VARDAAN LEARNING INSTITUTE

Explanation:
Any equivalence relation must contain the identity elements: \( (1, 1), (2, 2), (3, 3) \).
If it contains \( (1, 2) \), symmetry requires \( (2, 1) \).
• Smallest relation: \( R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} \), which is transitive and reflexive.
• If we add any other element such as \( (2, 3) \), transitivity forces all 9 elements of \( A \times A \): \( R_2 = A \times A \).
Hence, there are exactly 2 equivalence relations containing \( (1, 2) \).

Explanation:
• One-one: For \( x_1, x_2 \ge 0 \), \( x_1^2 = x_2^2 \implies x_1 = x_2 \) (since non-negative). Thus one-one.
• Onto: For any \( y \in [0, \infty) \), \( x = \sqrt{y} \ge 0 \) exists in the domain such that \( f(\sqrt{y}) = y \). Thus onto.
Therefore, \( f \) is one-one and onto (bijective).

Explanation:
\( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \).
\( \sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
Therefore, \( \tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \).

Explanation:
Product \( AB \) has order \( (2 \times 3) \times (3 \times 2) = 2 \times 2 \).
Transpose \( (AB)' \) has order \( 2 \times 2 \).

Explanation:
\( A + A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} + \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
\( 2\cos\alpha = 1 \implies \cos\alpha = \frac{1}{2} \implies \alpha = \frac{\pi}{3} \).

Explanation:
Area \( = \pm 35 = \frac{1}{2}\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} \).
Expanding: \( 2(4 - 4) - (-6)(5 - k) + 1(20 - 4k) = 0 + 30 - 6k + 20 - 4k = 50 - 10k \).
\( \frac{1}{2}(50 - 10k) = \pm 35 \implies 50 - 10k = \pm 70 \).
• \( 50 - 10k = 70 \implies -10k = 20 \implies k = -2 \).
• \( 50 - 10k = -70 \implies -10k = -120 \implies k = 12 \).
Thus, \( k = 12, -2 \).

Explanation:
Taking determinant on both sides: \( |A^2| = |3A| \implies |A|^2 = 3^3 |A| = 27|A| \).
Since \( A \) is non-singular, \( |A| \ne 0 \). Dividing by \( |A| \): \( |A| = 27 \).

Explanation:
The sum of two continuous functions is always continuous everywhere on \( \mathbb{R} \).
However, it has sharp corners and is not differentiable at the transition points \( x = 0 \) and \( x = 1 \).
Hence, it is continuous everywhere.

Explanation:
\( y = e^{\log(x^3)} = x^3 \).
\( \frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2 \).

Explanation:
Area \( A = \frac{\sqrt{3}}{4}x^2 \).
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \frac{dx}{dt} = \frac{\sqrt{3}}{2}x \frac{dx}{dt} \).
Given \( \frac{dx}{dt} = 2\text{ cm/s} \) and \( x = 10\text{ cm} \):
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}(10)(2) = 10\sqrt{3}\text{ cm}^2/\text{s} \).

Explanation:
\( \frac{dy}{dx} = 4x + 3\cos x \).
At \( x = 0 \): \( \left.\frac{dy}{dx}\right|_{x=0} = 4(0) + 3\cos(0) = 3 \).
Slope of the normal \( = -\frac{1}{\text{slope of tangent}} = -\frac{1}{3} \).

Explanation:
\( |\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta \implies \tan\theta = 1 \implies \theta = \frac{\pi}{4} \).

Explanation:
Projection \( = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2(1) + (-1)(2) + 1(2)}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{2 - 2 + 2}{\sqrt{9}} = \frac{2}{3} \).

Explanation:
Magnitude \( = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3 \).
Direction cosines: \( \left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) \).

Explanation:
Distance of point \( (x, y, z) \) from \( x \)-axis is \( \sqrt{y^2 + z^2} \).
Here, distance \( = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Explanation:
Standard theorem for unbounded regions in Linear Programming: \( M \) is the maximum value of \( Z \) if and only if the open half plane determined by \( ax + by > M \) has no common point with the feasible region.

