MATHEMATICS SECTION-A. sin x 1 cos x cos2x f(x) 1 sin x cos x cos2x, x R sin x cos x sin 2x. f(x) {x} {x}

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) MATHEMATICS

SECTION-A 1. The maximum value of is:

2 2

sin x 1 cos x cos2x f(x) 1 sin x cos x cos2x , x R

sin x cos x sin 2x

(1) 7 (2) ³

4 (3) 5 (4) 5

Ans. (3)

Sol. C1 C1 + C2 f(x) =

2 2 2

2 sin x cos2x 2 1 sin x cos2x 1 sin x sin 2x R2 R2 – R1

f(x) =

2

2 sin x cos2x

0 1 0

1 sin x sin 2x f(x) = 2 sin 2x – cos 2x f(x)max = 5

2. Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :

(1) ⁹

56 (2) ⁴

9 (3) ³

7 (4) ¹¹

27 Ans. (2)

Sol. n(S) = 6 × 6!

P(E) = 6! 2 (5 5!) 6 6!

= 6 10 36

= 4 9

3. Let R be such that the function

1 2 1

3

cos (1 {x} )sin (1 {x}),x 0

f(x) {x} {x}

, x 0

is continuous at x = 0, where {x} = x – [x], [x] is the greatest integer less than or equal to x. Then :

(1) 2 (2) = 0 (3) no such exists (4)

4 Ans. (3)

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II)

Sol. RHL =

xlim0

1 2 –1

2

cos (1 x )sin (1 x)

x(1 x )

=

2 ^xlim⁰ ^{cos (1 x )}¹ ² x

= 2 ^xlim ⁰

2 2

1

1 (1 x ) (–2x) (L Hospital Rule)

= xlim0

2 4

x 2x x

= xlim0

2

1 2 x =

2 LHL =

xlim0

1 2 –1

3

cos (1 (1 x) )sin ( x)

(1 x) (1 x) =

2 ^xlim⁰

1 2

sin x

(1 x)[(1 x) 1] =

2 ^xlim⁰ ^{sin x}₂ ¹ x 2x

= 2 1 2 =

4

As LHL RHL so f(x) is not continuous at x = 0

4. If (x, y, z) be an arbitrary point lying on a plane P which passes through the point (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of expression 3 ^{x 11}₂ ₂ ^{y 19}₂ ₂

(y 19) (z 12) (x 11) (z 12)

2 2

z 12 x y z

14(x 11)(y 19)(z 12) (x 11) (y 19)

(1) 0 (2) 3 (3) 39 (4) –45

Ans. (2)

Sol. equation of plane is x + y + z = 42 (x – 11) + (y – 19) + (z – 12) = 0 Let x – 11 = X, y – 19 = Y, z – 12 = Z

X + Y + Z = 0 X = Y = Z or X³ + Y³ + Z³ = 3XYZ 3 + _{2 2}^X ^Y₂ ₂ ₂^Z ₂ – ⁴²

14(X)(Y)(Z)

Y Z X Z X Y

3 +

3 3 3

2

X Y Z – 3XYZ XYZ

3 + 0 = 3

5. Consider the integral ¹⁰ _{x 1}^[x]

0

[x] e

I dx,

e where [x] denotes the greatest integer less than or equal to x.

Then the value of I is equal to:

(1) 9(e – 1) (2) 45(e + 1) (3) 45(e – 1) (4) 9(e + 1) Ans. (3)

Sol.

10 [x] 1 x

I [x].e dx

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2 3 4 10

2 x 3 x 4 x 10 x

1 2 3 9

e dx 2.e dx 3.e dx .... 9e dx

= –{(1 – e) +2 (1 – e) + 3 (1 – e) + ... + 9 (1 – e)}

= 45(e – 1)

6. Let C be the locus of the mirror image of a point on the parabola y² = 4x with respect to the line y

= x. Then the equation of tangent to C at P(2,1) is :

(1) x – y = 1 (2) 2x + y = 5 (3) x + 3y = 5 (4) x + 2y = 4 Ans. (1)

Sol.

y = x

Image of y² = 4x w.r.t. y = x is x² = 4y tangent from (2, 1)

xx1 = 2 (y + y1) 2x = 2(y + 1) x = y + 1

7. If y = y(x) is the solution of the differential equation ^dy (tanx) y sinx, 0 x

dx 3, with y(0) = 0,

theny

4 equal to : (1) ¹log 2_e

4 (2) ¹ log 2_e

2 2 (3) loge2 (4) ¹log 2_e

2 Ans. (2)

Sol. Integrating factor = e ^{tan x dx}= sec x solution of the equation is y sec x = sin x sec x dx

y

cosx = n(sec x) + c put x = 0, c = 0

y = cos x n (sec x) put x =

4 y = ¹ n 2

2 = ¹ n2

2 2

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) 8. Let A = {2, 3, 4, 5, .... , 30} and '~' be an equivalence relation on A × A, defined by

