Pioneer Education JEE (Mains) Mathematics Solution

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JEE (Mains) Mathematics Solution

26.2.2021 (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 a and b are perpendicular, then a a a a b is equal to:

(a) a b (b) 0 (c) 41

a b2

(d) 4

a b

Ans. (d)

Solution:

Let c be a unit vector in the direction of a b.

a b c, b c a & c a b

a b a b c

2

a a b a b b

3

a a a b a b c

4

a a a a b a b b

= 4

a b

2. The value of 22

x

2

cos xdx is :

1 3

(a) 4

(b) 2 (c)

2

(d) 4

Ans. (a)

Solution:

a a

a 0

f x dx f x f a x dx

22 22 2

x x x

0

2

cos xcos x cos xdx dx

1 3 1 3 1 3

= x2 2

2 2

x x

0 0

1 3cos x dx cos xdx

1 3 1 3

= 2

0

11 cos2x dx

2 4

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3. The value of

a 1 a 2 a 2 1

a 2 a 3 a 3 1 is :

a 3 a 4 a 4 1

(a) 0 (b) (a + 2) (a + 3) (a + 4)

(c) – 2 (d) (a + 1) (a + 2) (a + 3)

Ans. (c)

Solution:

Given determinant is 2

2

a 3a 2 a 2 1

D a 5a 6 a 3 1

a 7a 12 a 4 1

3 3 2 2 2 1R R R ; R R R

=

2a 3a 2 a 2 1

2a 4 1 0

2a 6 1 0

Expanding by C3

D = (2a + 4) – (2a + 6) = – 2

4. The maximum slope of the curve y = 4 3 21x 5x 18x 19x

2 occurs at the point:

(a) (2, 2) (b) (0, 0) (c) 21

3,2

(d) (2, 9)

Ans. (a)

Solution:

y = 4 3 21x 5x 18x 19x

2

Slope = y’ = 2x3 – 15x2 + 36x – 19 = g(x) say

g’(x) = 6x2 – 30x + 36 = 6(x – 2) (x – 3)

g’(x) = 0 x 2, 3

Slope g(x) has local maximum at x = 2

x = 2 y = 2

Local maximum at (2, 2)

[Note: Overall maximum (Absolute maximum) value of slope is far greater than that at (2, 2)].

5. In an increasing geometric series, the sum of the second and the sixth term is 25

2and the product

of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th term is equal to:

(a) 26 (b) 35 (c) 30 (d) 32

Ans. (b)

Solution:

2 6

25a a

2

23 5 2 6 4a a 25 a a a

24 4a 25 a 5





a2 & a6 are roots of 2 25x x 25 0

2

5x , 10

2

2 6

5a , a 10

2 GP is increasing

a4 = 5

2 2 24 2

5a a r 5 r r 2

2

28 6a a r 10 2 20

4 6 8a a a 5 10 20 35

6. The number of seven digit integers with sum of the digits equal to 10 and formed by using the

digits 1, 2 and 3 only is:

(a) 77 (b) 42 (c) 82 (d) 35

Ans. (a)

Solution:

Combination of digits

3, 2, 1, 1, 1, 1, 1 7!

425!

2, 2, 2, 1, 1, 1, 1 7!

354!3!

Total = 42 + 35 = 77 7. The sum of infinite series

2 3 4 5

2 7 12 17 221 .....

3 3 3 3 3 is equal to:

(a) 13

4 (b)

9

4 (c)

11

4 (d)

15

4 Ans. (a) Solution:

2 3 4 5

2 7 12 17 22S 1 .....

3 3 3 3 3

2 3 4 5

1 1 2 7 12 17S ......

3 3 3 3 3 3

2 3 4 5

2 1 5 5 5 5S 1 ......

3 3 3 3 3 3

=

5 54 59 9

1 23 313 3

2 4 5 13

S3 3 6 6

13S

4





8. Consider the three planes

P1: 3x + 15y + 21z = 9,

P2: x – 3y – z= 5, and

P3: 2x + 10y + 14z = 5

Then, which one of the following is true?

(a) P2 and P3 are parallel (b) P1 and P3 are parallel

(c) P1 and P2 are parallel (d) P1, P2 are P3 all are parallel

Ans. (b)

Solution:

Ratios of DRs of normals of P1 & P3 are 3 15 21

2 10 14

Normals are parallel

1 3P || P

9. Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of

A2 is 1, then the possible number of such matrices is:

(a) 6 (b) 1 (c) 4 (d) 12

Ans. (c)

Solution:

Let a c

Ac b

2 22

2 2

a c a c a c ac bcA

c b c b ac bc c b

2 2 2a b 2c 1 as a, b, c z

c = 0 and a, b = 1

Total 4 matrices are possible

10. The maximum value of the term independent of ‘t’ in the expansion of

1011 105

1 xtx

t

where x 0, 1 is :

(a)

2

2.10!

3 5! (b)

2

2.10!

3 3 5! (c)

2

10!

3 5! (d)

2

10!

2 5!