Explanation:
\( P(B|A) = \frac{P(A \cap B)}{P(A)} \implies P(A \cap B) = P(B|A) \cdot P(A) = 0.6 \times 0.4 = 0.24 \).
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.8 - 0.24 = 1.2 - 0.24 = 0.96 \).

Explanation:
Since \( A \) and \( B \) are independent, \( P(A \cap B) = P(A) \cdot P(B) \).
\( \frac{1}{6} = \frac{1}{2} \cdot P(B) \implies P(B) = \frac{2}{6} = \frac{1}{3} \).

Explanation:
Given \( A' = A \). Using Reason \( (A^{-1})' = (A')^{-1} \), we have \( (A^{-1})' = (A)^{-1} = A^{-1} \), which proves that \( A^{-1} \) is symmetric. Both A and R are true, and R correctly explains A.

Explanation:
\( \vec{b}_1 = \hat{i} + \hat{j} + \hat{k} \implies |\vec{b}_1| = \sqrt{3} \).
\( \vec{b}_2 = \hat{i} - \hat{j} + \hat{k} \implies |\vec{b}_2| = \sqrt{3} \).
\( \vec{b}_1 \cdot \vec{b}_2 = 1 - 1 + 1 = 1 \).
\( \cos\theta = \frac{1}{\sqrt{3}\sqrt{3}} = \frac{1}{3} \implies \theta = \cos^{-1}\left(\frac{1}{3}\right) \). Both A and R are true, and R is the correct formula and explanation.

Solution:
• \( \frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). We write \( \tan\left(\frac{3\pi}{4}\right) = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = \tan\left(-\frac{\pi}{4}\right) \).
Since \( -\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), \( \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = -\frac{\pi}{4} \).
• \( \frac{2\pi}{3} \in [0, \pi] \implies \cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{2\pi}{3} \).
Total value \( = -\frac{\pi}{4} + \frac{2\pi}{3} = \frac{-3\pi + 8\pi}{12} = \frac{5\pi}{12} \).

Alternative Solution:
\( \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) \).
Dividing numerator and denominator by \( \cos x \):
\( = \tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right) = \tan^{-1}\left(\frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4)\tan x}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right) = \frac{\pi}{4} - x \).

Solution:
\( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \).
\( A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \).
\( A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).

Multiply by \( A^{-1} \):
\( A - 5I + 7A^{-1} = O \implies 7A^{-1} = 5I - A = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \).
\( A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \).

Solution:
LHL \( = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx + 1) = 5k + 1 \).
RHL \( = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5) = 3(5) - 5 = 15 - 5 = 10 \).
Value of function \( f(5) = 5k + 1 \).
For continuity, \( \text{LHL} = \text{RHL} \implies 5k + 1 = 10 \implies 5k = 9 \implies k = \frac{9}{5} \).

Solution:
Area \( = |\vec{a} \times \vec{b}| \).
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2)) = 20\hat{i} + 5\hat{j} - 5\hat{k} \] \( |\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}\text{ sq units} \).

Solution:
Point \( \vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} \), Parallel vector \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \).
Vector equation: \( \vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) \).
Cartesian equation: \( \frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1} \).

Alternative Solution:
Sample space total outcomes \( = 6^3 = 216 \).
Event \( B \) ("6 on first and 5 on second"): \( B = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\} \implies n(B) = 6 \).
Event \( A \) ("4 on third throw"): Outcomes common to \( A \) and \( B \) is \( A \cap B = \{(6, 5, 4)\} \implies n(A \cap B) = 1 \).
\( P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{1}{6} \).

