a) Find the inverse of the function $y = x^3 - 2$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

1. Substitute: $u = x - 4 \Rightarrow x = u + 4 \Rightarrow dx = du.$

2. The integral becomes: $\int \sqrt{u} \, du.$

3. Now, evaluate the integral: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C$.

4. Substitute back: $= \frac{2}{3} (x - 4)^{3/2} + C.$

Using the chain rule:

1. Let $f(x) = 3 \tan^{-1}(2x)$.
2. The derivative is: $f'(x) = 3 \cdot \frac{1}{1 + (2x)^2} \cdot 2 = \frac{6}{1 + 4x^2}.$

1. Simplify the limit: $\lim_{x \to 0} \frac{\sin(2x)}{3x}$ because $2 \sin x \cos x = \sin(2x)$.

2. Use L'Hôpital's Rule since both the numerator and denominator approach 0: $= \lim_{x \to 0} \frac{2 \cos(2x)}{3} = \frac{2 \cdot 1}{3} = \frac{2}{3}.$

1. The inequality holds when the denominator is positive: $2x + 5 > 0 \Rightarrow 2x > -5 \Rightarrow x > -\frac{5}{2}.$

1. Let $p = \frac{2}{5}$, the probability of hitting the bullseye.
2. The probability of missing is: $q = 1 - p = \frac{3}{5}$.
3. The probability of hitting exactly one bullseye in three throws (using binomial distribution): $P(X = 1) = \binom{3}{1} p^1 q^2 = 3 \cdot \left(\frac{2}{5}\right)^1 \cdot \left(\frac{3}{5}\right)^2 = 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}.$

1. For at least two hits, calculate complements: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1).$
2. Calculate:
   $P(X = 0) = \binom{6}{0} p^0 q^6 = \left(\frac{3}{5}\right)^6 = \frac{729}{15625}.$
   $P(X = 1) = \binom{6}{1} p^1 q^5 = 6 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^5 = 6 \cdot \frac{2}{5} \cdot \frac{243}{3125} = \frac{2916}{15625}.$
3. Thus: $P(X \geq 2) = 1 - \left(\frac{729 + 2916}{15625}\right) = 1 - \frac{3645}{15625} = \frac{11980}{15625}.$