JEE Main 2022 June 28 – Shift 2 Maths Question Paper with Solutions

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JEE Main 2022 June 28th Maths Shift 2 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 R₁ = {(a, b) ∈ N × N : |a – b| ≤ 13} and R₂ = {(a, b) ∈ N × N : |a – b| ≠ 13}. Then on N:

(A) Both R₁ and R₂ are equivalence relations

(B) Neither R₁ nor R₂ is an equivalence relation

(C) R₁ is an equivalence relation but R₂ is not

(D) R₂ is an equivalence relation but R₁ is not

Answer (B)

Sol. R₁ = {(a, b) ∈ N × N : |a – b| ≤ 13} and

R₂ = {(a, b) ∈ N × N : |a – b| ≠ 13}

In R₁: ∵ |2 – 11| = 9 ≤ 13

∴ (2, 11) ∈ R₁ and (11, 19) ∈ R₁ but

(2, 19) ∉ R₁

∴ R₁ is not transitive

Hence R₁ is not equivalence

In R₂ : (13, 3) ∈ R₂ and (3, 26) ∈ R₂ but

(13, 26) ∉ R₂ (∵ |13 – 26| = 13)

∴ R₂ is not transitive

Hence R₂ is not equivalence.

2. Let f(x) be a quadratic polynomial such that f(–2) + f(3) = 0. If one of the roots of f(x) = 0 is –1, then the sum of the roots of f(x) = 0 is equal to:

Answer (A)

Sol. ∵ x = –1 be the roots of f(x) = 0

∴ let f(x) = A(x + 1)(x – b) …(i)

Now, f(–2) + f(3) = 0

⇒ A[–1(–2 – b) + 4(3 – b)] = 0

∴ Second root of f(x) = 0 will be 14/3.

∴ Sum of roots

3. The number of ways to distribute 30 identical candies among four children C₁, C₂, C₃ and C₄ so that C₂ receives atleast 4 and atmost 7 candies, C₃ receives atleast 2 and atmost 6 candies, is equal to:

(A) 205

(B) 615

(C) 510

(D) 430

Answer (D)

Sol. By multinomial theorem, no. of ways to distribute 30 identical candies among four children C₁, C₂ and C₃, C₄

= Coefficient of x³⁰ in (x⁴ + x⁵ + … + x⁷) (x² + x³ +…+ x⁶) (1 + x + x²…)²

= Coefficient of x²⁴ in (1 – x⁴ – x⁵ + x⁹) (1 – x)^–4

= ²⁷C₂₄ – ²³C₂₀ – ²²C₁₉ + ¹⁸C₁₅ = 430

4. The term independent of x in the expansion of

Answer (B)

Sol.

So, term independent from x in given expression

5. If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is:

(A) 21

(B) 22

(C) 23

(D) 24

Answer (C)

Sol. a, A₁, A₂ …….. A_n, 100

Let d be the common difference of above A.P. then

⇒ 7a + 8d = 100 …(i)

and a + n = 33 …(ii)

and 100 = a + (n + 1)d

⇒ 800 = 8a + 7a² – 338a + 3400

⇒ 7a² – 330a + 2600 = 0

∴ n = 23

6. Let f,g : R → R be functions defined by

Where [x] denotes the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :

(A) one point

(B) two points

(C) three points

(D) four points

Answer (B)

Sol.

Graph of y = g(x)

So, x = 0, 2 are the two points where fog is discontinuous.

7. Let f : R → R be a differentiable function such that

(A) 2

(B) 3

(C) 4

(D) –3

Answer (B)

Sol. Given :

Now,

8. Let f : R → R be a continuous function satisfying f(x) + f(x + k) = n, for all x ∈ R where k > 0 and n is a positive integer. If

Answer (C)

Sol. f : R → R and f(x) + f(x + k) = n ∀ x ∈ R

x → x + k

f(x + k) + f(x + 2k) = n

∴ f(x + 2k) = f(x)

So, period of f(x) is 2k

Now,

x = t + k ⇒ dx = dt (in second integral)

Now,

9. The area of the bounded region enclosed by the curve

Answer (C)

Sol.

