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 differentiate to find ( du = dx ) or ( dx = du ).
Then rewrite the integral:

$\int \sqrt{u} \, du$

Integrating this gives:

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

We apply the chain rule here. The derivative of $\tan^{-1}(u)$ is ( \frac{1}{1 + u^2} \frac{du}{dx} ):

Let $u = 2x$, then:

$\frac{du}{dx} = 2$

Thus:

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

Using the limit property of sine, we can simplify:

$\lim_{x \to 0} \frac{2\sin x \cos x}{3x} = \frac{2 \cdot 1 imes 1}{3} = \frac{2}{3}$

To solve this inequality, we consider the expression:

$\frac{3}{2x + 5} > x$

Cross-multiplying gives us:

$3 > x(2x + 5)$

Rearranging leads to:

$2x^2 + 5x + 3 < 0$

Finding the roots gives:

$x = -3, \; x = -\frac{1}{2}$

So the solution set is:

$-3 < x < -\frac{1}{2}$

The probability of hitting the bullseye is ( p = \frac{2}{5} ) and missing it is ( q = 1 - p = \frac{3}{5} ).
Using the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

For exactly one hit out of three throws (where $n = 3$, $k = 1$):

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

Here we can use the complement rule. The probabilities for getting at least 2 hits is:

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

Calculating these values:

$P(0) = \binom{6}{0} \left(\frac{2}{5}\right)^0 \left(\frac{3}{5}\right)^{6} = \frac{729}{15625}$

$P(1) = \binom{6}{1} \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^{5} = 6 \cdot \frac{2}{5} \cdot \frac{243}{3125} = \frac{2916}{15625}$

Thus,

$P(X \geq 2) = 1 - \left(\frac{729}{15625} + \frac{2916}{15625}\right) = 1 - \frac{3645}{15625} = \frac{11980}{15625}$