Question

Let \(f:[0, \pi] \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\begin{cases}x+\sqrt{2}\, a \sin x, & 0 \leq x<\frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2}<x \leq \pi\end{cases}\)
If \(f\) is continuous for \(0 \leq x \leq \pi\), then \(a^{2}+140 b^{2}\) is equal to

• A. \(0\)
• B. \(\pi^{2}\)
• C. \(2 \pi^{2}\)
• D. \(\frac{\pi^{2}}{2}\)

Hint:

Impose continuity at the break points \(x=\frac{\pi}{4}\) and \(x=\frac{\pi}{2}\) to get two equations in \(a, b\).

Answer 1

The Correct Option is B

At \(x=\frac{\pi}{4}\): left value \(=\frac{\pi}{4}+\sqrt{2}a\cdot\frac{1}{\sqrt{2}}=\frac{\pi}{4}+a\); right value \(=2\cdot\frac{\pi}{4}\cdot 1+b=\frac{\pi}{2}+b\). So \(a-b=\frac{\pi}{4}\).
At \(x=\frac{\pi}{2}\): middle value \(=2\cdot\frac{\pi}{2}\cdot 0+b=b\); right limit \(=a\cos\pi-b\sin\frac{\pi}{2}=-a-b\). So \(b=-a-b\), i.e. \(a=-2b\).
Substituting: \(-2b-b=\frac{\pi}{4}\), so \(b=-\frac{\pi}{12}\) and \(a=\frac{\pi}{6}\).
\(a^2+140b^2=\frac{\pi^2}{36}+\frac{140\pi^2}{144}=\frac{\pi^2}{36}+\frac{35\pi^2}{36}=\pi^2\).
Answer: (2) \(\pi^{2}\)