JEE Main 2022 July 29 – Shift 2 Maths Question Paper with Solutions

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JEE Main 2022 29th July Shift 2 Mathematics 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 z ≠ 0 be a complex number such that

$\begin{array}{l} | z - \frac{1}{z} | = 2, \end{array}$

then the maximum value of |z| is

(A) √2

(B) 1

(C) √2 – 1

(D) √2 + 1

Answer (D)

Sol.

\begin{array}{l} | z - \frac{1}{z} | \geq | | z | - \frac{1}{z} | \end{array}

\begin{array}{l} \Rightarrow | | z | - \frac{1}{| z |} | \leq 2 \end{array}

Let |z| = r

\begin{array}{l} | r - \frac{1}{r} | \leq 2 \end{array}

\begin{array}{l} - 2 \leq r - \frac{1}{r} \leq 2 \end{array}

\begin{array}{l} r - \frac{1}{r} \geq - 2 and r - \frac{1}{r} \leq 2 \end{array}

\begin{array}{l} r^{2} + 2 r - 1 \geq 0 and r^{2} - 2 r - 1 \leq 0 \end{array}

\begin{array}{l} r \in [- \infty, - 1 - \sqrt{2}] \cup [- 1 + \sqrt{2}, \infty] and r \in [1 - \sqrt{2}, 1 + \sqrt{2}] \end{array}

\begin{array}{l} Taking intersection r \in [\sqrt{2} - 1, \sqrt{2} + 1] \end{array}

2. Which of the following matrices can NOT be obtained from the matrix

$\begin{array}{l} [\begin{array}{c} - 1 & 2 \\ 1 & - 1 \end{array}] \end{array}$

by a single elementary row operation?

\begin{array}{l} (A) [\begin{array}{c} 0 & 1 \\ 1 & - 1 \end{array}] \end{array}

\begin{array}{l} (B) [\begin{array}{c} 1 & - 1 \\ - 1 & 2 \end{array}] \end{array}

\begin{array}{l} (C) [\begin{array}{c} - 1 & 2 \\ - 2 & 7 \end{array}] \end{array}

\begin{array}{l} (D) [\begin{array}{c} - 1 & 2 \\ - 1 & 3 \end{array}] \end{array}

Answer (C)

Sol.

\begin{array}{l} (1) By R_{1} \to R_{1} + R_{2}, [\begin{array}{c} 0 & 1 \\ 1 & - 1 \end{array}] is possible \end{array}

\begin{array}{l} (2) By R_{1} \leftrightarrow R_{2}, [\begin{array}{c} 1 & - 1 \\ - 1 & 2 \end{array}] is possible \end{array}

(3) This matrix can’t be obtained

\begin{array}{l} (4) By R_{2} \to R_{2} + 2 R_{1}, [\begin{array}{c} - 1 & 2 \\ - 1 & 3 \end{array}] is possible \end{array}

3. If the system of equations

$\begin{array}{l} x + y + z = 6 \\ 2 x + 5 y + α z = β \\ x + 2 y + 3 z = 14 \end{array}$

has infinitely many solutions, then α + β is equal to

(A) 8

(B) 36

(C) 44

(D) 48

Answer (C)

Sol.

\begin{array}{l} Δ = | \begin{array}{c} 1 & 1 & 1 \\ 2 & 5 & α \\ 1 & 2 & 3 \end{array} | = 1 (15 - 2 α) - 1 (6 - α) + 1 (- 1) \end{array}

\begin{array}{l} = 15 - 2 α - 6 + α - 1 \\ = 8 - α \end{array}

\begin{array}{l} For infinite solutions,~ Δ = 0 \Rightarrow α = 8 \end{array}

\begin{array}{l} Δ_{x} = | \begin{array}{c} 6 & 1 & 1 \\ β & 5 & 8 \\ 14 & 2 & 3 \end{array} | = 6 (- 1) - 1 (3 β - 112) + 1 (2 β - 70) \end{array}

\begin{array}{l} = - 6 - 3 β + 112 + 2 β - 70 \\ = 36 - β \end{array}

\begin{array}{l} Δ_{x} = 0 \Rightarrow f o r β = 36 \\ α + β = 44 \end{array}

4. Let the function

$\begin{array}{l} f (x) = {\begin{array}{c} \frac{\log_{e} (1 + 5 x) - \log_{e} (1 + α x)}{x} & ; if x \in 0 \\ 10 & ; if x = 0 \end{array} \end{array}$

be continuous at x = 0. Then α is equal to

(A) 10

(B) –10

(C) 5

(D) –5

Answer (D)

Sol.

\begin{array}{l} \underset{x \to 0}{It} \frac{\ln (1 + 5 x) - \ln (1 + α x)}{x} \\ = 5 - α = 10 \end{array}

\begin{array}{l} \Rightarrow α = - 5 \end{array}

5. If [t] denotes the greatest integer ≤ t, then the value of

$\begin{array}{l} \int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x \end{array}$

is

\begin{array}{l} (A) \frac{\sqrt{37} + \sqrt{13} - 4}{6} \end{array}

\begin{array}{l} (B) \frac{\sqrt{37} - \sqrt{13} - 4}{6} \end{array}

\begin{array}{l} (C) \frac{- \sqrt{37} - \sqrt{13} + 4}{6} \end{array}

\begin{array}{l} (D) \frac{- \sqrt{37} + \sqrt{13} + 4}{6} \end{array}

Answer (A)

