 – 4t It passes through (4, 4) 4mt – 2(4 + t) = m2t2 – 4t ⇒ m2t2 – 2(2m + 1)t + 8 = 0 For two such chords, we must have D > 0 ⇒ (2m + 1)2 – 8m2 > 0 1+ 2 1− 2 4m2 – 4m – 1 < 0 ⇒ <m< 2 2 23. (c): Product of roots < 0 24. (d) : Given points are co-normal points ⇒ p+q+r=0 Common root is 1. a 25. (c): ∆ = h g

h b f

g 1 0 1 f = 0 4 8 = −16 =/ 0 c 1 8 13

⇒ does not represent a pair of straight line. Again h2 = 0 and ab = 4 ⇒ h2 < ab ⇒ An ellipse. 26. (c): Any point on the ellipse

y2 + =1 (3 2 )2 (4 2 )2

can be taken as (3 2 cos θ, 4 2 sin θ) and the slope of 32(3 2 cos θ) 4 b2 x = − cot θ ... (i) the tangent = − 2 = − 3 18 (4 2 sin θ) a y 4 ... (ii) Given slope of the tangent = − 3 From equations (i) and (ii), we get π cot θ = 1 ⇒ θ = 4 16

Hence the equation of the tangent is 1 1 x⋅ y⋅ 2+ 2 = 1 i.e. x + y = 1 6 8 3 2 4 2

MATHEMATICS TODAY | SEPTEMBER ‘17

y2 + = 1 . Then, a 2 b2 A = Area of DPS1S2,

the ellipse

1 = 2

a cos θ b sin θ 1 1 ae 0 1 = b sin θ × 2ae = abe sinq 2 −ae 0 1

Area = abe sinq, which is maximum when q = p/2 \ Amax = abe

28. (d) : Q Centre of ellipse (0, 0) and centre of circle is (7, 2 30 ) and radius is 4 \ Maximum distance of any point on the circle from the centre of the ellipse = (7 − 0)2 + (2 30 − 0)2 + 4 = 17 29. (a) : Clearly a2 + 5 > a2 + 2  1 So (a2 + 2) = (a2 + 5) 1 −   3 3a2 + 6 = 2a2 + 10 ⇒ a2 = 4

x2 y2 + =1 6 9 2×6 \ Length of latus rectum = =4 3 2 2 x y 30. (a) : + =1 9 4 Let any point on ellipse be (3 cosq, 2 sinq) Since sinq and cosq can be rational for infinite many value of θ ∈[0, 2π] . 31. (b) : The circle through the points of intersection of the two curves will have centre at origin. \ Equation of ellipse is

2 2 Solving x2 – y2 = 0 and x + y = 1 , we get a 2 b2 2 2 ab x2 = y2 = 2 2 a +b 2a2b2 2ab Therefore radius of circle = 2 2 = a +b a 2 + b2

lx + my =1 −n By Theory of Homogenization, we can get the pairs of 2 x 2 y 2  lx + my  line 2 + 2 =    −n  a b

32. (a) : lx + my + n = 0 ⇒

n  2 n  2 2 2  2 − l  x +  2 − m  y − 2lmxy = 0 a  b  This represent a pair of coincident lines if  n2   n2  l 2m2 −  2 − l 2   2 − m2  = 0 a b  n4

n2m2 n2l 2 = + 2 ⇒ a2l 2 + b2m2 = n2 2 2 2 ab a b 33. (b) : Centre is (4, 3) and distance of focus from 5 centre is ae = 2 25 \ Locus is (x – 4)2 + (y – 3)2 = 4 34. (d) : Equation of a tangent at (at2, 2at) to y2 = 8x is ty = x + at2 where 4a = 8 i.e. a = 2 ⇒ ty = x + 2t2 which intersects the curve xy = –1 at x(x + 2t 2 ) = −1 clearly t =/ 0 the points given by t 2 2 or x + 2t x + t = 0 and will be a tangent to the curve if the roots of this quadratic equation are equal, for which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and an equation of a common tangent is y = x + 2. 35. (b) : For the ellipse, 2

e =

a 2 − b2

9 = 16

x2 y2 + =1 7 16 3 ⇒ e= 4

144 + a2 12 × 5 12 According to given condition, 12 144 + a2 × ⇒ 15 = 144 + a2 5 12 ⇒ a2 = 225 – 144 = 81 ⇒ a = 9 3=

1 1

3  dy  3 x ⇒  = ⋅ 1 =−  dx  4 y1 2

⇒ x1 = –2y1 which satisfy 3x2 – 4y2 = 72 ⇒ 12y12 – 4y12 = 72 ⇒ y12 = 9 ⇒ y1 = ± 3 ⇒ x1 = ± 6 Hence required points are (–6, 3) and (6, –3). 37. (c): Let P be (a, b) then PQ = 2b and OP = α2 + β2 Since OPQ is an equilateral triangle, OP = PQ a2 + b2 = 4b2 = a2 = 3b2 ⇒ α = ± 3β ∵ (a, b) is situated on the hyperbola ∴ b2

3β2 β2 3 1 1 β2 − 2 =1 ⇒ 2 − 2 = 2 > 0 − = 1 ⇒ 2 2 2 β a b a b a b 2 1 1 2 4 ⇒ e> > ⇒ e2 − 1 > ⇒ e > 3 3 3 3 α2

a2 38. (b) : x = 9 meets the hyperbola x2 – y2 = 9 at (9, 6 2 ) and (9, − 6 2 ) . The equations of the tangents to the hyperbola at these points are 3x − 2 2 y − 3 = 0 and 3x + 2 2 y − 3 = 0 \ Joint equation of the two tangents is (3x − 2 2 y − 3)(3x + 2 2 y − 3) = 0 ⇒ (3x − 3)2 − (2 2 y)2 = 0 ⇒ 9x2 – 8y2 – 18x + 9 = 0.

a2 3 \ ae = 4 × = 3 ... (i) 4 2 2 y x For the hyperbola − =1 144 a2 25 25 144 a2 2 2 2 + a +b 25 25 ⇒ e ′ = 144 + a e′2 = = 144 12 a2 25 ∴ ae ′ =

36. (b) : For the nearest point on the curve, tangent drawn to curve at that point should be parallel to the given line  dy  =0 ∴ 6x1 − 8 y1    dx (x , y )

(ii)

39. (a) : We have,

x 2 + ( y − 1)2 − x 2 + ( y + 1)2 = K

which is equivalent to |S1P – S2P| = const. Where S1 ≡ (0, 1), S2 ≡ (0, –1) and P ≡ (x, y) Now, 2a = K [where 2a is the transverse axis and e is the eccentricity] and 2ae = S1S2 = 2 2 Dividing, we have e = K Since, e > 1 for a hyperbola, therefore K < 2 Also, K must be a positive quantity. Hence, we have, K ∈ (0, 2). 40. (a) : The required point is point of intersection of internal angle bisectors \ P(x, y) = incentre of D.  2 + 6 +1 0 + 0 + 5   5  , 41. (c): Centroid ≡   ≡  3,   3 3   3 MATHEMATICS TODAY | SEPTEMBER‘17

and for orthocentre equation of line perpendicular to AB passing through C(1, 5) is x=1 ... (i) Eq. of line perpendicular to AC B(6, 0) and passing through B(6, 0) is y = (x – 6)1/5 ⇒ 5y = x – 6 ... (ii) Solving (i) and (ii) we get Orthocentre ≡ (1, –1) We known cent roid divides or t ho cent re and circumcentre in 2 : 1 (internally) \ Circumcentre is (4, 3) \ Distance between orthocentre and circumcentre = (1 − 4)2 + (−1 − 3)2 = 5 42. (a) : Median CD =

(2 − 2)2 + (2 − 1)2 = 1

43. (b) : Let the point (h, k) 1 2 2 Given, | k | = h +k 2 Squaring 4k2 = h2 + k2 ⇒ h2 – 3k2 = 0 \ Locus x2 – 3y2 = 0 44. (a) : Let, the coordinates of point A is (a, 0) Now –mAB = mAR If AR makes an angle q with +ve x-axis, then AB makes (p – q), therefore –mAB = mAR 0−2  0−3  = −  α − 1   α − 5 

13 ⇒ 2(α − 5) = −3(α − 1) ⇒ α = 5  13  \ A is  , 0  5  45. (d) : We have, x + 4y = 2a sinq x – y = a cosq

( x + 4 y )2 + (x − y)2 = a2

4 ⇒ 5x + 20y2 = 4a2

... (i) …..(ii) [From (i) and (ii)]

MATHEMATICS TODAY | SEPTEMBER ‘17



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

A number of the form x + iy, where x, y ∈ R and i = −1, is called a complex number and ‘i’ is called iota. A complex number is usually denoted by z and the set of complex numbers is denoted by C. i.e., C = {x + iy : x ∈ R, y ∈ R, i = −1} 5 + 3i, –1 + i, 0 + 4i, 4 + 0i etc. are complex numbers. z For any positive real number a, we have −a = −1 × a = −1 a = i a z

The property a b = ab is valid only if at least one of a and b is non-negative.

Integral powers of iota (i) : Since i = −1 hence we have i2 = –1, i3 = –i and i4 = 1. To find the value of in(n > 4). Let n = 4q + r where 0 r 3, \ in = i4q + r = (i4)q.(i)r = (1)q (i)r = ir In general we have the following results i4n = 1, i4n + 1 = i, i4n + 2 = –1, i4n + 3 = –i, where n is any integer.

REAL AND IMAGINARY PARTS OF A COMPLEX NUMBER If x and y are two real numbers, then a number of the form z = x + iy is called a complex number. Here ‘x’ is called the real part of z and ‘y’ is known as the imaginary part of z. The real part of z is denoted by Re(z) and the imaginary part by Im(z). If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4. A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0.

ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS Let z1 = a + ib and z2 = c + id be two complex numbers then, Addition (z1 + z2) : (a + ib) + (c + id) = (a + c) + i(b + d) Subtraction (z1 – z2) : (a + ib) – (c + id) = (a – c) + i(b – d) Multiplication (z1 z2) : (a + ib)(c + id) = (ac – bd) + i(ad + bc) a + ib Division (z1/z2) : c + id (where at least one of c and d is non-zero) (a + ib) (c − id ) (ac + bd ) i(bc − ad ) ⋅ = = + 2 . (c + id ) (c − id ) c 2 + d 2 c + d2 z Properties of algebraic operations on complex numbers : Let z1, z2 and z3 are any three complex numbers then their algebraic operations satisfy following properties : z z1 + z2 = z2 + z1 z (z1 + z2) + z3 = z1 + (z2 + z3) z z1z2 = z2z1 z (z1z2)z3 = z1(z2z3). z z1(z2 + z3) = z1z2 + z1z3 z (z2 + z3)z1 = z2z1 + z3z1. EQUALITY OF TWO COMPLEX NUMBERS Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are said to be equal if and only if their real and imaginary parts are separately equal. i.e., z1 = z2 x1 + iy1 = x2 + iy2 x1 = x2 and y1 = y2. Complex numbers do not possess the property of order i.e., (a + ib) < (or) > (c + id) is not defined. For example, the statement (9 + 6i) > (3 + 2i) makes no sense.

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants.

MATHEMATICS TODAY | SEPTEMBER‘17

CONJUGATE OF A COMPLEX NUMBER z Conjugate complex number : If there exists a complex number z = a + ib, (a, b) ∈R, then its conjugate is defined as z = a – ib.

In general |z1z2z3...zn| = |z1||z2||z3|...|zn|

P(z)

X z

Q(z)

Geometrically, the conjugate of z is the reflection or mirror image of z about real axis. Properties of conjugate : If z, z1 and z2 are existing complex numbers, then we have the following results: z (z ) = z z1 ± z2 = z1 ± z2 z

z1z 2 = z1 z 2 In general, z1 ⋅ z2 ⋅ z3 .....zn = z1 ⋅ z2 ⋅ z3 .....zn  z1  ( z1 )  z  = z ,[z2 ≠ 0] ( 2) 2

(z )n = (z n )

z + z = 2Re(z ) = 2Re(z ) = purely real z z − z = 2iIm(z ) = purely imaginary z z1z2 + z1z2 = 2Re( z1z2 ) = 2Re( z1z2 ) Reciprocal of a complex number : For an existing non-zero complex number z = a + ib, the reciprocal 1 z z is given by z −1 = = = . z z ⋅ z | z |2 z

MATHEMATICS TODAY | SEPTEMBER ‘17

| zn | = | z |n,n ∈ N |z1 ± z2|2 = |z1|2 + |z2|2 ± (z1z2 + z1z2)

or |z1|2 + |z2|2 ± 2Re(z1z2) z |z1 + z2|2 = |z1|2 + |z2|2 ⇒ 1 is purely real z2 |z1 + z2|2 + |z1 – z2|2 = 2{|z1|2 + |z2|2} (Law of parallelogram)

VARIOUS REPRESENTATIONS OF A COMPLEX NUMBER A complex number can be represented in the following form: z G e ome t r ic a l repre s ent at ion (C a r te si a n representation): The complex number z = a + ib = (a, b) is represented by a point P whose coordinates are referred to rectangular axes XOX′ and YOY′ which are called real and imaginary axis respectively. This plane is called argand plane or argand diagram or complex plane or Gaussian plane. Y

X′

Real axis O

Y′

MODULUS OF A COMPLEX NUMBER Modulus of a complex number z = a + ib is defined by a positive real number given by | z | = a2 + b2 , where a, b are real numbers. Geometrically |z| represents the distance of point P from the origin, i.e. |z| = OP. If | z | = 1 the corresponding Y P(z) complex number is known as unimodular complex number. Clearly z lies on a circle of unit q X O M radius having centre (0, 0). Properties of modulus z |z| ≥ 0 ⇒ |z| = 0 if z = 0 and |z| > 0 if z ≠ 0. z –|z| Re(z) |z| and –|z| Im(z) |z| z |z| = |z| = |–z| = |–z|=|zi| z zz = |z|2 ⇒ Purely real

z1 |z | = 1 , (z2 ≠ 0) z2 | z2 |

Imaginary axis

Imaginary

q -q Real

|z1z2| = |z1||z2|

P(a, b) b q a M

| z | = a 2 + b2 X

Angle of any complex number with positive direction of x–axis is called amplitude or argument of z.  b i.e., amp(z ) = arg (z ) = tan −1    a Trigonometrical (Polar) representation : In DOPM, let OP = r, then a = r cos q and b = r sin q. Hence z can be expressed as z = r(cos q + i sin q) where r = |z| and q = principal value of argument of z. For general values of the argument z = r[cos(2np + q) + isin(2np + q)] Sometimes (cos q + i sin q) is written in short as cisq. Vector representation : If P is the point (a, b) on the argand plane corresponding to the complex number z = a + ib.

Then OP = aiˆ + bjˆ , \ | OP | = a2 + b2 = | z | and  b arg(z) = direction of the vector OP = tan −1    a Eulerian representation (Exponential form) : Since we have eiq = cosq + i sinq and thus z can be expressed as z = reiq, where |z| = r and q = arg(z). Y Principal value of arg (z) : The value q of the a p–a argument, which satisfies (+, +) (–, +) X the inequality –p < q p X′ O (+, –) (–, –) is called the principal –a –(p – a) value of argument, where Y′ b α = tan −1 (acute angle) a and principal values of argument z will be a, p – a, –p + a and –a as the point z lies in the 1st , 2nd , 3rd and 4th quadrants respectively. Properties of arguments z arg(z1z2) = arg(z1) + arg(z2) + 2kp, K ∈ I In general arg(z1z2z3...zn) = arg(z1) + arg(z2) + arg(z3) + ... + arg(zn) + 2kp k ∈ I z arg(z1z2) = arg(z1) – arg(z2) z

z z z

z  arg  1  = arg z1 – arg z2 + 2kp k ∈ I  z2  z arg   = 2arg z + 2kp k ∈ I z arg(zn) = n arg z + 2kp k ∈ I z  z  If arg 2 = q, then arg  1  = 2kp – q, where k ∈ I  z   z2  1 1 arg z = –arg z = arg z

π arg(z – z) = (4k ± 1) , z not being purely 2 imaginary arg(–z) = arg(z) + (2n + 1)p arg(z) – a r g (z) = ±p (If z is purely imaginary) z1z2 + z1z2 = 2|z1| |z2| cos (q1 – q2) where q1 = arg(z1) and q2 = arg(z2) The general value of arg(z) is 2np – arg(z).

To find the square root of a – ib, replace i by –i in the above results.

LOGARITHM OF A COMPLEX NUMBER Let log(x + iy) = loge(reiq) = loger + loge eiq = loge r + iq  y = log e (x 2 + y 2 ) + itan −1    x

Hence, loge (z) = loge |z| + i ampz

GEOMETRY OF COMPLEX NUMBERS Q(z 2) Distance formula : The distance between two points P(z1) and Q(z2) is given by PQ = |z2 – z1| = |affix of Q – affix of P| P(z1) z Section formula : If R(z) divides the line segment joining P(z1) and Q(z2) in the ratio m1 : m2(m1, m2 > 0) then z

m1z2 + m2 z1 m1 + m2 m1z2 − m2 z1 z For external division z = m1 − m2 Equation of a straight line z Parametric form : Equation of a straight line joining the point having affixes z1 and z2 is z = tz1 + (1 – t)z2, when t ∈ R z Non-parametric form : Equation of a straight line joining the points having affixes z1 and z2 z z 1 is z1 z1 1 = 0 z2 z2 1 z

 z +a z −a  a + ib = ±  +i  , for b> 0 2 2    =± 

z +a z −a −i  , for b < 0. 2 2 

⇒ z(z1 – z2) – z(z1 – z2) + z1z2 – z2z1 = 0. Condition of Collinearity of three points : Three points A(z1), B(z2) and C(z3) are collinear z1 z1 1 if, z2 z2 1 = 0 z3 z3 1 z1 − z 2 z 2 − z3 z1 − z3 = = z1 − z 2 z 2 − z3 z1 − z3 General equation of a straight line : The general equation of a straight line is of the form az + az + b = 0, where a is non zero complex number and b is any real number. Slope of a line : The complex slope of the line a coeff.of z and real az + az + b = 0 is − = − a coeff. of z (a + a ) slope of the line az + az + b = 0 is − i . (a − a ) or

SQUARE ROOT OF A COMPLEX NUMBER Let z = a + ib be a complex number, Then

For internal division z =

MATHEMATICS TODAY | SEPTEMBER‘17

Length of perpendicular : The length of perpendicular from a point z 1 to the line | az1 + az1 + b | az + az + b = 0 is given by or |a | + |a | | az1 + az1 + b | 2|a | Equation of a circle : P(z) z The equation of a circle r whose centre is at point C(z 0) having affix z0 and radius r is |z – z0| = r z If the centre of the circle is at origin and radius r, then its equation is |z| = r. z |z – z0| < r represents interior of a circle, |z – z0| = r lie on the circle and |z – z0| > r represents exterior of the circle. z General equation of a circle : The general equation of the circle is zz + az + az + b = 0 (where a is complex number and b ∈ R) with z

Complex number as a rotating arrow in the argand plane : Let z = r(cosq + isinq) = reiq ...(i) i eiq be a complex number representing a point P in the argand plane. Y Q(zeif) Then OP = |z| = r P(z) f and∠POX = q Now consider complex X′ q X O number z1 = zeif Y′ or z1 = reiq eif i(q + f) {From (i)} = re Clearly the complex number z1 represents a point Q in the argand plane, when OQ = r and ∠QOX = q + f. Clearly multiplication of z with eif rotates the vector OP through angle f in anticlockwise sense. Similarly multiplication of z with e–if will rotate the vector OP in clockwise sense. z If z1, z2 and z3 are the affixes of the points A, B and C such that AC = AB and ∠CAB = q. C(z 3)

centre and radius as – a and | a |2 −b = aa − b respectively. Equation of circle in diametric form : If end points of diameter represented by A(z1) and B(z2) and P(z) be any point on the circle then, (z – z1)(z – z2) + (z – z2)(z – z1) = 0. which is required equation of circle in diametric form.

ROTATION THEOREM Rotational theorem is used to find the angle between two intersecting lines. This is also known as coni method. Let z1, z2 and z3 be the affixes Y C(z 3) of three points A, B and C B(z 2) a respectively taken on argand A(z 1) plane. f q Then we have AC = z3 – z1 X O and AB = z2 – z1 and let arg AC = arg(z3 – z1) = q and AB = arg(z2 – z1) = f Let ∠CAB = a, Q ∠CAB = a = q – f = arg AC – arg AB z −z  = arg(z3 – z1) – arg(z2 – z1) = arg  3 1   z2 − z1  or angle between AC and AB  affix of C − affix of A  = arg    affix of B − affix of A  22

MATHEMATICS TODAY | SEPTEMBER ‘17

A(z 1)

B(z 2)

Therefore, AB = z2 – z1, AC = z3 – z1. Then AC will be obtained by rotating AB through an angle q in anticlockwise sense, and therefore, iθ AC = ABe or (z3 – z1) = (z2 – z1)eiq z −z or 3 1 = eiq z2 − z1 z

If A, B and C are three points in argand plane such that AC = AB and ∠CAB = q then use the rotation about A to find eiq, but if AC ≠ AB use coni method. If four points z1, z2, z3 and z4 are concyclic then (z 4 − z1 )(z2 − z3 ) is real (z 4 − z2 )(z1 − z3 )

 (z − z )(z − z )  or arg  2 3 4 1  = ± π, 0  (z1 − z3 )(z 4 − z2 ) 

TRIANGLE INEQUALITIES In any triangle, sum of any two sides is greater than the third side and difference of any two sides is less than the third side. By applying this basic concept to the set of complex numbers we are having the following results.

z z

|z1 ± z2| |z1| + |z2| |z1 ± z2| ≥ ||z1| – |z2||

STANDARD LOCI IN THE ARGAND PLANE If z is a variable point and z1, z2 are two fixed points in the argand plane, then z |z – z1| = |z – z2| ⇒ Locus of z is the perpendicular bisector of the line segment joining z1 and z2. z |z – z1| + |z – z2| = constant > |z1 – z2| ⇒ Locus of z is an ellipse z |z – z1| + |z – z2| = |z1 – z2| ⇒ Locus of z is the line segment joining z1 and z2. z |z – z1| – |z – z2| = |z1 – z2| ⇒ Locus of z is a straight line joining z1 and z2 but z does not lie between z1 and z2. z |z – z1| – |z – z2| = constant (> 0 but ≠ |z1 – z2|) ⇒ Locus of z is a hyperbola. z |z – z1|2 + |z – z2|2 = |z1 – z2|2 ⇒ Locus of z is a circle with z1 and z2 as the extremities of diameter. z |z – z1| = k|z – z2|, (k ≠ 1) ⇒ Locus of z is a circle. z

 z − z1  arg  = a (fixed) ⇒ Locus of z is a segment  z − z2  of circle.  z − z1  arg  = ±p/2 ⇒ Locus of z is a circle with  z − z2  z1 and z2 as the vertices of diameter.  z − z1  arg  = 0 or p ⇒ Locus of z is a straight  z − z2  line passing through z1 and z2.

DE’ MOIVRE’S THEOREM z If n is any rational number, then (cos q + i sin q)n = cos nq + i sin nq. z If z = (cos q1 + i sin q1)(cos q2 + i sin q2) (cos q3 + i sin q3)...(cosqn + i sin qn) then z = cos(q1 + q2 + q3 + ... + qn) + i sin(q1 + q2 + q3 + ... + qn) where q1, q2, q3...qn ∈ R. z If z = r(cos q + i sin q) and n is a positive integer,   2k π + θ   2k π + θ   then z1/n = r1/n cos   + i sin  n   ,  n   where k = 0, 1, 2, 3,...(n – 1). Deductions: If n ∈ Q, then z (cos q – i sin q)n = cos nq – i sin nq z (cos q + i sin q)–n = cos nq – i sin nq z (cos q – i sin q)–n = cos nq + i sin nq

π  π  (sin θ + i cos θ)n = cos n  − θ + i sin n  − θ 2  2  This theorem is not valid when n is not a rational number or the complex number is not in the form of cos q + i sin q.

ROOTS OF A COMPLEX NUMBER z nth roots of complex number (z1/n) Let z = r(cos q + i sin q) be a complex number. By using De’moivre’s theorem nth roots having n distinct values of such a complex number are given by 2k π + θ   2k π + θ z1/n = r1/n cos + i sin , n n   where k = 0, 1, 2, ..., (n – 1). z Properties of the roots of z1/n z All roots of z1/n are in geometrical progression with common ratio e2pi/n. z Sum of all roots of z1/n is always equal to zero. z Product of all roots of z1/n = (–1)n – 1. z Modulus of all roots of z1/n are equal and each equal to r1/n or |z|1/n. z Amplitude of all the roots of z1/n are in A.P. with 2π common difference . n z All roots of z1/n lies on the circumference of a circle whose centre is origin and radius equal to |z|1/n. Also these roots divides the circle into n equal parts and forms a polygon of n sides. z The nth roots of unity : The nth roots of unity are given by the solution set of the equation xn = 1 = cos 0 + i sin 0 = cos 2kp + i sin 2kp x = [cos 2kp + i sin 2kp]1/n 2k π 2k π x = cos + i sin , where k = 0, 1, 2, ..., (n – 1). n n z Properties of nth roots of unity 2π 2π i(2 π/n) z Let α = cos + i sin =e , the nth roots n n of unity can be expressed in the form of a series i.e., 1, a, a2, ... an – 1. Clearly the series is G.P. with common ratio a i.e., ei(2p/n). z The sum of all n roots of unity is zero i.e., 1 + a + a2 + ... + an – 1 = 0. z Product of all n roots of unity is (–1)n–1. z Sum of pth power of n roots of unity 1 + ap + a2p + ... + a(n – 1)p 0,when p is not multiple of n = n, when p is a multiple of n MATHEMATICS TODAY | SEPTEMBER‘17

The n, nth roots of unity if represented on a complex plane locate their positions at the vertices of a regular polygon of n sides inscribed in a unit circle having centre at origin, one vertex on positive real axis. Cube roots of unity : Cube roots of unity are the solution set of the equation x3 – 1 = 0 ⇒ x = (1)1/3 ⇒ x = (cos0 + i sin0)1/3 2k π  2k π  ⇒ x = cos + i sin  , where k = 0, 1, 2  3  3 2π 2π Therefore roots are 1, cos + i sin , 3 3 4π 4π cos + i sin or 1, e 2 πi /3 , e 4 πi /3 3 3 z

Alternative : x = (1)1/3 ⇒ x3 – 1 = 0 ⇒ (x – 1)(x2 + x + 1) = 0 −1 + i 3 −1 − i 3 x = 1, , 2 2 If one of the complex root is , then other root will be 2 or vice-versa. Properties of cube roots of unity z 1+ + 2=0 3=1 z 0, if n isnot a multiple of 3 z 1 + ωn + ω 2n =  3, if n is a multiple of 3 z The cube roots of unity, when represented on complex plane, lie on vertices of an equilateral triangle inscribed in a unit circle having centre at origin, one vertex being on positive real axis. z Cube root of – 1 are –1, – , – 2.