(a, b) ~ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :

(1) 5 (2) 6 (3) 8 (4) 7

Ans. (4)

Sol. (a, b) R (c, d) ad = bc

(a, b) R (4, 3) 3a = 4b b = 3 4a a is a multiple of 4

(a, b) = (4, 3), (8, 6), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21) i.e., 7 ordered pairs

9. Let the lengths of intercepts on x-axis and y-axis made by the circle x² + y² + ax + 2ay + c = 0, (a < 0) be 2 2 and 2 5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is euqal to :

(1) 11 (2) 7 (3) 6 (4) 10

Ans. (3)

Sol. 2 ^a² c 2 2

4 , 2 a² c 2 5

a2

4 c 2

a2 c 5 a2

c 2

4 a² = 4c + 8 3c + 8 = 5 c = –1, a² = 4 Equation of tangent y = 2x + c'

D = c ' 5 = r

2 2

c ' a

a c

5 4

c ' 1 5

5 c' = 30

D = 6

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) 10. The least value of |z| where z is complex number which satisfies the inequality

exp | z | 3 | z | 1 _e ₂

log 2 log 5 7 9i

| z | 1 ,i 1, is equal to :

(1) 3 (2) 5 (3) 2 (4) 8

Ans. (1) Sol.

(|z| 3)(|z| 1) (|z| 1)

2 2³ (| z | 3)(| z | 1)

| z | 1 3 |z|² + 2|z| – 3 3|z| + 3

|z|² – |z| – 6 0 (|z| – 3) (|z| + 2) 0

|z|min = 3

11. Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let be the number of triangles having these points from different sides as vertices and be the number of quadrilaterals having these points from different sides as vertices.

Then ( – ) is equal to :

(1) 795 (2) 1173 (3) 1890 (4) 717

Ans. (4)

Sol. = ⁶C17C19C1 + ⁵C17C19C1 + ⁵C16C19C1 + ⁵C16C17C1 = 378 + 315 + 270 + 210 = 1173 = ⁵C16C17C1 9C1 = 1890

– = 1890 – 1173 = 717

12. If the point of intersections of the ellipse ^x² ^y²₂ 1

16 b and the circle x² + y² = 4b, b > 4 lie on the curve y² = 3x², then b is equal to:

(1) 12 (2) 5 (3) 6 (4) 10

Ans. (1) Sol. x² y²₂

16 b 1 ...(1)

2 2

x y 4b ...(2)

y² = 3x² ...(3)

From eq (2) and (3) x² = b and y² = 3b from equation (1) b 3b₂

16 b 1

b² + 48 = 16b b = 12

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) 13. Given that the inverse trigonometric functions take principal values only. Then, the number of

real values of x which satisfy sin ¹ ^3x sin ¹ ^4x sin x¹

5 5 is equal to:

(1) 2 (2) 1 (3) 3 (4) 0

Ans. (3)

Sol. Taking sine both sides

2 2

3x 16x 4x 9x

1 1

5 25 5 25 = x

3x 25 16x² = 25x – 4x 25 9x² x = 0 or 3 25 16x² = 25 – 4 25 9x²

9(25 – 16x²) = 625 – 200 25 9x² + 16 (25 – 9x²) 200 25 9x² = 800

25 9x2 = 4 x² = 1

x ± 1

Total number of solution = 3

14. Let A(–1, 1), B(3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of ABC and PQC respectively, such that A1 = 3A2, then the value of m is equal to :

(1) ⁴

15 (2) 1 (3) 2 (4) 3

Ans. (2)

Sol.

