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JEE Main Online Exam 2020

Questions & Answer 9th January 2020 | Shift - II

MATHEMATICS

Q.1 Given : ƒ(x) =

1x21;x1

21x;

21

21x0;x

and g(x) = 2

21x

, x R. Then the area (in sq. units) of the region

bounded by the curves y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = 3 , is -

(1) 43

31 (2)

43

21 (3)

43

21 (4)

31

43

Ans. [4]

Sol. Co-ordinates of P

0,

21 , Q

21,

21 , R

231,

23 and S

0,

23

O

y

21,

21Q

0,

21P

0,

23S

g(x) =2

21x

x R

Required area = Area of trapezium PQRS –

2/3

2/1

2

dx21x

= 2/3

2/1

3

21x

31

231

21

21

23

21

=

02

1331

233

213

21

3

= 31

43

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Q.2 Let a, b R, a 0 be such that the equation, ax2 – 2bx + 5 = 0 has a repeated root , which is also a root of

the equation, x2 – 2bx – 10 = 0. If is the other root of this equation, then 2 + 2 is equal to - (1) 26 (2) 24 (3) 25 (4) 28

Ans. [3] Sol. Roots of equation ax2 – 2bx + 5 = 0 are ,

+ = ab2 =

ab ...(i)

and 2 = a5 ...(ii)

from eq. (i) & (ii)

a5

ab

2

2 b2 = 5a ...(iii) (a 0)

Roots of equation x2 – 2bx – 10 = 0 are , + = 2b and = –10

Now, = ab is root of equation x2 – 2bx – 10 = 0

ab2

ab 2

2

2 – 10 = 0

a5 – 10 – 10 = 0 ( b2 = 5a)

a = 41 and b2 =

45

So, 2 = 2

2

ab = 20 and 2 = 5

2 + 2 = 25 Q.3 If A = {x R : |x| < 2} and B = {x R : |x – 2| 3}; then -

(1) A B = R – (2, 5) (2) A B = (–2, –1) (3) B – A = R – (–2, 5) (4) A – B = [–1, 2) Ans. [3] Sol. A = {x R : |x| < 2} A (–2, 2)

and B = {x R : |x – 2| 3} B (– , –1] [5, ) A B (– , 2) [5, )

A B (–2, –1] B – A (–, –2] [5, ) or R – (–2, 5) A – B (–1, 2)

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Q.4 Let an be the nth term of a G.P. of positive terms. If

100

1n1n2a = 200 and

100

1nn2a = 100, then

200

1nna is equal to-

(1) 175 (2) 225 (3) 300 (4) 150

Ans. [4] Sol. Let the G.P. is a, ar, ar2 ....

200a100

1n1n2

a3 + a5 + a7 + .... + a201 = 200

1r

)1r(ar2

2002

= 200 ...(1)

100a100

1nn2

a2 + a4 + .... + a200 = 100

1r

)1r(ar2

200

= 100 ...(2)

dividing (1) by (2) we get, r = 2 adding both eq. (1) & (2) a2 + a3 + a4 + a5 + ..... + a201 = 300 r(a1 + a2 + ..... + a200) = 300

a1 + a2 + ..... + a200 = r

300

2

300a200

1n1

= 150

Q.5 If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be -

(1) 10 (2) 8 (3) 7 (4) 2

17

Ans. [3] Sol. Let z = x + iy given that |Re(z)| + |Im(z)| = 4 |x| + |y| = 4

OA(4, 0)

D(0, –4)

(–4,0)C

B(0, 4)

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Maximum value of |z| = 4 Minimum value of |z| = perpendicular distance of line AB from (0, 0) = 22 So, |z| [ 22 , 4] So, |z| can not be 7

Q.6 If x =

0n

n)1( tan2n and y = ,cos0n

n2

for 0 < < 4 , then -

(1) y(1 – x) = 1 (2) x(1 – y) = 1 (3) y(1 + x) = 1 (4) x(1 + y) = 1 Ans. [1]

Sol. x =

0n

n2n tan)1(

x = 1 – tan2 + tan4 – tan6 ......

x = 2tan1

1 = cos2 ...(1)

and y =

0n

n2cos

y = 1 + cos2 + cos4 + cos6 + ....

y = 2cos1

1 = 2sin

1

sin2 = y1 ....(2)

Adding (1) & (2), we get,

x + y1 = 1

y(1 – x) = 1

Q.7 Let [t] denote the greatest integer t and 0x

lim

x4x = A. Then the function, ƒ(x) = [x2] sin (x) is

discontinuous, when x is equal to -

(1) 1A (2) A (3) 21A (4) 5A

Ans. [1]

Sol.

x4xlim

0x= A

x4

x4xlim

0x = A

x4x4lim

0x = A

A = 4 Now, f(x) = [x2] sin(x) is continuous at every integer point but discontinuous at non integer points then by options, 1A is correct answer.

