JEE Main 2021 March 18 – Shift 2 Maths Question Paper with Solutions

Section – A

Question 1: Let the system of linear equations

4x + λy + 2z = 0

2x – y + z = 0

𝝻x + 2y + 3z = 0, λ, 𝝻 ∈ R

Has a non-trivial solution. Then which of the following is true?

a. 𝝻 = 6, λ ∈ R
b. λ = 2, 𝝻 ∈ R
c. λ = 3, 𝝻 ∈ R
d. 𝝻 = – 6, λ ∈ R

Answer: (a)

For non-trivial solution

Δ = 0

4 (- 3 – 2) – λ (6 – 𝝻) + 2 (4 + 𝝻) = 0

– 20 – 6λ + λ𝝻 + 8 + 2𝝻 = 0

– 12 – 6λ + λ𝝻 + 2𝝻 = 0

– 6 (λ + 2) + 𝝻 (λ + 2) = 0

(λ + 2) (𝝻 – 6) = 0

𝝻 = 6, λ ∈ R

Question 2: A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be π / 3. If the radius of the circumcircle of △ABC is 2, then the height of the pole is equal to

a. 1 / √3
b. √3
c. 2√3
d. 2√3 / 3

Answer: (c)

tan 60^o = h / 2

h = 2√3

Question 3: Let in a series of 2n observations, half of them are equal to a and the remaining half are equal to -a. Also by adding a constant b in each of these observations, the mean and standard deviation of the new set become 5 and 20, respectively. Then the value of a² + b² is equal to:

a. 250
b. 925
c. 650
d. 425

Answer: (d)

Given series

(a,a,a…….n times), (-a, -a, -a,…… n times)

Now

As x_i → x_i + b, then

So,

No change in S.D. due to change in origin

σ = [∑x_i² / 2n] – (

20 = √a² ⇒ a = 20

a² + b² = 425

Question 4: Let g (x) =∫ ₀^t f (t) dt where f is a continuous function in [0, 3] such that (1 / 3) ≤ f (t) ≤ 1 for all t ∈ [0, 1] and 0 ≤ f (t) for all t ∈ (1, 3]. The largest possible interval in which g (3) lies is:

a. [1, 3]
b. [- 1, – 1 / 2]
c. [- 3 / 2, – 1]
d. [1 / 3, 2]

Answer: (d)

∫ ₀¹ (1 / 3) dt + ∫ ₁³ 0 dt ≤ g (3) ≤ ∫ ₀¹ (1) dt + ∫ ₁³ (1 / 2) dt

(1 / 3) ≤ g (3) ≤ 2

Question 5: If 15 sin⁴ θ + 10 cos⁴ θ = 6, for some θ ∈ R, then the value of 27 sec⁶ θ + 8 cosec⁶ θ is equal to:

a. 250
b. 500
c. 400
d. 350

Answer: (a)

15 sin⁴ θ + 10 cos⁴ θ = 6

⇒ 15 sin⁴ θ + 10 (1 – sin² θ)² = 6

⇒ 25 sin⁴ θ – 20 sin² θ + 4 = 0

(5 sin² θ – 2)² = 0 ⇒ sin² θ = 2 / 5, cos² θ = 3 / 5

Now 27 sec⁶ θ + 8 cosec⁶ θ = 27 (125 / 27) + 8 (125 / 8) = 250

Question 6: Let f : R – {3} → R – {1] be defined by f (x) = (x – 2) / (x – 3). Let g : R – R be given as g (x) = 2x – 3. Then, the sum of all the values of x for which f^-1 (x) + g^-1 (x) = 13 / 2 is equal to

a. 7
b. 5
c. 2
d. 3

Answer: (b)

f^-1 (x) + g^-1 (x) = 13 / 2

⇒ [3x – 2] / [x – 1] + [x + 3] / 2 = 13 / 2

⇒ 2 (3x – 2) + (x – 1) (x + 3) = 13 (x – 1)

⇒ x² – 5x + 6 = 0

⇒ x = 2 or 3

Question 7: Let S₁ be the sum of the first 2n terms of an arithmetic progression. Let S₂ be the sum of the first 4n terms of the same arithmetic progression. If (S₂ – S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

a. 3000
b. 7000
c. 5000
d. 1000

Answer: (a)

S_4n – S_2n = 1000

⇒ (4n / 2) (2a + (4n – 1)d) – (2n / 2) (2a + (2n – 1)d) = 1000

⇒ 2an + 6n²d – nd = 1000

⇒ (6n / 2) (2a + (6n – 1)d) = 3000

∴ S_6n = 3000

Question 8: Let S₁ = x² + y² = 9 and S₂ = (x – 2)² + y² = 1. Then the locus of the centre of a variable circle S which touches S₁ internally and S₂ externally always passes through the points:

a. (1 / 2, ± √5 / 2)
b. (2, ± 3 / 2)
c. (1, ± 2)
d. (0, ± √3)

Answer: (b)

C₁ : (0,0) , r₁ = 3

C₂ : (2, 0), r₂ = 1

Let the centre of the variable circle be C₃(h, k) and the radius is r.

