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Pioneer Education The Best Way To Success NTSE | Olympiad | JEE - Mains & Advanced Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 1 of 14 www.pioneermathematics.com JEE (Mains) Mathematics Solution 24.2.2021 (Shift-1) Section-I Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. 1. If 1 cos x sin x sin x cos x dx a sin c, b 8 sin2x where c is a constant of integration, then the ordered pair (a, b) is equal to: (a) (–1, 3) (b) (3, 1) (c) (1, – 3) (d) (1, 3) Ans. (d) Solution: 2 cos x sin x cos x sin x dx dx 8 sin2x 9 sin x cos x let sin x + cos x = t (cos x – sin x) dx = dt = 2 dt 9 t = 1 t sin c 3 = 1 sin x cos x sin c 3 Hence (a, b) = (1, 3) 2. The area (in sq. units) of the part of the circle x 2 + y 2 = 36, which is outside the parabola y 2 = 9x, is: (a) 12 33 (b) 24 33 (c) 12 33 (d) 24 33 Ans. (d) Solution: Area of the shaded region = 3 6 2 0 3 2 3 xdx 36 x dx = 3 6 3/2 2 1 0 3 1 x 2 2x x 36 x 18sin 2 6 = 9 3 26 3 9 3 33 12 2 Required area = 36 33 12 = 24 33

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JEE (Mains) Mathematics Solution

24.2.2021 (Shift-1)

Section-I

Multiple Choice Questions: This section contains 20 multiple choice questions. Each

question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

1. If 1cos x sin x sin x cos xdx a sin c,

b8 sin2x

where c is a constant of integration, then the

ordered pair (a, b) is equal to:

(a) (–1, 3) (b) (3, 1) (c) (1, – 3) (d) (1, 3)

Ans. (d)

Solution:

2

cosx sinx cosx sin xdx dx

8 sin2x 9 sinx cosx

let sin x + cos x = t

(cos x – sin x) dx = dt

= 2

dt

9 t

= 1 tsin c

3

= 1 sin x cos xsin c

3

Hence (a, b) = (1, 3)

2. The area (in sq. units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is:

(a) 12 3 3 (b) 24 3 3 (c) 12 3 3 (d) 24 3 3 Ans. (d)

Solution:

Area of the shaded region

= 3 6

2

0 32 3 xdx 36 x dx

=

3 6

3/2 2 1

0 3

1 x2 2x x 36 x 18sin

2 6

= 9 3

2 6 3 9 3 3 3 122

Required area = 36 3 3 12 = 24 3 3

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3. If 2 4 6

ecos x cos x cos x ... log 2e

satisfies the equation t2 – 9t + 8 = 0, then the value of

2sin x0 x is :

2sinx 3 cos x

(a) 3

2 (b) 3 (c) 2 3 (d)

1

2 Ans. (d)

Solution: 2

2 2

cos xln 2

1 cos x cot xe 2

t 1 or 8

So, 2cot x 0 3 22 2 or 2 cot x 0 or 3

x 0, then cot x 3 x2 6

12

2 sin x 12

2sinx 3 cos x 1 33.

2 2

4. The population P = P(t) at time ‘t’ of a certain species follows the differential equation

dP0.5P 450.

dt If P(0) = 850, then the time at which population becomes zero is:

(a) loge18 (b) e

1log 18

2 (c) loge 9 (d) 2loge 18

Ans. (d)

Solution:

dP 1

P 900dt 2

dP 1 1

dt ln P 900 t cP 900 2 2

When t = 0, P = 850 c ln 50

When P = 0, t = 2(ln 900 – ln 50) = 2ln 18

5. The statement among the following that is a tautology is:

(a) A A B (b) B [A A B ]

(c) [A A B ] B (d) A A B

Ans. (c)

Solution:

(a) A A B A

(b) A A B A ~ A B A B

So, B A B ~B A B ~B A

(c) A A B B A B B ~ A B

B ~ A ~B B (Tautology)

(d) A A B A

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6. Let p and q be two positive numbers such that p + q = 2 and p4 + q4 = 272.

