JEE Main 2021 March 16 – Shift 2 Maths Question Paper with Solutions

Section A

Question 1: The least value of |z| where z is a complex number which satisfies the inequality exp [{(|z| + 3) (|z| – 1)} / {||z| + 1}] log_e 2 ≥ log_√2 |5 √7 + 9i|, i = √-1, is equal to:

a. 2
b. 3
c. 8
d. √5

Answer: (b)

2^{[{(|z| + 3) (|z| – 1)} / {|z| + 1}]} ≥ 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

Question 2: Let f : S → S where S = (0, ∞) be a twice differentiable function such that f (x + 1) = x f (x). If g : S → R be defined as g (x) = log_e f (x), then the value of |g”(5) – g”(1)| is equal to:

a. 197 / 144
b. 187 / 144
c. 205 / 144
d. 1

Answer: (c)

f(x+1) = xf(x)

g(x+1) = log_e(f(x+1))

g(x+1) = log_e x + log_e f(x)

g(x+1) – g(x) = log_e x

g’’(x+1) – g’’(x) = – 1 / x²

g’’(2) – g’’(1) = –1

g’’(3) – g’’(2) = – 1 / 4

g’’(4) – g’’(3) = – 1 / 9

g’’(5) – g’’(4) = – 1 / 16

g’’(5) – g’’(1) =- [1 + (1 / 4) + (1 / 9) + (1 / 16)]

|g’’(5) – g’’(1)| = [144 + 36 + 16 + 9] / [16 * 9] = 205 / 144

Question 3: If y = y (x) is the solution of the differential equation (dy / dx) + (tan x) y = sinx, 0 ≤ x ≤ π / 3, with y(0) = 0, then y (π / 4) equal to:

a. log_e 2
b. (1 / 2) log_e 2
c. (1 / 2 √2) log_e 2
d. (1 / 4) log_e 2

Answer: (c)

I.F. = e^{∫tanx dx}

= e^{ln secx}

= sec x

Solution of the equation:

y(sec x) = ∫sin x sec x dx

y / (cosx) = ln (secx) + c

Put x = 0, c = 0

∴ y = cos x ln(sec x)

Put x = π / 4

y = (1 / √2) ln √2 = (1 / 2√2) ln 2

y = (ln 2) / (2√2)

Question 4: If the foot of the perpendicular from point (4, 3, 8) on the line L₁ : [x – a] / l = [y – 2] / 3 = [z – b] / 4, l ≠ 0 is (3, 5, 7), then the shortest distance between the line L₁ and line L₂ : [x – 2] / 3 = [y – 4] / 4 = [z – 5] / 5 is equal to:

a. √2 / 3
b. 1 / √3
c. 1 / 2
d. 1 / √6

Answer: (d)

(3, 5, 7) lies on given line L₁

A (4, 3, 8)

B (3, 5, 7)

DR’s of AB = (1, –2, 1)

AB ⊥ line L₁

(1)(l) + (–2)(3) + 4(1) = 0

⇒ l = 2

a = 1

a = 1, b = 3, l = 2

SD =

Question 5: If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression 3 + [x – 11] / [(y – 19)² (z – 12)²] + [y – 19] / [(x – 11)² (z – 12)²] + [z – 12] / [(z – 11)² (y – 19)²] – [x + y + z] / [14 (x – 11) (y – 19) (z – 12)] is equal to :

a. 3
b. 0
c. 39
d. – 45

Answer: (a)

Equation of plane is x + y + z = 42 or, (x − 11)+ (y − 19)+ (z −12) = 0.

Now, [x – 11] / [(y – 19)² (z – 12)²] + [y – 19] / [(x – 11)² (z – 12)²] + [z – 12] / [(z – 11)² (y – 19)²]

= [(x – 11)³ + (y – 19)³ + (z – 12)³] / [(x – 11)² (y – 19)² (z – 12)²]

= [3 (x – 11) (y – 19) (z – 12)] / [(x – 11)² (y – 19)² (z – 12)²] [If a + b + c = 0, then a³ + b³ + c³ = 3abc]

= 3 / [(x – 11) (y – 19) (z – 12)]

The given expression is equal to 3 + 3 / [(x – 11) (y – 19) (z – 12)] – 3 / [(x – 11) (y – 19) (z – 12)] = 3

Question 6: Consider the integral I = ∫₀¹⁰ {[x] e^[x]} / [e^x-1] dx, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :

a. 45 (e – 1)
b. 45 (e + 1)
c. 9(e – 1)
d. 9(e + 1)

Answer: (a)

I = ∫₀¹⁰ {[x] e^[x]} / [e^x-1] dx

= ∫₀¹⁰ [x] e^[x]+1-xdx

= ∫₁² e^2-x dx + ∫₂³ 2 e^3-x dx + ∫₃⁴ 3 e^4-x dx + ……… ∫₉¹⁰ 9 e^10-x dx

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

= 45(e – 1)