A(–1,1)

C(2,0) B(3,4)

P

Q y=mx

A1 = ABC =

1 1 1

1 2 0 1

2 3 4 1

A1 = 13 2

Equation of line AC is y – 1 = 1

3 (x + 1)

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solve it with line y = mx, we get P 2 2m 3m 1 3m 1, Equation of line BC is y – 0 = 4 (x – 2)

Solve it with line y = mx, we get Q –8 –8m m – 1 m – 4,

A2 = Area of PQC =

2 0 1

1 2 2m

2 3m 1 3m 1 1

–8 –8m

m 4 m 4 1

= A¹ 13

3 6

2 0 1

1 2 13

m 0 1

2 3m 1 6

0 –8 1

m 4

1 2 8 13

2 m(2) 3m 1 m 4 6

m 4 12m 4 13

2m (3m 1)(m 4) 6

26m2 13

(3m 1)(m 4) 6

m = 1

15. Let f be a real valued function, defined on R – {–1, 1} and given by f(x) = 3loge x 1 2

x 1 x 1 Then in which of the following intervals, function f(x) is increasing?

(1) ( , 1) ¹, {1}

2 (2) (– , ) – {–1, 1}

(3) 1,¹

2 (4) ,¹ { 1}

2 Ans. (1)

Sol. f (x) = 3 ^{x 1}

x 1 × ^{x 1 (x 1)}₂

(x 1) + ² ₂

(x 1) = ⁶

(x 1)(x 1) + ² ₂ (x 1)

= ² (x 1)

3 1

x 1 x 1 = ²₂

(x 1) (x 1) (4x – 2)

(9)

= ₂

8 x –1 2 (x 1) (x 1)

for increasing, y (x) 0

–1 1

₂^x

(x 1) (x 1) 0 ( , 1) ¹, {1}

2

16. Let f : S S where S = (0, ) be a twice differentiable function such that f(x + 1) = xf(x). If g : S R be defined as g(x) = logef(x), then the value of |g"(5) – g"(1)| is equal to :

(1) ²⁰⁵

144 (2) ¹⁹⁷

144 (3) ¹⁸⁷

144 (4) 1

Ans. (1)

Sol. g(x + 1) = n (f(x + 1)) = n (x f(x)) = nx + n(f(x)) g(x + 1) – g(x) = nx

g (x + 1) – g (x) = 1 x

g (x + 1) – g (x) = – 1

32 …..(i)

putting x = 1, 2, 3, 4 in (i), we get g (2) – g (1) = – 1

g (3) – g (2) = – 1 4 g (4) – g (3) = – 1 9 g (5) – g (4) = – 1

16 Adding,

g (5) – g (1) = – 1 – 1 4 – 1

9 – 1

16 = –205 144 |g (5) – g (1)| = 205

144

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17. Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that

1

0

P(x)dx 1 and P(x) leaves remainder 5 when it is divided by (x – 2). Then the value of 9(b + c) is equal to:

(1) 9 (2) 15 (3) 7 (4) 11

Ans. (3) Sol.

1 2 0

x bx c dx = 1

3 2 1

0

x bx

cx 1

3 2

b 2

2 c 3 …(i)

and P(2) = 5

2b + c = 1 …(ii)

by (i) and (ii) b = 2

9, c = 5 9 then 9 (b + c) = 7

18. If the foot of the perpendicular from point (4, 3, 8) on the line L :₁ ^{x a} ^{y 2} ^{z b}

3 4

l , l 0 is

(3, 5, 7), then the shortest distance between the line L1 and line L :₂ ^{x 2} ^{y 4} ^{z 5}

3 4 5 is equal to :

(1) ¹

2 (2) ¹

6 (3) ²

3 (4) ¹

3 Ans. (2)

Sol. Point (3,5,7) satisfy the given line L1 = 0

3 a 5 2 7 b

3 4

a = 3 – , b = 3

Direction ratio of line joining given points is (1, –2, – 1) which is perpendicular to L1 = 0 So, a1a2 + b1b2 + c1c2 = 0

– 6 + 4 = 0 = 2, a = 1 L1 = x 1 y 2 z 3

2 3 4 and L :₂ ^{x 2} ^{y 4} ^{z 5}

3 4 5 are skew lines

D = ² ¹ ¹ ²

1 2

a – a . b b

b b = i 2 j 2k . –i 2 j – k

6 = ¹

6

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) 19. Let C1 be the curve obtained by the solution of differential equation 2xy^dy y² x²

dx , x > 0. Let the curve C2 be the solution of ₂^2xy₂ ^dy

x y dx. If both the curves pass through (1,1), then the area enclosed by the curves C1 and C2 is equal to :

