Final JEE - Main Exam September, 2020/02-09-2020/Evening Session
FINAL JEE–MAIN EXAMINATION – SEPTEMBER, 2020 (Held On Wednesday 02nd SEPTEMBER, 2020)
TEST PAPER WITH SOLUTION
MATHEMATICS 1.
TIME : 3 PM to 6 PM
The area (in sq. units) of an equilateral triangle
1
inscribed in the parabola y2 = 8x, with one of
n
its vertices on the vertex of this parabola, is : (1) 64 3
(2) 256 3
(3) 192 3
(4) 128 3
2
Sol.
3
Official Ans. by NTA (3)
4
(2t2,4t) A y2=8x
Number of blue lines = Number of sides = n
Sol.
O (0,0)
30°
EN
Number of red lines = number of diagonals
nC
B 4t 2 Þ t =2 3 2 = 2t t
3.
AB = 8t = 16 3
Area = 256.3·
Let n > 2 be an integer. Suppose that there are
A
2.
3 = 192 3 4
n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are
(1) 199
(2) 101
(3) 201
(4) 200
n(n - 1) - n = 99 n 2
If the equation cos4q+sin4q + l = 0 has real solutions for q, then l lies in the interval : é 3 5ù (1) ê - , - ú ë 2 4û
æ 1 1ù (2) ç - , - ú è 2 4û
æ 5 ö (3) ç - , -1 ÷ 4 è ø
1ù é (4) ê -1, - ú 2û ë
Official Ans. by NTA (4) Sol. l = – (sin4q + cos4q) l = – (sin2q + cos2q)2 – 2sin2qcos2q
connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is :-
– n = 99 n Þ
n -1 - 1 = 99 Þ n = 201 2
LL
tan 30° =
2
= nC2 – n
l=
sin 2 2q -1 2
sin 2 2q é 1 ù Î ê0, ú 2 ë 2û
Official Ans. by NTA (3)
1ù é l Î ê-1, - ú 2û ë 1
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 4.
Let f(x) be a quadratic polynomial such that
6.
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
Let a, b, cÎR be all non-zero and satisfy a3 + b3 + c3 = 2. If the matrix
is 3, then its other root lies in : (1) (–3, –1)
(2) (1, 3)
(3) (–1, 0)
(4) (0, 1)
æa b cö ç ÷ A =çb c a÷ ç c a b÷ è ø
Official Ans. by NTA (3) satisfies ATA = I, then a value of abc can be :
Sol. f(x) = a(x – 3) (x – a) f(2) = a(a – 2)
(1)
2 3
(2) -
1 3
f(–1) = 4a(1 + a) f(–1) + f(2) = 0 Þ a(a – 2 + 4 + 4a) = 0
(3) 3
(4)
a ¹ 0 Þ 5a = – 2
Official Ans. by NTA (4)
1 3
a= -
EN
Sol. ATA = I
2 = – 0.4 5
Þ a2 + b2 + c2 = 1
and ab + bc + ca = 0
a Î (–1, 0)
Let f : R ® R be a function which satisfies f(x + y) = f(x) + f(y)"x,yÎR. If f(1) = 2 and g(n) =
(n -1)
å f(k), n Î N k=1
Þ a+b+c=±1
LL
5.
Now, (a + b + c)2 = 1
then the value of n, for
which g(n) = 20, is :
(2) 9
A
(1) 5 (3) 20
(4) 4
So, a3 + b3 + c3 – 3abc
= (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = ± 1 (1 – 0) = ± 1 Þ 3 abc = 2 ± 1 = 3, 1 Þ abc = 1,
Official Ans. by NTA (1) Sol. f(x + y) = f(x) + f(y) Þ f(n) = nf(1)
7.
Let f : (–1, ¥) ® R be defined by f(0) = 1 and f(x) =
f(n) = 2n
1 3
1 log e (1 + x), x ¹ 0 . Then the function f: x
(1) decreases in (–1, ¥) n -1
g(n) =
k=1
2
æ (n - 1)n ö = n(n – 1) 2 ÷ø
å 2n = 2 çè
(2) decreases in (–1, 0) and increases in (0, ¥) (3) increases in (–1, ¥)
g(n) = 20 Þ n(n – 1) = 20
(4) increases in (–1, 0) and decreases in (0, ¥)
n=5
Official Ans. by NTA (1)
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session x - l n(1 + x) Sol. f '(x) = 1 + x 2 x x - (1 + x) l n(1 + x) = x 2 (1 + x)
9.
