Question

Let $f\left(x\right)$ be defined on $[0,\pi ]$ by $f\left(x\right)=\{\begin{array}{cc}x+a\sqrt{2}\sin x& ,0\le x\le \pi /4\\ 2x\cot x+b& ,\frac{\pi }{4}<x\le \frac{\pi }{2}\\ a\cos 2x-b\sin x& ,\frac{\pi }{2}<x<\pi \end{array}$. If $f$ is continuous on $[0,\pi ]$ then

• A. $a=\frac{\pi }{6}$
• B. $b=-\frac{\pi }{12}$
• C. $a=\frac{\pi }{6}$ and $b=-\frac{\pi }{12}$
• D. $a=\frac{\pi }{3}$ and $b=-\frac{\pi }{12}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a=\frac{\pi }{6}$
B $b=-\frac{\pi }{12}$
C $a=\frac{\pi }{6}$ and $b=-\frac{\pi }{12}$

$f\left(\pi /4\right)=\frac{\pi }{4}+a$. Also
$f\left(\pi /4\right)=\underset{x\to \pi /4^{+}}{lim}(2x\cot x+b)$
$=\frac{\pi }{2}+b$
so $a-b=\pi /4$
Also $f\left(\pi /2\right)=b$ and $f\left(\pi /2\right)=\underset{x\to \pi /2^{+}}{lim}=\underset{x\to \pi /2}{lim}(a\cos 2x-b\sin x)$
$=-a-b$
so $2b=-a$ Hence $a=\pi /6$ and $b=-\pi /12$