VARDAAN LEARNING INSTITUTE

Explanation:
• Reflexive: \( a - a = 0 = 2(0) \) is even, so \( (a, a) \in R \).
• Symmetric: If \( (a, b) \in R \implies a - b = 2k \implies b - a = -2k = 2(-k) \) (even) \( \implies (b, a) \in R \).
• Transitive: If \( a - b = 2k \) and \( b - c = 2m \), then \( a - c = (a - b) + (b - c) = 2(k + m) \) (even) \( \implies (a, c) \in R \).
Hence, \( R \) is an equivalence relation.

Explanation:
• One-one: \( f(x_1) = f(x_2) \implies 3 - 4x_1 = 3 - 4x_2 \implies x_1 = x_2 \). Hence, one-one.
• Onto: For any \( y \in \mathbb{R} \), let \( y = 3 - 4x \implies x = \frac{3 - y}{4} \in \mathbb{R} \). Thus, \( f(x) = y \). Hence, onto.
Since \( f \) is both one-one and onto, it is bijective.

Explanation:
The principal value branch of \( \cos^{-1} \) is \( [0, \pi] \). Since \( \frac{7\pi}{6} \notin [0, \pi] \), we write:
\[ \cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{5\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \] Since \( \frac{5\pi}{6} \in [0, \pi] \), \( \cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6} \).

Explanation:
For a square matrix \( A \) of order \( n \), \( |kA| = k^n |A| \).
Here \( n = 3, k = 2 \implies |2A| = 2^3 |A| = 8 \times 4 = 32 \).

Explanation:
Given \( A' = A \) and \( B' = B \).
\( (AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA) \).
Hence, \( AB - BA \) is a skew-symmetric matrix.

Explanation:
LHS: \( 2(1) - 4(5) = 2 - 20 = -18 \).
RHS: \( (2x)(x) - 4(6) = 2x^2 - 24 \).
\( 2x^2 - 24 = -18 \implies 2x^2 = 6 \implies x^2 = 3 \implies x = \pm \sqrt{3} \).
(Correction: Option (b) is \( \pm\sqrt{3} \), which is the correct value).

Explanation:
For a square matrix \( A \) of order \( n \), \( |\text{adj } A| = |A|^{n-1} \).
Here \( n = 3 \implies |\text{adj } A| = |A|^{3-1} = |A|^2 = 5^2 = 25 \).

Explanation:
\( \lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} \). Let \( x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2}, h \to 0 \).
\( \lim_{h \to 0} \frac{k\cos(\pi/2 + h)}{\pi - 2(\pi/2 + h)} = \lim_{h \to 0} \frac{-k\sin h}{-2h} = \frac{k}{2} \).
Since \( f(x) \) is continuous at \( x = \frac{\pi}{2} \), \( \frac{k}{2} = f(\pi/2) = 3 \implies k = 6 \).

Explanation:
By chain rule: \( \frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x)) \cdot e^x = -e^x \frac{\sin(e^x)}{\cos(e^x)} = -e^x \tan(e^x) \).

Explanation:
\( V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dr} = 4\pi r^2 \).
\( S = 4\pi r^2 \implies \frac{dS}{dr} = 8\pi r \).
\( \frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4\pi r^2}{8\pi r} = \frac{r}{2} \).
At \( r = 4\text{ cm} \), \( \frac{dV}{dS} = \frac{4}{2} = 2\text{ cm} \).

Explanation:
\( f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2) \).
For strictly decreasing, \( f'(x) < 0 \implies (x - 1)(x - 2) < 0 \implies x \in (1, 2) \).

Explanation:
\( |\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \).
Unit vector \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}} \).

Explanation:
\( \vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (3)(4) = 2 - 2 + 12 = 12 \).

Explanation:
Let \( \alpha = \beta = \gamma \). Then \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \implies 3\cos^2\alpha = 1 \implies \cos\alpha = \pm \frac{1}{\sqrt{3}} \).
Direction cosines are \( \left(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right) \).

Explanation:
Two lines are perpendicular if \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \).
\( (-3)(3k) + (2k)(1) + (2)(-5) = 0 \implies -9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -\frac{10}{7} \).

Explanation:
Evaluating \( Z = 4x + 6y \) at corner points:
• \( (0, 2): Z = 0 + 12 = 12 \)
• \( (3, 0): Z = 12 + 0 = 12 \)
• \( (6, 0): Z = 24 + 0 = 24 \)
• \( (0, 5): Z = 0 + 30 = 30 \)
• \( (6, 8): Z = 4(6) + 6(8) = 24 + 48 = 72 \).
Maximum value is 72, which occurs uniquely at \( (6, 8) \).