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Q.8 If p (p ~q) is false, then the truth values of p and q are respectively -

(1) F, F (2) T, F (3) T, T (4) F, T Ans. [3] Sol. p (p ~q) will be false only when p is true and (p ~q) is false. So, p = T, q = T

Q.9 Let a function ƒ : [0, 5] R be continuous, ƒ(1) = 3 and F be defined as F(x) = x

1

2 dt)t(gt , where

g(t) = t

1

du)uƒ( . Then for the function F, the point x = 1 is

(1) not a critical point (2) a point of local maxima (3) a point of local minima (4) a point of inflection

Ans. [3]

Sol. F(x) = x

1

2 dt)t(gt

F (x) = x2g(x)

at F (1) = (1) g(1) = 0 { g(1) = 0} Now, F (x) = 2xg(x) + x2g(x) F (1) = 2g(1) + g(1) F (1) = 0 + g(1) {g(t) = f(t); gf F(1) = 3 So, at x = 1, F (1) = 0 but F (1) > 0 For the function f(x), x = 1 is a point of local minima. Q.10 A random variable X has the following probability distribution : X : 1 2 3 4 5 P(X) : K2 2K K 2K 5K2 Then P(X > 2) is equal to -

(1) 361 (2)

61 (3)

127 (4)

3623

Ans. [4]

Sol. 1)X(P5

1i

K2 + 2K + K + 2K + 5K2 = 1 6K2 + 5K – 1 = 0

K = 61 , K = –1 (rejected)

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So, K = 61

P(X > 2) = K + 2K + 5K2

= 365

62

61

= 3623

Q.11 If

)2sec2(tancosd

2 = tan + 2loge |ƒ()| + C where C is a constant of integration, then the ordered

pair (, ƒ()) is equal to - (1) (1, 1 – tan ) (2) (–1, 1 – tan ) (3) (–1, 1 + tan) (4) (1, 1 + tan)

Ans. [3]

Sol.

)2sec2(tancosd

2

=

2

2

22

tan1tan1

tan1tan2cos

d

=

d)tan1(

)tan1(sec2

22

Put tan = t sec2 d = dt

= dt

)t1()t1(2

2 =

dt

t1t1

=

dt

t121

= –t + 2 loge |1 + t| + C = – tan + 2loge |1 + tan| + C = –1 and f() = 1 + tan

Q.12 The length of the minor axis (along y-axis) of an ellipse in the standard form is 3

4 . If this ellipse touches

the line, x + 6y = 8, then its eccentricity is -

(1) 3

1121 (2)

311

31 (3)

35

21 (4)

65

Ans. [1]

Sol. Let the equation of ellipse 2

2

2

2

by

ax

= 1, (a > b)

Given that 2b = 3

4 b = 3

2

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Equation of tangent y = mx ± 222 bma ....(1) Given tangent is x + 6y = 8

y = –61 x +

68 ...(2)

from eq. (1) & (2)

m = –61 and a2m2 + b2 =

916

a2

361 +

34 =

916

a2 = 16

Now, e = 2

2

ab1 =

163/41 =

1211 =

311

21

Q.13 In the expansion of 16

sinx1

cosx

, if 1 is the least value of the term independent of x when

48

and 2 is the least value of the term independent of x when 816

, then the ratio 2 : 1 is equal to -

(1) 16 : 1 (2) 1 : 8 (3) 8 : 1 (4) 1 : 16

Ans. [1]

Sol. General term Tr+1 = rr16

r16

sinx1

cosxC

Tr+1 = r216rr16

r16 x

sin1

cos1C

term is independent of x 16 – 2r = 0 r = 8

T9 = 88

816

sin1

cos1C

T9 = 8

8

816

)2(sin2C

4,

8 2

2,

4

For least value sin2 should be maximum

2 = 2

So, 1 = 16C8(28) ...(1)

Again,

8,

16 2

4,

8

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Minimum value, 2 = 16C8 8

8

)2/1(2 = 16C8212 ...(2)

1

16124

1

2

Q.14 If x = 2 sin – sin2 and y = 2cos – cos2, [0, 2], then 2

2

dxyd at = is -

(1) 23 (2)

43 (3) –

43 (4) –

83

Ans. [Bonus] Sol. x = 2 sin – sin2

d

dx = 2cos – 2cos2

and y = 2 cos – cos 2

d

dy = – 2sin+ 2sin2

Now, dxdy =

)2cos(cos)sin2(sin2

=

2sin.

23sin2

2sin

23cos2

dxdy = cot

23

2

2

dxyd = – cosec2

23

dxd

23

2

2

dxyd = –

2cos2cos21

23eccos

23 2

at =

2

2

dxyd =

221)1(

23 =

83

Q.15 If dxdy = 22 yx

xy

; y(1) = 1; then a value of x satisfying y(x) = e is -

(1) 321 e (2) 2 e (3)

2e (4) 3 e

Ans. [4]

