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7340010333


| JEE MAIN-2020 | DATE : 05-09-2020 (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

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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. If and are the roots of the equation, 7x2 – 3x– 2 = 0, then the value of 22 11

is equal to :

(1) 24

1 (2)

32

27 (3)

16

27 (4)

8

3

Ans. (3)

Sol. + = 7

3. = –

7

2

22 –1–1

=

22 –1–1

–

=

222 –1–

2–1

–22

= 16

27

7

22

7

3–

7

21

7

3

7

2

7

3

22

2. The statement

(p (q p)) (p (p q) is :

(1) equivalent to ( p q ) (~ p) (2) a contradiction

(3) a tautology (4) equivalent to )q(~)qp(

Ans. (3)

Sol.

TTTFTFF

TTTTFTF

TTTTTFT

TTTTTTT

sr)qp(p:s)pq(p:rqppqqp

3. If the line y = mx + c is a common tangent to the hyperbola 164

y

100

x 22 and the circle x2 + y2 = 36, then

which one of the following is true ?

(1) 4c2 = 369 (2) 5m = 4 (3) c2 = 369 (4) 8m + 5 = 0

Ans. (1)

Sol. c2 = 36(1 + m2) …(1)

c2 = 100m2 – 64 …(2)

100m2 – 64 = 36 + 36m2






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64m2 = 100

m2 = 64

100

c2 = 36

64

1001 =

4

369

4. If the length of the chord of the circle, x2 + y2 = r2 (r > 0) along the line, y–2x = 3 is r, then r2 is equal to :

(1) 5

9 (2) 12 (3)

5

12 (4)

5

24

Ans. (3)

Sol. AB = r, AD = 2

r

A D B r

B C(0,0)

60º

r

B

r

B

CD = rsin60º = 2

r3

2

r3

21

3–00

22

r =

5

32 r2 =

5

12

5. If x = 1 is a critical point of the function f(x) = (3x2 + ax –2 – a) ex, then :

(1) x = 1 and x = 3

2 are local minima of f.

(2) x = 1 is a local maxima and x = 3

2 is a local minima of f.

(3) x = 1 is a local minima and x = 3

2 is a local maxima of f.

(4) x = 1 and x = 3

2 are local maxima of f.

Ans. (3)

Sol. f(x) = (3x2 + ax – 2 – a)ex

f'(x) = (3x2 + ax – 2 – a)ex + ex(6x + a) = ex(3x2 + (a + 6)x – 2)

x = 1 is a critical point f'(1) = 0

3 + a + 6 – 2 = 0

a = –7

f'(x) = ex(3x2 –x – 2) = ex(3x2 –3 x + 2x – 2) = ex(3x + 2)(x – 1)






PAGE # 3

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+ – +

–2/3 1

maxima at x = –2/3 minima at x = 1

6. The derivative of tan–1

x

1x1 2 with respect to

2

21

x21

x1x2tan at x =

2

1 is :

(1) 10

3 (2)

12

3 (3)

5

32 (4)

3

32

Ans. (1)

Sol. Let x = tan

y1 = tan–1

tan

1–sec = tan–1

2tan =

2

=

2

1 tan–1x

x = sin y2 = tan–1

2cos

cossin2 = tan–1 (tan2) = 2 = 2sin–1x

dx/dy

dx/dy

dy

dy

2

1

2

1 =

2

2

x–1

1.2

2

1.

x1

1

= 2

2

x14

x–1

=

4

114

4

1–1

= 10

3

7. If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

(1) )13(13

2 50 (2) )13(26

1 49 (3) )13(13

1 50 (4) )13(26

1 50

Ans. (4)

Sol. Let a, ar, ar2,…….G.P.