Explanation:
By definition, \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). Since \( P(B) = 0 \), division by zero is not defined.

Explanation:
If \( A \) and \( B \) are independent, then \( A' \) and \( B' \) are also independent.
\( P(A' \cap B') = P(A') \cdot P(B') = (1 - P(A))(1 - P(B)) = (1 - 0.3)(1 - 0.4) = (0.7)(0.6) = 0.42 \).

Explanation:
\( f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2 \ge 0 \) for all \( x \). Since \( f'(x) \) does not change sign across \( x = 1 \), it has neither local maximum nor minimum. Thus Assertion is true.
Reason states that \( f'(c) = 0 \) and \( f''(c) = 0 \) necessarily implies a point of inflection, which is false in general (e.g. \( f(x) = x^4 \) has local minimum at \( x = 0 \) where \( f'(0)=0, f''(0)=0 \)). Thus Reason is false.

Explanation:
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b} \).
Since \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \),
\( = \vec{0} - \vec{a} \times \vec{b} - \vec{a} \times \vec{b} - \vec{0} = -2(\vec{a} \times \vec{b}) \).
Both A and R are true, and R correctly explains A.

Solution:
For \( \sin^{-1}(t) \), the domain requires \( -1 \le t \le 1 \).
Here, \( -1 \le 2x - 3 \le 1 \)
Adding 3 to all parts: \( 2 \le 2x \le 4 \)
Dividing by 2: \( 1 \le x \le 2 \).
Domain: \( [1, 2] \).
The principal value branch is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

Alternative Solution:
Evaluate \( \tan^{-1}\left[ 2\cos\left(2\sin^{-1}\frac{1}{2}\right) \right] \).
We know \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \).
Then \( 2\sin^{-1}\frac{1}{2} = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \).
\( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
Expression becomes: \( \tan^{-1}\left[ 2 \times \frac{1}{2} \right] = \tan^{-1}(1) = \frac{\pi}{4} \).

Solution:
\( X = -(2A + B) \).
\( 2A = 2 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} \).
\( 2A + B = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} \).
\( X = -\begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix} \).

Solution:
\( x = a(\theta - \sin\theta) \implies \frac{dx}{d\theta} = a(1 - \cos\theta) \).
\( y = a(1 + \cos\theta) \implies \frac{dy}{d\theta} = -a\sin\theta \).
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-a\sin\theta}{a(1 - \cos\theta)} = \frac{-2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = -\cot(\theta/2) \).
At \( \theta = \frac{\pi}{2} \):
\( \frac{dy}{dx} = -\cot\left(\frac{\pi}{4}\right) = -1 \).

Solution:
Let \( \vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \).
\( |\vec{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
Unit vector in direction of \( \vec{a} \) is \( \hat{a} = \frac{2\hat{i} - \hat{j} + 2\hat{k}}{3} \).
Required vector of magnitude 6 is \( \pm 6\hat{a} = \pm 6 \left(\frac{2\hat{i} - \hat{j} + 2\hat{k}}{3}\right) = \pm (4\hat{i} - 2\hat{j} + 4\hat{k}) \).

Solution:
Direction ratios of line \( AB \): \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (3 - 1, 4 - (-1), -2 - 2) = (2, 5, -4) \).
Vector equation: \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 5\hat{j} - 4\hat{k}) \).
Cartesian equation: \( \frac{x - 1}{2} = \frac{y + 1}{5} = \frac{z - 2}{-4} \).

Alternative Solution:
Let \( E_1 \) be the event that the first card is an ace, and \( E_2 \) be the event that the second card is an ace.
Total cards = 52, Number of aces = 4.
\( P(E_1) = \frac{4}{52} = \frac{1}{13} \).
After drawing one ace without replacement, remaining cards = 51, remaining aces = 3.
\( P(E_2|E_1) = \frac{3}{51} = \frac{1}{17} \).
By multiplication theorem, \( P(E_1 \cap E_2) = P(E_1) \cdot P(E_2|E_1) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221} \).