Solution:
1. Reflexive: Take \( a = \frac{1}{2} \). Here \( a^3 = \frac{1}{8} \). Since \( \frac{1}{2} \le \frac{1}{8} \) is false, \( (1/2, 1/2) \notin R \). Hence, not reflexive.
2. Symmetric: Take \( a = 1, b = 2 \). Here \( 1 \le 2^3 = 8 \) is true \( \implies (1, 2) \in R \). But \( 2 \le 1^3 = 1 \) is false \( \implies (2, 1) \notin R \). Hence, not symmetric.
3. Transitive: Take \( a = 7, b = 2, c = 1.5 \).
\( 7 \le 2^3 = 8 \implies (7, 2) \in R \).
\( 2 \le (1.5)^3 = 3.375 \implies (2, 1.5) \in R \).
However, \( 7 \le (1.5)^3 = 3.375 \) is false \( \implies (7, 1.5) \notin R \). Hence, not transitive.
Conclusion: \( R \) is neither reflexive, nor symmetric, nor transitive.

Solution:
Let \( y = u + v \implies \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).
• \( u = x^x \implies \log u = x\log x \implies \frac{1}{u}\frac{du}{dx} = 1 + \log x \implies \frac{du}{dx} = x^x(1 + \log x) \).
• \( v = (\sin x)^{\cos x} \implies \log v = \cos x \log(\sin x) \).
\( \frac{1}{v}\frac{dv}{dx} = -\sin x \log(\sin x) + \cos x \cdot \frac{\cos x}{\sin x} = \cos x \cot x - \sin x \log(\sin x) \).
\( \frac{dv}{dx} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log(\sin x)] \).

Hence, \( \frac{dy}{dx} = x^x(1 + \log x) + (\sin x)^{\cos x}[\cos x\cot x - \sin x\log(\sin x)] \).

Alternative Solution:
\( \frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t \).
\( \frac{dy}{dt} = a(\cos t - \cos t + t\sin t) = at\sin t \).
\( \frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t \).
\( \frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at\cos t} = \frac{\sec^3 t}{at} \).
At \( t = \frac{\pi}{4} \):
\( \frac{d^2y}{dx^2} = \frac{(\sqrt{2})^3}{a(\pi/4)} = \frac{2\sqrt{2}}{\frac{\pi a}{4}} = \frac{8\sqrt{2}}{\pi a} \).

Solution:
Let \( x \) be the distance of foot from the wall and \( y \) be the height on the wall.
Ladder length: \( x^2 + y^2 = 5^2 = 25 \).
When \( x = 4\text{ m} \): \( 4^2 + y^2 = 25 \implies y^2 = 9 \implies y = 3\text{ m} \).
Given \( \frac{dx}{dt} = 2\text{ cm/s} = 0.02\text{ m/s} \).
Differentiating \( x^2 + y^2 = 25 \) w.r.t. \( t \):
\( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \).
Substituting values: \( \frac{dy}{dt} = -\frac{4}{3}(2) = -\frac{8}{3}\text{ cm/s} \).
The height on the wall is decreasing at the rate of \( \frac{8}{3}\text{ cm/s} \).

Solution:
\( A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \).
\( |A| = 1(3 - 0) - (-1)(2 - (-10)) + 2(0 - (-6)) = 3 + 12 + 12 = 27 \).
Cofactors of \( A \):
\( A_{11} = 3, \quad A_{12} = -12, \quad A_{13} = 6 \)
\( A_{21} = 1, \quad A_{22} = 5, \quad A_{23} = 2 \)
\( A_{31} = -11, \quad A_{32} = -1, \quad A_{33} = 5 \).
\( \text{adj } A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix} \).
Computing \( A(\text{adj } A) \):
\( = \begin{bmatrix} 1(3)+(-1)(-12)+2(6) & 1(1)+(-1)(5)+2(2) & 1(-11)+(-1)(-1)+2(5) \\ 2(3)+3(-12)+5(6) & 2(1)+3(5)+5(2) & 2(-11)+3(-1)+5(5) \\ -2(3)+0+1(6) & -2(1)+0+1(2) & -2(-11)+0+1(5) \end{bmatrix} = \begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix} = 27I = |A|I \). Verified.