Ans. (b)

Solution:

r1

1 1010 10 r5

r 1 r

1 xT C (tx )

t

For term independent of f

10 – r – r = 10 r 5

1

10 26 5T C x 1 x f x

Let





110 2

5 1

2

xf ' x C 1 x 0

2 1 x

2 – 2x = x 2

x3

2

f '' x 0 at x3

1

210

56 max 2

2 1 2.10!T C .

3 3 5! 3 2

11. A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to

probability of getting 9 heads, then the probability of getting 2 heads is:

(a) 13

15

2 (b)

12

15

2 (c)

8

15

2 (d)

14

15

2 Ans. (a)

Solution:

Let n number of tosses

Given, 7 n 7 9 n 9

n n7 9

1 1 1 1C C

2 2 2 2

n 16

Probability of getting 2 heads = 16

162

1C

2

= 13

15

2

12. The value of n100

x [x]

n 1 n 1

e dx,

where [x] is the greatest integer x, is :

(a) 100(e – 1) (b) 100(1 – e) (c) 100e (d) 100(1 + e)

Ans. (a)

Solution:

n 1

x x x

n 1 0

e dx e dx e 1

100

n 1

e 1 100 e 1





13. In the circle given below, let OA = 1 unit, OB = 13 unit and PQ OB. Then, the area of the triangle

PQB (in square units) is:

(a) 24 2 (b) 24 3 (c) 26 3 (d) 26 2 Ans. (b)

Solution:

Assume that OB is diameter of the given circle

Using Ptolemy’s Theorem,

OP, QB + OQ. PB = PQ OB

2OP. PB 13PQ

Also PA2 = OP2 – 1 = PB2 – 122

2 2PB OP 143

and OP2 + PB2 = 132

then PB2 = 156 and OP2 = 13

So, 2 13. 156

PQ 4 313

Area of 1

PQB .4 3. 12 24 32

14. The value of h 0

3 sin h cos h6 6

lim2 is3h 3cosh sinh

(a) 4

3 (b)

3

4 (c)

2

3 (d)

2

3 Ans. (a)

Solution:

h 0

3 1sin h cos h

2 6 2 6lim 2

3 13h cosh sin h

2 2

h 0

sin h 2 2 4lim 2 .

33 33h sin h3





15. Let R = {(P, Q) |P and Q are at the same distance from the origin} be a relation, then the

equivalence class of (1, – 1) is the set:

(a) S = 2 2{(x, y)| x y 2} (b) 2 2S {(x, y)|x y 1}

(c) 2 2S {(x,y)| x y 2} (d) S = 2 2{(x,y)| x y 4}

Ans. (a)

Solution:

R = {P, Q) | P and Q are at the same distance from the origin}.

Then equivalence class of (1, –1) will contain all such points which lies on circumference of the

circle of centre at origin and passing through point (1, – 1).

i.e., radius of circle = 2 21 1 2

Required equivalence class of (S)

= 2 2{(x, y)|x y 2}.

16. The rate of growth of bacteria in a culture is proportional to the number of bacteria present and

the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2

hours. If the population of bacteria is 2000 after

e

k

6log

5

hours, then

2

e

k

log 2

is equal to:

(a) 16 (b) 4 (c) 8 (d) 2

Ans. (b)

Solution:

At t = 0 Bo = 1000

dBB

dt

o

o

1.2B 2

B 0

dBkt

B [Given]

o

o

1.2Bln 2k

B

1

k ln 1.22

To find time when B = 2000

o

o

2B t

B 0

dB 1ln 1.2 dt

B 2

1

ln 2 ln 1.2 t2

ln4

t hrs.6

ln5

R ln 4

Thus 2

2K2 4

ln





17. Let f be any function defined on R and let it satisfy the condition:

2

f x f y x y , x, y R

If f(0) = 1, then:

(a) f(x) can take any value in R (b) f(x) < 0, x R

(c) f(x) = 0, x R (d) f(x) > 0, x R

Ans. (d)

Solution:

2

f x f y x y

f x f y

x yx y

x y x y

f x f yLim Lim x y

x y

f ' x 0

f ' x 0

f(x) is constant function.

f(0) = 1 then f(x) = 1

18. If 1 1 1sin x cos x tan y

;a b c

0 < x < 1, then the value of cosc

is :a b

(a) 1 – y2 (b) 21 y

y y

(c)

2

2

1 y

1 y

(d)

21 y

2y

Ans. (c)

Solution:

1 1 1sin x cos x tan y

k (say)a b c

sin–1 x = ak, cos–1x = bk and tan–1y = ck

Now,

1 1sin x cos x2

a b x2

k2 a b

Now

1 ctan y

2 a b

1ccos cos 2tan y

ab

= 2

1

2

1 ycos cos

1 y

[if y > 0]

= 2

2

1 y

1 y





19. The intersection of three lines x – y = 0, x + 2y = 3 and 2x + y = 6 is a:

(a) None of the above (b) Isosceles triangle

(c) Right angled triangle (d) Equilateral triangle

Ans. (b)

