JEE Main 2022 July 25 – Shift 2 Maths Question Paper with Solutions

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JEE Main 2022 25th 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. For

$\begin{array}{l} z \in C if the minimum value of (| z - 3 \sqrt{2} | + | z - p \sqrt{2} i |) \end{array}$

is 5√2, then a value of p is _________.

(A) 3

\begin{array}{l} (B) \frac{7}{2} \end{array}

(C) 4

\begin{array}{l} (D) \frac{9}{2} \end{array}

Answer (C)

Sol.

It is sum of distance of z from (3√2, 0) and (0, p√2).

For minimising, z should lie on AB and

\begin{array}{l} A B = 5 \sqrt{2} \end{array}

\begin{array}{l} (A B)^{2} = 18 + 2 p^{2} \end{array}

\begin{array}{l} p = \pm 4 \end{array}

2. The number of real values of λ, such that the system of linear equations

2x – 3y + 5z = 9

x + 3y – z = –18

3x – y + (λ² – | λ |)z = 16

has no solutions, is

(A) 0

(B) 1

(C) 2

(D) 4

Answer (C)

Sol.

\begin{array}{l} Δ = | \begin{array}{c} 2 & - 3 & 5 \\ 1 & 3 & - 1 \\ 3 & - 1 & λ^{2} - | λ | \end{array} | = 2 (3 λ^{2} - 3 | λ | - 1) + 3 (λ^{2} - | λ | + 3) + 5 (- 1 - 9) \end{array}

= 9λ² – 9 | λ | – 43

= 9 | λ |² – 9 | λ | – 43

Δ = 0 for 2 values of | λ | out of which one is –ve and other is +ve

So, 2 values of λ satisfy the system of equations to obtain no solution.

3. The number of bijective functions f : {1, 3, 5, 7, …, 99} → {2, 4, 6, 8, ….., 100} such that

$\begin{array}{l} f (3) \geq f (9) \geq f (15) \geq f (21) \geq \dots \geq f (99) \end{array}$

is_____.

\begin{array}{l} (A)^{50} P_{17} \end{array}

\begin{array}{l} (B)^{50} P_{33} \end{array}

\begin{array}{l} (C) 33! \times 17! \end{array}

\begin{array}{l} (D) \frac{50!}{2} \end{array}

Answer (B)

Sol. As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction

f(3) > f(9) > f(15) ….. > f(99)

So number of ways

\begin{array}{l} =^{50} C_{17} \cdot 1 \cdot 33! =^{50} P_{33} \end{array}

4. The remainder when (11)¹⁰¹¹ + (1011)¹¹ is divided by 9 is

(A) 1

(B) 4

(C) 6

(D) 8

Answer (D)

Sol.

\begin{array}{l} R e (\frac{(11)^{1011} + (1011)^{11}}{9}) = R e (\frac{2^{1011} + 3^{11}}{9}) \end{array}

\begin{array}{l} For R e (\frac{2^{1011}}{9}) \end{array}

\begin{array}{l} 2^{1011} = (9 - 1)^{337} =^{337} C_{0} 9^{337} (- 1)^{0} +^{337} C_{1} 9^{336} (- 1)^{1} +^{337} C_{2} 9^{335} (- 1)^{2} + \dots +^{337} C_{337} 9^{0} (- 1)^{337} \end{array}

so, remainder is 8.

and

\begin{array}{l} R e (\frac{3^{11}}{9}) = 0 \end{array}

So, remainder is 8.

5. The sum

$\begin{array}{l} \sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)} \end{array}$

is equal to

\begin{array}{l} (A) \frac{7}{87} \end{array}

\begin{array}{l} (B) \frac{7}{29} \end{array}

\begin{array}{l} (C) \frac{14}{87} \end{array}

\begin{array}{l} (D) \frac{21}{29} \end{array}

Answer (B)

Sol.

\begin{array}{l} \sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)} = \frac{3}{4} \sum_{n = 1}^{21} \frac{1}{4 n - 1} - \frac{1}{4 n + 3} \end{array}

\begin{array}{l} = \frac{3}{4} [(\frac{1}{3} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{11}) + \dots + (\frac{1}{83} - \frac{1}{87})] \end{array}

\begin{array}{l} = \frac{3}{4} [\frac{1}{3} - \frac{1}{87}] = \frac{3}{4} \frac{84}{3.87} = \frac{7}{29} \end{array}

6.

