Using definition of logarithm: \( \log_a b = m \Rightarrow a^m = b \)
\[ 3^4 = x + 5 \] \[ 81 = x + 5 \] \[ x = 81 - 5 \]

\[ \cos 75^{\circ} = \cos(45^{\circ} + 30^{\circ}) \] Using formula: \( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\[ = \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ} \] \[ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \] \[ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \]

Standard form: \( ax + by + c = 0 \Rightarrow 3x + 4y - 12 = 0 \)
Here \( a=3, b=4, c=-12 \)

Slope (m): \( -\frac{a}{b} = -\frac{3}{4} \)
X-intercept: \( -\frac{c}{a} = -\frac{-12}{3} = 4 \)
Y-intercept: \( -\frac{c}{b} = -\frac{-12}{4} = 3 \)

Differentiating w.r.t \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^x) + \frac{d}{dx}(e^x) + \frac{d}{dx}(a^a) \] Using formulas: \( \frac{d}{dx}(x^n) = nx^{n-1} \), \( \frac{d}{dx}(a^x) = a^x \log a \), \( \frac{d}{dx}(\text{const}) = 0 \)
\[ \frac{dy}{dx} = 10x^9 + 10^x \log 10 + e^x + 0 \]

Slope \( m = \frac{dy}{dx} \)
\( y = x^3 \Rightarrow \frac{dy}{dx} = 3x^2 \)
At \( x = 4 \):
\( m = 3(4)^2 = 3(16) = 48 \)

Step 1: Find \( f(0) \)
\( f(0) = (0)^4 - 2(0) + 7 = 7 \)

Step 2: Find \( f(2) \)
\( f(2) = (2)^4 - 2(2) + 7 = 16 - 4 + 7 = 19 \)

Step 3: Addition
\( f(0) + f(2) = 7 + 19 = 26 \)

Largest value (\( L \)) = 48
Smallest value (\( S \)) = 39

Range: \( L - S = 48 - 39 = 9 \)
Coefficient of Range:
\[ \frac{L-S}{L+S} = \frac{9}{48+39} = \frac{9}{87} \approx 0.103 \]

Step 1: Calculate \( A^2 \)
\[ A^2 = \begin{bmatrix}2&4&4\\ 4&2&4\\ 4&4&2\end{bmatrix} \begin{bmatrix}2&4&4\\ 4&2&4\\ 4&4&2\end{bmatrix} \] \[ = \begin{bmatrix} 4+16+16 & 8+8+16 & 8+16+8 \\ 8+8+16 & 16+4+16 & 16+8+8 \\ 8+16+8 & 16+8+8 & 16+16+4 \end{bmatrix} \] \[ = \begin{bmatrix} 36 & 32 & 32 \\ 32 & 36 & 32 \\ 32 & 32 & 36 \end{bmatrix} \]
Step 2: Calculate \( A^2 - 8A \)
\[ 8A = \begin{bmatrix} 16 & 32 & 32 \\ 32 & 16 & 32 \\ 32 & 32 & 16 \end{bmatrix} \] \[ A^2 - 8A = \begin{bmatrix} 36-16 & 32-32 & 32-32 \\ 32-32 & 36-16 & 32-32 \\ 32-32 & 32-32 & 36-16 \end{bmatrix} \] \[ = \begin{bmatrix} 20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20 \end{bmatrix} = 20I \]

Step 1: Simplify Bracket
\[ = \begin{bmatrix} 4 & 8 & 0 \\ 8 & -4 & 12 \end{bmatrix} - \begin{bmatrix} 2 & 6 & -2 \\ 4 & -6 & 8 \end{bmatrix} \] \[ = \begin{bmatrix} 4-2 & 8-6 & 0-(-2) \\ 8-4 & -4-(-6) & 12-8 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 2 \\ 4 & 2 & 4 \end{bmatrix} \]
Step 2: Multiply by Column Matrix
\[ \begin{bmatrix} 2 & 2 & 2 \\ 4 & 2 & 4 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} (2)(2)+(2)(0)+(2)(-1) \\ (4)(2)+(2)(0)+(4)(-1) \end{bmatrix} \] \[ = \begin{bmatrix} 4+0-2 \\ 8+0-4 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} \]

