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# 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.

1. If z ≠ 0 be a complex number such that

$$\begin{array}{l}\left|z-\frac{1}{z}\right|=2,\end{array}$$
then the maximum value of |z| is

(A) √2

(B) 1

(C) √2 – 1

(D) √2 + 1

Sol.

$$\begin{array}{l}\left|z-\frac{1}{z}\right|\ge \left||z| – \frac{1}{z}\right|\end{array}$$
$$\begin{array}{l}\Rightarrow \left||z| – \frac{1}{|z|}\right|\le 2\end{array}$$

Let |z| = r

$$\begin{array}{l}\left|r-\frac{1}{r}\right|\le 2\end{array}$$
$$\begin{array}{l}-2\le r-\frac{1}{r}\le 2\end{array}$$
$$\begin{array}{l}r-\frac{1}{r}\ge -2 \text{ and } r -\frac{1}{r}\le 2\end{array}$$
$$\begin{array}{l}r^2 + 2r – 1 \geq 0 ~\text{and} ~r^2 – 2r – 1 \le 0 \end{array}$$
$$\begin{array}{l}r\in [-\infty , -1 – \sqrt{2}]\cup [-1+\sqrt{2}, \infty] \text{ and } r \in [1-\sqrt{2}, 1+ \sqrt{2}]\end{array}$$
$$\begin{array}{l}\text{Taking intersection}\ r \in [\sqrt{2}-1, \sqrt{2}+1]\end{array}$$

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

$$\begin{array}{l}\begin{bmatrix} -1& 2 \\1 & -1 \\\end{bmatrix}\end{array}$$
by a single elementary row operation?

$$\begin{array}{l}(\text{A})\ \begin{bmatrix} 0& 1 \\1 & -1 \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}(\text{B})\ \begin{bmatrix} 1& -1 \\-1 & 2\\\end{bmatrix}\end{array}$$
$$\begin{array}{l}(\text{C})\ \begin{bmatrix} -1& 2 \\-2 & 7\\\end{bmatrix}\end{array}$$
$$\begin{array}{l}(\text{D})\ \begin{bmatrix} -1& 2 \\-1 & 3\\\end{bmatrix}\end{array}$$

Sol.

$$\begin{array}{l}(1) \text{By } R_1 \rightarrow R_1 + R_2, \begin{bmatrix}0 & 1 \\1 & -1 \\\end{bmatrix} \text{ is possible}\end{array}$$
$$\begin{array}{l}(2) \text{By } R_1 \leftrightarrow R_2, \begin{bmatrix}1 & -1 \\-1 & 2 \\\end{bmatrix} \text{ is possible}\end{array}$$

(3) This matrix can’t be obtained

$$\begin{array}{l}(4) \text{By } R_2 \rightarrow R_2 +2R_1, \begin{bmatrix}-1 & 2 \\-1 & 3 \\\end{bmatrix} \text{ is possible}\end{array}$$

3. If the system of equations

$$\begin{array}{l}x + y + z = 6\\ 2x + 5y + \alpha z = \beta\\ x + 2y + 3z = 14\end{array}$$
has infinitely many solutions, then α + β is equal to

(A) 8

(B) 36

(C) 44

(D) 48

Sol.

$$\begin{array}{l}\Delta = \begin{vmatrix}1 & 1 & 1 \\2 & 5 & \alpha \\1 & 2 & 3 \\\end{vmatrix} = 1(15-2\alpha) – 1 (6-\alpha)+1(-1)\end{array}$$
$$\begin{array}{l}= 15 – 2\alpha – 6 + \alpha -1\\ = 8 – \alpha\end{array}$$
$$\begin{array}{l}\text{For infinite solutions,~} \Delta = 0 \Rightarrow \alpha = 8\end{array}$$
$$\begin{array}{l}\Delta_x = \begin{vmatrix}6 & 1 & 1 \\\beta & 5 & 8 \\14 & 2 & 3 \\\end{vmatrix} = 6(-1)-1(3\beta – 112)+1(2\beta -70)\end{array}$$
$$\begin{array}{l}= -6 – 3\beta + 112 + 2\beta – 70\\ = 36 – \beta\end{array}$$
$$\begin{array}{l}\Delta_x = 0 \Rightarrow for ~\beta = 36\\ \alpha + \beta = 44\end{array}$$

4. Let the function

$$\begin{array}{l}f(x)= \left\{\begin{matrix}\frac{\log_e(1+5x)-\log_e(1+\alpha x)}{x} &; \text{if } x\in0 \\10 & ; \text{if } x=0 \\\end{matrix}\right.\end{array}$$
be continuous at x = 0. Then α is equal to

(A) 10

(B) –10

(C) 5

(D) –5

Sol.

