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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