Solution:
Given \( R = \{(a, b) : 3 \text{ divides } (a - b)\} \) on \( \mathbb{Z} \).
1. Reflexive: For all \( a \in \mathbb{Z} \), \( a - a = 0 = 3 \times 0 \), which is divisible by 3. Thus, \( (a, a) \in R \).
2. Symmetric: Let \( (a, b) \in R \implies a - b = 3k \) for some \( k \in \mathbb{Z} \). Then \( b - a = -(a - b) = 3(-k) \), which is divisible by 3. Thus, \( (b, a) \in R \).
3. Transitive: Let \( (a, b) \in R \) and \( (b, c) \in R \implies a - b = 3k \) and \( b - c = 3m \).
Then \( a - c = (a - b) + (b - c) = 3k + 3m = 3(k + m) \), which is divisible by 3. Thus, \( (a, c) \in R \).
Hence, \( R \) is an equivalence relation.
Equivalence class \( [0] \):
\( [0] = \{x \in \mathbb{Z} : (x, 0) \in R\} = \{x \in \mathbb{Z} : 3 \text{ divides } (x - 0)\} = \{ \dots, -6, -3, 0, 3, 6, \dots \} = \{3k : k \in \mathbb{Z}\} \).

Solution:
Let \( y = u + v \implies \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).
• For \( u = (\sin x)^x \): Take log on both sides: \( \log u = x \log(\sin x) \).
Differentiating w.r.t. \( x \):
\( \frac{1}{u}\frac{du}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x}\cos x = \log(\sin x) + x\cot x \)
\( \implies \frac{du}{dx} = (\sin x)^x [\log(\sin x) + x\cot x] \).

• For \( v = x^{\sin x} \): Take log on both sides: \( \log v = \sin x \log x \).
Differentiating w.r.t. \( x \):
\( \frac{1}{v}\frac{dv}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \cos x \log x + \frac{\sin x}{x} \)
\( \implies \frac{dv}{dx} = x^{\sin x} \left[\cos x \log x + \frac{\sin x}{x}\right] \).

Therefore, \( \frac{dy}{dx} = (\sin x)^x [x\cot x + \log(\sin x)] + x^{\sin x} \left[\frac{\sin x}{x} + \cos x \log x\right] \).

Alternative Solution:
Given \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \implies x\sqrt{1+y} = -y\sqrt{1+x} \).
Squaring both sides:
\( x^2(1 + y) = y^2(1 + x) \implies x^2 + x^2 y = y^2 + xy^2 \)
\( x^2 - y^2 + x^2 y - xy^2 = 0 \implies (x - y)(x + y) + xy(x - y) = 0 \)
\( (x - y)[x + y + xy] = 0 \).
Since \( x \ne y \) (otherwise \( 2x\sqrt{1+x}=0 \implies x=0 \) only), we have:
\( x + y(1 + x) = 0 \implies y = -\frac{x}{1 + x} \).
Differentiating w.r.t. \( x \):
\( \frac{dy}{dx} = -\frac{(1 + x)(1) - x(1)}{(1 + x)^2} = -\frac{1}{(1 + x)^2} \). (Hence proved).

Solution:
\( f'(x) = \cos x - \sin x \).
For critical points, \( f'(x) = 0 \implies \cos x = \sin x \implies \tan x = 1 \).
In \( [0, 2\pi] \), \( x = \frac{\pi}{4}, \frac{5\pi}{4} \).
The intervals are: \( \left[0, \frac{\pi}{4}\right), \left(\frac{\pi}{4}, \frac{5\pi}{4}\right), \left(\frac{5\pi}{4}, 2\pi\right] \).

• In \( \left(0, \frac{\pi}{4}\right) \): For \( x = 0 \), \( f'(0) = 1 > 0 \implies f'(x) > 0 \).
• In \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \): For \( x = \frac{\pi}{2} \), \( f'(\pi/2) = 0 - 1 = -1 < 0 \implies f'(x) < 0 \).
• In \( \left(\frac{5\pi}{4}, 2\pi\right) \): For \( x = \frac{3\pi}{2} \), \( f'(3\pi/2) = 0 - (-1) = 1 > 0 \implies f'(x) > 0 \).

(a) Strictly increasing in: \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \).
(b) Strictly decreasing in: \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).

Solution:
\( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \).
\( A^2 = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 1 & -2 - 2 \\ -2 - 2 & 1 + 4 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \).
\( -4A = \begin{bmatrix} -8 & 4 \\ 4 & -8 \end{bmatrix} \), \( 3I = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \).
\( A^2 - 4A + 3I = \begin{bmatrix} 5-8+3 & -4+4+0 \\ -4+4+0 & 5-8+3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \). Verified.