T2 + T3 + T4 = 3 ar (1 + r + r2) = 3 …..(i)

T6 + T7 + T8 = 243 ar5 (1 + r + r2) = 243 …..(ii)

by (i) and (ii)

r4 = 81 r = 3

a = 13

1

S50 = 1r

)1r(a 50

=

26

1350






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8. If the mean and the standard deviation of the data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :

(1) x2 – 20x + 18 = 0 (2) 2x2 – 20x + 19 = 0

(3) x2 – 10x + 19 = 0 (4) x2 – 10x + 18 = 0 Ans. (3)

Sol. 5 + 3 + 7 + a + b = 25 a + b = 10

S.D. = 25–2

ba735 222222

= 25–5

83ba 22 = 4 a2 + b2 = 62

(a + b)2 – 2ab = 62 ab = 19

so equation whose roots are a and b is x2 –10x + 19 = 0

9. If ,C)(BlogAdcos2sin75

cose2

where C is a constant of integration, then A

)(B can be:

(1) )3(sin5

1sin2

(2)

3sin

)1sin2(5

(3)

3sin

1sin2

(4)

1sin2

)3(sin5

Ans. (2)

Sol. I =

d3sin7sin2

cos2

sin = t cosd = dt

= dt

2

3t

2

7t

1

2

1

2

= dt

4

5

4

7t

1

2

122

= c3t

1t2ln

5

1

= c

3sin

1sin2ln

5

1

so A = 5

1

B() = 3sin

)1sin2(5

10. The value of

30

i1

3i1

is :

(1) 65 (2) 215 i (3) –215 (4) –215 i Ans. (4)






PAGE # 5

7340010333

Sol.

30

i–1

i31–

=

30

)4

sini–4

(cos2

3

2sini

3

2cos2

=

2

15sini–

2

15cos2

20sini20cos2

15

30

=

i2–

i0

i012 1515

11. 1xx1

1ex

lim42

x)1xx1(

0x

42

(1) does not exist (2) is equal to 1

(3) is equal to e (4) is equal to 0

Ans. (2)

Sol.

x

1xx1

1e

lim42

x

1xx1

0x

42

put x

1xx1 42 = t

clearly x 0 t 0

given limit = 0t

lim

t

1e t = 1

12. The area (in sq. units) of the region A = {x, y) : (x – 1) [x] y x2 , 0 x 2}, where [t] denotes the

greatest integer function, is :

(1) 2

12

3

4 (2) 12

3

8 (3) 12

3

4 (4)

2

12

3

8

Ans. (4)






PAGE # 6

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Sol. y = [x] (x – 1)

=

2x11x

1x00

x2y

0 1 2

1

1

Area = )1)(1(2

1dx.x2

2

0

= 2

0

2/3

3

x4

–

2

1 =

3

28 –

2

1

13. If the system of linear equations

x + y + 3z = 0

x + 3y + k2z = 0

3x + y + 3z = 0

has a non–zero solution (x, y, z) for some

k R, then x +

z

y is equal to :

(1) –9 (2) –3 (3) 9 (4) 3

Ans. (2)

Sol. So D = 0

313

k31

3112 = 0 k2 = 9

x + y + 3z = 0 …..(1)

x + 3y + 9z = 0 …..(2)

3x + y + 3z = 0 …..(3)

(1) – (3)

x = 0 y + 3z = 0






PAGE # 7

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3z

y

so x +

z

y = –3

14. If L =

8sin

16sin 22 and

M =

8sin

16cos 22

(1) M = 8

cos4

1

24

1 (2) M =

8cos

2

1

22

1

(3) L = 8

cos2

1

22

1 (4) L =

8cos

4

1

24

1

Ans. (2)

Sol. L =

816sin

816sin

16sin.

16

3sin

=

1616

3cos

1616

3cos

2

1 =

8cos

2

1

2

1

M =

816cos

816cos

16cos.

16

3cos

=

1616

3cos

1616

3cos

2

1 =

8cos

2

1

2

1

15. If for some R, the lines

L1 : 1

1z

1

2y

2

1x

and

L2 : 1

1z

5

1y2x

are coplanar, then the line L2 passes through the point :

(1) (2, –10, –2) (2) (10, 2, 2) (3) (–2, 10, 2) (4) (10, –2, –2)

Ans. (1)






PAGE # 8

7340010333

Sol. Lines are coplanar

so

231

112

15

=

–5 + ( – 5)3 + 7 = 0

–2 = 8 = –4

L2 : 1

1z

9

1y

4

2x

Now by cross checking option (1) is correct.

