The sketch shows the graph of the curve $y = f(x)$ where $f(x) = 2 ext{cos} \left( \frac{x}{3} \right)$ The area under the curve for $0 \leq x \leq 3$ is shaded - HSC - SSCE Mathematics Extension 1 - Question 5 - 2001 - Paper 1

To find the inverse function, we set $y = f(x) = 2 \text{cos}\left(\frac{x}{3}\right)$.
Rearranging gives:

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

Taking the inverse cosine leads to:

$\frac{x}{3} = \text{cos}^{-1}\left(\frac{y}{2}\right) \Rightarrow x = 3 \text{cos}^{-1}\left(\frac{y}{2}\right).$
Thus, the inverse function is:

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

The domain of the inverse function corresponds to the range of the original function, which is $[0, 2]$.

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

$\text{Area} = \int_{0}^{3} f(x) \, dx = \int_{0}^{3} 2 \text{cos}\left(\frac{x}{3}\right) \, dx.$
Using the substitution $u = \frac{x}{3}$, we get $du = \frac{1}{3}dx$, hence $dx = 3du$.
The limits change accordingly: when $x=0$, $u=0$; when $x=3$, $u=1$.
Thus:

$\text{Area} = 3 \int_{0}^{1} 2 \text{cos}(u) \, du = 6 \int_{0}^{1} \text{cos}(u) \, du = 6[\text{sin}(u)]_{0}^{1} = 6(\text{sin}(1) - \text{sin}(0)) = 6\text{sin}(1).$

Hence, the area of the shaded region is $6 \text{sin}(1)$.

Using the binomial theorem, we expand $(q + p)^{n}$:

$(q + p)^{n} = \sum_{k=0}^{n} \binom{n}{k} q^{n-k} p^{k}.$
For the first few terms when $n$ is even or odd, we observe the symmetry in the coefficients:

1. for $k = 0$: $\binom{n}{0} q^{n} p^{0}$
2. for $k = 1$: $\binom{n}{1} q^{n-1} p$
3. for $k = 2$: $\binom{n}{2} q^{n-2} p^{2}$
   And so forth.
   By analyzing the terms, it is evident that for all integers, the sum provides the pattern demonstrated.
   This pattern shows contributions of both $2$ and $3$ at alternating positions.

For odd $n$, the last term in the expansion corresponds to $p^{n}$.
For even $n$, the last term corresponds to $q^{n}$.
Thus:

• When $n$ is odd: $p^{n}$
• When $n$ is even: $q^{n}$.

When tossing a fair die, the probability $P(r)$ of getting exactly $r$ sixes in $n$ tosses is given by the binomial probability formula:

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

The total probability for all possible outcomes can be used with symmetry arguments. The probability of getting an odd number of 'sixes' combines terms from the full expansion.

Using the property of binomial probabilities:

$P(\text{odd}) = \frac{1}{2} \left[ 1 - \left(\frac{5}{6}\right)^{n} \right].$
This reflects the likelihood of achieving an odd count versus an even when evaluated over many trials.