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

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

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

This is a standard integral that can be evaluated using the sine substitution method. The integral evaluates to:

$\int_{-1}^{1} \frac{1}{\sqrt{4 - x^2}} \; dx = \frac{\pi}{2}.$

Using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^{k}$ We need $x^k$ and $y^4$, so $n - k = k$ for $x$ and $k = 4$ for $y$. Thus: [ k = 12 - 4 = 8. ] The coefficient is: [ {12 \choose 4} (2^8)(3^4) = 495 \cdot 256 \cdot 81 = 1028160. ]

Use the substitution $u = \sin \theta$ where $du = \cos \theta d\theta$. The limits change accordingly from 0 to $\frac{\sqrt{2}}{2}$. Thus: $\int_{0}^{\frac{\pi}{4}} \cos \theta \sin^{2} \theta \; d\theta = \int_{0}^{1/\sqrt{2}} u^2 \; du = \left[ \frac{u^3}{3} \right]_{0}^{1/{\sqrt{2}}} = \frac{1/8}{3} = \frac{1}{24}.$

For the function $f(x)$ to be defined, the argument of the logarithm must be positive: $[x - 3](5 - x) > 0.$ This occurs when:

1. Both factors are positive: $x - 3 > 0$ and $5 - x > 0$ leads to:

   • $x > 3$
   • $x < 5$
     Hence, $3 < x < 5$.

2. Both factors are negative simultaneously leads to:

   • $x < 3$
   • $x > 5$, which is not possible.
     Thus, the domain of $f(x)$ is: $(3, 5).$