IMPORTANT POINTS z 0 = 0 + i 0, is the identity element for addition. z 1 = 1 + i 0 is the identity element for multiplication. z The additive inverse of a complex number z = a + ib is –z (i.e. – a – ib). z For every non-zero complex number z, the 1 multiplicative inverse of z is . z z |z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |Im(z)| ≥ Im(z) z z is always a unimodular complex number if z ≠ 0. |z | z |Re(z)| + |Im(z)| 2 |z| 1 z If z + = a, the greatest and least values of |z| are z a − a2 + 4 a + a2 + 4 . respectively and 2 2 24

MATHEMATICS TODAY | SEPTEMBER ‘17

z z z

If |z1| 1, |z2| 1 then z |z1 – z2|2 (|z1| – |z2|)2 + (arg(z1) – arg(z2))2. z |z1 + z2|2 ≥ (|z1| + |z2|)2 – (arg(z1) – arg(z2))2. |z1 + z2|2 = |z1|2 + |z2|2 + 2|z1||z2| cos(q1 – q2). |z1 – z2|2 = |z1|2 + |z2|2 – 2|z1||z2| cos(q1 – q2). If |z1| = |z2| and amp(z1) + amp(z2) = 0, then z1 and z2 are conjugate complex numbers of each other The area of the triangle whose vertices are z, iz and 1 z + iz is | z |2. 2 The area of the triangle with vertices z, wz and 3 2 | z |. z + wz is 4 If z1, z2, z3 be the vertices of an equilateral triangle and z0 be the circumcentre, then z21 + z22 + z23 = 3z20. If z1, z2, z3...zn be the vertices of a regular polygon of n sides and z0 be its centroid, then z21 + z22 +...+ z2n = nz20. If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral iff (z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0 z21 + z22 + z23 = z1z2 + z2z3 + z3z1 1 1 1 + + =0 or z1 − z2 z2 − z3 z3 − z1 If z1, z2, z3 be the affixes of the vertices A, B, C respectively of a triangle ABC, then its orthocentre is a(sec A)z1 + b(sec B)z2 + (c sec C )z3 a sec A + b sec B + c sec C Re(iz) = –Im(z), Im(iz) = Re(z). If the complex numbers z1 and z2 are such that the sum z1 + z2 is a real number, then they are not necessarily conjugate complex. If z1 and z2 are two complex numbers such that the product z1z2 is a real number, then they are not necessarily conjugate complex. If and 2 are the complex cube roots of unity, then (a + b 2)(a 2 + b ) = a2 + b2 – ab (a + b)(a + b )(a 2 + b2 ) = a3 + b3 (a + b + c 2)(a + b 2 + c ) = a2 + b2 + c2 – ab – bc – ca (a + b + c)(a + b + c 2)(a + b 2 + c ) = a3 + b3 + c3 – 3abc If three points z1, z2, z3 connected by relation az1 + bz2 + cz3 = 0 where a + b + c = 0, then the three points are collinear. If z is a complex number, then ez is periodic. If three complex numbers are in A.P., then they lie on a straight line in the complex plane. or

z z

PROBLEMS Single Correct Answer Type

The value of i1 + 3 + 5 +...+ (2n + 1) is (a) i if n is even, – i if n is odd (b) 1 if n is even, – 1 if n is odd (c) 1 if n is odd, – 1 if n is even (d) i if n is even, – 1 if n is odd 1 = 2 cos θ, then x is equal to x (a) cos q + i sin q (b) cos q – i sin q (c) cos q ± i sin q (d) sin q ± i cos q

If x +

3 + 2i sin θ will be real, if q = 1 − 2i sin θ π (a) 2np (b) np + 2 (c) np (d) None of these [where n is an integer] The real part of (1 – cos q + 2i sin q)–1 is 1 1 (a) (b) 3 + 5 cos θ 5 − 3 cos θ 1 1 (c) (d) 3 − 5 cos θ 5 + 3 cos θ 3 , then x2 + y2 is equal to If x + iy = 2 + cos θ + i sin θ (a) 3x – 4 (b) 4x – 3 (c) 4x + 3 (d) None of these ( p + i)2 If = µ + iλ, then 2p − i (a) (c)

( p2 + 1)2 4 p2 − 1 ( p2 − 1)2 4 p2 + 1

(b) (d)

is equal to

( p2 − 1)2 4 p2 − 1 ( p2 + 1)2 4 p2 + 1

1 + iz If z(1 + a) = b + ic and a2 + b2 + c2 = 1, then = 1 − iz a + ib b − ic (a) (b) 1+ c 1+ a a + ic (c) (d) None of these 1+ b 1+ a If a = cos q + i sin q, then = 1− a θ (a) cot q (b) cot 2 θ θ (c) i cot (d) i tan 2 2

The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for (a) x = np

1  (b) x =  n +  π  2

(d) No value of x –

10. The number of solutions of the equation z2 + z = 0 is (a) 1 (b) 2 (c) 3 (d) 4 z −i (z ≠ −i) is a purely imaginary number, then z +i – z z is equal to (a) 0 (b) 1 (c) 2 (d) None of these

11. If

12. The maximum value of |z| where z satisfies the 2 condition z + = 2 is z (a)

3 −1

(b)

3 +1

(c)

(d)

2+ 3

13. If z1 and z2 be complex numbers such that z1 ≠ z2 and |z1| = |z2|. If z1 has positive real part and z2 has (z + z ) negative imaginary part, then 1 2 may be (z1 − z2 ) (a) Purely imaginary (b) Real and positive (c) Real and negative (d) None of these

14. The product of two complex numbers each of unit modulus is also a complex number, of (a) Unit modulus (b) Less than unit modulus (c) Greater than unit modulus (d) None of these 15. If |z1| = |z2| = ..... = |zn| = 1, then the value of |z1 + z2 + z3 +....+ zn| = (a) 1 (b) |z1| + |z2| +....+ |zn| 1 1 1 + + ......... + z1 z2 zn (d) None of these (c)

16. The values of z for which |z + i| = |z – i| are (a) Any real number (b) Any complex number (c) Any natural number (d) None of these MATHEMATICS TODAY | SEPTEMBER‘17

17. Let z be a complex number (not lying on X-axis) of 1 maximum modulus such that z + = 1 . Then z (a) Im(z) = 0 (b) Re(z) = 0 (c) amp(z) = p (d) None of these 18. The minimum value of |2z – 1| + |3z – 2| is (a) 0 (b) 1/2 (c) 1/3 (d) 2/3 19. If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2|, then arg (z1) – arg (z2) is equal to π π (a) –p (b) − (c) (d) 0 2 2 – 20. If z be the conjugate of the complex number z, then which of the following relations is false ? –

–

(b) z z = |z |2

(a) |z| = |z |

– (c) z1 + z2 = z1 + z2 (d) arg (z) = arg (z) 21. If for complex numbers z1 and z2, arg(z1/z2) = 0, then |z1 – z2| is equal to (a) |z1| + |z2| (b) |z1| – |z2|

(d) 0

22. |z1 + z2| = |z1| + |z2| is possible if

1 z1 (c) arg(z1) = arg(z2) (d) |z1| = |z2| –

(a) z2 = z 1

(b) z2 =

iθ 23. If −1 + −3 = re , then q is equal to 2π π π 2π (b) − (c) (d) − (a) 3 3 3 3 24. The value of (–i)1/3 is

(a)

1 + 3i 2

(b)

− 3 −i (c) 2

1 − 3i 2

3 3 +i (d) 2 − iθ

25. The amplitude of e e is equal to (a) sin q (b) –sin q (c) ecos q

(d) esin q

26. The real part of (1 – i)–i is 1  − π/ 4 cos  log 2 (a) e 2  1  − π/ 4 sin  log 2 (b) −e 2  1  1  π/ 4 − π/ 4 sin  log 2 (c) e cos  log 2 (d) e 2  2  26

MATHEMATICS TODAY | SEPTEMBER ‘17

 x −i is equal to 27. i log   x + i  (a) p + 2tan–1 x (b) p – 2tan–1 x –1 (c) –p + 2tan x (d) –p – 2tan–1 x –

–

28. The equation zz + az + az + b = 0, b ∈ R represents a circle if (a) |a|2 = b (b) |a|2 > b (d) None of these (c) |a|2 < b 29. Let the complex numbers z1, z2 and z3 be the vertices of an equilateral triangle. Let z0 be the circumcentre of the triangle, then z12 + z22 + z23 = (a) z20 (b) –z20 (c) 3z02 (d) –3z02

30. If the vertices of a quadrilateral be A = 1 + 2i, B = –3 + i, C = –2 – 3i and D = 2 – 2i, then the quadrilateral is (a) Parallelogram (b) Rectangle (c) Square (d) Rhombus 1 =2, ω then maximum distance of from origin is (a) 2 + 3 (b) 1 + 2 (c) 1 + 3 (d) None of these

31. If

is a complex number satisfying ω +

Multiple Correct Answer Type

32. Let z1, z2, z3, ...., zn are the complex numbers such  n  n 1  that |z1| = |z2| = .... = |zn| =1. If z =  ∑ z k   ∑   k =1   k =1 z k  then (a) z is purely imaginary (b) z is real (c) 0 < z n2 (d) z is a complex number of the form a + ib 33. Let x1, x2 are the roots of the quadratic equation x2 + ax + b = 0 where a,b are complex numbers and y1, y2 are the roots of the quadratic equation y2 + |a|y + |b| = 0. If |x1| = |x2| = 1 then (a) |y1| = 1 (b) |y2| = 1 (d) |y1| = |y2| = 2 (c) |y1| ≠ |y2| 34. If the equation z3 + (3 + i)z2 – 3z – (m + i) = 0 where m ∈ R, has atleast one real root then ‘m’ can have the value equal to (a) 1 (b) 2 (c) 3 (d) 5

35. Let z1, z2, z3 in G.P. be roots of the equation z3 – bz2 + 3z – 1 = 0 then (a) z2 = 1 (b) z2 = 2 (c) b = 3 (d) b can be –3

36. The complex numbers satisfying the equation (3z + 1) (4z + 1)(6z+ 1)(12z + 1) = 2 is /are 33 − 5 33 + 5 (b) (a) 24 24 −i 23 − 5 −i 23 + 5 (d) (c) 24 24 37. If all the three roots of az3 + bz2 + cz + d = 0 have negative real parts (a, b, c ∈ R) then (a) ab > 0 (b) bc > 0 (c) ad > 0 (d) bc – ad > 0 Comprehension Type

Paragraph for Q. No. 38 to 40 The complex slope M of a line joining two points z1 z −z and z2 in complex plane is defined as M = 1 2 .Its z1 − z2 real slope m is tanq, where q is inclination of the line . 38. M = (a)

1 + im 1 − im 2m

 1 − m2  (b) +i  1 + m2  1 + m2   1 + m2  (c) +i  1 + m2  1 − m2  2m

(d)

1 − im m−i

39. The inclination of a line whose complex slope is – 2 (where is a non real cube root of unity) is 2π π π π (a) (b) (c) (d) 3 3 6 12 40. Which of the following is false? (a) |M| = 1 (b) M = m is never possible  M − 1 (c) m = i   M + 1 (d) M = cis 2q Paragraph for Q. No. 41 to 43

cis2Kπ If a is any of 7th roots of unity, then a = 7 (K = 0 to 6) and

∑α i =1

= −1(α ≠ 1) and a7 = 1

41. The equation whose roots are a + a2 + a4 and a3 + a5 + a6 is (a ≠ 1) (a) x2 + x – 2 = 0 (b) x2 + x + 2 = 0 2 (c) x – x + 2 = 0 (d) x2 – x – 2 = 0

42. If f(x) = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + 7x6 then for a ≠ 1, f (x) + f (ax) + f (a2x) + f (a3x) +....+ f (a6x) = (a) 42 (b) 21 (c) 14 (d) 7 π 2π 3π 4π 5π 6π sin sin = sin sin sin = 7 7 7 7 7 7 7 7 7 7 (b) (c) (d) (a) 16 8 32 64

43. sin

Matrix-Match Type

44. Match the following. Column-I Column-II (p) 0 (A) The maximum value of – ||z – | – |z – || (where |z| = – 5 and , complex cube root of unity) is 2/3 (B) Tangent drawn to circle (x – 1)2 + (q) (y – 1)2 = 5 at a point P meets the line 2x + y + 6 = 0 at Q on the x-axis then the value of

(PQ)2 is 2 (C) One vertex of an equilateral (r) triangle is at the origin and the other two vertices are given by 2z2 + 2z + k = 0, then k is (s)

45. Match the following. Column –I

Column –II

z + i z2 (A) If z = 1 then |z| equals z2 + i z1

(p)

(q) z − 2i be purely z + 2i imaginary then |z| equals If |z + 6| = |2z + 3| then |z| (r) (C) equals Let ≠ 1 be a cube root of (s) (D) unity and

(B) I f z =

0 1 2 3

a + bω + cω2

+ c + aω + bω2 a + bω + cω2

then |z| equals

b + cω + aω2 (t)

MATHEMATICS TODAY | SEPTEMBER‘17

4 27

Integer Answer Type

46. Let l, z0 be two complex numbers. A(z1), B(z2), C(z3) be the vertices of a triangle such that z1 = z0 + l, z2 = z0 + leip/4, z3 = z0 + lei7p/11 and 3kπ ∠ABC = then the value of k is 22 – 47. If the argument of (z – a)(z – b) is equal to that ( 3 + i)(1 + 3 i) , where a, b are real numbers. of 1+ i 3+i then find If locus of z is a circle with centre 2 a + b. 48. If a1, a2, a3, a4, a5 are the roots of the equation x5 – 1 = 0, where ak = ak – 1, a = ei2p/5 and l = a1001 , 3 (669 + 1/3) (503 + 1/2) 2011 2011 2011 = a4 , = a5 , then [|l + + |] (where [ ] denotes the greatest integer function) is –

–

49. Two lines zi – zi + 2 = 0 and z(1 + i) + z(1 – i) + 2 = 0 intersect at a point P. There is a complex number a = x + iy at a distance of 2 units from the point – P which lies on line z(1 + i) + z (1 – i) + 2 = 0. Find [|x|] (where [ ] represents greatest integer function). 50. Suppose that w is the imaginary (2009)th roots of unity. If (22009 – 1)

2008

∑ 2 − wr = (a)(2b ) + c

where

r =1

a, b, c ∈ N and the least value of (a + b + c) is (2008)K. The numerical value of K is SOLUTIONS 1. (c) : Let z = i[1+ 3+ 5+....+(2n+1)] Clearly series is A.P. with common difference = 2 Q Tn = 2n − 1 and Tn+1 = 2n + 1 So, number of terms in A.P. = n + 1 n +1 Now, Sn+1 = [2 ⋅1 + (n + 1 − 1)2] 2 2 n +1 ⇒ Sn+1 = [2 + 2n] = (n + 1)2 i.e. z = i(n+1) 2 Now put n = 1, 2, 3, 4, 5, ..... n = 1, z = i4 = 1, n = 2, z = i6 = –1, n = 3, z = i8 = 1, n = 4, z = i10 = – 1, n = 5, z = i12 = 1, .......... 2. (c) : x + ⇒ x= 28

1 = 2 cos θ ⇒ x2 – 2x cos q + 1 = 0 x

2 cos θ ± 4 cos2 θ − 4 ⇒ x = cos q i sin q 2 MATHEMATICS TODAY | SEPTEMBER ‘17

3 + 2i sin θ (3 + 2i sin θ)(1 + 2i sin θ) = 1 − 2i sin θ (1 − 2i sin θ)(1 + 2i sin θ)  3 − 4 sin2 θ   8 sin θ  =   +i   1 + 4 sin2 θ   1 + 4 sin2 θ  Now, since it is real, therefore Im (z) = 0 8 sin θ ⇒ = 0 ⇒ sin q = 0 \ q = np 1 + 4 sin2 θ where n = 0, 1, 2, 3, ...... Remark : Check for (a), if n = 0, q = 0 the given number is absolutely real but (c) also satisfies this condition and in (a) and (c), (c) is most general value of q. 4. (d) : Let z = 0 {(1 – cosq) + i·2 sinq}–1 3. (c) : We have,

θ  =  2 sin   2

−1

θ  =  2 sin   2

−1

θ  θ sin + i ⋅ 2 cos  2  2

−1

θ θ sin − i ⋅ 2 cos 2 2 × θ θ θ θ sin + i ⋅ 2 cos sin − i ⋅ 2 cos 2 2 2 2 θ θ sin − i ⋅ 2 cos 2 2 = θ 2 θ θ 2 sin  sin + 4 cos2  2 2 2 θ sin 2 ∴ Re(z ) = θ θ 2 sin  1 + 3 cos2  2 2 1

5. (b) : If x + iy = =

3 2 + cos θ + i sin θ

3(2 + cos θ − i sin θ) (2 + cos θ)2 + sin2 θ

6 + 3 cos θ − 3i sin θ 4 + cos2 θ + 4 cos θ + sin2 θ

 6 + 3 cos θ   −3 sin θ  =  +i   5 + 4 cos θ   5 + 4 cos θ  3(2 + cos θ) −3 sin θ x= ,y= 5 + 4 cos θ 5 + 4 cos θ 9 [4 + cos2 q + 4 cos q + sin2 q] ∴ x2 + y2 = 2 (5 + 4 cos θ) =

9  6 + 3 cos θ  = 4  − 3 = 4x − 3 5 + 4 cos θ  5 + 4 cos θ 

6. (d) : We have, µ + iλ =

( p + i)2 ( p2 − 1 + 2 pi)(2 p + i) = 2p − i (2 p − i)(2 p + i)

MATHEMATICS TODAY | SEPTEMBERâ&#x20AC;&#x2DC;17

2 p( p2 − 2) + i(5 p2 − 1) 4 p2 + 1

\ µ2 + λ 2 =

4 p2 ( p2 − 2)2 + (5 p2 − 1)2 (4 p2 + 1)2

4 p6 + 6 p2 + 9 p 4 + 1

(4 p2 + 1)2 p 4 (4 p2 + 1) + 2 p2 (4 p2 + 1) + (4 p2 + 1)

p 4 + 2 p2 + 1

4 p2 + 1

(4 p2 + 1)2 =

( p2 + 1)2 4 p2 + 1

b + ic 7. (a) : Given, z(1 + a) = b + ic ⇒ z = 1+ a 1 + iz 1 + i(b + ic) / (1 + a) 1 + a − c + ib = = 1 − iz 1 − i(b + ic) / (1 + a) 1 + a + c − ib (1 + a − c + ib)(1 + a + c + ib) = (1 + a + c)2 + b2 =

1 + 2a + a2 − b2 − c 2 + 2ib + 2iab 1 + a2 + c 2 + b2 + 2ac + 2(a + c)

a2 + b2 + c 2 + 2a + a2 − b2 − c 2 + 2ib(1 + a) 1 + 1 + 2ac + 2(a + c) 2a(a + 1) + 2ib(1 + a) a + ib = = 2(1 + a)(1 + c) 1+ c =

9. (d) : sin x + i cos 2x and cos x – i sin 2x are conjugate to each other if sin x = cos x and cos 2x = sin 2x π 5π 9 π or tan x = 1 ⇒ x = , , ,...... ...(i) 4 4 4 π 5π 9 π and tan 2x = 1 ⇒ 2 x = , , ,........ 4 4 4 π 5π 9 π ⇒ x= , , ....... ...(ii) 8 8 8 There exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate. 10. (d) : Let z = x + iy, so, z = x − iy ,

\ z 2 + z = 0 ⇔ (x 2 − y 2 + x ) + i (2 xy − y ) = 0 Equating real and imaginary parts, we get x2 – y2 + x = 0 ...(i) 1 and 2xy – y = 0 ⇒ y = 0 or x = 2 If y = 0, then (i) gives x2 + x = 0 ⇒ x = 0 or x = –1

1 1 1 3 If x = , then x2 – y2 + x = 0 ⇒ y 2 = + = 2 4 2 4 3 ⇒ y=± 2 Hence, there are four solutions in all. 11. (b) : Here,

z − i x + i( y − 1) x − i( y + 1) = . z + i x + i( y + 1) x − i( y + 1)

(x 2 + y 2 − 1) + i(−2 x )

8. (c) : a = cos q + i sin q 1 + a (1 + cos θ) + i sin θ = ∴ . 1 − a (1 − cos θ) − i sin θ On Rationalizing the denominator, we get

1 + a (1 + cos θ) + i sin θ (1 − cos θ) + i sin θ = × 1 − a (1 − cos θ) − i sin θ (1 − cos θ) + i sin θ (1 + cos θ)(1 − cos θ) + (1 + cos θ) i sin θ + (1 − cos θ)i sin θ + i 2 sin2 θ = (1 − cos θ)2 − (i sin θ)2 2 1 − cos θ + i sin θ + i sin θ cos θ + i sin θ − i sin θ cos θ − sin2 θ = 1 + cos2 θ − 2 cos θ + sin2 θ

x2 + y2 – 1 = 0 ⇒ x2 + y2 = 1 ⇒ zz = 1

1 − (cos2 θ + sin2 θ) + 2i sin θ 2

1 + (cos θ + sin θ) − 2 cos θ

2i sin θ 2(1 − cos θ)

θ θ θ i ⋅ 2 sin cos i cos 2 2 2 = i cot θ = = θ θ 2 2 sin2 sin 2 2 30

MATHEMATICS TODAY | SEPTEMBER ‘17

x 2 + ( y + 1)2 z −i is purely imaginary, we get As z +i

12. (b) : z +

2 2 = 2 ⇒| z | − ≤ 2 ⇒ |z|2 – 2 |z| –2 z |z |

2± 4+8 ≤ 1± 3 . 2 Hence max. value of |z| is 1 + 3 |z |≤

13. (a) : Let z1 = a + ib = (a, b) and z2 = c – id = (c, –d) where a > 0 and d > 0 ...(i) Then |z1| = |z2| ⇒ a2 + b2 = c2 + d2 z +z (a + ib) + (c − id ) Now 1 2 = z1 − z2 (a + ib) − (c − id ) =

[(a + c) + i(b − d )][(a − c) − i(b + d )] [(a − c) + i(b + d )][(a − c) − i(b + d )]

= = \

(a2 + b2 ) − (c 2 + d 2 ) − 2(ad + bc)i

Since |z| = r is maximum, therefore

a2 + c 2 − 2ac + b2 + d 2 + 2bd 2

−(ad + bc )i

[using (i)]

a + b2 − ac + bd (z1 + z2 ) is purely imaginary. (z1 − z2 )

(z1 + z2 ) will be equal (z1 − z2 ) to zero. According to the conditions of the equation, we can have ad + bc = 0

However if ad + bc = 0, then

Remark : Assume any two complex numbers satisfying both conditions i.e., z1 ≠ z2 and |z1| = |z2| Let z1 = 2 + i, z2 = 1 – 2i, \ Hence the result.

z1 + z2 3 − i = = −i z1 − z2 1 + 3i

14. (a) 15. (c) : We have |zk| = 1, k = 1, 2, .... n

Therefore, | z1 + z2 + .... + zn | = | z1 + z2 + .... + zn | (Q | z | = | z |)

...(i)

Given |z + i| = |z – i| or |x + iy + i| = |x + iy – i| or |x + i(y + 1)| = |x + i(y – 1)| 2

x + ( y + 1) = x + ( y − 1)

⇒ ⇒

+ (y + 1)2 = x2 + (y – 1)2

⇒ y2 + 2y + 1 = y2 – 2y + 1 ⇒ 4y = 0 or y = 0 Hence from (i), we get z = x, where x is any real number. 17. (b) : Let z = r(cos q + i sin q) 2

1 1 =1 ⇒ z + =1 z z 2 1 ⇒ r (cos θ + i sin θ) + (cos θ − i sin θ) = 1 r

Then z +

 1  1 ⇒  r +  cos2 θ +  r −  sin2 θ = 1  r  r 1 2 ⇒ r + + 2 cos 2 θ = 1 r2

Differentiating (i) w.r.t. q, we get dr 2 dr 2r − − 4 sin 2θ = 0 dθ r 3 dθ dr π = 0, we get sin 2q = 0 ⇒ q = 0 or Putting dθ 2 \ z is purely imaginary. (Q q ≠ 0) 18. (c) : Given expression, |2z – 1| + |3z – 2|, minimum 1 value of |2z – 1| is 0 at z = . So value of given 2 1 1 expression = 0 + = , minimum value of |3z – 2| is 2 2 2 1 1 0 at z = . So value of given expression = + 0 = . 3 3 3 1 So minimum value of given expression is . 3 19. (d) : Let z1 = r1 (cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2) \ |z1 + z2| = [(r1 cos q1 + r2 cos q2)2 + (r1 sin q1 + r2 sin q2)2]1/2

1 ⇒ | zk |2 = 1 ⇒ zk zk = 1⇒ zk = zk

1 1 1 ⇒ | z1 + z2 + ..... + z n | = + + .... + z1 z2 zn 16. (a) : Let, z = x + iy

dr =0 dθ

...(i)

= [r12 + r12 + 2r1r2 cos (q1 – q2)]1/2 = [(r1 + r2)2]1/2 (Q |z1 + z2| = |z1| + |z2|) Therefore cos(q1 – q2) = 1 ⇒ q1 – q2 = 0 ⇒ q1 = q2 Thus arg(z1) – arg(z2) = 0 20. (d) : Let z = x + iy, z = x − iy y Since arg (z ) = θ = tan −1 x −1  − y  arg (z ) = θ = tan    x  Thus arg (z ) ≠ arg (z ) 21. (c) : We have, |z1 – z2|2 = |z1|2 + |z2|2 –2 |z1| |z2| cos(q1 – q2) where q1 = arg(z1) and q2 = arg(z2) Since arg(z1) – arg(z2) = 0 \ |z1 – z2|2 = |z1|2 + |z2|2 –2|z1| |z2| = (|z1| – |z2|)2 ⇒ |z1 – z2| = ||z1| – |z2|| 22. (c) : |z1 + z2| = |z1| + |z2| |z1 + z2|2 = |z1|2 + |z2|2 + 2 |z1| |z2| ⇒ |z1|2 + |z2|2 + 2Re | z1z2 | = |z1|2 + |z2|2 + 2 |z1| |z2| ⇒ 2Re | z1z2 | = 2|z1| |z2| ⇒ 2|z1| | z2 | cos(q1 – q2) = 2|z1| |z2| ⇒ cos(q1 – q2) = 1 or q1 – q2 = 0 \ arg(z1) = arg(z2) MATHEMATICS TODAY | SEPTEMBER‘17

[Q log(a + ib) = log(reiq) = log r + iq

23. (c) : Here, −1 + −3 = reiθ

= log a2 + b2 + i tan −1 (b / a) ]

⇒ −1 + i 3 = reiθ = r cos q + ir sin q Equating real and imaginary parts, we get ⇒

r cos q = –1 and r sin θ = 3 Hence, tan θ = − 3 ⇒ tan θ = tan

2π 2π . Hence θ = . 3 3

− 3 −i π π  24. (c) : Since = −  cos + i sin   2 6 6 3

π π  − 3 −i  ⇒  = −  cos + i sin  = −i    2  6 6 –iq

25. (b) : Let z = ee = ecosq – i sin q = ecos q e–i sin q z = ecosq [cos(sin q) – i sin(sin q)] z = ecosq cos(sin q) – iecosq sin(sin q)  e cos θ sin(sin θ)  amp(z ) = tan −1  −   e cos θ cos(sin θ)  = tan–1[tan(–sin q)] = – sin q

 2x  z x 4 + 1 − 2x 2 + 4 x 2 = log + i tan −1  2  1 − x 2  i ( x + 1)

⇒ z=

tan–1

= − i log

(

− i π/4

)

1 −i π / 4  = −i  log 2 + log e  2 

iπ  i π 1 = −i  log 2 −  = − log 2 − 4 2 4 2

⇒ |z + a|2 = |a|2 – b, {Q zz = | z |2 } This equation will represent a circle with centre z = –a, if |a|2 – b > 0, i.e. |a|2 > b since |a|2 = b represents point circle only. 29. (c) : Let r be the circumradius of the equilateral triangle and the cube root of unity. A(z1)

2π 3

C(z3)

Let ABC be the equilateral triangle with z1, z2 and z3 as its vertices A, B and C respectively with circumcentre O′(z0). iθ Then the vectors O ′ A = z1 − z0 = re



1  Re(z ) = e −π /4 cos  log 2 2   x −i z  x −i 27. (b) : Let z = i log  ⇒ = log    x + i  x + i  i  x 2 − 1 − 2ix  z x −i x −i log ⇒ = log  × =    i x +i x −i  x 2 + 1   x2 − 1 z 2x  ⇒ = log  −i  2 2 i  x + 1 x + 1 

z) 0

B(z2)



⇒ z Taking real part only, we get

O′(

2π 

i  θ+  iθ O ′ B = z2 − z0 = re  3  = rωe

−i (log 2) − π/4 2 =e e

z ⇒ = log i

= log 1 + i(2 tan–1 x) x = p –2tan–1 x.

28. (b) : By adding aa on both the sides of zz + az + az = −b we get, (z + a)(z + a ) = aa − b

26. (a) : Let z = (1 – i)–i. Taking log on both sides, π π  ⇒ log z = –i log(1 – i) = −i log 2  cos − i sin   4 4

–2tan–1

4π 

i θ + iθ 3  O ′C = z3 − z0 = re  = r ω 2e

\ z1 = z0 + reiθ , z2 = z0 + rωeiθ , z3 = z0 + rω2eiθ Squaring and adding z12 + z22 + z32 = 3z02 + 2(1 + ω + ω2 )z0reiθ

3z02,

since 1 +

+ (1 + ω2 + ω 4 )r 2ei 2θ =0=1+ 2+ 4

30. (c) : Given the vertices of quadrilateral A(1 + 2i), B(–3 + i), C(–2 – 3i) and D(2 – 2i) Now, AB = 16 + 1 = 17 , BC = 16 + 1 = 17

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

MATHEMATICS TODAY | SEPTEMBER ‘17

CD = 16 + 1 = 17, DA = 16 + 1 = 17 AC = 9 + 25 = 34 , BD = 25 + 9 = 34 Hence it is a square.

