JEE Main 2022 June 29 – Shift 1 Maths Question Paper with Solutions

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JEE Main 2022 June 29th Maths Shift 1 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. The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :

Answer (C)

Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Let E be the event that matrix of order 2 × 2 is singular

Case-I

All entries are same example

= ¹⁰C₁

Case-II

Matrix with two prime numbers only example

= ¹⁰C₂ × 2! × 2!

2. Let the solution curve of the differential equation

(A) 15

(B) 11

(C) 13

(D) 17

Answer (A)

Sol.

y = 4x tan θ

log |secθ + tanθ| = log |x| + C

y(1) = 3 ⇒ 3 = 4 tanθ

⇒ C = ln 2

∴ |secθ + tanθ| = 2|x|

To find y(2) put x = 2

(secθ + tanθ)² = 16

⇒ y = 15

3. If the mirror image of the point (2, 4, 7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to:

(A) 54

(B) 50

(C) –6

(D) –42

Answer (C)

Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is

(a, b, c) then

= –6

4. Let ƒ : R ⇒ R be a function defined by :

where [t] is the greatest integer less than or equal to t. Let m be the number of points where ƒ is not differentiable and

Answer (C)

Sol.

g(t) = t³ –3t ⇒ g′(t) = 3t² –3 = 3(t – 1)(t + 1)

Points of non-differentiability = {2, 3, 4, 5}

⇒ m = 4

5. Let

(A) 3

(B) 4

(C) 5

(D) 6

Answer (A)

Sol.

∴ α = 2

∴ β = 1

α + β = 2 + 1 = 3

6. The area enclosed by y² = 8x and y = √2x that lies outside the triangle formed by

Answer (C)

Sol.

7. If the system of linear equations

2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k, where δ, k ∈ R

has infinitely many solutions, then δ + k is equal to:

(A) –3

(B) 3

(C) 6

(D) 9

Answer (B)

Sol. 2x + y – z = 7

x – 3y + 2z = 1

x + 4y + δz = k

⇒ 6 – k = 0 ⇒k = 6

δ + k = -3 + 6 = 3

8. Let α and β be the roots of the equation x² + (2i – 1) = 0. Then, the value of |α² + β²| is equal to:

(A) 50

(B) 250

(C) 1250

(D) 1500

Answer (A)

Sol. x² + 2i – 1 = 0

α² = β² = 1 – 2i

α⁴ = (1 – 2i)² = 1 + (2i)² – 4i = –3 – 4i

α⁸ = (–3 – 4i)² = 9 – 16 + 24i = –7 + 24i

9. Let

Answer (C)

Sol.

Now

∴ Δ = ⇒

10. Let A = [a_ij] be a square matrix of order 3 such that a_ij = 2^j–i, for all i, j = 1, 2, 3. Then, the matrix A² + A³ + … + A¹⁰ is equal to :

Answer (A)

Sol.

A² = 3A

A³ = A.A² = A(3A) = 3A² = 3²A

A⁴ = 3³A

Now

A² + A³ + … + A¹⁰

A[3¹ + 3² + 3³ + … + 3⁹]

11. Let a set A = A₁ ⋃ A₂ ⋃ …⋃ A_k, where A_i ⋂ A_j = Φ for i ≠ j, 1 ≤ i, j ≤ k. Define the relation R from A to A by R = {(x, y) : y ∈ A_i if and only if x ∈ A_i, 1 ≤ i ≤ k}. Then, R is :

(A) reflexive, symmetric but not transitive

(B) reflexive, transitive but not symmetric

(C) reflexive but not symmetric and transitive

(D) an equivalence relation

Answer (D)

Sol. R = {(x, y) :y ∈ A_i, iff x ∈ A_i, 1 ≤ i ≤ k}

(1) Reflexive

(a, a) ⇒ a ∈ A_i iff a ∈ A_i

(2) Symmetric

(a, b) ⇒a ∈A_i iff b ∈ A_i

(b, a) ∈ R as b ∈ A_i iff a ∈ A_i

(3) Transitive

(a, b) ∈ R & (b, c) ∈ R.

⇒ a ∈ A_i iff b ∈ A_i & b ∈ A_i iff c ∈ A_i

⇒ a ∈ A_i iff c ∈ A_i

⇒ (a, c) ∈ R.

⇒ Relation is equivalence

12. Let

Answer (B)

Sol.

a_n+2 = 2a_n+1 – a_n + 1 & a₀ = a₁ = 0

a₂ = 2a₁ – a₀ + 1 = 1

a₃ = 2a₂ – a₁ + 1 = 3

a₄ = 2a₃ – a₂ + 1 = 6

a₅ = 2a₄ – a₃ + 1 = 10

13. The distance between the two points A and A′ which lie on y = 2 such that both the line segments AB and A′B (where B is the point (2, 3)) subtend angle π/4 at the origin, is equal to

(A) 10

(B) 48/5

(C) 52/5

(D) 3

Answer (C)

Sol. Let A(α, 2) Given B(2, 3)

4 – 3α = 2α + 6 & 4 – 3α = -2α – 6

14. A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is

Answer (B)

Sol.

