of 13 /13
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 1 JEE Main Online Exam 2020 Questions & Answer 9 th January 2020 | Shift - II MATHEMATICS Q.1 Given : ƒ(x) = 1 x 2 1 ; x 1 2 1 x ; 2 1 2 1 x 0 ; x and g(x) = 2 2 1 x , 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) 4 3 3 1 (2) 4 3 2 1 (3) 4 3 2 1 (4) 3 1 4 3 Ans.  Sol. Co-ordinates of P 0 , 2 1 , Q 2 1 , 2 1 , R 2 3 1 , 2 3 and S 0 , 2 3 O y 2 1 , 2 1 Q 0 , 2 1 P 0 , 2 3 S g(x) = 2 2 1 x x R Required area = Area of trapezium PQRS – 2 / 3 2 / 1 2 dx 2 1 x = 2 / 3 2 / 1 3 2 1 x 3 1 2 3 1 2 1 2 1 2 3 2 1 = 0 2 1 3 3 1 2 3 3 2 1 3 2 1 3 = 3 1 4 3

others
• Category

## Documents

• view

0

0

Embed Size (px)

### Transcript of JEE Main Online Exam 2020careerpoint.ac.in/answer-key-solutions/jee-main-jan-2020...2020/01/09  ·... CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 1

JEE Main Online Paper

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. 

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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 2

JEE Main Online Paper

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.  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.  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) CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 3

JEE Main Online Paper

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.  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.  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) CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 4

JEE Main Online Paper

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. 

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. 

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. CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 5

JEE Main Online Paper

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.  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. 

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. 

Sol. 1)X(P5

1i

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

K = 61 , K = –1 (rejected) CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 6

JEE Main Online Paper

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. 

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. 

Sol. Let the equation of ellipse 2

2

2

2

by

ax

= 1, (a > b)

Given that 2b = 3

4 b = 3

2 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 7

JEE Main Online Paper

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. 

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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 8

JEE Main Online Paper

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. 

Sol. 22 yxxy

dxdy

Put y = vx

then dxdy = v + x

dxdv CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 9

JEE Main Online Paper

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. 

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 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 10

JEE Main Online Paper

Ans.  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.  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.  Sol. y2 = 8x (a = 2)

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

2,

21

2at = –2 t = –1/2 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 11

JEE Main Online Paper

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.  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.  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.  Sol. C0 + 5·C1 + 9·C2 + …. + (101)·C25

=

25

0rr

25C)1r4( =

25

0rr

2525

0rr

25 CC.r4 CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 12

JEE Main Online Paper

= 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.  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.  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. 

Sol. |)cb(a|

= |)cb||a|

sin 2

= 3

sin|c||b||a| CAREER POINT

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-6630500 www.careerpoint.ac.in 13

JEE Main Online Paper

= 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