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COMPUTER BASED TEST (CBT) Questions & Solutions

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7340010333

FOR 2020–21



| JEE MAIN-2020 | DATE : 08-01-2020 (SHIFT-2) | PAPER-2 | OFFICIAL PAPER | MATHEMATICS




PAGE # 1

7340010333

PART : MATHEMATICS

SECTION – 1 : (Maximum Marks : 80)

Straight Objective Type (lh/ks oLrqfu"B izdkj)

This section contains 20 multiple 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 cgq&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 S be the set of all function f : [0,1] R , which are continuous on [0,1] and differentiable on (0,1) .

Then for every f in S, there exists a c (0,1) , depending on f, such that :

ekuk lHkh Qyuksa f : [0,1] R tks fd [0,1) ij larr gS rFkk (0,1] ij vodyuh; gSa] dk leqPp; S gSA rks S esa

izR;sd f ds fy, f ij fuHkZj ,d c (0,1) dk vfLrRo bl izdkj gS fd %

(1) |f(c) – f(1) | < (1 – c)| f'(c)| (2) |f(c) – f(1)| < |f'(c)|

(3) (1) – ( )

1–

f f c

c = f'(c) (4) |f(c) + f(1)| < (1 + c) |f'(c)|

Ans. Bonus

Sol. If we consider f(x) = constant function, then option (2) becomes 0 < 0 which is false so answer (2)

cannot be answer. But NTA given by Option (2)

2. The mirror image of the point (1,2,3) in a plane is 7 4 1

– ,– ,–3 3 3

. Which of the following points lies on

this plane ?

fcUnq (1,2,3) dk ,d lery esa izfrfcEc 7 4 1

– ,– ,–3 3 3

gSA fuEu esa ls dkSulk fcUnq bl lery ij fLFkr gS \

(1) (–1, –1, –1) (2) (1, –1,1)

(3) (–1, –1,1) (4) 1,1,1)

Ans. (2)

Sol. d.r of normal to the plane

lery ds yEcor~ f}dvuqikr

3

10,

3

10,

3

10

1, 1, 1

midpoint of P and Q is

3

4,

3

1,

3

2–

P rFkk Q dk e/; fcUnq

3

4,

3

1,

3

2–

equation of plane x + y + z = 1

lery dk lehdj.k x + y + z = 1







PAGE # 2

7340010333

3. If a hyperbola passes through the point P(10,16) and it has vertices at (±6,0) , then the equation of the

normal to it at P is :

;fn ,d vfrijoy; fcUnq P(10,16) ls gksdj tkrk gS rFkk blds 'kh"kZ (±6,0) ij gSa rks P ij blds vfHkyEc dk

lehdj.k gS &

(1) x + 3y = 58 (2) x + 2y = 42

(3) 3x + 4y = 94 (4) 2x + 5y = 100

Ans. (4)

Sol. Vertex is at (±6, 0)

a = 6

Let the hyperbola is 2

2

a

x –

2

2

b

y = 1

Putting point P(10, 16) on the hyperbola

36

100 –

2b

256 = 1 b2 = 144

hyperbola is 36

x2

– 144

y2

= 1

equation of normal is 1

2

x

xa +

1

2

y

yb = a2 + b2

putting we get 2x + 5y = 100

Sol. 'kh"kZ (±6, 0) gS

a = 6

ekuk vfrijoy;2

2

a

x –

2

2

b

y = 1

fcanq P(10, 16) vfrijoy; ij fLFkr gS

36

100 –

2b

256 = 1 b2 = 144

vfrijoy;36

x2

– 144

y2

= 1

vfHkyEc dk lehdj.k1

2

x

xa +

1

2

y

yb = a2 + b2

j[kus ij 2x + 5y = 100

4. Let S be the set of all real roots of the equation, 3x(3x – 1) + 2 = |3x –1| + |3x – 2|. Then S :

(1) contains exactly two elements (2) is a singleton

(3) contains at least four elements (4) is an empty set

ekuk lehdj.k 3x(3x – 1) + 2 = |3x –1| + |3x – 2| ds lHkh okLrfod ewyksa dk leqPp; S gSA rks S :

(1) ,d gh vo;o okyk leqPp; gSA (2) esa de ls de pkj vo;o gaaSA

(3) ,d fjDr leqPp; gSA (4) esa ek=k nks vo;o gSA

Ans. (2)