$\begin{array}{l} lim_{x \to \frac{x}{4}} \frac{8 \sqrt{2} - (\cos x + \sin x)^{7}}{\sqrt{2} - \sqrt{2} \sin 2 x} \end{array}$

is equal to

(A) 14

(B) 7

\begin{array}{l} (C) 14 \sqrt{2} \end{array}

\begin{array}{l} (D) 7 \sqrt{2} \end{array}

Answer (A)

Sol.

\begin{array}{l} lim_{x \to \frac{x}{4}} \frac{8 \sqrt{2} - (\cos x + s i n x)^{7}}{\sqrt{2} - \sqrt{2} \sin 2 x} (\frac{0}{0} form) \end{array}

\begin{array}{l} = lim_{x \to \frac{x}{4}} \frac{- 7 (\cos x + \sin x)^{6} (- \sin x + \cos x)}{- 2 \sqrt{2} \cos 2 x} \end{array}

using L–H Rule

\begin{array}{l} = lim_{x \to \frac{x}{4}} \frac{56 (\cos x - \sin x)}{2 \sqrt{2} \cos 2 x} (\frac{0}{0}) \end{array}

\begin{array}{l} = lim_{x \to \frac{x}{4}} \frac{- 56 (\sin x + \cos x)}{- 4 \sqrt{2} \sin 2 x} \end{array}

using L–H Rule

\begin{array}{l} = 7 \sqrt{2} \cdot \sqrt{2} = 14 \end{array}

7.

$\begin{array}{l} lim_{n \to \infty} \frac{1}{2^{n}} (\frac{1}{\sqrt{1 - \frac{1}{2^{n}}}} + \frac{1}{\sqrt{1 - \frac{2}{2^{n}}}} + \frac{1}{\sqrt{1 - \frac{3}{2^{n}}}} + \dots + \frac{1}{\sqrt{1 - \frac{2^{n} - 1}{2^{n}}}}) \end{array}$

is equal to

(A)

(B) 1

(C) 2

(D) –2

Answer (C)

Sol.

\begin{array}{l} I = lim_{n \to \infty} \frac{1}{2^{n}} (\frac{1}{\sqrt{1 - \frac{1}{2^{n}}}} + \frac{1}{\sqrt{1 - \frac{2}{2^{n}}}} + \frac{1}{\sqrt{1 - \frac{3}{2^{n}}}} + \dots + \frac{1}{\sqrt{1 - \frac{2^{n} - 1}{2^{n}}}}) \end{array}

Let 2ⁿ = t and if n → ∞ then t → ∞

\begin{array}{l} I = lim_{n \to \infty} \frac{1}{t} (\sum_{r = 1}^{t = 1} \frac{1}{\sqrt{1 - \frac{r}{t}}}) \end{array}

\begin{array}{l} l = \int_{0}^{1} \frac{d x}{\sqrt{1 - x}} = \int_{0}^{1} \frac{d x}{\sqrt{x}} \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{array}

\begin{array}{l} = {[2 x^{\frac{1}{2}}]}_{0}^{1} = 2 \end{array}

8. If A and B are two events such that

$\begin{array}{l} P (A) = \frac{1}{3}, P (B) = \frac{1}{5} and P (A \cup B) = \frac{1}{2}, then \end{array}$

$\begin{array}{l} P (A | B^{'}) + P (B | A^{'} |) \end{array}$
is equal to

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

\begin{array}{l} (B) \frac{5}{8} \end{array}

\begin{array}{l} (C) \frac{5}{4} \end{array}

\begin{array}{l} D) \frac{7}{8} \end{array}

Answer (B)

Sol.

\begin{array}{l} P (A) = \frac{1}{3}, P (B) = \frac{1}{5} and P (A \cup B) = \frac{1}{2} \end{array}

\begin{array}{l} ∴ P (A \cap B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{1}{30} \end{array}

JEE Main 2022 July 25 Shift 2 Maths A8

\begin{array}{l} Now, P (A | B^{'}) + P (B | A^{'}) = \frac{P (A \cap B^{'})}{P (B^{'})} + \frac{P (B \cap A^{'})}{P (A^{'})} \end{array}

\begin{array}{l} = \frac{\frac{9}{30}}{\frac{4}{5}} + \frac{\frac{5}{30}}{\frac{2}{3}} = \frac{5}{8} \end{array}

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

$\begin{array}{l} \int_{- 3}^{101} ([\sin (π x)] + e^{[\cos (2 π x)]}) d x \end{array}$

is equal to

\begin{array}{l} (A) \frac{52 (1 - e)}{e} \end{array}

\begin{array}{l} (B) \frac{52}{e} \end{array}

\begin{array}{l} (C) \frac{52 (2 + e))}{e} \end{array}

\begin{array}{l} (D) \frac{104}{e} \end{array}

Answer (B)

Sol.

\begin{array}{l} I = \int_{- 3}^{101} ([s i n (π x)] + e^{[c o s (2 π x)]}) d x \end{array}

\begin{array}{l} [\sin π x] is periodic with period 2 \end{array}

and

\begin{array}{l} e^{[\cos 2 π x]} is periodic with period 1. \end{array}

So,

\begin{array}{l} I = 52 \int_{0}^{2} ([\sin (π x)] + e^{[\cos 2 π x]}) d x \end{array}

\begin{array}{l} = 52 {\int_{1}^{2} - 1 d x + \int_{1 / 4}^{3 / 4} e^{- 1} d x + \int_{5 / 4}^{7 / 4} e^{- 1} d x + \int_{0}^{1 / 4} e^{0} d x + \int_{3 / 4}^{5 / 4} e^{0} d x + \int_{7 / 4}^{2} e^{0} d x} \end{array}

\begin{array}{l} = \frac{52}{e} \end{array}

10. Let the point P(α, β) be at a unit distance from each of the two lines L₁ : 3x – 4y + 12 = 0 and L₂ : 8x + 6y + 11 = 0. If P lies below L₁ and above L₂, then 100(α + β) is equal to

(A) –14

(B) 42

(C) –22

(D) 14

Answer (D)

Sol.