$$\begin{array}{l}\underset{x \to 0}{\text{It}}\frac{\ln(1+5x)-\ln(1+\alpha x)}{x}\\= 5 – \alpha = 10\end{array}$$
$$\begin{array}{l}\Rightarrow \alpha = -5\end{array}$$

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

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

$$\begin{array}{l}(\text{A})\ \frac{\sqrt{37} + \sqrt{13}-4}{6}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{\sqrt{37} – \sqrt{13}-4}{6}\end{array}$$
$$\begin{array}{l}(\text{C})\ \frac{-\sqrt{37} – \sqrt{13}+4}{6}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{-\sqrt{37} + \sqrt{13}+4}{6}\end{array}$$

Sol.

$$\begin{array}{l}I = \int_{0}^{1}[2x-|3x^2 – 5x + 2 |+1]dx\end{array}$$
$$\begin{array}{l}I = \int_{0}^{2/3} \left[ \underset{I_1}{\underbrace{-3x^2 + 7x -2}} \right]dx + \int_{2/3}^{1}\left[\underset{I_2}{\underbrace{3x^2-3x+2}} \right]dx + 1\end{array}$$
$$\begin{array}{l}I_1 = \int_{0}^{t_1}(-2)dx + \int_{t_1}^{1/3}(-1)dx + \int_{1/3}^{t_2}0.dx + \int_{t_2}^{2/3}dx\end{array}$$
$$\begin{array}{l}=-t_1 – t_2 + \frac{1}{3}, \text{ where } t_1 = \frac{7-\sqrt{37}}{6}, t_2 = \frac{7-\sqrt{13}}{6}\end{array}$$
$$\begin{array}{l}I_2=\int_{2/3}^{1}1dx =\frac{1}{3}\end{array}$$
$$\begin{array}{l}\therefore I = \frac{1}{3}-t_1 -t_2 + \frac{1}{3} + 1 = \frac{5}{3}-\left[\frac{7-\sqrt{37}}{6}+ \frac{7-\sqrt{13}}{6}\right]\end{array}$$
$$\begin{array}{l}=\frac{\sqrt{37}+\sqrt{13}-4}{6}\end{array}$$

6. Let

$$\begin{array}{l}\{a_n\}_{n=0}^\infty\ \text{be a sequence such that}\ a_0 = a_1 = 0\ \text{and}\end{array}$$
$$\begin{array}{l}a_{n+2}=3a_{n+1}-2a_{n} + 1, \forall \ n \ge 0.\ \text{Then}\ a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}\end{array}$$
is equal to

(A) 483

(B) 528

(C) 575

(D) 624

Sol.

$$\begin{array}{l}a_{n+2}=3a_{n+1}-2a_n+1, \forall \ n \ge 0 (a_0 = a_1 = 0)\end{array}$$
$$\begin{array}{l}\left(a_{n+2}-a_{n+1}\right)-2\left(a_{n+1}-a_n\right)-1 = 0\end{array}$$

Put n = 0

$$\begin{array}{l}\left(a_2-a_1\right)-2\left(a_1-a_0\right)-1=0\end{array}$$

n = 1

$$\begin{array}{l}\left(a_3-a_2\right)-2\left(a_2-a_1\right)-1=0\end{array}$$

n = 2

$$\begin{array}{l} \left(a_4-a_3\right)-2\left(a_3-a_2\right)-1=0\end{array}$$
$$\begin{array}{l}\vdots \\n = n \end{array}$$
$$\begin{array}{l}\left(a_{n+2 } – a_{n+1}\right)-2\left(a_{a+1}-a_n\right)-1=0\end{array}$$

$$\begin{array}{l}\left(a_{n+2 } – a_{1}\right)-2\left(a_{a+1}-a_0\right)-\left(n+1\right)=0\end{array}$$
$$\begin{array}{l}\therefore a_{n+2}-2a_{n+1}-\left(n+1\right)=0\end{array}$$
$$\begin{array}{l}n \rightarrow n – 2\end{array}$$
$$\begin{array}{l}a_n – 2a_{n-1}-n+1=0\end{array}$$

Now,

$$\begin{array}{l}a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}\end{array}$$
$$\begin{array}{l}=a_{25}\left(a_{23}-2a_{22}\right)-2a_{24}\left(a_{23}-2a_{22}\right)\end{array}$$
$$\begin{array}{l}=\left(a_{25}-2a_{24}\right)\left(a_{23}-2a_{22}\right) = 24 \cdot 22 = 528\end{array}$$

7.

