JEE Main 2022 June 28 – Shift 1 Maths Question Paper with Solutions

The JEE Main 2022 June 28 – Shift 1 Maths Question Paper with Solutions are provided on this page. The JEE Main 2022 answer keys are prepared by experts at BYJU’S. Students are recommended to revise and learn the JEE Main 2022 question paper solutions to score better marks in the JEE Main exams. Practising these solutions will definitely improve the confidence of the students. Find the JEE Main 2022 June 28 – Shift 1 Maths Question Paper with Solutions below.

JEE Main 2022 June 28th 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. If

(A) 1411

(B) 1320

(C) 1615

(D) 1855

Answer (A)

Sol.

= ⁶²C₃₂ – ⁶⁰C₃₁

2. Let a function ƒ : N →N be defined by

(A) One-one but not onto

(B) Onto but not one-one

(C) Neither one-one nor onto

(D) One-one and onto

Answer (D)

Sol. When n = 1, 5, 9, 13 then (n + 1)/2 will give all odd numbers.

When n = 3, 7, 11, 15 …

n – 1 will be even but not divisible by 4

When n = 2, 4, 6, 8, …

Then 2n will give all multiples of 4

So range will be N.

And no two values of n give same y, so function is one-one and onto.

3. If the system of linear equations

2x + 3y – z = –2

x + y + z = 4

x – y + |λ|z = 4λ – 4

where λ∈ R, has no solution, then

(A) λ = 7

(B) λ = –7

(C) λ = 8

(D) λ² = 1

Answer (B)

Sol.

But at λ = 7, D_x = D_y = D_z = 0

P₁ : 2x + 3y – z = –2

P₂ :x + y + z = 4

P₃ :x – y + |λ|z = 4λ – 4

So clearly, 5P₂ – 2P₁ = P₃, so at λ = 7, system of equation is having infinite solutions.

So λ = –7 is correct answer.

4. Let A be a matrix of order 3 × 3 and det (A) = 2. Then det (det (A) adj (5 adj (A³))) is equal to ______.

(A) 512 × 10⁶

(B) 256 × 10⁶

(C) 1024 × 10⁶

(D) 256 × 10¹¹

Answer (A)

Sol. |A| = 2

||A| adj(5 adjA³)|

= |25|A| adj(adjA³)|

= 25³ |A|³⋅ |adjA³|²

= 25³⋅ 2³⋅ |A³|⁴

= 25³⋅ 2³⋅ 2¹² = 10⁶⋅ 512

5. The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

(A) 36

(B) 48

(C) 60

(D) 72

Answer (D)

Sol. Number should be divisible by 6 and it should be even.

Total sum = 1 + 2 + 3 + 5 + 6 + 7 = 24

So number removed should be of type 3.

Total cases = 48 + 24 = 72

6. Let A₁, A₂, A₃, … be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = 1/1296 and A₂ + A₄ = 7/36 then, the value of A₆ + A₈ + A₁₀ is equal to

(A) 33

(B) 37

(C) 43

(D) 47

Answer (C)

Sol.

So A₆ + A₈ + A₁₀ = 1 + 6 + 36

= 43

7. Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral

(A) –1

Answer (C)

Sol.

8. Let f: ℝ → ℝ be defined as

Where a, b, c ∈  ℝ and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?

(A) There exists a, b, c ∈  ℝ  such that ƒiscontinuous on ∈  ℝ .

(B) If ƒ is discontinuous at exactly one point, then a + b + c = 1

(C) If ƒ is discontinuous at exactly one point, then a + b + c ≠ 1

(D) ƒ is discontinuous at atleast two points, for any values of a, b and c

Answer (C)

Sol.

To be continuous at x = 0

a – 1 = 0

to be continuous at x = 1

ae – 1 = b = b – 1 ⇒ not possible

to be continuous at x = 2

b – 1 = – c⇒b + c = 1

If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.

9. The area of the region

Answer (B)

Sol.