PAGE # 3

7340010333

Sol. Let 3x = t

t(t –1) + 2 = |t –1| + |t –2|

t2 – t + 2 = |t –1| + |t –2|

7/4

2

1 a 1 2

are positive solution

t = a

3x = a

x = alog3 so singleton set

Hindi. ekuk fd 3x = t

t(t –1) + 2 = |t –1| + |t –2|

t2 – t + 2 = |t –1| + |t –2|

7/4

2

1 a 1 2

/kukRed gy gSA

t = a

3x = a

x = alog3 ,dy leqPp;

5. Let a = i – ˆ2 j + k and b = i – j + k be two vectors. If c is a vector such that b c = b a and .c a

= 0, then .c b is equal to

ekuk nks lfn'k a = i – ˆ2 j + k rFkk b = i – j + k gSaA ;fn ,d lfn'k c bl izdkj gS fd b c = b a rFkk

.c a = 0 gSa] rks .c b cjkcj gS &

(1) 1

2 (2)

1–

2 (3) –1 (4)

3–

2

Ans. (2)

Sol. )cb(a

= )ab(a

– )b.a(

c

= ( a

. a

) b

– ( b.a

) a

– c4

= 6( kji ) – 4( kj2i )

– c4

= k2j2i2

c

= 2

1 ( kji )

c.b

= 2

1







PAGE # 4

7340010333

6. The system of linear equations

x + 2y + 2z = 5

2x + 3y + 5z = 8

4x + y + 6z = 10 has

(1) no solution when = 8 (2) a unique solution when = –8

(3) no solution when = 2 (4) infinitely many solution when = 2

jSf[kd lehdj.k fudk;

x + 2y + 2z = 5

2x + 3y + 5z = 8

4x + y + 6z = 10

(1) dk dksbZ gy ugha gS rc = 8 (2) dk dkbZ gy ugha gS tc = –8

(3) ds vuUr gy gSa tc = 2 (4) dk ek=k ,d gy gS tc = 2

Ans. (3)

Sol. D =

64

532

22

D = ( + 8) ( 2 – )

for = 2

D1 =

6210

538

225

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30] = 40 + 4 – 28 0

No solutions for = 2

7. The differential equation of the family of curves , x2 = 4b(y + b) , b R , is

oØksa x2 = 4b(y + b) , b R ds dqy dk vody lehdj.k gS &

(1) x(y')2 = x – 2yy' (2) xy'' = y' (3) x(y')2 = x + 2yy' (4) x(y')2 = 2yy' – x

Ans. (3)

Sol. 2x = 4by' 'y2

xb

So. differential equation is

2

''

2

y

xy.

y

x2x

vr% vody lehdj.k 2

''

2

y

xy.

y

x2x

x

dx

dyy2

dx

dyx

2

8. If the 10th term of an A.P. is 1

20 and its 20th term is

1

10, then the sum of its first 200 term is

;fn ,d lekUrj Js<+h dk 10 oka in 1

20 gS rFkk bldk 20 ok¡ in

1

10 gSa] rks blds izFke 200 inksa dk ;ksx gS &

(1) 100 (2) 1

504

(3) 1

1002

(4) 50

Ans. (3)







PAGE # 5

7340010333

Sol. T10 = 20

1 = a + 9d …………..(i)

T20 = 10

1 = a + 19d …………..(ii)

a = 200

1, d =

200

1 S200 =

2

200

200

199

200

2=

2

201 = 100

2

1

9. If a line, y = mx + c is a tangent to the circle, ( x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where

L1 is the tangent to the circle, x2 + y2 = 1 at the point 1 1

,2 2

; then :

;fn ,d js[kk y = mx + c, o`Ùk ( x – 3)2 + y2 = 1 dh ,d Li'kZ js[kk gS rFkk ;g ,d js[kk L1 ij yEc gSa] tgk¡ L1 o`Ùk

x2 + y2 = 1 ds fcUnq 1 1

,2 2

ij Li'kZ js[kk gS] rks %

(1) c2 + 7c + 6 = 0 (2) c2 + 6c + 7 = 0 (3) c2 – 6c + 7 = 0 (4) c2 – 7c + 6 = 0

Ans. (2)