C₃C₁ = 3 – r

C₂C₃ = 1 + r

C₃C₁ + C₂C₃ = 4

So locus is ellipse whose focii are C₁ & C₂

And major axis is 2a = 4 and 2ae = C₁C₂ = 2

⇒ e = 1 / 2

⇒ b² = 4 [1 – (1 / 4)] = 3

The centre of the ellipse is the midpoint of C₁ & C₂ is (1, 0).

The equation of the ellipse is [x – 1]² / 2² + [y²] / [√3]² = 1.

Now by cross-checking the option (2, ± 3 / 2) satisfied it.

Question 9: Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of ΔABC, then (R + r) is equal to

a. 2√2
b. 3√2
c. 7√2
d. 9 / √2

Answer: (d)

r = |(0 + 0 – 3) / √2| = 3 / √2

sin 30° = r / R = 1 / 2

R = 2r

So, r + R = 3r = 3 * (3 / √2) = 9 / √2

Question 10: In a triangle ABC, if vector BC = 8, CA = 7, AB = 10, then the projection of the vector AB on AC is equal to:

a. 25 / 4
b. 85 / 14
c. 127 / 20
d. 115 / 16

Answer: (b)

Projection of AB on AC is = AB cos A = 10 cos A

By cosine rule

cos A = [10² + 7² – 8²] / [2 . 10 . 7]

= 85 / 140

10 cos A = 10 (85 / 140) = 85 / 14

Question 11: Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:

a. 80 / 243
b. 32 / 625
c. 128 / 625
d. 40 / 243

Answer: (b)

⁵C₁ p¹ q⁴ = 0.4096 — (1)

⁵C₂ p² q³ = 0.2048 — (2)

(1) / (2) ⇒ q / 2p = 2 ⇒ q = 4p

p + q = 1 ⇒ p = 1 / 5 and q = 4 / 5

P (exactly 3) = ⁵C₃ (p)³ (q)² = ⁵C₃ (1 / 5)³ (4 / 5)²

= (10) * (1 / 125) * (16 / 25) = 32 / 625

Question 12: Let a and b be two non-zero vectors perpendicular to each other and |a| = |b|. If |a x b| = |a|, then the angle between the vectors (a + b + (a x b)) and a is equal to:

a. sin^-1 (1 / √3)
b. cos^-1 (1 / √3)
c. sin^-1 (1 / √6)
d. cos^-1 (1 / √2)

Answer: (b)

Question 13: Let a complex number be w = 1 – √3i. Let another complex number z be such that |zw| = 1 and arg (z) – arg (w) = π / 2. Then the area of the triangle with vertices origin, z and w is equal to:

a. 1 / 2
b. 4
c. 2
d. 1 / 4

Answer: (a)

w = 1 – √3i

|w| = 2

|zw| = 1 ⇒ |z| = 1 / |w| = 1 / 2

arg (z) – arg (w) = π / 2

Area of triangle = (1 / 2) * (1 / 2) * 2 = 1 / 2

Question 14: The area bounded by the curve 4y² = x² (4 – x) (x – 2) is equal to:

a. 3π / 2
b. π / 16
c. π / 8
d. 3π / 8

Answer: (a)

The domain of 4y² = x² (4 – x) (x – 2) is [2, 4] ⋃ {0}

Area of loop = 2 * (1 / 2) * ∫₂⁴ x √(4 – x) (x – 2) dx

Put x = 4 sin² θ + 2 cos² θ

dx = (8 sin θ cos θ – 4 cos θ sin θ) dθ

= 4 sin θ cos θ dθ

= ∫₀^π/2 (4 sin² θ + 2 cos² θ) √(2 cos² θ) (2 sin² θ) * [4 sin θ cos θ] dθ

= ∫₀^π/2 (4 sin² θ + 2 cos² θ) 8 (cos θ sin θ)² dθ

= ∫₀^π/2 32 sin⁴ θ cos² θ dθ + ∫₀^π/2 16 sin² θ cos⁴ θ dθ

Using wallis theorm

= 32 . {[3 . 1 . 1 π] / [6 . 4 . 2 . 2]} + 16 . {[3 . 1 . 1 π] / [6 . 4 . 2 . 2]}