4a + 3b = 22

Total area = A

15. The domain of the function

Answer (D)

Sol.

From (1) & (2)

16. If the constant term in the expansion of

(A) 6

(B) 7

(C) 8

(D) 9

Answer (D)

Sol. Constant term in

⇒ (3x⁸ – 2x⁷ + 5)¹⁰→x⁵⁰

General term of (3x⁸ – 2x⁷ + 5)¹⁰ is

Here 8p + 7q = 50 and p + q+r = 10

⇒ p = 1, q = 6, r = 3 is only valid solution

∴ k = 9

17.

(A) –3

(B) –2

(C) 2

(D) 0

Answer (D)

Sol.

= 0 + 0 + 0 = 0

18. Let PQ be a focal chord of the parabola y² = 4x such that it subtends an angle of π/2 at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse

(A) 1 + √2

(B) 3 + 2√2

(C) 1 + 2√3

(D) 4 + 5√3

Answer (B)

Sol.

As

⇒ t = ± 1

∴ P ≡ (1, 2) & Q(1, –2)

∴ for ellipse

⇒ (5 – e²)e² = 1 – e²

⇒ e⁴ – 6e² + 1 = 0

19. Let the tangent to the circle C₁: x² + y² = 2 at the point M(–1, 1) intersect the circle C₂: (x – 3)² + (y – 2)² = 5, at two distinct points A and B. If the tangents to C₂ at the points A and B intersect at N, then the area of the triangle ANB is equal to

Answer (C)

Sol. Tangent toC₁ at M: –x + y = 2 ≡T

Intersection of T with C₂⇒ (x – 3)² + x² = 5

⇒ x = 1, 2

A(1, 3) and B(2, 4)

Let N ≡ (α, β)

Then –x + y = 2 shall be chord of contact for x² + y² – 6x – 4y + 8 = 0

20. Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be 24/5 and 194/25, respectively. If the mean and variance of the first 4 observations are 7/2 and a, respectively, then (4a + x₅) is equal to

(A) 13

(B) 15

(C) 17

(D) 18

Answer (B)

Sol.

⇒ x₅ = 10

⇒ x₅ + 4a = 10 + 5 = 15

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. Let

Answer (26)

Sol.

S represents the shaded region shown in the diagram.

Clearly, z₁ will be the point of intersection of PA and given circle.

PA : 2x + y = 4 and given circle has equation

(x – 2)² + y² = 1.

On solving, we get

z₂ will be either B or C.

So

Clearly α = 30 and β = –4 ⇒α + β = 26

2. Let y = y(x) be the solution of the differential equation

Answer (2)

Sol.

When

When

So

3. Let d be the distance between the foot of perpendiculars of the point P(1, 2, –1) and Q(2, –1, 3) on the plane –x + y + z = 1. Then d² is equal to ____.

Answer (26)

Sol. Foot of perpendicular from P

and foot of perpendicular from Q

⇒ d² = 26

4. The number of elements in the set

Answer (32)

Sol.

⇒ 3cos²2θ + cos2θ = 0

⇒ cos2θ = 0 or cos2θ = -1/3

As θ ∈[0, π], cos2θ = -⅓ ⇒ 2 times

⇒ θ ∈[–4π, 4π], cos2θ = -1/3 ⇒ 16 times

Similarly, cos2θ = 0 ⇒ 16 times

∴ Total 32 solutions

5. The number of solutions of the equation

Answer (1)

Sol.

1 point of intersection = 1 solution.

6.

Answer (29)

Sol.

Where

7. Let c, k∈ R. If f(x) = (c + 1)x² + (1 – c²)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of

Answer (3395)

Sol. f(x) is polynomial

Put y = 1/x in given functional equation we get

⇒ 2(c + 1) = 2K – 1 …(1)

and put x = y = 0 we get

∴ k = 0 and 2c = –3 ⇒c = –3/2

8. Let H:

Answer (88)

Sol.

and

By (3) and (4)

⇒ a² + b² = 32 + 56 = 88

9. Let

Answer (28)

Sol. Direction ratio of normal to P₁≡< 2, 1, – 3 >

and that of

i.e.< –5, 5, 1 >

d.r’s of line of intersection are along vector

i.e.< 16, 13, 15 >

∴ α + β = 13 + 15 = 28

10. Let b₁b₂b₃b₄ be a 4-element permutation with b_i{1, 2, 3,…..,100} for 1 ≤i≤ 4 and b_i ≠b_j for i ≠ j, such that either b₁, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁b₂b₃b₄ is equal to _______.

Answer (18915)

Sol. There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.

Using the principle of inclusion and exclusion,

Number of permutations of b₁b₂b₃b₄ = Number of permutations when b₁b₂b₃ are consecutive + Number of permutations when b₂b₃b₄ are consecutive – Number of permutations when b₁b₂b₃b₄ are consecutive

=97 × 98 + 97 × 98 – 97 = 97 × 195 = 18915.

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