computer based test (cbt) questions & solutions...2020/09/06 · 14. 3consider the statement:...

RReessoonnaannccee EEdduuvveennttuurreess LLttdd.. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029

TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((MMAAIINN)) 22002200 SSoolluuttiioonn ppoorrttaall

7340010333

PPAAPPEERR--11 ((BB..EE..//BB.. TTEECCHH..))

JJEEEE ((MMaaiinn)) 22002200

COMPUTER BASED TEST (CBT)

Questions & Solutions

DDaattee:: 0066 SSeepptteemmbbeerr,, 22002200 ((SSHHIIFFTT--22)) || TTIIMMEE :: (03.00 p.m. to 06.00 p.m)

DDuurraattiioonn:: 33 HHoouurrss || MMaaxx.. MMaarrkkss:: 330000

SSUUBBJJEECCTT :: MMAATTHHEEMMAATTIICCSS

mailto:[email protected]://www.resonance.ac.in/reso/results/jee-main-2014.aspx




7340010333


| 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


This solution was download from Resonance JEE (MAIN) 2020 Solution portal

PAGE # 1

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)






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– =






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






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)






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)






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






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)






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






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






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





7340010333


computer based test (cbt) questions & solutions...2020/09/06 · 14. 3consider the statement:...

Documents