Then p and q are roots of the equation:

(a) x2 – 2x + 2 = 0 (b) x2 – 2x + 8 = 0

(c) x2 – 2x + 136 = 0 (d) x2 – 2x + 16 = 0

Ans. (d)

Solution:

44 4 2 2 2 2p q p 4 4pq p q 6p q

272 = 16 – 4pq(4 – 2pq) – 6p2q2

2p2q2 – 16pq – 256 = 0

pq = – 8 or 16

p, q > 0, so pq = 16

Required quadratic equation is

x2 – 2x + 16 = 0

7. The system of linear equations

3x – 2y – kz = 10

2x – 4y – 2z = 6

x + 2y – z = 5m

is inconsistent if:

(a) 4

k 3, m5

(b) k 3, m R (c) 4

k 3, m5

(d) 4

k 3, m5

Ans. (d)

Solution:

Here

3 2 k

2 4 2 0

1 2 1

3(4 + 4) + 2(–2 + 2) – k(k + 4) = 0

24 0 8k 0 k 3

Now,

1

10 2 3

6 4 2 10 4 4 2 6 10m 3 12 20m

5m 2 1

= 80 – 12 + 20m – 36 – 60m = 32 – 40m

2

3 10 3

2 6 2 3 6 10m 10 2 2 3 10m 6

1 5m 1

= –18 + 30m + 0 – 30m + 18 = 0

3

3 2 10

2 4 6 3 20m 12 2 10m 6 10 4 4

1 2 5m

= –60m – 36 + 20m – 12 + 80

= – 40m + 32

For inconsistent we have k = 3, & 32 – 40m 4

0 m5

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8. A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least

2 Indians and double the number of foreigners as Indians. Then the number of ways, the

committee can be formed, is:

(a) 560 (b) 1050 (c) 1625 (d) 575

Ans. (c)

Solution:

Indians = 6, Foreigners = 8

According to questions

The no. of ways to form the committee are

(2I, 4F) or (3I, 6F) or (4I, 8F)

6 8 6 8 6 82 4 3 6 4 8C C C C C C

= 15 70 20 28 15 1 = 1625

9. The equation of the plane passing through the point (1, 2, – 3) and perpendicular to the planes

3x + y – 2z= 5 and 2x – 5y – z = 7, is:

(a) 3x – 10y – 2z+ 11 = 0 (b) 6x – y – 2z – 2 = 0

(c) 6x – 5y + 2z+ 10 = 0 (d) 11x + y + 17z + 38 = 0

Ans. (d)

Solution:

The given planes are 3x + y – 2z = 5 ….(1)

2x – 5y – z = 7 ….(2)

Since the required plane passes through (1, 2, – 3)

So equation of this plane is

a(x – 1) + b(y – 2) + c(z +3) = 0 …(3)

Now this plane (3) is to the planes (1) & (2)

So, 3a + b – 2c = 0 & 2a – 5b – c = 0

a b c

11 1 17

So equation of plane is 11(x –1) + (y – 2) + 17(2 + 3) = 0

11x + y + 17z + 38 = 0

10. If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q,

then the ordinate of the point which divides PQ internally in the ratio 1 : 2 is:

(a) –2t3 (b) 2t3 (c) 0 (d) –t3

Ans. (a)

Solution:

Curve is y = x3 ….(1)

So equation of tangent at (t, t3)

(y – t3) = 3t2(x – t) …(2)

It meets again the curve at Q

So solving (1) & (2) we get

x = – 2t Q = (–2t, –8t3)

Now by section formula ordinate = 3 32t 8t

1 2

= 36t

3

= –2t3

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11. Let f: R R be defined as f(x) = 2x – 1 and g : R – {1} – R be defined as g(x) =

1x

2 .x 1

Then the composition function f(g(x)) is:

(a) neither one-one nor onto (b) onto but not one-one

(c) both one-one and onto (d) one-one but not onto

Ans. (d)

Solution:

Here f: R R, f x 2x 1

and g:

1x

2R {1} R g xx 1

So, f(g(x)) = 2 g(x) – 1

=

1x

22 1x 1

= 2x 1 x 1 x 1 1

x 1 x 1

= 1

1x 1

So clearly it is one-one but not onto

12.

