JEE Main 2021 March 17 – Shift 1 Maths Question Paper with Solutions

Section A

Question 1: Which of the following is true for y(x) that satisfies the differential equation (dy / dx) = xy – 1 + x – y; y (0) = 0

a. y (1) = 1
b. y (1) = e^1/2 – 1
c. y (1) = e^1/2 – e^-1/2
d. y (1) = e^-½ – 1

Answer: (d)

dy / dx = (x – 1) y + (x – 1)

dy / dx = (x – 1) (y + 1)

(dy / (y + 1)) = (x – 1) dx

ln (y + 1) = (x² / 2) – x + c

x = 0, y = 0 ⇒ c = 0

ln (y + 1) = (x² / 2) – x

Putting x = 1, ln (y + 1) = (1 / 2) – 1 = – 1 / 2

y + 1 = e^-½

y = e^-½ – 1

y (1) = e^-½ – 1

Question 2: The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k² has no solution if k is equal to

a. – 2
b. – 1
c. 1
d. 0

Answer: (a)

⇒ k(k² – 1) – (k – 1) + (1 – k) = 0

⇒ (k – 1) (k² + k – 1 – 1) = 0

⇒ (k – 1) (k² + k – 2) = 0

⇒ (k –1) (k –1) (k + 2) = 0

⇒k = 1, k = 2

For k = 1 equation identical so k = –2 for no solution.

Question 3: The value of is:

a. 2 + (4 / √5) (√30)
b. 4 + (4 / √5) (√30)
c. 2 + (2 / 5) (√30)
d. 5 + (2 / 5) √30

Answer: (c)

y = 4 + 1 / [5 + (1 / y)]

⇒ y = 4 + y / [5y + 1]

⇒ 5y²-20y-4 = 0

⇒ y = [20 ± √400 + 80] / 10

⇒ y = [20 ± 4√30] / 10

y = [10 + 2√30] / 5

Question 4: If the Boolean expression (p ⇒ q) ⇔ (q * (~ P)) is a tautology, then the Boolean expression p * (~q) is equivalent to:

a. p ⇒ ~ q
b. p ⇒ q
c. q ⇒ p
d. ~q ⇒ p

Answer: (c)

(p → q) ⇔ (q * ~ P)

∴ ~ (~ q ∧ p) = q ∨ ~ p = ~ p ∨ q

∴ * is equivalent to ∨

∴ p * ~ q = p ∨~ q

= ~ q ∨ p

= q p

Question 5: Choose the incorrect statement about the two circles whose equations are given below:

x² + y² – 10x – 10y + 41 = 0 and x² + y² – 16x – 10y + 80 = 0

a. Distance between two centres is the average radii of both the circles
b. Circles have two intersection points
c. Both circles’ centres lie inside the region of one another
d. Both circles pass through the centre of each other

Answer: (c)

C₁ (5, 5), C₂ (8, 5)

The position of C₁(5, 5) in S₂ = 0

= 25 + 25 – 80 – 50 + 80

= 0

The position of C₂(8, 5) in S₁ = 0

= 64 + 25 – 80 – 50 + 41

= 0

Question 6: The sum of possible values of x for tan^-1 (x + 1) + cot^-1 [1 / (x – 1)] = tan^-1 (8 / 31) is:

a. – 33 /4
b. – 32 / 4
c. – 31 / 4
d. – 30 / 4

Answer: (b)

Taking tan both sides

4x² + 31x – 8 = 0

x = – 8, 1 / 4

But at x = 1 / 4

LHS > π / 2 and RHS < π / 2

So, the only solution is x = – 8.

Question 7: Let

a. 10
b. 13
c. 12
d. 8

Question 8: The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is:

a. 3x + z = 6
b. 3x − z = 0
c. x + 3z = 10
d. x + 3z = 0

Answer: (b)

Let the equation of the plane is a(x – 1) + b(y – 2) + c (z – 3) = 0

Y–axis lies on it D.R.’s of y – axis are 0, 1, 0

0.a + 1.b + 0.c = 0 ⇒ b = 0

Equation of plane is a(x – 1) + c (z – 3) = 0

x = 0, z = 0 also satisfy it –a – 3c = 0 ⇒ a = –3c

–3c (x – 1) + c (z – 3) = 0

–3x + 3 + z – 3 = 0

3x – z = 0

Question 9: If A =

a. π / 6
b. π / 2
c. π / 3
d. π / 4

Question 10: The line 2x – y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x – 2y = 4. Then, the radius of the circle is:

a. 4√5
b. 3√5
c. 5√3
d. 5√4

Answer: (b)

m₁ * m₂ = – 1

a– 14 = 2 – a

2a = 16

a = 8

Centre (8, 2)

Radius = √(36 + 9)

= √45

= 3√5

Question 11: Team ‘A’ consists of 7 boys and n girls and Team ‘B’ has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to

a. 5
b. 6
c. 2
d. 4

Answer: (d)

7 × 4 + 6 × n = 52

6n = 24

n = 4

Question 12: In a triangle PQR, the coordinates of the points P and Q are (-2, 4) and (4, -2) respectively. If the equation of the perpendicular bisector of PR is 2x – y + 2 = 0, then the centre of the circumcircle of the △PQR is:

a. (-2, -2)
b. (0, 2)
c. (-1, 0)
d. (1, 4)

Answer: (a)

Perpendicular bisector of PR : 2x – y + 2 = 0 …..(1)

Midpoint of PQ is M (1, 1).