= π + (π / 2) = 3π / 2

Question 15: Define a relation R over a class of n × n real matrices A and B as “ARB iff there exists a non-singular matrix P such that PAP^-1 = B”. Then which of the following is true?

a. R is reflexive, symmetric but not transitive
b. R is symmetric, transitive but not reflexive
c. R is an equivalence relation
d. R is reflexive, transitive but not symmetric

Answer: (c)

For reflexive

(B, B) ∈ R ⇒ B = PBP^-1 which is true for P = I

∴ R is Reflexive

For symmetry

As (B, A) ∈ R for matrix P

B = PAP^-1 ⇒ P^-1B = P^-1PAP^-1

P^-1BP = IAP^-1P = IAI

P^-1BP = A ⇒ A = P^-1BP

(A, B) ∈ R for matrix P^-1

∴ R is symmetric

For transitivity

B = PAP^-1 and A = PCP^-1

B = P (PCP^-1) P^-1

B = P²C (P^-1)² ⇒ B = P²C (P²)^-1

(B, C) ∈ R for matrix P²

∴ R is transitive

So R is an equivalence relation.

Question 16: If P and Q are two statements, then which of the following compound statement is a tautology?

a. ((P ⇒ Q) ^ ~Q) ⇒ P
b. ((P ⇒ Q) ^ ~ Q) ⇒ ~ P
c. ((P ⇒ Q) ^ ~ Q)
d. ((P ⇒ Q) ^ ~ Q) ⇒ Q

Answer: (b)

(P ⇒ Q) ^ ~ Q

= (~ P ∨ Q) ^ ~ Q

≡ (~ P ∧ ~ Q) ∨ (Q ∧ ~ Q)

= (~ P ∨ ~ Q) ∨ (Q ^ ~ Q)

= ~ (P ∨ Q)

Now,

(1) ~ (P v Q) ⇒ P

= (P v Q) v P

= P v Q

(2)

~ (P v Q) ⇒ ~ P

= (P v Q) v ~ P

= T

(3) ~ (P v Q) ⇒ (P ^ Q)

= (P v Q) v (P ^ Q)

= P v Q

(4) ~ (P v Q) ⇒ Q

= (P v Q) v Q

= P v Q

Question 17: Consider a hyperbola H : x² – 2y² = 4. Let the tangent at a point P (4, √6) meet the x-axis at Q and latus rectum at R (x₁, y₁), x₁ > 0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR is equal to:

a. √6 – 1
b. 4√6 – 1
c. 4√6
d. (7 / √6) – 2

Answer: (d)

Tangent at P (4, √6)

4 (x) – 2 . √6y = 4

⇒ 2x – √6y = 2 …(1)

For Q, put y = 0

Q (1,0)

Equation of Latus rectum:

X = ae = 2 (√3 / 2) = √6 …(2)

Solving (1) & (2), we get R (√6, 2 – (2 / √6))

Area of ΔQFR = (1 / 2) (√6 – 1) (2 – (2 / √6))

= (7 / √6) – 2

Question 18: Let f : R → R be a function defined as

a. – 2
b. – 2 / 5
c. – 3 / 2
d. – 3

Answer: (c)

‘f’ is continuous at x = 0

⇒ f (0^–) = f (0) = f (0⁺)

f (0^–) = lim_x→0- [sin (a + 1) x + sin 2x] / 2x

lim_x→0- [sin (a + 1) x] / [(a + 1) x] * (a + 1) / 2 + (sin 2x / 2x)

= [(a + 1) / 2] + 1 …(1)

f (0⁺) = lim_x→0- [√(x + bx³) – √x] / bx^5/2

= lim_x→0+ [bx³] / [bx^5/2 (√(x + bx³) + √x)]

= lim_x→0+ [1 / (√(1 + bx²) + 1)]

= 1 / 2 …(2)

f(0) = b …(3)

From (1), (2) and (3)