To find \( A^{-1} \), multiply the equation by \( A^{-1} \):
\( A - 4I + 3A^{-1} = O \implies 3A^{-1} = 4I - A \)
\( 3A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
\( A^{-1} = \frac{1}{3}\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \).

Solution:
Line 1: \( x + 2y = 10 \). Points: \( (10, 0), (0, 5) \). Region: Above the line (\( \ge \)).
Line 2: \( 3x + 4y = 24 \). Points: \( (8, 0), (0, 6) \). Region: Below the line (\( \le \)).
Non-negative constraints: \( x \ge 0, y \ge 0 \) (first quadrant).

Intersection of \( x + 2y = 10 \) and \( 3x + 4y = 24 \):
Multiply first equation by 2: \( 2x + 4y = 20 \).
Subtracting: \( (3x + 4y) - (2x + 4y) = 24 - 20 \implies x = 4 \).
Then \( 4 + 2y = 10 \implies 2y = 6 \implies y = 3 \). Point is \( (4, 3) \).

Corner points of the feasible region are: \( A(0, 5), B(0, 6), C(4, 3) \).
Evaluating \( Z = 200x + 500y \):
• At \( A(0, 5) \): \( Z = 200(0) + 500(5) = 2500 \).
• At \( B(0, 6) \): \( Z = 200(0) + 500(6) = 3000 \).
• At \( C(4, 3) \): \( Z = 200(4) + 500(3) = 800 + 1500 = 2300 \).

Minimum value of \( Z \) is 2300 at \( x = 4, y = 3 \).

Solution:
Let \( E_1 \): Bag 1 is chosen, \( E_2 \): Bag 2 is chosen.
Let \( A \): The drawn ball is red.
\( P(E_1) = \frac{1}{2}, \quad P(E_2) = \frac{1}{2} \).
Bag 1 has 4 red, 4 black (total 8) \( \implies P(A|E_1) = \frac{4}{8} = \frac{1}{2} \).
Bag 2 has 2 red, 6 black (total 8) \( \implies P(A|E_2) = \frac{2}{8} = \frac{1}{4} \).

By Bayes' Theorem:
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \] \[ P(E_1|A) = \frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{4}\right)} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3} \].

Solution:
The system is \( AX = B \) where:
\( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \).

\( |A| = 1(12 - 5) - (-1)(9 - (-10)) + 2(-3 - 8) = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4 \ne 0 \).
Since \( |A| \ne 0 \), unique solution exists: \( X = A^{-1}B \).

Cofactors of \( A \):
\( A_{11} = 7, \quad A_{12} = -19, \quad A_{13} = -11 \)
\( A_{21} = -(-3 - (-2)) = 1, \quad A_{22} = (3 - 4) = -1, \quad A_{23} = -(-1 - (-2)) = -1 \)
\( A_{31} = (5 - 8) = -3, \quad A_{32} = -(-5 - 6) = 11, \quad A_{33} = (4 - (-3)) = 7 \).

\( \text{adj } A = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \implies A^{-1} = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \).

\( X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49 - 5 - 36 \\ -133 + 5 + 132 \\ -77 + 5 + 84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \).

Answer: \( x = 2, \quad y = 1, \quad z = 3 \).

Alternative Solution:
For \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \):
\( |A| = 1(1 + 3) - (-1)(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10 \ne 0 \).
Matrix of cofactors: \( C = \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix} \implies \text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \).
\( A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \).
The given system can be written as \( A' X = C \implies X = (A')^{-1}C = (A^{-1})' C \).
Computing gives: \( x = 1, \quad y = 1, \quad z = 1 \).

Solution:
Dimensions of rectangular sheet: \( 24\text{ cm} \times 9\text{ cm} \).
Square of side \( x \) is cut from each corner.
Dimensions of the open box: Length \( l = 24 - 2x \), Breadth \( b = 9 - 2x \), Height \( h = x \).
Here \( x > 0 \) and \( 9 - 2x > 0 \implies 0 < x < 4.5 \).

Volume \( V(x) = x(24 - 2x)(9 - 2x) = x(216 - 66x + 4x^2) = 4x^3 - 66x^2 + 216x \).
Differentiating w.r.t. \( x \):
\( \frac{dV}{dx} = 12x^2 - 132x + 216 = 12(x^2 - 11x + 18) = 12(x - 2)(x - 9) \).