Equation of perpendicular bisector of PQ : x – y = 0 ……(2)

POI of equation (1) & (2) is circumcentre.

So, circumcentre is (–2,–2).

Question 13: If cot^-1 (ɑ) = cot^-1 2 + cot^-1 8 + cot^-1 18 + cot^-1 32 + …….. upto 100 terms, then ɑ is:

a. 1.03
b. 1.00
c. 1.01
d. 1.02

Answer: (c)

RHS = ∑_n=1¹⁰⁰ cot^-1 2n² = ∑_n=1¹⁰⁰ tan^-1 (2 / 4n²)

= ∑_n=1¹⁰⁰ tan^-1 {[(2n + 1) – (2n – 1)] / [1 + (2n + 1) (2n – 1)]}

= ∑_n=1¹⁰⁰ tan^-1 (2n + 1) – tan^-1 (2n – 1)

= tan^-1 201 – tan^-11

= tan^-1 (200 / 202)

cot^-1 ɑ = cot^-1 [101 / 100]

ɑ = 1.01

Question 14: Which of the following statements is incorrect for the function g (ɑ) for ɑ ∈ R such that g (ɑ) = ∫_π/6^π/3 sin^ɑ x / [cos^ɑ x + sin^ɑ x] dx

a. g (ɑ) is a strictly decreasing function
b. g (ɑ) has an inflexion point at ɑ = – 1/2
c. g(ɑ) is an even function
d. g(ɑ) is a strictly increasing function

Answer: (a OR b OR c or Bonus)

g (x) = π/12 by applying prop. x → a + b – x.

Question 15: If the fourth term in the expansion of (x + x^{log_{2} x})⁷ is 4480, then the value of x where x ∈ N is equal to:

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

Answer: (c)

⁷C₃ x⁴ (x^{log₂ x})³ = 4480

35x⁴ (x^{log₂ x})³ = 4480

x⁴ (x^{log₂ x})³ = 128

Take log w.r.t base 2 we get 4 log₂ x + 3 log₂ (x^{log₂ x}) = log₂ 128

Let log₂ x = y

4y + 3y² = 7

y = 1, – 7 / 3

log₂ x = 1, – 7 / 3

x = 2, x = 2^–7/3

Question 16: Two dice are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is:

a. 17 / 36
b. 4 / 9
c. 5 / 12
d. 1/2

Answer: (a)

n (S) = 36

The possible ordered pairs are (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1)

Number of ordered pair = 17

Probability = 17 / 36

Question 17: The inverse of y = 5^logx is:

a. x = 5^logy
b. x= y^log5
c. x = y^1/log5
d. x = 5^1/logy

Answer: (c)

y = 5^logx

log₅ y = log_e x

x = e^log₅y

Question 18: In a school, there are three types of games to be played. Some of the students play two types of games, but none play all three games. Which Venn diagrams can justify the above statements.

a. P and R
b. P and Q
c. None of these
d. Q and R

Answer: (c)

In are the (A), (B), (C) there are some students which play all three games hence no Venn diagram is correct.

Question 19: The area of the triangle with vertices A (z), B (iz) and C (z + iz) is:

a. (1 / 2) |z + iz|²
b. 1
c. 1 / 2
d. (1 / 2) |z|²

Answer: (d)

Area of triangle = (1 / 2) (area of square) = (1 / 2) |z|²

Question 20: The value of lim_x→0+ [cos^-1 (x – [x]²) . sin^-1 (x – [x]²)] / [x – x³], where [x] denotes the greatest integer ≤ x is:

a. 0
b. π / 4
c. π / 2
d. π

Answer: (c)

lim_x→0+ [cos^-1 (x – [x]²) . sin^-1 (x – [x]²)] / [x – x³]

lim_x→0+ [cos^-1 (x – [x]²) . sin^-1 (x – [x]²)] / [x (1 – x²)]

lim_x→0+ [cos^-1 x sin^-1 x] / x = π / 2

Section B

Question 21: Let there be three independent events E₁, E₂ and E₃. The probability that only E₁ occurs is ɑ, only E₂ occurs is β and only E₃ occurs is 𝛾. Let ‘p’ denote the probability of none of the events that occur that satisfies the equations (ɑ – 2β) p = ɑβ and (β – 3𝛾) p = 2β𝛾. All the given probabilities are assumed to lie in the interval (0, 1) Then, [probability of occurrence of E₁] / [probability of occurrence of E₃] is equal to _______

Answer: 6

Let x, y, z be probability of B₁, B₂, B₃ respectively.