(3 + 2
Suppose h(x) = x – (1 + x) ln(1 + x) Þ h'(x) = 1 – ln(1 + x) – 1 = – ln(1 + x) Sol.
h'(x) < 0, " x Î (0, ¥) h(0) = 0 Þ h'(x) < 0 " x Î (–1, ¥)
)
2
) (
54 = 3 - 6 i
10.
A
Þ a1 + a11 = 0
where d is common difference Þ a1 = -5d
-54
)
1/ 2
)
2
(
+ 3 - 2 -54
) (
)
1/ 2
)
= 6, –6, 2 6i, – 2 6i,
11 =0 2
Þ a1 + a1 + 10d = 0
can be :
)
(
Official Ans. by NTA (2) Sol. a1 + a2 + a3+.....+a11 = 0 Þ (a1 + a11) ×
1/2
-54 = 3 + 2 ´ 3 ´ 6 i
EN
121 (4) 10
(3 + 2
= ± 3+ 6 i ± 3- 6 i
LL
72 (3) 5
)
(4) - 6
6
(3 + 2
is 0 (a 1 ¹ 0), then the sum of the A.P., a1, a3, a5,...,a23 is ka1, where k is equal to : 72 5
(
– 3 - 2 -54
(3)
(3 - 2
If the sum of first 11 terms of an A.P., a1 a2, a3,...
(2) -
1/2
(2) 6
(
Þ f(x) is a decreasing function for all xÎ(–1, ¥)
121 10
)
(1) -2 6
= 3+ 6 i
Þ f'(x) < 0 " xÎ(–1, ¥)
(1)
-54
Official Ans. by NTA (1)
h'(x) > 0, " x Î (–1, 0)
8.
The imaginary part of
1/x
æ æp öö lim ç tan ç + x ÷ ÷ x®0 øø è è4
is equal to :
(1) 2
(2) e
(3) 1
(4) e2
Official Ans. by NTA (4) 1/x
Sol.
ì æp öü lim í tan ç + x ÷ ý x®0 øþ î è4
=
e
1ì æp ö ü lim ítan ç + x ÷ -1ý x®0 x î è4 ø þ
a1 + a3 + a5 +......+a23 12 = (a1 + a 23 ) ´ = (a1 + a1 + 22d) ´ 6 2
æ 1+ tan x -1+ tan x ö ÷ x(1 - tan x) ø
ç ®0 è = e xlim
2 tan x
æ æ -a1 ö ö = ç 2a1 + 22 ç ÷÷ ´ 6 è 5 øø è
= -
= e xlim ®0 x(1- tan x) = e2
72 -72 a1 Þ K = 5 5 3
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 11.
The equation of the normal to the curve y = (1+x)2y + cos2(sin–1x) at x = 0 is :
12.
p For some qÎæç 0, ö÷ , if the eccentricity of the è 2ø
(1) y = 4x + 2
(2) x + 4y = 8
hyperbola, x2–y 2sec2q = 10 is
(3) y + 4x = 2
(4) 2y + x = 4
eccentricity of the ellipse, x2sec2q + y2 = 5, then the length of the latus rectum of the ellipse, is:
Official Ans. by NTA (2) Sol. Given equation of curve y = (1 + x)2y + cos2(sin–1x)
5 times the
30
(2)
4 5 3
(3) 2 6
(4)
2 5 3
(1)
at x = 0 y = (1 + 0)2y + cos2(sin–10) y=1+1 Official Ans. by NTA (2) y=2
So we have to find the normal at (0, 2) 2 y ln(1+ x) + cos2 cos -1 1 - x 2 Now y = e
y = e 2y ln(1+ x ) +
(
1 - x2
y = e2yln(1+x) + (1–x2)
)
)
equation of hyperbola Þ x2 – y2sec2q = 10
EN
(
æ pö Sol. Given qÎç 0, ÷ è 2ø
Þ
2
…(1)
LL
Now differentiate w.r.t. x
x2 y2 =1 10 10 cos2 q
Hence eccentricity of hyperbola
(e H ) = 1 +
é ù æ 1 ö y ' = e 2y ln(1+x) ê 2y × ç + ln(1 + x).2y' ú - 2x ÷ è1+ x ø ë û
10 cos2 q 10
…(1)
ìï b 2 üï íe = 1 + 2 ý a ïþ îï
Put x = 0 & y = 2
A
é ù æ 1 ö y ' = e 2´2 l n1 ê 2 ´ 2 ç + ln(1 + 0).2y 'ú - 2 ´ 0 ÷ è1+ 0 ø ë û
y' = e0[4 + 0] – 0
Now equation of ellipse Þ x2sec2q + y2 = 5