For maximum volume, \( \frac{dV}{dx} = 0 \implies x = 2 \) or \( x = 9 \).
Since \( x < 4.5 \), we have \( x = 2\text{ cm} \).

Second derivative test:
\( \frac{d^2V}{dx^2} = 24x - 132 \).
At \( x = 2 \): \( \frac{d^2V}{dx^2} = 24(2) - 132 = 48 - 132 = -84 < 0 \).
Hence, volume is maximum at \( x = 2\text{ cm} \).

Dimensions of the box:
Length \( = 24 - 2(2) = 20\text{ cm} \)
Breadth \( = 9 - 2(2) = 5\text{ cm} \)
Height \( = 2\text{ cm} \).
Maximum volume: \( V = 20 \times 5 \times 2 = 200\text{ cm}^3 \).

Alternative Solution:
Given surface area \( S = \pi r l + \pi r^2 \) is constant \( \implies l = \frac{S - \pi r^2}{\pi r} \).
Volume \( V = \frac{1}{3}\pi r^2 h \implies V^2 = \frac{1}{9}\pi^2 r^4 (l^2 - r^2) \).
Substituting \( l \) and maximizing \( Z = V^2 \) with respect to \( r \):
Setting \( \frac{dZ}{dr} = 0 \) yields \( S = 4\pi r^2 \).
Then \( \pi r l + \pi r^2 = 4\pi r^2 \implies \pi r l = 3\pi r^2 \implies l = 3r \).
Let semi-vertical angle be \( \alpha \). Then \( \sin\alpha = \frac{r}{l} = \frac{r}{3r} = \frac{1}{3} \implies \alpha = \sin^{-1}\left(\frac{1}{3}\right) \). (Hence proved).

Solution:
Line 1: \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) where \( \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k} \).
Line 2: \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \) where \( \vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k} \) and \( \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k} \).

1. Vector \( \vec{a}_2 - \vec{a}_1 = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \).

2. Cross product \( \vec{b}_1 \times \vec{b}_2 \):
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k} \] \( |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19} \).

3. Dot product \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\( (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = 3(-9) + 3(3) + 3(9) = -27 + 9 + 27 = 9 \).

4. Shortest Distance \( d \):
\[ d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}\text{ units} \] Since \( d \ne 0 \), the two lines do not intersect.

Solution:
Given \( y = (x + \sqrt{x^2 + 1})^m \).
Differentiating w.r.t. \( x \):
\[ \frac{dy}{dx} = m(x + \sqrt{x^2 + 1})^{m-1} \cdot \left(1 + \frac{2x}{2\sqrt{x^2 + 1}}\right) = m(x + \sqrt{x^2 + 1})^{m-1} \cdot \left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}\right) \] \[ \frac{dy}{dx} = \frac{m(x + \sqrt{x^2 + 1})^m}{\sqrt{x^2 + 1}} = \frac{my}{\sqrt{x^2 + 1}} \] Cross multiplying:
\[ \sqrt{x^2 + 1} \frac{dy}{dx} = my \] Squaring both sides:
\[ (x^2 + 1)\left(\frac{dy}{dx}\right)^2 = m^2 y^2 \] Differentiating both sides w.r.t. \( x \):
\[ (2x)\left(\frac{dy}{dx}\right)^2 + (x^2 + 1) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} = m^2 \cdot 2y \frac{dy}{dx} \] Dividing throughout by \( 2\frac{dy}{dx} \) (since \( \frac{dy}{dx} \ne 0 \)):
\[ x\frac{dy}{dx} + (x^2 + 1)\frac{d^2y}{dx^2} = m^2 y \] \[ (x^2 + 1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m^2 y = 0 \] (Hence proved).

(i) [1 Mark]
Fuel cost per hour \( F \propto v^2 \implies F = k v^2 \).
At \( v = 40 \), \( F = 800 \implies 800 = k(40)^2 = 1600k \implies k = \frac{800}{1600} = \frac{1}{2} \).
Thus, fuel cost per hour \( = \frac{1}{2}v^2 \).
Fixed charges per hour \( = 3200 \). Total cost per hour \( = \frac{1}{2}v^2 + 3200 \).
Time taken for \( 400\text{ km} \) journey is \( t = \frac{400}{v}\text{ hours} \).
Total Cost \( C(v) = t \times (\text{Cost per hour}) = \frac{400}{v}\left(\frac{1}{2}v^2 + 3200\right) = 200v + \frac{1280000}{v} \).

