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

Using the substitution $u = x - 4$ gives us:

1. Then, $x = u + 4$ and $dx = du$.

2. The integral becomes:

   $\int \sqrt{u} \, du = \int u^{1/2} \, du$

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

Using the chain rule, we differentiate:

$$\frac{d}{dx}[3 \tan^{-1}(2x)] = 3 \cdot \frac{1}{1 + (2x)^2} \cdot (2) = \frac{6}{1 + 4x^2}.$

Use the identity $\sin(2x) = 2 \sin x \cos x$:

1. Rewrite the limit:

   $\lim_{x \to 0} \frac{\sin(2x)}{3x}$

2. This can be evaluated as:

   $$\frac{1}{3} \cdot 2 = \frac{2}{3}.$

To solve the inequality:

1. Set $2x + 5 > 0 \Rightarrow 2x > -5 \Rightarrow x > -\frac{5}{2}$.
2. Thus, the solution is $x > -\frac{5}{2}$.

Using the binomial probability formula:

1. The probability of exactly one hit:

   $P(X = 1) = \binom{3}{1} \left(\frac{3}{5}\right)^1 \left(\frac{2}{5}\right)^2$
   $$= 3 \cdot \frac{3}{5} \cdot \frac{4}{25} = \frac{36}{125}.$

Using the cumulative distribution function:

1. We want $P(X \geq 2) = 1 - P(X < 2)$:

   $P(X < 2) = P(X = 0) + P(X = 1)$

   2. Calculate:

   $P(X = 0) = \binom{6}{0} \left(\frac{3}{5}\right)^0 \left(\frac{2}{5}\right)^6 = \frac{64}{15625}$ $P(X = 1) = \binom{6}{1} \left(\frac{3}{5}\right)^1 \left(\frac{2}{5}\right)^5 = 6 \cdot \frac{3}{5} \cdot \frac{32}{3125} = \frac{576}{15625}$.

   3. Therefore:

   $$P(X \geq 2) = 1 - \left(\frac{64 + 576}{15625}\right) = \frac{12485}{15625}.$