⇒ a = -2 & b = ½

Thus, a + b = -3/2

Question 19: Let y = y (x) be the solution of the differential equation dy / dx = (y + 1) [(y + 1) e^{x^2/2} – x], 0 < x < 2.1, with y (2) = 0. Then the value of dy / dx at x = 1 is equal to:

a. [e^5/2] / [1 + e²]²
b. [5e^1/2] / [e² + 1]²
c. – 2e² / (1 + e²)²
d. [-e^3/2] / [e² + 1]²

Answer: (d)

dy / dx = (y + 1) [(y + 1) e^{x^2/2} – x]

⇒ [- 1 / (y + 1)²] (dy / dx) – x [1 / (y + 1)] = – e^{x^2/2}

Put, 1 / (y + 1) = z

IF = e^{∫-x dx} = e^{-x^2/2}

z (e^{-x^2/2}) = – ∫ e^{-x^2/2} . e^{x^2/2} dx = – ∫ 1 dx = – x + c

⇒ (e^{x^2/2} / (y + 1)) = – x + c …(1)

Given y = 0 at x = 2

Put in (1)

e^-2 / (0 + 1) = – 2 + c

C = e^-2 + 2 …(2)

From (1) and (2)

y + 1 = e^{-x^2/2} / (e^-2 + 2 – x)

Again, at x = 1

y + 1 = e^3/2 / [e² + 1] [y + 1] = e^3/2 / [e² + 1]

dy / dx|_x=1 = {e^3/2 / [e² + 1]} {[e^3/2 / (e² + 1)] * [e^1/2 – 1]}

= – [e^3/2] / [e² + 1]²

Question 20: Let a tangent be drawn to the ellipse (x² / 27) + y² = 1 at (3√3 cos θ, sin θ) where θ ∈ (0, π / 2). Then the value of θ such that the sum of intercepts on axes made by a tangent is minimum is equal to:

a. π / 8
b. π / 6
c. π / 3
d. π / 4

Answer: (b)

Equation of tangent [x / 3√3] cos θ + y sin θ = 1

A [(3√3 / cos θ), 0], B [(0, 1 / sin θ)]

Now sum of intercept = [3√3 / cos θ] + [1 / sin θ]

Let y = 3√3 sec θ + cosec θ

y’ = 3√3 sec θ tan θ – cosec θ cot θ

y’ = 0

tan θ = 1 / √3

⇒ θ = π / 6

SECTION – B

Question 21: Let P be a plane containing the line [x – 1] / 3 = [y + 6] / 4 = [z + 5] / 2 and parallel to the line [x – 3] / 4 = [y – 2] / – 3 = [z + 5] / 7. If the point (1, -1, α) lies on the plane P, then the value of |5α| is equal to __________ .

Answer: (38)

DR’s of normal

(34, -13, -25)

P ≡ 34 (x – 1) – 13 (y + 6) – 25 (z + 5)

Q (1, -1, α) lies on P.

⇒ 3 (1 – 1) – 13 (- 1 + 6) – 25 (α + 5) = 0

⇒ – 25 (α + 5) = 65

⇒ +5α = – 38

⇒ |5α| = 38

Question 22: ∑_r=1¹⁰ r! (r³ + 6r² + 2r + 5) = α (11!). Then the value of α is equal to _______ .

Answer: (160)

T_r = r! ((r + 1) (r + 2) (r + 3) – 9r – 1)

= (r + 3) ! – 9r . r! – r!

= (r + 3) ! – 9 (r – 1 + 1)) r! – r!

= (r + 3) ! – 9 (r + 1) ! + 8r!

= {(r + 3) ! – (r + 1) ! – 8 (r + 1) ! – r !

Now,

∑_r=1¹⁰ T_r = {13! + 12! – 3! – 2!} – 8 {11! – 1!}

= 13! + 12! − 8 (11!)

= (13 * 12 + 12 – 8) * 11!

= 160 * 11!

α = 160

Question 23: The term independent of x in the expansion of [(x + 1) / (x^2/3 – x^1/3 + 1) – (x – 1) / (x – x^1/2)]¹⁰, x ≠ 1, is equal to ________.

Answer: (210)

General term, T_r+1 = ¹⁰C_r (x^1/3)^10-r (- x^-½)^r

For term independent of x

20 – 2r – 3r = 0

r = 4

Therefore required term, T₅ = ¹⁰C₄ = [10 * 9 * 8 * 7] / [4 * 3 * 2 * 1] = 210

Question 24: Let ⁿC_r denote the binomial coefficient of x^r in the expansion of (1 + x)ⁿ. If ∑_k=0¹⁰ [2² + 3k] ⁿC_k = ɑ . 3¹⁰ + β . 2¹⁰ then ɑ + β is equal to _______.

