Use the principle of mathematical induction to show that $$2 \cdot 1! + 5 \cdot 2! + 10 \cdot 3! + \ldots + (n^2 + 1)n! = n(n + 1)!$$ for all positive integers $n$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2002 - Paper 1

From the geometry of the cone, we use similar triangles:

If $h$ is the height of the water, then the radius at this height, $r$, can be expressed as:

$\frac{r}{4} = \frac{h}{12}$

Cross-multiplying gives:

$12r = 4h$

Rearranging yields:

$r = \frac{1}{3}h$

Given that the volume $V$ of water in the cone is:

$V = \frac{1}{3}\pi r^2 h$

Using the relationship found previously, substituting for $r$, we have:

$V = \frac{1}{3}\pi \left(\frac{1}{3}h\right)^2 h = \frac{1}{27}\pi h^3$

To find the rate of height change, differentiate both sides with respect to time:

$\frac{dV}{dt} = \frac{1}{9} \pi h^2 \frac{dh}{dt}$

Given that $\frac{dV}{dt} = 3$ cm³/s, set:

$3 = \frac{1}{9} \pi (9^2) \frac{dh}{dt}$

Simplifying, we find:

$3 = \frac{1}{9} \pi (81) \frac{dh}{dt} \implies 3 = 9\pi \frac{dh}{dt}$

Thus:

$\frac{dh}{dt} = \frac{1}{3\pi}$

To find $f'(x)$, we differentiate:

f'(x) = \frac{d}{dx}igg[2 \sin^{-1}(\sqrt{x})\bigg] - \frac{d}{dx}\bigg[\sin^{-1}(2x - 1)\bigg]

Using the derivative of $\sin^{-1}(u)$, we have:

$f'(x) = \frac{2}{\sqrt{x}(2\sqrt{x})} - \frac{2}{\sqrt{1 - (2x - 1)^2}}$

Simplifying gives:

$f'(x) = \frac{1}{\sqrt{x}} - \frac{2}{\sqrt{1 - (2x - 1)^2}}$

Setting $f'(x) = 0$ to find critical points:

$\frac{1}{\sqrt{x}} = \frac{2}{\sqrt{1 - (2x - 1)^2}}$

After solving, we find that $f'(x) = 0$ in the interval as required.

To sketch the graph of $y = f(x)$, observe the behavior at the endpoints, notably,

• At $x = 0$, $f(0) = 0$.
• At $x = 1$, $f(1) = \pi$.

Evaluate the critical points found in the previous step. From the calculation, $f'(x)$ changes signs, indicating local maxima or minima. The graph is continuous in $[0, 1]$ and returns satisfactory smooth behavior with careful attention at critical points.