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

$$\begin{array}{l}\left(A\right) 22! – 21!\\ \left(B\right) 22! – 2(21!)\\ \left(C\right) 21! – 2(20!)\\ \left(D\right) 21! – 20!\end{array}$$

Sol.

$$\begin{array}{l}\sum_{r=1}^{20}\left(r^2+1+2r-2r\right)r! = \sum_{r=1}^{20}\left((r+1)^2-2r\right)r!\end{array}$$
$$\begin{array}{l}=\sum_{r=1}^{20}\left[(r+1)(r+1)! -rr!\right]-\sum_{r=1}(r+1)r!=r!\end{array}$$
$$\begin{array}{l}=(2\cdot 2! – 1!)+(3\cdot 3! – 2\cdot 2!)+…+ (21\cdot21! – 20\cdot 20!) -\left[(2!-1!)+(3! -2!)+…+(21!-20!)\right]\end{array}$$
$$\begin{array}{l}=(21\cdot 21! – 1)-(21! -1)\end{array}$$
$$\begin{array}{l}=20 \cdot 21! = (22-2)21! = 22! – 2(21!)\end{array}$$

8. For

$$\begin{array}{l}I(x)=\int\frac{\sec^2 x – 2022}{\sin^{2022}x} dx,\ \text{if}\ I\left(\frac{\pi}{4}\right)=2^{1011},\end{array}$$
then

$$\begin{array}{l}(\text{A}) \ 3^{1010}I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0\end{array}$$
$$\begin{array}{l}(\text{B}) \ 3^{1010}I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0\end{array}$$
$$\begin{array}{l}(\text{C}) \ 3^{1011}I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0\end{array}$$
$$\begin{array}{l}(\text{C}) \ 3^{1011}I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0\end{array}$$

Sol.

$$\begin{array}{l}I(x)=\int\frac{\sec^2x – 2022}{\sin^{2022}x}dx\end{array}$$
$$\begin{array}{l}=\int \left(\sec^2x \cdot \sin^{-2022}x – 2022\sin^{-2022}x\right)dx\end{array}$$
$$\begin{array}{l}=\sin^{-2022}x \tan x + \int 2022\sin^{-2023}x \cos x \cdot \tan x \ dx – \int2022 \sin^{-2022}x \ dx + c\end{array}$$
$$\begin{array}{l}I(x)=\sin^{-2022}x \tan x + c\end{array}$$
$$\begin{array}{l}\because I\left(\frac{\pi}{4}\right)=2^{1011}\Rightarrow c = 2^{1011}-2^{1011}=0\end{array}$$
$$\begin{array}{l}\therefore I\left(\frac{\pi}{3}\right)=\left(\frac{2}{\sqrt{3}}\right)^{2022}\sqrt{3},I\left(\frac{\pi}{6}\right)=2^{2022}\frac{1}{\sqrt{3}}\end{array}$$
$$\begin{array}{l}\text{So, option (A)} : \frac{3^{1010}2^{2022}}{3^{1011}}\cdot \sqrt{3}-\frac{2^{2022}}{\sqrt{3}}=0\end{array}$$

∴ Option (A) is correct

9. if the solution curve of the differential equation

$$\begin{array}{l}\frac{dy}{dx}=\frac{x+y-2}{x-y} \end{array}$$
passes through the points (2, 1) and (k + 1, 2), k > 0, then

$$\begin{array}{l}(\text{A})\ 2\tan^{-1}\left(\frac{1}{k}\right)=\log_e(k^2+1)\end{array}$$
$$\begin{array}{l}(\text{B})\ \tan^{-1}\left(\frac{1}{k}\right)=\log_e(k^2+1)\end{array}$$
$$\begin{array}{l}(\text{C})\ 2\tan^{-1}\left(\frac{1}{k+1}\right)=\log_e(k^2+2k +2)\end{array}$$
$$\begin{array}{l}(\text{D})\ 2\tan^{-1}\left(\frac{1}{k}\right)=\log_e\left(\frac{k^2+1}{k^2}\right)\end{array}$$

Sol.

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

Let x – 1 = X, y – 1 = Y

$$\begin{array}{l}\frac{dY}{dX}=\frac{X+Y}{X-Y}\end{array}$$
$$\begin{array}{l}\text{Let } Y = tX \Rightarrow \frac{dY}{dX}=t + X\frac{dt}{dX}\end{array}$$
$$\begin{array}{l}t+X\frac{dt}{dX}=\frac{1+t}{1-t}\end{array}$$
$$\begin{array}{l}X\frac{dt}{dX}=\frac{1+t}{1-t}-t=\frac{1+t^2}{1-t}\end{array}$$
$$\begin{array}{l}\int\frac{1-t}{1+t^2}dt = \int \frac{dX}{X}\end{array}$$
$$\begin{array}{l}\tan^{-1}t -\frac{1}{2}\ln (1+t^2)=\ln |X|+c\end{array}$$
$$\begin{array}{l}\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2}\ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln|x-1|+c\end{array}$$

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

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

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

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

Sol.

