(a) The sketch shows a polar graph with \(\theta = 0\) at the pole and the curve extending to \(\theta = \pi\). The point furthest from the pole is \((1, 0)\).

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

\(= \frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi}\)

\(= \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\).

Differentiate and set \(\frac{dy}{d\theta} = 0\):

\((\theta^2 + 1)^{\frac{1}{2}} \cos \theta - \theta (\theta^2 + 1)^{-\frac{1}{2}} \sin \theta = 0\)

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Check for root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.