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

To find the inverse function, we first write:

$y = 2 \cos \left( \frac{x}{3} \right).$

Solving for $x$ gives:

$y = 2 \cos \left( \frac{x}{3} \right) \Rightarrow \cos \left( \frac{x}{3} \right) = \frac{y}{2} \Rightarrow \frac{x}{3} = \cos^{-1} \left( \frac{y}{2} \right) \Rightarrow x = 3 \cos^{-1} \left( \frac{y}{2} \right).$

Thus, the inverse function is:

$f^{-1}(y) = 3 \cos^{-1} \left( \frac{y}{2} \right).$

The domain $D$ of this inverse function is derived from the range of $f(x)$, which is $[ -2, 2 ]$. Hence, $D = [ -2, 2 ].$

To find the area under the curve from $0$ to $3$, we need to evaluate the integral:

$\text{Area} = \int_{0}^{3} f(x) \, dx = \int_{0}^{3} 2 \cos \left( \frac{x}{3} \right) \, dx.$

To solve this, we use substitution. Let:

$u = \frac{x}{3} \\ dx = 3 \, du.$

Changing the limits accordingly: when $x = 0$, $u = 0$; when $x = 3$, $u = 1$. Then we have:

$\text{Area} = 6 \int_{0}^{1} \cos(u) \, du = 6 \left[ \sin(u) \right]_{0}^{1} = 6\left( \sin(1) - \sin(0) \right) = 6 \sin(1).$

Using the binomial expansion, we expand both $(q + p)^{n}$ and $(q - p)^{n}$:

$(q + p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} p^{k},$

and

$(q - p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} (-p)^{k} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} (-1)^{k} p^{k}.$

Subtracting these two expansions gives:

$(q + p)^{n} - (q - p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} p^{k} - \sum_{k=0}^{n} \binom{n}{k} q^{n-k} (-1)^{k} p^{k}.$

When we combine like terms, we notice that the terms where $k$ is odd will be doubled:

$= 2 \sum_{\text{odd } k} \binom{n}{k} q^{n-k} p^{k}.$

Hence, the formula is verified.

When $n$ is odd, the last term in the expansion is given by the term where $k = n - 1$:

$2 \binom{n}{n-1} q^{1} p^{n-1} = 2n qp^{n-1}.$

When $n$ is even, the last term occurs at $k = n$, which gives:

$2 \binom{n}{n} q^{0} p^{n} = 2p^{n}.$

The probability of exactly $r$ sixes appearing in the uppermost position can be calculated using the binomial probability formula:

$P(X = r) = \binom{n}{r} \left(\frac{1}{6}\right)^{r} \left(\frac{5}{6}\right)^{n - r}.$

Using the result from part (b), we know the probability for odd occurrences can be derived from the total:

$P(\text{odd}) = \frac{1}{2} \left[ 1 - \left( \frac{2}{3} \right)^{n} \right].$

This is due to symmetry in the distribution of probabilities for the number of sixes coming up.