Solution:
Line 1: \( 3x + 5y = 15 \implies (0, 3), (5, 0) \).
Line 2: \( 5x + 2y = 10 \implies (0, 5), (2, 0) \).
Intersection: Multiply (1) by 2 and (2) by 5:
\( 6x + 10y = 30 \)
\( 25x + 10y = 50 \)
Subtracting: \( 19x = 20 \implies x = \frac{20}{19} \).
Then \( y = \frac{15 - 3(20/19)}{5} = 3 - \frac{12}{19} = \frac{45}{19} \).

Corner points: \( O(0, 0), A(2, 0), B\left(\frac{20}{19}, \frac{45}{19}\right), C(0, 3) \).
Evaluating \( Z \):
• At \( O(0, 0): Z = 0 \)
• At \( A(2, 0): Z = 5(2) + 0 = 10 \)
• At \( B\left(\frac{20}{19}, \frac{45}{19}\right): Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100 + 135}{19} = \frac{235}{19} \approx 12.37 \)
• At \( C(0, 3): Z = 0 + 3(3) = 9 \).

Maximum value of \( Z = \frac{235}{19} \) at \( x = \frac{20}{19}, y = \frac{45}{19} \).

Solution:
Total insured drivers \( = 2000 + 4000 + 6000 = 12000 \).
\( P(E_1) \) (scooter) \( = \frac{2000}{12000} = \frac{1}{6} \).
\( P(E_2) \) (car) \( = \frac{4000}{12000} = \frac{1}{3} \).
\( P(E_3) \) (truck) \( = \frac{6000}{12000} = \frac{1}{2} \).
Probabilities of accident \( A \):
\( P(A|E_1) = 0.01 = \frac{1}{100}, \quad P(A|E_2) = 0.03 = \frac{3}{100}, \quad P(A|E_3) = 0.15 = \frac{15}{100} \).

By Bayes' Theorem:
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)} \] \[ = \frac{\frac{1}{6} \times \frac{1}{100}}{\left(\frac{1}{6} \times \frac{1}{100}\right) + \left(\frac{1}{3} \times \frac{3}{100}\right) + \left(\frac{1}{2} \times \frac{15}{100}\right)} = \frac{\frac{1}{6}}{\frac{1}{6} + 1 + \frac{15}{2}} = \frac{\frac{1}{6}}{\frac{1 + 6 + 45}{6}} = \frac{1}{52} \].

Solution:
\( A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} \).

\( |A| = 2(4 - (-1)) - 3(-2 - 3) + 3(-1 - (-6)) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \ne 0 \).
Unique solution \( X = A^{-1}B \).

Cofactor matrix \( C \):
\( A_{11} = 5, \quad A_{12} = 5, \quad A_{13} = 5 \)
\( A_{21} = -(-6 - (-3)) = 3, \quad A_{22} = -4 - 9 = -13, \quad A_{23} = -(-2 - 9) = 11 \)
\( A_{31} = 3 - (-6) = 9, \quad A_{32} = -(2 - 3) = 1, \quad A_{33} = -4 - 3 = -7 \).

\( \text{adj } A = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \implies A^{-1} = \frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \).

\( X = \frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \).

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

Alternative Solution:
Compute \( AB \):
\( AB = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} = \begin{bmatrix} 2+4+0 & 2-2+0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I \).
Therefore, \( A^{-1} = \frac{1}{6}B \).
The given system is \( AX = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \implies X = A^{-1}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} = \frac{1}{6}B\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \).
\( X = \frac{1}{6}\begin{bmatrix} 2(3) + 2(17) - 4(7) \\ -4(3) + 2(17) - 4(7) \\ 2(3) - 1(17) + 5(7) \end{bmatrix} = \frac{1}{6}\begin{bmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{bmatrix} = \frac{1}{6}\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \).
Answer: \( x = 2, \quad y = -1, \quad z = 4 \).