⇒ x(1 – y)(1 – z) = ɑ

⇒ y(1 – x)(1 – z) = β

⇒ z(1 – x)(1 – y) = 𝛾

⇒ (1 – x) (1 – y) (1 – z) = ⍴

Putting in the given relation we get x = 2y and y = 3z

⇒ x = 6z

⇒ x / z = 6

Question 22: If the equation of the plane passing through the line of intersection of the planes 2x – 7y + 4z – 3 = 0, 3x – 5y + 4z + 11 = 0 and the point (-2, 1, 3) is ax + by + cz – 7 = 0, then the value of 2a + b + c – 7 is __________

Answer: 4

The equation of plane can be written using the family of planes: P₁ + λP₂ = 0.

(2x –7y + 4z – 3) + λ (3x – 5y + 4z + 11) = 0

It passes through (-2, 1, 3).

(- 4 – 7 + 12 – 3) + λ (- 6 – 5 + 12 + 11) = 0

– 2 + λ (12) = 0

λ = 1 / 6

12x – 42y + 24z –18 + 3x – 5y + 4z + 11 = 0

15x – 47y + 28z – 7 = 0

a = 15, b = –47, c = 28

2a + b + c –7 = 30 – 47 + 28 – 7

= 4

Question 23: If

Question 24: The minimum distance between any two points P₁ and P₂ while considering point P₁ on one circle and point P₂ one the other circle for the given circles’ equations x² + y² – 10x – 10y + 41 = 0 and x² + y² – 24x – 10y + 160 = 0 ________

Answer: 1

S₁ : (x – 5)² + (y – 5)² = 9 centre (5, 5), r₁ = 3

S₂ : (x – 12)² + (y – 5)² = 9 centre (12, 5), r₂ = 3

So (P₁P₂)_min = 1

Question 25: If (2021)³⁷⁶² is divided by 17, then the remainder is______

Answer: 4

(2021)³⁷⁶² = (2023 – 2)³⁷⁶² = multiple of 17 + 2³⁷⁶²

= 17λ + 2² (2⁴)⁹⁴⁰

= 17λ + 4 (17 – 1)⁹⁴⁰

= 17λ + 4 (17μ + 1)

17k + 4; (k ∈ I)

Remainder = 4

Question 26: If [ ^. ] represents the greatest integer function, then the value of |∫₀^√π/2 [[x²] – cosx] dx| is ________

Answer: 1

∫₀^√π/2 [[x²] – cosx] dx

= ∫₀¹ [- cosx] dx + ∫₁^√π/2 [1 – cosx] dx

= ∫₀¹ – 1 – [cosx] dx + ∫₁^√π/2 dx + ∫₁^√π/2 – 1 – [cosx] dx

= – ∫₀¹ dx + ∫₁^√π/2 dx – ∫₁^√π/2 dx

= – (x)₀¹

= – 1

The answer is 1.

Question 27: If f (x) = sin [cos^-1 (1 – 2^2x) / (1 + 2^2x)] and its first derivative with respect to x is (- b / a) log_e 2 when x = 1, where a and b are integers, then the minimum value of |a² – b²| is_______

Answer: 481

cos^-1 [1 – 4^x] / [1 + 4^x]

Let 2^x = t > 0

cos^-1 [(1 – t²) / (1 + t²)], t > 0 and t = tan θ

cos^-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)

⇒ 2θ

sin {cos^-1 [[1 – 4^x] / [1 + 4^x]]}

= sin2θ

y = [2 tanθ / (1 + tan² θ)] = 2t / (1 + t²) = [2 * 2^x] / [1 + 4^x]

dy / dx = [20 ln 2 – 32 ln 2] / 25 = – 12 ln 2 / 25

a =25, b= 12

|a² – b²|_min = |25² – 12²| = 481

Question 28: If the function f (x) = [cos (sinx) – cosx] / x⁴ is continuous at each point in its domain and f (0) = 1 / k, then k is ___________

Answer: 6

(1 / k) = lim_x→0 [{2 sin (sinx + x) / 2} sin {x – sinx} / 2] / x⁴

= lim_x→0 [(sinx + x) (x – sinx)] / 2x⁴ = {lim_x→0 [sinx + x] / 2x} {lim_x→0 (x – sinx) / x³}

= 1 * lim_x→0 [1 – cosx] / 3x² = 1 / 6

Question 29: The maximum value of z in the following equation z = 6xy + y², where 3x + 4y ≤ 100 and 4x + 3y ≤ 75 for x ≥ 0 and y ≥ 0 is _________

Answer: 625

x ≥ 0, y ≥ 0

4x + 3y ≤ 75

3x + 4y ≤ 100

(0, 0), z = 0 + 0 = 0

z_max = 625 at x = 0, y = 25