Sol. Slope of tangent to x2 + y2 = 1 at

2

1,

2

1

x2 + y2 = 1 dh

2

1,

2

1ij Li'kZ js[kk dh izo.krk

x2 + y2 = 1

2x + 2yy = 0

y = – y

x = –1

y = mx + c is tangent of x2 + y2 = 1

y = mx + c o`Ùk x2 + y2 = 1 dh Li'kZ js[kk gS

so blfy;s m = 1

y = x + c

now distance of (3, 0) from y = x + c is

vc (3, 0) dh js[kk y = x + c ls nwjh gS

2

3c = 1

c2 + 6c + 9 = 2

c2 + 6c + 7 = 0







PAGE # 6

7340010333

10. 0

limx

x

dt)t10sin(t

X

0

is equal to

0

limx

x

dt)t10sin(t

X

0

cjkcj gS &

(1) 1

–10

(2) 1

10 (3)

1–

5 (4) 0

Ans. (4)

Sol. Using L’Hospital

L gkWfLiVy ls

0x

l im 1

)x01sin(x = 0

11. The area (in sq. units) of the region (x,y) R2 ; x2 y 3 – 2x}, is

{ks=k (x,y)R2 ; x2 y 3 – 2x} dk {ks=kQy ¼oxZ bZdkb;ksa esa½ gS &

(1) 29

3 (2)

34

3 (3)

32

3 (4)

31

3

Ans. (3)

Sol. Point of intersection of y = x2 & y = – 2x + 3 is

y = x2 rFkk y = – 2x + 3 dk izfrPNsnu fcUnq

obtained by izkIr gksrk gS] x2 + 2x – 3 = 0

x = – 3, 1

So, Area vr% {ks=kQy =

1

3

2 dx)xx23( = 3(4) – 2

2

31 22

–

3

31 33

= 12 + 8 – 3

28 =

3

32

–3 1

12. Let = –1 3

2

i if a = (1 + )

1002

0

k

k

and b =

1003

0

k

k

, then a and b are the roots of the quadratic

equation :

ekuk = –1 3

2

i gSA ;fn a = (1 + )

1002

0

k

k

rFkk b =

1003

0

k

k

, rks a rFkk b fuEu esa ls fdl f}?kkr lehdj.k

ds ewy gSa \

(1) x2 + 101x + 100 = 0 (2) x2 – 102x + 101 = 0

(3) x2 –101x + 100 = 0 (4) x2 + 102x + 101 = 0







PAGE # 7

7340010333

Ans. (2)

Sol. = b = 1 + 3 + 6 + ……= 101

a = (1 + ) (1 + 2 + 4 + …..198 + 200)

11

11

1

1

122

1012

Equation : lehdj.k x2 – (101 +1)x + (101) × 1 = 0 x2 – 102x + 101 = 0

13. If = 2

3 21 2 – 9 12 4

dx

x x x , then

;fn = 2

3 21 2 – 9 12 4

dx

x x x , rc

(1) 1

6< 2 <

1

2 (2)

1

16 < 2 <

1

9 (3)

1

8 < 2 <

1

4 (4)

1

9 < 2 <

1

8

Ans. (4)

Sol. f(x) = 4x12x9–x2

1

23

f'(x) = 2

1–

2

323

2

)4x12x9–x2(

)12x18–x6(

=

23

23 4x12x9–x22

2–x1–x6–

f(1) = 3

1, f(2) =

8

1

3

1 < I <

8

1

14. If and be the coefficient of x4 and x2 respectively in the expansion of

62 – 1x x + 62– –1x x , then :

;fn 62 – 1x x + 62– –1x x ds izlkj esa x4 rFkk x2 ds xq.kkad Øe'k% rFkk gSa ] rks

(1) + = –30 (2) – = 60 (3) – = – 132 (4) + = 60

Ans. (3)

Sol. 2[6C0 x6 + 6C2 x4 (x2 – 1) + 6C4 x2 (x2 – 1)2 + 6C6 (x2 – 1)3]

= 2[x6 + 15(x6 – x4) + 15x2(x4 – 2x2 + 1) + (–1 + 3x2 – 3x4 + x6)]

= 2(32x6 – 48x4 + 18x2 – 1)

= – 96 and vkSj = 36 – = –132







PAGE # 8

7340010333

15. Let f : (1,3) R be a function defined by f(x) = 2

[ ]

1

x x

x, where [x] denotes the greatest integer x. Then

the range of f is :

ekuk f : (1,3) R ,d Qy gS] tks f(x) = 2

[ ]

