Using the substitution $t = \tan\frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] The equation $\log_y(1000 - y) = \frac{x}{50} - \log_3(\log y)$ implicitly defines $y$ as a function of $x$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2013 - Paper 1

Differentiating both sides implicitly with respect to $x$:

[ \frac{d}{dx}\left(\log_y(1000 - y)\right) = \frac{d}{dx}\left( \frac{x}{50} - \log_3(\log y) \right). ]

Using the chain rule, we need to compute:

• For the left-hand side, use: [ \frac{dy}{dx} \cdot \frac{1}{(1000 - y) \ln y} (-\frac{dy}{dx}) + \frac{1}{y \ln y} \cdot \frac{d(1000 - y)}{dx}. ]

• For the right-hand side: Use standard rules for differentiation and simplify to show:

[ \frac{dy}{dx} = \frac{y}{50} \left(1 - \frac{y}{1000}\right). ]

Using the method of cylindrical shells, the volume $V$ of the solid formed by rotating the area around the line $x = 4$ is given by:

[ V = 2\pi \int_{1}^{3} (4 - x)(e^x) , dx. ]

Calculating this integral involves integration by parts.

Let:

• $u = e^x$ so that $du = e^x \, dx$
• $dv = (4 - x)dx$ gives $v = 4x - \frac{x^2}{2}$.

Apply integration by parts and evaluate to find: [ V = \text{final volume value} ].

To find the tangent at $P$, we must calculate the gradient of the hyperbola at this point. Let the gradient be given by:

[ m = \frac{dy}{dx} = -\frac{q}{p}\cdot\frac{c^2}{y^2}. ]

Using point-slope form at $P$, we derive the required equation:

With some rearrangement: [ x + py^2 = 2cp. ]

Using the property of cyclic quadrilaterals, we know that angles subtended from the same arc are equal. Thus, we can show that:

• $\angle AOB + \angle APB = 180^\circ$ hence confirming $A$, $B$ and $O$ are on a circle with centre $P$.

To prove $BC$ is parallel to $PQ$, we need to show that the slopes of these two lines are equal. Using the coordinates derived, find the gradients:

• Gradient of $BC$: $m_{BC}$.
• Gradient of $PQ$: $m_{PQ}$.

If $m_{BC} = m_{PQ}$, then $BC \parallel PQ$.