y' = 4 = slope of tangent to the curve 1 so slope of normal to the curve = - {m1m2=–1} 4
Hence eccenticity of ellipse
Hence equation of normal at (0, 2) is 1 y - 2 = - (x - 0) 4
Þ 4y – 8 = –x Þ x + 4y = 8
4
a 2 ïü ïì e = 1 í ý b 2 ïþ ïî
x2 y2 + =1 Þ 5cos2 q 5
(e E ) = 1 -
5cos2 q 5
(e E ) = 1 - cos 2 q = | sin q |= sin q
…(2)
ì æ p öü íQ qÎ ç 0, ÷ ý è 2 øþ î
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session given Þ eH =
Sol. Hence normal is ^r to both the lines so normal vector to the plane is r ˆ ´ (2iˆ + 3ˆj - k) ˆ n = (iˆ - 2 ˆj + 2k)
5 ee
Hence 1 + cos2q = 5sin2q 1 + cos2q = 5(1 – cos 2q) 1 + cos2q = 5 – 5cos 2q
ˆi ˆj kˆ r ˆ + 4) n = 1 -2 2 = ˆi(2 - 6) - ˆj( -1 - 4) + k(3 2 3 -1
6cos2q = 4 cos2q =
2 3
…(3)
r n = -4iˆ + 5ˆj + 7kˆ
Now length of latus rectum of ellipse
Now equation of plane passing through 2a 10 cos q 20 4 5 = = = b 3 5 3 5 2
=
(3,1,1) is Þ –4(x–3) + 5(y – 1) + 7(z – 1) = 0 Þ –4x + 12 + 5y – 5 + 7z – 7 = 0
Which of the following is a tautology ?
EN
13.
2
(1) (~p) Ù (pÚq)®q
(2) (q®p)Ú~(p®q)
Þ –4x + 5y + 7z = 0
(3) (p®q)Ù(q®p)
(4) (~q)Ú(pÙq)®q
Plane is also passing through (a, –3, 5) so this
Official Ans. by NTA (1) Sol. Option (1) is ~pÙ(pÚq) ® q
point satisfies the equation of plane so put in equation (1)
–4a + 5 × (–3) + 7 × (5) = 0
LL
º (~pÙp)Ú(~pÙq) ® q º CÚ(~pÙq) ® q º (~pÙq)®q º ~(~pÙq)Úq º (pÚ~q)Úq
15.
A
º (pÚq)Ú(~qÚq) º (pÚq) Ú t
14.
Þ –4a –15 + 35 = 0
Þ a= 5 Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1ÇE2ÇE3) = 0. Then P( E C2 Ç E 3C /E1) is equal to :
so ~pÙ(pÚq) ® q is a tautology
(1) P( E 3C ) – P(E2)
A plane passing through the point (3, 1,1) contains two lines whose direction ratios are 1,
(3) P( E 3C ) – P( E C2 )
–2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (a, –3, 5), then a is equal to: (1) –10
(2) 5
(3) 10
(4) –5
Official Ans. by NTA (2)
…(1)
(2) P( E C2 ) + P(E3) (4) P(E3) – P( E C2 )
Official Ans. by NTA (1) Sol. Given E1, E2, E3 are pairwise indepedent events so P(E1ÇE2) = P(E1).P(E2) and P(E2ÇE3) = P(E2).P(E3) and P(E3ÇE1) = P(E3).P(E1) & P(E1ÇE2ÇE3) = 0
5
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session By solving these equation
æ E 2 Ç E3 ö P ëé E1 Ç (E 2 Ç E 3 ) ûù ÷= Now P ç P(E1 ) è E1 ø
=
we get x =
P(E1 ) - [ P(E1 Ç E2 ) + P(E1 Ç E3 ) - P(E1 Ç E2 Ç E3 )]
-11l 7l ; y = l; z = 2 2
Also given, x2 + y2 + z2 = 1
P(E1) 2
=
P(E1 ) - P(E1 ).P(E 2 ) - P(E1 )P(E 3 ) - 0 P(E1 )
Þ l=±
= 1 – P(E2) – P(E3) = [1 – P(E3)] – P(E2)
1 121 49 +1+ 4 4
= P(E C3 ) - P(E 2 )
so, there are 2 values of l.
Let A = {X = (x, y, z)T : PX = 0 and
\ so, there are 2 solution set of (x,y,z).
EN
16.