1

x x

x, }kjk ifjHkkf"kr gS tgk¡ [x] egÙke iw.kk±d x dks n'kkZrk gSA rks f

dk ifjlj gS &

(1) 3 4

,4 5

(2) 2 1

,5 2

3 4

,5 5

(3) 2 3

,5 5

3 4

,4 5

(4) 2 4

,5 5

Ans. (2)

Sol. f(x)

)3,2[x;1x

x2

)2,1(x;1x

x

2

2

f(x) is a decreasing function

f(x) âkleku Qyu gSA

y

2

1,

5

2

5

4,

10

6

y

2

1,

5

2

5

4,

5

3

16. The length of the perpendicular from the origin on the normal to the curve, x2 + 2xy – 3y2 = 0 at the

point (2,2) is

oØ x2 + 2xy – 3y2 = 0 ds fcUnq (2,2) [khpsa x;s vfHkyEc ij ewy fcUnq ls Mkys x;s yEc dh yEckbZ gS &

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

Ans. (3)

Sol. x2 + 2xy – 3y2 = 0

x2 + 3xy – xy – 3y2 = 0

(x – y) (x + 3y) = 0

x – y = 0 x + 3y = 0

(2, 2) satisfy x – y = 0 larq"V djrk gS

Normal vfHkyEc :

x + y = = 4

Hence bl izdkj x + y = 4

perpendicular distance from origin = 222

400

ewy fcanq ls yEcor~ nwjh = 222

400







PAGE # 9

7340010333

17. Let A and B be two events such that the probability that exactly one of them occurs is 2

5 and the

probability that A or B occurs is 1

2, then the probability of both of them occur together is

ekuk A rFkk B nks ?kVuk;sa blizdkj gS fd nksuksa esa ls ek=k ,d ds gksus dh izk;fdrk 2

5 gS rFkk A ;k B ds gksus dh

izkf;drk 1

2 gSa] rks nksuksa ds ,d lkFk gksus dh izkf;drk gS &

(1) 0.02 (2) 0.10 (3) 0.01 (4) 0.20

Ans. (2)

Sol. P(exactly one Bhd ,d) = 5

2

P(A) + P(B) – 2 BAP5

2

2

1BAP

P(A) + P(B) – BAP2

1

BAP10

1

10

4–5

5

2–

2

1

18. The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it

was found that an observation 9 was incorrect and the correct observation was 11. Then the correct

variance is :

20 izs{k.kksa ds ek/; rFkk izlj.k Øe'k% 10 rFkk 4 ik;s x;sA iqu% tk¡p djus ij ik;k x;k fd ,d izs{k.k 9 xyr Fkk

rFk lgh izs{k.k 11 FkkA rks lgh izlj.k gS &

(1) 3.99 (2) 4.02 (3) 4.01 (4) 3.98

Ans. (1)

Sol. 20

x i= 10 …….(i)

100–20

x2

i= 4 …….(ii)

2

ix = 104 × 20 = 2080

Actual mean lgh ek/; = 20

119–200 =

20

202

Variance pfjrk = 20

12181–2080 –

2

20

202

= 20

2120– (10.1)2 = 106 – 102.01 = 3.99







PAGE # 10

7340010333

19. If A = 2 2

9 4

and = 1 0

0 1

, then 10A–1 is equal to :

;fn A = 2 2

9 4

rFkk = 1 0

0 1

gSa] rks 10A–1 cjkcj gS &

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

Ans. (2)

Sol. Characteristics equation of matrix ‘A’ is

vkO;wg ‘A’ dh vfHkyk{kf.kd lehdj.k gS &

0x–49

2x–2 x2 – 6x – 10 = 0

A2 – 6A – 10 I = 0

10A–1 = A – 6I

20. Which of the following statement is a tautology ?

fuEu esa ls dkSulk dFku ,d iqu:fDr gS \

(1) ~(p ~q) p q (2) ~(p ~q) p q

(3) ~ (p ~q) p q (4) p (~q) p q

Ans. (2)

Sol. (~p q) (p q)

~{(~p q) (~p ~q)}

~{~p f}

SECTION – 2 : (Maximum Marks : 20)

This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit.

If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places.

Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 20)

bl [kaM esa ik¡p (05) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad rFkk n'keyo ,dy&vadu eas gSA

;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM vkWQ (truncate/round-off) djsaA

vadu ;kstuk : iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

21. Let a line y = mx (m > 0) intersect the parabola, y2 = x at a point P, other than the origin. Let the tangent

to it at P meet the x-axis at the point Q. If area (OPQ) = 4 sq. units, then m is equal to

ekuk ,d js[kk y = mx (m > 0) ijoy; y2 = x dks ewy fcUnq ds vfrfjDr ,d fcUnq P ij dkVrh gSA ekuk P ij

bldh Li'kZ js[kk x v{k dks fcUnq Q ij feyrh gSA ;fn OPQ dk {ks=kQy 4 oxZ bdkbZ gS rks m cjkcj gS &

Ans. 0.5







PAGE # 11

7340010333

Sol.

P(t2, t)

Q O

2ty = x + t2

Q(–t2, 0)

4

10t

1tt

100

2

1

2

2

8t3

t = ± 2 (t > 0) m = 2

1

22. Let f(x) be a polynomial of degree 3 such that f(–1) = 10, f(1) = – 6, f(x) has a critical point at x = –1 and

f'(x) has a critical point at x = 1. Then f(x) has a local minima at x =

ekuk ?kkr 3 dk ,d cgqin f(x) bl izdkj gS fd f(–1) = 10, f(1) = – 6, f(x) dk ,d Økafrd fcUnq x = –1 gSa rFkk

f'(x) dk ,d Økafrd fcUnq x = 1 gSA rks f(x) dk ,d LFkkuh; fuEufu"B gS x =

Ans. 3

Sol. f''(x) = (x –1)

f'(x) = 2( – 1)

2

x– 2

f(x) = 3( – 1)

6

x –2x + µ

f(1) = – 6 –2 + µ = – 6

f(–1) = 10 2

3

+µ = 10

= 6 and µ = 6

f'(x) = 0 x = – 1 or x = 3

x = 3 has a local minima

23. The sum, 7

1

( 1)(2 1)

4n

n n n

is equal to

;ksxQy 7

1

( 1)(2 1)

4n

n n n

cjkcj gS &

Ans. 504







PAGE # 12

7340010333

Sol.

7

1n

23 nn3n24

1

2

8.7

6

15.8.73

2

8.72

4

12

4

1[2 × 49 × 16 + 28 × 15 + 28]

4

1[1568 + 420 + 28] = 504

24. If 2 sin

1 cos2

= 1

7 and

1– cos2

2

=

1

10, , 0,

2

, then tan( + 2) is equal to

;fn 2 sin

1 cos2

= 1

7 rFkk

1– cos2

2

=

1

10, , 0,

2

, gSa] rks tan( + 2) cjkcj gS &

Ans. 1

Sol.

cos2

sin2=

7

1 and rFkk

2

sin2 =

10

1

tan = 7

1

sin = 10

1

tan = 3

1

tan2 =

9

1–1

3

1.2

=

9

83

2

= 4

3

tan ( + 2) =

2tantan1

2tantan=

4

3.

7

1–1

4

3

7

1

= 1

28

2528

214

25. The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the

word 'EXAMINATION' is

'kCn 'EXAMINATION' ds X;kjg v{kjksa ls cu ldus okys 4 v{kjksa ds 'kCnksa ¼vFkZ okys rFkk vFkZfoghu½ dh la[;k gS

&

Ans. 2454







PAGE # 13

7340010333

Sol. EXAMINATION

2N, 2A, 2I, E, X, M, T, O

Case I All are different so 16805.6.7.8!4

!8P4

8

Case II 2 same and 2 different so 75612.21.3!2

!4.C.C 2

71

3

Case III 2 same and 2 same so 186.3!2!.2

!4.C2

3

Total = 1680 + 756 + 18 = 2454

Hindi. EXAMINATION

2N, 2A, 2I, E, X, M, T, O

fLFkfr - I lHkh fHkUu&fHkUu gS 16805.6.7.8!4

!8P4

8

LFkfr -II 2 leku rFkk 2 fHkUu&fHkUu gS 75612.21.3!2

!4.C.C 2

71

3

LFkfr -III 2 leku rFkk 2 leku 186.3!2!.2

!4.C2

3

dqy = 1680 + 756 + 18 = 2454






7340010333



computer based test (cbt) questions & solutions · 2020-01-23 · this solution was download...

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