Sol.

\begin{array}{l} I = \int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x \end{array}

\begin{array}{l} I = \int_{0}^{2 / 3} [\underset{I_{1}}{\underset{⏟}{- 3 x^{2} + 7 x - 2}}] d x + \int_{2 / 3}^{1} [\underset{I_{2}}{\underset{⏟}{3 x^{2} - 3 x + 2}}] d x + 1 \end{array}

\begin{array}{l} I_{1} = \int_{0}^{t_{1}} (- 2) d x + \int_{t_{1}}^{1 / 3} (- 1) d x + \int_{1 / 3}^{t_{2}} 0. d x + \int_{t_{2}}^{2 / 3} d x \end{array}

\begin{array}{l} = - t_{1} - t_{2} + \frac{1}{3}, where t_{1} = \frac{7 - \sqrt{37}}{6}, t_{2} = \frac{7 - \sqrt{13}}{6} \end{array}

\begin{array}{l} I_{2} = \int_{2 / 3}^{1} 1 d x = \frac{1}{3} \end{array}

\begin{array}{l} ∴ I = \frac{1}{3} - t_{1} - t_{2} + \frac{1}{3} + 1 = \frac{5}{3} - [\frac{7 - \sqrt{37}}{6} + \frac{7 - \sqrt{13}}{6}] \end{array}

\begin{array}{l} = \frac{\sqrt{37} + \sqrt{13} - 4}{6} \end{array}

6. Let

$\begin{array}{l} {a_{n}}_{n = 0}^{\infty} be a sequence such that a_{0} = a_{1} = 0 and \end{array}$

$\begin{array}{l} a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0. Then a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24} \end{array}$

is equal to

(A) 483

(B) 528

(C) 575

(D) 624

Answer (B)

Sol.

\begin{array}{l} a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0 (a_{0} = a_{1} = 0) \end{array}

\begin{array}{l} (a_{n + 2} - a_{n + 1}) - 2 (a_{n + 1} - a_{n}) - 1 = 0 \end{array}

Put n = 0

\begin{array}{l} (a_{2} - a_{1}) - 2 (a_{1} - a_{0}) - 1 = 0 \end{array}

n = 1

\begin{array}{l} (a_{3} - a_{2}) - 2 (a_{2} - a_{1}) - 1 = 0 \end{array}

n = 2

\begin{array}{l} (a_{4} - a_{3}) - 2 (a_{3} - a_{2}) - 1 = 0 \end{array}

\begin{array}{l} ⋮ \\ n = n \end{array}

\begin{array}{l} (a_{n + 2} - a_{n + 1}) - 2 (a_{a + 1} - a_{n}) - 1 = 0 \end{array}

Adding,

\begin{array}{l} (a_{n + 2} - a_{1}) - 2 (a_{a + 1} - a_{0}) - (n + 1) = 0 \end{array}

\begin{array}{l} ∴ a_{n + 2} - 2 a_{n + 1} - (n + 1) = 0 \end{array}

\begin{array}{l} n \to n - 2 \end{array}

\begin{array}{l} a_{n} - 2 a_{n - 1} - n + 1 = 0 \end{array}

Now,

\begin{array}{l} a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24} \end{array}

\begin{array}{l} = a_{25} (a_{23} - 2 a_{22}) - 2 a_{24} (a_{23} - 2 a_{22}) \end{array}

\begin{array}{l} = (a_{25} - 2 a_{24}) (a_{23} - 2 a_{22}) = 24 \cdot 22 = 528 \end{array}

7.

$\begin{array}{l} \sum_{r = 1}^{20} (r^{2} + 1) (r!) \end{array}$

is equal to

\begin{array}{l} (A) 22! - 21! \\ (B) 22! - 2 (21!) \\ (C) 21! - 2 (20!) \\ (D) 21! - 20! \end{array}

Answer (B)

Sol.

\begin{array}{l} \sum_{r = 1}^{20} (r^{2} + 1 + 2 r - 2 r) r! = \sum_{r = 1}^{20} ((r + 1)^{2} - 2 r) r! \end{array}

\begin{array}{l} = \sum_{r = 1}^{20} [(r + 1) (r + 1)! - r r!] - \sum_{r = 1} (r + 1) r! = r! \end{array}

\begin{array}{l} = (2 \cdot 2! - 1!) + (3 \cdot 3! - 2 \cdot 2!) + \dots + (21 \cdot 21! - 20 \cdot 20!) - [(2! - 1!) + (3! - 2!) + \dots + (21! - 20!)] \end{array}