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

Equation of angle bisector of L₁ and L₂ of angle containing origin

2(3x – 4y + 12) = 8x + 6y + 11

2x + 14y – 13 = 0 …(i)

\begin{array}{l} \frac{3 α - 4 β + 12}{5} = 1 \end{array}

\begin{array}{l} \Rightarrow 3 α - 4 β + 7 = 0 \dots (i i) \end{array}

Solution of 2x + 14y – 13 = 0 and 3x – 4y + 7 = 0 gives the required point

$\begin{array}{l} P (α, β), α = \frac{- 23}{25}, β = \frac{53}{50} \end{array}$

$\begin{array}{l} 100 (α + β) = 14 \end{array}$

11. Let a smooth curve y = f(x) be such that the slope of the tangent at any point (x, y) on it is directly proportional to (-y/x). If the curve passes through the points (1, 2) and (8, 1), then

$\begin{array}{l} | y (\frac{1}{8}) | is equal to \end{array}$

(A) 2 log_e2

(B) 4

(C) 1

(D) 4 log_e2

Answer (B)

Sol.

\begin{array}{l} \frac{d y}{d x} \propto \frac{- y}{x} \end{array}

\begin{array}{l} \frac{d y}{d x} = \frac{- k y}{x} \Rightarrow \int \frac{d y}{y} = - K \int \frac{d x}{x} \end{array}

ln | y | = -Kln | x | + C

If the above equation satisfies (1, 2) and (8, 1)

\begin{array}{l} l n 2 = - K \times 0 + C \Rightarrow C = l n 2 \end{array}

\begin{array}{l} \ln 1 = - K \ln 8 + \ln 2 \Rightarrow K = \frac{1}{3} \end{array}

\begin{array}{l} So, at x = \frac{1}{8} \end{array}

\begin{array}{l} \ln | y | = - \frac{1}{3} \ln (\frac{1}{8}) + \ln 2 = 2 \ln 2 \end{array}

|y| = 4

12. If the ellipse

$\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 meets the line \frac{x}{7} + \frac{y}{2 \sqrt{6}} = 1 \end{array}$

on the x-axis and the line
$\begin{array}{l} \frac{x}{7} - \frac{y}{2 \sqrt{6}} = 1 \end{array}$
on the y-axis, then the eccentricity of the ellipse is

\begin{array}{l} (A) \frac{5}{7} \end{array}

\begin{array}{l} (B) \frac{2 \sqrt{6}}{7} \end{array}

\begin{array}{l} (C) \frac{3}{7} \end{array}

\begin{array}{l} (D) \frac{2 \sqrt{5}}{7} \end{array}

Answer (A)

Sol.

\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 meets the line \frac{x}{7} + \frac{7}{2 \sqrt{6}} = 1 on the x-axis. \end{array}

So, a = 7

and

\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 meets the line \frac{x}{7} - \frac{y}{2 \sqrt{6}} = 1 on the y-axis. \end{array}

\begin{array}{l} So, b = 2 \sqrt{6} \end{array}

\begin{array}{l} ∴ e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{24}{49} \end{array}

\begin{array}{l} e = \frac{5}{7} \end{array}

13. The tangents at the points A(1, 3) and B(1, –1) on the parabola y² – 2x – 2y = 1 meet at the point P. Then the area (in unit²) of the triangle PAB is :

(A) 4

(B) 6

(C) 7

(D) 8

Answer (D)

Sol. Given curve : y² – 2x – 2y = 1.

Can be written as

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

JEE Main 2022 July 25 Shift 2 Maths A13

And, the given information can be plotted as shown in figure.