(1) – 1 (2) 1

2 (3) + 1 (4) 1

4 Ans. (2)

Sol. dy y² x²

dx 2xy

Put y = vx v + x dv

dx =

2 2 2 2

2

v x – x v – 1 2vx 2v

xdv

dx = ^{v – 1 – 2v}² ² ^(v² ¹⁾

2v 2v

2

2v dx

dv x

v 1

n(v² + 1) = – nx + nc v² + 1 = c x

2 2

y c

1 x

x x² + y² = cx If pass through (1, 1)

x² + y² – 2x = 0

similarly for second differential equation ^dx

dy = x² y² 2xy equation of curve is x² + y² – 2y = 0

Now required area is

C(1, 0) x

(1, 1) y

(0, 1)

O

= 1 ² 1

1 – 1 1

4 2 × 2

= – 1 sq. units

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) 20. Let a i 2 j 3k^ˆ ^ˆ ^ˆ and b 2i 3j 5k^ˆ ^ˆ ^ˆ. If r a b r, r. i 2 j k^ˆ ^{ˆ ˆ} 3and r. 2i 5j^ˆ ^ˆ k^ˆ 1,

R, then the value of r² is equal to :

(1) 9 (2) 15 (3) 13 (4) 11

Ans. (2)

Sol. r (a b) 0 r (a b) r (3i – j 2k)

r.( i 2j k) 3

(3 – 2 2) = 3 = 1 and r.(2i 5j – k) –1

(6 – 5 – 2 ) = –1 1

(1 – 2 ) –1 = 1, = 1 ˆ ˆ ˆ

r 3i – j 2k

| r | 14, = 1 r2 = 15

SECTION-B

1. If the distance of the point (1, –2, 3) from the plane x + 2y – 3z + 10 = 0 measured parallel to the line, ^{x 1 2 y} ^{z 3}

3 m 1 is ⁷

2 , then the value of |m| is equal to ______.

Ans. 2

Sol. Line parallel to given line and passes through (1, –2, 3) is

x 1 y 2 z 3

3 m 1 =

(x, y, z) = (1 + 3 , –2 – m, 3 + ) satisfy plane x + 2y – 3z + 10 = 0

+3 – 4 – 2 m – 9 – 3 + 10 = 0 m = 1

So, point of intersection 1 ³ , –3,3 ¹

m m

Distance = 9₂ 1₂ 7

1 2

m m

m² = 4 or |m| = 2

2. Consider the statistics of two sets of observations as follows :

Size Mean Variance

Observation I 10 2 2

Observation II n 3 1

If the variance of the combined set of these two observations is 17

9 , then the value of n is equal to __________.

Ans. 5

(13)

Sol. For group-1 : xⁱ

10 = 2 x_i = 20

2i

x

10 – (2)² = 2 x²_i = 60 for group-2 : yⁱ

n = 3 y_i = 3n

2

yi

n – 3² = 1 y²_i = 10n Now, combined variance

2 =

2 2

i i

(x y )

10 n –

2

i i

(x y ) 10 n

2 2

17 60 10n (20 3n)

9 10 n – (10 n)

17(n² + 20n + 100) = 9(n² + 40n + 200) 8n² – 20n – 100 = 0

2n² – 5n – 25 = 0 n = 5 3. Let ¹

2

A a

a and ¹

2

B b

b be two 2 × 1 matrices with real entries such that A = XB, where 1 1

X 1

1 k

3 , and k R. If a₁² a²₂ ² b₁² b²₂

3 and (k² 1)b²₂ 2b b_{1 2}, then the value of k is _______.

Ans. 1

Sol. A = XB

1 1

2 2

a 1 1 1 b

a 3 1 k b

1 1 2

a 1 (b b )

3 and a₂ ¹ (kb₂ b )₁ 3

Given that

2 2 2 2

1 2 1 2

a a 2 b b

3

2 2 2 2

1 2 2 1 1 2

1 1 2

b b kb b b b

3 3 3

(1 + k²) b22 + 2b12 + 2 (k –1) b1b2 = 2b12 + 2b22 (1 + k²) b22 + 2(k – 1) b1b2 = 2b22

this is only possible when k = 1

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4. For real numbers , , and , if

2 1 2

4 2 1 2

x 1 (x 1) tan

x dx

x 1 (x 3x 1)tan

x

= log tan_e ¹ ^x² ¹ x

1 (x2 1)

tan x

where C is an arbitrary constant, then the value of 10( ) is equal to __________.