Solution:
Total wire length \( = 28\text{ m} \).
Let length of wire for square be \( x \). Then length of wire for circle is \( 28 - x \).
Side of square \( s = \frac{x}{4} \implies \text{Area of square } A_1 = s^2 = \frac{x^2}{16} \).
Circumference of circle \( 2\pi r = 28 - x \implies r = \frac{28 - x}{2\pi} \implies \text{Area of circle } A_2 = \pi r^2 = \frac{(28 - x)^2}{4\pi} \).

Total Area \( A(x) = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi} \).
\( \frac{dA}{dx} = \frac{2x}{16} - \frac{2(28 - x)}{4\pi} = \frac{x}{8} - \frac{28 - x}{2\pi} \).
For minimum area, \( \frac{dA}{dx} = 0 \implies \frac{x}{8} = \frac{28 - x}{2\pi} \implies \pi x = 4(28 - x) = 112 - 4x \).
\( (\pi + 4)x = 112 \implies x = \frac{112}{\pi + 4}\text{ m} \).

Second derivative: \( \frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0 \). Hence area is strictly minimum.
Lengths of the two pieces:
• Piece for square: \( x = \frac{112}{\pi + 4}\text{ m} \).
• Piece for circle: \( 28 - x = 28 - \frac{112}{\pi + 4} = \frac{28\pi}{\pi + 4}\text{ m} \).

Alternative Solution:
Given volume \( V = \pi r^2 h \) is constant \( \implies h = \frac{V}{\pi r^2} \).
Total surface area \( S = 2\pi r h + 2\pi r^2 = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2 \).
\( \frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r = 0 \implies 4\pi r = \frac{2V}{r^2} \implies 2\pi r^3 = V \).
Since \( V = \pi r^2 h \), we have \( 2\pi r^3 = \pi r^2 h \implies h = 2r = \text{diameter} \).
\( \frac{d^2S}{dr^2} = \frac{4V}{r^3} + 4\pi > 0 \implies S \) is minimum when \( h = 2r \). (Hence proved).

Solution:
\( \vec{a}_1 = \hat{i} + \hat{j} \), \( \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \).
\( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k} \), \( \vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k} \).
1. \( \vec{a}_2 - \vec{a}_1 = (2 - 1)\hat{i} + (1 - 1)\hat{j} + (-1 - 0)\hat{k} = \hat{i} - \hat{k} \).
2. Cross product \( \vec{b}_1 \times \vec{b}_2 \):
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2 - (-5)) - \hat{j}(4 - 3) + \hat{k}(-10 - (-3)) = 3\hat{i} - \hat{j} - 7\hat{k} \] \( |\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \).
3. Dot product: \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 1(3) + 0(-1) + (-1)(-7) = 3 + 7 = 10 \).
4. Shortest distance: \( d = \left|\frac{10}{\sqrt{59}}\right| = \frac{10}{\sqrt{59}}\text{ units} \).

Solution:
Given \( y = (\tan^{-1} x)^2 \).
Differentiating w.r.t. \( x \):
\[ \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1 + x^2} \] \[ (1 + x^2)\frac{dy}{dx} = 2\tan^{-1} x \] Differentiating again w.r.t. \( x \):
\[ (1 + x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} \] Multiplying throughout by \( (1 + x^2) \):
\[ (1 + x^2)^2 \frac{d^2y}{dx^2} + 2x(1 + x^2)\frac{dy}{dx} = 2 \] (Hence proved).

(i) [1 Mark]
Given volume \( V = \pi r^2 h = 128\pi \implies r^2 h = 128 \implies h = \frac{128}{r^2} \).

(ii) [1 Mark]
Total surface area \( S = 2\pi r h + 2\pi r^2 = 2\pi r \left(\frac{128}{r^2}\right) + 2\pi r^2 = \frac{256\pi}{r} + 2\pi r^2 \).
\( \frac{dS}{dr} = -\frac{256\pi}{r^2} + 4\pi r \).