2x

03x 0

sin t dt

limx

is equal to:

(a) 0 (b) 1

15 (c)

2

3 (d)

3

2 Ans. (c)

Solution:

2x

03x 0

sin t dt0

lim formx 0

By D, L Hospital rule

2x 0 x 0

2x sin x 2 sin xlim lim

3x 3 x

= 2 2

13 3

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13. The ordinary dice is rolled for a certain number of times. If the probability of getting an odd

number 2 times is equal to the probability of getting an even number 3 times, then the probability

of getting an odd number for odd number of times is:

(a) 3

16 (b)

1

2 (c)

1

32 (d)

5

16 Ans. (b)

Solution:

Let number of trials be ‘n’

given 2 n 2 3 n 3

n n2 3

1 1 1 1C C

2 2 2 2

n = 5

Probability of getting odd number for odd number of times is

= 1 3 5 5

15C 5C 5C

2

= 4

5

2 1

2 2

14. Two vertical poles are 150m apart and the height of one is three times that of the other. If from

the middle point of the line joining their feet, an observed finds the angles of elevation of their

tops to be complementary, then the height of the shorter pole (in meters) is:

(a) 25 3 (b) 30 (c) 25 (d) 20 3 Ans. (a)

Solution:

Given 3h

tan75

….(i)

and h

tan 9075

….(ii)

Multiplying (i) and (ii) we get,

2

2

3h1

75

h 25 3

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15. If f: R R is a function defined by f(x) = [x – 1] cos 2x 1

,2

where [.] denotes the greatest

integer function, then f is:

(a) discontinuous only at x = 1

(b) continuous for every real x

(c) discontinuous at all integral values of x except at x = 1

(d) continuous only at x = 1

Ans. (b)

Solution:

2x 1

f x x 1 cos2

at x = 1

x 1

2x 1lim x 1 cos 0

2

x 1

2x 1lim x 1 cos 0

2

f(1) = 0

at any general integer x = k

x K

2k 1lim x 1 cos 0

2

x K

2k 1lim x 1 cos 0

2

f(k) = 0

f(x) is continuous x R

16. The distance of the point (1, 1, 9) from the point of intersection of the line x 3 y 4 z 5

1 2 2

and the plane x + y + z = 17 is:

(a) 38 (b) 19 2 (c) 2 19 (d) 38 Ans. (d)

Solution:

Let a point P on the line

x 3 y 4 z 5

1 2 2

P 3, 2 4, 2 5

as P also satisfies the given plane

3 2 4 2 5 17

5 5 1

P 4, 6,7

Distance from (1, 1, 9) is

2 2 2

4 1 6 1 7 9

= 9 25 4 38

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17. The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a

moving point of the parabola, is another parabola whose directrix is

(a) x = a (b) x = a

2 (c) x = 0 (d) x =

a

2

Ans. (c)

Solution:

Let the moving point be P(at2, 2at)

Focus of given parabola is (a, 0)

Let point of required locus (h, k)

2at a

h2

…(i)

and 2at 0

k2

…(ii)

2at 1 h

2 …(iii)

and k

ta

…(iv)

By (iii) and (iv) we have 2

2

a k1 h

2 a

Locus of k2 + a2 = 2ah

2 ay 2a x

2

Equation of directrix x – a a

02 2

x 0

18. A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this

line on the coordinate axes is 1

.4

Three stones A, B and C are placed at the points (1, 1), (2, 2) and

(4, 4) respectively. Then which of these stones is/are on the path of the man? (a) C only (b) B only (c) All the three (d) A only Ans. (b) Solution:

Let line be x y

1a b …(i)

given

1 11a b

2 4

1 1 1

a b 2 …(ii)

By (i) and (ii), we get x 1 1

y 1a 2 a

y

x y 1 02

Represents family of line passing through (2, 2)

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19. The function

3 24x 3x

f x 2 sinx 2x 1 cos x :6

(a) Decreases in 1

,2

(b) Decreases in

1,2

(c) Increases in 1

,2

(d) Increases in 1

,2

Ans. (d)

Solution:

3 24x 3x

f x 2 sinx 2x 1 cos x6

On differentiating both sides w. r. t. x we get f’(x) = 212x 6x

2 cos x 2x 1 sin x 2 cos x6

2f ' x 2x x 2x 1 sin x

f ' x 2x 1 x sin x

When 1

x , 2x 1 0 and x sin x 0.2

1

f ' x 0 if x2

1

f x is increasing in ,2

20. The value of

–15C1+ 2. 15C2 – 3. 15C3 + …. –15. 15C15 + 14C1 + 14C3 + 14C5 + … + 14C11 is

(a) 214 (b) 216 – 1 (c) 213 – 13 (d) 213 – 14

Ans. (d)

Solution:

14 14 14 14 2 14 14

0 1 2 141 x C C x C x ... C x …(i)

14 14 14 14 140 1 2 142 C C C .... C …(ii)

14 14 14 140 1 2 140 C C C ... C …(iii)

14 14 14 131 3 13C C .... C 2 …(iv)

and (1 – x)15 = 15C0 – 15C1x + 15C2x2 + …15C15x15

Differentiate w. r. t. x we get

–15(1 – x)14 = –15C1 + 2. 15C2x + ….–15. 15C15x14

Put x = 1, we get

–15C1 + 2. 15C2 – 3. 15C3 + …. –15.15C15 = 0 …(v)

From equation (iv) + equation (v) we get

–15C1 + 2. 15C2 + ….–1515C15 + 14C1 + 14Cr + … + 14C11

= 213 – 14C13

= 213 – 14

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Section-II

Numerical Value Type Questions: This section contains 10 questions. In section II, attempt any

five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question,

enter the correct numerical value (in decimal notation, truncated/.rounded-off to the second

decimal place; e.g. 06.25, 07.00, -00.33, -00.30, 30.27, -27.30) using the mouse and the on-screen

virtual numeric keypad in the place designated to enter the answer.

1. Let A = { n N: n is a 3-digit number}

B = {9k + 2: k N}

and C = {9k + I: k N} for some I (0 < I < 9}

If the sum of all the elements of the set A B C is 274 400, then I is equal to _______ .

Ans. 5

Solution:

Sum of all elements of A B C is

99

k 0

274 400 99 9k I 99 9k 2

99 100

274 400 200 100 100I 18.2

274 4 200 I 9 99

I 5

2. If a

a

(| x| | x 2|)dx 22, (a 2)

and [x] denotes the greatest integer x, then a

a

(x [x])dx is equal to

__________ .

Ans. 3

Solution:

a

2 22 2

a

1 1 1 1x x 2 dx a a a 2 a 2

2 2 2 2

222 2a 4 a 3

Now, 3 3 2 2

3 3 3 3

(x [x]dx [x]dx [x]dx ( 3)dx

3. Let P =

3 1 2

2 0 ,

3 5 0

where R. Suppose ijQ [a ] is a matrix satisfying PQ = KI3 for some

non-zero k R. If 23

kq

8 and

2k|Q| ,

2 then 2 2k is equal to _________ .

Ans. 17

Solution:

2

1 3 kQ k.P and |P||Q| k , |Q|

2 then |P| 2k

3223

kCq

|P| (Where ijC is co-factor of ijP of P)

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3 4 kk k3 4

8 2k 4

…(1)

Also |P| 2k 12 20 2k

k 6 10 …(2)

From (1) and (2) we get

k = 4 and 1

then 2 2k 17

4. If one of the diameters of the circle x2 + y2 – 2x – 6y + 6 = 0 is a chord of another circle ‘C’, whose

center is at (2, 1), then its radius is __________ .

Ans. 3

Solution:

Circle x2 + y2 – 2x– 6y + 6 = 0 has centre O, (1, 3) and radius r1 = 2.

Let centre O2(2, 1) of required circle and its radius being R.

So 2 2 21 2R O O r

2R 5 4

R = 3

5. Let Bi (i = 1, 2, 3) be three independent events in a sample space. The probability that only B1

occurs is , only B2 occurs is and only B3 occurs is . Let p be the probability that none of the

events Bi occurs and these 4 probabilities satisfy the equations 2 p and 3 p 2

(All the probabilities are assumed to lie in the interval (0, 1)). Then

1

3

P B

P Bis equal to _______ .