2
æ -11l ö æ 7l ö + (l )2 + ç ÷ = 1 Þ ç ÷ è 2 ø è 2 ø
17.
é1 2 1ù x 2 + y 2 + z 2 = 1} where P = ê -2 3 -4 ú , ê ú ëê 1 9 -1úû
then the set A :
LL
(1) is a singleton
Consider a region R = {(x, y) ÎR2 : x2£y £2x}. If a line y = a divides the area of region R into two equal parts, then which of the following is true? (1) a3–6a2 + 16 = 0
(2) 3a2 – 8a + 8 = 0
(3) a3 – 6a3/2–16 = 0
(4) 3a2 –8a3/2+8 = 0
Official Ans. by NTA (4)
(2) contains exactly two elements
(3) contains more than two elements
y=x
A
6
C A
y=a
O(0,0)
* y ³ x2 Þ upper region of y = x2 y £ 2x Þ lower region of y = 2x According to ques, area of OABC = 2 area of OAC
x + 2y + z = 0
infinite many solutions,
B
a
é1 2 1ù Sol. Given P = ê -2 3 -4 ú , Here |P| = 0 & also ê ú êë 1 9 1 úû
ü ï Þ -2x + 3y - 4z = 0 ý Q D = 0, so system have ï x + 9y - z = 0 þ
(2,4)
4
Official Ans. by NTA (2)
é 1 2 1 ù éxù Þ ê -2 3 -4 ú ê y ú = 0 ê úê ú êë 1 9 1 úû êë z úû
y=2x
Sol.
(4) is an empty set
given PX = 0
2
4
Þ
a
yö yö æ æ ò0 çè y - 2 ÷ø dy = 2 ò0 çè y - 2 ÷ø .dy
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session Þ
4 1 é2 ù = 2 ê a 3/2 - .a 2 ú 3 4 û ë3
Þ -
Þ 3a2 - 8a3/ 2 + 8 = 0 18.
Þ
If a curve y = f(x), passing through the point
ì Put x = 1& y = 2 ü 2x = lnx + C í ý y î then we get C = -1þ
-2x = l n(x) - 1 y
(1,2), is the solution of the differential equation, Þ y=
æ1ö 2x2dy= (2xy + y2)dx, then f ç ÷ is equal to : è2ø 1 (1) 1 - log 2 e
EN 19.
{x4+(k + 6)a}+..... where a ¹ 0 and x ¹ 1. If
Þ t+x
dt t2 = t+ dx 2
dt t = dx 2
Þ 2ò
dt = t2
ò
x10 - x + 45a(x - 1) , then k is equal to : x -1
(1) –5
(2) 1
(3) –3
(4) 3
Official Ans. by NTA (3) Sol. S = [x + ka + 0] + [x2 + ka + 2a] + [x3 + ka + 4a] + [x4 + ka + 6a] +......9 terms Þ S = (x + x2 + x3 + x4+.....9 terms) + (ka + ka + ka + ka +.......9 terms) + (0 + 2a + 4a + 6a + .......9 terms)
é x9 - 1 ù = S x ê ú + 9ka + 72a Þ ë x -1 û
2
Þ x
S=
LL
dt 2x 2 t + x 2 t 2 = dx 2x 2
A
Þ t+x
Let S be the sum of the first 9 terms of the series: {x + ka} + {x2 + (k + 2)a} + {x3+(k+4)a}+
dy 2xy + y 2 = {Homogeneous D.E.} dx 2x 2
ì let y = xt ü ï ï í dy dt ý ïî Þ dx = t + x dx ïþ
2x 1 - log e x
1 æ1ö so, f ç ÷ = è 2 ø 1 + loge 2
(4) 1+ loge2
Official Ans. by NTA (2) Sol. 2x2dy = (2xy + y2) dx Þ
Þ f(x) =
1 (2) 1 + log 2 e
-1 (3) 1 + log 2 e
2x 1- lnx
dx x
yü ì æ 1ö Þ 2 ç - ÷ = l n(x) + C í Put t = ý xþ è tø î
(x10 - x) + (9k + 72)a(x - 1) ÞS= (x - 1) Compare with given sum, then we get, (9k + 72) = 45 Þ k = -3
7
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session 20.