(iii) [2 Marks]
For minimum surface area, \( \frac{dS}{dr} = 0 \implies 4\pi r = \frac{256\pi}{r^2} \implies r^3 = 64 \implies r = 4\text{ cm} \).
Second derivative: \( \frac{d^2S}{dr^2} = \frac{512\pi}{r^3} + 4\pi \).
At \( r = 4 \): \( \frac{d^2S}{dr^2} = \frac{512\pi}{64} + 4\pi = 8\pi + 4\pi = 12\pi > 0 \).
Hence, surface area is strictly minimum at \( r = 4\text{ cm} \).

At \( r = 4\text{ cm} \), height is \( h = \frac{128}{4^2} = \frac{128}{16} = 8\text{ cm} = 2(4) = 2r = \text{base diameter} \).
Minimum surface area \( S = \frac{256\pi}{4} + 2\pi(4^2) = 64\pi + 32\pi = 96\pi\text{ cm}^2 \approx 301.59\text{ cm}^2 \).

(i) [1 Mark]
\( T(1, 2, 3) \) and \( P(4, 6, 8) \).
\( \vec{TP} = (4 - 1)\hat{i} + (6 - 2)\hat{j} + (8 - 3)\hat{k} = 3\hat{i} + 4\hat{j} + 5\hat{k} \).

(ii) [1 Mark]
Distance \( |\vec{TP}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}\text{ meters} \approx 7.07\text{ m} \).

(iii) [2 Marks]
\( \vec{TP} = 3\hat{i} + 4\hat{j} + 5\hat{k} \), \( \vec{s} = 3\hat{i} + 4\hat{j} + 12\hat{k} \).
\( |\vec{s}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \).
\( \vec{TP} \cdot \vec{s} = (3)(3) + (4)(4) + (5)(12) = 9 + 16 + 60 = 85 \).
\( \cos\theta = \frac{\vec{TP} \cdot \vec{s}}{|\vec{TP}||\vec{s}|} = \frac{85}{5\sqrt{2} \times 13} = \frac{17}{13\sqrt{2}} = \frac{17\sqrt{2}}{26} \approx 0.9248 \implies \theta \approx 22.36^\circ \).

Projection of \( \vec{TP} \) on \( \vec{s} = \frac{\vec{TP} \cdot \vec{s}}{|\vec{s}|} = \frac{85}{13}\text{ meters} \approx 6.538\text{ meters} \).

Events defined:
\( E_1 \): Bolt manufactured by Machine 1.
\( E_2 \): Bolt manufactured by Machine 2.
\( E_3 \): Bolt manufactured by Machine 3.
\( A \): Bolt is defective.
Given:
\( P(E_1) = 0.25, \quad P(E_2) = 0.35, \quad P(E_3) = 0.40 \).
\( P(A|E_1) = 0.05, \quad P(A|E_2) = 0.04, \quad P(A|E_3) = 0.02 \).

(i) [1 Mark]
Probability bolt manufactured by Machine 2: \( P(E_2) = 0.35 \) (or \( 35\% \)).

(ii) [1 Mark]
Total probability that a bolt is defective:
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3) \] \[ P(A) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345 \text{ (or } 3.45\% \text{)} \].

(iii) [2 Marks]
Probability defective bolt came from Machine 3:
\[ P(E_3|A) = \frac{P(E_3)P(A|E_3)}{P(A)} = \frac{0.0080}{0.0345} = \frac{80}{345} = \frac{16}{69} \approx 0.2319 \text{ (or } 23.19\% \text{)} \].

Probability defective bolt came from Machine 1:
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)} = \frac{0.0125}{0.0345} = \frac{125}{345} = \frac{25}{69} \approx 0.3623 \text{ (or } 36.23\% \text{)} \].