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

Integrating factor I.F

$$\begin{array}{l}=e^{\int \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11 x + 6}dx}\end{array}$$
$$\begin{array}{l}\text{Let } \frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\end{array}$$

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

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

Curve passes through (0, 1)

$$\begin{array}{l}1\times \frac{1\times 2}{3}=0+c \Rightarrow c = \frac{2}{3}\end{array}$$
$$\begin{array}{l}\text{So, }y(1)=\frac{\frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} }{\frac{(2^2 \times 3)}{4}} = \frac{3}{2}\end{array}$$

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

$$\begin{array}{l}a^2 + 11a + 3(m_1^2 + m_2^2) = 220. \end{array}$$
If one vertex of the square is
$$\begin{array}{l}\left(10\left(cos~\alpha – sin~\alpha\right),10\left(sin~\alpha + cos~ \alpha\right)\right),\ \text{where}\ \alpha \in \left(0, \frac{\pi}{2}\right)\end{array}$$
and the equation of one diagonal is
$$\begin{array}{l}(\cos \alpha – \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10,\ \text{then}\ 72(\sin^4 \alpha + \cos^4 \alpha) + a^2 -3a + 13 \end{array}$$
is equal to :

(A) 119

(B) 128

(C) 145

(D) 155

Sol. One vertex of square is

$$\begin{array}{l}\left(10\left(cos\alpha – sin\alpha\right), 10\left(sin\alpha + cos\alpha\right)\right) \end{array}$$

and one of the diagonal is

$$\begin{array}{l}\left(cos\alpha – sin\alpha\right) x + \left(sin\alpha + cos\alpha\right)y = 10\end{array}$$

So the other diagonal can be obtained as

$$\begin{array}{l}\left(cos\alpha + sin\alpha\right)x – \left(cos\alpha – sin\alpha\right)y = 0\end{array}$$

So, point of intersection of diagonal will be

$$\begin{array}{l}\left(5\left(cos\alpha – sin\alpha\right), 5\left(cos\alpha + sin\alpha\right)\right).\end{array}$$

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

$$\begin{array}{l}a^2+11a+3\left(m_1^2 + m_2^2\right) = 220\end{array}$$
$$\begin{array}{l}m_1^2 + m_2^2 = \frac{220 – 100 -110}{3} = \frac{10}{3}\end{array}$$

and m1 m2 = – 1

Slopes of the sides are tanα and – cotα

$$\begin{array}{l}\tan^2 \alpha = 3 \text{ or } \frac{1}{3}\end{array}$$
$$\begin{array}{l}72\left(sin^4\alpha + cos^4\alpha\right) + a^2 – 3a + 13 \end{array}$$
$$\begin{array}{l}=72 \cdot \frac{\tan^4 \alpha + 1}{(1+\tan^2 \alpha)^2} + a^2 -3a + 13 =128\end{array}$$

12. The number of elements in the set

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

(A) 1

(B) 3

(C) 0

(D) infinite

Sol.

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

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

Sol.

Circumcentre of ΔABC

$$\begin{array}{l}=\left(\frac{\alpha + \frac{\alpha}{4}}{2}, \frac{6-2}{2}\right)\end{array}$$
$$\begin{array}{l}=\left(\frac{5\alpha}{8},2\right)\end{array}$$
$$\begin{array}{l}=\left(5, \frac{\alpha}{4}\right)\end{array}$$
$$\begin{array}{l}\Rightarrow \alpha = 8\end{array}$$
$$\begin{array}{l}area(\Delta ABC)=\frac{1}{2}\cdot \frac{3\alpha}{4}\times 8 = 24~ \text{sq. units}\end{array}$$
$$\begin{array}{l}\text{Perimeter } = 8 + \frac{3\alpha}{4}+\sqrt{8^2 + \left(\frac{3\alpha}{4}\right)^2}\\=8+6+10=24\end{array}$$
$$\begin{array}{l}\text{Circumradius} = \frac{10}{2}=5\end{array}$$
$$\begin{array}{l}r=\frac{\Delta}{s}=\frac{24}{12}=2\end{array}$$

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 :

$$\begin{array}{l}(\text{A})\ \frac{\sqrt{3}}{2}\end{array}$$
$$\begin{array}{l}(\text{B})\ \sqrt{3}\end{array}$$
$$\begin{array}{l}(\text{C})\ 2\sqrt{3}\end{array}$$
$$\begin{array}{l}(\text{D}) 3\end{array}$$

Sol.

$$\begin{array}{l}PQ = \left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}\end{array}$$
$$\begin{array}{l}QR = \frac{PQ}{\tan 60^{\circ}} = \frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\end{array}$$
$$\begin{array}{l}\text{Area}(\Delta PQR) = \frac{1}{2}\cdot PQ \cdot QR = \sqrt{3}\end{array}$$