Tangent at A : 2y – x – 5 = 0 {using T = 0}

Intersection with y = 1 is x = –3

Hence, point P is (–3, 1)

Taking advantage of symmetry

\begin{array}{l} Area of Δ P A B = 2 \times \frac{1}{2} \times (1 - (- 3)) \times (3 - 1) = 8 sq. units \end{array}

14. Let the foci of the ellipse

$\begin{array}{l} \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 and the hyperbola \frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25} \end{array}$

coincide. Then the length of the latus rectum of the hyperbola is :

\begin{array}{l} (A) \frac{32}{9} \end{array}

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

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

\begin{array}{l} (D) \frac{27}{10} \end{array}

Answer (D)

Sol.

\begin{array}{l} Ellipse : \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 \end{array}

\begin{array}{l} Eccentricity = \sqrt{1 - \frac{7}{16}} = \frac{3}{4} \end{array}

\begin{array}{l} Foci \equiv (\pm a e, 0) \equiv (\pm 3, 0) \end{array}

\begin{array}{l} Hyperbola : \frac{x^{2}}{(\frac{144}{25})} - \frac{y^{2}}{(\frac{α}{25})} = 1 \end{array}

\begin{array}{l} Eccentricity = \sqrt{1 + \frac{α}{144}} = \frac{1}{12} \sqrt{144 + α} \end{array}

\begin{array}{l} Foci \equiv (\pm a e, 0) \equiv (\pm \frac{12}{5} \cdot \frac{1}{12} \sqrt{144 + α}, 0) \end{array}

If foci coincide then

\begin{array}{l} 3 = \frac{1}{5} \sqrt{144 + α} \Rightarrow α = 81 \end{array}

Hence, hyperbola is

\begin{array}{l} \frac{x^{2}}{{(\frac{12}{5})}^{2}} - \frac{y^{2}}{{(\frac{9}{5})}^{2}} = 1 \end{array}

Length of latus rectum

\begin{array}{l} = 2 \cdot \frac{\frac{81}{25}}{\frac{12}{5}} = \frac{27}{10} \end{array}

15. A plane E is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4, and passes through the point P(1, –1, 1). If the distance of the plane E from the point Q(a, a, 2) is 3√2, then (PQ)² is equal to

(A) 9

(B) 12

(C) 21

(D) 33

Answer (C)

Sol. First plane, P₁ = 2x – 2y + z = 0,

\begin{array}{l} normal vector \equiv {\overset{―}{n}}_{1} = (2, - 2, 1) \end{array}

Second plane, P₂ ≡ x – y + 2z = 4,

\begin{array}{l} normal vector \equiv {\overset{―}{n}}_{2} = (1, - 1, 2) \end{array}

\begin{array}{l} Plane perpendicular to P_{1} and P_{2} will have normal vector {\overset{―}{n}}_{3} \end{array}

Where

\begin{array}{l} {\overset{―}{n}}_{3} = ({\overset{―}{n}}_{1} \times {\overset{―}{n}}_{2}) \end{array}

Hence,

\begin{array}{l} {\overset{―}{n}}_{3} = (- 3, - 3, 0) \end{array}

\begin{array}{l} Equation of plane E through P (1, - 1, 1) and {\overset{―}{n}}_{3} as normal vector \end{array}

\begin{array}{l} (x - 1, y + 1, z - 1) \cdot (- 3, - 3, 0) = 0 \end{array}

\begin{array}{l} \Rightarrow x + y = 0 \equiv E \end{array}

Distance of PQ(a, a, 2) from E = |2a/√2|

as given,

\begin{array}{l} | \frac{2 a}{\sqrt{2}} | = 3 \sqrt{2} \Rightarrow a = \pm 3 \end{array}

\begin{array}{l} Hence, Q \equiv (\pm 3, \pm 3, 2) \end{array}

Distance PQ

\begin{array}{l} = \sqrt{21} \Rightarrow (P Q)^{2} = 21 \end{array}

16. The shortest distance between the lines

$\begin{array}{l} \frac{x + 7}{- 6} = \frac{y - 6}{7} = z and \frac{7 - x}{2} = y - 2 = z - 6 \end{array}$

is

\begin{array}{l} (A) 2 \sqrt{29} \end{array}

(B) 1

\begin{array}{l} (C) \sqrt{\frac{37}{29}} \end{array}

\begin{array}{l} (D) \sqrt{\frac{29}{2}} \end{array}

Answer (A)

Sol.

\begin{array}{l} L_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1} \end{array}

Any point on it

\begin{array}{l} {\vec{a}}_{1} (- 7, 6, 0) \end{array}

and L₁ is parallel to

\begin{array}{l} {\vec{b}}_{1} (- 6, 7, 1) \end{array}

\begin{array}{l} L_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1} \end{array}

Any point on it,

\begin{array}{l} {\vec{a}}_{2} (7, 2, 6) \end{array}

and L₂ is parallel to

\begin{array}{l} {\vec{b}}_{2} (- 2, 1, 1) \end{array}

Shortest distance between L₁ and L₂

\begin{array}{l} = | \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} | = | \frac{(- 14, 4, - 6) \cdot (3, 2, 4)}{\sqrt{9 + 4 + 16}} | \end{array}

\begin{array}{l} = 2 \sqrt{29} \end{array}

17. Let

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

$\begin{array}{l} and let \vec{b} be a vector such that \end{array}$

$\begin{array}{l} \vec{a} \times \vec{b} = 2 \hat{i} - \hat{k} and \vec{a} \cdot \vec{b} = 3 \end{array}$
Then the projection of
$\begin{array}{l} \vec{b} on the vector \vec{a} - \vec{b} is : \end{array}$

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

\begin{array}{l} (B) 2 \sqrt{\frac{3}{7}} \end{array}

\begin{array}{l} (C) \frac{2}{3} \sqrt{\frac{7}{3}} \end{array}

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

Answer (A)

Sol.

