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7340010333
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COMPUTER BASED TEST (CBT)
Questions & Solutions
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7340010333
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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7340010333
PART : MATHEMATICS
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer,
out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Let = 5
and A =
cos sin
sin cos
. If B = A + A4, then det (B) :
(1) is zero (2) is one (3) lies in (2, 3) (4) lies in (1, 2)
Ans. (4)
Sol. A2 =
cossin–
sincos
cossin–
sincos
A2 =
2cos2sin–
2sin2cos
A4 =
4cos4sin–
4sin4cos
B=
cossin–
sincos
4cos4sin–
4sin4cos
=
cos4cossin4sin–
sin4sincos4cos
22 sin4sincos4cosB
= 2 + 2 (cos4.cos + sin4. sin)
= 2 + 2cos(4 – )
= 2 + 2.cos3
|B| = 2 + 2 cos 5
3
= 2 –
2
1–5 =
2
5–5 (1, 2)
2. The area (in sq. units) of the region enclosed by the curves y = x2 – 1 and y = 1 – x2 is equal to :
(1) 8
3 (2)
7
2 (3)
4
3 (4)
16
3
Ans. (1)
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PAGE # 2
7340010333
Sol.
1
1
–1
–1
Given curves are y = x2 – 1 and y = 1 – x2 so intersection point are (±1, 0)
bounded area = 41
2
0
(1 x )dx = 1
3
0
x4 x
3
=
14 1
3
=
8
3 sq. units
3. Let L denote the line in the xy-plane with x and y intercepts as 3 respectively. Then the image of the point
(–1, –4) in the line is:
(1) 8 29
,5 5
(2) 11 28
,5 5
(3) 29 8
,5 5
(4) 29 11
,5 5
Ans. (2)
Sol. Equation of line is
y
x+
1
y = 1 x + 3y – 3 = 0
If image is (x1, y1) then
10
3–12–1–2–
3
4y
1
1x 11
5
16
3
4y1x 11
5
281y,
5
11x 11
4. If and are the roots of the equation 2x(2x + 1) = 1, then is equal to :
(1) 2( – 1) (2) 2( + 1) (3) 22 (4) –2( + 1)
Ans. (4)
Sol Given equation is 2x (2x + 1) = 1 4x2 + 2x – 1 = 0 ………..(1)
roots of equation (1) are and
+ = 2
1– = –
2
1– ………..(2)
and
42 + 2 – 1 = 0 2 =2
–4
1 ………..(3)
now
–2 ( + 1) = – 22 – 2
= – 2
2–
4
1– 2= –
2
1– =
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
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PAGE # 3
7340010333
5. For a suitably chosen real constant a, let a function, f : R – {–a} R be defined by f(x)= a x
a x
.Further
suppose that for any real number x –a and f(x) –a, (fof)(x) = x. Then f1
2
is equal to :
(1) 3 (2) –3 (3) 1
3 (4)
1
3
Ans. (1)
Sol. fof(x) = x)x(fa
)x(f–a
)x(fx1
ax–a
xa
x–a
x1
x–1a
a = 1
so f(x) = x1
x–1
32
1–f
6. If the tangent to the curve, y = f(x)=xlogex, (x>0) at a point (c, f(c)) is parallel to the line-segment joining
the points (1, 0) and (e, e), then c is equal to:
(1) e 1
e
(2)
1
e 1 (3)
1
e 1e
(4)
1
1 ee
Ans. (3)
Sol. f'(c) = 1 + nc = 1–e
e
nc = 1–e
1
c = 1–e1
e
7. If the constant term in the binomial expansion of
10
2
kx
x
is 405, then |k| equals:
(1) 3 (2) 2 (3) 1 (4) 9
Ans. (1)
Sol. Tr+1 = 10Cr. r–10r
2x
x
K–
= 10Cr . .K– r 2r5
–5
x
for constant term 5 – 02
r5 r = 2
T3 = 10C2. K2. = 405
405K2
910 2 K2 = 9 3K
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
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PAGE # 4
7340010333
8. The probabilities of three events A, B and C are given P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(AB)=0.8, P(AC) = 0.3, P(ABC)=0.2, P(BC)= and P(ABC) = , where 0.85 0.95, then lies in the interval: (1) [0.36, 0.40] (2) [0.35, 0.36] (3) [0.25, 0.35] (4) [0.20, 0.25] Ans. (3)
Sol. P(A B C) = P(A) + P(B) + P(C) – P(A B) – P(B C) – P(C A) + P(A B C)
= 1.4 – P(A B) – + = 1.4 – P (A B) …………(1)
again
P(A B) = P(A) + P(B) – P(A B) P(A B) = .2 ………….. (2)
by (1) and (2) = 1.2 –
now 0.85 0.95
0.85 1.2 – 0.95 [0.25, 0.35]
9. The integral
2
x 2
e
1
e .x (2 log x)dx equals:
(1) e(2e – 1) (2) e(4e + 1) (3) 4e2 – 1 (4) e(4e–1) Ans. (4)
Sol. Let y = (ex)x
ny = [1 + nx]
nx2dx
dy
y
1
dy = (ex)x (2 + nx) dx
2
x 2
e
1
e .x (2 log x)dx = e–e4exy 22
1x2
1
10. The common difference of the A.P. b1,b2,….,bm is 2 more than common difference of A.P. a1,a2,…,an. If a40 = –159,a100 = – 399 and b100 = a70, then b1 is equal to : (1) 127 (2) 81 (3) – 127 (4) – 81 Ans. (4)
Sol. Let a1 a1 + d, a1 + 2d …… first A.P.