\begin{array}{l} = (21 \cdot 21! - 1) - (21! - 1) \end{array}

\begin{array}{l} = 20 \cdot 21! = (22 - 2) 21! = 22! - 2 (21!) \end{array}

8. For

$\begin{array}{l} I (x) = \int \frac{\sec^{2} x - 2022}{\sin^{2022} x} d x, if I (\frac{π}{4}) = 2^{1011}, \end{array}$

then

\begin{array}{l} (A) 3^{1010} I (\frac{π}{3}) - I (\frac{π}{6}) = 0 \end{array}

\begin{array}{l} (B) 3^{1010} I (\frac{π}{6}) - I (\frac{π}{3}) = 0 \end{array}

\begin{array}{l} (C) 3^{1011} I (\frac{π}{3}) - I (\frac{π}{6}) = 0 \end{array}

\begin{array}{l} (C) 3^{1011} I (\frac{π}{6}) - I (\frac{π}{3}) = 0 \end{array}

Answer (A)

Sol.

\begin{array}{l} I (x) = \int \frac{\sec^{2} x - 2022}{\sin^{2022} x} d x \end{array}

\begin{array}{l} = \int (\sec^{2} x \cdot \sin^{- 2022} x - 2022 \sin^{- 2022} x) d x \end{array}

\begin{array}{l} = \sin^{- 2022} x \tan x + \int 2022 \sin^{- 2023} x \cos x \cdot \tan x d x - \int 2022 \sin^{- 2022} x d x + c \end{array}

\begin{array}{l} I (x) = \sin^{- 2022} x \tan x + c \end{array}

\begin{array}{l} ∵ I (\frac{π}{4}) = 2^{1011} \Rightarrow c = 2^{1011} - 2^{1011} = 0 \end{array}

\begin{array}{l} ∴ I (\frac{π}{3}) = {(\frac{2}{\sqrt{3}})}^{2022} \sqrt{3}, I (\frac{π}{6}) = 2^{2022} \frac{1}{\sqrt{3}} \end{array}

\begin{array}{l} So, option (A) : \frac{3^{1010} 2^{2022}}{3^{1011}} \cdot \sqrt{3} - \frac{2^{2022}}{\sqrt{3}} = 0 \end{array}

∴ Option (A) is correct

9. if the solution curve of the differential equation

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

passes through the points (2, 1) and (k + 1, 2), k > 0, then

\begin{array}{l} (A) 2 \tan^{- 1} (\frac{1}{k}) = \log_{e} (k^{2} + 1) \end{array}

\begin{array}{l} (B) \tan^{- 1} (\frac{1}{k}) = \log_{e} (k^{2} + 1) \end{array}

\begin{array}{l} (C) 2 \tan^{- 1} (\frac{1}{k + 1}) = \log_{e} (k^{2} + 2 k + 2) \end{array}

\begin{array}{l} (D) 2 \tan^{- 1} (\frac{1}{k}) = \log_{e} (\frac{k^{2} + 1}{k^{2}}) \end{array}

Answer (A)

Sol.

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

Let x – 1 = X, y – 1 = Y

\begin{array}{l} \frac{d Y}{d X} = \frac{X + Y}{X - Y} \end{array}

\begin{array}{l} Let Y = t X \Rightarrow \frac{d Y}{d X} = t + X \frac{d t}{d X} \end{array}

\begin{array}{l} t + X \frac{d t}{d X} = \frac{1 + t}{1 - t} \end{array}

\begin{array}{l} X \frac{d t}{d X} = \frac{1 + t}{1 - t} - t = \frac{1 + t^{2}}{1 - t} \end{array}

\begin{array}{l} \int \frac{1 - t}{1 + t^{2}} d t = \int \frac{d X}{X} \end{array}

\begin{array}{l} \tan^{- 1} t - \frac{1}{2} \ln (1 + t^{2}) = \ln | X | + c \end{array}

\begin{array}{l} \tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} \ln (1 + {(\frac{y - 1}{x - 1})}^{2}) = \ln | x - 1 | + c \end{array}

Curve passes through (2, 1)

\begin{array}{l} 0 - 0 = 0 + c \Rightarrow c = 0 \end{array}

If (k + 1, 2) also satisfies the curve

\begin{array}{l} \tan^{- 1} (\frac{1}{k}) - \frac{1}{2} \ln (\frac{1 + k^{2}}{k^{2}}) = \ln k \end{array}

\begin{array}{l} 2 \tan^{- 1} (\frac{1}{k}) = \ln (1 + k^{2}) \end{array}

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

$\begin{array}{l} \frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1 \end{array}$

which passes through the point (0, 1). Then y(1) is equal to

(A) 1/2

(B) 3/2

(C) 5/2

(D) 7/2

Answer (B)

Sol.

\begin{array}{l} \frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1 \end{array}

Integrating factor I.F

\begin{array}{l} = e^{\int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d x} \end{array}

\begin{array}{l} Let \frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3} \end{array}

A = 2, B = 1, C = –1

\begin{array}{l} I . F . = e^{(2 \ln | x + 1 | + \ln | x + 2 | - \ln | x + 3 |)} \end{array}