31. (b) : Since maximum distance of any complex number from origin is given by | | 1 1 1 1 1 therefore, | ω | = ω + − ≤ ω+ + =2+ |ω| ω ω ω ω ⇒ | |2 –2 | | –1

0 ⇒ | ω |≤

2±2 2 2

Hence, max | | is 1 + 2 32. (b, c) : We have, 1 1 1 z = ( z1 + z2 + ... + zn )  + + ... +  zn   z1 z2 = |z1 + z2 + ... + zn|2

|z1|2 + |z2|2 + |z3|2 + ... + |zn|2 = n2

33. (a, b) : x2 + ax + b = 0 ⇒ |a| y=

−a±

\ |y| = 1

a −4 b 2

− a ±i 4− a

1 + i tan θ 1 + im = 1 − i tan θ 1 − im

 2m   1 − m2  M =i +   1 + m2   1 + m2  \ M = cos 2q + i sin 2q = cis 2q ⇒ |M| = 1 π 1+ i 3 = cis 2θ ⇒ θ = Also, −ω2 = 2 6 41. (b) 42. (d)

2π 43. (b) : Let a = cis then a = a + a2 + a4, 7 3 5 6 b=a +a +a ⇒ a + b = a + a2 + a3 + a4 + a5 + a6 = –1 and ab = (a + a2 + a4)(a3 + a5 + a6) = 2 \ Equation whose roots are a, b is x2 – x(–1) + 2 = 0 f(x) + f(ax) + f(a2x) + f(a3x) + .... + f(a6x) = (1 + 1 + 1 + .... + 1) + 2x (1 + a + a2 + a3 + .... + a6) + 3x2 (1 + a2 + a4 + .... + a12) + .... + 7x6(1 + a6 + a12 + a18 + .... + a36) = 7

35. (a, c) : We have z1, z2, z3 in G.P. \ z22 = z1z3 ⇒ z23 = 1 So, z2 = 1, , 2 Hence, 1 – b + 3 – 1 = 0 ⇒ b = 3 36. (a, c) : (3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2 ⇒ 8(3z + 1) 6 (4z + 1) 4 (6z + 1) 2 (12z + 1) =2×8×6×4×2 (24z + 8)(24z + 6)(24z + 4)(24z + 2) = 768 Let 24z + 5 = U (U + 3)(U + 1)(U – 1)(U – 3) = 768 ⇒ (U2 – 9)(U2 – 1) = 768 ⇒ U4 – 10U2 – 759 = 0 ⇒ U2 = 33 or –23 ⇒ 24 z + 5 = ± 33 or ± i 23 ± 33 − 5 ±i 23 − 5 z= or 24 24 37. (a, b, c, d) : Let z1 = x1; z2, z3 = x2 ± iy2 b ⇒ z1 + z2 + z3 = − a b ⇒ x1 + 2 x2 = − < 0 ⇒ ab > 0 a d Also, z1 z2 z3 = x1  x22 + y22  = − ⇒ ad > 0   a bc 2 2 Also − < x1 x2 + y2 ⇒ bc > ad a2 38. (c)

)

( x − x ) + i ( y1 − y2 ) z −z 40. (c) : Given M = 1 2 = 1 2 z1 − z2 ( x1 − x2 ) − i ( y1 − y2 )

2 and |b| = 1

34. (a, d) : Let a is a real root then a3 + (3 + i)a2 – 3a = m + i ⇒ a3 + 3a2 – 3a – m = 0 or a2 – 1 = 0 ⇒ a = 1 or –1 ⇒ m = 1 or 5

(

39. (c)

Also,

x7 − 1 = (x – a)(x – a2)(x – a3)(x – a4)(x – a5) x −1 (x – a6)

Putting a = cis

2π of applying x 7

1 gives

2π   4π   12π   7 =  1 − cis   1 − cis  ....  1 − cis   7  7   7  π  2π   6π   7 =  2 sin   2 sin  ....  2 sin   7 7   7  π 2π 3π 4π 5π 6π 7 7 ⇒ sin sin sin = sin sin sin = = 7 7 7 7 7 7 64 8 44. (A) → (s), (B) → (q), (C) → (r) (A) ||z –

| – |z – ω ||max = | – ω | =

(B) Q (–3, 0) \ PQ = S1 = 12 (C) 2z2 + 2z + k = 0

−2 ± 4 − 8k 4 Since 'z' is a complex number 4 – 8k will be negative 1 ⇒ k> 2 \ z=

MATHEMATICS TODAY | SEPTEMBER‘17

 −1 2k − 1   −1 −1  2k − 1  \ Points are (0, 0),  ,   ,  2  2   2 2 Since triangle is equilateral

z1 + i z2

z2 + i z1

| i || −iz1 + z2 | 1 × z2 − iz1 = =1 z2 − iz1 | z2 + iz1 |

π  z − 2i  π (B) Here, arg  = or − , So it represents a  z + 2i  2 2 semicircle with diametric ends at points (0, 2) and (0, –2) So, |z| = radius = 2 (C) (x + 6)2 + y2 = (2x + 3)2 + (2y)2 ⇒ x2 + y2 + 12x + 36 = 4x2 + 4y2 + 12x + 9 ⇒ 3x2 + 3y2 = 27 ⇒ x2 + y2 = 9, |z| = 3 (D) z =

2 3 2 3 4 1  aw + bw + cw  1  aw + bw + cw  +     w  c + aw + bw 2  w 2  b + cw + aw 2 

1 1 ×1+ × 1 = w 2 + w = −1 ⇒ | z | = 1 2 w w

46. (5) : |z1 – z0| = |z2 – z0| = |z3 – z0| = |l| Now,

z3 − z0 ei7 π/11 i17 π /44 = =e i π / 4 z2 − z0 e

⇒ ∠BSC = 17

\ l = a1001 = (a2)1001 = a2002 = a5 × 400 + 2 = a2 3 = (a5)503+1/2 = (a4)503+1/2 = a2014 = a5·402+4 = a4

45. (A) → (q), (B) → (r), (C) → (s), (D) → (q) (A) | z | =

where a = ei2p/5

= (a4)669+1/3 = (a3)(669+1/3) = a2008 = a3

1 1 (2k − 1) + = (2k − 1) ⇒ k = 2 / 3 4 4

48. (1) : Clearly, a1 = 1 ; a2 = a; a3 = a2 ; a4 = a3 ; a5 = a4;

π (where S represents z0) 44

Also sum of 2011th power of roots of unity is 0 So, 1 + a2011 + l2011 + 2011 + v2011 = 0 l2011 +

2011

+ v2011 = –(1 + a2011)

l2011 +

2011

+ v2011 = –(1 + a)

|l2011 +

2011

+ v2011| = |–(1 + ei2p/5)|

= |1 + cos 2p/5 + i sin 2p/5| = |2 cos p/5 (cos p/5 + i sin p/5)| =2

5 +1 5 +1 = 4 2

\ [|l2011 +

iπ

α = i ± 2e 4 \ x = ± 2 ⇒ [|x|] = 1 50. (2) : Let x be the (2009)th root of unity ≠ 1, then x2009 – 1 = (x – 1)(x – w)....(x – w2008) Taking log on both sides, we get ln (x2009 – 1) = ln (x – 1) + ln (x – w) + ln (x – w2) +....+ ln (x – w2008) \ Differentiating both the sides w.r.t. x, we get (2009)x 2008

2008 1 1 + ∑ x − 1 r =1 x − w r

x 2009 − 1 Putting x = 2 in equation (i), we get

z −z π Similarly 2 0 = eiπ /4 ⇒ ∠ACB = z1 − z0 8

47. (3) : tan −1

(a − b) y 2

x + y − (a + b)x + ab

π 4

⇒ x2 + y2 – (a + b)x – (a – b)y + ab = 0 Centre =

3+i ⇒a+b = 3 2

MATHEMATICS TODAY | SEPTEMBER ‘17

2008

∑ 2 − wr

r =1

π 17 π 15π − = 22 8 88

+ v2011|] = 1

49. (1) : Solving the equation of the lines we get z = −z ⇒ z = i |a – i| = 2; a = i ± 2eiq , put it in the equation of the second line, we get cos q – sin q = 0

π ⇒ ∠BAC = 17 88

∴ ∠ ABC = π −

2011

...(i)

2009 (22008 ) 22009 − 1

Multiplying both sides of above equation by (22009 – 1), we get 2008 1 \ ( 22009 − 1) ∑ = 2009 ⋅ 22008 − 22009 + 1 r r =1 2 − w = 22008 (2009 – 2) + 1 = 22008 2007 + 1 = [(a)(2b) + c] \ a = 2007, b = 2008, c = 1 Hence a + b + c = 4016 

CLASS XI

Series 5

CBSE Permutations and Combinations | Binomial Theorem

PERMUTATIONS AND COMBINATIONS X X X

n factorial = n or n! = n(n − 1)(n − 2) ... 2.1 0 = 0! = 1 nP = P(n, r) = n(n – 1)(n – 2) ... (n – r + 1) r

n! ( r < n) = (n − r )! X X

nP = n

n(n – 1) ... 2 .1 = n!

Number of permutations of n different things taken r at a time when

nC = nC r n–r

nC + nC n+1C r r–1 = r

(i) repetition is allowed = nr (ii) a particular thing is included = r · n–1Pr–1

(ii) If 'r' particular things are identical then no. of permutations = (n – r + 1)! X

No. of ways of arranging 'n' things when 'p' things are n! p !q !

Cr =

n n −1 Cr −1 r

No. of combinations of n different things taken 'r' at

No. of combinations when 'p' particular things are always excluded = n–pCr

No. of ways of selecting zero or more things out of 'n' identical things = 1 + 1 + 1 + ... upto (n + 1) terms = n + 1

of one kind, 'q' things are of second kind and rest are all different =

nC ⇒ x = y or x + y = n y

= n–pCr–p

(i) No. of permutations when 'r' particular things

are together = (n – r + 1)! r!

a time when 'p' particular things are always included

(iii) a particular thing is not included = n–1Pr X

n  n P n! (r < n) = r Cr =   = C ( n , r ) = r !(n − r )! r! r

No. of ways of selecting zero or more things out of n different things = nC0 + nC1 + ... + nCn = 2n MATHEMATICS TODAY | SEPTEMBER‘17

BINOMIAL THEOREM X

(a + x)n = nC0 anx0 + nC1an–1x1 + nC2 an–2x2 + ......

....+

h X

nC n

a0x n

n∈

I+

b) n

General term in the expansion of (a + Tr + 1 i.e., (r + 1)th term = nCr an – r . br

Middle term : In the expansion of (a + b)n, the middle term is given by n   + 1 2  n +1   2 

th th

Some Important Expansions

(1 + x)n = 1 + nC1 x + nC2 x2 + ..... + nCnxn n

= ∑ nCr x r h

(1 – x)n = 1 – nC1 x + nC2 x2 – ..... + (–1)n nCnxn n = ( −1)r .n Cr xr r =0

∑

term, if n is odd

Number of terms in the binomial expansion of (a + x)n = n + 1

(1 + x)–1 = 1 – x + x2 – x3+ ..... + (–1)rxr + ...... ∞

In the expansion of (x + y)n (i) rth term from end = (n – r + 2)th term from beginning

(1 + x)–2 = 1 – 2x + 3x2 – ..... + (–1)r(r + 1) xr + ..... ∞

(1 – x)–1 = 1 + x + x2 + x3 + ..... + xr + ..... ∞

(1 – x)–2 = 1 + 2x + 3x2 + 4x3 + ..... + (r + 1) xr + ..... ∞

the same sequence of answers. What is the maximum number of students in the class for this to be possible?

WORK IT OUT VERY SHORT ANSWER TYPE

1. If (n + 2)! = 2550 (n!), find n. 2. Prove that : (2n)! = 2n (n!) [1 3 5 ... (2n–1)] 3. Find the (n+1)th term from the end in the expansion 3n

10. Find (i) Coefficient of x2 in (1 – 2x + 3x2) (1 – x)14 (ii) Coefficient of x11 in the expansion of (1 – 2x + 3x2) (1 + x)11. LONG ANSWER TYPE - I

1  of  x −  .  x 4. How many 4-letter words, with or without meaning, can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not allowed? 11

1  5. Expand  x −  , y ≠ 0 y  SHORT ANSWER TYPE

6. Find the middle terms in the expansion of 7  x3   .  3x − 6 

7. Find the greatest term in (5 + 2y)17 if y = 8. Find r, if 5P(4, r) = 6P(5, r – 1).

1. 2

9. For a set of five true or false questions, no student has written all correct answer and no two students have given 36

in the expansion of (y + x)n

r =0

term, if n is even n +1  term and  + 1  2 

(ii) rth term from end = rth term from beginning

MATHEMATICS TODAY | SEPTEMBER ‘17

11 If C(n, r) : C(n, r + 1) = 1 : 2 and C(n, r + 1) : C(n, r + 2) = 2 : 3, determine the values of n and r. 12. Assuming that x is so small that its second and higher powers may be neglected, simplify the 2 /3 3/ 2 following: (1 − 2 x ) (4 + 5x ) 1− x

1  13. If xp occurs in the expansion of  x 2 +  , prove  x (2n)! that its coefficient is 4 n − p    2n + p    !  ! 3 3  14. The letters of the word 'OUGHT' are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word 'TOUGH' in this dictionary. 15. In how many ways can 9 examination papers be arranged so that the best and the worst papers are never together?

LONG ANSWER TYPE - II

16. In how many ways can 3 prizes be distributed among 4 boys, when (i) no boy gets more than 1 prize (ii) a boys may get any number of prizes (iii) no boy gets all the prizes? 17. Find a, b and n in the expansion of (a – b)n if the first three terms of the expansion are 729, 7290 and 30375 respectively. 18. How many numbers are there between 100 and 1000 such that at least one of their digits is 7? 19. If a1, a2, a3 and a4 be any four consecutive coefficients in the expansion of (1 + x)n, prove that a3 a1 2a2 + = a1 + a2 a3 + a4 a2 + a3 20. If p is nearly equal to q and n > 1, show that (n + 1) p + (n − 1) q  p 1/n = . (n − 1) p + (n + 1) q  q 

1/ 6

1 C0 x   − 11C1x10   y y

1 1 1 + 11C2 x 9   − 11C3 x 8   + 11C4 x 7   y y y 5

1 1 1 − 11C5 x 6   + 11C6 x 5   − 11C7 x 4   y y y 8

1 1 1 + 11C8 x 3   − 11C9 x 2   + 11C10 x   y y     y

1 − 11C11   y

(n + 2)! = 2550 (n!) (n + 2) (n + 1) (n!) = 2550 (n!) (n + 2) (n + 1) = 2550 ⇒ n2 + 3n – 2548 = 0 (n – 49) (n + 52) = 0 n = 49 as n = –52 is rejected being n ∈ N. n = 49 (2n)! = (2n)(2n – 1)(2n – 2)(2n – 3) ... 4 . 3 . 2 . 1 = [(2n)(2n – 2)(2n – 4) ... 4 . 2][(2n – 1)(2n – 3) ... 3 . 1] = 2n × [n(n – 1)(n – 2) ... 2 . 1] × [(2n – 1)(2n – 3) ... 3 . 1] = 2n (n!) [1 3 5 ... (2n – 1)]. 3n

1  3. Given expansion is  x −   x th (n + 1) term from the end in the given expansion = (3n – n + 1)th i.e., (2n + 1)th term from the beginning in the same expansion .  1 \ Required term = T2n + 1 = 3nC2n x 3n − 2n  −   x

+ 462

x5 y

− 330

x4 y

+ 165

x3 y

− 55

x2 y

11x 10

−

y11

6. Clearly, the given expansion contains 8 terms

SOLUTIONS

11  1 

x10 x9 x8 x7 x6 = x11 − 11 + 55 − 165 + 330 − 462 y y2 y3 y4 y5

 99  . Hence, find the approximate value of   101  1. ⇒ ⇒ ⇒ ⇒ \ 2.

1   x − y 

3n ! n (−1) 3n ! 1 = x . 2n 2 (2n)! n ! ( n )! n ! x n x

4. The word 'LOGARITHMS' contains 10 different letters. \ Nu mb e r of re qu i re d word s = nu mb e r of arrangements of 10 letters, taken 4 at a time = 10P4 = 10 × 9 × 8 × 7 = 5040

8 \ Middle terms are   2 i.e,, 4th and 5th terms 37

8  and  + 1 2  (7 − 3)

Now, t 4 = t3 + 1 = (−1) C3 ⋅ (3x )

 x3  ⋅   6 

 x 9  −105x13 =  −35 × 81 × x 4 × . =  216  8 4

(7 − 4 )

And, t5 = t 4 + 1 = (−1) ⋅ C4 ⋅ (3x )

terms, 3

 x3  ⋅   6 

 x12  35x15 =  35 × 27 × x 3 × . =  1296  48 7. Here,

Tr + 1 Tr

1 17 − r + 1 2 y 18 − r  ⋅ = Q y =  2 r 5 5r

\ Tr + 1 > , = or < Tr according as

18 − r 5r

>, = or < 1,

Now, 18 – r = 5r ⇒ 18 = 6r ⇒ r = 3 Hence the 3rd and the 4th terms are equal when 1 y = \ T3 = T4 = 17C2(5)15 = 17 × 8 × (5)15. 2 MATHEMATICS TODAY | SEPTEMBER‘17

8. 5P(4, r) = 6P(5, r – 1) 4! 5! ⇒ 5⋅ = 6⋅ (4 − r )! [5 − (r − 1)]!

⇒

12. Using binomial expansion, we have (1 − 2 x )2 /3 (4 + 5x )3/ 2

1 6 = (4 − r )! (6 − r )(5 − r )(4 − r )!

1− x

⇒ (6 – r) (5 – r) = 6 ⇒ – 11r + 30 = 6 ⇒ (r – 3) (r – 8) = 0 ⇒ r = 3 or 8 Since P(n, r) is defined only when r n, we reject r = 8. Hence r = 3. 9. Clearly, there are 2 ways of answering each of the 5 questions, i.e., true for false. \ Total number of different sequences of answers = 2 × 2 × 2 × 2 × 2 = 32. There is only one all-correct answer. So, the maximum number of sequences leaving allcorrect answer is (32 – 1) = 31. Since different students have given different sequences of answers, so the maximum possible number of students = 31. 10. (i) We have, (1 – 2x + 3x2) (1 – x)14 = (1 – 2x + 3x2) × (1 – 14C1x + 14C2x2 – .... to 15 terms) \ Co-efficient of x2 = 14C2 – 2 × 14C1 + 3 = 66 (ii) we have (1 – 2x + 3x2) (1 + x)11 = (1 – 2x + 3x2) × (1 + 11C1x + 11C2x2 + ... + 11C9x9 + 11C10x10 + x11) \ Coefficient of x11 = 1 × 1 – 2 × 11C10 + 3 × 11C9 = 1 − 2 × 11 + 3 × 11. C(n, r) : C(n, r + 1) = 1 : 2

11 × 10 = 144 2!

⇒

n! n! : =1: 2 r !(n − r )! (r + 1)!(n − r − 1)!

⇒

(r + 1) r !(n − r − 1)! 1 n! × = n! r !(n − r )(n − r − 1)! 2

r +1 1 = ⇒ n − 3r − 2 = 0 n−r 2 Also C(n, r + 1) : C(n, r + 2) = 2 : 3

⇒

n! n! : =2:3 (r + 1)!(n − r − 1)! (r + 2)!(n − r − 2)!

⇒

n! × (r + 1)!(n − r − 1) (n − r − 2)!

(r + 2)(r + 1)!(n − r − 2)! n! MATHEMATICS TODAY | SEPTEMBER ‘17

2 /3

= (1 − 2 x )

⋅4

5  1 + x  4

3/ 2 

 5  = 8(1 − 2 x )2 /3 ⋅ 1 + x   4 

{

3/ 2

⋅ (1 − x )−1/ 2

}{

}

 2 3 5 = 8 ×  1 + × (−2 x ) + ... × 1 + × x + ... 3 2 4    1  × 1 +  −  (− x ) + ....    2     4   15   1   = 8 ×  1 − x  1 + x  1 + x   3  8   2   [neglecting x2 and higher powers of x]  4 15  = 8 ×  1 − x + x   3 8  

 1  1 + x   [neglecting x2] 2 

  13   1   = 8 ×  1 + x  1 + x   = 8 ×   24   2  

13   1 1 + 2 x + 24 x  [neglecting x2]

25   25   = 8 × 1 + x  =  8 + x  .  24   3  13. The general term in the expansion of  x 2 + 1   x r 1   2 n − r Tr + 1 = 2nCr ( x 2 ) ⋅  x 4 n − 2r 4n − 3r = 2nCr x ⋅ x −r = 2nCr x

...(i)

⇒

...(ii)

Solving (i) and (ii) we get, n = 14, r = 4

6 × 5! 5! ⇒ = (4 − r )! (6 − r )! ⇒

r +2 2 = ⇒ 2n − 5r − 8 = 0 n − r −1 3

2 3

is :

4n − p 3 4n − p \ Co-efficient of xp = 2nCr where r = 3 (2n)! (2n)! = = 4n − p   4n − p    4n − p   2n + p  ! 2n − !  ! !  3   3  3   3  14. Total number of letters in the word OUGHT is 5 and all the five letters are different. Alphabetical order of letters is G, H, O, T, U. This contains xp if 4n − 3r = p ⇒ r =

Number of words beginning with G = 4! = 24 Number of words beginning with H = 4! = 24 Number of words beginning with O = 4! = 24 Number of words beginning with TG = 3! = 6 Number of words beginning with TH = 3! = 6 Number of words beginning with TOG = 2! = 2 Number of words beginning with TOH = 2! = 2 There will be two words beginning with TOU and TOUGH is first among them \ Rank of word 'TOUGH' in the dictionary = 24 + 24 + 24 + 6 + 6 + 2 + 2 + 1 = 89th 15. The number of ways of arranging 9 papers = 9P9 = 9 ! Let us first consider the number of permutations, each containing the best and the worst papers together. Let us tie these two papers together and consider them as one paper. Then, these 8 papers can be arranged in 8P8 = 8 ! ways. Also, these two papers may be arranged among themselves in 2! i.e., 2 ways. So, the number of permutations with the worst and the best papers together = (2 × 8!). \ The number of arrangements with best and worst papers never together = [(9!) –(2 × 8!)] = (9 – 2) × 8! = 7 × 8 ! = 282240 16. (i) The first prize can be given away in 4 ways as it may be give to anyone of the 4 boys. So, the second prize can be given away in 3 ways, and third prize can be given away to anyone of the remaining 2 boys in 2 ways. \ The of ways in which all the prizes can be given away = (4 × 3 × 2) = 24. (ii) The first prize can be given away in 4 ways as it may be given to anyone of the 4 boys. The second prize can be given away in 4 ways, because there is no restriction as to the number of prizes a student gets. Similarly, the third prize can be given away in 4 ways. Hence, the number of ways in which all the prizes can be given away = (4 × 4 × 4) = 64. (iii) The number of ways in which a boy gets all the 3 prizes is 4, as anyone of the 4 boys may get all the prizes. \ The number of ways in which a boy does not get all the prizes = 64 – 4 = 60 b)n

= = 729 17. T1 of (a + n n T2 of (a + b) = C1an – 1 b = 7290 T3 of (a + b)n = nC2a n – 2 b2 = 30375

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

Dividing (i) by (ii), we get an n

n −1

C1a

a 1 a n 729 1 ...(iv) ⇒ = ⇒ = = nb 10 b 10 7290 10

Dividing (ii) by (iii), we get n n

n −1

C1a

C2 a

⇒

n−2 2

7290 30375

n −1

na b 7290 6 = = n(n − 1) n − 2 2 30375 25 a b 2 a 6 2 × = n − 1 b 25

...(v)

2 n 6 × = (using (iv )) ⇒ n = 6 n − 1 10 25 Putting n = 6 in (i) we get, a6 = 729 ⇒ a = 3 Putting n = 6, a = 3 in (iv), we get b = 5 18. We have to form 3-digit numbers such that at least one of their digits is 7. Now, (i) 3-digit numbers with 7 at the unit's place: The number of ways to fill the hundred's place = 9 [by any digit from 1 to 9]. The number of ways to fill the ten's place = 10 [by any digit from 0 to 9]. The number of ways to fill the unit's place = 1 [by 7 only] \ Number of such numbers = 9 × 10 × 1 = 90. (ii) 3-digit numbers with 7 at the ten's place but not at unit's place : The number of ways to fill the hundred's place = 9 [by any digit from 1 to 9]. The number of ways to fill the ten's place = 1 [by 7 only] The number of ways to fill the unit's place = 9 [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. \ Number of such numbers = 9 × 1 × 9 = 81.

ANSWER KEY

MPP-5 CLASS X I 1.

(b)

6. 11. 15. 20.

(d) 7. (a,b,c) 8. (b,c) (b, c) 12. (a,b,c,d) (b) 16. (a) 17. (4) (8)

(c)

(d)

(c)

9 . (a,b,c) 10. (d) 13. (a,d) 14. (c) 18. (5) 19. (3 )

MATHEMATICS TODAY | SEPTEMBER‘17

(iii) 3-digit numbers with 7 at the hundred's place but neither at the unit's place nor at ten's place: The number of ways to fill the hundred's place = 1. [by 7 only]. The number of ways to fill the ten's place = 9. [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. The number of ways to fill the unit's place = 9 [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. \ Number of such numbers = 1 × 9 × 9 = 81. Hence, the number of required type of numbers = (90 + 81 + 81) = 252. 19. Given, a1, a2, a3 and a4 are the coefficients of the rth, (r + 1)th, (r + 2)th and (r + 3)th terms respectively in the expansion of (1 + x)n Now, a1 = rth coefficient = nCr – 1 ...(i) a2 = (r + 1)th coefficient = nCr ...(ii) a3 = (r + 2)th coefficient = nCr + 1 ...(iii) th n a4 = (r + 3) coefficient = Cr + 2 ...(iv)

n (r − 1)!(n − r + 1)! Cr a2 n! = = ⋅ n a1 n! Cr − 1 r !(n − r )! n − r +1 = r a2 + a1 a2 n − r +1 n +1 ∴ = +1 = +1 = a1 a1 r r a1 r ...(v) = ∴ a1 + a2 n + 1 Putting r + 1 in place of r in (v), we get a2 r +1 ...(vi) = a2 + a3 n + 1 a3 r+2 ...(vii) Similarly, = a3 + a4 n + 1

Now,

∴

a3 a1 r r+2 + = + a1 + a2 a3 + a4 n + 1 n + 1 =

2a2 2(r + 1) = a2 + a3 n +1

20. Let p = q + h, where h is so small that its square and higher powers may be neglected. Then, (n + 1) p + (n − 1) q (n − 1) p + (n + 1) q

(n + 1)(q + h) + (n − 1) q (n − 1)(q + h) + (n + 1)q

  n +1  1 +   h 2nq + (n + 1)h   2nq   = = 2nq + (n − 1)h   n − 1   1 +  2nq  h    [on dividing num. and denom. by 2nq] 40

MATHEMATICS TODAY | SEPTEMBER ‘17

  n +1  = 1 +   h   2nq  

  n − 1  1 +  2nq  h   

−1

  n +1    n − 1  = 1 +   h  h  1 −    2nq     2nq   [expanding the 2nd expression and neglecting h2, h3, etc]  n +1  n − 1  [neglecting h2] = 1 +   h −  2nq  h  2 nq    h   = 1 +  .  nq 

1/n

h   p q+h  h  = 1 +  = 1 +  . Also,   =   q  q   q  nq  2 [Expanding and neglecting h and higher powers of h] (n + 1) p + (n − 1) q

1/n

 p =  (n − 1) p + (n + 1) q  q  Deduction : Taking p = 99, q = 101 and n = 6 in the above result, we get ∴

1/ 6

 99    101 

(6 + 1) × 99 + (6 − 1) × 101 1198 599 = = . (6 − 1) × 99 + (6 + 1) × 101 1202 601 

MADHYA PRADESH at • • • • • • • • • • • • • • • • • •

Telang Book Corner - Bhopal Ph: 0755-2741446; Mob: 9827058587, 9425301772 Books N Books - Bhopal Ph: 0755-2673602, 2574603; Mob: 9826086858, 9826431103 UBS Publishers Distributors Pvt Ltd. - Bhopal Ph: 0755-4203183, 4203193, 2555285; Mob: 9425013150 Krishna Pustak Kendra - Gwalior Ph: 0751-2625982; Mob: 9425118370, 9425110809, 8989475382 New Bhartiya Prakashan - Gwalior Ph: 0751-2625534, 2423004; Mob: 9407079998 Chelawat Book Depot - Indore Ph: 0731-3027430; Mob: 9926500362, 9479723775 Shiv Shakti Book Centre - Indore Ph: 0731-2459956, 2459955; Mob: 9425063647 Arun Prakashan - Indore Ph: 0731-2454372, 4054544; Mob: 9826369234, 9425313294 Bhaiya Store - Indore Ph: 0731-2531207, 2540166; Mob: 9425318103 Case Traders - Indore Mob: 9827259249, 9827620395 Chelawat Book Centre - Indore Mob: 9406666147 Jain Book Palace - Indore Ph: 0371-4054817, 4054839; Mob: 9926636333 Kitab Kunj - Indore Ph: 0731-4054442; Mob: 9826325258, 9926058881 Sushil Prakashan - Indore Ph: 0731-2535892; Mob: 9425322330 New Radhika Book Palace - Jabalpur Mob: 9425411566, 7614083315 Shankar Competition Book House - Rewa Ph: 07662-403161, 421128; Mob: 9407041342 M.P. Publishers & Suppliers - Ujjain Ph: 0734-2515680, 2514262; Mob: 9826099912 O.K. Book Depot - Ujjain Mob: 9425091418

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.