Area of shaded region (required area)

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

(A) e log_e(2)

(B) -e log_e(2)

(C) e² log_e(2)

(D) -e² log_e(2)

Answer (D)

Sol. Given differential equation

Now, using x(1) = 0, c = 2

So, for x(e), Put y = e in (i)

11. Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tanx(cosx – y). If the curve passes through the point (π/4, 0) then the value of

Answer (B)

Sol.

∴ Solution of D.E. will be

∵ Curve passes through

12. Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : –4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to

Answer (B)

Sol. L₁ : 2x + 5y = 10

L₂ : – 4x + 3y = 12

Solving L₁ and L₂ we get

Now, Let

and

So, 3x₁ + x₂ = 8 and 10 x₁ – x₂ = – 5

So,

13. Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola

(A) 100

(B) 110

(C) 120

(D) 130

Answer (D)

Sol. H :

and

(I′ be the length of LR of conjugate hyperbola)

By (i) and (ii)

7a = 4b

then by (i)

⇒ 44b = 65 and 77a = 65

∴ 77a + 44b = 130

14. Let,

(A) 10

(B) 7

(C) 9

(D) 14

Answer (D)

Sol.

Now

⇒ (2 + α)² + (α – 2)² + (α² + 4)² = 15(α² + 4)

⇒ α⁴ – 5α² – 36 = 0

∴ α = ±3

Now,

15. If vertex of a parabola is (2, –1) and the equation of its directrix is 4x – 3y = 21, then the length of its latus rectum is :

(A) 2

(B) 8

(C) 12

(D) 16

Answer (B)

Sol. Vertex of Parabola : (2, –1)

and directrix : 4x – 3y = 21

Distance of vertex from the directrix

∴ length of latus rectum = 4a = 8

16. Let the plane ax + by + cz = d pass through (2, 3, –5) and is perpendicular to the planes 2x + y – 5z = 10 and 3x + 5y – 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :

(A) 18

(B) 20

(C) 24

(D) 22

Answer (D)

Sol. Equation of plane through point (2, 3, –5) and perpendicular to planes 2x + y – 5z = 10 and

3x + 5y – 7z = 12 is

∴ Equation of plane is (x – 2) (– 7 + 25) – (y – 3) (– 14 + 15) + (z + 5) · 7 = 0

∴ 18x – y + 7z + 2 = 0

⇒ 18x – y + 7z = – 2

∴ – 18x + y – 7z = 2

On comparing with ax + by + cz = d where d > 0 is

a = – 18, b = 1, c = – 7, d = 2

∴ a + 7b + c + 20d = 22

17. The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) – f(c) = f(d) is :

Answer (D)

Sol. Number of one-one function from {a, b, c, d} to set {1, 2, 3, 4, 5} is

The required possible set of value

(f(a), f(b), f(c), f(d)) such that f(a) + 2f(b) – f(c) = f(d) are (5, 3, 2, 1), (5, 1, 2, 3), (4, 1, 3, 5), (3, 1, 4, 5), (5, 4, 3, 2) and (3, 4, 5, 2)

∴ n(E) = 6

∴ Required probability

18. The value of

(A) 1

(B) 2

(C) 3

(D) 6

Answer (C)

Sol.

19. Let

(A) 1/3

(B) 1

(C) 5/3

(D) 7/3

Answer (C)

Sol. Let

and

and

∴ a₂ = 2 …(ii)

and a₁ – 2a₃ = 13 …(iii)

From eq. (i) and (iii) : a₁ = 3 and a₃ = –5

20. If

(A) -1/7 and IV^th quadrant

(B) 7 and I^st quadrant

(C) – 7 and IV^th quadrant

(D) 1/7 and I^st quadrant

Answer (A)

Sol. ∵ cot α = 1,

then tan α = 1

and

then

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 the image of the point P(1, 2, 3) in the line

Answer (125)

Sol. The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.