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

\begin{array}{l} \vec{a} \times \vec{b} = 2 \hat{i} - \hat{k} \end{array}

\begin{array}{l} \vec{a} \cdot \vec{b} = 3 \end{array}

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

\begin{array}{l} \Rightarrow 5 + 9 = 6 {| \vec{b} |}^{2} \end{array}

\begin{array}{l} \Rightarrow {| \vec{b} |}^{2} = \frac{7}{3} \end{array}

\begin{array}{l} | \vec{a} - \vec{b} | = \sqrt{{| \vec{a} |}^{2} + {| \vec{b} |}^{2} - 2 \vec{a} \cdot \vec{b}} = \sqrt{\frac{7}{3}} \end{array}

\begin{array}{l} projection of | \vec{b} | on \vec{a} - \vec{b} = \frac{\vec{b} \cdot (\vec{a} - \vec{b})}{| \vec{a} - \vec{b} |} \end{array}

\begin{array}{l} = \frac{\vec{b} \cdot \vec{a} - | \vec{b} |^{2}}{| \vec{a} - \vec{b} |} = \frac{3 - \frac{7}{3}}{\sqrt{\frac{7}{3}}} \end{array}

\begin{array}{l} = \frac{2}{\sqrt{21}} \end{array}

18. If the mean deviation about median for the number 3, 5, 7, 2k, 12, 16, 21, 24 arranged in the ascending order, is 6 then the median is

(A) 11.5

(B) 10.5

(C) 12

(D) 11

Answer (D)

Sol.

\begin{array}{l} Median = \frac{2 k + 12}{2} = k + 6 \end{array}

\begin{array}{l} Mean deviation = \sum \frac{| x_{i} - M |}{n} = 6 \end{array}

\begin{array}{l} \Rightarrow \frac{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)}{8} \end{array}

\begin{array}{l} ∴ \frac{58 - 2 k}{8} = 6 \\ k = 5 \end{array}

\begin{array}{l} Median = \frac{2 \times 5 + 12}{2} = 11 \end{array}

19.

$\begin{array}{l} 2 \sin (\frac{π}{22}) \sin (\frac{3 π}{22}) \sin (\frac{5 π}{22}) \sin (\frac{7 π}{22}) \sin (\frac{9 π}{22}) \end{array}$

is equal to :

\begin{array}{l} (A) \frac{3}{16} \\ (B) \frac{1}{16} \\ (C) \frac{1}{32} \\ (D) \frac{9}{32} \end{array}

Answer (B)

Sol.

\begin{array}{l} 2 \sin (\frac{π}{22}) \sin (\frac{3 π}{22}) \sin (\frac{5 π}{22}) \sin (\frac{7 π}{22}) \sin (\frac{9 π}{22}) \end{array}

\begin{array}{l} = 2 \sin (\frac{11 π - 10 π}{22}) \sin (\frac{11 π - 8 π}{22}) \sin (\frac{11 π - 6 π}{22}) \sin (\frac{11 π - 4 π}{22}) \sin (\frac{11 π - 2 π}{22}) \end{array}

\begin{array}{l} = 2 \cos \frac{π}{11} \cos \frac{2 π}{11} \cos \frac{3 π}{11} \cos \frac{4 π}{11} \cos \frac{5 π}{11} \end{array}

\begin{array}{l} = \frac{2 \sin \frac{32 π}{11}}{2^{5} \sin \frac{π}{11}} \end{array}

\begin{array}{l} = \frac{1}{16} \end{array}

20. Consider the following statements :

P : Ramu is intelligent.

Q : Ramu is rich.

R : Ramu is not honest.

The negation of the statement “Ramu is intelligent and honest if and only if Ramu is not rich” can be expressed as :

(A) ((P ∧ (~ R)) ∧ Q) ∧ ((~ Q) ∧ ((~ P) ∨ R))

(B) ((P ∧ R) ∧ Q) ∨ ((~ Q) ∧ ((~ P) ∨ (~ R)))

(C) ((P ∧ R) ∧ Q) ∧ ((~ Q) ∧ (( ~ P) ∨ (~ R)))

(D) ((P ∧ (~ R)) ∧ Q) ∨ ((~ Q) ∧ ((~ P) ∧ R))

Answer (D)

Sol. P : Ramu is intelligent

Q : Ramu is rich

R : Ramu is not honest

Given statement, “Ramu is intelligent and honest if and only if Ramu is not rich”

