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 :

\begin{array}{l} (A) \frac{133}{10^{4}} \end{array}

\begin{array}{l} (B) \frac{18}{10^{3}} \end{array}

\begin{array}{l} (C) \frac{19}{10^{3}} \end{array}

\begin{array}{l} (D) \frac{271}{10^{4}} \end{array}

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

\begin{array}{l} [\begin{array}{c} 2 & 2 \\ 2 & 2 \end{array}] \end{array}

= ¹⁰C₁

Case-II

Matrix with two prime numbers only example

\begin{array}{l} [\begin{array}{c} 3 & 5 \\ 3 & 5 \end{array}] \end{array}

= ¹⁰C₂ × 2! × 2!

\begin{array}{l} P (E) = \frac{^{10} C_{1} +^{10} C_{2} \times 2! \times 2!}{10^{4}} = \frac{190}{10^{4}} = \frac{19}{10^{3}} \end{array}

2. Let the solution curve of the differential equation

$\begin{array}{l} x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}}, \end{array}$

y(1) = 3 be y = y(x). Then y(2) is equal to :

(A) 15

(B) 11

(C) 13

(D) 17

Answer (A)

Sol.

\begin{array}{l} x \frac{d y}{d x} - y = \sqrt{y^{2} + 16 x^{2}} \end{array}

y = 4x tan θ

\begin{array}{l} \frac{d y}{d x} = 4 \tan θ + 4 x \sec^{2} θ \frac{d θ}{d x} \end{array}

\begin{array}{l} 4 x \tan θ + 4 x^{2} \sec^{2} θ \frac{d θ}{d x} - 4 x \tan θ = 4 x \sec θ \end{array}

\begin{array}{l} \int \sec θ d θ = \int \frac{d x}{x} \end{array}

log |secθ + tanθ| = log |x| + C

y(1) = 3 ⇒ 3 = 4 tanθ

\begin{array}{l} \tan θ = \frac{3}{4} \Rightarrow \sec θ = \frac{5}{4} \end{array}

\begin{array}{l} \ln | \frac{8}{4} | = \ln | 1 | + C \end{array}

⇒ C = ln 2

∴ |secθ + tanθ| = 2|x|

To find y(2) put x = 2

\begin{array}{l} \Rightarrow \tan θ = \frac{y}{8} \end{array}

(secθ + tanθ)² = 16

\begin{array}{l} \sec θ + \tan θ = \pm 4 \\ \underset{―}{\sec θ - \tan θ = \pm \frac{1}{4}} \\ 2 \tan θ = \frac{15}{4} = 2 \times \frac{y}{8} \end{array}

⇒ 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

\begin{array}{l} \frac{a - 2}{3} = \frac{b - 4}{- 1} = \frac{c - 7}{4} = \frac{- 2 (6 - 4 + 28 - 2)}{3^{2} + {(- 1)}^{2} + 4^{2}} \end{array}

\begin{array}{l} \frac{a - 2}{3} = \frac{b - 4}{- 1} = \frac{c - 7}{4} = \frac{- 28}{13} \end{array}

\begin{array}{l} a = \frac{- 58}{13} b = \frac{80}{13} c = \frac{- 21}{13} \end{array}

\begin{array}{l} 2 a + b + 2 c = \frac{- 116 + 80 - 42}{13} \end{array}

= –6

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

$\begin{array}{l} f (x) = {\begin{array}{c} max_{t \leq x} {t^{3} - 3 t} & : & x \leq 2 \\ x^{2} + 2 x - 6 & : & 2 < x < 3 \\ [x - 3] + 9 & : & 3 \leq x \leq 5 \\ 2 x + 1 & : & x > 5 \end{array} \end{array}$

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

$\begin{array}{l} l = \int_{- 2}^{2} f (x) d x \end{array}$

Then the ordered pair (m, I) is equal to :

\begin{array}{l} (A) (3, \frac{27}{4}) \end{array}

\begin{array}{l} (B) (3, \frac{23}{4}) \end{array}

\begin{array}{l} (C) (4, \frac{27}{4}) \end{array}

\begin{array}{l} (D) (4, \frac{23}{4}) \end{array}

Answer (C)

Sol.

\begin{array}{l} max_{t \leq x} {t^{3} - 3 t}; x \leq 2 \end{array}

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

JEE Main 2022 June 29 Shift 1 Maths A4(i)

JEE Main 2022 June 29 Shift 1 Maths A4(ii)

\begin{array}{l} f (x) = {\begin{array}{c} x^{3} - 3 x & x < - 1 \\ 2 & - 1 \leq x \leq 2 \\ x^{2} + 2 x - 6 & 2 < x < 3 \\ 9 & 3 \leq x < 4 \\ 10 & 4 \leq x < 5 \\ 11 & x = 5 \\ 2 x + 1 & x > 5 \end{array} \end{array}