Ans. 6

Sol. I = ² ₂

4 2 1

(x 1) dx

x 1 (x 3x 1)tan

x

+ ₄ ¹ ₂

x 3x 1dx

= n

x2 1

x + ²

2 2

2

1 x dx

2 x 1 c

x

n ² ₂²

1 1

x 1 1 x dx

x 2 1

x 5

x

– ²₂

1 1

1 x

2 1

x 1

x

dx

= n

2 2 2

–1 –1

x 1 1 1 x 1 1 x 1

tan – tan

x 2 5 5x 2 x

= 1, = ¹

2 5, = ¹

5, = –1 2 10( + y + ) = 1 ¹ –¹ 10

10 2 = 6

5. Let f : R R and g : R R be defined as f(x) ^{x a,} ^{x 0}

| x 1 |, x 0 and g(x) ^{x 1,}₂ ^{x 0} (x 1) b, x 0 where a, b are non-negative real numbers. If (gof)(x) is continuous for all x R, then a + b is equal to ___________.

Ans. 1

Sol. f(x) ^{x a,} ^{x 0}

| x 1 |, x 0, g(x) ^{x 1,}₂ ^{x 0} (x 1) b, x 0 Now gof(x) is continues for x R then (i) f(0) = f(0^–)

1 = a

and f(x) = 0 ^{x 1} x a at x =1

g(f(x)) g(f(1)) = g(f(1⁺)) = b + 1

(15)

JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) at x = – 1, g(f(x)) g(f(–1)) = g(f(–1^–))

b + 1 = 1 b = 0 a + b = 1

6. Let ¹

16, a and b be in G.P. and ^{1 1}, ,6

a b be in A.P., where a, b > 0. Then 72(a + b) is equal to ________.

Ans. 14 Sol. ² b

a 16 and 2 1 b a 6 Solving, we get a = 1

12 or a = 1

4 [rejected]

If a = 1

12 b = 1 9 72 (a + b) = 72 1 1

12 9 = 14

7. In ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of ABC is 30 cm² and R and r are respectively the radii of circumcircle and incircle of ABC, then the

value of 2R + r

(in cm) is equal to _______.

Ans. 15 Sol. C = 90º

A

13

C 5 B 12

R = 13

2 , S = 12 5 13 2 = 15 r = S = 30

15 = 2 2R + r = 13 + 2 = 15

8. Let n be a positive integer. Let

k

k k k k k

n k

C k 0

1 3 7 15 31

A ( 1) n

2 4 8 16 32 If 63A =

30

1 1

2 , then n is equal to _______.

Ans. 6

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) Sol. Use ⁿ ^{r n} _r ^r

r 0

( 1) C x = (1 – x)ⁿ

A = 1 n

1 2 + 3 n

1 4 + 7 n

1 8 +

15 n

1 16 +

31 n

1 32 =

1 n

2 + 1 n

4 + 1 n

8 + 1 n

16 + 1 n

32

=

n 5

n

1 1

2 1 2

1 1 2 63A = 1 ¹₃₀

2 n = 6

9. Let c be a vector perpendicular to the vectors a i j k^{ˆ ˆ ˆ} and b i 2 j k^ˆ ^{ˆ ˆ}. If c. i j 3k^{ˆ ˆ} ^ˆ 8 then the value of c. a b is equal to _________.

Ans. 28

Sol. c = ( a × b )

c =

ˆ ˆi j kˆ 1 1 –1

1 2 1

= i 1 2 – j 1 1 k 2 – 1

c = (3 ˆi – 2ˆj + ˆk ) given c .( ˆi + ˆj + 3 ˆk ) = 8

(3 ˆi – 2ˆj + ˆk ).( ˆi + ˆj + 3 ˆk ) = 8 (3 – 2 + 3) = 8

= 2

c = 2 ( a × b )

c .( a × b ) = 2| a × b |² = 28

10. Let S (x) logn a1/2x loga1/3x loga1/6x log_a_1/11x log_a_1/18x log_a_{1/ 27}x ... up to n-terms, where a > 1. If S24 (x) = 1093 and S12 (2x) = 265, then value of a is equal to _______.

Ans. 16

Sol. S_n(x) = (log_ax)

7

2 3 6 11 ...

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JEE (MAIN) MARCH-2021 DATE-16/03/2021 (SHIFT-II) tn = an² + bn + c

a + b + c = 2 ... (i) 4a + 2b + c = 3 ... (ii) 9a + 3b + c = 6 ... (iii)

a = 1; b = – 2; c = 3

Tn = n² – 2n + 3 = (n – 1)² + 2 S₂₄(x) = 1093 = log_ax.(4372) x = a^1/4 ... (i)

Also, S₁₂(2x) = 265 = (log_a2x).(530) 2x = a^1/2 ... (ii) from (i) and (2)

a^1/2 = 2a^1/4

a² = 16a a = 16