= (P ∧ ~ R) ⇔ ~ Q

So, negation of the statement is

~ [(P ∧ ~ R) ⇔ ~ Q]

= ~ [{~ (P ∧ ~ R) ∨ ~ Q} ∧ {Q ∨ (P ∧ ~ R)}]

= ((P ∧ ~ R) ∧ Q) ∨ (~ Q ∧ (~ P ∨ R))

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 A = {1, 2, 3, 4, 5, 6, 7}. Define

$\begin{array}{l} B = {T \subseteq A : either 1 \notin T o r 2 \in T} \end{array}$

and
$\begin{array}{l} C = {T \subseteq A : T the sum of all the elements of T is a prime number} \end{array}$
. Then the number of elements in the set B ∪ C is ______.

Answer (107)

Sol. ∵ (B ∪ C)′ = B′ ∩ C′

B′ is a set containing sub sets of A containing element 1 and not containing 2.

And C′ is a set containing subsets of A whose sum of elements is not prime.

So, we need to calculate number of subsets of

{3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.

Number of such 5 elements subset = 1

Number of such 4 elements subset = 3 (except selecting 3 or 7)

Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})

Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})

Number of such 1 elements subset = 3 (except selecting {4} or {6})

Number of such 0 elements subset = 1

n(B′∩C′) = 21 ⇒ n(B∪C) = 2⁷ – 21 = 107

2. Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p ≠ 0, and f(1) = 1/3 . If the equations f(x) = 0 and fo fo fo f(x) = 0 have a common real root, then f(–3) is equal to ______.

Answer (25)

Sol. Let f(x) = (x – α) (x – β)

It is given that f(0) = p ⇒ αβ = p

and

\begin{array}{l} f (1) = \frac{1}{3} \Rightarrow (1 - α) (1 - β) = \frac{1}{3} \end{array}

Now, let us assume that α is the common root of f(x) = 0 and fofofof(x) = 0

fofofof(x) = 0

⇒ fofof(0) = 0

⇒ fof(p) = 0

So, f(p) is either α or β.

(p – α) (p – β) = α

(αβ – α) (αβ – β) = α ⇒ (β – 1) (α – 1) β = 1

So, β = 3 (∵ α ≠ 0)

\begin{array}{l} (1 - α) (1 - 3) = \frac{1}{3} \end{array}

\begin{array}{l} α = \frac{7}{6} \end{array}

\begin{array}{l} f (x) = (x - \frac{7}{6}) (x - 3) \end{array}

\begin{array}{l} f (- 3) = (- 3 - \frac{7}{6}) (3 - 3) = 25 \end{array}

3. Let

$\begin{array}{l} A = [\begin{array}{c} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}], a, b \in R \end{array}$

. If for some
$\begin{array}{l} n \in N, A^{n} = [\begin{array}{c} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{array}] \end{array}$
then n + a + b is equal to __________.

Answer (24)

Sol.

\begin{array}{l} A = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] + [\begin{array}{c} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}] = I + B \end{array}

\begin{array}{l} B^{2} = [\begin{array}{c} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}] + [\begin{array}{c} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}] = [\begin{array}{c} 0 & 0 & a b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{array}

B³ = 0

∴ Aⁿ = (1 + B)ⁿ = ⁿC₀I + ⁿC₁ B + ⁿC₂B² + ⁿC₃ B³ + ….

\begin{array}{l} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] + [\begin{array}{c} 0 & n a & n a \\ 0 & 0 & n b \\ 0 & 0 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & \frac{n (n - 1) a b}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} 1 & n a & n a + \frac{n (n - 1)}{2} a b \\ 0 & 1 & n b \\ 0 & 0 & 1 \end{array}] = [\begin{array}{c} 1 & 48 & 2160 \\ 0 & 1 & 48 \\ 0 & 0 & 1 \end{array}] \end{array}

On comparing we get na = 48, nb = 96 and

\begin{array}{l} n a + \frac{n (n - 1)}{2} a b = 2160 \end{array}

⇒ a = 4, n = 12 and b = 8

n + a + b = 24

4. The sum of the maximum and minimum values of the function f(x) = |5x – 7| + [x² + 2x] in the interval [5/4, 2], where [t] is the greatest integer ≤ t, is ____.

Answer (15)

Sol. f(x) = |5x – 7| + [x² + 2x]

= |5x – 7| + [(x + 1)²] – 1

\begin{array}{l} Critical points of f (x) = \frac{7}{5}, \sqrt{5} - 1, \sqrt{6} - 1, \sqrt{7} - 1, \sqrt{8} - 1, 2 \end{array}

∴ Maximum or minimum value of f(x) occur at critical points or boundary points

\begin{array}{l} ∴ f (\frac{5}{4}) = \frac{3}{4} + 4 = \frac{19}{4} \end{array}

\begin{array}{l} f (\frac{7}{5}) = 0 + 4 = 4 \end{array}

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5.