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 :

$$\begin{array}{l}(\text{A})\ \frac{21}{2}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{59}{8}\end{array}$$
$$\begin{array}{l}(\text{C})\ \frac{57}{8}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{95}{8}\end{array}$$

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

$$\begin{array}{l}\therefore \begin{vmatrix}\lambda – 2 & -1 & -6 \\-1 & \lambda-3 & -1 \\3 & -1 & -8 \\\end{vmatrix} =0\end{array}$$
$$\begin{array}{l}\therefore 8\lambda^2 – 67\lambda + 95 = 0\end{array}$$
$$\begin{array}{l}\therefore\text{Product of all values of }\lambda = \frac{95}{8}\end{array}$$

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 :

$$\begin{array}{l}(\text{A})\ \frac{4}{9}\end{array}$$
$$\begin{array}{l}(\text{B})\ \frac{5}{18}\end{array}$$
$$\begin{array}{l}(\text{C})\ \frac{1}{6}\end{array}$$
$$\begin{array}{l}(\text{D})\ \frac{3}{10}\end{array}$$

Sol.

$$\begin{array}{l} \text{Let} E \rightarrow \text{Ball drawn from Bag II is black.} \\ E_R \rightarrow \text{Bag I to Bag II red ball transferred.}\\ E_B \rightarrow \text{Bag I to Bag II black ball transferred.}\\ E_W \rightarrow \text{Bag I to Bag II white ball transferred.}\end{array}$$
$$\begin{array}{l}P\left(E_R/E\right) =\frac{P(E/E_R)\cdot P(E_R)}{P(E/E_R)P(E_R) + P(E/E_B)\cdot P(E_B) + P(E/E_W)P(E_W)}\end{array}$$

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

$$\begin{array}{l}P\left(\dfrac{E}{E_R}\right)=\frac{5}{10}, P\left(\dfrac{E}{E_B}\right)=\frac{6}{10}, P\left(\dfrac{E}{E_W}\right)=\frac{5}{10}\end{array}$$
$$\begin{array}{l}\therefore P\left(\dfrac{E_R}{E}\right)= \frac{\frac{15}{100}}{\frac{15}{100} + \frac{24}{100} + \frac{15}{100}}\end{array}$$
$$\begin{array}{l}=\frac{15}{54}=\frac{5}{18}\end{array}$$

17. Let

$$\begin{array}{l}S = \{z = x + iy : \left|z – 1 + i\right| \geq \left|z\right|, \left|z\right| < 2, \left|z + i\right| = \left|z – 1\right|\}.\end{array}$$
Then the set of all values of x, for which w = 2x + iy ∈ S for some y ∈ R is

$$\begin{array}{l}(\text{A})\ \left(-\sqrt{2}, \frac{1}{2\sqrt{2}}\right]\end{array}$$
$$\begin{array}{l}(\text{B})\ \left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]\end{array}$$
$$\begin{array}{l}(\text{C})\ \left(-\sqrt{2}, \frac{1}{2}\right]\end{array}$$
$$\begin{array}{l}(\text{D})\ \left(-\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\right]\end{array}$$

Sol.

$$\begin{array}{l}S : \{z = x + iy : \left|z – 1 + i\right| \geq \left|z\right|, \left|z\right| < 2, \left|z – i\right| = \left|z – 1\right|\} \left|z – 1 + i\right| \geq \left|z\right| \end{array}$$

|z| < 2

$$\begin{array}{l}\left|z – i\right| = \left|z – 1\right|\end{array}$$

$$\begin{array}{l}\because w \in S ~\text{and} ~w = 2x + iy\end{array}$$
$$\begin{array}{l}2x< \frac{1}{2}\ \ \ \therefore x < \frac{1}{4}\end{array}$$
$$\begin{array}{l}(2x)^2 + (-2x)^2 < 4\end{array}$$
$$\begin{array}{l}4x^2 + 4x^2 < 4\end{array}$$
$$\begin{array}{l}x^2 < \frac{1}{2}\Rightarrow x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\end{array}$$
$$\begin{array}{l}\therefore x \in \left(-\frac{1}{2},\frac{1}{4}\right]\end{array}$$

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
$$\begin{array}{l}(\vec{a}\times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c})\cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a})\cdot (\vec{a} \times \vec{b}) = 168,\ \text{then}\ |\vec{a}| + |\vec{b}| + |\vec{c}|\end{array}$$
is equal to :

(A) 10

(B) 14

(C) 16

(D) 18

Sol.

