State the difference between "physical change" and "chemical change" and provide examples for each.

(1) (ক)

Step 1: $\sin^4 x = (\sin^2 x)^2 = \left( \frac{1 - \cos 2x}{2} \right)^2$

$$\sin^4 x = \left( \frac{1 - \cos 2x}{2} \right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}$$

Step 2: $\cos^2 2x = \frac{1 + \cos 4x}{2}$

$$\sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{1 - 2\cos 2x + \frac{1}{2} + \frac{\cos 4x}{2}}{4} = \frac{\frac{3}{2} - 2\cos 2x + \frac{\cos 4x}{2}}{4}$$ $$= \frac{3}{8} - \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x$$

Step 3: Integrate term by term from 0 to $\frac{\pi}{2}$

$$\int_0^{\pi/2} \sin^4 x \, dx = \int_0^{\pi/2} \left( \frac{3}{8} - \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x \right) dx$$ = \left[ \frac{3}{8} x - \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x \right]_0^{\pi/2} = \frac{3}{8} \cdot \frac{\pi}{2} - 0 = \frac{3\pi}{16} ** \boxed{\dfrac{3\pi}{16}} (1) (খ) Step 1: $\cos^5 x = \cos^4 x \cos x = (1 - \sin^2 x)^2 \cos x$ Let $u = \sin x$, $du = \cos x \, dx$

\int \cos^5 x , dx = \int (1 - u^2)^2 , du

$$Step 2: Expand$$

(1 - u^2)^2 = 1 - 2u^2 + u^4

$$$$

\int (1 - 2u^2 + u^4) , du = u - \frac{2}{3} u^3 + \frac{1}{5} u^5 + C

$$Step 3: Substitute back$$

\sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C
**

\sin x - \dfrac{2}{3} \sin^{3} x + \dfrac{1}{5} \sin^{5} x + C

(1) (গ)

Step 1: Complete the square in denominator

$$x^2 + 4x + 3 = (x+2)^2 - 4 + 3 = (x+2)^2 - 1$$ $$\int \frac{dx}{\sqrt{(x+2)^2 - 1}}$$

Step 2: Let $u = x+2$, $du = dx$

\int \frac{du}{\sqrt{u^2 - 1}} = \cosh^{-1} u + C \quad \text{or} \quad \ln |u + \sqrt{u^2 - 1}| + C ** \boxed{\ln \left| x + 2 + \sqrt{x^{2} + 4x + 3} \right| + C} (2) (ক) (i) Step 1: $y = \frac{\sin(x + a)}{\cos(x - a)}$ Use quotient rule: $\frac{dy}{dx} = \frac{ u' v - u v' }{v^2}$ where $u = \sin(x+a)$, $v = \cos(x-a)$ $u' = \cos(x+a)$, $v' = \sin(x-a)$ Step 2: Numerator: $\cos(x+a) \cos(x-a) - \sin(x+a) (-\sin(x-a))$

= \cos(x+a)\cos(x-a) + \sin(x+a)\sin(x-a) = \cos( (x+a) - (x-a) ) = \cos 2a

Step 3: Denominator: $\cos^2 (x-a)$

\frac{dy}{dx} = \frac{\cos 2a}{\cos^2 (x - a)}
**

\dfrac{\cos 2a}{\cos^{2}(x - a)}

(2) (ক) (ii)

Step 1: $y = x^{\sin x}$

Take $\ln$: $\ln y = \sin x \ln x$

Step 2: Differentiate: $\frac{1}{y} y' = \cos x \ln x + \sin x \cdot \frac{1}{x}$

y' = y \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) ** x^{\sin x} \left( \cos x \ln x + \dfrac{\sin x}{x} \right) (2) (ক) (iii) Step 1: $y = \log (\sec x + \tan x)$ $\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} (\sec x + \tan x)$ Step 2: Derivative of $\sec x + \tan x = \sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$ Step 3: So $\frac{dy}{dx} = \frac{ \sec x (\sec x + \tan x) }{ \sec x + \tan x } = \sec x$ ** \sec x (3) (ক) Step 1: $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$ $|A| = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5$ 5 (3) (খ) Assuming inverse of same A. Step 1: $A^{-1} = \frac{1}{|A|} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$ \dfrac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} (3) (গ) Assuming product of two matrices, say P = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, Q = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} Step 1: PQ_{11} = 1\cdot5 + 2\cdot7 = 5+14=19 PQ_{12} = 1\cdot6 + 2\cdot8 = 6+16=22 PQ_{21} = 3\cdot5 + 4\cdot7 = 15+28=43 PQ_{22} = 3\cdot6 + 4\cdot8 = 18+32=50 \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}