(ii) [2 Marks]
\( \frac{dC}{dv} = 200 - \frac{1280000}{v^2} \).
For minimum cost, \( \frac{dC}{dv} = 0 \implies 200 = \frac{1280000}{v^2} \implies v^2 = \frac{1280000}{200} = 6400 \implies v = 80\text{ km/h} \).
Critical speed is \( 80\text{ km/h} \).

\( \frac{d^2C}{dv^2} = \frac{d}{dv}\left(200 - 1280000v^{-2}\right) = \frac{2560000}{v^3} \).
At \( v = 80 \), \( \frac{d^2C}{dv^2} = \frac{2560000}{512000} = 5 > 0 \).
Since \( \frac{d^2C}{dv^2} > 0 \), the total cost is strictly minimized at \( v = 80\text{ km/h} \).

(iii) [1 Mark]
Minimum total cost at \( v = 80\text{ km/h} \):
\( C(80) = 200(80) + \frac{1280000}{80} = 16000 + 16000 = 32000 \), i.e., ₹32,000.

(i) [1 Mark]
Coordinates: \( P(2, 3, 4) \) and \( Q(4, 1, 2) \).
Position vector \( \vec{PQ} = (4 - 2)\hat{i} + (1 - 3)\hat{j} + (2 - 4)\hat{k} = 2\hat{i} - 2\hat{j} - 2\hat{k} \).

(ii) [1 Mark]
Distance between drones is \( |\vec{PQ}| = \sqrt{2^2 + (-2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}\text{ km} \approx 3.464\text{ km} \).

(iii) [2 Marks]
\( \vec{d}_1 = \hat{i} + 2\hat{j} + 2\hat{k} \implies |\vec{d}_1| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
\( \vec{d}_2 = 2\hat{i} - \hat{j} + 2\hat{k} \implies |\vec{d}_2| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
\( \vec{d}_1 \cdot \vec{d}_2 = (1)(2) + (2)(-1) + (2)(2) = 2 - 2 + 4 = 4 \).
\( \cos\theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1||\vec{d}_2|} = \frac{4}{3 \times 3} = \frac{4}{9} \).

Cross product \( \vec{d}_1 \times \vec{d}_2 \):
\[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(4 - (-2)) - \hat{j}(2 - 4) + \hat{k}(-1 - 4) = 6\hat{i} + 2\hat{j} - 5\hat{k} \] Magnitude \( = \sqrt{6^2 + 2^2 + (-5)^2} = \sqrt{36 + 4 + 25} = \sqrt{65} \).
Unit vector perpendicular to both is \( \hat{n} = \pm \frac{6\hat{i} + 2\hat{j} - 5\hat{k}}{\sqrt{65}} \).

Events defined:
\( E_1 \): Person has the disease.
\( E_2 \): Person is disease-free.
\( A \): Test result is positive.
Given:
\( P(E_1) = 0.2\% = \frac{0.2}{100} = 0.002 \).
\( P(A|E_1) = 99\% = 0.99 \).
\( P(A|E_2) = 0.5\% = 0.005 \).

(i) [1 Mark]
Probability person is disease-free: \( P(E_2) = 1 - P(E_1) = 1 - 0.002 = 0.998 \) (or \( 99.8\% \)).

(ii) [1 Mark]
Total probability that a person tests positive:
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = (0.002)(0.99) + (0.998)(0.005) = 0.00198 + 0.00499 = 0.00697 \].

(iii) [2 Marks]
Probability that a person testing positive actually has the disease (Bayes' Theorem):
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)} = \frac{0.00198}{0.00697} = \frac{198}{697} \approx 0.2841 \text{ (or } 28.41\% \text{)} \].

Probability that a person testing negative is genuinely disease-free:
\( P(A'|E_1) = 1 - 0.99 = 0.01 \).
\( P(A'|E_2) = 1 - 0.005 = 0.995 \).
\( P(A') = P(E_1)P(A'|E_1) + P(E_2)P(A'|E_2) = (0.002)(0.01) + (0.998)(0.995) = 0.00002 + 0.99301 = 0.99303 \).
\[ P(E_2|A') = \frac{P(E_2)P(A'|E_2)}{P(A')} = \frac{0.99301}{0.99303} \approx 0.99998 \text{ (or } 99.998\% \text{)} \].