$$\begin{array}{l}|\vec{a}| |\vec{b}| |\vec{c}| = 14\end{array}$$
$$\begin{array}{l}\vec{a} \wedge \vec{b} = \vec{b} \wedge \vec{c} = \vec{c} \wedge \vec{a} = \theta = \frac{2\pi}{3}\end{array}$$
$$\begin{array}{l}\vec{a} \cdot \vec{b} = -\frac{1}{2}|\vec{a}||\vec{b}|\end{array}$$
$$\begin{array}{l}\vec{b} \cdot \vec{c} = -\frac{1}{2}|\vec{b}||\vec{c}|\end{array}$$
$$\begin{array}{l}\vec{c} \cdot \vec{a} = -\frac{1}{2}|\vec{c}||\vec{a}|\end{array}$$

Now,

$$\begin{array}{l}(\vec{a}\times\vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c})\cdot(\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a})\cdot (\vec{a} \times\vec{b}) = 168\ \ \ …(i)\end{array}$$
$$\begin{array}{l}(\vec{a}\times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a}\cdot \vec{b}) (\vec{b} \cdot \vec{c}) – (\vec{a} \cdot \vec{c})|\vec{b}|^2\end{array}$$
$$\begin{array}{l}=\frac{1}{4}|\vec{b}|^2|\vec{a}||\vec{c}| + \frac{1}{2}|\vec{a}||\vec{b}|^2|\vec{c}|\end{array}$$
$$\begin{array}{l}=\frac{3}{4}|\vec{a}||\vec{b}|^2|\vec{c}|\ \ \ …(ii)\end{array}$$
$$\begin{array}{l}\text{Similarly } (\vec{b} \times \vec{c})\cdot (\vec{c} \times \vec{a}) = \frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|^2\ \ \ …(iii)\end{array}$$
$$\begin{array}{l}(\vec{c} \times \vec{a}) \cdot (\vec{a}\times \vec{b}) = \frac{3}{4}|\vec{a}|^2 |\vec{b}||\vec{c}|\ \ \ …(iv)\end{array}$$

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

$$\begin{array}{l}\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|\left[|\vec{a}|+|\vec{b}|+|\vec{c}|\right] = 168\end{array}$$
$$\begin{array}{l}\frac{3}{4}\times 14\left[|\vec{a}|+|\vec{b}|+|\vec{c}|\right] = 168\end{array}$$
$$\begin{array}{l}|\vec{a}|+|\vec{b}|+|\vec{c}|= 16\end{array}$$

19. The domain of the function

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

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

Sol.

$$\begin{array}{l}f(x)=\sin^{-1}\left(\frac{x^2-3x+2}{x^2+2x+7}\right)\end{array}$$
$$\begin{array}{l}-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1\end{array}$$
$$\begin{array}{l}\frac{x^2-3x+2}{x^2+2x+7}\le 1\end{array}$$
$$\begin{array}{l}x^2-3x+2 \le x^2+2x+7\end{array}$$
$$\begin{array}{l}5x \ge -5\end{array}$$
$$\begin{array}{l}x\ge -1\ \ \ …(i)\end{array}$$
$$\begin{array}{l}\frac{x^2-3x+2}{x^2 +2x +7}\ge -1\end{array}$$
$$\begin{array}{l}x^2-3x+2\ge -x^2 -2x -7\end{array}$$
$$\begin{array}{l}2x^2- x + 9 \ge 0\end{array}$$
$$\begin{array}{l}x \in R\ \ \ \ …(ii) \end{array}$$
$$\begin{array}{l}\left(i\right)\cap \left(ii\right)\end{array}$$
$$\begin{array}{l}\text{Domain} \in \left[-1, \infty\right) \end{array}$$

20. The statement

$$\begin{array}{l}\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)\end{array}$$
is NOT equivalent to

$$\begin{array}{l}(\text{A})\ (p \wedge (\sim r))\Rightarrow q\end{array}$$
$$\begin{array}{l}(\text{B})\ (\sim q)\Rightarrow ((\sim r)\vee p)\end{array}$$
$$\begin{array}{l}(\text{C})\ p\Rightarrow (q\vee r)\end{array}$$
$$\begin{array}{l}(\text{D})\ (p\wedge (\sim q))\Rightarrow r\end{array}$$

Sol.

$$\begin{array}{l}\left(A\right) \left(p \wedge \left(\sim r\right)\right) \Rightarrow q\\ \sim \left(p \wedge \sim r\right) \vee q\\ \equiv \left(\sim p \vee r\right) \vee q \\ \equiv \sim p \vee \left(r \vee q\right)\\ \equiv p \rightarrow \left(q \vee r\right)\\ \equiv \left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)\end{array}$$
$$\begin{array}{l}\left(C\right) p \Rightarrow \left(q \vee r\right)\\ \equiv \sim p \vee \left(q \vee r\right)\\ \equiv \left(\sim p \vee q\right) \vee \left(\sim p \vee r\right)\\ \equiv \left(p \rightarrow q\right) \vee \left(p \rightarrow r\right)\end{array}$$
$$\begin{array}{l}\left(D\right) \left(p \wedge \sim q\right) \Rightarrow r\\ \equiv p \Rightarrow \left(q \vee r\right)\\ \equiv \left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right) \end{array}$$

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 _______.