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

Solution of differential equation

\begin{array}{l} y \cdot \frac{(x + 1)^{2} (x + 2)}{x + 3} = \int (x + 1) (x + 2) d x \end{array}

\begin{array}{l} y \cdot \frac{(x + 1)^{2} (x + 2)}{x + 3} = \frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c \end{array}

Curve passes through (0, 1)

\begin{array}{l} 1 \times \frac{1 \times 2}{3} = 0 + c \Rightarrow c = \frac{2}{3} \end{array}

\begin{array}{l} So, y (1) = \frac{\frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3}}{\frac{(2^{2} \times 3)}{4}} = \frac{3}{2} \end{array}

11. Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that

$\begin{array}{l} a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220. \end{array}$

If one vertex of the square is
$\begin{array}{l} (10 (c o s α - s i n α), 10 (s i n α + c o s α)), where α \in (0, \frac{π}{2}) \end{array}$
and the equation of one diagonal is
$\begin{array}{l} (\cos α - \sin α) x + (\sin α + \cos α) y = 10, then 72 (\sin^{4} α + \cos^{4} α) + a^{2} - 3 a + 13 \end{array}$
is equal to :

(A) 119

(B) 128

(C) 145

(D) 155

Answer (B)

Sol. One vertex of square is

\begin{array}{l} (10 (c o s α - s i n α), 10 (s i n α + c o s α)) \end{array}

and one of the diagonal is

\begin{array}{l} (c o s α - s i n α) x + (s i n α + c o s α) y = 10 \end{array}

So the other diagonal can be obtained as

\begin{array}{l} (c o s α + s i n α) x - (c o s α - s i n α) y = 0 \end{array}

So, point of intersection of diagonal will be

\begin{array}{l} (5 (c o s α - s i n α), 5 (c o s α + s i n α)) . \end{array}

Therefore, the vertex opposite to the given vertex is (0, 0).

So, the diagonal length

\begin{array}{l} = 10 \sqrt{2} \end{array}

Side length (a) = 10

It is given that

\begin{array}{l} a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220 \end{array}

\begin{array}{l} m_{1}^{2} + m_{2}^{2} = \frac{220 - 100 - 110}{3} = \frac{10}{3} \end{array}

and m₁m₂ = – 1

Slopes of the sides are tanα and – cotα

\begin{array}{l} \tan^{2} α = 3 or \frac{1}{3} \end{array}

\begin{array}{l} 72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13 \end{array}

\begin{array}{l} = 72 \cdot \frac{\tan^{4} α + 1}{(1 + \tan^{2} α)^{2}} + a^{2} - 3 a + 13 = 128 \end{array}

12. The number of elements in the set

$\begin{array}{l} S = {x \in R : 2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} \end{array}$

(A) 1

(B) 3

(C) 0

(D) infinite

Answer (A)

Sol.

\begin{array}{l} S = {x \in R : 2 \cos (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} \end{array}

LHS is less than or equal to 2 and RHS is greater than or equal to 2.

So equality holds only if LHS = RHS = 2

RHS is 2 when x = 0

and at x = 0, LHS is also 2.

So, only one solution exist.

13. Let A(α, -2), B(α, 6) and C(α/4, -2) be vertices of a ΔABC. If (5, α/4) is the circumcentre of ΔABC, then which of the following is NOT correct about ΔABC?

(A) Area is 24

(B) Perimeter is 25

(C) Circumradius is 5

(D) Inradius is 2

Answer (B)

Sol.

Circumcentre of ΔABC

\begin{array}{l} = (\frac{α + \frac{α}{4}}{2}, \frac{6 - 2}{2}) \end{array}

\begin{array}{l} = (\frac{5 α}{8}, 2) \end{array}

\begin{array}{l} = (5, \frac{α}{4}) \end{array}

\begin{array}{l} \Rightarrow α = 8 \end{array}

\begin{array}{l} a r e a (Δ A B C) = \frac{1}{2} \cdot \frac{3 α}{4} \times 8 = 24 sq. units \end{array}

\begin{array}{l} Perimeter = 8 + \frac{3 α}{4} + \sqrt{8^{2} + {(\frac{3 α}{4})}^{2}} \\ = 8 + 6 + 10 = 24 \end{array}

\begin{array}{l} Circumradius = \frac{10}{2} = 5 \end{array}

\begin{array}{l} r = \frac{Δ}{s} = \frac{24}{12} = 2 \end{array}

14. Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :

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

\begin{array}{l} (B) \sqrt{3} \end{array}

\begin{array}{l} (C) 2 \sqrt{3} \end{array}

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

Answer (B)

Sol.

\begin{array}{l} P Q = | \frac{1 + 4 + 3 - 14}{\sqrt{6}} | = \sqrt{6} \end{array}

\begin{array}{l} Q R = \frac{P Q}{\tan 60^{\circ}} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2} \end{array}

\begin{array}{l} Area (Δ P Q R) = \frac{1}{2} \cdot P Q \cdot Q R = \sqrt{3} \end{array}