∴ f(2) = 3 + 8 = 11

\begin{array}{l} ∴ f {(\frac{7}{5})}_{min} = 4 and f (2)_{max} = 11 \end{array}

Sum is 4 + 11 = 15

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

$\begin{array}{l} \frac{d y}{d x} = \frac{4 y^{3} + 2 y x^{2}}{3 x y^{2} + x^{3}}, y (1) = 1. \end{array}$

If for some n ∈ N, y(2) ∈ [n – 1, n), then n is equal to _________.

Answer (3)

Sol.

\begin{array}{l} \frac{d y}{d x} = \frac{y}{x} \frac{(4 y^{2} + 2 x^{2})}{(3 y^{2} + x^{2})} \end{array}

Put y = vx

\begin{array}{l} \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}

\begin{array}{l} \Rightarrow v + x \frac{d v}{d x} = \frac{v (4 v^{2} + 2)}{(3 v^{2} + 1)} \end{array}

\begin{array}{l} \Rightarrow x \frac{d v}{d x} = v (\frac{(4 v^{2} + 2 - 3 v^{2} - 1)}{3 v^{2} + 1}) \end{array}

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

⇒ ln|v³ +v| = lnx + c

\begin{array}{l} \Rightarrow \ln | {(\frac{y}{x})}^{3} + (\frac{y}{x}) | = \ln x + C \end{array}

↓ y(1) = 1

⇒ C = ln2

∴ for y(2)

\begin{array}{l} \ln (\frac{y^{3}}{8} + \frac{y}{2}) = 2 \ln 2 \Rightarrow \frac{y^{3}}{8} + \frac{y}{2} = 4 \end{array}

⇒ [y(2)] = 2

⇒ n = 3

6. Let f be a twice differentiable function on R. If f’(0) = 4 and

$\begin{array}{l} f (x) + \int_{0}^{x} (x - 1) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) \cos 2 x + \frac{2}{a} x, \end{array}$

then (2a + 1)⁵ a² is equal to ________________.

Answer (8)

Sol.

\begin{array}{l} ∵ f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) \cos 2 x + \frac{2 x}{a} \dots (i) \end{array}

Here f(0) = 2 …(ii)

On differentiating equation (i) w.r.t. x we get :

\begin{array}{l} f^{'} (x) + \int_{0}^{x} f^{'} (t) d t + x f^{'} (x) - x f^{'} (x) \\ = 2 (e^{2 x} - e^{- 2 x}) \cos 2 x - 2 (e^{2 x} + e^{- 2 x}) \sin 2 x + \frac{2}{a} \end{array}

⇒ f′(x) + f(x) – f(0) = 2(e²^x – e^–2^x)cos2x – 2 (e²^x + e^–2^x) sin 2x + (2/a)

Replace x by 0 we get :

\begin{array}{l} \Rightarrow 4 = \frac{2}{a} \Rightarrow a = \frac{1}{2} \end{array}

\begin{array}{l} ∴ (2 a + 1)^{5} \cdot a^{2} = 2^{5} \cdot \frac{1}{2^{2}} = 2^{3} = 8 \end{array}

7. Let

$\begin{array}{l} a_{n} = \int_{- 1}^{n} (1 + \frac{x}{2} + \frac{x^{2}}{3} + \dots + \frac{x^{n - 1}}{n}) d x \end{array}$

for every n ∈ N. Then the sum of all the elements of the set {n ∈ N : a_n ∈ (2, 30)} is ________________ .

Answer (5)

Sol.

\begin{array}{l} ∵ a_{n} = \int_{- 1}^{n} (1 + \frac{x}{2} + \frac{x^{2}}{3} + \dots + \frac{x^{n - 1}}{n}) d x \end{array}

\begin{array}{l} = {[x + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{3^{2}} + \dots + \frac{x^{n}}{n^{2}}]}_{- 1}^{n} \end{array}

\begin{array}{l} a_{n} = \frac{n + 1}{1^{2}} + \frac{n^{2} - 1}{2^{2}} + \frac{n^{3} + 1}{3^{2}} + \frac{n^{4} - 1}{4^{2}} + \dots + \frac{n^{n} + (- 1)^{n + 1}}{n^{2}} \end{array}

\begin{array}{l} Here a_{1} = 2, a_{2} = \frac{2 + 1}{1} + \frac{2^{2} - 1}{2} = 3 + \frac{3}{2} = \frac{9}{2} \end{array}

\begin{array}{l} a_{3} = 4 + 2 + \frac{28}{9} = \frac{100}{9} \end{array}

\begin{array}{l} a_{4} = 5 + \frac{15}{4} + \frac{65}{9} + \frac{255}{16} > 31 \end{array}

∴ The required set is {2, 3}. ∵ a_n ∈(2, 30)

∴ Sum of elements = 5.

