Answer: (i) At \(P\), \(2\theta\tan\theta=1\), with a root between \(0.6\) and \(0.7\).

(ii) \(\theta=\dfrac{\pi}{4}\).

(iii) \(C_1\) starts at \(O\), lies outside \(C_2\), and meets \(C_2\) again at \(Q\).

(iv) The enclosed area is \(\dfrac14\ln 2+\dfrac{\pi}{8}\left(\dfrac{\pi}{4}-1\right)\).

(i) The line \(\theta=\dfrac{\pi}{2}\) is the positive \(y\)-axis. For a point with polar coordinates \((r,\theta)\), the perpendicular distance from this line is

\(x=r\cos\theta\).

On \(C_1\), \(r^2=2\theta\), so \(r=\sqrt{2\theta}\). We maximise

\(x=\sqrt{2\theta}\cos\theta\).

Differentiate:

\(\dfrac{\mathrm d}{\mathrm d\theta}\left(\sqrt{2\theta}\cos\theta\right)=\sqrt2\left(-\theta^{1/2}\sin\theta+\dfrac12\theta^{-1/2}\cos\theta\right)\).

At the furthest point this is zero, so

\(-\theta^{1/2}\sin\theta+\dfrac12\theta^{-1/2}\cos\theta=0\).

Multiplying by \(2\theta^{1/2}\) gives

\(\cos\theta=2\theta\sin\theta\),

and hence

\(2\theta\tan\theta=1\).

Let \(f(\theta)=2\theta\tan\theta-1\). Then

\(f(0.6)=2(0.6)\tan(0.6)-1\approx -0.179\),

whereas

\(f(0.7)=2(0.7)\tan(0.7)-1\approx 0.179\).

Since the signs are different, the equation has a root between \(0.6\) and \(0.7\).

(ii) At \(Q\), the two curves have the same value of \(r\), and \(\theta\neq0\). Therefore

\(2\theta=\theta\sec^2\theta\).

Since \(\theta\neq0\), divide by \(\theta\):

\(\sec^2\theta=2\).

Thus \(\cos^2\theta=\dfrac12\). In the interval \(0\leq\theta\leq\dfrac{\pi}{4}\), this gives

\(\theta=\dfrac{\pi}{4}\).

(iii) The curve \(C_1\) starts at \(O\) when \(\theta=0\). For \(0<\theta<\dfrac{\pi}{4}\),

\(2\theta>\theta\sec^2\theta\),

so \(C_1\) has the larger radius and lies outside \(C_2\). It then meets \(C_2\) at \(Q\), where \(\theta=\dfrac{\pi}{4}\). The sketch should therefore show a smooth curve from \(O\) to \(Q\), outside \(C_2\), with the second intersection at \(Q\).

(iv) The enclosed area is

\(\dfrac12\int_0^{\pi/4}\left(r_1^2-r_2^2\right)\,\mathrm d\theta\).

Using \(r_1^2=2\theta\) and \(r_2^2=\theta\sec^2\theta\),

\(\text{Area}=\dfrac12\int_0^{\pi/4}\left(2\theta-\theta\sec^2\theta\right)\,\mathrm d\theta\).

Write the integrand as \(\theta(2-\sec^2\theta)\). Integrating by parts with \(u=\theta\) and \(\mathrm dv=(2-\sec^2\theta)\,\mathrm d\theta\), we have \(v=2\theta-\tan\theta\). Hence

\(\text{Area}=\dfrac12\left[\theta(2\theta-\tan\theta)\right]_0^{\pi/4}-\dfrac12\int_0^{\pi/4}(2\theta-\tan\theta)\,\mathrm d\theta\).

Now

\(\int(2\theta-\tan\theta)\,\mathrm d\theta=\theta^2+\ln(\cos\theta)\).

Therefore

\(\text{Area}=\dfrac12\left[\theta(2\theta-\tan\theta)-\theta^2-\ln(\cos\theta)\right]_0^{\pi/4}\).

Substitute \(\theta=\dfrac{\pi}{4}\), \(\tan\dfrac{\pi}{4}=1\), and \(\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}\):

\(\text{Area}=\dfrac12\left(\dfrac{\pi}{4}\left(\dfrac{\pi}{2}-1\right)-\dfrac{\pi^2}{16}+\dfrac12\ln2\right)\).

So

\(\text{Area}=\dfrac14\ln2+\dfrac{\pi}{8}\left(\dfrac{\pi}{4}-1\right)\).