JEE Main 2022 June 25 – Shift 2 Maths Question Paper with Solutions

The JEE Main 2022 June 25 – Shift 2 Maths Question Paper with Solutions are provided on this page. JEE Main 2022 answer keys are prepared by subject matter experts at BYJU’S. These are prepared in a step-by-step manner, to help students easily understand the tricks used in solving the problems. Students can also download the JEE Main 2022 June 25 – Shift 2 Maths Question Paper with Solutions in PDF format. Practising the JEE Main 2022 question papers will help students to prepare effectively for the upcoming exams. It will also help them to evaluate the difficulty level of the JEE Main 2022 exam.

JEE Main 2022 June 25th Shift 2 Maths Question Paper and Solutions

SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. Let A = {x ∈ R : | x + 1 | < 2} and B = {x ∈ R : | x – 1| ≥ 2}. Then which one of the following statements is NOT true?

(A) A – B = (–1, 1)

(B) B – A = R – (–3, 1)

(C) A ⋂ B = (–3, –1]

(D) A U B = R – [1, 3)

Answer (B)

Sol. A = (–3, 1) and B = (–∞, –1] ∪ [3, ∞)

So, A – B = (–1, 1)

B – A = (–∞, –3] ∪ [3, ∞) = R – (–3, 3)

A ∩ B = (–3, –1]

and A ∪ B = (–∞, 1) ∪ [3, ∞) = R – [1, 3)

2. Let a, b ∈ R be such that the equation ax² – 2bx + 15 = 0 has a repeated root α. If α and β are the roots of the equation x² – 2bx + 21 = 0, then α² + β² is equal to

(A) 37

(B) 58

(C) 68

(D) 92

Answer (B)

Sol. ax² – 2bx + 15 = 0 has repeated root so b² = 15a

and

∵α is a root of x² – 2bx + 21 = 0

So

Now α² + β² = (α + β)² – 2αβ = 4b² – 42 = 100 – 42

= 58

3. Let z₁ and z₂ be two complex numbers such that

Answer (C)

Sol.

Also

⇒ arg(z₁) + arg(z₂) = π …(ii)

From (i) and (ii), we get

4. The system of equations

–kx + 3y – 14z = 25

–15x + 4y – kz = 3

–4x + y + 3z = 4

is consistent for all k in the set

(A) R

(B) R – {–11, 13}

(C) R – {13}

(D) R – {–11, 11}

Answer (D)

Sol. The system may be inconsistent if

Hence if system is consistent then

5.

(A) 1/12

(B) -1/18

(C) -1/12

(D) 1/6

Answer (A)

Sol.

6. The area of the region enclosed between the parabolas y² = 2x – 1 and y² = 4x – 3 is

(A) ⅓

(B) ⅙

(C) ⅔

(D) ¾

Answer (A)

Sol. Area of the shaded region

7. The coefficient of x¹⁰¹ in the expression (5 + x)⁵⁰⁰ + x(5 + x)⁴⁹⁹ + x²(5 + x)⁴⁹⁸ + ……+ x⁵⁰⁰, x > 0, is

(A) ⁵⁰¹C₁₀₁ (5)³⁹⁹

(B) ⁵⁰¹C₁₀₁ (5)⁴⁰⁰

(C) ⁵⁰¹C₁₀₀ (5)⁴⁰⁰

(D) ⁵⁰⁰C₁₀₁ (5)³⁹⁹

Answer (A)

Sol.

8. The sum 1 + 2 ⋅ 3 + 3 ⋅ 3² + …. + 10 ⋅ 3⁹ is equal to

Answer (B)

Sol. Let

9. Let P be the plane passing through the intersection of the planes

(A) X and X + Y are on the same side of P

(B) Y and Y – X are on the opposite sides of P

(C) X and Y are on the opposite sides of P

(D) X + Y and X – Y are on the same side of P

Answer (C)

Sol. Let the equation of required plane

⇒ λ = 1

Hence,

Clearly, X and Y are on opposite sides of plane π.

10. A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is

Answer (D)

Sol. Let the centre be (h, k)

So,

⇒ 2h² = h² + k² + 2hk

Locus will be x² – y² = 2xy

11. Water is being filled at the rate of 1 cm³/sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm²/sec) at which the wet conical surface area of the vessel increase, is

(A) 5

Answer (C)

Sol.