Sol. 22 yxxy

dxdy

Put y = vx

then dxdy = v + x

dxdv

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v + xdxdv = 222 xvx

)vx(x

xdxdv = 2v1

v

– v

dvv

v13

2 = – x

dx

2v21

+ log v = – log x + C

xylog

yx

21

2

2 = –logx + C ...(1)

put x = 1, y = 1

C = – 21

from eq. (1)

xylog

yx

21

2

2 = –logx –

21

Put y = e xlogxelog

e2x

2

2

= –

21

x2 = 3e2 x = ± 3 e

x = 3 e

Q.16 Let a – 2b + c = 1. If ƒ(x) = 3x4xcx2x3xbx1x2xax

, then -

(1) ƒ(–50) = – 1 (2) ƒ(50) = 1 (3) ƒ(50) = – 501 (4) ƒ(–50) = 501 Ans. [2]

Sol. ƒ(x) = 3x4xcx2x3xbx1x2xax

R1 R1 – 2R2 + R3

f(x) = 3x4xcx2x3xbx

001

f(x) = 1 f(50) = 1 Q.17 If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these

boxes contain exactly 2 and 3 balls is -

(1) 102945 (2) 112

965 (3) 112945 (4) 102

965

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Ans. [1] Sol. Total ways n = 410

Number of ways placing exactly 2 and 3 balls in two of these boxes

= 4C2 × !3!2

!5 × 2! × 10C5 × 25

Required probability = 102945

Q.18 The following system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0, has

(1) no solution (2) infinitely many solutions, (x, y, z) satisfying y = 2z (3) only the trivial solution (4) infinitely many solutions, (x, y, z) satisfying x = 2z

Ans. [4] Sol. 7x + 6y – 2z = 0 ....(1) 3x + 4y + 2z = 0 ....(2) x – 2y – 6z = 0 ....(3)

= 621

243267

= 7(–24 + 4) – 6(–18 – 2) – 2(–6 – 4) = 0 = 0 infinite non-trival solution exist to eliminate y we operate eq. (1) – (2) + (3) 5x = 10z x = 2z

Q.19 If one end of a focal chord AB of the parabola y2 = 8x is at

2,21A , then the equation of the tangent to it

at B is - (1) 2x + y – 24 = 0 (2) x + 2y + 8 = 0 (3) x – 2y + 8 = 0 (4) 2x – y – 24 = 0

Ans. [3] Sol. y2 = 8x (a = 2)

Let one end of focal chord is (at2, 2at) =

2,

21

2at = –2 t = –1/2

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other end of focal chord will be

ta2,

ta2 (8, 8)

Now, tangent at B(8, 8)

y(8) = 8

28x

x – 2y + 8 = 0 Q.20 Let ƒ and g be differentiable functions on R such that fog is the identity function. If for some a, b R, g(a) = 5

and g(a) = b, then ƒ(b) is equal to -

(1) 51 (2)

52 (3) 5 (4) 1

Ans. [1] Sol. fog is an identity function fog(x) = x

f {g(x)} g(x) = 1 put x = a f {g(a)} g(a) = 1 f (b) (5) = 1

f (b) = 51

Q.21 If the distance between the plane, 23x – 10y – 2z + 48 = 0 and the plane containing the lines

31z

43y

21x

and

1z

62y

23x ( R) is equal to

633k , then k is equal to ____________.

Ans. [3] Sol. Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)

222 )2()10()23(

4823023

= 633k

6333 =

633k

k = 3 Q.22 If Cr = 25Cr and C0 + 5·C1 + 9·C2 + …. + (101)·C25 = 225·k, then k is equal to ___________. Ans. [51] Sol. C0 + 5·C1 + 9·C2 + …. + (101)·C25

=

25

0rr

25C)1r4( =

25

0rr

2525

0rr

25 CC.r4

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= 2525

0r1r

24 2Cr

25.r4

= 100 . 224 + 225 = 225(50 + 1) k = 51 Q.23 The number of terms common to the two A.P.’s 3, 7, 11, ….. 407 and 2, 9, 16, …. 709 is ___________. Ans. [14] Sol. First A.P. is 3, 7, 11, 15, 19, 23, ..... 407 Second A.P. is 2, 9, 16, 23, ..... 709 First common term = 23 Common difference d = L.C.M. (4, 7) = 28 Last term 407

23 + (n – 1) (28) 407 n 14.7 So, n = 14 Q.24 If the curves, x2 – 6x + y2 + 8 = 0 and x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the

largest value of k is ______________. Ans. [36] Sol. C1 = (3, 0), C2(0, 4)

r1 = 809 = 1; r2 = k1616 = k

Two circles touch each other C1C2 = |r1 ± r2|

5 = |1 ± k |

1 + k = 5 or k – 1= 5

k = 16 or k = 36 Maximum value of k = 36

Q.25 Let candb,a be three vectors such that |

a | = 3 ,|

b | = 5,

c·b = 10 and the angle between

candb

is 3 . If

a is perpendicular to the vector

cb , then | )cb(a

| is equal to _______.

Ans. [30]

Sol. |)cb(a|

= |)cb||a|

sin 2

= 3

sin|c||b||a|

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= 23.|c|)5()3(

= |c|2

15 ...(1)

cos 3 =

|c||b|

c.b

21 =

|c|5

10 |c|

= 4

from eq. (1)

|)cb(a|

= 2

15 × 4 = 30