Question 11 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 11 - 2016 - Paper 1

To show that $z^6$ is real, we compute:

For $z^n$ to be purely imaginary, the cosine part must equal zero. This occurs when: $-\frac{\pi}{6} \cdot n = -\frac{\pi}{2} + k\pi \, (k \in \mathbb{Z}).$
Solving for $n$, we find: $n = 3 + 6k.$
Choosing the smallest positive integer, we set $k = 0$, yielding $n = 3$.

To solve the integral, we employ integration by parts where we let:

• $u = x$ and $dv = e^{-2x} dx$. Then, we differentiate and integrate to find:
• $du = dx$ and $v = -\frac{1}{2} e^{-2x}.$ Now applying the integration by parts formula: $\int u \, dv = uv - \int v \, du,$
  we obtain: $\int xe^{-2x} dx = -\frac{1}{2}xe^{-2x} - \int -\frac{1}{2} e^{-2x} dx = -\frac{1}{2}xe^{-2x} + \frac{1}{4}e^{-2x} + C.$

To differentiate the equation implicitly with respect to $x$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(2xy).$
Applying the chain rule and product rule yields: $3x^2 + 3y^2 \frac{dy}{dx} = 2\left(y + x \frac{dy}{dx}\right).$
Rearranging terms leads to: $3y^2 \frac{dy}{dx} - 2x \frac{dy}{dx} = 2y - 3x^2.$
Factoring out $\frac{dy}{dx}$ gives: $\frac{dy}{dx}(3y^2 - 2x) = 2y - 3x^2.$
Thus, $\frac{dy}{dx} = \frac{2y - 3x^2}{3y^2 - 2x}.$