Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2008 - Paper 1

To differentiate $\cos^{-1}(3x)$ with respect to $x$, we use the chain rule. The derivative of $\cos^{-1}(u)$ is $-\frac{1}{\sqrt{1-u^2}}$ where $u = 3x$.

Therefore, applying the chain rule, we have:

$\frac{d}{dx}[\cos^{-1}(3x)] = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot \frac{d}{dx}(3x) = -\frac{3}{\sqrt{1 - 9x^2}}.$

To evaluate the integral $\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} dx,$ we recognize that this is a standard integral. We can substitute $x = 2\sin\theta$, which gives $dx = 2\cos\theta d\theta$ and transforms the limits:

When $x = -1$, $heta = \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$; and when $x = 1$, $heta = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}.$

The integral becomes:

$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{2\cos\theta}{\sqrt{4 - (2\sin\theta)^2}} d\theta =\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{2\cos\theta}{\sqrt{4(1 - \sin^2\theta)}} d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{2\cos\theta}{\sqrt{4\cos^2\theta}} d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 1 d\theta = \frac{\pi}{3}.$

To find the coefficient of $x^8y^4$ in the expansion of $(2x + 3y)^{12}$, we use the Binomial Theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.$

For our expression, let $a = 2x$ and $b = 3y$, thus:

We need $n - k = 8$ and $k = 4$. Therefore, $n = 12 = k + (n - k) = 4 + 8 = 12$.

The coefficient of $x^8y^4$ is:

$\binom{12}{4} (2^8)(3^4) = \frac{12!}{4!8!} \cdot 256 \cdot 81 = 495 \cdot 256 \cdot 81 = 10205760.$

To evaluate the integral $\int_0^{\frac{\pi}{4}} \cos \theta \sin^2 \theta d\theta,$ we can use the substitution method:

Let $u = \sin \theta$, hence $du = \cos \theta d\theta$. The limits change from:

When $heta = 0$, $u = \sin(0) = 0$;

When $heta = \frac{\pi}{4}$, $u = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$

Thus, the integral becomes:

$\int_0^{\frac{\sqrt{2}}{2}} u^2 du = \left. \frac{u^3}{3} \right|_0^{\frac{\sqrt{2}}{2}} = \frac{1}{3} \left(\frac{\sqrt{2}}{2}\right)^3 = \frac{1}{3} \cdot \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{12}.$

To determine the domain of the function $f(x) = \log_e \left[ (x-3)(5-x) \right],$ we must ensure that the argument of the logarithm is positive:

$(x-3)(5-x) > 0.$

This inequality holds when:

1. $x < 3$ and $x > 5$, which is impossible, or
2. $x > 3$ and $x < 5$.

Thus, the domain of $f(x)$ is:

$\boxed{(3, 5)}.$