Question 7: 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 ΔABC and ΔPQC respectively, such that A₁ = 3A₂, then the value of m is equal to:

a. 4 / 15
b. 1
c. 2
d. 3

Answer: (b)

A₁ = Area of ΔABC =

A₁ = 13 / 2

Equation of line AC is y – 1 = (- 1 / 3) (x + 1)

Solving it with line y = mx, we get P [2 / (3m + 1), 2m / (3m + 1)]

Equation of line BC is y – 0 = 4(x – 2)

Solving it with line y = mx, we get Q [- 8 / (m – 4), – 8m / (m – 4)]

A₂ = Area of ΔPQC =

⇒ 12m² = ± (3m² – 11m – 4)

Taking +ve sign,

9m² + 11m + 4 = 0 (Rejected ∵ m is imaginary)

Taking –ve sign,

15m² – 11m – 4 = 0

m = 1, – 4 / 15

m = 1 as m > 0

Question 8: Let f be a real-valued function, defined on R – {–1, 1} and given by f(x) = 3 log_e |(x – 1) / (x + 1)| – 2 / (x – 1). Then in which of the following intervals, function f (x) is increasing?

a. (- ∞, – 1) ⋃ ([1 / 2, ∞) – {1})
b. (- 1, 1 / 2]
c. (–∞, ∞) – {–1, 1}
d. (- ∞, 1 / 2] – {- 1}

Answer: (a)

f’ (x) = [(x + 1) / (x – 1) * (x + 1 – (x – 1)) / (x + 1)²] 3 + 2 / (x – 1)² = 6 / (x – 1) (x + 1) + 2 / (x – 1)²

= [4(2x – 1)]/[(x + 1)(x – 1)²]

x ∈ (- ∞, – 1) ⋃ ([1 / 2, ∞) – {1})

Question 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 equal to:

a. √10
b. √6
c. √11
d. √7

Answer: (b)

2 √(a² / 4 – c) = 2√2

√(a² – 4c) = 2√2

a² – 4c = 8 — (1)

2√(a² – c) = 2√5

a² – c = 5 — (2)

(2) – (1), we get

3c = –3 ⇒ c = –1

a² = 4 ⇒ a = –2

x² + y² – 2x – 4y – 1 = 0

Equation of tangent 2x – y + λ = 0

∴ p = r

|(0 – 0 + λ) / √5| = √6

λ = ± √30

∴ Tangents are 2x – y ± √30 = 0

Distance from origin = √30 / √5 = √6

Question 10: 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 the probability of event A is equal to:

a. 4 / 9
b. 9 / 56
c. 3 / 7
d. 11 / 27

Answer: (a)

Total case = 6∟6

Fav. case = (0, 1, 2, 3, 4, 5) + (0, 1, 2, 4, 5, 6) + (1, 2, 3, 4, 5, 6)

= 5 ∟5 + 5 ∟5 + ∟6

= 1920

Probability = 1920 / 6∟6 = 4 / 9

Question 11: Let α ∈ R be such that the function is continuous at x = 0, where {x} = x – [x], [x] is the greatest integer less than or equal to x. Then:

a. α = π / 4
b. No such α exists
c. α = 0
d. α = π / √2

Answer: (b)

Question 12: The maximum value of f (x) =

a. √7
b. √5
c. 5
d. 3 / 4

Answer: (b)

C₁ → C₁ + C₂

R₁ → R₁ – R₂

= (–1)[2sin2x – cos2x] = cos2x – 2sin2x

Maximum value = √5

Question 13: Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β – α) is equal to:

a. 1890
b. 795
c. 717
d. 1173

Answer: (c)

α = ⁶C₁ ⁷C₁ ⁹C₁ + ⁵C₁ ⁷C₁ ⁹C₁ + ⁵C₁ ⁶C₁ ⁹C₁ + ⁵C₁ ⁶C₁ ⁷C₁

= 378 + 315 + 270 + 210

= 1173

β = ⁵C₁ ⁶C₁ ⁷C₁ ⁹C₁ = 1890

⇒ β – α = 1890 – 1173 = 717

Question 14: 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:

a. 2x + y = 5
b. x + 2y = 4
c. x + 3y = 5
d. x – y = 1

Answer: (d)

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

Question 15: Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin^-1 (3x / 5) + sin^-1 (4x / 5) = sin^-1 x =

a. 1
b. 2
c. 3
d. 0

Answer: (c)

Taking sin on both sides, (3x / 5) √[1 – (16x² / 25)] + (4x / 5) √[1 – (9x² / 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 solutions = 3