15. If (2, 3, 9), (5, 2, 1), (1, λ, 8) and (λ, 2, 3) are coplanar, then the product of all possible values of λ is :

\begin{array}{l} (A) \frac{21}{2} \end{array}

\begin{array}{l} (B) \frac{59}{8} \end{array}

\begin{array}{l} (C) \frac{57}{8} \end{array}

\begin{array}{l} (D) \frac{95}{8} \end{array}

Answer (D)

Sol. ∵ (2, 3, 9), (5, 2, 1), (1, λ, 8) and (λ, 2, 3) are coplanar.

\begin{array}{l} ∴ | \begin{array}{c} λ - 2 & - 1 & - 6 \\ - 1 & λ - 3 & - 1 \\ 3 & - 1 & - 8 \end{array} | = 0 \end{array}

\begin{array}{l} ∴ 8 λ^{2} - 67 λ + 95 = 0 \end{array}

\begin{array}{l} ∴ Product of all values of λ = \frac{95}{8} \end{array}

16. Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

\begin{array}{l} (A) \frac{4}{9} \end{array}

\begin{array}{l} (B) \frac{5}{18} \end{array}

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

\begin{array}{l} (D) \frac{3}{10} \end{array}

Answer (B)

Sol.

\begin{array}{l} Let E \to Ball drawn from Bag II is black. \\ E_{R} \to Bag I to Bag II red ball transferred. \\ E_{B} \to Bag I to Bag II black ball transferred. \\ E_{W} \to Bag I to Bag II white ball transferred. \end{array}

\begin{array}{l} P (E_{R} / E) = \frac{P (E / E_{R}) \cdot P (E_{R})}{P (E / E_{R}) P (E_{R}) + P (E / E_{B}) \cdot P (E_{B}) + P (E / E_{W}) P (E_{W})} \end{array}

Here,

\begin{array}{l} P (E_{R}) = \frac{3}{10}, P (E_{B}) = \frac{4}{10}, P (E_{W}) = \frac{3}{10} \end{array}

and

\begin{array}{l} P (\frac{E}{E_{R}}) = \frac{5}{10}, P (\frac{E}{E_{B}}) = \frac{6}{10}, P (\frac{E}{E_{W}}) = \frac{5}{10} \end{array}

\begin{array}{l} ∴ P (\frac{E_{R}}{E}) = \frac{\frac{15}{100}}{\frac{15}{100} + \frac{24}{100} + \frac{15}{100}} \end{array}

\begin{array}{l} = \frac{15}{54} = \frac{5}{18} \end{array}

17. Let

$\begin{array}{l} S = {z = x + i y : | z - 1 + i | \geq | z |, | z | < 2, | z + i | = | z - 1 |} . \end{array}$

Then the set of all values of x, for which w = 2x + iy ∈ S for some y ∈ R is

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

\begin{array}{l} (B) (- \frac{1}{\sqrt{2}}, \frac{1}{4}] \end{array}

\begin{array}{l} (C) (- \sqrt{2}, \frac{1}{2}] \end{array}

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

Answer (B)

Sol.

\begin{array}{l} S : {z = x + i y : | z - 1 + i | \geq | z |, | z | < 2, | z - i | = | z - 1 |} | z - 1 + i | \geq | z | \end{array}

JEE Main 2022 July 29 Shift 2 Maths A17 (i)

|z| < 2

JEE Main 2022 July 29 Shift 2 Maths A17 (ii)

\begin{array}{l} | z - i | = | z - 1 | \end{array}

JEE Main 2022 July 29 Shift 2 Maths A17 (iii)

JEE Main 2022 July 29 Shift 2 Maths A17 (iv)

\begin{array}{l} ∵ w \in S and w = 2 x + i y \end{array}

\begin{array}{l} 2 x < \frac{1}{2} ∴ x < \frac{1}{4} \end{array}

\begin{array}{l} (2 x)^{2} + (- 2 x)^{2} < 4 \end{array}

\begin{array}{l} 4 x^{2} + 4 x^{2} < 4 \end{array}

\begin{array}{l} x^{2} < \frac{1}{2} \Rightarrow x \in (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \end{array}

\begin{array}{l} ∴ x \in (- \frac{1}{2}, \frac{1}{4}] \end{array}

18. Let

$\begin{array}{l} \vec{a}, \vec{b}, \vec{c} \end{array}$

be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and
$\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168, then | \vec{a} | + | \vec{b} | + | \vec{c} | \end{array}$
is equal to :

(A) 10

(B) 14

(C) 16

(D) 18

Answer (C)

Sol.

\begin{array}{l} | \vec{a} | | \vec{b} | | \vec{c} | = 14 \end{array}

\begin{array}{l} \vec{a} \land \vec{b} = \vec{b} \land \vec{c} = \vec{c} \land \vec{a} = θ = \frac{2 π}{3} \end{array}

\begin{array}{l} \vec{a} \cdot \vec{b} = - \frac{1}{2} | \vec{a} | | \vec{b} | \end{array}

\begin{array}{l} \vec{b} \cdot \vec{c} = - \frac{1}{2} | \vec{b} | | \vec{c} | \end{array}

\begin{array}{l} \vec{c} \cdot \vec{a} = - \frac{1}{2} | \vec{c} | | \vec{a} | \end{array}

