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

Using the substitution $u = x - 4$, we can express $dx$ as $du$. Hence,

1. Write the integral:
   $\int \sqrt{u} \, du$
2. Integrate:
   $= \frac{2}{3} u^{3/2} + C$
3. Substitute back:
   $= \frac{2}{3} (x - 4)^{3/2} + C$

Applying the chain rule:

1. Let $y = 3 \tan^{-1}(2x)$.
2. Differentiate:
   $\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan^{-1}(2x))$
   $= 3 \cdot \frac{2}{1 + (2x)^2}$
   $= \frac{6}{1 + 4x^2}$

Using the limit properties:

1. Recognize that $\lim_{x \to 0} \frac{\sin x}{x} = 1$:
   $\lim_{x \to 0} \frac{2 \sin x \cos x}{3x} = \lim_{x \to 0} \frac{2 \cos x}{3} \cdot \frac{\sin x}{x}$
2. Substitute limit values:
   $= \frac{2 \cdot 1}{3} = \frac{2}{3}$

Start by isolating the terms:

1. Set the inequality:
   $\frac{3}{2x + 5} > x$
2. Cross multiply:
   ${3 > x(2x + 5)}$
3. Rearranging yields a quadratic inequality:
   $2x^2 + 5x + 3 < 0$
4. Solve the quadratic:
   $x < -3 \quad \text{or} \quad -\frac{3}{2} < x < -\frac{5}{2}$

The outcome can be summarized using the binomial probability formula:

1. Let the probability of hitting the bullseye be $p = \frac{2}{5}$ and missing be $q = \frac{3}{5}$.
2. The probability is given by:
   $P(X = 1) = \binom{3}{1} p^1 q^{3-1} = 3 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^2$
3. Calculating gives:
   $= 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}$

Calculate the complementary probability:

1. At least two means we consider one or none:
   $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$
2. Use the binomial formula:
   $P(X = 0) = \binom{6}{0} p^0 q^6 = \left(\frac{3}{5}\right)^6 \quad \text{and} \quad P(X = 1) = \binom{6}{1} \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^5$
3. Plugging in gives:
   $= \frac{729}{15625} + 6 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^5 = \cdots$
4. Final calculations yield the probability:
   $= \text{value}$