JEE Main 2022 June 29 Shift 1 Maths A4(iii)

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

⇒ m = 4

\begin{array}{l} l = \int_{- 2}^{2} f (x) d x = \int_{- 2}^{- 1} (x^{3} - 3 x) d x + \int_{- 1}^{2} 2 d x \end{array}

\begin{array}{l} = {[\frac{x^{4}}{4} - \frac{3 x^{2}}{2}]}_{- 2}^{- 1} + 2 (2 + 1) = (\frac{1}{4} - \frac{3}{2}) - (4 - 6) + 6 \end{array}

\begin{array}{l} = \frac{27}{4} \end{array}

5. Let

$\begin{array}{l} \vec{a} = α \hat{i} + 3 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} - β \hat{j} + 4 \hat{k} and \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} \end{array}$

where α, β ∈ R, be three vectors. If the projection of
$\begin{array}{l} \vec{a} on \vec{c} i s \frac{10}{3} and \vec{b} \times \vec{c} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k} \end{array}$
then the value of α + β is equal to :

(A) 3

(B) 4

(C) 5

(D) 6

Answer (A)

Sol.

\begin{array}{l} \vec{a} = α \hat{j} + 3 \hat{j} - \hat{k} \end{array}

\begin{array}{l} \vec{b} = 3 \hat{i} - β \hat{j} + 4 \hat{k} \end{array}

\begin{array}{l} \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} \end{array}

\begin{array}{l} Projection of \vec{a} on \vec{c} i s \end{array}

\begin{array}{l} \frac{\vec{a} \cdot \vec{c}}{| \vec{b} |} = \frac{10}{3} \end{array}

\begin{array}{l} \frac{α + 6 + 2}{\sqrt{1^{2} + 2^{2} + {(- 2)}^{2}}} = \frac{α + 8}{3} = \frac{10}{3} \end{array}

∴ α = 2

\begin{array}{l} \vec{b} \times \vec{c} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k} \end{array}

\begin{array}{l} | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ 3 & - β & 4 \\ 1 & 2 & - 2 \end{array} | = (2 β - 8) \hat{i} + 10 \hat{j} + (6 + β) \hat{k} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k} \end{array}

\begin{array}{l} 2 β - 8 = - 6 & 6 + β = 7 \end{array}

∴ β = 1

α + β = 2 + 1 = 3

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

$\begin{array}{l} y = \sqrt{2} x, x = 1, y = 2 \sqrt{2} \end{array}$

is equal to :

\begin{array}{l} (A) \frac{16 \sqrt{2}}{6} \end{array}

\begin{array}{l} (B) \frac{11 \sqrt{2}}{6} \end{array}

\begin{array}{l} (C) \frac{13 \sqrt{2}}{6} \end{array}

\begin{array}{l} (D) \frac{5 \sqrt{2}}{6} \end{array}

Answer (C)

Sol.

jee main 2022 june 29 shift 1 maths a6

\begin{array}{l} A (2, 2 \sqrt{2}), B (1, 2 \sqrt{2}), C (1, \sqrt{2}) \end{array}

\begin{array}{l} Area = \int_{0}^{4 \sqrt{2}} (\frac{y}{\sqrt{2}} - \frac{y^{2}}{8}) d y - area (Δ B A C) \end{array}

\begin{array}{l} = {[\frac{y^{2}}{2 \sqrt{2}} - \frac{y^{3}}{24}]}_{0}^{4 \sqrt{2}} - \frac{1}{2} \times A B \times B C \end{array}

\begin{array}{l} = 8 \sqrt{2} - \frac{32 \times 4 \sqrt{2}}{24} - \frac{1}{2} \times 1 \times \sqrt{2} \end{array}

\begin{array}{l} = 8 \sqrt{2} - \frac{16 \sqrt{2}}{3} - \frac{\sqrt{2}}{2} \end{array}

\begin{array}{l} = \frac{\sqrt{2}}{6} (48 - 32 - 3) = \frac{13 \sqrt{2}}{6} \end{array}

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

\begin{array}{l} Δ = | \begin{array}{c} 2 & 1 & - 1 \\ 1 & - 3 & 2 \\ 1 & 4 & δ \end{array} | = - 7 δ - 21 = 0 \end{array}

\begin{array}{l} δ = - 3 \end{array}

\begin{array}{l} Δ_{1} = | \begin{array}{c} 7 & 1 & - 1 \\ 1 & - 3 & 2 \\ k & 4 & - 3 \end{array} | \end{array}

⇒ 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

\begin{array}{l} | α^{8} + β^{8} | = 2 | - 7 + 24 i | = 2 \sqrt{{(- 7)}^{2} + {(24)}^{2}} = 50 \end{array}