8. If the circles x² + y² + 6x + 8y + 16 = 0 and

$\begin{array}{l} x^{2} + y^{2} + 2 (3 - \sqrt{3}) x + 2 (4 - \sqrt{6}) y = k + 6 \sqrt{3} + 8 \sqrt{6}, k > 0, \end{array}$

touch internally at the point P(α, β), then
$\begin{array}{l} (α + \sqrt{3})^{2} + (β + \sqrt{6})^{2} \end{array}$
is equal to ______.

Answer (25)

Sol. The circle x² + y² + 6x + 8y + 16 = 0 has centre

(–3, –4) and radius 3 units.

The circle

\begin{array}{l} x^{2} + y^{2} + 2 (3 - \sqrt{3}) x + 2 (4 - \sqrt{6}) y = k + 6 \sqrt{3} + 8 \sqrt{6}, k > 0 \end{array}

\begin{array}{l} has centre (\sqrt{3} - 3, \sqrt{6} - 4) \end{array}

\begin{array}{l} and radius \sqrt{k + 34} \end{array}

∵ These two circles touch internally hence

\begin{array}{l} \sqrt{3 + 6} = | \sqrt{k + 34} - 3 | \end{array}

Here, k = 2 is only possible (∵ k > 0)

Equation of common tangent to two circles is

\begin{array}{l} 2 \sqrt{3} x + 2 \sqrt{6} y + 16 + 6 \sqrt{3} + 8 \sqrt{6} + k = 0 \end{array}

∵ k = 2 then equation is

\begin{array}{l} x + \sqrt{2} y + 3 + 4 \sqrt{2} + 3 \sqrt{3} = 0 \dots (i) \end{array}

∵ (α, β) are foot of perpendicular from (–3, –4) to line (i) then

\begin{array}{l} \frac{α + 3}{1} = \frac{β + 4}{\sqrt{2}} = \frac{- (- 3 - 4 \sqrt{2} + 3 + 4 \sqrt{2} + 3 \sqrt{3})}{1 + 2} \end{array}

\begin{array}{l} ∴ α + 3 = \frac{β + 4}{\sqrt{2}} = - \sqrt{3} \end{array}

\begin{array}{l} \Rightarrow (α + \sqrt{3})^{2} = 9 and (β + \sqrt{6})^{2} = 16 \end{array}

\begin{array}{l} ∴ (α + \sqrt{3})^{2} + (β + \sqrt{6})^{2} = 25 \end{array}

9. Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0 at the point (–2, 3) be A. Then 8A is equal to _______.

Answer (170)

Sol. 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0 differentiating both sides we get

12x² – 3y² – 6xyy’ + 12x – 5y -5xy’ – 16yy’ + 9 = 0

\begin{array}{l} ↓ (- 2, 3) \end{array}

⇒ 48 – 27 + 36y’ – 24 – 15 + 10y’ – 48y’ + 9 = 0

⇒ 2y’ = – 9

\begin{array}{l} \Rightarrow m_{T} = \frac{- 9}{2} & m_{N} = \frac{2}{9} \end{array}

\begin{array}{l} ∴ Area = \frac{1}{2} \times Base \times Hight \end{array}

\begin{array}{l} A = \frac{1}{2} \times (\frac{- 4}{3} + \frac{31}{2}) (3) = \frac{1}{2} (\frac{85}{6}) \cdot 3 = \frac{85}{4} \\ = 8 A = 170 \end{array}

10. Let x = sin(2tan^–1 α) and

$\begin{array}{l} y = \sin (\frac{1}{2} \tan^{- 1} \frac{4}{3}) . \end{array}$

$\begin{array}{l} If S = {α \in R : y^{2} = 1 - x}, then \sum_{a \in S} 16 α^{3} \end{array}$
is equal to _______.

Answer (130)

Sol.

\begin{array}{l} ∵ x = \sin (2 \tan^{- 1} α) = \frac{2 α}{1 + α^{2}} \dots (i) \end{array}

and

\begin{array}{l} y = \sin (\frac{1}{2} \tan^{- 1} \frac{4}{3}) = \sin (\sin^{- 1} \frac{1}{\sqrt{5}}) = \frac{1}{\sqrt{5}} \end{array}

Now,

\begin{array}{l} y^{2} = 1 - x \end{array}

\begin{array}{l} \frac{1}{5} = 1 - \frac{2 α}{1 + α^{2}} \end{array}

\begin{array}{l} \Rightarrow 1 + α^{2} = 5 + 5 α^{2} - 10 α \end{array}

\begin{array}{l} \Rightarrow 2 α^{2} - 5 α + 2 = 0 \end{array}

\begin{array}{l} ∴ α = 2, \frac{1}{2} \end{array}

\begin{array}{l} ∴ \sum_{a \in S} 16 α^{3} = 16 \times 2^{3} + 16 \times \frac{1}{2^{3}} \end{array}

= 130

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JEE Main 2022 July 25th Shift 2 Paper Analysis

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