Let \( \frac{x+3}{(x+1)(x+5)} = \frac{A}{x+1} + \frac{B}{x+5} \)
\( x + 3 = A(x+5) + B(x+1) \)

Put \( x = -1 \):
\( -1 + 3 = A(-1+5) + 0 \Rightarrow 2 = 4A \Rightarrow A = 1/2 \)

Put \( x = -5 \):
\( -5 + 3 = 0 + B(-5+1) \Rightarrow -2 = -4B \Rightarrow B = 1/2 \)

Formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
\[ = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2})(\frac{1}{3})} \] \[ = \frac{\frac{3+2}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \]

\[ L.H.S. = \frac{(\sin 4\theta + \sin 2\theta)}{(1 + \cos 4\theta) + \cos 2\theta} \] Using formulas: \( \sin C + \sin D \) and \( 1 + \cos A = 2\cos^2(A/2) \)
Num: \( \sin 4\theta = 2\sin 2\theta \cos 2\theta \)
Denom: \( 1 + \cos 4\theta = 2\cos^2 2\theta \)
\[ = \frac{2\sin 2\theta \cos 2\theta + \sin 2\theta}{2\cos^2 2\theta + \cos 2\theta} \] \[ = \frac{\sin 2\theta(2\cos 2\theta + 1)}{\cos 2\theta(2\cos 2\theta + 1)} \] \[ = \frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta \]

Let \( \cos^{-1}(4/5) = A \Rightarrow \cos A = 4/5, \sin A = 3/5 \)
Let \( \cos^{-1}(12/13) = B \Rightarrow \cos B = 12/13, \sin B = 5/13 \)
Formula: \( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\[ = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \] \[ = \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \]

Slope of given line (\( m_1 \)) = \( -A/B = -2/-4 = 1/2 \)
Slope of perpendicular line (\( m_2 \)): \( m_1 m_2 = -1 \Rightarrow m_2 = -2 \)
Equation using Slope-Point form \( y - y_1 = m(x - x_1) \):
\( y - 4 = -2(x - 3) \)
\( y - 4 = -2x + 6 \)
\( 2x + y - 10 = 0 \)

1. Find Mean (\( \bar{x} \)):
Sum = 76, N = 8
\( \bar{x} = 76/8 = 9.5 \)

2. Find Deviations \( |x_i - \bar{x}| \):
\( |3-9.5|=6.5, |6-9.5|=3.5, |5-9.5|=4.5, |7-9.5|=2.5 \)
\( |10-9.5|=0.5, |12-9.5|=2.5, |15-9.5|=5.5, |18-9.5|=8.5 \)
Sum \( \Sigma |d| = 34 \)

3. Mean Deviation:
\( M.D. = \frac{\Sigma |d|}{N} = \frac{34}{8} = 4.25 \)

\( \frac{dx}{d\theta} = -a \sin \theta \)
\( \frac{dy}{d\theta} = b \cos \theta \)
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta \)
At \( \theta = \pi/4 \):
\( \frac{dy}{dx} = -\frac{b}{a} \cot(45^{\circ}) = -\frac{b}{a}(1) \)

Take log on both sides:
\( y \log x = (x-y) \log e = x - y \)
\( y \log x + y = x \Rightarrow y(1 + \log x) = x \Rightarrow y = \frac{x}{1+\log x} \)
Differentiate using Quotient Rule:
\[ \frac{dy}{dx} = \frac{(1+\log x)(1) - x(\frac{1}{x})}{(1+\log x)^2} \] \[ = \frac{1 + \log x - 1}{(1+\log x)^2} \]

Using half-angle formulas:
\( 1-\cos 2x = 2\sin^2 x \), \( 1+\cos 2x = 2\cos^2 x \)
\[ y = \tan^{-1}\sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \tan^{-1}\sqrt{\tan^2 x} \] \[ y = \tan^{-1}(\tan x) = x \] Differentiating w.r.t x:
\( \frac{dy}{dx} = 1 \)