Binomial Theorem Total Marks : 80

Time Taken : 60 Min.

Only One Option Correct Type

One or More Than One Option(s) Correct Type

1. If O be the sum of odd terms and E that of even terms in the expansion of (x + a)n, then O2 – E2 = (b) (x2 – a2)n (a) (x2 + a2)n (c) (x – a)2n

(d) none of these

2. (115)96 –(96)115 is divisible by (a) 15 (b) 17 (c) 19 (d) 21 3. The number of terms in the expansion of (a + b + c)10 is (a) 11 (b) 21 (c) 55 (d) 66 4. The coefficient of the term independent of x in the 9

1  3 expansion of (1 + x + 2 x 3 )  x 2 −  , is 2 3x  (a)

1 3

5. The sum

(b) m

19 54

∑  i   m − i  ,

i =0

17 54

(d)

7. If (1 + x –2x2)6 = 1 + a1x + a2x2 + a12x12, then (a) a2 + a4 + a6 + ...... + a12 = 31 (b) a1 + a3 + a5 + ...... + a11 = –32 (c) a1 + a2 + a3 + ...... + a12 = –1 (d) none of these 8. Which of the following is an even positive integer? (a) ( 3 + 1)2 n + 1 + ( 3 − 1)2 n +1 (b) ( 3 + 1)2 n + ( 3 − 1)2 n (c) ( 3 + 1)2 n +1 − ( 3 − 1)2 n +1 (d) none of these 9. If (1 + x)n = C0 + C1x + C2x2 + ... + Cnxn, then (a) C1 + 2C2 + 3C3 + ...+ nCn = n2n–1 (b) C0 + 2C1 + 3C2 + ... + (n + 1) Cn = 2n–1 (n + 2) C C C 2n +1 − 1 (c) C0 + 1 + 2 + ... + n = 2 3 n +1 n +1 (d) none of these

1 4

 p where  q  = 0 if p < q,

is maximum when m is (a) 5 (b) 10 (c) 15 (d) 20 6. If C0, C1, C2, ....., Cn denote the coefficients in the expansion of (1 + x)n, then C0 + 3·C1 + 5·C2 + ..... + (2n + 1)Cn = (b) (n – 1)2n (a) n·2n (c) (n + 1)2n–1 (d) (n + 1)2n

10.

 r −1 n r  ∑  ∑ Cr C p 2 p  is equal to  r =1  p = 0 n

(a) 4n – 3n + 1 (c) 4n – 3n + 2

(b) 4n – 3n –1 (d) 4n –3n

1   11. In the expansion of  3 4 + 4   6 (a) (b) (c) (d)

the number of rational terms = 4 the number of irrational terms = 19 the middle term is irrational the number of irrational terms = 17 MATHEMATICS TODAY | SEPTEMBER‘17

12. Coefficient of xn in the expansion of (1 + x)2n is equal to (a)

P (2n, n) n

(b)

(n + 1)(n + 2)(n + 3).....(2n) n

(d)

where a is an integer and 0 < b < 1, then (a) a is an even integer (b) (a + b)2 is divisible by 22n+1 (c) the integer just below (3 3 + 5)2n+1 divisible by 3 (d) a is divisible by 10 Comprehension Type Consider (1 + x + x ) = ∑ ar x , where r

r =0

a0, a1, a2, ...., a4n are real numbers and n is a positive integer. 14. The value of

n −1

∑ a2r

r =0

9n + 2a2n + 1 (b) 4

9n − 2a2n + 1 (c) 4

9n + 2a2n − 1 (d) 4

15. The value of

∑ a2r −1

(a) 2

(b) 3

(d) 2

3 Integer Answer Type

17. The remainder when 9103 is divided by 25 is equal to

9n − 2a2n − 1 (a) 4

 9n + 1  (d)   2 

Column I Column II 1. If last digit of the number l+ =6 P. 9 100 99 is l and last digit of 2l is , then Q. If last digit of the number 2. l + = 11 2 999 is l and last digit of 3lll is , then 3. l – = 4 72 ! 1 − is divisible R. Let a = (36 !)2 by 10l + , then 4. l + l = 9

13. If n is a positive integer and (3 3 + 5)2n+1 = α + β,

 32n + 1  (c)   4  16. Match the following :

2 2n

 32n − 1  (b)   4 

Matrix Match Type

 9n − 1  (a)   2 

18. If (2r + 3)th and (r – 1)th terms in the expansion of (1 + x)15 have equal coefficients then r is 19. If the coefficient of the middle term in the binomial expansions of (1 + ax)4 and (1 – ax)6 is same, then –10a equals 20. If n–1C3 + would be

r =1

n–1

C4 > nC3, then, the least value of n 

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 | SEPTEMBER ‘17

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

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

FUNCTIONS Functions can be easily defined with the help of concept mapping. Let X and Y be any two non-empty sets. “A function from X to Y is a rule or correspondence that assigns to each element of set X, one and only one element of set Y”. Let the correspondence be ‘f ’ then mathematically we write f : X Y where y = f(x), x ∈ X and y ∈ Y. We say that ‘y’ is the image of ‘x’ under f (or x is the pre image of y). A mapping f : X Y is said to be a function if each element in the set X has its image in set Y. It is also possible that there are few elements in set Y which are not the images of any element in set X. Every element in set X should have one and only one image. That means it is impossible to have more than one image for a specific element in set X. Functions can not be multi-valued (A mapping that is multivalued is called a relation from X and Y) e.g. Set X

Set Y

Set X

Set Y

1 2 3

a b c

Function

Set X

Set Y

Set X

Set Y

1 2 3

a b c

1 2 3

b c Not Function

Not Function

Testing for a function by vertical line test : A relation f : A B is a function or not, it can be checked by a graph of the relation. If it is possible to draw a vertical line which cuts the given curve at more than one point then the given relation is not a function and when this vertical line means line parallel to Y-axis cuts the curve at only one point then it is a function. Figure (iii) and (iv) represents a function. Y

X′

Y X

X′ Y′

(i)

X′

X Y′

(ii)

X Y′

(iii)

X′ O

X Y′

(iv)

Number of functions : Let X and Y be two finite sets having m and n elements respectively. Then each element of set X can be associated to any one of n elements of set Y. So, total number of functions from set X to set Y is nm. Real valued function : If R, be the set of real numbers and A, B are subsets of R, then the function f:A B is called a real function or real valued function.

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants.

MATHEMATICS TODAY | SEPTEMBER‘17

D OM A I N , C O - D OM A I N A N D R A NG E OF FUNCTION If a function f is defined from a set A to set B then (f : A B) set A is called the domain of f and set B is called the co-domain of f. The set of all f-images of the elements of A is called the range of f. In other words, we can say Domain = All possible values of x for which f(x) exists. Range = For all values of x, all possible values of f(x).

Equal functions : Two function f and g are said to be equal functions, if and only if (i) Domain of f = Domain of g (ii) Co-domain of f = Co-domain of g (iii) f(x) = g(x) x ∈ their common domain KINDS OF FUNCTION One-one function (injection) : A function f : A B is said to be a one-one function or an injection, if different elements of A have different images in B. e.g. Let f : A B and g : X Y be two functions represented by the following diagrams. A

Range Domain A a b c d

Co-domain B p q r s

Domain = (a, b, c, d) = A Co-domain = (p, q, r, s) = B Range = (p, q, r)

Methods for finding domain and range of function Domain : Expression under even root (i.e., square root, fourth root etc.) ≥ 0. Denominator ≠ 0. If domain of y = f(x) and y = g(x) are D1 and D2 respectively then the domain of (i) f(x) g(x) or f(x) g(x) is D1 D2. f (x ) is D1 D2 – {g(x) = 0} (ii) g (x ) (iii)

(

)

f ( x ) = D1

{x : f(x) ≥ 0}

Range : Range of y = f(x) is collection of all outputs f(x) corresponding to each real number in the domain. If domain ∈ finite number of points ⇒ range ∈ set of corresponding f(x) values. If domain ∈ R or R – [some finite points]. Then express x in terms of y. From this find y for x to be defined (i.e., find the values of y for which x exists). If domain ∈ a finite interval, find the least and greatest value for range using monotonicity. Algebra of functions : For functions f : x R; g : x R, we have x ∈ X, (cf )(x) = cf(x), where c is a scalar. (f g)(x) = f(x) g(x). (fg)(x) = (gf )(x) = f(x) g(x). f (x ) f  g  (x ) = g (x ) , g (x ) ≠ 0 44

a1 a2 a3 a4

MATHEMATICS TODAY | SEPTEMBER‘17

b1 b2 b3 b4 b5

x1 x2 x3 x4

Y y1 y2 y3 y4 y5

Clearly, f : A B is a one-one function. But g:X Y is not one-one function because two distinct elements x1 and x3 have the same image under function g. Method to check the injectivity of a function (i) Take two arbitrary elements x, y (say) in the domain of f. (ii) Solve f(x) = f(y). If f(x) = f(y) gives x = y only, then f : A B is a one-one function (or an injection). Otherwise not. If function is given in the form of ordered pairs and if two ordered pairs do not have same second element then function is one-one. If the graph of the function y = f(x) is given and each line parallel to x-axis cuts the given curve at maximum one point then function is one-one. e.g. Y

(0, 1) X′

O Y′

X f(x) = ax + b

X′

X O f(x) = ax (0 < a < 1) Y′

Number of one-one functions (injections) : If A and B are finite sets having m and n elements respectively, then number of one-one functions  n P , if n ≥ m from A to B =  m  0 , if n < m Many-one function : A function f : A B is said to be a many-one function if two or more elements of set A have the same image in B.

Thus, f : A B is a many-one function if there exist x, y ∈ A such that x ≠ y but f(x) = f(y). In other words, f : A B is a many-one function if it is not a one-one function. A a1 a2 a3 a4 a5

b1 b2 b3 b4 b5 b6

x1 x2 x3 x4 x5

y1 y2 y3 y4 y5

If function is given in the form of set of ordered pairs and the second element of atleast two ordered pairs are same then function is many-one. If the graph of y = f(x) is given and the line parallel to x-axis cuts the curve at more than one point then function is many-one. Y

Method to find onto or into function : (i) Solve f(x) = y by taking x as a function of y i.e., g(y)(say). (ii) Now if g(y) is defined for each y ∈ co-domain and g(y) ∈ domain then f(x) is onto and if any one of the above requirements is not fulfilled, then f(x) is into. One-one onto function (bijection) : A function f : A B is a bijection if it is one-one as well as onto. In other words, a function f : A B is a bijection if (i) It is one-one i.e., f(x) = f(y) ⇒ x = y for all x, y ∈ A. (ii) It is onto i.e., for all y ∈ B, there exists x ∈ A such that f(x) = y.

a1 a2

b1 b2 b3 b4

X′

O f(x) = x2 Y′

X′

O f(x) = |x| Y′

Onto function (surjection) : A function f : A B is onto if each element of B has its pre-image in A. In other words, Range of f = Co-domain of f. e.g. The following arrow-diagram shows onto function. A a1

x1 x2 x3 x4

Y y1 y2 y3

Number of onto function (surjection) : If A and B are two sets having m and n elements respectively such that 1 n m, then number of onto functions from A to B is

Into function : A function f : A B is an into function if there exists an element in B having no pre-image in A. In other words, f : A B is an into function if it is not an onto function e.g., The following arrowdiagram shows into function. B

Clearly, f is a bijection since it is both injective as well as surjective. Number of one-one onto function (bijection) : If A and B are finite sets and f : A B is a bijection, then A and B have the same number of elements. If A has n elements, then the number of bijection from A to B is the total number of arrangements of n items taken all at a time i.e. n!. Algebraic functions : Functions consisting of finite number of terms involving powers and roots of the independent variable and the four fundamental operations +, –, × and are called algebraic functions. 3 2 x

∑ (−1)n−r nCr r m .

r =1

Y y1 y2 y3 y4

+ 5x e.g., (i) 4 (iii) 3x – 5x + 7

(ii)

x +1 ,x ≠1 x −1

Transcendental function : A function which is not algebraic is called a transcendental function. e.g., trigonometric; inverse trigonometric , exponential and logarithmic functions are all transcendental functions. Trigonometric function : A function is said to be a trigonometric function if it involves circular functions (sine, cosine, tangent, cotangent, secant, cosecant) of variable angles. MATHEMATICS TODAY | SEPTEMBER‘17

Sine function

Cotangent function Y

(–3p/2, 1)

(p/2, 1)

(–3p/2, 0)

(–p/2, 0)

X′

–p

(–p/2, –1) Y′ Domain = R ; Range = [–1, 1]

X′

Function

(0, 1)

p 2

Y′ (–p, –1) Domain = R; Range = [–1, 1]

(p, –1)

Tangent function Y

–3p/2

–p

–p/2

p/2

3p/2

Y′ Domain = R – {(2n + 1) p/2 |n ∈ I}; Range = R Y

Domain

Definition of the function [–1, 1] y = sin–1 x sin–1 x [–p/2, p/2] x = sin y –1 [–1, 1] y = cos–1 x cos x [0, p] x = cos y –1 y = tan–1 x tan x (–∞, ∞) or R (–p/2, p/2) x = tan y y = cot–1 x cot–1 x (–∞, ∞) or R (0, p) x = cot y y = cosec–1 x cosec–1 x R – (–1, 1) [–p/2, p/2] x = cosec y – {0} –1 R – (–1, 1) [0, p] – {p/2} y = sec–1 x sec x x = sec y

(p/2, 1)

X′

y=1 X y = –1

(–3p/2, –1) (–p/2, –1) Y′ x = 2p x = –2p x = –p x=p Domain = R – {(np|n ∈ I}; Range = (–∞, –1] [1, ∞)

Secant function (–2p, 1)

X′

Y (2p, 1)

(0, 1)

a>1

(0, 1) X′

f(x) = ax X

Y′ Graph of f(x) = ax, when a > 1

MATHEMATICS TODAY | SEPTEMBER‘17

(0, 1) O

[1, ∞)

Y′ Graph of f(x) = ax, when a < 1

f(x) = logax O

(1, 0)

(p, –1)

f(x) = ax X′

X′

(–p, –1) Y′ x = –3p/2 x = –p/2 x = p/2 x = 3p/2 Domain = R – {(2n + 1) p/2 |n ∈ I}; Range = (–∞, –1]

a<1

Logarithmic function : Let a ≠ 1 be a positive real number. Then f : (0, ∞) R defined by f(x) = logax is called logarithmic function. Its domain is (0, ∞) and range is R.

Range

Exponential function : Let a ≠ 1 be a positive real number. Then f : R (0, ∞) defined by f(x) = ax called exponential function. Its domain is R and range is (0, ∞).

Cosecant function (–3p/2, 1)

x = 2p

Inverse trigonometric functions

pO 2

X′

(3p/2, 0)

Y′ x = –2p x = –p x=p Domain = R – {np | n ∈ I}; Range = R

(3p/2, –1)

Cosine function Y

(p/2, 0) O

Y′ Graph of f(x) = logax, when a > 1

X′

(1, 0) O

f(x) = logax

Y′ Graph of f(x) = logax, when a < 1

Explicit and implicit functions : A function is said to be explicit if it can be expressed directly in terms of the independent variable otherwise is an implicit. e.g., y = sin–1 x + log x is explicit function, while x2 + y2 = xy and x3y2 = (a – x)2 (b – y)2 are implicit functions. Constant function : Let Y k be a fixed real number. f(x) = k k Then a function f(x) X′ X O given by f(x) = k for all x ∈ R is called a constant Y′ function. The domain of the constant function f(x) = k is the complete set of real numbers and the range of f is the singleton set {k}. The graph of a constant function is a straight line parallel to x-axis as shown in figure and it is above or below the x-axis according as k is positive or negative. If k = 0, then the straight line coincides with x-axis. Identity function : The Y f u nc t i on d e f i ne d by f(x) = x f(x) = x for all x ∈ R, X′ X O is called the identity function on R. Clearly, Y′ the domain and range of the identity function is R. The graph of the identity function is a straight line passing through the origin and inclined at an angle of 45° with positive direction of x-axis. Modulus function : The function defined by  x , when x ≥ 0 f (x ) =| x |=  is called the modulus − x , when x < 0 Y function. The domain of the modulus function f(x) =–x f(x) = x X is the set R of all real X′ O numbers and the range is Y′ the set of all non-negative real numbers. Greatest integer function : Let f(x) = [x] where [x] denotes the greatest integer less than or equal to x. The domain is R and the range is I. e.g. [1.1] = 1, [2.2] = 2, [– 0.9] = –1, [– 2.1] = – 3 etc. The function f defined by f(x) = [x] for all x ∈ R, is called the greatest integer function.

Y 3

X′

2 1 –3 –2 –1 O

1 –1

–2 –3 Y′

Signum function : The f u nc t i on d e f i ne d by | x | X′  ,x ≠ 0 or f (x ) =  x  0, x = 0

Y (0, 1) X

O (0, –1) Y′

 1, x > 0  f (x ) =  0, x = 0 is called the signum function. −1, x < 0  The domain is R and the range is the set {–1, 0, 1}. Reciprocal function : The function that associates 1 is each non-zero real number x to be reciprocal x called the reciprocal function. The domain and range of the reciprocal Y f(x) = 1/x function are both equal to R – {0} i.e., the set of X′ X O all non-zero real numbers. The graph is Y′ as shown. Power function : A function f : R R defined by, f(x) = xa , a ∈ R is called a power function. EVEN AND ODD FUNCTION Even function : If f(–x) = f(x), x ∈ domain then f(x) is called even function. e.g., f(x) = ex + e–x, f(x) = x2, f(x) = x sin x, f(x) = cos x, f(x) = x2 cos x, all are even functions. Odd function : If we put (–x) in place of x in the given function and if f(–x) = –f(x), x ∈ domain then f(x) is called odd function. e.g., f(x) = ex – ex, f(x) = sin x, f(x) = x3, f(x) = xcos x, f(x) = x2 sin x, all are odd functions. Properties of even and odd function (i) The graph of even function is always symmetric with respect to y-axis. The graph of odd function is always symmetric with respect to origin. (ii) The product of two even functions is an even function. MATHEMATICS TODAY | SEPTEMBER‘17

f –1 : B A, f –1(b) = a ⇒ f (a) = b In terms of ordered pairs inverse function is defined as f –1 = (b, a) if (a, b) ∈ f. For the existence of inverse function, it should be one-one and onto. Properties of Inverse function (i) Inverse of a bijection is also a bijection and is unique. (ii) (f –1)–1 = f (iii) If f : A B is bijection and g : B A is inverse of f. Then fog = IB and gof = IA. where, IA and IB are identity functions on the sets A and B respectively. (v) If f : A B and g : B C are two bijections, then gof : A C is bijection and (gof )–1 = ( f –1og–1). (iv) fog ≠ gof but if, fog = gof then either f –1 = g or g –1 = f also, ( fog)(x) = (gof )(x) = (x).

(iii) The sum and difference of two even functions is an even function. (iv) The sum and difference of two odd functions is an odd function. (v) The product of two odd functions is an even function. (vi) The product of an even and an odd function is an odd function. It is not essential that every function is even or odd. It is possible to have some functions which are neither even nor odd function. e.g., f(x) = x2 + x3, f(x) = loge x, f(x) = ex . (vii) The sum of even and odd function is neither even nor odd function. (viii) Zero function f(x) = 0 is the only function which is even and odd both. PERIODIC FUNCTION A function is said to be periodic function if its each value is repeated after a definite interval. So a function f(x) will be periodic if a positive real number T exist such that, f(x + T) = f(x), x ∈ domain. Here the least positive value of T is called the period of the function. COMPOSITE FUNCTION If f : A B and g : B C are two functions then the composite function of f and g, gof : A C will be defined as gof(x) = g(f(x)), x ∈ A Properties of composition of function (i) f is even, g is even ⇒ fog is even function. (ii) f is odd, g is odd ⇒ fog is odd function. (iii) f is even, g is odd ⇒ fog is even function. (iv) f is odd, g is even ⇒ fog is even function. (v) Composite of functions is not commutative i.e., fog ≠ gof. (vi) Composite of functions is associative i.e., (fog)oh = fo(goh) (vii) Function gof will exist only when range of f is the subset of domain of g. (viii) If both f and g are one-one, or onto then fog and gof are also one-one or onto.

PROBLEMS Single Correct Answer Type 1.

MATHEMATICS TODAY | SEPTEMBER‘17

x −3 , then f [f {f (x)}] equals x +1 1 x (a) x (b) –x (c) (d) − x 2 If f (x) = cos(logx), then the value of  1 x f ( x ) ⋅ f ( 4 ) −  f   + f ( 4 x ) 2 4  If f (x ) =

(a) 1

(b) –1

(d)

Let f : R R be defined by f(x) = 2x + |x|, then f(2x) + f (–x) – f(x) = (a) 2x (b) 2|x| (c) –2x (d) –2|x|

If f(x + ay, x – ay) = axy, then f(x, y) is equal to (a) xy (b) x2 – a2y2 (c)

INVERSE FUNCTION If f : A B be a one-one onto (bijection) function, A which associates then the mapping f –1 : B each element b ∈ B with element a ∈ A, such that f(a) = b, is called the inverse function of the function f : A B.

 2x  1+ x  If f (x ) = log  , then f  is equal to  1− x   1 + x 2  (a) [f(x)]2 (b) [f(x)]3 (c) 2f(x) (d) 3f(x)

x2 − y2 4

(d)

x2 − y2

a2 If f(x) = cos[p2]x + cos[–p2]x, then

π (a) f   = 2 4

(b) f (–p) = 2

(d)

π f   = −1 2

10 + x , x ∈(−10, 10) and If e f ( x ) = 10 − x

 200 x  f (x ) = kf  ,then k =  100 + x 2  (a) 0.5 8.

(b) 0.6

(d) 0.8

The graph of the function y = f(x) is symmetrical about the line x = 2, then (a) f(x) = –f (–x) (b) f(2 + x) = f(2 – x) (c) f(x) = f (–x) (d) f(x + 2) = f(x – 2) Mapping f : R R which is defined as f(x) = cosx, x ∈ R will be (a) Neither one-one nor onto (b) One-one (c) Onto (d) One-one onto

10. Let f : N N defined by f(x) = x2 + x + 1, x ∈ N, then f is (a) One-one onto (b) Many one onto (c) One-one but not onto (d) None of these 11. Set A has 3 elements and set B has 4 elements. The number of injection that can be defined from A to B is (a) 144 (b) 12 (c) 24 (d) 64 x −m 12. Let f : R R be a function defined by f (x ) = , x −n where m ≠ n. Then (a) f is one-one onto (b) f is one-one into (c) f is many one onto (d) f is many one into 13. Which one of the following is a bijective function on the set of real numbers (d) x2 + 1 (a) 2x – 5 (b) |x| (c) x2 14. Let the function f : R R be defined by f (x) = 2x + sin x, x ∈ R. Then f is (a) One-to-one and onto (b) One-to-one but not onto (c) Onto but not one-to-one (d) Neither one-to-one nor onto x 15. If f : [0, ∞) [0, ∞) and f (x ) = , then f is 1 + x (a) One-one and onto (b) One-one but not onto (c) Onto but not one-one (d) Neither one-one nor onto 16. If f : R S defined by f (x ) = sin x − 3 cos x + 1 is onto, then the interval of S is (a) [–1, 3] (b) [1, 1] (c) [0, 1] (d) [0, –1]

17. If R denotes the set of all real numbers then the function f : R R defined f(x) = [x] (a) One-one only (b) Onto only (c) Both one-one and onto (d) Neither one-one nor onto 18. f (x ) = x + x 2 is a function from R R , then f(x) is (a) Injective (b) Surjective (c) Bijective (d) None of these 19. The period of f (x) = x – [x], if it is periodic, is 1 (a) f (x) is not periodic (b) 2 (c) 1 (d) 2 20. If f (x) = ax + b and g (x) = cx + d, then f(g(x)) = g(f(x)) is equivalent to (a) f (a) = g (c) (b) f (b) = g (b) (c) f (d) = g (b) (d) f (c) = g (a) 21. The domain of the function f (x ) = (a) [2, 4] (c) [2, ∞)

sin −1 (3 − x ) is log(| x | − 2)

(b) (2, 3) (3, 4] (d) (– ∞, –3) [2, ∞)

22. The function f (x ) =

sec −1 x

, where [.] denotes x − [x] the greatest integer less than or equal to x is defined for all x belonging to (a) R (b) R – {(–1, 1) (n | n ∈ Z)} (c) R+ – (0, 1) (d) R+ – {n | n ∈ N} 1/2

  5x − x 2   23. Domain of the function f (x ) =  log10     4   (a) –∞ < x < ∞ (b) 1 x 4 (c) 4 x 16 (d) –1 x 1

24. f : [0, 1] R is a differentiable function such that f (0) = 0 and |f ′(x)| k |f (x)| for all x ∈ [0, 1], (k > 0), then which of the following is/are always true ? (a) f (x) = 0, x ∈ R (b) f (x) = 0, x ∈ [0, 1] (c) f (x) ≠ 0, x ∈ [0, 1] (d) f (1) = k MATHEMATICS TODAY | SEPTEMBER‘17

25. Domain of definition of the function 3 f (x ) = + log10 (x 3 − x ) , is 2 4−x (a) (1, 2) (b) (–1, 0) (1, 2) (c) (1, 2) (2, ∞) (d) (–1, 0) (1, 2) (2, ∞)

35. The function f (x ) = sin  log(x + x 2 + 1)  is   (a) Even function (b) Odd function (c) Neither even nor odd (d) Periodic function ex − e−x 36. The inverse of the function f (x ) = + 2 is x −x e + e given by

26. The domain of the function f (x ) = x − x 2 + 4 + x + 4 − x is (a) [–4, ∞) (b) [–4, 4] (c) [0, 4] (d) [0, 1] 27. The domain of the function log( x 2 − 6 x + 6) is (a) (–∞, ∞) (b) (−∞, 3 − 3 ) ∪ (3 + 3 , ∞) (c) (–∞, 1]

[5, ∞)

(d) [0, ∞)

28. The natural domain of the real valued function defined by f (x ) = x 2 − 1 + x 2 + 1 is (a) 1 < x < ∞ (b) –∞ < x < ∞ (c) –∞ < x < –1 (d) (–∞, ∞) – (–1, 1) 29. The domain of the function f (x ) = (a) [1, 2)

−1

sin (x − 3)

9 − x2 (b) [2, 3) (c) [1, 2] (d) [2, 3]

π  30. The range of f (x ) = sec  cos2 x  , − ∞ < x < ∞ is 4  (a) [1, 2 ]

(b) [1, ∞)

31. If f (x) = a cos(bx + c) + d, then range of f (x) is (a) [d + a, d + 2a] (b) [a – d, a + d] (c) [d + a, a – d] (d) [d – a, d + a] 32. Range of f (x ) = (a) [5, 9] (c) (5, 9)

x 2 + 34 x − 71

is x 2 + 2x − 7 (b) (–∞, 5] [9, ∞) (d) None of these

33. The function f : R R is defined by f(x) = cos2 x + sin4 x for x ∈ R, then f (R) = 3  3  3  3  (a)  , 1 (b)  , 1 (c)  , 1 (d)  , 1  4  4  4 4  2 x + 14 x + 9 34. If x is real, then value of the expression 2 x + 2x + 3 lies between (a) 5 and 4 (b) 5 and –4 (c) – 5 and 4 (d) None of these 52

MATHEMATICS TODAY | SEPTEMBER‘17

1/2

 x −1 (d) log e   x + 1 

−2

37. If the function f : [1, ∞) [1, ∞) is defined by x(x – 1) –1 f (x) = 2 , then f (x) is 1 (a)   2

x ( x −1)

(b)

1 (1 + 1 + 4 log 2 x ) 2

1 (1 − 1 + 4 log 2 x ) (d) Not defined 2

2x − 1 (x ≠ −5) , then f –1(x) is equal to x +5 x +5 1 5x + 1 (a) (b) ,x≠ ,x ≠2 2x − 1 2 2−x