a40 = a1 + 39d = – 159 ……….(1)
a100 = a1 + 99d = – 399 ..……..(2)
from equation (1) and (2)
d = – 4, a1 = – 3
now
b100 = a70
b1 + 99D = a1 + 69d
b1 + 99 × – 2 = – 3 + 69 × –4 (According to question D = d + 2)
b1 = – 81
11. If the normal at an end of latus rectum of an ellipse passes through an extremity of the minor axis, the
eccentricity e of the ellipse satisfies:
(1) e4 + 2e2 – 1 = 0 (2) e2 + e – 1 = 0
(3) e2 + 2e – 1 = 0 (4) e4 + e2 – 1 = 0
Ans. (4)
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PAGE # 5
7340010333
Sol. Equation of normal at
a
b,ae
2
22
2
22
b–aa/b
yb–
ae
xa
It passes through (0, – b)
ab = a2e2
a2b2 = a4e4 (b2 = a2 (1–e2)
1 – e2 = e4
12. Let f : R R be a function defined by f(x) = max {x, x2}. Let S denote the set of all points in R, where f is
not differentiable. Then:
(1) (an empty set) (2) {1} (3) {0} (4) {0, 1}
Ans. (4)
Sol.
0,0
• y = x
y = x2
• • • • • • • • • • • • • • • • • • • • • • • • •
• • • •
• • • •
• • • • • • • • • • •
k
13. The set of all real values for which the function f(x) = (1 – cos2x).( + sinx), x ,2 2
, has exactly one
maxima and exactly one minima, is :
(1) 1 1
,2 2
– {0} (2) 3 3
,2 2
(3) 1 1
,2 2
(4) 3 3
,2 2
– {0}
Ans. (4)
Sol. f(x) = sin2x ( + sinx)
f'(x) = sinx cosx (2 + 3sinx)
sinx = 0 (one point)
sinx = –3
2 (–1,1) – {0} …………(i)
3 3
,2 2
– {0}
14. Consider the statement: “For an integer n, if n3 – 1 is even, then n is odd”. The contrapositive statement
of this statement is:
(1) For an integer n, if n is even, then n3 – 1 is odd.
(2) For an integer n, if n is even, then n3 – 1 is even.
(3) For an integer n, if n3 – 1 is not even, then n is not odd.
(4) For an integer n, if n is odd, then n3 – 1 is even.
Ans. (1)
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PAGE # 6
7340010333
Sol. P : n3 – 1 is even, q : n is odd
contrapositive of p q = ~ q ~p
" If n is not odd then n3 – 1 is not even"
For an integer n, if n is even, then n3 – 1 is odd.
15. The centre of the circle passing through the point (0, 1) and touching the parabola y = x2 at the point (2,
4) is:
(1) 3 16
,10 5
(2) 53 16
,10 5
(3) 16 53
,5 10
(4) 6 53
,5 10
Ans. (3)
Sol. y = x2 , (2,4)
tangent at (2,4) is
x24y2
1
y + 4 = 4x 4x – y – 4 = 0
Equation of circle (x – 2)2 + (y – 4)2 + (4x – y – 4) = 0
it passes through (0,1)
4 + 9 + (0 –1 – 4) = 0
13 = 5 5
13
circle is x2 – 4x + 4 + y2 – 8y + 16 + 5
13(4x – y – 4) = 0
x2 + y2 + 05
52–20y
5
138–x4–
5
52
x2 + y2 + 05
48y
5
53–x
5
32
centre is
10
53,
5
16–
16. If y = 2
x 1
cosec x is the solution of the differential equation, dy
dx+p(x)y=
2
cosec x, 0 < x <
2
, then
the function p(x) is equal to:
(1) tan x (2) cosec x (3) cot x (4) sec x
Ans. (3)
Sol. y = 2
x 1
cosec x
xcotecxcos1–x2
–ecxcos2
dx
dy
ecxcos2
xcotecxcos1–x2
dx
dy
ecxcos2
xcotydx
dy
P(x) = cotx
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
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PAGE # 7
7340010333
17. Let z = x + iy be a non-zero complex number such that z2 = i |z|2, where i = 1 , then z lies on the:
(1) real axis (2) line, y = x (3) line, y = –x (4) imaginary axis
Ans. (2)
Sol. 222 yxiiyx x2 – y2 + 2ixy = i (x2 + y2)
compare real and imaginary parts
x = y
18. A plane P meets the coordinate axes at A, B and C respectively. The centroid of ABC is given to be
(1,1,2). Then the equation of the line through this centroid and perpendicular to the plane P is:
(1) x 1
2
=
y 1
1
=
z 2
1
(2)
x 1
1
=
y 1
2
=
z 2
2
(3) x 1
1
=
y 1
1
=
z 2
2
(4)
x 1
2
=
y 1
2
=
z 2
1
Ans. (4)
Sol. Let A(,0,0) , B (0,,0), C(0,0,) then
3,
3,
3G (1,1,2)
B(0,,0)
A(,0,0)
G(1,1,2)
C(0,0,) = 3, = 3, = 6
equation of plane is 1zyx
16
z
3
y
3
x
2x + 2y + z = 6
required line 1
2–z
2
1–y
2
1–x
19. The angle of elevation of the summit of a mountain from a point on the ground is 45º. After climbing up
one km towards the summit at an inclination of 30º from the ground, the angle of elevation of the summit
is found to be 60º. Then the height (in km) of the summit from the ground is:
(1) 3 1
3 1
(2)
1
3 1 (3)
1
3 1 (4)
3 1
3 1
Ans. (3)
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PAGE # 8
7340010333
Sol.
