- Let \(a > 0\). Sketch the graph of

\[y=\dfrac{a+x}{a-x} \quad \text{for} \quad -a < x < a.\]

Now \(y=\dfrac{a+x}{a-x} = \dfrac{2a-(a-x)}{a-x} = \dfrac{2a}{a-x}-1\), so as \(x\) increases from \(-a\) towards \(a, y\) increases from \(0\) towards \(\infty\). Also \((0,1)\) is on the curve.

If we use calculus, \(\dfrac{dy}{dx} = \dfrac{2a}{(a-x)^2} > 0\), and so this function is always increasing (with a discontinuity at \(x = a\)).

- Let \(0 < \theta < \dfrac{\pi}{2}\). In the diagram below is the half-disc given by \(x^2 + y^2 \leq 1\) and \(y \geq 0\).

The shaded region \(A\) consists of those points with \(-\cos \theta \leq x \leq \sin \theta\). The region \(B\) is the remainder of the half-disc.

Find the area of \(A\).

We can see from the diagram that A consists of a quarter of the circle with two additional congruent triangles.

Thus the area \(A = \dfrac{\pi}{4}+2\left[\dfrac{1}{2}\sin\theta\cos\theta\right] = \dfrac{\pi}{4}+\sin\theta\cos\theta\).

- Assuming
*only*that \(\sin^2\theta + \cos^2\theta = 1\), show that \(\sin\theta\cos\theta \leq \dfrac{1}{2}\).

We know that \(0\leq (\sin\theta-\cos\theta)^2 = \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta.\)

Thus \(2\sin\theta\cos\theta \leq \sin^2\theta + \cos^2\theta = 1 \implies \sin\theta\cos\theta \leq \dfrac{1}{2}.\)

- What is the largest \(\dfrac{\text{area of } A}{\text{area of } B}\) can be, as \(\theta\) varies?

We know \(\dfrac{\text{area of } A}{\text{area of } B} = \dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+\sin\theta\cos\theta\right)}= \dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}\).

Now from part (i), we know that \(\dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}\) increases as \(\sin\theta\cos\theta\) increases.

But the maximum value \(\sin\theta\cos\theta\) can take is \(\dfrac{1}{2}\), and so the maximum value of \(\dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}\) is \(\dfrac{\dfrac{\pi}{4}+\dfrac{1}{2}}{\dfrac{\pi}{4}-\dfrac{1}{2}} = \dfrac{\pi+2}{\pi-2}\).