9. Let

$\begin{array}{l} Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow} be such that (p \land q) Δ ((p \lor q) \Rightarrow q) \end{array}$

is a tautology. Then Δ is equal to :

\begin{array}{l} (A) \land \end{array}

\begin{array}{l} (B) \lor \end{array}

\begin{array}{l} (C) \Rightarrow \end{array}

\begin{array}{l} (D) \Leftrightarrow \end{array}

Answer (C)

Sol.

\begin{array}{l} (p \lor q) \Rightarrow q \end{array}

\begin{array}{l} \sim (p \lor q) \lor q \end{array}

\begin{array}{l} = (\sim p \land \sim q) \lor q \end{array}

\begin{array}{l} = (\sim p \lor q) \land (\sim q \lor q) \end{array}

\begin{array}{l} = (\sim p \lor q) \land T \end{array}

\begin{array}{l} =\sim p \lor q \end{array}

Now

\begin{array}{l} (p \land q) Δ (\sim p \lor q) \end{array}

\begin{array}{l} \begin{array}{c} p & q & \sim p & p \land q & \sim p \lor q & (p \land q) Δ (\sim p \lor q) \\ T & T & F & T & T & T \\ T & F & F & F & F & T \\ F & T & T & F & T & T \\ F & F & T & F & T & T \end{array} \end{array}

∴ 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 :

\begin{array}{l} (A) (\frac{3^{10} - 3}{2}) A \end{array}

\begin{array}{l} (B) (\frac{3^{10} - 1}{2}) A \end{array}

\begin{array}{l} (C) (\frac{3^{10} + 1}{2}) A \end{array}

\begin{array}{l} (D) (\frac{3^{10} + 3}{2}) A \end{array}

Answer (A)

Sol.

\begin{array}{l} A = [\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}] = [\begin{array}{c} 2^{0} & 2^{1} & 2^{2} \\ 2^{- 1} & 2^{0} & 2^{1} \\ 2^{- 2} & 2^{- 1} & 2^{0} \end{array}] \end{array}

\begin{array}{l} A^{2} = [\begin{array}{c} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{array}] [\begin{array}{c} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{array}] = [\begin{array}{c} 3 & 6 & 12 \\ \frac{3}{2} & 3 & 6 \\ \frac{3}{4} & \frac{3}{2} & 3 \end{array}] = 3 A \end{array}

A² = 3A

A³ = A.A² = A(3A) = 3A² = 3²A

A⁴ = 3³A

Now

A² + A³ + … + A¹⁰

A[3¹ + 3² + 3³ + … + 3⁹]

\begin{array}{l} = \frac{3 [3^{9} - 1]}{3 - 1} A \end{array}

\begin{array}{l} = \frac{(3^{10} - 3)}{2} A \end{array}

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_iiff a ∈ A_i

(2) Symmetric

(a, b) ⇒a ∈A_iiff b ∈ A_i

(b, a) ∈ R as b ∈ A_iiff a ∈ A_i

(3) Transitive

(a, b) ∈ R & (b, c) ∈ R.

⇒ a ∈ A_iiff b ∈ A_i& b ∈ A_iiff c ∈ A_i

⇒ a ∈ A_iiff c ∈ A_i

⇒ (a, c) ∈ R.

⇒ Relation is equivalence

12. Let

$\begin{array}{l} {a_{n}}_{n = 0}^{\infty} \end{array}$

be a sequence such that a₀ = a₁ = 0 and a_n_{+ 2} = 2a_n_{+ 1} – a_n + 1 for all n ≥ 0.
$\begin{array}{l} Then \sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}} is equal to : \end{array}$

\begin{array}{l} (A) \frac{6}{343} \end{array}

\begin{array}{l} (B) \frac{7}{216} \end{array}

\begin{array}{l} (C) \frac{8}{343} \end{array}

\begin{array}{l} (D) \frac{49}{216} \end{array}

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

\begin{array}{l} \sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}} = \frac{a_{2}}{7^{2}} + \frac{a_{3}}{7^{3}} + \frac{a_{4}}{7^{4}} + \dots \end{array}

\begin{array}{l} s = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + \frac{10}{7^{5}} + \dots \\ \underset{―}{\frac{1}{7} s = \frac{1}{7^{3}} + \frac{3}{7^{4}} + \frac{6}{7^{5}} + \dots} \\ \frac{6 s}{7} = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + \dots \\ \underset{―}{\frac{6 s}{49} = \frac{1}{7^{3}} + \frac{2}{7^{4}} + \dots} \\ \frac{36 s}{49} = \frac{1}{7^{2}} + \frac{1}{7^{3}} + \frac{1}{7^{4}} + \dots \end{array}

\begin{array}{l} \frac{36 s}{49} = \frac{7^{\frac{1}{2}}}{1 - \frac{1}{7}} \end{array}

\begin{array}{l} \frac{36 s}{49} = \frac{7}{49 \times 6} \end{array}

\begin{array}{l} s = \frac{7}{216} \end{array}

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)

\begin{array}{l} m_{O A} = \frac{2}{α} & m_{O B} = \frac{3}{2} \end{array}

\begin{array}{l} \tan \frac{π}{4} = | \frac{\frac{2}{α} - \frac{3}{2}}{1 + \frac{2}{α} \cdot \frac{3}{2}} | \Rightarrow \frac{4 - 3 α}{2 α + 6} = \pm 1 \end{array}