38. If f (x ) =

 x −1  (b) log e   3 − x 

 x  (c) log e   2 − x 

1/2

x −2 (a) log e   x − 1 

5x − 1 ,x ≠2 2−x

(d)

x −5 1 ,x≠ 2x + 1 2

Multiple Correct Answer Type

39. Let f(x) be a non constant polynomial satisfying the relation f (x) f (y) = f (x) + f (y) + f (xy) – 2 for all real x and y and f (0) ≠ 1, suppose f (4) = 65. Then (a) f 1(x) is a polynomial of degree 2 (b) roots of f 1(x) = 2x + 1 are real (c) xf 1(x) = 3[f (x) – 1] (d) f 1 (–1) = 3 40. If a function satisfies (x – y) f (x + y) – (x + y) f (x – y) = 2(x2y – y3) x, y ∈ R and f (1) = 2, then (a) f (x) must be polynomial function (b) f (3) = 12 (c) f (0) = 0 (d) f (x) may not be differentiable

41. Solutions of the equations [x] + [y] = [x] [y] is/are where [.] denotes the greater integer function (a) 2 x < 3 and 2 y < 3 (b) 0 x < 1 and 0 y < 1 (c) 0 < x 2 and 0 < y 2 (d) None 42. Let f (x) = a1cos (a1 + x) + a2 cos(a2 + x) + … + an cos(an + x). If f (x) vanishes for x = 0 and x = x 1 (where x1 ≠ kp, k ∈ Z), then (a) a1cos a1 + a2 cos a2 + … + an cos an = 0 (b) a1sin a1 + a2 sin a2 + … + an sin an = 0 (c) f (x) = 0 has only two solutions 0, x1 (d) f (x) is identically zero x 43. Consider the real valued function satisfying 2f (sin x) + f (cos x) = x, then (a) domain of f (x) is R (b) domain of f (x) is [–1, 1]  −2 π π  (c) range of f (x) is  ,   3 6  −2 π π  (d) range of f (x) is  ,   3 3  x 2 − 4 x + 3 x < 3 44. Let f (x ) =  and x ≥3  x − 4  x − 3 g (x ) =  2 (x + 1) + 1 7 (a) ( f − g )   = −1 2

x<4 x≥4

then

46. The set A is equal to (a) [–5, –2] (c) [–5, 2]

(b) [2, 5] (d) [–3, –2]

Matrix-Match Type

47. Match the following. Column I Column II (A) The interval containing (p)  1  t h e c ompl e te s e t of 0, 2    solution of the equation 1 − x2 1 are +x= x x (B) The interval containing (q) [–1, 1]– {0} t h e c ompl e te s e t of values of ‘a’, for which (a + 1) x + ay – 1 = 0 is a normal to the curve xy =1, are (C) Complete set of values (r) [0, 1] of ‘a’ for which equation a sin2 x + |cos x| – 2a = 0 has atleast one solution belongs to the interval (D) The interval containing (s) [–1, 0] the range of the function 1 f (x ) = 2 1 + 2 cos x + 3 cos 4 x + 4 cos6 x + .....∞

(b) fog(3) = 3

(t)

Paragraph for Q. No. 45 and 46 4 2 Let f : [2, ∞) [1, ∞) defined by f (x) = 2x – 4x and π  sin x + 4 g :  , π  → A defined by g (x ) = be two 2  sin x − 2 invertible function, then 45. f –1(x) is equal to (a)

2 + 4 − log 2 x

(b)

(c)

2 − 4 + log 2 x

(d) None of these

2 + 4 + log 2 x

 1  −1, 2   

Integer Answer Type

48. Find n

the

natural

number

for

which

∑ f (c + r ) = 16(2n − 1) where the function satisfies

r =1

the relation f(x + y) = f (x) . f (y) for all x, y ∈ N (natural numbers) and f (1) = 2. 49. For non negative integers m, n define a function as follows  n +1 if m=0  f (m − 1,1) f (m, n) =  if m ≠ 0, n = 0  f (m − 1, f (m, n − 1)) if m ≠ 0, n ≠ 0  Then the value of f (1, 1) is MATHEMATICS TODAY | SEPTEMBER‘17

50. If function f satisfies the relation f(x) f ′(–x) = f(–x). f ′(x) for all x and f (0) = 3, now if f (3) = 3, then the value of f (–3) is 51. Number of real values of x, satisfying the equation [x]2 – 5[x] + 6 – sin x = 0, [.] denoting the greatest integer function is 52. If f (x + 2) – 5f (x + 1) + 6f (x) = 0 for all, and x x f (x) = Aa1 + Ba 2, then a1 + a2 is

53. Let f be a function from the set of positive integers to the set of real numbers i.e., f : N R such that (i) f (1) = 1 (ii) f (1) + 2f (2) + 3f (3) + ... + nf(n) = n(n + 1) f (n); for n ≥ 2. Find the value of 800·f (200).

54. If for all real values of u and v, 2f (u) cos v = f (u – v) + f (u + v) prove that for all real values of x, (i) f(x) + f(–x) = 2a cos x (ii) f (p – x) + f(–x) = 0 (iii) f (p – x) + f(x) = 2b sin x Then f(x) = a cos mx + b sin nx where a and b are arbitrary constants, find m + n.

(

)

55. Let f(x + y + 1) = f (x ) + f ( y ) for all x, y ∈ R and f (0) = 1. Then f(x) = (x + 1)m. Find m. SOLUTIONS 1+ x  1. (c) : Given, f(x) = log   1 − x  \

2x   2  1 +  2x  1 + x 2  = log  x + 1 + 2 x   log = f  1 + x 2  1 − 2 x   x 2 + 1 − 2x   1 + x 2  2 1 + x  1 + x  = log   = 2 log 1 − x  = 2f(x) 1 − x    

 x − 3  x + 1  − 3 f (x ) − 3 2. (a) : f [f(x)] = == f (x ) + 1  x − 3   +1 x + 1 x − 3 − 3x − 3 3 + x = = x − 3 + x +1 1− x  3+ x  1 − x  − 3  3+ x = Now, f [f(f(x))] = f   1 − x   3+ x  1 − x  + 1 3 + x − 3 + 3x =x = 3 + x +1− x 54

MATHEMATICS TODAY | SEPTEMBER‘17

3. (c) : f(x) = cos(log x)

 1   x f   + f ( 4 x ) 2   4   ⇒ y = cos(log x) cos(log 4)  1   x – cos  log    + cos(log 4 x )  4 2   1 ⇒ y = cos(log x) cos(log 4) – [cos(log x – log 4) 2 + cos(log x + log 4)] 1 ⇒ y = cos(log x) cos(log 4) – [2 cos(log x) cos (log 4)] 2 ⇒ y = 0. 4. (b) : f(2x) = 2(2x) + |2x| = 4x + 2|x|, f (–x) = –2x + |–x| = – 2x + |x|, f (x) = 2x + |x| ⇒ f (2x) + f (–x) – f (x) = 4x + 2|x| + |x| – 2x – 2x – |x| = 2|x|. 5. (c) : Given f(x + ay, x – ay) = axy Let x + ay = u and x – ay = v ...(i) u+v u−v and y = Then x = 2 2a Substituting the value of x and y in (i), we obtain Now let y = f(x) f(4) –

u2 − v 2 x2 − y2 . ⇒ f(x, y) = 4 4 6. (d) : f(x) = cos[p2]x + cos[–p2]x f(x) = cos(9x) + cos(–10x) = cos(9x) + cos(10x)  19 x   x cos   = 2 cos   2   2 f(u, v) =

 π  19π   π cos   ; f   = 2 cos   2  4   4 −1 1  π f   =2× × = −1 .  2 2 2 10 + x 7. (a) : e f(x) = , x ∈ (–10, 10) 10 − x  10 + x  ⇒ f(x) = log   10 − x  ⇒

 200 x  f  100 + x 2 

200 x   10 +  100 + x 2  = log  10 − 200 x   100 + x 2  2

 10 + x  10(10 + x )  = 2 log  = 2f(x) = log    10 − x  10(10 − x )  \

f(x) =

1 2

 200 x  1 ⇒k= = 0.5. f  2  100 + x  2

8. (b) : f(x) = f(–x) ⇒ f(0 + x) = f(0 – x) is symmetrical about x = 0. \ f(2 + x) = f(2 – x) is symmetrical about x = 2.

−2 ≤ (sin x − 3 cos x ) ≤ 2

9. (a) : Let x1, x2 ∈ R, then f(x1) = cos x1, f(x2) = cos x2, so f(x1) = f(x2) ⇒ cos x1 = cos x2 ⇒ x1 = 2np x2 ⇒ x1 ≠ x2, so it is not one-one. Again the value of f-image of x lies in between –1 to 1 ⇒ f (R) = { f(x) : – 1 f(x) 1} So other numbers of co-domain is not f-image. f (R) ∈ R, so it is also not onto. So this mapping is neither one-one nor onto.

−1 ≤ (sin x − 3 cos x + 1) ≤ 3 i.e., range = [–1, 3] \ For f to be onto S = [–1, 3].

10. (c) : Let x, y ∈ N such that f(x) = f(y) Then f(x) = f(y) ⇒ x2 + x + 1 = y2 + y + 1 ⇒ (x – y)(x + y + 1) = 0 ⇒ x = y or x = –(y + 1) N (Rejected) \ f is one-one. Now, Let f(x) = 1 ⇒ x2 + x + 1 = 1 ⇒ x(x + 1) = 0 ⇒ x = 0 or x = –1 (Not possible) \ f is not onto. 11. (c) : The total number of injective functions from a set A containing 3 elements to a set B containing 4 elements is equal to the total number of arrangements of 4 by taking 3 at a time i.e., 4 P3 = 24 . 12. (b) : For any x, y ∈ R, we have x −m y −m = ⇒x=y f(x) = f(y) ⇒ x −n y −n \ f is one-one. x −m Let a ∈ R such that f(x) = a ⇒ =α x −n m − nα ⇒ x= 1− α Clearly x R for a = 1. So, f is not onto.

13. (a) : |x|, x2 and x2 + 1 is not one-one. But 2x – 5 is one-one as f(x) = f(y) ⇒ 2x – 5 = 2y – 5 ⇒ x = y Also, f(x) = 2x – 5 is onto. \ f(x) = 2x – 5 is bijective. 14. (a) : f ′(x) = 2 + cos x > 0. So, f(x) is strictly monotonic increasing so, f(x) is one-to-one and onto. 15. (b) : f ′(x) =

1 (1 + x )2

> 0,

x ∈ [0 ∞) and

range ∈[0, 1) ⇒ Function is one-one but not onto. 16. (a) : − 1 + (− 3 )2 ≤ (sin x − 3 cos x ) ≤ 1 + (− 3 )2

−2 + 1 ≤ (sin x − 3 cos x + 1) ≤ 2 + 1

17. (d) : Let f(x1) = f(x2) ⇒ [x1] = [x2] ≠ x1 = x2 {For example, if x1= 1 4, x2 = 1 5, then [1 4] = [1 5] =1} \ f is not one-one. Also, f is not onto as its range I (set of integers) is a proper subset of its co-domain R. 18. (d) : We have f(x) = x + x 2 = x + |x| Clearly f is not one-one as f(–1) = f(–2) = 0 but – 1 ≠ 2. Also f is not onto as f(x) ≥ 0 x ∈ R, Also range of f = [0, ∞) R. 19. (c) : Let f(x) be periodic with period T. Then, f(x + T) = f(x) for all x ∈ R ⇒ x + T – [x + T] = x – [x], for all x ∈ R ⇒ x + T – x = [x + T] – [x] ⇒ [x + T] – [x] = T for all x ∈ R ⇒ T = 1, 2, 3, 4,... The smallest value of T satisfying f(x + T) = f(x) for all x ∈ R is 1. Hence f(x) = x – [x] has period 1. 20. (c) : We have f(x) = ax + b, g(x) = cx + d and f(g(x)) = g(f(x)) ⇒ f(cx + d) = g(ax + b) ⇒ a(cx + d) + b = c(ax + b) + d ⇒ ad + b = cb + d ⇒ f(d) = g(b). 21. (b) : f(x) =

sin −1 (3 − x ) log(| x | −2)

Let g(x) = sin–1(3 – x) ⇒ –1 3 – x 1 Domain of g(x) is [2, 4] and let h(x) = log(|x| – 2) ⇒ |x| – 2 > 0 ⇒ |x| > 2 ⇒ x < –2 or x > 2 ⇒ (–∞, –2) (2, ∞) We know that f (x ) ∀x ∈D1 ∩ D2 − {x ∈R : g (x ) = 0} (f/g)(x) = g (x ) \ Domain of f(x) = (2, 4] – {3} = (2, 3) (3, 4]. 22. (b) : The function sec –1 x is defined for all 1 is defined for x ∈ R – (–1, 1) and the function x −[x] all x ∈ R – Z. So the given function is defined for all x ∈ R – {(–1, 1) (n | n ∈ Z)}. MATHEMATICS TODAY | SEPTEMBER‘17

1/2

  5x − x 2   ...(i) 23. (b) : We have f(x) =  log10     4   From (i), clearly f(x) is defined for those values of x  5x − x 2  for which log10  ≥0  4   5x − x 2   5x − x 2  0 ≥ 10 ⇒ ⇒     ≥1  4   4  ⇒ x2 – 5x + 4 0 ⇒ (x – 1)(x – 4) 0 Hence domain of the function is [1, 4]. 24. (b) : (f ′(x))2 – k2(f(x))2 0 ⇒ (f ′(x) – kf(x)) (f ′(x) + kf(x)) 0 ⇒ (f(x)e–kx)′ (f(x)ekx)′ 0 ⇒ Exactly one of the functions 2 e 3 5 g1(x) = f(x)e–kx or g2(x) = f(x)ekx is non-decreasing. But f(0) = 0 ⇒ both function g1 and g2 have a value zero at x = 0 x ∈ [0, 1], g1(0) = 0 and g1 increasing ⇒ g1(x) ≥ 0 ⇒ f(x) ≥ 0 g2(0) = 0 and g2 decreasing ⇒ g2(x) 0 ⇒ f(x) 0 ⇒ f(x) = 0 x ∈ [0, 1] 3

+ log10(x3 – x). 4 − x2 So, 4 – x2 ≠ 0 ⇒ x ≠ ± 4 and x3 – x > 0 ⇒ x(x2 – 1) > 0 25. (d) : f(x) =

– \

D = (–1, 0) D = (–1, 0)

–

–1 0 1 (1, ∞) – { 4 } i.e., (1, 2) (2, ∞).

26. (d) : f(x) = x − x 2 + 4 + x + 4 − x Clearly f(x) is defined, if 4 + x ≥ 0 ⇒ x ≥ – 4 4–x≥0⇒x 4 x(1 – x) ≥ 0 ⇒ x ≥ 0 and x 1 \ Domain of f = (–∞, 4] [–4, ∞) [0, 1] = [0, 1]. 27. (c) : log(x2 – 6x + 6) ≥ 0 i.e., x2 – 6x + 6 ≥ 1 x2 – 6x + 5 ≥ 0 (x – 1)(x – 5) ≥ 0 i.e., x ∈ (–∞, 1] [5, ∞) 28. (d) : f(x) =

x 2 − 1 + x 2 + 1 ⇒ f(x) = y1 + y2

Domain of y1 = x 2 − 1 ⇒ x2 – 1 ≥ 0 ⇒ x2 ≥ 1 x ∈ (–∞, ∞) – (–1, 1) and Domain of y2 is real number, \ Domain of f(x) = (–∞, ∞) – (–1, 1). 56

MATHEMATICS TODAY | SEPTEMBER‘17

29. (b) : To define f(x), 9 – x2 > 0 ⇒ –3 < x < 3 ...(i) –1 (x – 3) 1 ⇒ 2 x 4 ...(ii) From (i) and (ii), 2 x < 3 i.e., [2, 3). π 2  30. (a) : f(x) = sec  cos x  4  2 We know that, 0 cos x 1 Also, at cos x = 0, f(x) = 1 and 2 ⇒ x ∈ [1, 2 ]. at cos x = 1, f(x) = 2 \ 1 x 31. (d) : f(x) = a cos(bx + c) + d For minimum cos(bx + c) = –1 from (i), f(x) = –a + d = (d – a) For maximum cos(bx + c) = 1 from (i), f(x) = a + d = (d + a) \ Range of f(x) = [d – a, d + a].

...(i)

x 2 + 34 x − 71

=y x 2 + 2x − 7 ⇒ x2(1 – y) + 2(17 – y)x + (7y – 71) = 0 For real value of x, discriminant, D ≥ 0 ⇒ y2 – 14y + 45 ≥ 0 ⇒ y ≥ 9, y 5. 32. (b) : Let

33. (c) : y = f(x) = cos2x + sin4 x ⇒ y = f(x) = cos2x + sin2x(1 – cos2x) ⇒ y = cos2x + sin2x – sin2x cos2x 1 ⇒ y = 1 – sin2x cos2x ⇒ y = 1 – sin22x 4 3 f(x) 1, (∵ 0 sin22x \ 4 ⇒ f(R) ∈ [3/4, 1]. 34. (c) :

x 2 + 14 x + 9

=y x 2 + 2x + 3 ⇒ x2(y – 1) + 2x(y – 7) + (3y – 9) = 0 Since x is real, \ 4(y – 7)2 – 4(3y – 9)(y – 1) ≥ 0 ⇒ 4(y2 + 49 – 14y) –4(3y2 + 9 – 12y) ≥ 0 ⇒ 4y2 + 196 – 56y – 12y2 – 36 + 48y ≥ 0 ⇒ 8y2 + 8y – 160 0 ⇒ y2 + y – 20 0 ⇒ (y + 5)(y – 4) 0 \ y lies between – 5 and 4. 35. (b) : f (x ) = sin  log (x + 1 + x 2 )   ⇒ f(–x) = sin  log( − x + 1 + x 2 )    ( 1 + x2 + x) ⇒ f(–x) = sin log  ( 1 + x 2 − x )   ( 1 + x 2 + x )    1  ⇒ f(–x) = sin log   (x + 1 + x 2 ) 

 2 −1  ⇒ f(–x) = sin  log(x + 1 + x )     2  ⇒ f(–x) = sin  − log(x + 1 + x )   ⇒ f(–x) = − sin  log(x + 1 + x 2 ) ⇒ f(–x) = –f(x)   \ f(x) is odd function. ex − e−x e2 x − 1 +2 36. (b) : y = x − x + 2 ⇒ y = 2 x e +e e +1 ⇒

 y − 1 1− y y −1 1 e2 x = = ⇒ x = log e  y −3 3− y 2  3 − y 

⇒ f

–1(y)

 y − 1 = loge   3 − y 

1/2

⇒f

–1(x)

 x − 1 = loge   3 − x 

Only x =

1 + 1 + 4 log 2 f (x ) 2

1/2

1 ± 1 + 4 log 2 f (x ) 2

lies in the domain

1 [1 + 1 + 4 log 2 x ] . 2 38. (b) : Let f(x) = y ⇒ x = f –1(y) 2x − 1 Now, y = ,(x ≠ −5) x+5 xy + 5y = 2x – 1 ⇒ 5y + 1 = 2x – xy 5y +1 ⇒ x(2 – y) = 5y + 1 ⇒ x = 2− y 5y +1 − 1 ⇒ f ( y) = 2− y 5x + 1 \ f −1 (x ) = , x≠2 2−x 39. (a, b, c, d) : Given f(x) f(y) = f(x) + f(y) + f(xy) – 2 x, y ∈ R. Put x = y = 1 then f(1)2 – 3f(1) + 2 = 0 ⇒ f(1) = 1 or f(1) = 2 If f(1) = 1, then f(x) = 1 x ∈ R. A contradiction ∵ degree of f(x) is positive. \ f(1) ≠ 1. hence, f(1) = 2. 1 Replace 'y' with then x  1  1 f(x) f   = f(x) + f   (∵ f(1) = 2)  x  x \ f(x) must be in the form xn + 1 or –xn + 1. ∵ f(4) = 65, f(x) = x3 + 1 ⇒ f 1(x) = 3x2. \

–1(x)

f (x ) –x=l x ⇒ f(x) = (lx + x2) f(1) = 2 l + 1 = 2 ⇒ l = 1 f(x) = x2 + x

Let

37. (b) : Given f(x) = 2x(x – 1) ⇒ x(x – 1) = log2 f(x) ⇒ x2 – x – log2 f(x) = 0 ⇒ x =

40. (a, b, c) : (x – y)f(x + y) – (x + y)f(x – y) = 2y((x – y)(x + y)) Let x – y = u; x + y = v 2uv(v − u) uf(v) – vf(u) = 2 f (v ) f (u) =v–u − v u  f (v )   f (u)   v − v  =  u − u = constant

41. (a, b) : Let a = [x] + [y] = [x] [y] Then from the given equation, we have a + b = a b ⇒ ab – a – b = 0 ⇒ ab – a – b + 1 = 1 ⇒ (a – 1)(b – 1) = 1. This is possible if (i) a – 1 = 1, b – 1 = 1 or (ii) a – 1 = – 1, b – 1 = – 1 Now, for (i), a = 2 and b = 2 And for (ii) a = 0 and b = 0 Thus [x] = 2 and [y] = 2 or [x] = 0, [y] = 0 But [x] = 2 ⇒ 2 x < 3 and [y] = 2 ⇒ 2 y < 3. Again [x] = 0 ⇒ 0 x < 1 and [y] = 0 ⇒ 0 y < 1 Thus, the solution sets are 0 x < 1 and 0 y < 1 ; 2 x < 3 and 2 y < 3. 42. (a, b, d) : f(0) = 0 ⇒ a1cos a1 + a2cos a2 +....+ ancos an = 0 f(x1) = 0 ⇒ (a1cos a1 + a2cos a2 +....+ ancos an)cos x1 – (a1 sin a1 + a2 sin a2 +....+ an sin an)sin x1 = 0 ⇒ a1 sin a1 + a2 sin a2 +...+ an sin an = 0 (∵ x1 ≠ np) \ a1 cos a1 + a2 cos a2 +....+ an cos an = 0 and a1 sin a1 + a2 sin a2 +...+ an sin an = 0 ⇒ f(x) = 0 x 43. (b, d) : Given 2f(sin x) + f(cos x) = x π Replace x by – x, 2 π ⇒ 2f(cos x) + f(sin x) = –x 2 π π ⇒ f(sin x) = x – ⇒ f(x) = sin–1 x – 6 6 \ Domain of f(x) is [–1, 1] and range of f  −π π π π   −2π π   2 − 6 , 2 − 6  i.e.,  3 , 3      MATHEMATICS TODAY | SEPTEMBER‘17

–1(x)

 7  7 44. (a, b, c, d) : f   = –0.5, g   = 0.5, g(3) = 0,  2  2 f(0) = 3; f(2) = –1, g(2) = –1; f(4) = 0; g(4) = 26 45. (b) : ∵ ff –1(x) = x –1

–1

2(f (x)) – 4(f (x)) = x ⇒ (f –1(x))4 – 4(f –1(x))2 – log2x = 0

\ \

(f –1(x))2 = 2 + 4 + log 2 x Range of f –1(x) is [2, ∞).

\ \

f –1(x) = 2 + 4 + log 2 x f –1(x) > 0

46. ⇒ ⇒ \

Putting x = 1, f(1) = ek ⇒ 2 = ek. Hence from (i), f(x) = 2x. This can also be obtained by putting y = 1 so that F(x + 1) = f(x) f(1) = 2f(x) ⇒ f(x) = 2f(x – 1). Putting successively x – 1, x – 2, x – 3, ...., 2 for x in the above and multiplying them, we get f(x) = 2x. Now,

∑ f (c + r ) = f(c + 1) + f(c + 2) +...+ f(c + n)

r =1

= 2c + 1 + 2c + 2 +...+ 2c + n = 2c 2 + 2c 22 +...+ 2c 2n = 2c 2{1 + 2 + 22 +... to n terms}

sin x + 4 (a) : g(x) = sin x − 2 −6 cos x π  g′(x) = ≥ 0 ∵ x ∈  , π 2 2  (sin x − 2) g(x) is increasing function, hence one-one function.   π  Range is  g   , g (π) lie [–5, –2].   2  

47. (A) (q), (B) (q, s, t), (C) (p, q, r, t), (D) (q, r) 1 − x2 1− xx 1 + | x |= +x = x x x 2 1− x ⇒ ⋅ x ≥ 0 ⇒ x 2 − 1 ≤ 0, x ≠ 0 x ⇒ x ∈ [–1, 1] – {0} −(a + 1) (B) Slope > 0 ⇒ > 0 ⇒ a ∈(−1, 0) a (C) a – a cos2 x + |cos x| – 2a = 0 | cos x | 1 = ⇒ a= 2 1 + cos x | cos x | + 1 | cos x | 1  1 | cos x | ∈[0, 1] ⇒ | cos x | + ∈[2, ∞) ⇒ a ∈ 0,   2 | cos x |

(A)

1(2n − 1) = 2c 2(2n – 1) 2 −1 ⇒ 16(2n – 1) = 2c + 1(2n – 1) ⇒ 2c + 1 = 16 = 24 ⇒ c + 1 = 4 \ c = 3. = 2c 2

49. (3) : f(1, 1) = f(0, f(1, 0)) = f(0, f(0, 1)) = f(0, 2) = 3 50. (3) : f(x) × f ′(–x) = f(–x) × f ′(x) ⇒ f ′(x) × f(–x) – f(x) × f ′(–x) = 0 d ⇒ [ f(x)f(–x)] = 0 dx ⇒ f(x) f(–x) = k Given, ( f(0))2 = k = 9 ⇒ k = 9 Then f(3) f(–3) = 9 ⇒ f(–3) = 3 5 ± 25 + 4 sin x − 24 5 ± 1 + 4 sin x = 2 ⋅1 2 1; –4 4 sin x 4; –3 1 + 4 sin 5

51. (1) : [x] = –1

sin x

1 + 4 sin x 5 0 ⇒ [x] is an integer sin x = 0 ⇒ [x] = 3 ⇒ x = p

For x = np, f (x) is not difined, Range of f (x) = (0, 1]

52. (5) : Let f(x) = emx be a solution. Then f(x + 2) = em(x + 2) = emx e2m, f(x + 1) = emx em Therefore, f(x + 2) – 5f(x + 1) + 6f(x) = 0 ⇒ emx e2m – 5emx em + 6emx = 0 ⇒ emx[e2m – 5em + 6] = 0 ⇒ e2m – 5em + 6 = 0; emx ≠ 0 Let em = u. Then from the above equation, u2 – 5u + 6 = 0 ⇒ (u – 2)(u – 3) = 0 u = 2 or u = 3. ⇒ em = 2 or em = 3 ⇒ emx = 2x or emx = 3x Hence, f(x) = A 2x + B 3x .

48. (3) : From the given equation f(x + y) = f(x) f(y), we have f(x) = ekx Where k is a constant.

53. (2) : Given, f(1) + 2f(2) + 3f(3) +...+ nf(n) = n(n + 1)f(n) Replacing n by (n + 1), we get

(D) If x ≠ np S = 1 + 2 cos2x + 3 cos4 x + ........... (1) cos2 xS = cos2 x + cos4 x + ........... (2), Subtracting eq. (1) and eq. (2), we get sin2 xS = 1 + cos2 x + cos4 x + ........... 1 1 sin2 xS = ⇒S= ⇒ f (x ) = sin 4 x 2 4 1 − cos x sin x

MATHEMATICS TODAY | SEPTEMBER‘17

...(i)

f(1) + 2f(2) + 3f(3) +...+ nf(n) + (n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1) ...(ii) Subtracting (i) from (ii), we get (n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1) – n(n + 1) f(n) ⇒ f(n + 1) = (n + 2)f(n + 1) – nf(n) ⇒ nf(n) = (n + 2)f(n + 1) – f(n + 1) = (n + 2 – 1)f(n + 1) = (n + 1)f(n + 1). Thus, we have 2f(2) = 3f(3) = ... = nf(n). Hence, from (1), f(1) + (n – 1)nf(n) = n(n + 1)f(n) ⇒ f(n){n(n + 1) – n(n – 1)} = f(1) = 1 ⇒ (n2 + n – n2 + n)f(n) = 1 ⇒ 2nf(n) = 1 1 1 1 \ f(200) = ⇒ f(n) = = 2n 2 × 200 400 54. (2) : Given f(u + v) + f(u – v) = 2f(u) cos v ...(i) Putting u = 0 and v = x in (i), we get f(x) + f(–x) = 2f(0) cos x ...(ii) = 2a cos x where a = f(0) π π – x and v = in (i), we get Now, putting u = 2 2 π f(p – x) + f(–x) = 0 ∵ cos = 0. 2

π π and v = – x in (i), we get 2 2  π π  f(p – x) + f(x) = 2f   cos  − x  = 2b sin x ...(iii)  2 2 

Also, on putting u =

 π where b = f    2 Now, adding (ii) and (iii), we get 2f(x) – f(–x) + f(–x) = 2a cos x + 2b sin x ⇒ 2f(x) = 2a cos x + 2b sin x (from (iii)) Hence f(x) = a cos x + b sin x. 55. (2) : Given, f(x + y + 1) =

(

f (x ) +

(

Putting y = 0, we get f(x + 1) =

f ( y)

)

f (x ) + 1 ;

Putting x = 0, we have f(1) = (1 + 1)2 = 22; Putting x = 1, f(2) = (2 + 1)2 = 32;

(

)

f (2) + 1 = (3 + 1)2 = 42 and Putting x = 2, f(3) = so on. Proceeding in this way, we get f(x) = (x + 1)2. 