Ans. 6

Solution:

Let P(B1) = x, P(B2) = y, P(B3) = z

2 3 2 31 1P B B B P B P B P B

x 1 y 1 z …(i)

Similarly 1 x y 1 z …(ii)

1 x 1 y z …(iii)

p 1 x 1 y 1 z …(iv)

(i) & (iv) x

x1 x p p

(iii) & (iv) z

z1 z p p

1

3

p p1

P B x pp pP B z 1

p

…(v)

Given that,

2 p 2 …(vi)

3 p 2 3 p p 2 …(vii)

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(vi) & (vii) 2p

3 p 2

p 6p 5

p 6p5

p p1 6 1

…(viii)

(v) & (viii)

1

3

P B6

P B

6. If the least and the largest real values of , for which the equation

z + z 1 2i 0 z C and i 1 has a solution, are p and q respectively, then 2 24 p q

is equal to _____________

Ans. 10

Solution:

Let z = x + iy, x, R

z z 1 2i 0.

2 2x i y 2 x 1 y 0.

2

y 2 and x x 1 4 0.

Now 2

x 1 4 x *

Squaring both sides : 22 2x x 1 4

2 2 2 21 x 2 x 5 0.

For real x, D 0.

4 2 24 4.5 1 0.

2 45 4 0.

5 5

,2 2

min max.

5 5p and q

2 2

2 24 p q 10.

*The step is not permissible as squaring may include extraneous roots.

No maximum value of exists as is shown in the below procedure.

Let z = x + iy

2 2x x 1 y i y 2 0

2

xy 2 and

x 2x 5

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2 2

3 32 22 2

x x x 2x 5d x 5

dxx 2x 5 x 2x 5

So is decreasing in , 5 and increasing in 5,

min

5 51 p at x 5

420

and max 2x

xlim 1 q 1

x 2x 5

(however this value is not achievable.)

7. The minimum value of for which the equation 4 1

sin x 1 sin x

has at least one solution in

0, is_____ .2

Ans. 9

Solution:

Let f(x) = 4 1

sinx 1 sin x

where sin x 0, 1

2 2 13sinx sinx 1 sinx

sin x sinx3 1 sinx2 2

f x 9

So least value of is 9.

8. Let M be any 3 3 matrix with entries from the set (0, 1, 2). The maximum number of such

matrices, for which the sum of diagonal elements of MTM is seven is ___________ .

Ans. 540

Solution:

Let ij 3 3a

3 3

T 2r ij

i 1 j 1

T M .M a 7

So there will be two cases.

Case I: Any seven ija s are 1 and remaining two elements are zero.

Number of such matrices M = 9

367|2

Case II: Any one elements is 2, any three elements are 1 and reaming elements are 0.

Number of such matrices = 9

5041 3 5

Total number of possible matrices M = 540.

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9. n

1

2nr 1

1lim tan tan

1 r r

is equal to _____________.

Ans. 1

Solution:

1 1

2

r 1 r1tan tan

1 r r 1 r 1 r

= tan–1(r + 1) –tan–1 r

So n

1 1 1

2r 1

1tan tan 2 tan 1

1 r r

+ (tan–13 – tan–12) + …… + (tan–1(n + 1) – tan–1n)

= tan–1(n + 1) – tan–11

n

1

2nr 1

1lim tan tan

1 r r

1 1

nlim tan tan n 1 tan 1

= tan 12 4

10. Let three vectors a, b and c be such that c is coplanar with a and b, a. c 7 and b is

perpendicular to c, where a i j k and b 2i k , then the value of 2

2 a b c is ____ .

Ans. 75

Solution:

Let c a b a, b and c are coplanar

a. c 7, b. c 0 and a. b 1

So 2

7 a a. b 3 7

and 2

0 a. b b 5 0

Clearly 5 1 5a b

and c2 2 2

So

22 7a 3b

2 a b c 22

= 21

i 7j 10k2

= 1 49 100

752