Now,

\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168 \dots (i) \end{array}

\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c}) | \vec{b} |^{2} \end{array}

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

\begin{array}{l} = \frac{3}{4} | \vec{a} | | \vec{b} |^{2} | \vec{c} | \dots (i i) \end{array}

\begin{array}{l} Similarly (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} | \vec{a} | | \vec{b} | | \vec{c} |^{2} \dots (i i i) \end{array}

\begin{array}{l} (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} | \vec{a} |^{2} | \vec{b} | | \vec{c} | \dots (i v) \end{array}

Substitute (ii), (iii), (iv) in (i)

\begin{array}{l} \frac{3}{4} | \vec{a} | | \vec{b} | | \vec{c} | [| \vec{a} | + | \vec{b} | + | \vec{c} |] = 168 \end{array}

\begin{array}{l} \frac{3}{4} \times 14 [| \vec{a} | + | \vec{b} | + | \vec{c} |] = 168 \end{array}

\begin{array}{l} | \vec{a} | + | \vec{b} | + | \vec{c} | = 16 \end{array}

19. The domain of the function

$\begin{array}{l} f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) \end{array}$

is :

\begin{array}{l} (A) [1, \infty) \\ (B) [- 1, 2] \\ (C) [- 1, \infty) \\ (D) (- \infty, 2] \end{array}

Answer (C)

Sol.

\begin{array}{l} f (x) = \sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) \end{array}

\begin{array}{l} - 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1 \end{array}

\begin{array}{l} \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1 \end{array}

\begin{array}{l} x^{2} - 3 x + 2 \leq x^{2} + 2 x + 7 \end{array}

\begin{array}{l} 5 x \geq - 5 \end{array}

\begin{array}{l} x \geq - 1 \dots (i) \end{array}

\begin{array}{l} \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 \end{array}

\begin{array}{l} x^{2} - 3 x + 2 \geq - x^{2} - 2 x - 7 \end{array}

\begin{array}{l} 2 x^{2} - x + 9 \geq 0 \end{array}

\begin{array}{l} x \in R \dots (i i) \end{array}

\begin{array}{l} (i) \cap (i i) \end{array}

\begin{array}{l} Domain \in [- 1, \infty) \end{array}

20. The statement

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

is NOT equivalent to

\begin{array}{l} (A) (p \land (\sim r)) \Rightarrow q \end{array}

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

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

\begin{array}{l} (D) (p \land (\sim q)) \Rightarrow r \end{array}

Answer (B)

Sol.

\begin{array}{l} (A) (p \land (\sim r)) \Rightarrow q \\ \sim (p \land \sim r) \lor q \\ \equiv (\sim p \lor r) \lor q \\ \equiv\sim p \lor (r \lor q) \\ \equiv p \to (q \lor r) \\ \equiv (p \Rightarrow q) \lor (p \Rightarrow r) \end{array}

\begin{array}{l} (C) p \Rightarrow (q \lor r) \\ \equiv\sim p \lor (q \lor r) \\ \equiv (\sim p \lor q) \lor (\sim p \lor r) \\ \equiv (p \to q) \lor (p \to r) \end{array}

\begin{array}{l} (D) (p \land \sim q) \Rightarrow r \\ \equiv p \Rightarrow (q \lor r) \\ \equiv (p \Rightarrow q) \lor (p \Rightarrow r) \end{array}

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. The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is _______.

Answer (96)

Sol. Given np + npq = 82.5 … (1)

and np (npq) = 1350 … (2)

∴ Mean and Vairance be the roots of x² – 82.5x + 1350 = 0

⇒ x² – 22.5 x – 60x + 1350 = 0

⇒ x – (x – 22.5) – 60 (x – 22.5) = 0

Mean = 60 and Variance = 22.5

np = 60, npq = 22.5

\begin{array}{l} \Rightarrow q = \frac{9}{24} = \frac{3}{8}, p = \frac{5}{8} \end{array}

\begin{array}{l} ∴ n \frac{5}{8} = 60 \\ \Rightarrow n = 96 \end{array}

2. Let α, β(α > β) be the roots of the quadratic equation x² – x – 4 = 0.

$\begin{array}{l} If P_{n} = α^{n} - β^{n}, n \in N then \frac{P_{15} P_{16} - P_{14} P_{16} - P_{15}^{2} + P_{14} P_{15}}{P_{13} P_{14}} \end{array}$

is equal to _______.

Answer (16)

Sol.

