a) Function: \( f(x) = x\sqrt{3 – x^2} \), domain: \( -\sqrt{3} \leq x \leq \sqrt{3} \).
Key points:
– At \( x = 0 \): \( f(0) = 0 \).
– At \( x = \sqrt{3} \): \( f(\sqrt{3}) = \sqrt{3} \sqrt{3 – 3} = 0 \).
– At \( x = -\sqrt{3} \): \( f(-\sqrt{3}) = -\sqrt{3} \sqrt{3 – 3} = 0 \).
– Test \( x = 1 \): \( f(1) = 1 \sqrt{3 – 1} = \sqrt{2} \approx 1.414 \).
Symmetry: Odd function, symmetric about origin.
Graph: As shown below.
[2]

b) (i) Volume formula: \( V = \pi \int_a^b [f(x)]^2 \,dx \).
\( [f(x)]^2 = (x\sqrt{3 – x^2})^2 = x^2 (3 – x^2) = 3x^2 – x^4 \).
Limits: \( -\sqrt{3} \) to \( \sqrt{3} \).
Since \( [f(x)]^2 \) is even, \( V = 2\pi \int_0^{\sqrt{3}} (3x^2 – x^4) \,dx \).
Integral: \( V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (x\sqrt{3 – x^2})^2 \,dx \) or \( 2\pi \int_0^{\sqrt{3}} (x\sqrt{3 – x^2})^2 \,dx \). [2]
(ii) Evaluate: \( V = 2\pi \int_0^{\sqrt{3}} (3x^2 – x^4) \,dx \).
Antiderivative: \( \int (3x^2 – x^4) \,dx = x^3 – \frac{x^5}{5} \).
At limits: \( \left[ x^3 – \frac{x^5}{5} \right]_0^{\sqrt{3}} = \left( (\sqrt{3})^3 – \frac{(\sqrt{3})^5}{5} \right) – 0 = 3\sqrt{3} – \frac{9\sqrt{3}}{5} = \frac{6\sqrt{3}}{5} \).
Volume: \( V = 2\pi \cdot \frac{6\sqrt{3}}{5} = \frac{12\pi \sqrt{3}}{5} \approx 13.1 \). [3]

c) Transformation:
– Stretch by 2 along \( x \)-axis: \( f\left(\frac{x}{2}\right) = \frac{x}{2} \sqrt{3 – \left(\frac{x}{2}\right)^2} \).
– Stretch by 0.5 along \( y \)-axis: \( g(x) = \frac{1}{2} f\left(\frac{x}{2}\right) \).
New domain: \( -2\sqrt{3} \leq x \leq 2\sqrt{3} \).
Volume: \( V_g = \pi \int_{-2\sqrt{3}}^{2\sqrt{3}} [g(x)]^2 \,dx \).
\( [g(x)]^2 = \left( \frac{1}{2} f\left(\frac{x}{2}\right) \right)^2 = \frac{1}{4} \left( f\left(\frac{x}{2}\right) \right)^2 \).
Substitute \( u = \frac{x}{2} \), \( dx = 2 \,du \), limits \( -\sqrt{3} \) to \( \sqrt{3} \):
\( V_g = \pi \int_{-\sqrt{3}}^{\sqrt{3}} \frac{1}{4} [f(u)]^2 \cdot 2 \,du = \frac{\pi}{2} \int_{-\sqrt{3}}^{\sqrt{3}} [f(u)]^2 \,du \).
Since \( \pi \int_{-\sqrt{3}}^{\sqrt{3}} [f(x)]^2 \,dx = \frac{12\pi \sqrt{3}}{5} \),
\( V_g = \frac{1}{2} \cdot \frac{12\pi \sqrt{3}}{5} = \frac{6\pi \sqrt{3}}{5} \approx 6.53 \). [3]