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

$$\begin{array}{l}\Rightarrow q = \frac{9}{24}=\frac{3}{8}, p = \frac{5}{8}\end{array}$$
$$\begin{array}{l}\therefore n\frac{5}{8}=60\\\Rightarrow n =96\end{array}$$

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

$$\begin{array}{l}\text{If}\ P_n = \alpha^n – \beta^n, n \in \mathbb{N}\ \text{then}\ \frac{P_{15}P_{16} – P_{14}P_{16} – P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}\end{array}$$
is equal to _______.

Sol.

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

$$\begin{array}{l}P_n = \alpha^n – \beta^n, \end{array}$$
$$\begin{array}{l}\therefore I = \frac{(P_{15}-P_{14})P_{16}-P_{15}(P_{15}-P_{14})}{P_{13}P_{14}} = \frac{(P_{16}-P_{15})(P_{15}-P_{14})}{P_{13}P_{14}}\end{array}$$
$$\begin{array}{l}\Rightarrow I = \frac{(\alpha^{16} – \beta^{16} – \alpha ^{15} + \beta^{15})(\alpha^{15} – \beta^{15} – \alpha ^{14} + \beta^{14})}{(\alpha^{13} – \beta^{13})(\alpha^{14} – \beta^{14})}\end{array}$$
$$\begin{array}{l}\Rightarrow I = \frac{(\alpha^{15}(\alpha – 1)-\beta^{15}(\beta -1))(\alpha^{14}(\alpha-1)-\beta^{14} (\beta-1))}{(\alpha^{13}-\beta^{13})(\alpha^{14}-\beta^{14})}\end{array}$$
$$\begin{array}{l}\text{As } \alpha^2 – \alpha = 4 \Rightarrow \alpha -1 = \frac{4}{\alpha} \text{ and } \beta -1 = \frac{4}{\beta}\end{array}$$
$$\begin{array}{l}\Rightarrow I = \frac{\left(\alpha^{15}\cdot \frac{4}{\alpha} – \beta^{15}\cdot \frac{4}{\beta}\right)\left(\alpha^{14}\cdot \frac{4}{\alpha} – \beta^{14}\cdot \frac{4}{\beta}\right)}{(\alpha^{13}-\beta^{13})(\alpha^{14}-\beta^{14})}\end{array}$$
$$\begin{array}{l}= \frac{16(\alpha^{14}-\beta^{14})(\alpha^{13}-\beta^{13})}{(\alpha^{14}-\beta^{14})(\alpha^{13}-\beta^{13})} = 16\end{array}$$

3. Let

$$\begin{array}{l}x=\begin{bmatrix}1 \\ 1\\1\end{bmatrix} and\ A = \begin{bmatrix}-1 & 2 & 3 \\0 & 1 & 6 \\0 & 0 & -1 \\\end{bmatrix}\end{array}$$
For k ∈ N, if X’AkX = 33, then k is equal to _______.

Sol. Given

$$\begin{array}{l}A=\begin{bmatrix}-1 & 2 & 3 \\0 & 1 & 6 \\0 & 0 & -1 \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}A^2=\begin{bmatrix}1 & 0 & 6 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}, A^4 = \begin{bmatrix}1 & 0 & 12 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}\Rightarrow A^k=\begin{bmatrix}1 & 0 & 3k \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}\end{array}$$
$$\begin{array}{l}\therefore X’A^kX=\begin{bmatrix}1 & 1 & 1 \\\end{bmatrix}\begin{bmatrix}1 & 0 & 3k \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix}1 \\1 \\1\end{bmatrix} = \begin{bmatrix}3k + 3\end{bmatrix}\end{array}$$

⇒ [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 _______.

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

$$\begin{array}{l}\left(\overline {a\ b\ c\ d}\right)\end{array}$$
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

$$\begin{array}{l}\sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 = 22000L,\end{array}$$
then L is equal to _____.

Sol.

$$\begin{array}{l}\sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 =1^2\ ^{10}C_1^2 + 2^2\ ^{10}C_2^2 + … + 10^2\ ^{10}C_{10}\end{array}$$

Let

$$\begin{array}{l}\left(1 + x\right)^{10} = ^{10}C_0 + ^{10}C_1 x + ^{10}C_2 x^2 + ….+ ^{10}C_{10} x^{10} \end{array}$$
$$\begin{array}{l}\Rightarrow 10\left(1 + x\right)^9 = ^{10}C_1 + 2\cdot ^{10}C_2 x +… + 10\cdot ^{10}C_{10} x^9 …\left(1\right)\end{array}$$

Similarly,

$$\begin{array}{l}10\left(x + 1\right)^9 = 10\cdot ^{10}C_0 x^9 + 9\cdot ^{10}C_1 x^8 + … + 1\cdot ^{10}C_9\end{array}$$
$$\begin{array}{l}100\left(1+ x\right)^{18}\text{has required term with coefficient of} ~x^9 \end{array}$$
$$\begin{array}{l}i.e. ^{18}C_9 100 = 22000 L\\ \Rightarrow L = 221\end{array}$$

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

$$\begin{array}{l}f(x)=4|2x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]\end{array}$$
is not differentiable in the open interval (–20, 20), is ________.

