JEE Main 2022 June 24 – Shift 1 Maths Question Paper with 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 = {z ∈ C : 1 ≤ |z – (1 + i)| ≤ 2} and B = {z ∈ A : |z – (1 – i)| = 1}. Then, B :

(A) Is an empty set

(B) Contains exactly two elements

(C) Contains exactly three elements

(D) Is an infinite set

Answer (D)

Sol.

Set A represents region 1 i.e.R₁ and clearly set B has infinite points in it.

2. The remainder when 3²⁰²² is divided by 5 is :

(A) 1

(B) 2

(C) 3

(D) 4

Answer (D)

Sol. 3²⁰²² = (10 – 1)¹⁰¹¹

= ¹⁰¹¹C₀(10)¹⁰¹¹ (–1)⁰ + ¹⁰¹¹C₁(10)¹⁰¹⁰ (–1)¹ + ….. + ¹⁰¹¹C₁₀₁₀(10)¹ (–1)¹⁰¹⁰ + ¹⁰¹¹C₁₀₁₁(10)⁰ (–1)¹⁰¹¹

= 5k –1, where k ∈ I

So when divided by 5, it leaves remainder 4.

3. The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

(A) 9

(B) 10

(C) 11

(D) 12

Answer (A)

Sol. S = 4πr²

at t = 0, r = 3

at t = 5,

At t = 9,

So, r = 9

4. Bag A contains 2 white, 1 black and 3 red balls and bas B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is

(A) 13

(B) 6

(C) 4

(D) 3

Answer (C)

Sol.

⇒ n² + 9n – 52 = 0

⇒ n = 4 is only possible value

5. Let

(A) 16

(B)

(C) 72

(D) –8

Answer (A)

Sol. For tangent to parabola y = x² at (2, 4)

Equation of tangent is

y – 4 = 4(x – 2)

⇒ 4x – y – 4 = 0

Family of circle can be given by

(x – 2)² + (y – 4)² + λ(4x – y – 4) = 0

As it passes through (0, 6)

2² + 2² + λ(–10) = 0

Equation of circle is

So, A + C = 16

6. The number of values of α for which the system of equations :

x + y + z = α

αx + 2αy + 3z = –1

x + 3αy + 5z = 4

is inconsistent, is

(A) 0

(B) 1

(C) 2

(D) 3

Answer (B)

Sol.

= 1(10α – 9α) – 1 (5α – 3) + 1 (3α² – 2α)

= α – 5α + 3 + 3α² – 2α

= 3α² – 6α + 3

For inconsistency Δ = 0 i.e. α = 1

Now check for α = 1

x + y + z = 1 …(i)

x + 2y + 3z = –1 …(ii)

x + 3y + 5z = 4 …(iii)

By (ii) × 2 – (i) × 1

x + 3y + 5z = –3

so equations are

inconsistent for α = 1

7. If the sum of the squares of the reciprocals of the roots α and β of the equation 3x² + λx – 1 = 0 is 15, then 6(α³ + β³)² is equal to :

(A) 18

(B) 24

(C) 36

(D) 96

Answer (B)

Sol.

α³ + β³ = (α + β) (α² + β² – αβ)

8. The set of all values of k for which

(A)

(B)

(C)

(D)

Answer (A)

Sol. Let

Max will occur around

Range of

9. Let

Let

If

(A) 218

(B) 221

(C) 663

(D) 1717

Answer (B)

Sol. |A| = a² + 1

|adj A| = (a² + 1)²

= 2² (1² + 2² + 3² + … + 25²)

λ = 221

10. For the function

f(x) = 4 log_e(x – 1) – 2x² + 4x + 5, x > 1, which one of the following is NOT correct?

(A) f is increasing in (1, 2) and decreasing in (2, )

(B) f(x) = –1 has exactly two solutions

(C) f′(e) – f′′(2) < 0

(D) f(x) = 0 has a root in the interval (e, e + 1)

Answer (C)

Sol.

So maxima occurs at x = 2

f(2) = 4⋅0 – 2⋅2² + 4⋅2 + 5 = 5

so clearly f(x) = –1 has

exactly 2 solutions

so f′(e) – f ′′(2) > 0

so option c is not correct

11. If the tangent at the point (x₁, y₁) on the curve y = x³ + 3x² + 5 passes through the origin, then (x₁, y₁) does NOT lie on the curve :

(A)

(B)

(C) y = 4x² + 5

(D)

Answer (D)

Sol.

m_op – m_Tangent

So, (x₁, y₁) = (1, 9)

12. The sum of absolute maximum and absolute minimum values of the function f(x) = |2x² + 3x – 2| + sinx cosx in the interval [0, 1] is :

(A)

(B)

(C)

(D)

Answer (B)

Sol.

f‘(x) = –4x – 3 + cos2x <0

For

So, minima occurs at

So, maxima is possible at x = 0 or x = 1

Now checking for x = 0 and x = 1, we can see it attains its maximum value at x = 1

Sum of absolute maximum and minimum value

13. If

(A) 48

(B) 96

(C) 92

(D) 104

Answer (B)

Sol.

a₂ + a₄ + a₆ + … + a_n = 120

⇨ n = 96

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

(A) e³(e^e – 1)

(B) e^e(e³ – 1)

(C) e²(e^e + 1)

(D) e^e(e² – 1)

Answer (A)

Sol.