B
60º
30º 45º
E
A F x y C
h
D
z
In CDF
sin30º = 1
z [CD = 1 km(given)]
z = 2
1 ………….(1)
cos30º = 1
y =
2
3
now in ABC
tan45º = yx
h
h = x + y
x = h – 2
3 ………….(2)
now
In BDE,
tan60º = x
z–h
2
1–hx3
2
1–h
2
3–h3
1h1–3
h = km1–3
1
20. For all twice differentiable functions f : R R, with f(0) = f(1) = f(0) = 0,
(1) f(x) = 0, for some x (0, 1) (2) f(x) = 0, at every point x (0, 1)
(3) f(0) = 0 (4) f(x) 0, at every point x (0, 1)
Ans. (1)
Sol. Applying Rolle's theorem in [0,1] for function f(x)
f'(c) = 0, c (0,1)
again applying Rolle's theorem in [0,c] for function f'(x)s
f''(c1) = 0, c1 (0,c)
option (1) is correct
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
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PAGE # 9
7340010333
Numerical Value Type (la[;kRed izdkj)
This section contains 5 Numerical value type questions.
bl [k.M esa 5 l[;kRed izdkj ds iz'u gSaA
21. The number of words (with or without meaning) that can be formed from all the letters of the word
“LETTER” in which vowels never come together is……..
Ans. 120
Sol. Consonants are L, T, T, R
Vowels are E,E,
Total number of words (with or without meaning) from letters of word 'LETTER' = 180!2!2
!6
Total number of words (with or without meaning) from letters of word 'LETTER' if vowels are together
= 60!2
!5
Required = 180 – 60 = 120
22. Consider the data on x taking the values 0, 2, 4, 8, …., 2n with frequencies nC0, nC1, nC2,…, nCn
respectively. If the mean of this data is n
728
2, then n is equal to……..
Ans. 6
Sol. n
n2
n1
n0
ni
n2i
CCCC)frequency(f
2220)nobservatio(x
i
ii
f
xfx
n
n
nn
2n
1n
0n
nnn
2n2
1n
0n
2
1–3
C......CCC
C2......C2C2C0
=
n2
728
3n = 36
n = 6
23. Suppose that a function f : R R satisfies f(x + y) = f(x)f(y) for all x, y R and f(1) = 3. If n
i 1
f(i) 363
,
then n is equal to……..
Ans. 5
Sol. f(x) = ax
f(1) = a = 3
so f(x) = 3x
n
i 1
f(i) 363
3 + 32 + ….. 3n = 363
3632
)1–3(3 n
3n = 243 n = 5
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| JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | MATHEMATICS
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PAGE # 10
7340010333
24. If x and y be two non-zero vectors such that x y = x and 2x y is perpendicular to y , then the
value of is…….
Ans. 1
Sol. 22
xyx
0y.x2y2
………..(1)
and 0y.yx2
s
0y.x2y2
…………..(2)
by (1) and (2) = 1
25. The sum of distinct values of l for which the system of equations:
( – 1)x + (3 + 1)y + 2z = 0
( – 1)x + (4 – 2)y + ( + 3)z = 0
2x + (3 + 1)y + 3( – 1)z = 0,
Has non-zero solutions, is..........
Ans. 3
Sol. 0
)1–(3132
32–41–
2131–
R2 R2 – R1
R3 R3 – R1
0
3–0–3
3–3–0
2131–
C1 C1 + C3
0
3–00
–33––3
2131–3
(– 3)2 [6] = 0 = 0, 3
sum of values of = 3
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