∴ Let a point on line be λ

as P′ is foot of perpendicular

(3λ + 5)3 + (2λ – 1)2 + (3λ – 1)3 = 0

⇒ 22λ + 15 – 2 – 3 = 0

Mid-point of PP’

⇒ 22(α + β + γ) = 62 + 23 + 40 = 125

2. Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is __________.

Answer (0)

Sol. According to given data

So for any x_i , (x_i – 62)² ≤ 140

⇒ x_i > 50 ∀ i = 1, 2, 3, …7

So no student is going to score less than 50.

3. If one of the diameters of the circle

Answer (10)

Sol. For

Radius

⇒ Diameter = 2√6

If this diameter is chord to

⇒ r² = 6 + 4 = 10

⇒ r² = 10

4. If

Answer (11)

Sol.

So f(x) = 2x³ – 7x² + ax +b = 0 has x = 1 as repeated root, therefore f(1) = 0 and f ′(1) = 0 gives

a + b + 5 and a = 8

So, a – b = 11

5. Let for n = 1, 2, …, 50, S_n be the sum of the infinite geometric progression whose first term is n² and whose common ratio is

Answer (41651)

Sol.

Now

= 1 + 25 × 17 (101 – 3)

= 41651

6. If the system of linear equations

2x – 3y = γ + 5,

αx + 5y = β + 1, where α, β, γ ∈ R has infinitely many solutions, then the value of |9α + 3β + 5γ| is equal to _______.

Answer (58)

Sol. If 2x – 3y = γ + 5 and αx + 5y = β + 1 have infinitely many solutions then

So |9α + 3β + 5γ| = |–30 – 28| = 58

7. Let

Answer (25)

Sol.

So A⁵ = A, A⁹ = A and so on.

Clearly n = 1, 5, 9, ….., 97

Number of values of n = 25

8. Sum of squares of modulus of all the complex numbers z satisfying

Answer (2)

Sol. Let z = x + iy

So 2x = (1 + i)(x² – y² + 2xyi)

⇒ 2x = x² – y² – 2xy …(i) and

x² – y² + 2xy = 0 …(ii)

From (i) and (ii) we get

x = 0 or

When x = 0 we get y = 0

When

So there will be total 3 possible values of z, which are

Sum of squares of modulus

= 2

9. Let S = {1, 2, 3, 4}. Then the number of elements in the set {f : S × S → S : f is onto and f(a,b)=f(b,a) ≥ a ∀ (a, b)∈  S × S is _____.

Answer (37)

Sol. There are 16 ordered pairs in S × S. We write all these ordered pairs in 4 sets as follows.

A = {(1, 1)}

B = {(1, 4), (2, 4), (3, 4) (4, 4), (4, 3), (4, 2), (4, 1)}

C = {(1, 3), (2, 3), (3, 3), (3, 2), (3, 1)}

D = {(1, 2), (2, 2), (2, 1)}

All elements of set B have image 4 and only element of A has image 1.

All elements of set C have image 3 or 4 and all elements of set D have image 2 or 3 or 4.

We will solve this question in two cases.

Case I : When no element of set C has image 3.

Number of onto functions = 2 (when elements of set D have images 2 or 3)

Case II : When atleast one element of set C has image 3.

Number of onto functions = (2³ – 1)(1 + 2 + 2)

= 35

Total number of functions = 37

10. The maximum number of compound propositions, out of p ∨ r ∨ s, p ∨ r ∨ ~s, p ∨ ~q ∨ s, ~p ∨ ~r ∨ s, ~p ∨ ~r ∨ ~s, ~p ∨ q ∨ ~s, q ∨ r ∨ ~s, q ∨ ~r ∨ ~s, ~p ∨ ~q ∨ ~s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to ____________ .

Answer (9)

Sol. There are total 9 compound propositions, out of which 6 contain ~s. So if we assign s as false, these 6 propositions will be true.

In remaining 3 compound propositions, two contain p and the third contains ~r. So if we assign p and r as true and false respectively, these 3 propositions will also be true.

Hence maximum number of propositions that can be true are 9.

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