If

Solution is given by

⇒ x = y² (ye^y + c)

at, y = 1, x = 0

⇒ 0 = 1(1.e¹ + c) ⇒ c = –e

at y = e,

x = e²(e.e^e – e)

15. Let λx – 2y = μ be a tangent to the hyperbola a²x² – y² = b². Then

(A) –2

(B) –4

(C) 2

(D) 4

Answer (D)

Sol.

Tangent in slope form ⇒

i.e., same as

Comparing coefficients,

Eliminating

16. Let

(A)

(B)

(C)

(D)

Answer (C)

Sol.

17. If a random variable X follows the Binomial distribution B(33, p) such that 3P(X = 0) = P(X = 1), then the value of

(A) 1320

(B) 1088

(C)

(D)

Answer (A)

Sol. 3P(X = 0) = P(X = 1)

⇒ 1 – P = 11P

⇒ 11³ – 11 = 1320

18. The domain of the function

(A)

(B) (2, ∞)

(C)

(D)

Answer (D)

Sol.

Solution to this inequality is

for x² – 3x + 2 > 0 and ≠ 1

Combining the two solution sets (taking intersection)

19. Let

(A)

(B) 9

(C)

(D) 10

Answer (B)

Sol. tanθ (sinθ + 1) – sin2θ = 0

tanθ (sinθ + 1 – 2cos²θ) = 0

⇒ tanθ = 0 or 2sin²θ + sinθ – 1 = 0

⇒ (2sinθ + 1)(sinθ – 1) = 0

⇒

But, sinθ = 1 not possible

n(S) = 5

= 4

20. The number of choices for

(A) 1

(B) 2

(C) 3

(D) 4

Answer (B)

Sol. Let x : (pΔq) ⇒ (pΔ ~ q) ∨ (~pΔq)

Case-I

When Δ is same as ∨

Then (pΔ~q) ∨ (~pΔq) becomes

(p ∨ ~q) ∨ ( ~p ∨ q) which is always true, so x becomes a tautology.

Case-II

When Δ is same as ∧

Then (p ∧ q) ⇒ (p ∧ ~q) ∨ (~p ∧ q)

If p ∧ q is T, then (p ∧ ~ q) ∨ (~ p ∧ q) is F

so x cannot be a tautology.

Case-III

When Δ is same as ⇒

Then (p ⇒ ~ q) ∨ (~ p ⇒ q) is same at (~ p ∨ ~ q) ∨ (p ∨ q), which is always true, so x becomes a tautology.

Case-IV

When Δ is same as ⇔

Then (p ⇔ q) ⇒ (p ⇔ ~ q) ∨ (~ p ⇔ q)

p ⇔ q is true when p and q have same truth values, then p ⇔ ~ q and ~p ⇔ q both are false. Hence x cannot be a tautology.

So finally x can be ∨ or ⇒.

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 andw the on-screen virtual numeric keypad in the place designated to enter the answer.

1. The number of one-one functions ƒ : {a, b, c, d} → {0, 1, 2, …, 10} such that 2ƒ(a) – ƒ(b) + 3ƒ(c) + ƒ(d) = 0 is ________.

Answer (31)

Sol. 3ƒ(c) + 2ƒ(a) + ƒ(d) = ƒ(b)

2. In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, –2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is _____.

Answer (40)

Sol. Let student marks x correct answers and y incorrect. So

3x – 2y = 5 and x + y 5 where x, y W

Only possible solution is (x, y) = (3, 2)

Student can mark correct answer by only one choice but for incorrect answer, there are two choices. So total number of ways of scoring 5 marks = ⁵C₃(1)³. (2)² = 40

3. Let

Answer (8)

Sol. Clearly B is

Area of

⇒

4. Let a line having direction ratios 1, –4, 2 intersect the lines

Answer (84)

Sol. Let A(3λ + 7, –λ + 1, λ – 2) and B (2μ, 3μ + 7, μ)

So, DR’s of AB ∝ 3λ – 2μ + 7, – (λ + 3μ + 6), λ – μ – 2

Clearly

⇒ 5λ – 3μ = –16 …(i)

And λ – 5μ = 10 …(ii)

From (i) and (ii) we get λ = –5, μ = –3

So, A is (–8, 6, –7) and B is (–6, –2, –3)

5. The number of points where the function

Answer (7)

Sol. f (–1) = 2 and f (1) = 3

For x ∈ (–1, 1), (4x² – 1) ∈ [–1, 3)

hence f(x) will be discontinuous at x = 1 and also

whenever 4x² – 1 = 0, 1 or 2

So there are total 7 points of discontinuity.

6. Let

Answer (1)

Sol.

Clearly f(θ) = asinθ + bcosθ

Where

and