Sol.

$$\begin{array}{l}f(x)=4|2x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]\end{array}$$
$$\begin{array}{l}=4|2x+3|+9\left[x+\frac{1}{2}\right]-12[x]-240\end{array}$$

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

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

as well as

$$\begin{array}{l}\left\{-\frac{39}{2}, -\frac{37}{2},…,-\frac{3}{2},-\frac{1}{2},\frac{1}{2},…,\frac{39}{2}\right\}\end{array}$$
,

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 = x3x2 + 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 ________.

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

$$\begin{array}{l}=m=\left(\frac{dy}{dx}\right)_{\text{at}(2,-1)} = 22\end{array}$$

∴ 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

$$\begin{array}{l}\therefore a = 3 \text{ or } -\frac{7}{3}\end{array}$$

Also b = a3a2 + a

$$\begin{array}{l}\text{Then} \left(a, b\right) = \left(3, 21\right) or \left(-\frac{7}{3}, – \frac{151}{9}\right)\end{array}$$
$$\begin{array}{l}\left(-\frac{7}{3}, – \frac{151}{9}\right)\ \text{does not satisfy the equation of tangent}\end{array}$$
$$\begin{array}{l}\therefore a = 3, b = 21\\ \therefore \left|2a + 9b\right| = 195\end{array}$$

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 _______.

Sol. Here AM = BM = 6

$$\begin{array}{l}OM = \sqrt{\left(\frac{13}{2}\right)^2-6^2}=\frac{5}{2}\end{array}$$

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

In ΔPAO,

$$\begin{array}{l}\frac{PO}{OA}=\sec \theta\end{array}$$
$$\begin{array}{l}PO = \frac{13}{2}\cdot \frac{13}{5}=\frac{169}{10}\end{array}$$
$$\begin{array}{l}\therefore PM = \frac{169}{10}-\frac{5}{2}=\frac{144}{10}=\frac{72}{5}\end{array}$$
$$\begin{array}{l}\therefore 5PM = 72\end{array}$$

9. Let

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

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

Sol.

$$\begin{array}{l}\because |\vec{a} + \vec{b}|^2 = |\vec{a}|^2+2|\vec{b}|^2\end{array}$$
$$\begin{array}{l}\text{ or } |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot \vec{b} = |\vec{a}|^2 + 2|\vec{b}|^2\end{array}$$
$$\begin{array}{l}\therefore |\vec{b}|^2 = 6 \ \ \ …(i)\end{array}$$
$$\begin{array}{l}\text{Now} |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 -\left(\vec{a} \cdot \vec{b}\right)^2\end{array}$$
$$\begin{array}{l}75 = |\vec{a}|^2 \cdot 6-9\end{array}$$
$$\begin{array}{l}\therefore |\vec{a}|^2 =14\end{array}$$

10. Let

$$\begin{array}{l}S =\left\{(x,y)\in \mathbb{N}\times \mathbb{N} : 9(x-3)^2 + 16(y-4)^2 \le 144\right\}\end{array}$$
and
$$\begin{array}{l}T =\left\{(x,y)\in \mathbb{R}\times \mathbb{R} : (x-7)^2 + (y-4)^2 \le 36\right\}.\end{array}$$
Then n(S ⋂ T) is equal to ______.

Sol.

$$\begin{array}{l}S = \left\{(x,y) \in \mathbb{N} \times \mathbb{N}: \frac{(x-3)^2}{16} + \frac{(y-4)^2}{9}\le 1\right\}\end{array}$$
represents all the integral points inside and on the ellipse
$$\begin{array}{l}\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9}=1\end{array}$$

$$\begin{array}{l}\text{and }T = \left\{(x,y) \in \mathbb{R}\times \mathbb{R} : (x-7)^2 + (y-4)^2 \le 36\right\}\end{array}$$
represents all the points on and inside the circle
$$\begin{array}{l}(x-7)^2+(y-4)^2=36\end{array}$$
.

$$\begin{array}{l}\therefore \left(S\cap T \right)=\{\left(3,1\right)\left(2,2\right)\left(3,2\right)\left(4,2\right)\left(5,2\right)\left(2,3\right)……….\left(6,5\right)\}\end{array}$$

Total number of points = 27