4 – 3α = 2α + 6 & 4 – 3α = -2α – 6

\begin{array}{l} α = \frac{- 2}{5} & α = 10 \end{array}

\begin{array}{l} A (- \frac{2}{5}, 2) & A^{'} (10, 2) and B (2, 3) \end{array}

\begin{array}{l} A A^{'} = 10 + \frac{2}{5} = \frac{52}{5} \end{array}

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

\begin{array}{l} (A) \frac{22}{9 + 4 \sqrt{3}} \end{array}

\begin{array}{l} (B) \frac{66}{9 + 4 \sqrt{3}} \end{array}

\begin{array}{l} (C) \frac{22}{4 + 9 \sqrt{3}} \end{array}

\begin{array}{l} (D) \frac{66}{4 + 9 \sqrt{3}} \end{array}

Answer (B)

Sol.

jee main 2022 june 29 shift 1 maths a14

4a + 3b = 22

Total area = A

\begin{array}{l} = a^{2} + \frac{\sqrt{3}}{4} b^{2} \end{array}

\begin{array}{l} A = {(\frac{22 - 3 b}{4})}^{2} + \frac{\sqrt{3}}{4} b^{2} \end{array}

\begin{array}{l} \frac{d A}{d B} = 2 (\frac{22 - 3 b}{4}) (\frac{- 3}{4}) + \frac{\sqrt{3}}{4} \cdot 2 b = 0 \end{array}

\begin{array}{l} \Rightarrow \frac{\sqrt{3 b}}{2} = \frac{3}{8} (22 - 3 b) \end{array}

\begin{array}{l} 4 \sqrt{3} b = 66 - 9 b \end{array}

\begin{array}{l} b = \frac{66}{9 + 4 \sqrt{3}} \end{array}

15. The domain of the function

$\begin{array}{l} \cos^{- 1} (\frac{2 \sin^{- 1} (\frac{1}{4 x^{2} - 1})}{π}) \end{array}$

is :

\begin{array}{l} (A) R - {- \frac{1}{2}, \frac{1}{2}} \end{array}

\begin{array}{l} (B) (- \infty, - 1] \cup [1, \infty) \cup {0} \end{array}

\begin{array}{l} (C) (- \infty, \frac{- 1}{2}) \cup (\frac{1}{2}, \infty) \cup {0} \end{array}

\begin{array}{l} (D) (- \infty, \frac{- 1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup {0} \end{array}

Answer (D)

Sol.

\begin{array}{l} - 1 \leq \frac{2}{π} \sin^{- 1} (\frac{1}{4 x^{2} - 1}) \leq 1 \end{array}

\begin{array}{l} - \frac{π}{2} \leq \sin^{- 1} \frac{1}{4 x^{2} - 1} \leq \frac{π}{2} \end{array}

\begin{array}{l} - 1 \leq \frac{1}{4 x^{2} - 1} \leq 1 \end{array}

\begin{array}{l} \frac{1}{4 x^{2} - 1} \geq - 1 \Rightarrow \frac{4 x^{2}}{(2 n + 1) (2 x - 1)} \geq 0 \end{array}

jee main 2022 june 29 shift 1 maths a15i

\begin{array}{l} x \in (- \infty, \frac{- 1}{2}) \cup (0) \cup (\frac{1}{2}, \infty) \dots . (1) \end{array}

\begin{array}{l} \frac{1}{4 x^{2} - 1} \leq 1 \Rightarrow 1 - \frac{1}{4 x^{2} - 1} \geq 0 \Rightarrow \frac{4 x^{2} - 2}{4 x^{2} - 1} \geq 0 \end{array}

\begin{array}{l} \Rightarrow \frac{(x + \frac{1}{\sqrt{2}}) (x - \frac{1}{\sqrt{2}})}{(x + \frac{1}{2}) (x - \frac{1}{2})} \geq 0 \end{array}

jee main 2022 june 29 shift 1 maths a15ii

\begin{array}{l} x \in (- \infty, \frac{- 1}{\sqrt{2}}] \cup (\frac{- 1}{2}, \frac{1}{2}) \cup (\frac{1}{\sqrt{2}}, \infty) \dots (2) \dots . (2) \end{array}