MATHEMATICS TODAY | SEPTEMBER‘17

CLASS XII

Series 5

CBSE Integrals | Application of Integrals IMPORTANT FORMULAE INTEGRALS

∫ dx = x + C , where 'C ' is an arbitrary constant xn + 1 n x dx = + C ,where n ≠ –1 ∫ n+1 x x ∫ e dx = e + C ax x a dx = + C , where a > 0 ∫ log e a 1 ∫ x dx = log e | x| + C , where x ≠ 0 ∫ sinx dx = − cos x + C ∫ cosx dx = sin x + C 2 ∫ sec x dx = tan x + C 2 ∫ cosec x dx = −cot x + C ∫ sec x tan x dx = sec x + C ∫ cos e c x cot x dx = −cosec x + C ∫ tan x dx = log|sec x| + C or –log |cos x| + C ∫ cot x dx = log|sin x| + C or –log cosec x + C ∫ sec x dx = log|sec x + tan x| + C ∫

x π = log tan  +  + C 2 4 cosecx dx = log|cosec x – cot x| + C x = log tan + C 2

MATHEMATICS TODAY | SEPTEMBER ‘17

∫

dx 1− x

= sin −1 x + C or –cos– 1 x + C, where |x| < 1

∫ 1 + x 2 = tan– 1 x + C or – cot– 1 ∫

1 2

x x −1

x+C

dx = sec– 1 x + C or – cosec– 1 x + C

x−a

a+x

∫ x 2 + a2 = a tan

−1  x 

 a  + C

∫ x 2 − a2 = 2a log x + a + C ∫ a2 − x 2 = 2a log a − x + C ∫ ∫ ∫

dx 2

x + a2 dx 2

x −a dx

= log x + x 2 + a2 + C = log x + x 2 − a2 + C

x = sin −1   + C ,| x |< a a a2 − x 2

∫

a2 + x 2 dx =

x 2 a2 a + x 2 + log| x + a2 + x 2 | +C 2 2

∫

x 2 − a 2 dx =

x 2 a2 x − a2 − log| x + x 2 − a 2 | +C 2 2

∫

a2 − x 2 dx =

a2 x 2 x a − x 2 + sin −1   + C 2 2 a

Properties of Definite Integrals

By parts (ILATE)   d ∫ I ⋅ II dx = I ⋅ ∫ II dx − ∫  dx (I ) ⋅ ∫ II dx  dx (Here I and II functions are choosen on the basis of ILATE) IN T E G R A T IO N

B Y P A R T IA L F R A C T IO N S

Rational form

Partial form

px + q (x − a)(x − b)

A B + (x − a) (x − b)

A B C px + qx + r + + (x − a)(x − b)(x − c) (x − a) (x − b) (x − c) A B + (x − a) (x − a)2

px + q

(x − a)2 px 2 + qx + r

A B C + + (x − a) (x − a)2 (x − b)

(x − a) (x − b)

∫

f ( x )dx = f (t )dt

∫

f ( x )dx = −

a b

∫

A B C D + + + (x − a) (x − a)2 (x − a)3 (x − b)

(x − a)3 (x − b)

(x − a)(x 2 + bx + c)

A Bx + C + 2 (x − a) x + bx + c

where x2 + bx + c cannot be factorised further

f ( x )dx . In particular,

∫

∫ f ( x) dx = 0

∫

f ( x )dx = f ( x )dx +

∫ f ( x)dx c

∫

f ( x )dx = f ( a + b − x )dx

a a

∫ f ( x)dx = ∫ f ( a − x )dx 0

∫

f ( x )dx = f ( x )dx +

∫ f (2a − x )dx 0

 a  2a 2 f ( x )dx , if f ( 2 a − x ) = f ( x ) f ( x )dx =   0   0   0 , if f ( 2 a − x ) = − f ( x )

∫

 a  2 f ( x ), if f ( − x ) = f ( x ) [even function] f ( x )dx =   0   −a ( x ) [ odd function ] 0 , if f ( x ) f − = −   a

px 2 + qx + r

∫

∫ e [ f ( x) + f ′( x)]dx = e x

f ( x) + C

( f ( x ))n +1 +C n+1

∫

( f ( x ))n f ′( x )dx =

∫

f ′( x ) dx = log( f ( x )) + C f ( x)

APPLICATION OF INTEGRALS

The area of a region bounded by y2 = 4ax and 16ab sq. units. 3 The area of a region bounded by y 2 = 4ax and 8a2 y = mx is sq. units. 3m3 The area of a region bounded by y2 = 4ax and its x2 = 4by is

latus rectum is

8a2 sq. units. 3

The area of a region bounded by one arc of sinax or 2 sq. units. cosax and x-axis is a Area of an ellipse

x2 a2

y2 b2

= 1 is pab sq. units.

The area of a region bounded by y = ax2 + bx + c and x-axis is

3 2 − 4ac)

6a2

sq. units.

MATHEMATICS TODAY | SEPTEMBER ‘17

WORK IT OUT VERY SHORT ANSWER TYPE

1. Evaluate ∫ e 2. Find

(ii)

(x ) dx

 px + qx + r  ∫ log  px 2 − qx + r  dx − π /2 dx

(a x + b x )2 x x

a b

f ′( x ) + f ′( − x )

10. Find the area of the region bounded by the curve x2 = y, and the line y = 4. LONG ANSWER TYPE - I

2x − 3

∫ (x 2 − 1)(2x + 3) dx e

∫−1

log x

13. Evaluate

∫ 3 sin x + 4 cos x

14. Evaluate

∫−1 e

2 sin x + 3 cos x

dx as the limit of a sum.

15. Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 2), (1, 5) and (3, 4). LONG ANSWER TYPE - II

16. Evaluate

∫

x 2 + 1(log(x 2 + 1) − 2 log x )

17. Show that π /2

(i)

∫ 0

f (sin 2 x ) sin x dx =

−1

(1 − x + x 2 )dx

SOLUTIONS

dx 8. Find the value of ∫−2008 (2008)x + 1 dx 9. Evaluate ∫ 1 − 2x − x 2

12. Find the value of

∫ cot

a2 − x 2

20. Find the area enclosed between the curves y = sinx and y = cos x that lies between the lines x = 0 and π x= . 2

2008

0 1

6. Find the area bounded by the curves y = x and y = x3.

11. Find

f (cos 2 x )cos x dx

∫x+

(ii)

SHORT ANSWER TYPE

∫

(i)

5. Evaluate ∫ (2 x + 2− x )2 dx

7. Evaluate

∫ 0

3x − 2

19. Evaluate the following definite integrals :

π /2

f (sin 2 x ) sin x dx =

π/ 4

18. Find the smaller of the two areas in which the circles x2 + y2 = 4 is divided by the parabola y2 = 3(2x – 1).

∫−1| x | dx

∫

∫ 0

3. Evaluate 4. Solve

3 log x

π /2

∫

dx , x > 0

f (sin 2 x ) cos x dx

MATHEMATICS TODAY | SEPTEMBER ‘17

∫e

3 log x

(x 4 )dx = ∫ x 3 ⋅ x 4 dx (

elog x = x)

= x 7 dx = x + C ∫ 8 2. We have,

∫−1| x | dx = ∫−1 −x dx + ∫0 xdx 0

 x2   x2  1  1   = −   +   = −  0 −  +  − 0 = 1  2  2   2  −1  2  0  px 2 + qx + r  3. Let f (x ) = log    px 2 − qx + r   p(− x )2 + q(− x ) + r  ⇒ f (− x ) = log    p(− x )2 − q(− x ) + r   px 2 − qx + r  = log    px 2 + qx + r  −1  px 2 + qx + r   px 2 + qx + r  = log  log = −   2  = − f (x )  px 2 − qx + r   px − qx + r  ⇒ \

f (x) is an odd function. π /2

 px 2 + qx + r  log ∫  px 2 − qx + r  dx = 0 − π /2

4. We have, ∫ = 3⋅

3 3x − 2

dx = 3∫ (3x − 2)−1/2 dx

( 3x − 2 )1/2 1 ⋅3 2

+ C = 2 3x − 2 + C.

5. We have, ∫ (2 x + 2− x )2 dx = ∫ (22 x + 2−2 x + 2) dx −2 x

2 2 + + 2⋅ x + C (log 2) × 2 (log 2) (− 2 ) 1 = (22 x − 2 −2 x ) + 2 x + C 2 log 2 6. The given curves are y = x ...(i) and y = x3 ...(ii) Solving (i) and (ii), we get Y x3 = x ⇒ x(x2 – 1) = 0 P (1, 1) O ⇒ x = 0, 1, –1 3 X x When x = 0, y = 0; = y when x = 1, y = 1 and when x = –1, y = –1. Q (–1, 1) So, O(0, 0), P(1, 1) and Q(–1, –1) are points of intersection of (i) and (ii).

\ Required area = 2 area OABO

= 2∫ xdy 0 4

= 2∫

ydy

= 2 ⋅ 2  y3/ 2    3

y=4

x2 = y B

O 4 0

4 32 ( 8 − 0) = sq. units 3 3

10. Given curves are x2 = y and y = 4

 x2 x 4  \ Required area = 2∫ (x − x 3 )dx = 2  −  2 4 0 0  1 1   1 1 = 2  −  − (0 − 0) = 2 × = sq. units 4 2  2 4   7. We have,

(a x + b x )2

∫

axbx

dx = ∫

  a   b = ∫    +   + 2 dx =  a  b 

 a   b

a2 x + b2 x + 2a x b x axbx

 b   a

+ + 2x + C, a ≠ b b a log logg a b 2008 f ′(− x ) + f ′( x ) 8 . I=∫ dx −2008 (2008)− x + 1 2008 f ′(− x ) + f ′( x ) =∫ × (2008)x dx −2008 1 + (2008)x \ 2I = ∫

2008

−2008 2008

= 2∫

f ′(− x ) + f ′(x )dx f ′(x ) + f ′(− x )dx

I = [ f (x ) − f (− x )]2008 = f (2008) − f (−2008) 0 9.

I=∫

=∫

1 − (x 2 + 2 x ) 2 − (x 2 + 2 x + 1) dx dx =∫ = ∫ ...(i) 2 2 2 − (1 + x ) ( 2 ) − (1 + x )2 Let z = 1 + x, then dz = dx From (i), dz z 1+ x  +c I=∫ = sin −1 + c = sin −1   2  2 ( 2 )2 − z 2

11. Let

2x − 3 (x − 1) (x + 1) (2 x + 3) =

A B C + + x − 1 x + 1 2x + 3

⇒ 2x – 3 = A (x + 1) (2x +3) + B(x – 1) (2x + 3) + C (x –1) (x + 1) ⇒

2x – 3 = A (2x2 + 5x + 3) + B (2x2 + x – 3) + C (x2 –1)

Equating the coefficients of x2, x and constant terms, we get 2A + 2B + C = 0, 5A + B = 2 and 3A – 3B – C = –3 Solving these equations, we get 1 5 24 A = − , B = and C = − 10 2 5 1 24 5 − − 2x − 3 10 5 2 + + ∴ = 2 1 1 + − + 2 3 x x x (x − 1)(2 x + 3) 2x − 3 1 dx ∫ (x 2 − 1)(2x + 3) dx = − 10 ∫ x − 1 +

5 2

∫ x + 1−

24 5

∫ 2x + 3

1 5 24 log 2 x + 3 log x − 1 + log x + 1 − +C 10 2 5 2 5 12 1 = − log x − 1 + log x + 1 − log 2 x + 3 + C 10 2 5 =−

12. The value of

e2 e

∫−1

log x dx x

1  log x = 0 when x = 1 and hence negative in  , 1 and e  positive in (1, e2]

MPP-5 CLASS XII 1. (a) 6. (b) 11. (a,c,d) 16. (b)

2. 7. 12. 17.

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

3. 8. 13. 18.

ANSWER KEY (c) (a,b,c) (a,c) (1)

4. 9. 14. 19.

MATHEMATICS TODAY | SEPTEMBER‘17

5. 10. 15. 20.

(b) (a) (c) (1) 63

log x

∴ I=∫ 1 e 1

∫

dx +

log x

∫

− log x

1/e

dx +

∫

log x

eh e h  enh − 1  h ⋅ h  = lim h →0 e  e − 1  h →0 e

= lim

1 4 5  1  1  =  − (log x )2  +  (log x )2  = (−1)2 + − 0 = 1 2 2 2 2 2    1 e

13. Let 2 sin x + 3 cos x = l (3 sin x + 4 cos x) + m (3 cos x – 4 sin x). Equating the coefficients of sinx and cosx on both sides, we get 2 = 3l – 4m and 3 = 4l + 3m. 18 1 On solving, we get l = and m = 25 25 2 sin x + 3 cos x \ ∫ dx 3 sin x + 4 cos x

∫

dt = lx + m log | t | + C t 18 1 = x + log | 3 sin x + 4 cos x | + C 25 25 = lx + m∫

14.

∑h ∫a f (x) dx = hlim →0

f (a + rh)

15. Let the vertices of the given triangle be A (–1, 2), B (1, 5) and C (3, 4). The equations of the sides AB, BC and CA respectively are: − x + 11 3x + 7 y= ...(i); ...(ii) y= 2 2 x+5 and y = ...(iii) 2

r =1

From (i), n

∫−1 e

= lim ∑ h e −1 ⋅ e rh = lim e −1 ⋅ h h→0 r =1

h→0

∑ erh

r =1

1 = lim ⋅ h(e h + e 2h + e 3h + ... + enh ) h →0 e

h  e h {1 − (e h )n }  e h  h(1 − enh )  = lim  = lim    h →0 e  1 − eh  h →0 e  1 − e h  MATHEMATICS TODAY | SEPTEMBER ‘17

∫

3x + 7 2

−1

dx + 1

 1  3x 2 =  + 7x  2  2 

dx = lim ∑ h e a ⋅ e rh h→0 r =1

\ Required area = Area of region bounded by trap. ADEB + Area of region bounded by trap. BEFC – Area of region bounded by trap. ADFC =

...(i)

Where, nh = b – a, and n ∞ Here, f (x) = ex, a = –1, b = 1 \ f (a + rh) = ea + rh = ea erh Also, nh = b – a =1 – (–1) = 2

1 0 e2 − 1  1 eh − 1  ⋅e ⋅ = 1 = e − Q lim e 1 e  h →0 h 

l (3 sin x + 4 cos x ) + m (3 cos x − 4 sin x )

dx 3 sin x + 4 cos x 3 cos x − 4 sin x = l ∫ 1 dx + m∫ dx 3 sin x + 4 cos x (put 3 sin x + 4 cos x = t ⇒ (3 cos x – 4 sin x) dx = dt) =

 e2 − 1   h   e −1   h 

∫

− x + 11 2

dx −

∫

x+5

−1

2 3

dx 3

  1  x2 1  x2 + − + 11x  −  + 5x  2  2  −1 1 2  2 −1

1  3  3   1  9   1   + 7 −  − 7  + 2  − 2 + 33 −  − 2 + 11   2 2 2    1  9  1  −  + 15  −  − 5   2  2   2

1 1 = (14 + 22 − 4) − (4 + 20) 2 2 1 1 = (32) − (24) = 16 − 12 = 4 sq. units. 2 2 16.

∫

x 2 + 1(log(x 2 + 1) − 2 log x )

x 1+

x4 1 x

(log(x 2 + 1) − log x 2 ) x4

1  = ∫ 1 +   x2 

1/2

(Q f (cos 2t ) cos t is an even function and f(cos 2t) sin t is an odd function)

1 1  log  1 +  ⋅ dx  x2  x3

1 2 1 1    Put 1 + 2 = t ⇒ − 3 dx = dt ⇒ 3 dx − 2 dt  x x x 1 1   = ∫ t1/2 log t  −  dt = − ∫ log t ⋅ t1/2dt  2 2 3 / 2 3 t 1 1 t /2  = −  log t ⋅ −∫ ⋅ dt  (Integrate by parts) 2 3/2 t 3/2 

1 1 1 1 t 3/2 +C = − t 3/2 log t + ∫ t1/2dt = − t 3/2 log t + ⋅ 3 3 3 3 3/2

1 3/2 t (2 − 3 log t ) + C 9

1 1 1+   9  x2  π /2

3/2

 1   2 − 3 log  1 + 2   + C x   π /2

= 2

π/ 4

∫

f (cos 2 x ) cos x dx

18. The given circle is x2 + y2 = 4 Its centre is (0, 0) and radius = 2. The given parabola is y2 = 3(2x – 1)

π /2

∫

f (sin (π − 2 x )) cos x dx =

π /2

∫

( ) 1 2 ,0

f (sin 2 x ) cos x dx

π π (ii) Put x = − t ⇒ t = − x ⇒ dx = − dt 4 4 When x = 0, t=

P(1, 3 )

π π π π π π − 0 = and when x = , t = − = − 4 2 4 2 4 4 π /2

− π/4

...(iii)

∫

f (sin 2 x ) sin x dx =

...(ii)

1  ⇒ y2 = 6  x −   2 Solving (i) and (ii), we get x2 + 3(2x – 1) = 4 ⇒ x2 + 6x – 7 = 0 ⇒ (x + 7) (x – 1) = 0 ⇒ x = –7, 1 1 but x ≥ , from (iii) we get, x = 1 2 When x = 1, y2 = 3 ⇒ y = ± 3

 π  f  sin 2  − x   2   0 0 π/2   π π  n 2 x ) sin x dx = ∫ f  sin 2  − x   sin  − x  dx 2    2  0

∫

17. (i)

...(i)

B X (2, 0)

Q(1, − 3 )

\ P (1, 3 ), and Q (1, − 3 ) are points of intersection of (i) and (ii) So, required area = 2 . area of the region PAB 1

 π   ∴ ∫ f (sin 2x)sin x dx = ∫ f  sin 2  4 − t   0 π/4 π  sin  − t  (−dt ) 4  − π/ 4 π π = − ∫ f (cos 2t ) (sin cos t − cos sin t ) dt 4 4

= 2∫

 1  1 = ∫ f (cos 2t )  cos t − sin t  dt 2  2  − π/ 4

 2 3  1  (1 − 0) + (0 + 4 sin −1 (1)) −  3 + 4 sin −1     2   3

2 3 π  π + 4⋅ − 3 + 4⋅  3 2  6

π/ 4 π/ 4

π/4

∫

2 − π/4 1 2

⋅2

f (cos 2t ) cos t dt −

π/ 4

∫ 0

π/4

∫

2 − π/4

f (cos 2t ) cos t dt −

1 2

⋅0

f (cos 2t ) sin t dt

1/2

3(2 x − 1) dx + 2 ∫ 4 − x 2 dx 1

  x 4 − x2  (2 x − 1)3/2  4 x  + 2⋅  + sin −1  = 2⋅ 3   2 2 2 1  3 ⋅2  1/2  2

2 3 4π 2  3  − 1 + 2π − π = − + 3  3 3 3 1 = (4 π − 3 ) sq. units 3

MATHEMATICS TODAY | SEPTEMBER‘17

19. (i) Let I =

∫ 0

20. Given, curves are y = sin x and y = cos x;

dx x + a2 − x 2

Y y = cos x y = sin x

Put x = a sin t ⇒ dx = a cos t dt π 2 a π /2 a cos t dx \ I=∫ = ∫ dt 2 2 a sin t + a cos t + − x a x 0 0

When x = 0, t = 0 and when x = a, t =

π /2

∫

cos t sin t + cos t

Then by using

∫

...(i)

dt a

f (x ) dx = ∫ f (a − x ) dx , we get 0

π  cos  − t  π /2 π /2 sin t 2  dt = ∫ dt I= ∫ cos t + sin t π  π  0 sin  − t  + cos  − t  0 ...(ii) 2  2  On adding (i) and (ii), we get 2I =

π /2

∫

cos t + sin t dt = sin t + cos t

π /2

∫ 1dt 0

= [t ]0π /2

π π = −0= 2 2

π 4 1   (ii) We have, cot–1 (1 – x + x2) = tan −1  2 1− x + x  −1  x + (1 − x )  –1 –1 == tan   = tan x + tan (1 – x)  1 − x(1 − x )  ⇒ I=

⇒

∫ cot 1

−1

= ∫ tan 0 1

−1

(1 − x + x )dx = ∫ tan 1

−1

0 1 0

(integrate by parts)

  1 1 1 = 2  (tan −1 x ⋅ x ) 0 − ∫ ⋅ x dx  2   0 1+ x 2x

π = 2(1 ⋅ tan −1(1) − 0) − ∫ dx = 2 ⋅ − | log(1 + x 2 )|10 2 4 0 1+ x π π π = − (log 2 − log 1) = − (log 2 − 0) = − log 2 2 2 2 66

\ Required area =

MATHEMATICS TODAY | SEPTEMBER ‘17

π /2

∫ | cos x − sin x |dx

π/ 4

0 π /2

0 π/ 4

π/ 4 π /2

∫ | cos x − sin x |dx + ∫ | cos x − sin x |dx ∫ 0

(cos x − sin x ) dx + π/ 4

= sin x + cos x  0

∫

(sin x − cos x ) dx

π/ 4

π /2

+  − cos x − sin x  π/ 4

 1   1  =  + − (0 + 1) −  (0 + 1) −  2  2   2 2 = −1−1 + = 2 2 − 2 = 2 2 −1 2 2

(

 1 1  +   2 2 

) sq. units. 

JAMMU & KASHMIR at

x dx + ∫ tan (1 − (1 − x )) dx

The curves cross each other at sin x = cos x i.e., π at tan x = 1 i.e., at x = . 4

x dx + ∫ tan (1 − x )dx

−1

= 2∫ tan −1 x ⋅1 dx

−1

= tan −1 x dx + tan −1 x dx ∫ ∫ 0

• • • • • • • • • • • •

Kabli Book Stall - Anantnag Mob: 8803043296, 9906460029 New Valley Book Depot - Anantnag Mob: 9469183540 Sharjha Book Depot - Anantnag Mob: 9419040456 Harnam Dass & Bro’S - Jammu Ph: 0191-2542175, 2574428; Mob: 9419664141 Sahitya Sangam - Jammu Ph: 0191-2579593, 2562191, 2579593; Mob: 9419190177 Shiela Book Centre - Jammu Ph: 0191-2574912; Mob: 9419146803 Kapoor Sons - Srinagar Ph: 0194-2456458; Mob: 9419425757, 9419069199 Kashmir Book Depot - Srinagar Ph: 2450262, 474440, 2475973; Mob: 9906726231, 9419761773 Shah Book Centre - Srinagar Mob: 9906763627, 9419062444 Abdullah News Agency - Srinagar Ph: 0194-2472621, 2435057; Mob: 9419074859 Highway Book Shop - Srinagar Mob: 9858304786 Paradise Book House - Srinagar Mob: 9419067856

Class XII

Application of Derivatives Total Marks : 80

Only One Option Correct Type 1. The coordinates of the point P(x, y) lying in the first quadrant on the ellipse x2/8 + y2/18 = 1 so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, are given by (a) (2, 3) (b) ( 8 , 0) (d) none of these (c) ( 18 , 0) 3/2 2. If f(x) = x (3x – 10), x > 0, then f(x) is increasing in (a) (–∞, –1) (1, ∞) (b) [2, ∞) (c) (–∞, –1) [1, ∞) (d) (–∞, 0] (2, ∞) 3. The set of all values of a for which the function  a+4  5 − 1 x – 3x + log 5 decreases for all f(x) =   1− a  real x is  5 − 27  ∪ (2, ∞) (a)  −3, 2    3 − 21  (b)  −4,  ∪ (1, ∞) 2   (c) (–∞, ∞) (d) [1, ∞). 4. If the function f(x) = cos|x| – 2ax + b increases along the entire number scale, the range of values of a is given by (a) a b (b) a = b/2 (c) a –1/2 (d) a ≥ 3/2 5. The image of the interval [–1, 3] under the maping f(x) = 4x3 – 12x is (a) [–2, 0] (b) [–8, 72] (c) [–8, 0] (d) none of these

Time Taken : 60 Min.

6. By LMVT, which of the following is true for x > 1? (a) 1 + x ln x < x < 1 + ln x (b) 1 + ln x < x < 1 + x ln x (c) x < 1 + x ln x < 1 + ln x (d) 1 + ln x < 1 + x ln x < x One or More Than One Option(s) Correct Type

1 7. For the function f(x) = x cos , x ≥ 1, which of the x following is (are) correct? (a) for at least one x in the interval [1, ∞), f(x + 2) – f(x) < 2 (b) lim f ′(x) = 1 x→∞

(c) for all x in the interval [1, ∞], f(x + 2) – f(x) > 2 (d) f ′(x) is strictly decreasing in the interval [1, ∞) 8. Let F : R R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = – 4 and F′(x) < 0 for all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R. Then which of the following statements is (are) correct? (a) f ′(1) < 0 (b) f(2) < 0 (c) f ′(x) ≠ 0 for all x ∈ (1, 3) (d) f ′(x) = 0 for some x ∈ (1, 3) 9. If f(x) = f(x) + f(2a – x) and f ′′(x) > 0, a > 0, 0 x 2a, then (a) f(x) increases in (a, 2a) (b) f(x) increases in (0, a) (c) f(x) decreases in (a, 2a) (d) f(x) decreases in (0, a)  x 3 + x 2 − 10 x −1 ≤ x < 0  sin x 0 ≤ x < π/2 10. Let f(x) =   1 + cos x π/2 ≤ x ≤ π  then f(x) has MATHEMATICS TODAY | SEPTEMBER‘17

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

15. If L is the total light transmitted, then critical points of L are p 3p (a) y = (b) y = 6 + 3π / 2 12 + 3π / 2

local maxima at x = p/2 local minima at x = p/2 absolute minima at x = 0 absolute maxima at x = p/2

11. Consider the function f : R f(x) =

x − ax + 1 x 2 + ax + 1

If g(x) =

R given by

, 0 < a < 2.

3p 12 + 5π/2

(d) y =

2p 11 + 5π/2

Matrix Match Type

f ′(t )

∫ 1 + t 2 dt , which of the following is (are)

16. Match the following.

not true? (a) g′(x) is positive on (–∞, 0) and negative on (0, ∞) (b) g′(x) is negative on (–∞, 0) and positive on (0, ∞) (c) g′(x) changes sign on both (–∞, 0) and (0, ∞) (d) g′(x) does not change sign on (–∞, ∞) 12. For x > 1, y = loge x satisfies the inequality (a) x – 1 > y (b) x2 – 1 > y x −1 (c) y > x – 1 (d) <y x 13. Suppose f ′(x) exists for each x and h(x) = f(x) – (f(x))2 + (f(x))3 for every real number x. Then (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general. Comprehension Type A window of fixed perimeter (including the base of the arch) is in the form of a rectangle surmounted by a semi-circle. The semi-circular portion is fitted with coloured glass, while the rectangular portion is fitted with clear glass. The clear glass transmits three times as much light per square metre as the coloured glass. Suppose that y is the length and x is the breadth of the rectangular portion and p the perimeter. 14. The ratio of the sides y : x of the rectangle so that the window transmit the maximum light is (a) 3 : 2 (b) 6 : 6 + p (c) 6 + p : 6 (d) 1 : 2

Column I Column II 3 2 The function f(x) = 2x – 9x 1. (3, ∞) – 24x + 7 decreases on

Q. f(x) = R.

1 + 3x 4 + 5x 2

increases on 2.

(–1, 4)

3 2 x 3. 2

(–∞, 2]

f(x) = (x2 – 2x) log x – + 4x increases on

4. (a) (b) (c) (d)

P 2 2 1 4

Q 3 3 2 3

(e, ∞)

R 1 4 3 1

Integer Answer Type

17. If the point on y = x tan α −

ax 2

(a > 0), 2u2 cos2 α where the tangent is parallel to y = x has an ordinate u2/4a, then 4 sin2 a is equal to 18. If the greatest and least values of the functions 1   1 , 3  are G and f(x) = arc tan x − ln x on  2  3  L respectively, then [G + L] = (where [·] is greater integer) 19. If f(x) = |x – 1| + |x – 3| + |5 – x|, x ∈ R is symmetrical about the line x = l, then l = 20. If A > 0, B > 0 and A + B = p/3 then the maximum value of 3 tan A tan B is  Keys are published in this issue. Search now!