16. If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non–zero distinct real numbers,

then

czycz

byyby

axyax

is equal to :

(1) y (a – b) (2) 0 (3) y(b – a) (4) y(a – c)

Ans. (1)

Sol. Given x + a = y + b + 1 = z + c

Now

zcycz

ybyby

xayax

=

cycz

byby

ayax

(C3 C3 – C1)

= 322 C–CCcyz

byy

ayx

= y

c1z

b1y

a1x

R2 R2 – R1 and R3 R3 – R1

y.

a–c0x–z

a–b0x–y

a1x

= y

a–c0x–z

)ba(–0ba

a1x

= y(a – b)

a–c0x–z

1–01

a1x

= –y(a – b)(c – a + z – x) = y(a – b)






PAGE # 9

7340010333

17. Let y = y(x) be the solution of the differential equation

2,0x,x2sinxsiny2

dx

dyxcos

If y(/3) =0, then y(/4) is equal to :

(1) 12

1 (2) 22 (3) 22 (4) 22

Ans. (3)

Sol. dx

dy + 2tanx.y = 2sinx

I.F. = xdxtan2

e = sec2x

solution is

y.sec2 x = Cxdxsec.xsin22

y sec2 x = 2secx + C

0 = 2.2 + c c = –4

ysec2x = 2secx – 4

224

y

18. There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :

(1) 3000 (2) 2250 (3) 2255 (4) 1500 Ans. (2)

Sol. A 5Q B 5Q C 5Q

A1, A2, A3, A4, A5 B1, B2, B3, B4, B5 C1, C2, C3, C4, C5

A1A2A3 B1C1 3C1 × 5C3 × 5C1 × 5C1 = 750

A1A2B1B2C1 3C2 × 5C2 × 5C2 × 5C1 = 1500

total = 2250






PAGE # 10

7340010333

19. If the sum of the first 20 terms of the series log ...xlogxlogxlog)7()7()7( 4/13/12/1

is 460, then x is equal

to :

(1) e2 (2) 746/21 (3) 72 (4) 71/2 Ans. (3)

Sol. Given times20......xlogxlogxlog4

1

3

1

2

1

777

= 460

xlog21.....432 7 = 460

xlog)212(2

207 = 460

xlog7 = 2

x = 49

20. Which of the following points lies on the tangent to the curve 31y2ex y4 at the point

(1, 0) ?

(1) (–2, 4) (2) (2, 2) (3) (–2, 6) (4) (2, 6)

Ans. (3)

Sol. eyy'x4 + 4x3 ey + 2y' 1y2

1

= 0

at (1, 0)

y' + 4 + y' = 0 y' = –2

equation of tangent at (1, 0) is 2x + y – 2 = 0

so option (3) is correct

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. In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is …..

Ans. 11

Sol. let probability of hitting the target = p p = 2

1

Let n be the minimum number of bombs






PAGE # 11

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According to given condition

1 – (nC0 P0(1 – P)n + nC1 P1(1 – P)n–1) 100

99

2n (n + 1)100

n = 10 210 1100 Reject

n = 11 211 1200 Select

22. Let the vectors c,b,a

be such that 2a

, 4b

and 4c

. If the projection of b

on a

is equal to the

projection of c

on a

and b

is perpendicular to c

, then the value of cba

is …….

Ans. 6

Sol. a.ca.b

c.a–c.b–b.a2cbac–ba 2222

= 4 + 16 + 16 + 2 b.a–0–b.a

= 36

6c–ba

23. Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the number of elements in the set C = {f : A B | 2 f(A) and f is not one–one } is ….

Ans. 19

Sol. only '2' in range 1 function

1 2

3

4

a

b

c

one element out of 1,3,4, is in range with '2'

number of ways = 3C1. 18!2.!1!.2

!3

(Select one from 1, 3, 4 and distribute among a, b, c)

Total function = 1 + 18 = 19

24. If the lines x + y = a and x – y = b touch the curve y = x2 – 3x + 2 at the points where the curve intersects

the x–axis, then b

a is equal to ….

Ans. 0.5






PAGE # 12

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Sol. y = x2 – 3x + 2 , x + y = a, x – y = b

2x1 – 0 = 3 1 2x2 – 3 = – 1

x1 = 2 x2 = 1

x1 = 4 – 6 + 2 = 0 x2 = 0

(2,0) (1,0)

b = 2 a = 1

5.02

1

b

a

25. The coefficient of x4 in the expansion of (1 + x + x2 + x3)6 in powers of x, is …..

Ans. 49

Sol. Given times20......xlogxlogxlog4

1

3

1

2

1

777

= 460

xlog21.....432 7 = 460

xlog)212(2

207 = 460

xlog7 = 2

x = 49





7340010333


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