Answer: Bonus

n must be equal to 10

∑_k=0¹⁰ [2² + 3k] ⁿC_k

= ∑_k=0¹⁰ [4 + 3k] ⁿC_k

= 4 ∑_k=0¹⁰ⁿC_k + 3 ∑_k=0¹⁰ k ⁿC_k

= 4 (2¹⁰) + 3 * 10 * 2⁹

= 19 × 2¹⁰

∴ α = 0 and β = 19

Thus, α + β = 19

Question 25: Let P (x) be a real polynomial of degree 3 which vanishes at x = – 3. Let P(x) have local minima at x = 1, local maxima at x = -1 and ∫_-1¹ P (x) dx = 18, then the sum of all the coefficients of the polynomial P (x) is equal to _______________ .

Answer: (8)

P’ (x) = a (x + 1) (x – 1)

P (x) = (ax³ / 3) – ax + c

P (-3) = 0 (given)

⇒ a(- 9 + 3) + C = 0

⇒ 6a = C …(i)

Also, ∫_-1¹ P (x) dx = 18 ⇒ ∫_-1¹ [a ((x³ / 3) – x) + c] dx = 18

⇒ 0 + 2 C = 18 ⇒ C = 9

From (i), a = 3 / 2

P (x) = (x³ / 2) – (3 / 2) x + 9

Sum of co-efficient = – 1 + 9 = 8

Question 26: Let the mirror image of the point (1, 3, a) with respect to the plane r . (2i – j + k) – b = 0 be (-3, 5, 2). Then, the value of |a + b| is equal to ______.

Answer: (1)

Plane : 2x -y + z = b

R ≡ [- 1, 4, (a + 2) / 2] on plane

– 2 – 4 + (a + 2) / 2 = b

⇒ a + 2 = 2b + 12 ⇒ a = 2b + 10 …(1)

PQ < 4, -2, a -2>

(2 / 4) = – 1 / – 2 = 1 / (a – 2)

a- 2 = 2

a = 4, b = – 3

|a + b| = 1

Question 27: If f (x) and g (x) are two polynomials such that the polynomial P (x) = f (x³) + x g (x³) is divisible by x² + x + 1, then P (1) is equal to _______.

Answer: 0

Roots of x² + x + 1 are ω and ω² now

Q (ω) = f (1) + ω g (1) = 0 …(1)

Q (ω²) = f (1) + ω² g (1) = 0 …(2)

Adding (1) and (2)

2 f (1) – g (1) = 0

g (1) = 2 f (1)

f (1) = g (1) = 0

Therefore , Q (1) = f (1) + g (1)

= 0 + 0

= 0

Question 28: Let I be an identity matrix of order 2 × 2 and

Answer: (6)

P⁶ = 5I – 8P

Thus, n = 6

Question 29: Let f : R → R satisfy the equation f (x + y) = f (x) . f (y) for all x, y ∈ R and f (x) ≠ 0 for any x ∈ R. If the function f is differentiable at x = 0 and f’ (0) = 3, then lim_h→0 (1 / h) [f (h) – 1] is equal to _______.

Answer: (3)

f (x + y) = f (x) . f (y) then

f (x) = a^kx

f’(x) = (a^kx ) k ln a

f’(0) = k ln a = 3 (given f’(0) = 3)

a = e^3/k

f (x) = (e^3/k)^kx = e^3x

Now, lim_h→0 [f (h) – 1] / h = lim_h→0 {[e^3h – 1] / 3h} * 3 = 1 * 3 = 3

Question 30: Let y = y (x) be the solution of the differential equation xdy – ydx = √(x² – y²) dx, x ≥ 1, with y (1) = 0. If the area bounded by the line x = 1, x = e^π, y = 0 and y = y (x) is αe^2π + β, then the value of 10 (α + β) is equal to ________.

Answer: (4)

xdy – ydx = √(x² – y²) dx

⇒ sin^-1 (y / x) = ln |x| + c

At x = 1, y = 0 ⇒ c = 0

y = x sin (l nx)

A = ∫₁^{e^π} x sin (ln x) dx

x = e^t, dx = e^t dt = ∫₀^π e^2t sin (t) dt

αe^2π + β = [(e^2t / 5) [2 sin t – cos t])₀^π = [1 + e^2π] / 5

α = 1 / 5, β = 1 / 5

Thus, 10 (α + β) = 4