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 find that $du = dx$ and $x = u + 4$.

The integral becomes:

$\int \sqrt{x - 4} \, dx = \int \sqrt{u} \, du$

Now, we can integrate:

$= \frac{2}{3} u^{3/2} + C$

Substituting back $u = x - 4$ gives:

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

Using the chain rule, we differentiate:

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

We can use the identity $\sin(2x) = 2 \sin x \cos x$. Thus, we rewrite the limit:

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

Using L'Hôpital's Rule:

$= \lim_{x \to 0} \frac{2 \cos(2x)}{3} = \frac{2 \cdot 1}{3} = \frac{2}{3}$

We start by combining the fractions:

$\frac{3 - x(2x + 5)}{2x + 5} > 0$

This simplifies to:

$\frac{3 - 2x^2 - 5x}{2x + 5} > 0$

Finding the roots of the numerator $-2x^2 - 5x + 3 = 0$, we can use the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-2)(3)}}{2(-2)} = \frac{5 \pm \sqrt{49}}{-4}$

This gives $x = -1$ and $x = \frac{3}{2}$. We test intervals to find:

$x < -1 ext{ or } -\frac{5}{2} < x < -1$

The probability of hitting the bullseye in one throw is $\frac{2}{5}$, and not hitting it is $\frac{3}{5}$.

The probability of hitting exactly once in three throws is given by:

$P = \binom{3}{1}\left(\frac{2}{5}\right)^1\left(\frac{3}{5}\right)^2 = 3 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^2 = 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}$

We can find the probabilities of hitting at least two throws using the complement rule:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Calculating these:

1. For $P(X = 0)$: $= \left(\frac{3}{5}\right)^6 = \frac{729}{15625}$
2. For $P(X = 1)$: $= \binom{6}{1}\left(\frac{2}{5}\right)^1\left(\frac{3}{5}\right)^5 = 6 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^5 = 6 \cdot \frac{2}{5} \cdot \frac{243}{3125} = \frac{2916}{15625}$

So: $P(X \geq 2) = 1 - \frac{729}{15625} - \frac{2916}{15625} = 1 - \frac{3645}{15625} = \frac{11980}{15625}$