α, β are the roots of x² – x – 4 = 0 and

\begin{array}{l} P_{n} = α^{n} - β^{n}, \end{array}

\begin{array}{l} ∴ I = \frac{(P_{15} - P_{14}) P_{16} - P_{15} (P_{15} - P_{14})}{P_{13} P_{14}} = \frac{(P_{16} - P_{15}) (P_{15} - P_{14})}{P_{13} P_{14}} \end{array}

\begin{array}{l} \Rightarrow I = \frac{(α^{16} - β^{16} - α^{15} + β^{15}) (α^{15} - β^{15} - α^{14} + β^{14})}{(α^{13} - β^{13}) (α^{14} - β^{14})} \end{array}

\begin{array}{l} \Rightarrow I = \frac{(α^{15} (α - 1) - β^{15} (β - 1)) (α^{14} (α - 1) - β^{14} (β - 1))}{(α^{13} - β^{13}) (α^{14} - β^{14})} \end{array}

\begin{array}{l} As α^{2} - α = 4 \Rightarrow α - 1 = \frac{4}{α} and β - 1 = \frac{4}{β} \end{array}

\begin{array}{l} \Rightarrow I = \frac{(α^{15} \cdot \frac{4}{α} - β^{15} \cdot \frac{4}{β}) (α^{14} \cdot \frac{4}{α} - β^{14} \cdot \frac{4}{β})}{(α^{13} - β^{13}) (α^{14} - β^{14})} \end{array}

\begin{array}{l} = \frac{16 (α^{14} - β^{14}) (α^{13} - β^{13})}{(α^{14} - β^{14}) (α^{13} - β^{13})} = 16 \end{array}

3. Let

$\begin{array}{l} x = [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}] a n d A = [\begin{array}{c} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{array}] \end{array}$

For k ∈ N, if X’A^kX = 33, then k is equal to _______.

Answer (10*)

Sol. Given

\begin{array}{l} A = [\begin{array}{c} - 1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & - 1 \end{array}] \end{array}

\begin{array}{l} A^{2} = [\begin{array}{c} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], A^{4} = [\begin{array}{c} 1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

\begin{array}{l} \Rightarrow A^{k} = [\begin{array}{c} 1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

\begin{array}{l} ∴ X^{'} A^{k} X = [\begin{array}{c} 1 & 1 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}] = [\begin{array}{c} 3 k + 3 \end{array}] \end{array}

⇒ [3k + 3] = 33 (here it shall be [33] as matrix can’t be equal to a scalar)

i.e. [3k + 3] = 33

3k + 3 = [33] ⇒ k = 10

If k is odd and apply above process, we don’t get odd value of k

∴ k = 10

4. The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2, 3, 4, 5, 6 (repetition of digits is not allowed) and divisible by 55 is _______.

Answer (6)

Sol. Case-I When number is 4-digit number

\begin{array}{l} (\overset{―}{a b c d}) \end{array}

here d is fixed as 5

So, (a, b, c) can be (6, 4, 3), (3, 4, 6), (2, 3, 6),
(6, 3, 2), (3, 2, 4) or (4, 2, 3)

⇒ 6 numbers

Case-II No number possible

5. If

$\begin{array}{l} \sum_{K = 1}^{10} K^{2} {(^{10} C_{K})}^{2} = 22000 L, \end{array}$

then L is equal to _____.

Answer (221)

Sol.

\begin{array}{l} \sum_{K = 1}^{10} K^{2} {(^{10} C_{K})}^{2} = 1^{2}^{10} C_{1}^{2} + 2^{2}^{10} C_{2}^{2} + \dots + 10^{2}^{10} C_{10} \end{array}

Let

\begin{array}{l} {(1 + x)}^{10} =^{10} C_{0} +^{10} C_{1} x +^{10} C_{2} x^{2} + \dots . +^{10} C_{10} x^{10} \end{array}

\begin{array}{l} \Rightarrow 10 {(1 + x)}^{9} =^{10} C_{1} + 2 \cdot^{10} C_{2} x + \dots + 10 \cdot^{10} C_{10} x^{9} \dots (1) \end{array}

Similarly,

\begin{array}{l} 10 {(x + 1)}^{9} = 10 \cdot^{10} C_{0} x^{9} + 9 \cdot^{10} C_{1} x^{8} + \dots + 1 \cdot^{10} C_{9} \end{array}

\begin{array}{l} 100 {(1 + x)}^{18} has required term with coefficient of x^{9} \end{array}

\begin{array}{l} i . e .^{18} C_{9} 100 = 22000 L \\ \Rightarrow L = 221 \end{array}

6. If [t] denotes the greatest integer ≤ t, then the number of points, at which the function

$\begin{array}{l} f (x) = 4 | 2 x + 3 | + 9 [x + \frac{1}{2}] - 12 [x + 20] \end{array}$

is not differentiable in the open interval (–20, 20), is ________.

Answer (79)

Sol.

\begin{array}{l} f (x) = 4 | 2 x + 3 | + 9 [x + \frac{1}{2}] - 12 [x + 20] \end{array}

\begin{array}{l} = 4 | 2 x + 3 | + 9 [x + \frac{1}{2}] - 12 [x] - 240 \end{array}

f(x) is non differentiable at x = -3/2

and f(x) is discontinuous at {–19, –18, ….., 18, 19}

as well as

\begin{array}{l} {- \frac{39}{2}, - \frac{37}{2}, \dots, - \frac{3}{2}, - \frac{1}{2}, \frac{1}{2}, \dots, \frac{39}{2}} \end{array}

,

at same point they are also non differentiable

∴ Total number of points of non differentiability

= 39 + 40

= 79

7. If the tangent to the curve y = x³ – x² + x at the point (a, b) is also tangent to the curve y = 5x² + 2x – 25 at the point (2, –1), then |2a + 9b| is equal to ________.