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 | SEPTEMBER ‘17

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

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.

JEE MAIN

1. If x2 – y2 – 84y = 2012, x, y ∈ N, then x – 3y = (a) 2 (b) 3 (c) 6 (d) 7 2. Let g (x ) =

(x − 1)n

log cosm (x − 1) integers, m ≠ 0, n > 0, and let p be the left hand derivative of |x – 1| at x = 1. If lim g (x ) = p, then (a) n = 1, m = 1 (c) n = 2, m = 2

Let

; 0 < x < 2, m and n are

x →1+

(b) n = 1, m = –1 (d) n > 2, m = n

3. Given, x = a cos t cos 2t and

(1 + (dy / dx )2 )3/2 y = a sin t cos 2t (a > 0), then at d 2 y / dx 2 π is given by 6 a 2 2a (a) (b) a 2 (c) (d) 3a 3 3 sin3 x / 2 dx is equal to 4. ∫ x cos cos3 x + cos2 x + cos x 2 (a) cos −1 sec x + tan x + 1 + c (b) tan −1 sin x + cos x + 1 + c (c) sin −1 tan x + sec x + 1 + c (d) tan −1 cos x + sec x + 1 + c

5. The limiting points of the coaxial system of circles given by x2 + y2 + 2gx + c + l(x2 + y2 + 2fy + k) = 0 subtend a right angle at the origin, if c k c k (b) (a) − − 2 =2 + 2 = −2 2 2 g f g f c k c k (d) 2 + 2 = 2 (c) 2 − 2 = 2 g f g f JEE ADVANCED

6. Five balls are to be placed in three boxes. Each can hold all the five balls. In how many different ways can we place the balls so that no box remains empty,

if balls are different but boxes are identical? (a) 25 (b) 15 (c) 10 (d) 35 COMPREHENSION f be a function satisfying a = g a (x )(a > 0) f (x ) = x a + a 1995  r   Let f(x) = g9(x), then the value of  ∑ f    r =1 1996   is (where [.] denotes the greatest integer function) (a) 995 (b) 996 (c) 997 (d) 998 2n 1  r  = + 987, then If the value of ∑ f   2n + 1  1 + a r =0 the value of n is (a) 493 (b) 494 (c) 987 (d) 988 INTEGER TYPE The remainder when 22003 is divided by 17 is MATRIX MATCH

10. Match the following. List-I 1/ x

 tan x  lim   x →0  x 

1. e

The number of points at which 2. 1 f ( x ) = | x | −1 is not differentiable, is

tan x

∫

1/e

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

List-II

4t dt 1+ t

cot x

∫

1/e

4 dt t (1 + t 2 )

The degree of the differential equation satisfied by all circles of radius r is P 3 1 3 2

Q 2 2 2 4

R 5 4 4 5

S 4 3 5 3

3. 2 4. 3 5. 4

See Solution Set of Maths Musing 176 on page no 85

MATHEMATICS TODAY | SEPTEMBER ‘17

TWO DIMENSIONAL GEOMETRY Time : 1 hr.

Marks : 60

MULTIPLE CORRECT CHOICE 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. If e1 and e2 are the eccentricities of the conic sections 16x2 + 9y2 = 144 and 9x2 – 16y2 = 144,then (a) e12 + e22 = 3 (c) e2 + e2 < 3 1 2

(b) e12 + e22 > 3 (d) e12 − e22 < 0

2. The equation(s) to the tangent(s) to the conic x2 + 4xy + 3y2 – 5x – 6y + 3 = 0, which are parallel to x + 4y = 0 are (a) x + 4y – 1 = 0 (b) x + 4y – 3 = 0 (c) x + 4y – 5 = 0 (d) x + 4y – 8 = 0 3. Consider the parabola y2 = 4ax and x2 = 4by. The straight line b1/3 y + a1/3x + a2/3 b2/3 = 0 (a) touches y2 = 4ax (b) touches x2 = 4by (c) intersects both parabolas in real points (d) touches first and intersects other 4. The co-ordinates of a point on the parabola y2 = 8x whose distance from the circle x2 + (y + 6)2 = 1 is minimum is (a) (2, 4) (b) (2, –4) (c) (18, –12) (d) 8, 8 5. The angle between the asymptotes of the hyperbola x2

−

= 1 is

1 1 (b) 2 cos−1   (a) cos−1   e  e  1 (c) sin−1   (d) None of these e  6. The circle x2 + y2 + 4x – 6y + 3 = 0 is one of the circles of a coaxial system of circles having as radical axis the line 2x – 4y + 1 = 0. Then the equation of the circle of the system which touches the line x + 3y – 2 = 0 is (a) x2 + y2 + 2x – 2y + 2 = 0 (b) x2 + y2 + 2x + 6y = 0 (c) x2 + y2 – 2x + 6y = 0 (d) x2 + y2 + 2x – 6y = 0 7. If a circle of constant radius 3k passes through the origin and meets the axes at ‘A’ and ‘B’, the locus of the centroid of DOAB is (b) x2 + y2 = 2k2 (a) x2 + y2 = k2 2 2 2 (c) x + y = 3k (d) None of these 8.

= 1 will represent the P 2 − P − 6 P 2 − 6P + 5 ellipse if P lies in the interval (a) (−∞, −2) (b) (1, ∞) (d) (5, ∞) (c) (3, ∞) +

9. If the eccentric angles of the extremities of a focal 2 2 chord of an ellipse x + y = 1 are a and b , then a2 b2 cos α + cos β sin α + sin β (b) e = (a) e = cos(α + β) sin(α + β)

By : Vidyalankar Institute, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367

MATHEMATICS TODAY | SEPTEMBER ‘17

Area (∆AMN ) 3 on OB and AB. If the ratio of = , Area (∆OAB) 8 AN . then find the value of BN

α −β α +β (c) cos  = e cos   2   2  (d) tan

α β e −1 tan = 2 2 e +1

10. If b and c are the lengths of the segments of any focal chord of a parabola y2 = 4ax, then the length of the semi-latus rectum is b+c bc (b) (a) 2 b+c 2bc (c) (d) bc b+c

19. A(0, 0), B(4, 2) and C(6, 0) are the vertices of a triangle ABC and BD is its altitude. The line through D parallel to the side AB intersects the side BC at a point E. Find the product of areas of DABC and DBDE. 20. Find the number of integral values of l if (l , 2) is an interior point of DABC formed by, x + y = 4, 3x – 7y = 8, 4x – y = 31.

ONE INTEGER VALUE CORRECT TYPE

ANSWER KEY

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]

(c, d)

(a, b)

(b)

(a, c)

(d)

(a, d)

11. The locus of the centre of the circle for which one end of diameter is (3, 3) while the other end lies on the line x + y = 4 is x + y = k, then k equals_____

(b, c, d) 10. (c)

13. (3) 17. (6)

x2 y2 + = 1, then |c| equals_____ 8 4 x2 y2 15. The foci of the ellipse + = 1 and the 16 b2 2 2 hyperbola x − y = 1 coincide, then value of b2 144 81 25 equals_____ ellipse

16. If the angle between the two lines represented by 2x2+ 5xy +3y2+ 7y + 4 = 0 is tan–1(m), then the value of 10 m must be_____ 17. From a point, common tangents are drawn to the circle x2 + y2 = 8 and parabola y2 = 16x. If the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola is 10k, then find k.

12. (3)

14. (6)

15. (7)

16. (2)

18. (3)

19. (8)

20. (1)

For detailed solution to the Sample Paper, visit our website www. vidyalankar.org 

12. The greatest distance of the point (10, 7) from the circle x2 + y2– 4x –2y – 20 = 0 is 5a , then a equals____ 13. Angle between the tangents drawn from (1, 4) to π the parabola y2 = 4 x is , where m equals_____ m 14. If the straight line y = 2x + c is a tangent to the

11. (5)

HARYANA at • • • • • • • • • • • • • • • • • • •

Bharat Book Depot - Ambala Ph: 0171-2643568; Mob: 9896316290 Gulati Book Palace - F aridabad Ph: 0129-2295960, 4007870; Mob: 9871450002 Book Care Centre - Gurgaon Ph: 0124-6521670, 4251670; Mob: 9873621670, 9873401670 Brilliant E ducational Aid - Gurgaon Ph: 0124-406725600; Mob: 9811092425, 9873092425 Mangla And Company Book Seller - Gurgaon Ph: 0172-2725216, 4642255; Mob: 9814006655 Modern Book Shop - Hissar Mob: 9416346209 New Gulati Book Depot - Jind Ph: 01681-252586; Mob: 9416947565, 9416166838 Books Ally Incorporation - Karnal Mob: 9812440050 Balbir And Brothers - Kurukshetra Mob: 9996400044 Hanuman Book Depot - Narnaul Mob: 9416178859, 9813194705 Holkaran Dass Hemraj - Narnaul Ph: 01282-252008; Mob: 9355552008, 9416268011 Kamal Book Depot - Narnaul Ph: 252305; Mob: 9416382533, 9050632762 Anamika Trading - Panchkula Ph: 2579143, 5011377; Mob: 9316002471 Y adav Book Depot - Rewari Ph: 1281245052; Mob: 9896126652, 9813074160 Manish Traders - Rohtak Ph: 01262-268715; Mob: 9812556687 Sunrise - Rohtak Mob: 8199944420, 8950350733 Tinku Book Depot - Rohtak Ph: 01262-211532; Mob: 9812375783 F air Deal Book Centre - Sirsa Mob: 9896054814, 9255102978 Sachdeva News Agency - Y amuna Nagar Mob: 9896377026

18. The line x + y = a meets the x-axis at A and y-axis at B. A DAMN is inscribed in the DOAB, O being the origin,with right angle at N; M and N lie respectively MATHEMATICS TODAY | SEPTEMBER‘17

Series-3 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. 3

Topic

Syllabus In Details

Permutations and Combinations Trigonometry Co-ordinate Geometry-2D

Fundamental principle of counting, permutation as an arrangement and combination as selection, meaning of P(n, r) and C(n, r), simple applications. General solution and Properties of Triangle. Circles: Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equations of a circle when the end points of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent.

1. The number of positive terms in the sequence xn = (a) 2 (c) 4

195 n

4 ⋅ Pn

−

n+3 n+1

Pn+1

, n ∈ N is (b) 3 (d) none of these.

2. The number of numbers greater than 50,000 that can be formed by using the digits 3, 5, 6, 6, 7 is (a) 36 (b) 48 (c) 54 (d) none of these. 3. The rank of the word 'FLOWER' is (a) 165 (b) 155 (c) 145 (d) none of these. 4. A person write letters to his 4 friends and addressed the corresponding envelopes. The number of ways at least two of them are in the wrong envelopes is (a) 23 (b) 19 (c) 17 (d) 14. 5. A test consists of 6 multiple choice questions each having 4 alternative answers of which only one is correct. Also only one of the alternatives must be marked by each candidate. The number of ways

of getting exactly 4 correct answers by a candidate answering all the questions is (b) 135 (a) 46 – 32 (c) 55 (d) 120. 6. Number of ways in which 3 boys and 3 girls(all are of different heights) can be arranged in a line so that boys as well as girls among themselves are in decreasing order to their height (from left to right) is (a) 720 (b) 72 (c) 10 (d) 20. 7. The number of permutations of the letters of the word HINDUSTAN such that neither the pattern 'HIN' nor 'DUS' nor 'TAN' appears, are (a) 166674 (b) 169194 (c) 166680 (d) 181434. 8. The number of ways in which four letters of the word MATHEMATICS can be arranged is given by 8! (a) (b) 2454 4!4! (c) 2464 (d) 2474.

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

MATHEMATICS TODAY | SEPTEMBER ‘17

9. The number of ways of distributing of 50 identical things among 8 persons in such a way that three of them get 8 things each, two of them get 7 things each, and remaining three of them get 4 things each, is equal to (50 !)(8 !) (a) 3 (8 !) (3 !)2 (7 !)2 (4 !)3 (2 !) (b) (c)

(50 !)(8 !) (8 !)3 (7 !)3 (4 !)3 (50 !)

(d)

⋅1 (8 !) (7 !) (4 !) (3 !) ⋅ 2 ! 3 ! 10. Consider seven digit number x1, x2, ...., x7 where x1 , x2 , ..., x7 =/ 0 having the property that x4 is the greatest digit and digits towards the left and right of x4 are in decreasing order. Then total number of such numbers in which all digits are distinct is (a) 9C7 · 6C3 (b) 9C6 · 5C3 10 6 (c) C7 · C3 (d) none of these. 3

11. The most general solution of the equation logcosqtanq +logsinqcotq = 0, is π π (b) nπ − (a) nπ + 4 4 π π (c) 2nπ − (d) 2nπ + 4 4 x 12. The general solution of 2 − cos x = 2 tan is 2 π π (b) ( 4n + 1) (a) (2n + 1) 2 2 (c) 2np (d) (4n + 1)p. 13. The general solution of 2 sinx + cosx = min{1, a − 4a + 6} is a∈R

(a)

nπ π + (−1)n 4 2

(b) 2nπ + (−1)n π 4

(d) nπ + (−1)n

π 4

π π − 4 4

14. The equation cos4q + sin4q + l = 0 has real solution for q, if 1 3 (a) (b) −1 < λ < − < λ <1 4 2 (c) 0 < l < 1 d) l < –1 π π 15. In a triangle ABC, if ∠B = , ∠C = and D divides 3 4

sin ∠BAD BC internally in the ratio 1:3, then sin ∠CAD equals 1 1 2 1 (a) (b) (c) (d) 3 3 3 6 16. The ratio of the sides of a triangle is 19:16:5, then A B C cot : cot : cot equals 2 2 2 (a) 1:15:4 (c) 4:1:15

(b) 15:1:4 (d) 1:4:15.

17. If the sides a, b, c of a triangle ABC are the roots of the equation x3 – 15x2 + 47x – 82 = 0 then the value cos A cos B cos C equals of + + a b c 131 169 225 (b) (c) 131 (d) (a) 164 82 164 82 18. If two sides of a triangle are 2 3 − 2 and 2 3 + 2 and their included angle is 60°, then the other angles are (a) 75°, 45° (b) 105°, 15° (c) 60°, 60° (d) 90°, 30° 19. The maximum value of (a) sin 2A (c) sin

C 2

a cos A + b cos B + c cos C is a +b+c A B C (b) 4 sin sin sin 2 2 2 (d) none of these.

20. Let a, b, c are the sides of a triangle and (sinA + sinB + sinC)(sinA + sinC – sinB) = sinA sinC where sinA = ak, sinB = bk, sinC = ck then the range of is (a) [0, 1] (b) [–4, 4] (c) (0, 4) (d) [0, 4] 21. A line passing through the point (11, –2) and touching the circle x2 + y2 = 25 are (a) 4x + 3y = 38, 7x – 24y = 125 (b) 3x + 4y = 25, 7x – 24y = 125 (c) 3x – 4y = 41, 7x + 24y = 125 (d) 7x – 24y = 125, 4x – 3y = 38 22. Two circles each of radius 5, have a common tangent at (1, 1) whose equation is 4x + 3y – 7 = 0 The centres are (a) (–4, 4), (6, 2) (b) (–3, 4), (5, –2) (c) (5, 4), (–3, –2) (d) (4, 2), (–2, 0) MATHEMATICS TODAY | SEPTEMBER‘17

23. The equation of the circles touching the line x + 2y = 0 and passing through the points of intersection of the circle x2 + y2 = 4 and x2 + y2 – 2x – 4y + 4 = 0 is (a) x2 + y2 + x + 2y = 0 (b) x2 + y2 – x + 2y = 0 (c) x2 + y2 + x – 2y = 0 (d) x2 + y2 – x – 2y = 0. 24. A circle which passes through the point (1, 1) and cuts orthogonally the two circles x2 + y2 – 8x – 2y + 16 = 0 and x2 + y2 –4x – 4y + 1 = 0 . If its centre (a, b), then a + b = 1 1 (a) 0 (b) –1 (c) (d) − 2 2 2 2 25. The x + y + 1 = 0 meets the circle x + y + 3x + y – 6 = 0 at the points A and B. If C is a point on the circle, then the locus of the orthocentre of the triangle ABC is (a) x2 + y2 – x + y – 8 = 0 (b) x2 + y2 – x + y = 0 (c) x2 + y2 + x – y – 8 = 0 (d) x2 + x – y = 0 26. The circle C1 touches the line x – y = 0 at the origin and the circle C2 touches the line x + y = 0 at the origin. Let C be the circle x2 + y2 – 4x + 2y – 6 = 0. The common chord of C and C1 pass through A and common chords of C and C2 pass through B. Then AB = (c) 2 5 (d) 3 5 (a) 5 (b) 5 27. The triangle PQR is inscribed in the circle x2 + y2 = 25 if Q and R have co-ordinates (3, 4) and (–4, 3), then ∠QPR is equal to π π π π (a) (b) (c) d) 2 3 4 6 28. In an equilateral triangle, 3coins of radii 1 are kept n so that they touch each other and also the=sides ∴D Cn−of n r Dr the triangle. Area of the triangle is (b) 6 + 4 3 (a) 4 + 2 3 7 3 7 3 (c) 12 + (d) 3 + 4 4 29. Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. If (h, k) is the centre of circle, then 1 1 1 (a) k ≥ (b) − ≤ k ≤ 2 2 2 1 1 (d) 0 < k < (c) k < 2 2 74

MATHEMATICS TODAY | SEPTEMBER ‘17

30. If (–4, 3) and (12, –1) are the ends of diameter of a circle which makes an intercept of 2l on the y-axis, then l = (a) 13 (b) 4 13 (c) 3 13 (d) 2 13 SOLUTIONS n+3 P3 − , n ∈N n n+1 Pn+1 4 ⋅ Pn 195 (n + 2)(n + 1) 195 (n + 3)(n + 2) = − = − 4 ⋅ n! (n + 1)! 4 ⋅ n! n!

1. (c): Given that xn =

195

195 − 4n2 − 20n − 24 171 − 4n2 − 20n = 4n ! 4 ⋅ n! 2 171 − 4n − 20n Q xn is positive∴ >0 4 ⋅ n! ⇒ 4n2 + 20n − 171 < 0 The above inequality is true for n = 1, 2, 3 and 4. 2. (b) : Here, we have 5 digits among which 6 is repeated twice and number greater than 50,000 are consists of 5 digits. 5 ! 120 \ Number of permutation = = = 60 2! 2 But there are some numbers among these 60 numbers which are started with 3 and those are less than 50,000, so they are to be rejected. 4! Such number of numbers = = 12 2! \ The required number of numbers = 60 – 12 = 48 3. (b) : Words before FLOWER are 1 × 5! + 1 × 4! + 1 × 3! + 2 × 2! = 154 \ Rank of the word FLOWER is 154 + 1 = 155. 4. (a) : Let Dr denotes the number of ways in which r things goes to wrong places. If r goes to wrong place out of n then (n – r) goes to correct places. \ Dn = nCn – rDr =

1 1 1 1 1 where WhereDDrr == r! r !  − + − + ... + (−1)r   2 3! 4 ! 5! r! ∴

i =1

∑ Dn = ∑ nCn−r Di (r < n)

Now for our problem 1 1 1  (Dr )r= 4 = 4 !  − +  = 12 − 4 + 1 = 9 = D4  2 3! 4 !  1 1  (Dr )r=3 = 3 !  −  = 3 − 1 = 2 = D3  2 3!  1 (Dr )r=2 = 2 !   = 1 = D2  2! 

Now no. of ways at least two goes to the wrong places, n = 4, r > 2 4

∑ nCn−r Dr = 4C2 D2 + 4C1D3 + 4C0 D4

i =2

= 6D2 + 4D3 + D4 = 6 + 8 + 9 = 23 So, 23 are the number of ways in which at least two things goes to wrong place. 5. (b) : The number of ways by which 4 questions with correct answers can be chosen in 6C4. Since only one of the four alternatives is correct, the wrong answers can be given in 3 different ways for each of the two remaining questions attempted unsuccessfully by the candidate. The desired number of ways = 6C4 · 32 = 135 6. (d) : Since order of boys and girls are to be maintained in any of the different arrangements, the 6! required number = 3!3! 9! 7. (b) : Total number of permutations = ; no. of 2! permutations where 'HIN' are always together = 7!; no. of permutations where 'DUS' are always together 7! and no. of permutations where 'TAN' are always = 2! together= 7!; Now the number of permutations where 'HIN' and 'DUS' are always together = 5! Number of permutations where 'HIN' and 'TAN' are always together = 5! Number of permutations where 'TAN' and 'DUS' are always together = 5! Number of permutations where 'HIN', 'DUS' and 'TAN' are always together = 3! Required number of permutations 9!  7!  = −  7 !+ 7 !+  + 3 × 5 !− 3 ! = 169194 2!  2!  8. (b) : The word 'MATHEMATICS' consists of 11 letters in which there are 2M's, 2A's and 2 T's The following cases are possible. Case I: Four letters having a pair of similar letters. This can be done in 3C2 ways and no. of permutation 4! of such four letters is 2!2! Case II: one similar pair and 2 different letters. This can be done in 3C1 × 7C2 ways and then no. of possible 4! permutations = 2! Case III: All different letters are taken. This can be done in 8C4 ways and then no. of permutation = 4!

Therefore the required no. is 4! 3 4! 3 C2 × + C1 × 7C2 × + 8C4 × 4 ! = 2454. 2!2! 2! 9. (d) : Number of ways of dividing 8 persons in three groups first having 3 persons, second having 2 persons 8! and third having 3 persons = 3!2 !3! Since all the 50 things are identical 8! ⋅1 \ Required number = 2 (3!) ⋅ 2 !3! 10. (a) : Number of selections of 7 digits out of digits 1, 2, 3, ..., 9 = 9C7 Out of these 7 selected digits excluding the greatest digit = 6, the 6 digits can be divided in two groups 6! 1 each having 3 digits = = 6C3 × 3 !× 3 !× 2 ! 2! But the three digits on one side can go on the other side 1 \ Required number = 9C7 ⋅ 6C3 .2 ! = 9C7 ⋅ 6C3 2! 11. (a) : The given equation is logcosqtanq + logsinqcotq = 0 ⇒ logcos θ tan θ − logsin θ tan θ = 0 log tan θ log tan θ log sin θ ⇒ = ⇒ =1 log cos θ log sin θ log cos θ π ⇒ logcos θ sin θ = 1 ⇒ cos θ = sin θ ⇒ tan θ = tan 4 π ∴ θ = nπ + , n ∈ Z . 4 x 12. (b) : The given equation is 2 − cos x = 2 tan 2 x x   ⇒ 2 1 − tan  = cos x ⇒ 2 1 − tan  =   2 2 x   1 + tan x   2 ⇒ 1 − tan   2 − =0 x  2 2  1 + tan   2 x 1 + tan x 2 =0 Either 1 − tan = 0 or 2 − 2 2 x 1 + tan 2 x x 2 x ⇒ tan = 1 or 2 tan − tan + 1 = 0 2 2 2 x π x π ⇒ tan = 1 = tan ⇒ = nπ + 2 4 2 4 π π ⇒ x = 2nπ + ⇒ x = (4n + 1) 2 2 x 2 x and 2 tan − tan + 1 = 0 2 2

x 2 2 x 1 + tan 2 1 − tan2

MATHEMATICS TODAY | SEPTEMBER‘17

x 1 ± 12 − 4 ⋅ 2 ⋅1 1 ± i 7 (no solution) ⇒ tan = = 2⋅2 4 2 13. (d) : Here, a2 – 4a + 6 = (a – 2)2 + 2 > 2 ∴ min{1, a2 − 4a + 6} = 1 a∈R

Now, sin x + cos x = 1 ⇒

sinx +

cosx =

2 2 2 π π π  nπ ⇒ sin  x +  = sin ∴ x + = nπ + (−1)  4 4 4 4 nπ π ∴ x = nπ + (−1) − 4 4 14. (b) : We have, cos4 q + sin4 q + l = 0 2

⇒ 1 − 2 sin θ cos θ = − λ 1 1 ⇒ 1 − sin2 2θ = − λ ⇒ 1 − (1 − cos 4θ) = − λ 2 4 3 1 ⇒ + cos 4θ = − λ 4 4 1 1 1 Q − 1 < cos 4θ < 1 ∴ − < cos 4θ < 4 4 4 3 1 3 1 3 1 ⇒ − < + cos 4θ < + 4 4 4 4 4 4 1 1 ⇒ < −λ < 1 ⇒ − 1 < λ < − 2 2 BD AD 15. (c): From ∆ABD we have, = sin ∠BAD sin B CD AD From ∆ACD we have, = . sin ∠CAD sin C AD sin ∠BAD AD sin ∠CAD : = 1: 3 sin B sin C sin ∠BAD sin ∠CAD : = 1: 3 π π sin sin 3 4 2 sin ∠BAD 1 sin ∠BAD 1 = ⇒ = sin ∠CAD 3 sin ∠CAD 3 6

QBD : CD = 1 : 3 ∴ ⇒

⇒

16. (d) : Given a : b : c = 19 : 16 : 5 \ 2s = a + b + c = 40 A B C \ cot : cot : cos 2 2 2 s( s − a ) s(s − b) s( s − c ) = : : (s − b)(s − c) (s − a)(s − c) (s − a)(s − b) = (s – a) : (s – b) : (s – c) a+b+c  a +b+c  a +b+c  = − a :  − b :  − c  2   2   2   40   40   40  =  − 19  :  − 16  :  − 15  = 1 : 4 : 15 2  2  2  76

MATHEMATICS TODAY | SEPTEMBER ‘17

17. (b) : We are given that, a, b, c are the roots of the equation x3 – 15x2 + 47x – 82 = 0 \ a + b + c = 15, ab + bc + ca = 47, abc = 82 cos A cos B cos C Now + + a b c =

b2 + c2 − a2

c2 + a2 − b2

a2 + b2 − c2

a2 + b2 + c2

2abc 2abc 2abc 2abc 2 2 (a + b + c) − 2(ab + bc + ca) (15) − 2(47) 131 = = = 2abc 2(82) 164 18. (b) : Let b = 2 3 + 2, c = 2 3 − 2 and A = 60° A 4  B −C  b−c ∴ tan  = cot = cot 30° = 1 = tan 45°  2  b + c 2 4 3 ⇒ B − C = 90° Also, B + C = 120° (Q A = 60°) Therefore the other two angles are B = 105°andC = 15° a cos A + b cos B + c cos C a +b+c 2 sin A cos A + 2 sin B cos B + 2 sin C cos C = 2(sin A + sin B + sin C ) a b c   Q sin A = sin B = sin C = 2R    sin 2 A + sin 2 B + sin 2C = 2(sin A + sin B + sin C ) 4 sin A sin B sin C A B C = = 4 sin sin sin A B C 2 2 2  2  4 cos cos cos    2 2 2 20. (c) : (sinA + sinB + sinC)(sinA + sinC – sinB) = sinAsinC (given) (a + b + c) (a + c – b) = ac a b c   = = = 2R  Q  sin A sin B sin C ⇒ (a + c)2 – b2 = ac ⇒ a2 + c2 – b2 + 2ac = ac 19. (b) :

a2 + c2 − b2 µ +1 = 2ac 2  a2 + c2 − b2  µ ⇒ 1 + cos B = cos B = Q   2  2ac B µ µ ⇒ cos2 = ⇒ 0 < < 1 (∵ ∠B ≠ 0 and ∠B ≠ 180°) 2 4 4 \ 0< <4 21. (b) : Let the equation of the line be y + 2 = m(x – 11) or mx – y – (11m + 2) = 0 ...(i) Since the line (i) touches the circle x2 + y2 = 25 − (11m + 2) 3 7 ∴ =5⇒m= − , 4 24 m2 + 1 Therefore (i) becomes 3x + 4y = 25 or 7x – 24y = 125 .

