Here are the solutions to the questions: 1. (i) a) Differentiate $\ln\left(\frac{3x^4}{\sin x}\right)$ with respect to $x$. Step 1: Use logarithm properties to simplify the expression. $$y = \ln\left(\frac{3x^4}{\sin x}\right) = \ln(3x^4) - \ln(\sin x)$$ $$y = \ln 3 + \ln(x^4) - \ln(\sin x)$$ $$y = \ln 3 + 4\ln x - \ln(\sin x)$$ Step 2: Differentiate each term with respect to $x$. $$\frac{dy}{dx} = \frac{d}{dx}(\ln 3) + \frac{d}{dx}(4\ln x) - \frac{d}{dx}(\ln(\sin x))$$ $$\frac{dy}{dx} = 0 + 4\left(\frac{1}{x}\right) - \left(\frac{1}{\sin x} \cdot \cos x\right)$$ $$\frac{dy}{dx} = \frac{4}{x} - \frac{\cos x}{\sin x}$$ $$\frac{dy}{dx} = \frac{4}{x} - \cot x$$ The derivative is $\boxed{\frac{4}{x} - \cot x}$. 1. (i) b) A curve is defined parametrically by $x = t^3$, $y = t\left(1 - \frac{1}{4}t^3\right)$. Find the equation of the tangent to the curve at the point where $t = -1$. Step 1: Find the coordinates $(x_0, y_0)$ of the point of tangency when $t = -1$. $$x_0 = (-1)^3 = -1$$ $$y_0 = (-1)\left(1 - \frac{1}{4}(-1)^3\right) = (-1)\left(1 - \frac{1}{4}(-1)\right) = (-1)\left(1 + \frac{1}{4}\right) = (-1)\left(\frac{5}{4}\right) = -\frac{5}{4}$$ The point is $\left(-1, -\frac{5}{4}\right)$. Step 2: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$. $$x = t^3 \implies \frac{dx}{dt} = 3t^2$$ $$y = t - \frac{1}{4}t^4 \implies \frac{dy}{dt} = 1 - \frac{1}{4}(4t^3) = 1 - t^3$$ Step 3: Find the gradient $\frac{dy}{dx}$ using the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $$\frac{dy}{dx} = \frac{1 - t^3}{3t^2}$$ Step 4: Evaluate the gradient at $t = -1$. $$m = \left.\frac{dy}{dx}\right|_{t=-1} = \frac{1 - (-1)^3}{3(-1)^2} = \frac{1 - (-1)}{3(1)} = \frac{1+1}{3} = \frac{2}{3}$$ Step 5: Use the point-slope form of a line $y - y_0 = m(x - x_0)$. $$y - \left(-\frac{5}{4}\right) = \frac{2}{3}(x - (-1))$$ $$y + \frac{5}{4} = \frac{2}{3}(x + 1)$$ Step 6: Clear denominators and rearrange into general form. Multiply by 12: $$12\left(y + \frac{5}{4}\right) = 12\left(\frac{2}{3}(x + 1)\right)$$ $$12y + 15 = 8(x + 1)$$ $$12y + 15 = 8x + 8$$ $$8x - 12y + 8 - 15 = 0$$ $$8x - 12y - 7 = 0$$ The equation of the tangent is $\boxed{8x - 12y - 7 = 0}$. 1. (ii) a) Resolve $\frac{6}{(1+2x)(3-x)}$ into partial fractions. Step 1: Set up the partial fraction decomposition. $$\frac{6}{(1+2x)(3-x)} = \frac{A}{1+2x} + \frac{B}{3-x}$$ Step 2: Multiply both sides by $(1+2x)(3-x)$ to clear denominators