Required area

10. Let the solution curve y = y(x) of the differential equation

Answer (A)

Sol.

Putting y = tx

⇒ sin^–1t + e^t = lnx + C

atx = 1, y = 0

So, 0 + e⁰ = 0 + C⇒C = 1

at (2α, α)

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

(A) –18

(B) –12

(C) –6

(D) –3

Answer (A)

Sol.

Solution of D.E. can be given by

at x = 2, y = –2

12. The number of real solutions of x⁷ + 5x³ + 3x + 1 = 0 is equal to ______.

(A) 0

(B) 1

(C) 3

(D) 5

Answer (B)

Sol.

f(x) is always increasing.

So clearly, it intersects the x-axis at only one point.

13. Let the eccentricity of the hyperbola

(A) 18

(B) 20

(C) 24

(D) 32

Answer (B)

Sol.

14. If the tangents drawn at the points O(0, 0) and P(1 + √5, 2) on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

Answer (C)

Sol.

= 5cot θ.cos²θ

15. If two distinct points Q, R lie on the line of intersection of the planes –x + 2y – z = 0 and 3x – 5y + 2z = 0 and

Answer (B)

Sol.

Line L is x = y = z

⇒ (α – 3) + α + 2 + α – 1 = 0

So, Area

16. The acute angle between the planes P₁ and P₂, when P₁ and P₂ are the planes passing through the intersection of the planes 5x + 8y + 13z – 29 = 0 and 8x – 7y + z – 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is

Answer (A)

Sol. Family of Plane’s equation can be given by

(5 + 8λ)x + (8 – 7λ)y + (13 + λ) z – (29 + 20λ) = 0

P₁ passes through (2, 1, 3)

⇒ (10 + 16λ) + (8 – 7λ) + (39 + 3λ) – (29 + 20λ) = 0

⇒ –8λ + 28 = 0

d.r,s of normal to P₁

P₂ passes through (0, 1, 2)

• 8 – 7λ + 26 + 2λ – (29 + 20λ) = 0

⇒ 5 – 25λ = 0

d.r, s of normal to P₂

Angle between normals

17. Let the plane

(A) 90

(B) 93

(C) 95

(D) 97

Answer (B)

Sol. P₁: x + 3y – z = 6

P₂: –6x + 5y – z = 7

Family of planes passing through line of intersection of P₁ and P₂ is given by x(1 – 6λ) + y(3 + 5λ) + z (–1 – λ) – (6 + 7λ) = 0

It passes through (2, 3, 1/2)

So,

Required plane is

–5x + 8y – 2z – 13 = 0

Or

18. The probability, that in a randomly selected 3-digit number at least two digits are odd, is

Answer (A)

Sol. Required cases = Total – all digits even – exactly one digit even

Total = 900 ways

19. Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let π/8 and θ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan²θ is equal to

Answer (C)

Sol.

From (i) and (ii)

20. Let p, q, r be three logical statements. Consider the compound statements

S₁ : ((~p) ∨q) ∨ ((~p) ∨r) and

S₂ :p→ (q∨r)

Then, which of the following is NOT true?

(A) If S₂ is True, then S₁ is True

(B) If S₂is False, then S₁ is False

(C) If S₂ is False, then S₁ is True

(D) If S₁ is False, then S₂ is False

Answer (C)

Sol. S₁: (~p∨q) ∨ (~ p∨r)

≅ (~ p∨q∨r)

S₂: ~p ∨ (q∨r)

Both are same

So, option (C) is incorrect.

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 R₁ and R₂ be relations on the set {1, 2, ….., 50} such that R₁ ={(p, pⁿ) :p is a prime and n≥ 0 is an integer} and R₂ = {(p, pⁿ) : p is a prime and n = 0 or 1}. Then, the number of elements in R₁ – R₂ is ______.

