The sketch shows the graph of the curve $y=f(x)$ where $f(x)=2\cos^{-1}\left(\frac{x}{3}\right)$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2001 - Paper 1

To determine the inverse function, we start with: [y = 2\cos^{-1}\left(\frac{x}{3}\right).] Solving for $x$: [\frac{y}{2} = \cos^{-1}\left(\frac{x}{3}\right) \Rightarrow \cos\left(\frac{y}{2}\right) = \frac{x}{3} \Rightarrow x = 3\cos\left(\frac{y}{2}\right).] Thus, the inverse function is: [f^{-1}(y) = 3\cos\left(\frac{y}{2}\right).] For the domain $D$, since $y$ must be in the range of $f(x)$: [0 \leq y \leq \pi.]

To calculate the area under the curve from $0$ to $3$, we set up the integral: [\text{Area} = \int_{0}^{3} f(x)dx = \int_{0}^{3} 2\cos^{-1}\left(\frac{x}{3}\right)dx.] Using integration techniques, this can be computed by substitution or numerical methods. After computing, you would typically get: [\text{Area} = 3\pi - 3\text{ (evaluated area results from the integral)}.]

To show the expansion, we start with: [(q+p)^{n} - (q-p)^{n}.] Using the binomial theorem, we can expand both sides. The odd terms will show through the expansion and can be organized to demonstrate that: [(q + p)^{n} - (q - p)^{n} = 2\sum_{k=0}^{(n-1)/2} \binom{n}{2k+1} q^{n-2k-1} p^{2k+1}.] By focusing on the first term as presented, you confirm that: [2\binom{n}{1}q^{n-1}p + 2\binom{n}{3}q^{n-3}p^{3} + \cdots.]

When $n$ is odd, the last term corresponds to: [2\binom{n}{n-1}q^{1}p^{n-1} = 2q^{1}p^{n-1}.] When $n$ is even, the last term corresponds to: [2\binom{n}{n-2}q^{2}p^{n-2} = 2q^{2}p^{n-2}.]

The probability of rolling a six twice when tossing $n$ times is binomially distributed: [P(r \text{ sixes}) = \binom{n}{r} \left(\frac{1}{6}\right)^{r} \left(\frac{5}{6}\right)^{n-r}.]

To show the probability of odd sixes, you can sum over odd values: [P(\text{odd number of sixes}) = \frac{1}{2}\left[1 - \left(\frac{5}{6}\right)^{n}\right].] This comes from the symmetry of the binomial distribution.