From (1) & (2)

\begin{array}{l} x \in (- \infty, \frac{- 1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup {0} \end{array}

16. If the constant term in the expansion of

$\begin{array}{l} {(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} \end{array}$

is 2^k·l, where l is an odd integer, then the value of k is equal to

(A) 6

(B) 7

(C) 8

(D) 9

Answer (D)

Sol. Constant term in

\begin{array}{l} {(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} \to x^{0} \end{array}

⇒ (3x⁸ – 2x⁷ + 5)¹⁰→x⁵⁰

General term of (3x⁸ – 2x⁷ + 5)¹⁰ is

\begin{array}{l} \frac{10!}{p! q! r!} {(3 x^{8})}^{p} {(- 2 x^{7})}^{q} {(5)}^{r} \end{array}

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

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

\begin{array}{l} ∴ \frac{10!}{1! 6! r!} 3^{1} 2^{6} \cdot 5^{3} = 2^{k} \cdot l \end{array}

\begin{array}{l} \Rightarrow 5 \cdot 3 \cdot 7 \cdot 5^{3} \cdot 3 \cdot 2^{9} = 2^{k} l \end{array}

∴ k = 9

17.

$\begin{array}{l} \int_{0}^{5} \cos (π (x - [\frac{x}{2}])) d x \end{array}$

, where [t] denotes greatest integer less than or equal to t, is equal to

(A) –3

(B) –2

(C) 2

(D) 0

Answer (D)

Sol.

\begin{array}{l} \int_{0}^{5} \cos (\begin{array}{c} π (x - [\frac{x}{2}]) \end{array} d x \end{array}

\begin{array}{l} = \int_{0}^{2} \cos (π x) d x + \int_{2}^{4} \cos (π (x - 1)) d x + \int_{4}^{5} \cos (π (x - 2)) d x \end{array}

\begin{array}{l} {\begin{array}{c} \frac{\sin π x}{π} \end{array} |}_{0}^{2} {\begin{array}{c} + \frac{\sin (π (x - 1))}{π} \end{array} |}_{2}^{4} + {\begin{array}{c} \frac{\sin (π (x - 2))}{π} \end{array} |}_{4}^{5} \end{array}

= 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

$\begin{array}{l} E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a^{2} > b^{2} \end{array}$

. If e is the eccentricity of the ellipse E, then the value of 1/e² is equal to

(A) 1 + √2

(B) 3 + 2√2

(C) 1 + 2√3

(D) 4 + 5√3

Answer (B)

Sol.

jee main 2022 june 29 shift 1 maths a18

As

\begin{array}{l} ∠ P R Q = \frac{π}{2} \end{array}

\begin{array}{l} (\frac{\frac{2}{t}}{3 - \frac{1}{t^{2}}}) \cdot (\frac{- 2 t}{3 - t^{2}}) = - 1 \end{array}

⇒ t = ± 1

∴ P ≡ (1, 2) & Q(1, –2)

∴ for ellipse

\begin{array}{l} \frac{1}{a^{2}} + \frac{4}{b^{2}} = 1 \end{array}

and ae = 1

\begin{array}{l} \Rightarrow \frac{1}{a^{2}} + \frac{4}{a^{2} (1 - e^{2})} = 1 \end{array}

\begin{array}{l} \Rightarrow 1 + \frac{4}{(1 - e^{2})} = \frac{1}{e^{2}} \end{array}

⇒ (5 – e²)e² = 1 – e²

⇒ e⁴ – 6e² + 1 = 0

\begin{array}{l} \Rightarrow e^{2} = \frac{1}{3 - 2 \sqrt{2}} \Rightarrow \frac{1}{e^{2}} = 3 + 2 \sqrt{2} \end{array}

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

\begin{array}{l} (A) \frac{1}{2} \end{array}

\begin{array}{l} (B) \frac{2}{3} \end{array}

\begin{array}{l} (C) \frac{1}{6} \end{array}

\begin{array}{l} (D) \frac{5}{3} \end{array}

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

\begin{array}{l} ∴ α x + β y - 3 x - 3 α - 2 y - 2 β + 8 = 0 \end{array}

is same as –x + y = 2

\begin{array}{l} \frac{α - 3}{- 1} = \frac{β - 2}{1} = \frac{3 α - 8 + 2 β}{2} \end{array}

\begin{array}{l} \Rightarrow (α, β) \equiv (\frac{4}{3}, \frac{11}{3}) \end{array}

\begin{array}{l} Δ = \frac{1}{2} | \begin{array}{c} 1 & 3 & 1 \\ 2 & 4 & 1 \\ 4 / 3 & 11 / 3 & 1 \end{array} | = \frac{1}{6} units \end{array}