Question 16: Let C₁ be the curve obtained by the solution of the differential equation 2xy (dy / dx) = y² – x², x > 0. Let curve C₂ be the solution of 2xy / (x² – y²) = dy / dx. If both the curves pass through (1, 1), then the area enclosed by the curves C₁ and C₂ is equal to :

a. (π / 2) – 1
b. (π / 4) + 1
c. π – 1
d. π + 1

Answer: (a)

dy / dx = [y² – x²] / 2xy

Put y = vx

v + x (dv / dx) = [v²x² – x²] / 2vx² = [v² – 1] / 2v

x (dv / dx) = [v² – 1 – 2v²] / 2v = – (v² + 1) / 2v

2v / (v² + 1) dv = – dx / x

ln (v² + 1) = – ln x + ln c ⇒ v² + 1 = c / x

(y² / x²) + 1 = c / x ⇒ x² + y² = cx

It passes through (1, 1)

∴ x² + y² – 2x=0

Similarly, for second differential equation dy / dx = [x² – y²] / 2xy

equation of curve is x² + y² – 2y = 0

Now the required area is

= [(1 / 4) * π * 1² – (1 / 2) (1) (1)] * 2

= [(π / 2) – 1] square units

Question 17: Let a = i + 2j – 3k and b = 2i – 3j + 5k. If r x a = b x r, r . (ɑi + 2j + k) = 3 and r . (2i + 5j – ɑk) = – 1, ɑ ∈ R, then the value of ɑ + |r|² =

a. 11
b. 15
c. 9
d. 13

Answer: (b)

r x a = – r x b

r x (a + b) = 0 (a + b = 3i – j + 2k)

r || (a + b)

r = λ (a + b)

∵ r . (2i + 5j – ɑk) = – 1

λ [3i – j + 2k] . [2i + 5j – ɑk] = – 1

⇒ λ(6 – 5 – 2α) = –1

λ(1 – 2α) = –1 …(1)

r (ɑi + 2j + k) = 3

λ (3i – j + 2k) . (ɑi + 2j + k) = 3

⇒ λ[3α – 2 + 2] = 3 ⇒ λα = 1 …(2)

Solving (1) & (2)

λ [1 – (2 / λ)] = – 1

λ – 2 = –1 ⇒ λ = 1 and α = 1

r = 3i – j + 2k

α + |r|² = 1 + 14 = 15

Question 18: Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P (x) dx and P(x) leaves remainder 5 when it is divided by (x – 2). Then the value of 9 (b+c) is equal to :

a. 7
b. 11
c. 15
d. 9

Answer: (a)

(x – 2) Q (x) + 5 = x² + bx + c

Put x = 2

5 = 2b + c + 4 …(1)

∫₀¹ (x² + bx + c) dx = 1

(1 / 3) + (b / 2) + c = 1

(b / 2) + c = 2 / 3 …(2)

Solving (1) & (2)

b = 2 / 9

c = 5 / 9

9 (b + c) = 7

Question 19: If the points of intersections of the ellipse (x² / 16) + (y² / b²) = 1 and the circle x² + y² = 4b, b > 4 lie on the curve y² = 3x², then b is equal to :

a. 5
b. 6
c. 12
d. 10

Answer: (c)

(x² / 16) + (y² / b²) = 1 …(1)

x² + y² = 4b …(2)

y² = 3x² …(3)

From eqns. (2) and (3), x² = b and y² = 3b

From eqn. (1), (b / 16) + (3b / b² ) = 1

b² + 48 = 16b

b = 12

Question 20: 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 :

a. 7
b. 5
c. 6
d. 8

Answer: (a)

a = bc

(a, b) R (4, 3) ⇒ 3a = 4b

a = (4 / 3) b

b must be multiple of 3

b can be 3, 6, 9, ……, 30

Also, a must be less than or equal to 30.

(a, b) = (4, 3), (8, 6), (12, 9) (16, 12), (20, 15), (24, 18), (28, 21)

⇒ 7 ordered pairs

Section B

Question 21: Let

Answer: (28)

Question 22: In ΔABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of ΔABC is 30 cm² and R and r are respectively the radii of circumcircle and incircle of ΔABC, then the value of 2R + r (in cm) is equal to ______

Answer: (15)

Area = (1 / 2) * 5 * 12 * sin θ = 30

sin θ = 1 ⇒ θ = π / 2

ABC is right-angled at A.

r = (s – a) tan (A / 2)

r = (s – a)

2R + r = s ( a = 2R)

2R + r = 30 / 2 = 15

Question 23: Consider the statistics of two sets of observations as follows:

If the variance of the combined set of these two observations is 17 / 9, then the value of n is equal to ______

Answer: (5)

For group-1: ∑x_i / 10 = 2 ⇒ ∑x_i = 20

∑x_i² / 10 – (2²) = 2

∑x_i² = 60

For group-2 : ∑y_i / n = 3

∑y_i = 3n

∑y_i² / n – 3² = 1

∑y_i² = 10n

Now, combined variance

𝜎² = [∑(x_i² + y_i²) / (10 + n)] – [∑ (x_i + y_i) / (10 + n)]²

(17 / 9) = [60 + 10n] / [10 + n] – [20 + 3n]² / [10 + n]²

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

8n² – 20n – 100 = 0

2n² – 5n – 25 = 0

n = 5

Question 24: Let S_n(x) = up to n-terms, where a >1. If S₂₄(x) = 1093 and S₁₂(2x) = 265, then value of a is equal to______

Answer: (16)

S_n (x) = 2 log_a x + 3 log_a x + 6 log_a x + 11 log_a x + ……..

