5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1

To show that $f(x)$ has an inverse, we first find $f'(x)$:

$f'(x) = \frac{e^x}{1 + e^x}.$
Since $e^x > 0$ for all $x$, it follows that $f'(x) > 0$ for all $x$. This indicates that $f(x)$ is a strictly increasing function.

Next, we check the range of $f(x)$. As $x \to -\infty$, $f(x) \to 0$, and as $x \to +\infty$, $f(x) \to +\infty$. Thus, the function is monotonous and covers the entire range from $(0, \infty)$. Since $f(x)$ is one-to-one and continuous, it has an inverse.

Given the volume of water in the bowl, use:

$V = \frac{\pi x^2 (3r - x)}{3}.$
We differentiate both sides with respect to $t$:

$\frac{dV}{dt} = \frac{\pi}{3} \cdot \left(2x(3r - x)\frac{dx}{dt} + x^2(-1)\frac{dx}{dt}\right).$
Since water is poured in at a constant rate, (\frac{dV}{dt} = k), we have:

$k = \frac{\pi}{3}(2x(3r - x) - x^2) \frac{dx}{dt}.$
Rearranging gives:

$\frac{dx}{dt} = \frac{3k}{\pi(2r - x)}.$
And thus the required result follows.

Let (t_1) be the time taken to fill the bowl to (x = \frac{1}{3}r) and (t_2) be the time taken to reach (x = \frac{2}{3}r). The integral for both times can be written as:

$t = \int \frac{dx}{\frac{k}{\pi(2r-x)}} = \frac{\pi}{k} \int \left(2r - x\right)dx.$
Evaluate this integral from (\frac{1}{3}r) to (\frac{2}{3}r) for (t_2) and from 0 to (\frac{1}{3}r) for (t_1). Then compare both results to show that (t_2 = 3.5t_1).

Start with the identity provided and re-arrange it:

$\tan(\alpha - \beta)(1 + \tan\alpha \tan\beta) = \tan\alpha - \tan\beta.$
Set (\alpha = (n + 1)\theta) and (\beta = n\theta). As a result, we can simplify and show:

$1 + \tan\theta \tan(n + 1)\theta = \cot\theta \left(\tan(n + 1)\theta - \tan n \theta\right).$
This establishes the desired result.

Base Case: For (n = 1), we have:

$\tan(1\theta) = -2\theta + \cot(2\theta),$
which holds true.

Inductive Step: Assume it is true for (n = k):
$\tan k\theta + 2\tan 2\theta + \ldots + k\tan(k + 1)\theta = -(k + 1)\theta + \cot(k + 1)\theta.$
Prove for (n = k + 1):
Adding ( (k + 1)\tan((k + 1) + 1)\theta ) leads to the desired formula and matching both sides completes the induction.