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 set $y = f(x)$: $y = 2 \text{cos} \left( \frac{x}{3} \right).$ To find the inverse, we need to solve for $x$: $y = 2 \text{cos} \left( \frac{x}{3} \right) \Rightarrow \text{cos} \left( \frac{x}{3} \right) = \frac{y}{2}.$ Thus, we have: $\frac{x}{3} = \text{cos}^{-1} \left( \frac{y}{2} \right) \Rightarrow x = 3 \cdot \text{cos}^{-1} \left( \frac{y}{2} \right).$ So, the inverse function is: $y = f^{-1}(x) = 3 \cdot \text{cos}^{-1} \left( \frac{x}{2} \right).$

Domain D: Since the cosine function ranges from $-1$ to $1$, we have: $\frac{y}{2} \in [-1, 1], \text{ thus, } y \in [-2, 2].$

To find the area under the curve from $x = 0$ to $x = 3$, we calculate the integral: $A = \int_{0}^{3} f(x) \; dx = \int_{0}^{3} 2 \cos \left( \frac{x}{3} \right) \; dx.$ Using substitution, let $u = \frac{x}{3} \Rightarrow dx = 3 \, du$. Changing the limits accordingly: When $x=0, u=0$ and when $x=3, u=1$: $A = 2 \int_{0}^{1} 3 \cos(u) \; du = 6 \int_{0}^{1} \cos(u) \; du = 6 [\sin(u)]_{0}^{1} = 6(\sin(1) - \sin(0)) = 6\sin(1).$ So, the area of the shaded region is: $A = 6\sin(1).$

Using the binomial theorem: $(q + p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} p^{k},$ $(q - p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} (-p)^{k}.$

Subtracting these gives: $(q + p)^{n} - (q - p)^{n} = 2 \sum_{k=0, k \text{ odd}}^{n} \binom{n}{k} q^{n-k} p^{k}.$

This shows that the coefficients of odd powers of $p$ yield: $2 \binom{n}{1} q^{n-1} p + 2 \binom{n}{3} q^{n-3} p^{3} + \cdots.$

When $n$ is odd, the last term corresponds to $k = n$ which is: $2 \binom{n}{n} q^{0} p^{n} = 2 p^{n}.$ When $n$ is even, the last term corresponds to $k = n - 1$ which is: $2 \binom{n}{n - 1} q^{1} p^{n - 1} = 2n q p^{n - 1}.$

The probability of rolling a six on a fair die is rac{1}{6}, and of not rolling a six is rac{5}{6}. Thus, the probability of having exactly r sixes in n rolls is given by: $P(X = r) = \binom{n}{r} \left( \frac{1}{6} \right)^{r} \left( \frac{5}{6} \right)^{n-r}.$

Using the binomial expansion from part (b), we can see that the total probability of getting an odd number of successes is: $P(Odd) = \frac{1}{2}\left( (q+p)^{n} - (q-p)^{n} \right).$ Substituting $q = \frac{5}{6}$ and $p = \frac{1}{6}$: $P(Odd) = \frac{1}{2} \left[ \left( \frac{5}{6} + \frac{1}{6} \right)^{n} - \left( \frac{5}{6} - \frac{1}{6} \right)^{n} \right] = \frac{1}{2} \left( 1 - \left( \frac{2}{3} \right)^{n} \right).$