S_n (x) = log_a x (2 + 3 + 6 + 11 + …… )

S_r = 2 + 3 + 6 + 11

General term, T_r = r² – 2r + 3

S_n (x) = ∑_r=1ⁿ x (r² – 2r + 3)

S₂₄ (x) = x ∑_r=1²⁴ (r² – 2r + 3)

1093 = 4372 log_a x

log_a x = 1 / 4

x = a^1/4

S₁₂ (2x) = log_a (2x) ∑_r=1¹² (r² – 2r + 3)

265 = 530 log_a(2x)

log_a(2x) = 1 / 2

2x = a^1/2

After solving (i) and (ii), we get

a^1/4 = 2

a = 16

Question 25: Let n be a positive integer. Let A = ∑_k=0ⁿ (-1)^knC_k [(1 / 2)^k + (3 / 4)^k + (7 / 8)^k + (15 / 16)^k + (31 / 32)^k]. If 63A = 1 – (1 / 2³⁰), then n is equal to ______

Answer: (6)

A = (1 / 2)ⁿ + (1 / 4)ⁿ + (1 / 8)ⁿ + (1 / 16)ⁿ + (1 / 32)ⁿ

= (1 / 2ⁿ) + (1 / 2²ⁿ) + (1 / 2³ⁿ) + (1 / 2⁴ⁿ) + (1 / 2⁵ⁿ)

= (1 / 2ⁿ) [1 – (1 / 2ⁿ)]⁵ / [1 – (1 / 2ⁿ)]

A = [2⁵ⁿ – 1] / [2⁵ⁿ (2ⁿ – 1)]

63A = [63 (2⁵ⁿ – 1)] / [2⁵ⁿ (2ⁿ – 1)]

= (63 / (2ⁿ – 1)) [1 – (1 / 2⁵ⁿ)] = 1 – (1 / 2³⁰)

n = 6

Question 26: Let f : R → R and g : R → R be defined as f (x) =

Answer: (1)

Given g(f(x)) is continuous.

So, at x = –a & at x = 0

1 = b + 1 & (a–1)² + b = b

b = 0 & a = 1

⇒ a + b = 1

Question 27: If the distance of the point (1, – 2, 3) from the plane x + 2y – 3z + 10 = 0 measured parallel to the line, [x – 1] / 3 = [2 – y] / m = [z + 3] / 1 is √7 / 2, then the value of |m| is equal to_____

Answer: (2)

Point Q (3λ + 1, – mλ – 2, λ + 3) lies on the plane.

(3λ + 1) + 2 (- mλ – 2) – 3 (λ + 3) + 10 =0

3λ – 2mλ – 3λ + 1 – 4 – 9 + 10 = 0

– 2mλ = 2

mλ = –1 ⇒ λ = – 1 / m

Q [(- 3 / m) + 1, – 1, (- 1 / m) + 3]

PQ = √(7/2)

√(- 3 / m)² + 1 + (- 1 / m)² = √(7/2)

⇒ 20 + 2m² = 7m²

m² = 4 ⇒ |m| = 2

Question 28: Let 1 / 16, a and b be in G.P. and 1/a, 1/b, 6 be in A.P., where a, b, > 0. Then 72 (a + b) is equal to ______

Answer: (14)

a² = b / 16 and 2 / b = (1 / 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 / 12) + (1 / 9)] = 14

Question 29: Let

Answer: (1)

XB = A

(1 / √3)

b₁ – b₂ = √3a₁

3a₁² = b₁² + b₂² – 2b₁b₂

b₁ + kb₂ = √3a₂

3a₂² = b₁² + k²b₂² + 2kb₁b₂

Adding both equations, 3 (a₁² + a²) = 2b₁² + (k² + 1) b₂² + 2b₁b₂ (k – 1)

(k² – 1) b₂² + 2b₁b₂ (k – 1) = 0

(k – 1) [(k + 1) b₂² + 2b₁b₂ = 0

k = 1

Question 30: For real numbers α, β, γ and δ, if where C is an arbitrary constant, then the value of 10 (α + βγ + δ) is equal to______.

Answer: (6)