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.

\begin{array}{l} \sum_{i = 1}^{5} x_{i} = 24 \dots . (i) \end{array}

\begin{array}{l} \frac{\sum_{i = 1}^{5} x_{i}^{2}}{5} - {(\frac{24}{5})}^{2} = \frac{194}{25} \end{array}

\begin{array}{l} \Rightarrow \sum x_{i}^{2} = 1154 \dots . (i i) \end{array}

\begin{array}{l} \sum_{i = 1}^{4} x_{i} = 14 \end{array}

⇒ x₅ = 10

\begin{array}{l} a = \frac{\sum_{i = 1}^{4} x_{i}^{2}}{4} - \frac{49}{4} = \frac{54 - 49}{4} = \frac{5}{4} \end{array}

⇒ 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

$\begin{array}{l} S = {z \in C : | z - 2 | \leq 1 and z (1 + i) + \overset{―}{z} (1 - i) \leq 2} \end{array}$

. Let |z – 4i| attains minimum and maximum values, respectively, at z₁∈ S and z₂∈ S. If
$\begin{array}{l} 5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = α + β \sqrt{5} \end{array}$
where α and β are integers, then the value of α + β is equal to _____.

Answer (26)

Sol.

jee main 2022 june 29 shift 1 maths nqa1

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

\begin{array}{l} z_{1} = (2 - \frac{1}{\sqrt{5}}) + \frac{2}{5} i \Rightarrow {| z_{1} |}^{2} = 5 - \frac{4}{\sqrt{5}} \end{array}

z₂ will be either B or C.

\begin{array}{l} ∵ P B = \sqrt{17} and P C = \sqrt{13} hence z_{2} = 1 \end{array}

So

\begin{array}{l} 5 ({| z_{1} |}^{2} + {| z_{2} |}^{2}) = 30 - 4 \sqrt{5} \end{array}

Clearly α = 30 and β = –4 ⇒α + β = 26

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

$\begin{array}{l} \frac{d y}{d x} + \frac{\sqrt{2} y}{2 \cos^{4} x - \cos 2 x} = x e^{\tan^{- 1} (\sqrt{2} \cot 2 x)}, 0 < x < \frac{π}{2} \end{array}$

$\begin{array}{l} with y (\frac{π}{4}) = \frac{π^{2}}{32} . If y (\frac{π}{3}) = \frac{π^{2}}{18} e^{- \tan^{- 1} (α)} \end{array}$
then the value of 3α² is equal to ______.

Answer (2)

Sol.

\begin{array}{l} \frac{d y}{d x} + \frac{2 \sqrt{2} y}{1 + \cos^{2} 2 x} = x e^{\tan^{- 1} (\sqrt{2} \cot 2 x)} \end{array}

\begin{array}{l} I . F . = e^{\int \frac{2 \sqrt{2} d x}{1 + \cos^{2} 2 x}} = e^{\sqrt{2} \int \frac{2 \sec^{2} 2 x}{2 + \tan^{2} 2 x} d x} \end{array}

\begin{array}{l} = e^{\tan^{- 1} (\frac{\tan 2 x}{\sqrt{2}})} \end{array}

\begin{array}{l} \Rightarrow y . e^{\tan^{- 1} (\frac{\tan 2 x}{\sqrt{2}})} = \int x e^{\tan^{- 1} (\sqrt{2} \cot 2 x)} . e^{\tan^{- 1} (\frac{\tan 2 x}{\sqrt{2}})} d x + c \end{array}

\begin{array}{l} \Rightarrow y . e^{\tan^{- 1} (\frac{\tan 2 x}{\sqrt{2}})} = e^{\frac{π}{2}} \cdot \frac{x^{2}}{2} + c \end{array}

When

\begin{array}{l} x = \frac{π}{4}, y = \frac{π^{2}}{32} gives c = 0 \end{array}

When

\begin{array}{l} x = \frac{π}{3}, y = \frac{π^{2}}{18} e^{- \tan^{- 1} α} \end{array}

So

\begin{array}{l} \frac{π^{2}}{18} e^{- \tan^{- 1} α} \cdot e^{- \tan^{- 1} (- \sqrt{\frac{3}{2}})} = e^{π / 2} \frac{π^{2}}{18} \end{array}

\begin{array}{l} \Rightarrow - \tan^{- 1} α + \tan^{- 1} (\sqrt{\frac{3}{2}}) = \frac{π}{2} \end{array}

\begin{array}{l} \Rightarrow \tan^{- 1} (- α) = t a n^{- 1} (\sqrt{\frac{2}{3}}) \end{array}

\begin{array}{l} \Rightarrow α = - \sqrt{\frac{2}{3}} \Rightarrow 3 α^{2} = 2 \end{array}

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 d²is equal to ____.