⇒

22. (c) : The line joining the centres passing through (1, 1) and perpendicular to the tangent 4x + 3y – 7 = 0 is 3x – 4y + 1 = 0 ...(i) Let the abscissa of the centre be a and its ordinate can be obtained from (i)  3α + 1  ∴ Centre C  α,  4  The distance between the centre C and (1, 1) is 5 2

 3α + 1  ∴ (α − 1) +  − 1 = 25  4  On solving the above equation we get a = 5, –3 therefore C = (5, 4), (–3, –2) 23. (d) : The equation of the circle passing through the intersection of the circle x2 + y2 = 4 and x2 + y2 – 2x – 4y + 4 = 0 is (x2 + y2 – 2x – 4y + 4) + l(x2 + y2 – 4) = 0 2

⇒ (1 + λ)x2 + (1 + λ) y2 − 2 x − 4 y + 4(1 − λ) = 0

4 y 4(1 − λ) 2x ...(i) − + =0 1+ λ 1+ λ 1+ λ Since the line x + 2y = 0 touches the circle (i) 1 4 2 2 +  1   2  4(1 − λ) ∴ 1+ λ 1+ λ =  + −  1 + λ   1 + λ  1+ λ 12 + 22 ⇒ x2 + y2 −

⇒ ⇒

5 2 1 + λ = 5 − (4)(1 − λ ) 5 (1 + λ)2

5 1 + 4 λ2 5 1 + 4 λ2 = ⇒ = 1+ λ (1 + λ)2 (1 + λ)2 (1 + λ)2

⇒ 1 + 4 λ2 = 5 ⇒ λ2 = 1 ⇒ λ = 1

⇒

h2 + k2 + h − k − 8 = 0

Therefore the locus is x2 + y2 + x – y – 8 = 0 26. (c) : Here C1 = x2 + y2 + l(x – y) = 0, l = R, C2 : x2 + y2 + λ(x + y ) = 0, λ ∈ R

and C : x2 + y2 – 4x + 2y – 6 = 0 The common chord of C and C1 are l(x – y) + 4x – 2y + 6 = 0 which are concurrent at x − y = 0, ⇒ 4 x − 2 x + 6 = 0 ⇒ x = −3 ∴ A = (−3, −3) The common chord of C and C2 are l(x + y) + 4x – 2y + 6 = 0 ⇒ x = − y , − 6 y + 6 = 0 ⇒ y = 1 ∴ B = (−1, 1) ∴ AB = 42 + 22 = 2 5 27. (c) : The centre of the circle x2 + y2 = 25 is (0, 0). Now slope of QO × slope of RO 4−0 3−0 4 3 = × = × − = −1 3 − 0 −4 − 0 3 4 π Therefore angle at the centre is ∠QOR = , 2 1 π \ Angle at circumference ∠QPR = × ∠QOR = 2 4 28. (b) : Length of the side π = 2 + 2 cot = 2(1 + 3 ) 6

...(ii)

...(iii) 7 17 Solving (i), (ii) and (iii) we get g = − , f = , c = −3 3 6 7 17 1  7 17  \ The centre is  , −  = (a, b) ⇒ a = , b = − , a + b = − 3 6  3 6 2

((

))

2 3 2 1+ 3 = 6 + 4 3 4 29. (a) : Circle with centre (h, k) and touching x-axis is x2 + y2 – 2hx – 2ky + h2 = 0 ....(i)

Area =

Now put the value of l = 1 in equation (i) we get x2 + y2 – x – 2y = 0 24. (d) : Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0. It cuts the two given circles orthogonally \ –8g – 2 f = c + 16 and –4g – 4f = c + 1 ...(i) ⇒ 8 g + 2 f + c = −16 and 4 g + 4 f + c = −1 The circle passes through (1, 1) \ 2g + 2f + c = –2

25. (c) : Let (h, k) be the orthocentre. The image of (h, k) on the line x + y + 1 = 0 is given by x − h y − k −2(h + k + 1) = = 1 1 1+1 The image is (–1 –k, –1 – h) It lies on the circumcircle of triangle ABC. \ (1 + k)2 + (1 + h)2 – 3(1 + k) – (1 + h) – 6 = 0

Q (−1,1) lies on (i) ∴ 2 + 2h − 2k + h2 = 0

⇒ (h2 + 2h + 1) + 1 − 2k = 0 ⇒ 2k = (h + 1)2 + 1 1 1 1 ⇒ k = + (h + 1)2 ⇒ k > 2 2 2 30. (d) : The equation of circle having extremities of the diameter (–4, 3) and (12, –1) is (x + 4)(x – 12) + (y – 3)(y + 1) = 0

⇒ x2 + y2 − 8 x − 2 y − 51 = 0 Now, length of the chord intercepted on y-axis is 2 f 2 − c = 2 (−1)2 − (−51) = 2 52 = 4 13 ∴ 2 λ = 4 13 ⇒ λ = 2 13 MATHEMATICS TODAY | SEPTEMBER‘17

 77

hallen ing PROBLEMS CALCULUS ON

1. Consider a function f : (0, ∞) → R and a real number a > 0 such that f (a) = 1 and a a f (x) ⋅ f ( y) + f   ⋅ f   = 2 f (xy) x  y for all x, y ∈(0, ∞) and f(2) = 1 then f(3) = (a) 1 (b) 6 (c) 9 (d) 16 2. Consider the polynomial P(x) with integral coefficients such that P(P ′(x)) = P ′(P(x)) and P(4) = 4 then P(2) = (a) 2 (b) 3 (c) 4 (d) 5 3. All polynomials P of degree n having only real zeroes n2 1 = x1, x2, x3, ..., xn such that ∑ for xp′(x) i =1 P(x) − xi n

all non-zero real numbers x is of the form P(x) = (a) 2xn + 3xn – 1 (b) 2xn + 3xn – 2 n (c) 2x (d) 2xn – 1 + 3xn – 2 + 4xn

4. Consider the polynomial with real coefficients P(x) = a0xn + a1xn–1 + .... + an, an 0. If the equation P(x) = 0 has all its roots real and distinct then the equation x 2P ′′(x) + 3xP ′(x) + P(x) = 0 has (a) real and distinct roots (b) real and equal roots (c) not real roots (d) not all real roots

5. Let f, g : R → R be periodic functions of periods 'a' f (x) g (x) and 'b' such that lim = l and lim =m, x →0 x x →0 x l ∈ R, m ∈ R − {0}. Then lim

n→∞

(a) 0 (c) –1

f ((3 + 7 )n a)

g ((2 + 2 )nb)

(b) 1 (d) does not exist

6. Let f, g : R → R be two periodic functions with respective periods T1 and T2 and lim ( f (x) − g (x)) = 0 then x →∞

(a) T1 = T2 (c) 2T1 = T2

(b) T1 = 2T2 (d) T1 = 1 + T2

7. Let f :[1, ∞] → R be the function defined by f (x) =

[x] + {x}

. The smallest number k such x that f(x) < k for all x > 1 is (a) 1

(b)

MATHEMATICS TODAY | SEPTEMBER ‘17

(d) 2

8. The real parameter m such that the graph of 3

the function f (x) = 8x 3 + mx 2 − nx has the horizontal asymptote y = 1 is (a) 10 (b) 12 (c) 14 (d) 16 9. f is continuous function R → R, f (0) = 1 and f(2x) – f(x) = x x ∈ R then f(2) = (a) 1 (b) 2 (c) 3 (d) 4 10. Real functions f, g, h : R → R are such that for all real x, y, (x – y)f(x) + h(x) – xy + y2 < h(y) < (x – y) g(x) + h(x) – xy + y2 then h(x) is (a) a linear function (b) a quadratic function (c) a reciprocal function (d) an exponential function 11. Let f(x) be a function which contains element 2 in its domain and range. Suppose that f(f(x)) · (1 + f(x)) = –f(x) for all numbers x in the domain of f, then f(2) = (a) 1 (b) 2/3 (c) –1 (d) –2/3

By : Tapas Kr. Yogi, Visakhapatnam Mob : 09533632105

(c)

12. Let a0

∈ (–1, 1) and defined recursively,

1 + an−1 an = , n > 0 . Let An = 4n(1 – an) then 2 lim An =

n→∞

(cos −1 a0 )2 (b) 2 (d) 3(cos–1a0)2

(a) (cos–1a0)2 (c) 2(cos–1a0)2

1 n  2n  n ⋅ ∑   − 2   = ___ k  n→∞ n k =1  k 

13. lim

([·] denotes the greatest integer function) (a) log4 (b) log4 + 1 (c) log4 – 1 (d) 2log2 – 2 14. Let f(x) be a continuous function on (a, b) and lim f (x) = +∞ , lim f (x) = −∞ and n →b −

n→ a +

f ′(x) + f 2(x) ≥ −1 for x ∈(a, b) then minimum value of (b – a) is (a) p (b) p/2 (c) 2p (d) p/4 15. Let f(x, y) be a function satisfying the functional equation f(x, y) = f(2x + 2y, 2y – 2x) for all real number x, y. Define g(x) by g(x) = f(2x, 0). Then period of g(x) is (a) 3 (b) 4 (c) 6 (d) 12 16. The number of real solutions to the equation x 3 5 = + (x − 3) is/are x + 1 2 16 (a) 1 (b) 2 (c) 3 (d) 5 17. Consider the function, log(1 + x + x 2), x ≤ b f (x) =  x >b  ax + c, If the graph of f(x) is concave in R then (a) b > (c) a ≤

−(1 + 3) 2 2b + 1

b + b +1

(b) c + ab = log(b + b2) 2b + 1 (d) a > 2 b + b +1

18. Consider an = sin(np/2) and the three sequences a  nπ  bn = nan (n > 0), cn = n (n ≥ 1) and dn = an cos    2  n (n > 1) then (a) lim bn does not exist n→∞

(b) lim cn does not exist n→∞

(d) lim (cn + dn ) does not exist n→∞

SOLUTIONS

1. (a) : Putting x = 1 = y gives f(1) = 1 a Putting y = 1 gives, f (x) = f   , x > 0 x a   = 1 x

a gives, f (x)⋅ f x so that, gives, f 2(x) = 1 Putting y =

 a Putting x = y = t gives f 2( t ) + f 2   t As L.H.S. is positive so R.H.S is positive. Hence f(x) = 1 for all x.

  = 2 f (t )

2. (a) : Consider a polynomial P(x) = a0xn + a1xn – 1 + ... + an, a0 =/ 0 where a0, a1, a2, ...., an ∈ I, then

P ′(x) = na0 x n−1 + (n − 1) a1x n−2 + ... + an−1

By comparing coefficient of xn(n – 1) in the relation

n+1 n n P(P ′(x)) = P ′(P(x)) , we have a0 ⋅ n = a0 ⋅ n 1 i.e., a0 = n−1 . n Now, since a0 is an integer ⇒ n = 1 and a0 = 1 So, P(x) = x + a1 Putting back this value of P(x) in the given relation, we have 1 + a1 = 1, i.e. a1 = 0 So, P(x) = x

P ′(x) n2 = 3. (c) : The given equation is ∑ x i =1 P(x) − xi n

Integrating, or, log n

∑ log | P(x) − xi | = n2 log c | x |, c > 0

i =1

∏ | P(x) − x | = log cn i =1

| x |n

i.e. ∏ | P(x) − xi | = k | x |n , k > 0 i =1

i.e. |P(P(x)| = k|x|n Hence, P(x) = axn

4. (a) : Define, Q(x) = xP(x). As an =/ 0, the polynomial Q has distinct real roots. So, Q'(x) has distinct real roots as well. MATHEMATICS TODAY | SEPTEMBER‘17

Define f(x) = x Q′(x). Again, we observe that H' has distinct real roots and

since H ′(x) = x 2P ′′(x) + 3xP ′(x) + P(x) has distinct real roots. n 5. (a) : Numerator = f ((3 + 7 ) a)

= f (3 + 7 )n a + (3 − 7 )n a − (3 − 7 )n a  = f [Ma − (3 − 7 )na] where (3 + 7 )n + (3 − 7 )n = integer (M ) = f (−(3 − 7 )n a) [as f is periodic] n Similarly, denominator = g ((2 + 2 ) b)

= g (−(2 − 2 ) b) Hence, given limit  f (−(3 − 7 )n) −(2 − 2 )n a (3 − 7 )n  = lim  × ×  b n→∞  (−(3 − 7 )n) g (−(2 − 2 )n) (2 − 2 )n  3− 7 2− 2

6. (a) : lim f (x0 + nT1 + T2 ) − g (x0 + nT1 + T2 ) = 0 n→∞

lim f (x0 + T2 ) − g (x0 + nT1) = 0

n→∞

i.e. f(x0 + T2) = g(x + nT1) But according to the question, lim g (x0 + nT1) − f (x0 + nT1) = 0

... (1)

n→∞

So, lim g (x0 + nT1) − f (x0 ) = 0 n→∞

i.e. f(x0 + T2) = f(x0) i.e. T2 is period of f(x) as well.

(By (1))

7. (b) : Let [x] = a, {x} = b then x = a + b a+ b

and f (x) =

a +b

Squaring: ( f (x))2 = Using A.M.

a + b + 2 ab ab = 1+ a +  b a +b  2 

G.M., we have f (x) ≤ 2

8. (b) : lim = 1 (given) x 3(8 − n3) + mx 2 3

(8x 3 + mx 2 )2 + nx ⋅ 8x 3 + mx 2 + n2 x 2

(8 – n3) must be zero and then 80

As | x | → ∞, f (x) = So, m = 12

MATHEMATICS TODAY | SEPTEMBER ‘17

1/3

 m + 2 8 +   x

m = 1 (A.T.Q) 12

9. (c) : f(x) – f(x/2) = x/2 f(x/2) – f(x/4) = x/4  x  x x f  n−1  − f  n  = n 2  2  2

x x + + ... + ∞ 2 4

Hence, f(x) = x + 1 10. (b) : Putting y = x – 1 in L.H.S. and R.H.S, we have f (x) ≤ g(x). Putting y = x + 1 in L.H.S. and R.H.S., f (x) ≥ g(x) Hence, for all x, f(x) = g(x) and so inequality becomes equality and (x – y)f(x) + h(x) – xy + y2 = h(y) For x = 0, h(y) = y2 – y f(0) + h(0) for all y ∈ R. Hence, h(x) is a quadratic function. 11. (d) : Let f(x0) = 2 then at x = x0, f(f(x0))·(1 + f(x0)) = –f(x0) i.e. f(2)(1 + 2) = –2 −2 Hence, f (2) = 3 12. (b) : Let a0 = cosq, q ∈ (0, p) 1 + cos θ θ = cos   2 2 Similarly, a2, a3, ..., an = cos(q/2n)   θ  Hence, An = 4n 1 − cos  n    2   So, a1 =

 sin(θ / 2n )  θ2 =    θ   θ / 2n  1 + cos  n  2  So, as n → ∞ , An =

x →+∞ x →−∞

f (x) =

 m  8 +  x

2 /3

Adding and n → ∞, f (x) − f (0) =

= zero as

f (x) =

θ2 θ2 1 (1) = = (cos −1 a0 )2 2 2 2

Solution Sender of Maths Musing SET-176 1. Devjit Acharjee (West Bengal)

MATHEMATICS TODAY | SEPTEMBERâ&#x20AC;&#x2DC;17

13. (c) : The given limit =

2

1

∫  x  − 2  x  ⋅ dx

2   2 ,  0 if x ∈    2 1  2n + 1 2n  f (x) =   − 2   =  x x  2   2 , 1 if x ∈     2n + 2 2n + 1  So, required integral 2 2 2 2 2 2 =  −  +  −  +  −  + .... 3 4 5 6 7 8 1  1 1 1  = 2  − + − .... = 2  log 2 − 1 +   3 4 5  2 = log4 – 1 14. (a) : Given inequation becomes d f ′(x) (x + tan −1 f (x)) = +1 ≥ 0 dx 1 + f 2(x) So, x + tan–1f(x) is non-dec. π π Hence, + a ≤ − + b, so b − a ≥ π 2 2 Notice that a = 0, b = x, f (x) = cotx gives the minimum condition. 15. (d) : g(x) = f(2x, 0) = f(2x + 1, –2x + 1) = f(0, –2x + 3) = f(–2x + 4, –2x + 4) = f(–2x + 6, 0) = f(–2x + 7, 2x + 7) = f(0, 2x + 9) = f(2x + 10, 2x + 10) = f(2x + 12, 0) = g(x + 12) Hence, g(x) is periodic with period = 12

MATHEMATICS TODAY | SEPTEMBER ‘17

16. (a) : Let f (x) =

. Domain (−1, − ∞) x +1 Notice, f ′(x) > 0 and f ′′(x) < 0 . So, f is strictly increasing and concave in its domain. Now, f(3) = 3/2 and f '(3) = 5/16. So, the tangent line 3 5 to graph of f at (3, 3/2) is y = + (x − 3). Since, f is 2 16 3 5 strictly concave, so f (x) < + (x − 3) 2 16 Hence, the only solution is x = 3. 17. (c) : Let g(x) = log(1 + x + x2). Domain is x ∈ R. g ′′(x) < 0 ⇒ x >

−(1 + 3) 3 −1 or < 2 2

(1 + 3) 2 and lim f (x) = f (b) for continuity of f gives So, f is concave, when b < x →b

log(1 + b + b2) = ab + c and if f is concave then 2b + 1 f ′(b − ) ≥ f ′(b + ) ⇒ ≥a 1 + b + b2 18. (a) : The subsequence of bn corresponding to odd integers is b2n + 1 = (–1)n (2n + 1), which does not have a fixed limit. Hence bn also does not have a limit. The sequence cn is product of bounded sequence an and (1/n). So limit tends to zero.  nπ  1 And dn = an cos   = sin(nπ) = 0, n ∈ N . 2 2 

10 BEST

PROBLE

Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE Main & Advanced Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE Main & Advanced. In every issue of MT, challenging problems are offered with detailed solution. The readers’ & comments and suggestions regarding the problems and solutions offered are always welcome. 1.

The eccentricity of the ellipse which meets the x y straight line + = 1 on the axis of x and the straight 7 2 x y line − = 1 on the axis of y and whose axes lie along 3 5 the axes of coordinates is 3 2 4 2 6 (c) (d) 7 7 7 x −3 x −1 y +1 z −1 2. If the lines and = = 1 2 3 4 y−k z = = intersect, then k is equal to 2 1 (a) 9/2 (c) 0 (d) –1 (d) 2/9 (a)

2 6 7

(b)

3. The vectors a and b are not perpendicular and c and d are the vectors satisfying b × c = b × d and a ⋅ d = 0. Then the vector d is equal to:  a ⋅c  (a) c −  b  a ⋅ b 

 b ⋅c  c (b) b −   a ⋅ b 

 a ⋅c  (c) c +  b  a ⋅ b 

 b ⋅c  c (d) b +   a ⋅ b 

4. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (a) 55 (b) 66 (c) 77 (d) 88 5. The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard B y : Prof. Shyam Bhushan, D i r e ct o r , N a r a ya

deviations were 3 and 4 respectively. The variance of combined sample of size 500 is (a) 64 (b) 65.2 (c) 67.2 (d) 64.2 6. The mean and variance of random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is 1 1 1 1 (b) (c) (d) (a) 32 8 16 4 7. The only statement among the following that is a tautology is: (a) A ∧ ( A ∨ B) (b) A ∨ ( A ∧ B) (c) [ A ∧ ( A → B)] → B (d) B → [ A ∧ ( A → B)] 8. In a DPQR, if 3sinP + 4cosQ = 6 and 4sinQ + 3cosP = 1, Then the angle R is equal to (a) p/4 (b) 3p/4 (c) 5p/6 (d) p/6 9. The equation 2cos–1x = sin–1  2 x 1 − x2  is valid   for all values of x satisfying (a) –1 < x < 1 (b) 0 < x < 1 (c) 0 < x < 1 / 2 10. (a) (b) (c) (d) n a I I T A ca

(d) 1 / 2 < x < 1

Let f(x) = (x – 4)(x – 5)(x – 6) then f ′(x) = 0 has four real roots three roots of f ′(x) = 0 lie in (4, 5) ∪ (5, 6) ∪(6, 7) the equation f ′(x) has only two roots three roots of f ′(x) lie in (3, 4) ∪ (4, 5) ∪ (5, 6) d e m y , Ja

m sh

e d p u r . Mo b . : 0 9 3 3 4 8 7 0 0 2 1

MATHEMATICS TODAY | SEPTEMBER‘17

SOLUTIONS

6. (d) : We are given, np = 4, npq = 2

x2 y2 ...(i) 1. (a) : 2 + 2 = 1 a b x y x y + = 1 meets x-axis at A(7,0), line − = 1 meets 7 2 3 5 y-axis at B(0,–5) Eq. (i) passes through A and B 49 25 ⇒ + 0 = 1, 0 + = 1 2 a b2 2

⇒ a = 49, b = 25; b = a (1 − e ) ⇒ 25 = 49(1 − e ) 2 6 24 ⇒e= 49 7 2. (a) : Any point on the first line is (2r1 + 1, 3r1 – 1, 4r1 + 1) And on the second line is (r2 + 3, 2r2 + k, r2) The lines will intersect when 2r1 + 1 = r2 + 3, 3r1 – 1 = 2r2 + k, 4r1 + 1 = r2 ⇒ e2 =

⇒ q=

2 1 1 = ∴ n  = 4 ⇒ n = 8 2 4 2

1 1 ∴ P ( X = 1) = nC1 pqn−1 = 8 pq7 = 8   =  2  32 7. (c) : Note that A ∧ ( A ∨ B) is F when A = F A ∨ ( A ∧ B) is F when A = F, B = F, and B → [ A ∧ ( A → B)] is F when A = F, B = T \ We check only (c) [ A ∧ ( A → B)] → B ≡ [ A ∧ (~ A ∨ B)] → B ≡ [( A ∧ ( A))] ∨ ( A ∧ B) → B ≡ A ∧ B → B ≡~ ( A ∧ B) ∨ B ≡~[( A ∧ B) ∧ ( B)] ≡~[ A ∧ (B∧ B)] ≡~[ A ∧ F ] ≡~ F ≡ T Thus [ A ∧ ( A → B)] → B is a tautology.

And k = 3r1 – 2r2 – 1 = –9/2 + 10 – 1 = 9/2

8. (d) : Squaring and adding the given relations we get 16 + 9 + 24 sin(P + Q) = 37 ⇒ sin(P + Q) = 1 / 2 ⇒ sin R = 1 / 2

3. (a) : b × c = b × d ⇒ b × (c − d ) = 0

⇒ R = π / 6 or 5π / 6

⇒ c − d || b ⇒ c − d = αb for some α ∈ R ⇒ d = c − αb Also, a ⋅ d = a ⋅ c − αa ⋅ b ⇒ 0 = a ⋅ c − αa ⋅ b

If R = 5p/6, then P < p/6

a ⋅c a ⋅c ⇒ α= Thus, d = c − b a ⋅b a ⋅b

So, R ≠ 5π / 6

⇒ 2r1 − r2 = 2, 4r1 − r2 = −1 ⇒ r1 = −3 / 2, r2 = −5

4. (c) : There are only two possibilities for sum of the digits equal to 10. Case (i): 1,1,1,1,1,2,3 7! Number of seven digit integers = = 42 5! Case (ii): 1,1,1,1,2,2,2

7! = 35 4 !3! \ Total number of integers = 42 + 35 = 77 n x +n x 5. (c) : Combined mean x = 1 1 2 2 n1 + n2 200 × 25 + 300 × 10 = = 16 500 Number of seven digit integers =

Now we know that σ2 = =

(

+ d12

) (

+ n2 σ22

n1 + n2

200(9 + 81) + 300(16 + 36) 33600 = = 67.2 500 500

MATHEMATICS TODAY | SEPTEMBER ‘17

9. (d) : If we denote cos–1x by y, then since 0 < cos−1 x < π ⇒ 0 < 2 y < 2π,

...(i)

Also since − π / 2 < sin−1  2 x 1 − x2  < π / 2   ⇒ − π / 2 < sin−1 sin(2 y ) < π / 2, −

...(ii)

π π < 2y < 2 2

From (i) and (ii) we find 0 < 2 y < π / 2 ⇒ 0 < y < π / 4 ⇒ 0 < cos−1 x < π / 4

Let d1 = x1 − x = 25 − 16 = 9, d2 = x2 − x = 10 − 16 = −6 n1 σ12

⇒ 3 sin P < 3 / 2 ⇒ 3 sin P + 4 cos Q < 3 / 2 + 4 < 6

+ d22

)

Which holds if 1 / 2 < x < 1 10. (b) : Since f(4) = f(5) = f(6) = f(7) = 0, so by Rolle’s theorem applied to the intervals [4, 5], [5, 6], [6, 7] there exist x1 ∈(4, 5), x2 ∈(5, 6), x3 (6, 7) such that f ′(x1) = f ′(x2) = f1(x3) = 0. Since f ′ is a polynomial of degree 3 so cannot have four roots. 

SOLUTION SET-176

a+b = ab + 2 ⇒ b = a + 2 2 (a, b) = (1, 9), (4, 16), (9, 25), (16, 36), (25, 49), (36, 64), (49, 81), (64, 100), (81, 121) = 9 pairs. 1. (b) :

2. (d) : The nth term, tn n2(n + 1)2 13 + 23 + .... + n3 (n + 1)2 4 = = = 1 + 3 + ..... + (2n − 1) 4 n2 9 (n + 1)2 10 9   1  ∑ tn = ∑ 4 = 4  ∑ n2 − 1 n=1  n =1 n =1

{

}

1 1  ⇒ x( x + 1)  x −   x −  < 0    3 2 1 1  ∴ x ∈ (−1, 0) ∪  ,  3 2 7. (a) : z1 = cosq + isinq z2 = cos2q + isin2q – (cosq + isinq) = (cos2q – cosq) + i(sin2q – sinq) |z2|2 = (cos2q – cosq)2 + (sin2q – sinq)2 2 θ = 2 – 2(cos2qcosq + sin2q sinq) = 2 – 2cosq = 4 sin 2 θ \ |z2| = 2 sin 2 8. (c) : z2 = (cos2q – cosq) + i(sin2q – sinq) 3θ θ 3θ θ = −2 sin sin + i 2 cos sin 2 2 2 2 3θ 3θ  θ = 2i sin  cos + i sin  2 2 2 θ 3θ 3θ  π 3θ   arg z2 = arg  2 i sin  + arg  cos + i sin  = +   2 2 2 2 2

1 1 1 10 ⋅11 ⋅ 21 = − 1 = {385 − 1} = × 384 = 96 4 4 6 4 3. (a) : The area of the pentagon = Area of BCDE + Area of DEAB = 4ab + a2 – ab = a(3b + a) = 451 = 11 × 41 ⇒ a = 11, b = 10 \ a–b=1

Since, 4np < q < (4n + 2)p θ θ \ 2nπ < < (2n + 1) π ⇒ sin > 0 2 2 2013 2011 =42 9. (5) : m = 2

4. (c) : y2 = 4ax ⇒ yy′ = 2a Eliminating a, 2x y′ = y 1 we get, 2x + yy′ = 0 Replacing y′ by − y′

2m – 1 = (24)2 – 1 = 16 –1 Last digit is 6 – 1 = 5 f ( x ) − f (0) 1 10. (b) : P. f ′(0) = lim = lim x α −1 sin   = 0 x x x →0 x →0

2.2011

2011

f ( x ) − f (0) 1 f ′(0) = lim = lim x α −1 sin   = 0 if a > 1 x x1→ 0 1 x x →0 These are ellipses with eccentricity = 1 − = 1 1 2 x ≠ 0 ⇒ f ′(x ) = αx α −1 sin   − x α − 2 cos   2 x x 5. (c) : lim f ′( x ) = 0 if α > 2 \ a = 3 Solving, x2 + 2 = c, c ∈ R

x →0

Q. y(1) = 2 ⇒ 2 = a + b + c, y(0) = 0 ⇒ c = 0 Also, y′(0) = 1 ⇒ b = 1 \ a = 1 Hence, y = x2 + x ⇒ y(–1) = 0 BM = 2, C1M = 6 − 4 = 2 = MC2 AM = 4 cos 30° = 2 3 ∴ b = 2 3 + 2 1 Area = bc sin A = b = 2 3 + 2 2 6. (b, c, d) :  1  2 ⇒

2( x −1)/(2 x 2 + x −1)

1/ x

1 <  2 1 3 x − 2( x − 1) 1 >0 ⇒ > x ( 2 x − 1) ( x + 1) 2x2 + x − 1 x

⇒ (1 – 3x)x(2x – 1)(x + 1) > 0

R. f(x) = sinx(1 + a), where a = ∴

f (x ) = 2 sin x ,

π /2

∫

π /2

∫

cos tdt = 1

f (x )dx = 2.

x2 y2 2 − = 1 and − x 2 = 1 have y S. The hyperbolas 2 2 common tangents with slopes ±1. The four tangents are x + y ± 1 = 0, x – y ± 1 = 0. They form a square of area 2 × 2 = 2  MATHEMATICS TODAY | SEPTEMBER‘17

MATHEMATICS TODAY | SEPTEMBER â&#x20AC;&#x2DC;17

Registered R.N.I. No. 40700/83 Published on 1st of every month Postal Regd. No. DL-SW-1/4045/15-17 Lic. No. U(SW)-29/2015-17 to post without prepayment of postage N.D.P.S.O.N.D-1 on 2-3rd same month