Answer (8)

Sol. R₁ – R₂ = {(2, 2²), (2, 2³), (2, 2⁴), (2, 2⁵), (3, 3²)

(3, 3³), (5, 5²), (7, 7²)}

So number of elements = 8

2. The number of real solutions of the equation is _____.

Answer (2)

Sol. Dividing by e^2x

e^2x + 4e^x – 58 + 4e^–x + e^–2x = 0

⇒ (e^x + e^–x)² + 4(e^x + e^–x) – 60 = 0

Let e^x + e^–x= t∈[2, ∞)

⇒ t² + 4t – 60 = 0

⇒ t = 6 is only possible solution

e^x + e^–x = 6 ⇒e^2x – 6e^x + 1 = 0

Let e^x = p,

p² – 6p + 1 = 0

So

3. The mean and standard deviation of 15 observations are found to be 8 and 3, respectively. On rechecking, it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _______.

Answer (17)

Sol.

∑x_c²1095 – 25 + 400 = 1470

Correctvariance

4. If

Answer (150)

Sol. 2C₁ + C₂ + 3C₃ = 5 …(i)

3C₁ + 3C₂ + C₃ = 0 …(ii)

= 2(3C₃ – C₂) – 1(3C₃ – C₁) + 3(3C₂ – 3C₁)

= 3C₃ + 7C₂ – 8C₁

⇒ 8C₁ – 7C₂ – 3C₃ = 0 …(iii)

So 122(C₁ + C₂ + C₃) = 150

5. A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (α, β). Then, the value of 7α + 3β is equal to ___________.

Answer (31)

Sol.

Bisector of angle PAQ is X = 23/7.

So, 7α + 3β = 31

6. Let l be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line l and O is origin, then the area of the triangle OPQ is equal to ___________.

Answer (13)

Sol.

⇒ 6 – x₁ = 8x₁³ + 6x₁² – 7x₁ – 2

⇒ 8x₁³ + 6x₁² – 6x₁ – 8 = 0

So x₁ = 1 ⇒y₁ = 5

7. Let A = {1, a₁, a₂…a₁₈, 77} be a set of integers with 1 <a₁<a₂<….<a₁₈< 77. Let the set A + A = {x + y :x, y∈A} contain exactly 39 elements. Then, the value of a₁ + a₂ +…+a₁₈ is equal to _____.

Answer (702)

Sol. If we write the elements of A + A, we can certainly find 39 distinct elements as 1 + 1, 1 + a₁, 1 + a₂,…..1 + a₁₈, 1 + 77, a₁ + 77, a₂ + 77,……a₁₈ + 77, 77 + 77.

It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.

Let the common difference be ‘d’.

77 = 1 + 19d ⇒d = 4

So,

8. The number of positive integers k such that the constant term in the binomial expansion of

Answer (2)

Sol.

For constant term 36 – 3r – kr = 0

So, k can be 1, 3, 6, 9, 15, 33

In order to get 2⁸, check by putting values of k and corresponding in general term. By checking, it is possible only where k = 3 or 6

9. The number of elements in the set

Answer (40)

Sol.

at line y = –2, we have (5, –2) (6, –2) (1, –2) (0, –2) ⇒ 4 points

at line y = –1, we have (4, –1) (5, –1) (6, –1) (2, –1) (1, –1) (0, –1) ⇒ 6 points

at line y = 0, we have (0, 0) (1, 0) (2, 0) (3, 0) (4, 0) (5, 0) (6, 0) ⇒ 7 points

at line y = 1, we have (1, 1), (2, 1), (3, 1), (4, 1), (5, 1) i.e. 5 points

symmetrically

at line y = –5, we have 5 points

at line y = –4, we have 7 points

at line y = –3, we have 6 points

So Total integral points = 2(5 + 7 + 6) + 4

= 40

10. Let the lines

Answer (816)

Sol. L₁ : y + 2x = √11 + 7√7

L₂ : 2y + x = 2√11 + 6√7

Point of intersection of these two lines is centre of circle i.e.

So (5h – 8K)² + 5r²

= 64 × 11 + 112 = 816.

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