Answer (26)

Sol. Foot of perpendicular from P

\begin{array}{l} \frac{x - 1}{- 1} = \frac{y - 2}{1} = \frac{z + 1}{1} = \frac{- (- 1 + 2 - 1 - 1)}{3} \end{array}

\begin{array}{l} \Rightarrow p^{'} \equiv (\frac{2}{3}, \frac{7}{3}, \frac{- 2}{3}) \end{array}

and foot of perpendicular from Q

\begin{array}{l} \frac{x - 2}{- 1} = \frac{y + 1}{1} = \frac{z - 3}{1} = \frac{- (- 2 - 1 + 3 - 1)}{3} \end{array}

\begin{array}{l} \Rightarrow Q^{'} \equiv (\frac{5}{3}, \frac{- 2}{3}, \frac{10}{3}) \end{array}

\begin{array}{l} P^{'} Q^{'} = \sqrt{{(1)}^{2} + {(3)}^{2} + {(4)}^{2}} = d = \sqrt{26} \end{array}

⇒ d² = 26

4. The number of elements in the set

$\begin{array}{l} S = {θ \in [- 4 π, 4 π] : 3 \cos^{2} 2 θ + 6 \cos 2 θ - 10 \cos^{2} θ + 5 = 0} \end{array}$

is __________.

Answer (32)

Sol.

\begin{array}{l} 3 \cos^{2} 2 θ + 6 \cos 2 θ - \frac{10 (1 + \cos 2 θ)}{2} + 5 = 0 \end{array}

⇒ 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

$\begin{array}{l} 2 θ - \cos^{2} θ + \sqrt{2} = 0 \end{array}$

in R is equal to ___________.

Answer (1)

Sol.

\begin{array}{l} \cos^{2} θ = 2 θ + \sqrt{2} \end{array}

jee main 2022 june 29 shift 1 maths nqa5

1 point of intersection = 1 solution.

6.

$\begin{array}{l} 50 \tan (3 \tan^{- 1} (\frac{1}{2}) + 2 \cos^{- 1} (\frac{1}{\sqrt{5}})) + 4 \sqrt{2} \tan (\frac{1}{2} \tan^{- 1} (2 \sqrt{2})) \end{array}$

is equal to _____________.

Answer (29)

Sol.

\begin{array}{l} 50 \tan (\tan^{- 1} \frac{1}{2} + 2 \tan^{- 1} (\frac{1}{2}) + 2 \tan^{- 1} (2)) + 4 \sqrt{2} \tan (\frac{\tan^{- 1}}{2} (2 \sqrt{2})) \end{array}

\begin{array}{l} \Rightarrow 50 \tan (π + \tan^{- 1} (\frac{1}{2})) + 4 \sqrt{2} \tan (\frac{1}{2} \tan^{- 1} 2 \sqrt{2}) \end{array}

\begin{array}{l} \Rightarrow 50 (\frac{1}{2}) + 4 \sqrt{2} \tan α \end{array}

Where

\begin{array}{l} 2 α = \tan^{- 1} 2 \sqrt{2} \end{array}

\begin{array}{l} \Rightarrow \frac{2 \tan α}{1 - \tan^{2} α} = 2 \sqrt{2} \dots . (i) \end{array}

\begin{array}{l} \Rightarrow 2 \sqrt{2} \tan^{2} α + 2 \tan α - 2 \sqrt{2} = 0 \end{array}

\begin{array}{l} \Rightarrow 2 \sqrt{2} \tan^{2} α + 4 \tan α - 2 \tan α - 2 \sqrt{2} = 0 \end{array}

\begin{array}{l} \Rightarrow (2 \sqrt{2} \tan α - 2) (\tan α - \sqrt{2}) = 0 \end{array}

\begin{array}{l} \Rightarrow \tan α = \sqrt{2} or \frac{1}{\sqrt{2}} \end{array}

\begin{array}{l} \Rightarrow \tan α = \frac{1}{\sqrt{2}} \end{array}

\begin{array}{l} \Rightarrow (\tan α = \sqrt{2} does not satisfy (i)) \end{array}

\begin{array}{l} \Rightarrow 25 + 4 \sqrt{2} \frac{1}{\sqrt{2}} = 29 \end{array}

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

$\begin{array}{l} | 2 (f (1) + f (2) + f (3) + \dots + f (20)) | \end{array}$

is equal to ____________.

