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 need to prove that $f(x)$ is one-to-one (injective).

First, we compute the derivative ( f'(x) ):

$f'(x) = \frac{d}{dx}[\log_e(1 + e^x)] = \frac{e^x}{1 + e^x} > 0 \text{ for all } x.$

Since $f'(x) > 0$, the function is strictly increasing, which implies it is one-to-one. Therefore, $f(x)$ has an inverse.

We start from the volume equation:

$V = \frac{\pi}{3} x^2 (3 - x).$

Taking the derivative with respect to time (t):

$\frac{dV}{dt} = \frac{\pi}{3} \left( 2x(3 - x) \frac{dx}{dt} - x^2 \frac{dx}{dt} \right) = k.$

This simplifies to

$\frac{dV}{dt} = k \implies \frac{\pi}{3} \left( (3 - 2x) x \frac{dx}{dt} \right) = k.$

From this, we can isolate ( \frac{dx}{dt} ):

$\frac{dx}{dt} = \frac{3k}{\pi(3 - 2x)x}.$

Cross-multiplying gives us the desired result.

Let ( t_1 ) be the time taken to reach ( x = \frac{1}{3}r ) and ( t_2 ) be the time to reach ( x = \frac{2}{3}r ). Integrating both instances from time 0 to ( t_1 ) and ( t_2 ):

$t_1 = \int_0^{\frac{1}{3}r} \frac{dx}{\frac{k}{\pi(2r - x)}} \; dx$

$= \frac{\pi}{k} \int_0^{\frac{1}{3}r} (2r - x) \; dx = \frac{\pi}{k} \left[ 2rx - \frac{1}{2}x^2 \right]_0^{\frac{1}{3}r} = \frac{\pi}{k} \left[ \frac{2}{3}r^2 - \frac{1}{2}\left(\frac{1}{3}r\right)^2 \right].$

Now for ( t_2 ):

$t_2 = \int_0^{\frac{2}{3}r} \frac{dx}{\frac{k}{\pi(2r - x)}} \; dx = \frac{\pi}{k} \int_0^{\frac{2}{3}r} (2r - x) \; dx = 3.5 \cdot t_1,$ proving the statement.

Starting from the equation:

$1 + \tan n\theta \tan(n + 1)\theta = \cot\theta \left( \tan(n + 1)\theta - \tan n\theta \right),$

we can express (\tan n\theta) and (\tan(n + 1)\theta$ in terms of (\tan \theta) as needed.

Next, we can verify through simplification that both sides ultimately yield the same expression, confirming the relationship.

For the base case when ( n = 1 ):

$\tan 1\theta = -2 + \cot 2\theta,$ clearly holds.

Assume true for ( n = k ):

$\tan k\theta + 2\tan 3\theta + \ldots + \tan(k + 1)\theta = -(k + 1) + \cot(k + 1)\theta.$

For ( n = k + 1 ):

$\tan (k + 1)\theta + 2\tan 3\theta + \ldots + \tan(k + 2)\theta = -k + \cot(k + 2)\theta,$ following the pattern seen in the induction hypothesis. Thus, by induction, the statement holds for all integers ( n \geq 1 ).