Answer (195)

Sol. Slope of tangent to curve y = 5x² + 2x – 25

\begin{array}{l} = m = {(\frac{d y}{d x})}_{at (2, - 1)} = 22 \end{array}

∴ Equation of tangent: y + 1 = 22(x – 2)

∴ y = 22x – 45

Slope of tangent to y = x³ – x² + x at point (a, b) = 3a² – 2a + 1

3a² – 2a + 1 = 22

3a² – 2a – 21 = 0

\begin{array}{l} ∴ a = 3 or - \frac{7}{3} \end{array}

Also b = a³ – a² + a

\begin{array}{l} Then (a, b) = (3, 21) o r (- \frac{7}{3}, - \frac{151}{9}) \end{array}

\begin{array}{l} (- \frac{7}{3}, - \frac{151}{9}) does not satisfy the equation of tangent \end{array}

\begin{array}{l} ∴ a = 3, b = 21 \\ ∴ | 2 a + 9 b | = 195 \end{array}

8. Let AB be a chord of length 12 of the circle

$\begin{array}{l} (x - 2)^{2} + (y + 1)^{2} = \frac{169}{4} . \end{array}$

If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to _______.

Answer (72)

Sol. Here AM = BM = 6

\begin{array}{l} O M = \sqrt{{(\frac{13}{2})}^{2} - 6^{2}} = \frac{5}{2} \end{array}

JEE Main 2022 July 29 Shift 2 Maths NQA 8

\begin{array}{l} \sin θ = \frac{12}{13} \end{array}

In ΔPAO,

\begin{array}{l} \frac{P O}{O A} = \sec θ \end{array}

\begin{array}{l} P O = \frac{13}{2} \cdot \frac{13}{5} = \frac{169}{10} \end{array}

\begin{array}{l} ∴ P M = \frac{169}{10} - \frac{5}{2} = \frac{144}{10} = \frac{72}{5} \end{array}

\begin{array}{l} ∴ 5 P M = 72 \end{array}

9. Let

$\begin{array}{l} \vec{a} and \vec{b} be two vectors such that \end{array}$

$\begin{array}{l} | \vec{a} + \vec{b} |^{2} = | \vec{a} |^{2} + 2 | \vec{b} |^{2}, \vec{a} \cdot \vec{b} = 3 and | \vec{a} \times \vec{b} |^{2} = 75. Then | \vec{a} |^{2} \end{array}$
is equal to _____.

Answer (14)

Sol.

\begin{array}{l} ∵ | \vec{a} + \vec{b} |^{2} = | \vec{a} |^{2} + 2 | \vec{b} |^{2} \end{array}

\begin{array}{l} or | \vec{a} |^{2} + | \vec{b} |^{2} + 2 \vec{a} \cdot \vec{b} = | \vec{a} |^{2} + 2 | \vec{b} |^{2} \end{array}

\begin{array}{l} ∴ | \vec{b} |^{2} = 6 \dots (i) \end{array}

\begin{array}{l} Now | \vec{a} \times \vec{b} |^{2} = | \vec{a} |^{2} | \vec{b} |^{2} - {(\vec{a} \cdot \vec{b})}^{2} \end{array}

\begin{array}{l} 75 = | \vec{a} |^{2} \cdot 6 - 9 \end{array}

\begin{array}{l} ∴ | \vec{a} |^{2} = 14 \end{array}

10. Let

$\begin{array}{l} S = {(x, y) \in N \times N : 9 (x - 3)^{2} + 16 (y - 4)^{2} \leq 144} \end{array}$

and
$\begin{array}{l} T = {(x, y) \in R \times R : (x - 7)^{2} + (y - 4)^{2} \leq 36} . \end{array}$
Then n(S ⋂ T) is equal to ______.

Answer (27)

Sol.

\begin{array}{l} S = {(x, y) \in N \times N : \frac{(x - 3)^{2}}{16} + \frac{(y - 4)^{2}}{9} \leq 1} \end{array}

represents all the integral points inside and on the ellipse

\begin{array}{l} \frac{(x - 3)^{2}}{16} + \frac{(y - 4)^{2}}{9} = 1 \end{array}

, in first quadrant.

\begin{array}{l} and T = {(x, y) \in R \times R : (x - 7)^{2} + (y - 4)^{2} \leq 36} \end{array}

represents all the points on and inside the circle

\begin{array}{l} (x - 7)^{2} + (y - 4)^{2} = 36 \end{array}

.

JEE Main 2022 July 29 Shift 2 Maths NQA 10

\begin{array}{l} ∴ (S \cap T) = {(3, 1) (2, 2) (3, 2) (4, 2) (5, 2) (2, 3) \dots \dots \dots . (6, 5)} \end{array}

Total number of points = 27

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