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

To show that $z^6$ is real, we can use the modulus–argument form derived earlier:

$z^6 = \left(2\right)^6 \left(\cos\left(-\frac{6\pi}{6}\right) + i\sin\left(-\frac{6\pi}{6}\right)\right) = 64 \left(\cos(-\pi) + i\sin(-\pi)\right) = 64 (-1 + 0i) = -64.$

Since $-64$ is a real number, we confirm that $z^6$ is indeed real.

For $z^n$ to be purely imaginary, the argument of $z^n$ must be an odd multiple of $\frac{\pi}{2}$. From the argument of $z$:

$\text{Arg}(z) = -\frac{\pi}{6},$

We have:

$\text{Arg}(z^n) = n\left(-\frac{\pi}{6}\right).$

Setting this equal to an odd multiple of $\frac{\pi}{2}$, we can write:

$n\left(-\frac{\pi}{6}\right) = (2k + 1)\frac{\pi}{2}$

where $k$ is an integer. Solving for $n$ gives:

$n = -3(2k + 1).$

Taking the smallest positive integer, let $k = 0$, so:

$n = -3.$

However, because we require a positive integer, we can choose $n = 3$, as it gives:

$z^3 = -\sqrt{3} - i,$

which is purely imaginary.

To solve the integral $\int xe^{-2x} dx$, we can use integration by parts.

Let:

• $u = x$ thus $du = dx$.
• $dv = e^{-2x}dx$ thus $v = -\frac{1}{2}e^{-2x}$.

Using integration by parts, we have:

$\int u \, dv = uv - \int v \, du,$

Substituting in our values gives:

$\int xe^{-2x} dx = -\frac{x}{2}e^{-2x} - \int\left(-\frac{1}{2}e^{-2x}\right)dx = -\frac{x}{2}e^{-2x} + \frac{1}{4} e^{-2x} + C.$

Combining these terms yields:

$\int xe^{-2x} dx = -\frac{e^{-2x}}{2}(x - \frac{1}{2}) + C.$

To find $\frac{dy}{dx}$ from the implicit equation $x^3 + y^3 = 2xy$, we differentiate both sides 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 leads to:

$3x^2 + 3y^2\frac{dy}{dx} = 2\left(y + x\frac{dy}{dx}\right).$

Rearranging gives:

$3y^2\frac{dy}{dx} - 2x\frac{dy}{dx} = 2y - 3x^2,$

factoring out $\frac{dy}{dx}$,

$\frac{dy}{dx}(3y^2 - 2x) = 2y - 3x^2.$

Thus, we find:

$\frac{dy}{dx} = \frac{2y - 3x^2}{3y^2 - 2x}.$