Now,

12. If

(A) b₃ – b₂, b₄ – b₃, b₅ – b₄ are in an A.P. with a common difference –2

(C) b₃ – b₂, b₄ – b₃, b₅ – b₄ are in a G.P.

Answer (D)

Sol.

So, b₃ – b₂, b₄ – b₃, b₅ – b₄ are in H.P.

13. If y = y(x) is the solution of the differential equation

(A) ⅓

(B) ⅔

(C) 3/2

(D) 3

Answer (B)

Sol.

Now,at

14. If the angle made by the tangent at the point (x₀, y₀) on the curve x = 12(t + sin t cos t),

(C) 27

(D) 48

Answer (C)

Sol.

So,

15. The value of 2 sin(12°) – sin(72°) is :

Answer (D)

Sol.

16. A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is 1/n. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

Answer (D)

Sol. There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)

So, the required probability

17. The negation of the Boolean expression ((~ q) ∧ p) ⇒ ((~ p) ∨ q) is logically equivalent to :

(A) p ⇒ q

(B) q ⇒ p

(C) ~ (p ⇒ q)

(D) ~ (q ⇒ p)

Answer (C)

Sol. Let S : ((~ q) ∧ p) ⇒ ((~ p) ∨ q)

⇒ S : ~ ((~ q) ∧ p) ∨ ((~ p) ∨ q)

⇒ S : (q ∨ (~ p)) ∨ ((~ p) ∨ q)

⇒ S : (~ p) ∨ q

⇒ S : p ⇒ q

So, negation of S will be ~ (p ⇒ q)

18. If the line y = 4 + kx, k > 0, is the tangent to the parabola y = x – x² at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :

Answer (C)

Sol. ∵ Line y = kx + 4 touches the parabola y = x – x².

So, kx + 4 = x – x² ⇒ x² + (k – 1) x + 4 = 0 has only one root

(k – 1)² = 16 ⇒ k = 5 or – 3 but k > 0

So, k = 5.

And hence x² + 4x + 4 = 0 ⇒ x = – 2

So, P(–2, –6) and V is

Slope of PV

19. The value of

Answer (B)

Sol.

20. The line y = x + 1 meets the ellipse

(A) 20

(B) 12

(C) 11

(D) 8

Answer (A)

Sol. Let point (a, a + 1) as the point of intersection of line and ellipse.

So,

If the roots of this equation are α and β.

So, P(α, α + 1) and Q(β, β + 1)

PQ² = 4r² = (α – β)² + (α – β)²

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, 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.

Answer (1)

Sol.

nA^n + mB^m = nA + mB

So, 2n – m = 1, –n + m = 0, 2m – n = 1

So, (m, n) = (1, 1)

2. Let f(x) = [2x² + 1] and

Answer (62)

Sol.

The possible points where fog(x) may be discontinuous are

2(2x – 3)² ∈ I & x ∈ (–1, 0)

2(2x + 3)² ∈ I & x ∈ [0, 1)

So, total = 62

3. The value of b > 3 for which

Answer (6)

Sol.

⇒ b = 6

4. If the sum of the co-efficients of all the positive even powers of x in the binomial expansion of

Answer (83)

Sol.

So, r ≠ 8, 9, 10

Sum of required Coeff.

5. If the mean deviation about the mean of the numbers 1, 2, 3, …. n, where n is odd, is 5(n + 1)/n, then n is equal to ___________.

Answer (21)

Sol. Mean

So, n = 21

6.

Answer (14)

Sol. Let

So,

and x + y + λz = –21

⇒ clearly, λ = 3, x = –2, y = 2 and z = –7

So,

and

7. The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

Answer (243)

Sol. C-1 : All digits are non-zero

C-2 : One digit is 0

Total = 216 + 27 = 243

8. Let f(x) = |(x – 1)(x² – 2x – 3)| + x – 3, x ∈ R. If m and M are, respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to

Answer (3)

Sol.

f(x) = |(x – 1)(x² – 2x – 3)| + x – 3

So, 3 points

9. Let the eccentricity of the hyperbola

Answer (85)

Sol.

So,

Also, (8/√5, 12/5) lies on the given hyperbola.

So,

Equation of normal

So,

Gives

10. Let l₁ be the line in xy-plane with x and y intercepts ⅛ and

Answer (51)

Sol.

Equation of L₂

d^–2 = 51

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