Solution:

The given three lines are x – y = 0, x + 2y = 3 and 2x + y = 6 then point of intersection

lines x – y = 0 and x + 2y = 3 is (1, 1)

lines x – y = 0 and 2x + y = 6 is (2, 2)

and lines x +2y = 3 and 2x + y = 0 is (3, 0)

The triangle ABC has vertices A(1, 1), B(2, 2) and C(3, 0)

AB 2, BC 5 and AC 5

ABC is isosceles

20. If (1, 5, 35), (7, 5, 5), (1, , 7 ) and (2 , 1, 2) are coplanar, then the sum of all possible values of

is

(a) 44

5 (b) –

44

5 (c)

39

5 (d) –

39

5 Ans. (a)

Solution:

Four points (1, 5, 35), (7, 5, 5), (1, , 7) and (2 , 1, 2) are coplanar then

6 0 30

0 5 28 0

2 1 4 33

3 3 1

6 0 0

0 5 28 R C 5 C 0

2 1 4 33

6 5 10 38 112 0

210 88 78 0

25 44 39 0

Sum of all possible values of 44

5





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. The number of integral values of ‘k’ for which the equation 3 sin x + 4 cos x = k + 1 has a solution,

k R is ________ . Ans. 11

Solution:

3 sin x + 4 cos x = k + 1 has a solution then

k 1 [ 5, 5]

k [ 6, 4]

Number of possible integral values of k = 11.

2. The value of the integral 0

sin2xdx is________ .

Ans. 2

Solution:

0

sin2x dx

= 2

0

2

sin2xdx sin2xdx

= 2

02

cos 2x cos 2x

2 2

= 1 1 1 1

2 2 2 2

= 2

3. If 23 cos x 3 1 cos x 1, the number of solutions of the given equation when

x 0, is____ .2

Ans. 1

Solution:

23cos x 3 1 cosx 1

23cos x 3cosx cos x 1 0

3cos x cosx 1 cos x 1 0

cosx 1 3 cosx 1 0

1

cos x 1 or3

Number of solution in x 0, is 1.2





4. The sum of 162th power of the roots of the equation x3 – 2x2 + 2x – 1 = 0 is ____________ .

Ans. 3

Solution:

x3 – 1 + 2x – 2x2 = 0

2x 1 x x 1 0

2x 1, ,

162

162 162 2S 1

= 1 + 1 + 1 = 3

5. The difference between degree and order of a differential equation that represents the family of

curves given by y2 = a a

x , a 0 is ____ .2

Ans. 2

Solution:

2 ay a x

2

…(1)

dy2y. a

dx …(2)

From (1) and (2)

2 dy 1 dyy 2y x 2y

dx 2 dx

2 dy dy dyy 2x y . 2y

dx dx dx

2 3

3dy dyy 2x 2y

dx dx

Order 1 and degree 3.

6. If y = y(x) is the solution of the equation sin y sin ydye cos y e cos x cos x, y(0) 0;

dx

then 3 1

1 y y y6 2 3 42

is equal to ___________.

Ans. 1

Solution:

sin y sin ydye .cosx e .cosx cos x

dx

Let esin y = Y

dY

Ycos x cos xdx

I.F. = sin xe

sin x sin xY. e e .cos dx c

siny sinx sinxe .e e c

When x = 0, y = 0 then c = 0

sinx sin y sinx sin ye e 0





y 0

y x 0

hence, 3 1

1 y y y 16 2 3 42

7. Let , 2, 1 be a point on the plane which passes through the point (4, –2, 2). If the plane is

perpendicular to the line joining the points (–2, –21, 29) and (–1, –16, 23), then 2

44

11 11

is

equal to ____________ .

Ans. 8

Solution:

Vector perpendicular to the plane is

n i 5j 6k.

Given A ,2, 1 and B 4, 2, 2

AB n, so

4 5 4 6 1 0

4 20 6 0

22

211

hence 2

4 4 811 11

8. The number of solutions of the equation log4(x – 1) = log2(x – 3) is ____________ .

Ans. 1

Solution:

Domain: x – 1 > 0 and x – 3 > 0

x 3,

log4 (x – 1) = log2 (x – 3)

2

x 1 x 3

2x 7x 8 0

7 17

x2

but only 7 17

2

is the correct answer.





9. Let m, n N and gcd (2, n) = 1, if

m30 30 30 3030 29 ...... 2 1 n.2 ,

0 1 28 29

then n + m is equal to _________ . n

k

nHere C

k

Ans. 45

Solution:

29

30r

r 0

30 r . C

= 30 30

30 3030 r r

r 1 r 1

r. C r. C

= 30

29 29r 1

r 1

30 C 30.2

= 3015.2

Clearly n = 15, m = 30

and m + n = 45

10. The area bounded by the lines y x 1 2 is _____ .

Ans. 4

Solution:

Area of the shaded region = 1

4 2 42

* As per given answer key the equation in the question should be y x 1 2


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Pioneer Education JEE (Mains) Mathematics Solution