Lower boundary of first class (S) = \( 10 - 0.5 = 9.5 \)
Upper boundary of last class (L) = \( 69 + 0.5 = 69.5 \)
Range: \( L - S = 69.5 - 9.5 = 60 \)
Coeff. of Range:
\[ \frac{L-S}{L+S} = \frac{60}{79} = 0.759 \]

Formula: \( C.V. = \frac{\sigma}{\bar{x}} \times 100 \)
\( C.V._A = \frac{5.1}{44} \times 100 = 11.59\% \)
\( C.V._B = \frac{6.31}{54} \times 100 = 11.68\% \)
Since \( C.V._B > C.V._A \), Batsman B is more variable.

Matrix \( A = \begin{bmatrix}1&3&3\\ 1&4&4\\ 1&3&4\end{bmatrix} \)
Determinant |A|: \( 1(16-12) - 3(4-4) + 3(3-4) = 4 - 0 - 3 = 1 \)
Cofactors:
\( C_{11}=4, C_{12}=0, C_{13}=-1 \)
\( C_{21}=-3, C_{22}=1, C_{23}=0 \)
\( C_{31}=0, C_{32}=-1, C_{33}=1 \)
Adjoint A: \( \begin{bmatrix}4&-3&0\\ 0&1&-1\\ -1&0&1\end{bmatrix} \)
Solution X = A^-1B:
\[ \begin{bmatrix}4&-3&0\\ 0&1&-1\\ -1&0&1\end{bmatrix} \begin{bmatrix}12\\ 15\\ 13\end{bmatrix} = \begin{bmatrix}48-45\\ 15-13\\ -12+13\end{bmatrix} = \begin{bmatrix}3\\ 2\\ 1\end{bmatrix} \]

Part (i):
LHS: \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
RHS: \( 2 \sin 30^{\circ} \cos 30^{\circ} = 2(\frac{1}{2})(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} \)
Verified.

Part (ii):
\( \tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}\frac{2+3}{1-6} \) (since xy>1)
\( = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
Total = \( \tan^{-1}1 + \frac{3\pi}{4} = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \)

Part (i):
\( m_1 = -3/2, m_2 = 2/3 \).
\( m_1 m_2 = -1 \), so lines are perpendicular. Angle = \( 90^{\circ} \).

Part (ii):
Line: \( 2x + y - 34 = 0 \). Point (5,4).
\( p = \left| \frac{2(5) + 4 - 34}{\sqrt{2^2+1^2}} \right| = \left| \frac{10+4-34}{\sqrt{5}} \right| \)
\( p = \left| \frac{-20}{\sqrt{5}} \right| = 4\sqrt{5} \)

1. Mean (\( \bar{x} \)):
\[ \bar{x} = \frac{\Sigma fx}{N} = \frac{440}{20} = 22 \]
2. Standard Deviation (\( \sigma \)):
\[ \sigma = \sqrt{\frac{\Sigma fx^2}{N} - (\bar{x})^2} \] \[ \sigma = \sqrt{\frac{11900}{20} - (22)^2} = \sqrt{595 - 484} = \sqrt{111} \approx 10.536 \]
3. Coefficient of Variance (C.V.):
\[ C.V. = \frac{\sigma}{\bar{x}} \times 100 = \frac{10.536}{22} \times 100 = 47.89\% \]

Perimeter \( 2(x+y) = 100 \Rightarrow y = 50-x \)
Area \( A = x(50-x) = 50x - x^2 \)
\( \frac{dA}{dx} = 50 - 2x = 0 \Rightarrow x = 25 \)
\( y = 50 - 25 = 25 \)

\( y_1 = a \cdot \frac{1}{\sec(x/a)} \cdot \sec(x/a)\tan(x/a) \cdot \frac{1}{a} = \tan(x/a) \)
\( y_2 = \sec^2(x/a) \cdot \frac{1}{a} \)
\( \rho = \frac{(1+\tan^2)^{3/2}}{y_2} = \frac{(\sec^2)^{3/2}}{\frac{1}{a}\sec^2} = a \sec(x/a) \)