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

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

$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot 3 = -\frac{3}{\sqrt{1 - 9x^2}}.$

This integral can be evaluated by recognizing it as part of the inverse sine function. The evaluation goes as follows:

We can rewrite the integral as:

Now, evaluating from $-1$ to $1$ gives: $$\left[\sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}\left(-\frac{1}{2}\right)\right] = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}.$$

To find the coefficient of $x^8 y^4$, we use the Binomial Theorem:

In our case, let $a = 2x$, $b = 3y$, and $n = 12$. We need the terms where $x$ has a power of $8$ and $y$ has a power of $4$. Thus, we set: - $n - k = 8 \Rightarrow k = 4.$ Substituting: $${12 \choose 4}(2x)^8(3y)^4 = {12 \choose 4} \cdot 2^8 \cdot 3^4 \cdot x^8 \cdot y^4.$$ Therefore, the coefficient is: $$ {12 \choose 4} \cdot 2^8 \cdot 3^4.$$

Using the identity for integration, we can simplify this integral:

$\int 4 \cos \theta \sin^2 \theta d\theta = \int 2 \sin 2\theta d\theta,$

This leads to:

$= 2 \left[ \frac{\sin^2 2\theta}{2} \right]_{0}^{\frac{\pi}{4}} = [\sin^2(\frac{\pi}{2}) - \sin^2(0)] = 1 - 0 = 1.$

The function $f(x)$ is defined as long as the argument of the logarithm is positive:

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

This product is positive when:

1. $x - 3 > 0$ and $5 - x > 0$ (i.e., $3 < x < 5$)
2. $x - 3 < 0$ and $5 - x < 0$, which is not possible as it would require $x < 3$ and $x > 5$ simultaneously.

Thus, the domain of $f(x)$ is: $\text{Domain: } (3, 5).$