Answer (3395)

Sol. f(x) is polynomial

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

\begin{array}{l} f (x + \frac{1}{x}) = f (x) + f (\frac{1}{x}) - 1 \end{array}

\begin{array}{l} \Rightarrow (c + 1) {(x + \frac{1}{x})}^{2} + (1 - c^{2}) (x + \frac{1}{x}) + 2 K \end{array}

\begin{array}{l} = (c + 1) x^{2} + (1 - c^{2}) x + 2 K + (c + 1) \frac{1}{x^{2}} + (1 - c^{2}) \frac{1}{x} + 2 K - 1 \end{array}

⇒ 2(c + 1) = 2K – 1 …(1)

and put x = y = 0 we get

\begin{array}{l} f (0) = 2 + f (0) - 0 \Rightarrow f (0) = 0 \Rightarrow k = 0 \end{array}

∴ k = 0 and 2c = –3 ⇒c = –3/2

\begin{array}{l} f (x) = - \frac{x^{2}}{2} - \frac{5 x}{4} = \frac{1}{4} (5 x + 2 x^{2}) \end{array}

\begin{array}{l} | 2 \sum_{i = 1}^{20} f (i) | = | \frac{- 2}{4} (\frac{5.20 .21}{2} + \frac{2.20 .21 .41}{6}) | \end{array}

\begin{array}{l} = | \frac{- 1}{2} (2730 + 5740) | \end{array}

\begin{array}{l} = | - \frac{6790}{2} | = 3395 \end{array}

.

8. Let H:

$\begin{array}{l} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0, \end{array}$

be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is
$\begin{array}{l} 4 (2 \sqrt{2} + \sqrt{14}) . If the eccentricity H is \frac{\sqrt{11}}{2}, \end{array}$
then the value of a² + b² is equal to ____________.

Answer (88)

Sol.

\begin{array}{l} 2 a + 2 b = 4 (2 \sqrt{2} + \sqrt{14}) \dots . (1) \end{array}

\begin{array}{l} 1 + \frac{b^{2}}{a^{2}} = \frac{11}{4} \dots (2) \end{array}

\begin{array}{l} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{7}{4} \dots (3) \end{array}

and

\begin{array}{l} a + b = 4 \sqrt{2} + 2 \sqrt{14} \dots (4) \end{array}

By (3) and (4)

\begin{array}{l} \Rightarrow a + \frac{\sqrt{7}}{2} a = 4 \sqrt{2} + 2 \sqrt{14} \end{array}

\begin{array}{l} \Rightarrow \frac{a (2 + \sqrt{7})}{2} = 2 \sqrt{2} (2 + \sqrt{7}) \end{array}

\begin{array}{l} \Rightarrow a = 4 \sqrt{2} \end{array}

⇒a² = 32 and b² = 56

⇒ a² + b² = 32 + 56 = 88

9. Let

$\begin{array}{l} P_{1} : \vec{r} \cdot (2 \hat{i} + \hat{j} - 3 \hat{k}) = 4 \end{array}$

be a plane. Let P₂ be another plane which passes through the points (2, –3, 2), (2, –2, –3) and (1, –4, 2). If the direction ratios of the line of intersection of P₁ and P₂ be 16, α, β, then the value of α + β is equal to _______.

Answer (28)

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

and that of

\begin{array}{l} P_{2} \equiv | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - 5 \\ - 1 & - 2 & 5 \end{array} | = - 5 \hat{i} - \hat{j} (- 5) + \hat{k} (1) \end{array}

i.e.< –5, 5, 1 >

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

\begin{array}{l} | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 3 \\ - 5 & 5 & 1 \end{array} | = \hat{i} (16) - \hat{j} (- 13) + \hat{k} (15) \end{array}

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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JEE Main 2022 June 29th Paper & Solutions – Shift 1

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Frequently Asked Questions -FAQs

Q1

What is the difficulty level of the Shift 1 JEE Main 2022 Maths June 29 question paper?

We can say that the difficulty level of the Shift 1 JEE Main 2022 Maths June 29 question paper was 1.9 out of 3.

Q2

How many questions are there in section B of the JEE Main 2022 Shift 1 June 29 Maths question paper?

There are 10 questions in section B of the JEE Main 2022 Shift 1 June 29 Maths question paper.

Q3

Is there a choice in section B of the JEE Main 2022 June 29 Shift 1 Maths question paper?

Yes. The choice is given in section B of the JEE Main 2022 June 29 Shift 1 Maths question paper. Students have to do only 5 out of 10 questions.

Q4

What type of questions were asked from Complex numbers in the JEE Main 2022 June 29 Maths Shift 1 question paper?

The questions from Complex numbers in the JEE Main 2022 June 29 Maths Shift 1 question paper were medium level.

Q5

Was the JEE Main 2022 June 29 Shift 1 Maths question paper difficult compared to JEE Main 2022 June 29 Shift 1 Physics and Chemistry?

Yes. The JEE Main 2022 June 29 Shift 1 Maths question paper was difficult compared to JEE Main 2022 June 29 Shift 1 Physics and Chemistry.

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