The set of all possible values of q in the interval (0, p) for which the points (1, 2) and (sin q,
Sol. Let a be the first term and d be the common difference of the given A.P. Where d > 0
cosq) lie on the same side of the line x + y =
X= a+
1 is : æ pö (1) ç 0, ÷ è 4ø
æ 3p ö (2) ç 0, ÷ è 4 ø
æ p 3p ö (3) ç , ÷ è4 4 ø
æ pö (4) ç 0, ÷ è 2ø
= a + 5d S(X - x i )2 Þ varience = 11
Þ 90 × 11 = (25d 2 + 16d2 + 9d2 + 4d2) × 2
Official Ans. by NTA (4) Sol. Given that both points (1, 2) & (sinq, cosq) lie on same side of the line x + y – 1 = 0 n q, B(si
) co sq
Þ d = ±3 Þ d = 3 22.
x+y–1=0
2)
6 4 ü 1 ì3 If y = å k cos - í cos kx - sin kx ý , 5 î5 þ k =1
then
dy at x = 0 is________. dx
EN
A (1,
0 + d + 2d + ... + 10d 11
Official Ans. by NTA (91)
3 4 p 0<a< Sol. Put cos a = ,sin a = 5 5 2
æ Put (1,2) in ö æ Put (sin q, cos q in ö So, ç ÷ ç ÷>0 è given line ø è given line ø
Þ (1 + 2 – 1) (sin q + cos q – 1) > 0 Þ sin q + cos q > 1 ¸ by 2
1 1 1 sin q + cos q > 2 2 2
A
Þ
pö 1 æ Þ sin ç q + ÷ > 4ø 2 è
Þ
3p 4
p 4
p 2
If the variance of the terms in an increasing A.P., b 1 , b 2 , b 3 ,....b 11 is 90, then the common difference of this A.P. is_______. Official Ans. by NTA (3.00)
3 4 cos kx - sin kx 5 5
= cos a . cos kx – sin a . sin kx = cos(a + kx) As we have to find derivate at x = 0 We have cos–1 (cos(a + kx)) = (a + kx) Þ y=
p p 3p < q+ < 4 4 4
Þ 0<q< 21.
}
LL
{
Now
6
å (a + kx) k =1
dy = Þ dx at x = 0
23.
6
åk = k= x
6 ´ 7 ´ 13 = 91 6
Let the position vectors of points 'A' and 'B' be ˆi + ˆj + kˆ and 2iˆ + ˆj + 3kˆ , respectively. A point
'P' divides the line segment AB internally in the ratio l : 1 (l > 0). If O is the origin and uuur uuur uuur uuur OB × OP - 3 | OA ´ OP |2 = 6 , then l is equal to______. Official Ans. by NTA (0.8)
8
Final JEE - Main Exam September, 2020/02-09-2020/Evening Session l A(i + j + k)
n
B(2i + j + 3k)
Using section formula we get
OP =
2l + 1 ˆ l + 1 ˆ 3l + 1 ˆ i+ j+ k l +1 l +1 l +1
Now OB × OP =
=
4 l + 2 + l + 1 + 9l + 3 l +1
24.
æ 1ö For a positive integer n, ç1 + ÷ is expanded è xø
in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2 : 5 : 12, then n is equal to________. Official Ans. by NTA (118) Sol. nCr–1 : nCr : nCr+1 = 2:5:12 n
Now
14l + 6 l +1
Cr -1 2 = Cr 5
n
Þ 7r = 2n + 2
…(1)
n n
Þ 17r = 5n – 12
Þ n = 118
| OA ´ OP |2 =
(2l + 1)2 + l 2 + l 2 (l + 1) 2
6l 2 + 1 = (l + 1)2
Let [t] denote the greatest integer less than or equal to t. Then the value of
Þ l = 0,
8 = 0.8 10
2
1
| 2x - [3x] | dx
5 < 3x < 6 ; 2
Now
ò | 2x - [3x] | dx 1
14l + 6 (6l 2 + 1) 3 ´ =6 Þ l +1 (l + 1) 2 Þ 10l2 – 8l = 0
ò
is_______. Official Ans. by NTA (1.0) Sol. 3 < 3x < 6 Take cases when 3 < 3x < 4, 4 < 3x < 5,
LL
2l + 1 ˆ -l ˆ -l ˆ i+ j+ k l +1 l +1 l +1
…(2)
On solving (1) & (2)
25.
=
Cr 5 = C r +1 12
EN
ˆi ˆj kˆ OA ´ OP = 1 1 1 2l + 1 3l + 1 1 l +1 l +1
A
Sol.
1
4/3
=
ò 1
=
5/3
(3 - 2x)dx + ò (4 - 2x) dx + 4/3
2
ò (5 - 2x)dx
5/3
2 3 4 + + =1 9 9 9
Þ l = 0.8
9
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