Question

If the function $f(x)=\begin{cases} x+a\sqrt{2}\sin x & 0\leq x < \frac{\pi }{4} \\ 2x\cot x+b & \frac{\pi }{4}\leq x\leq \frac{\pi }{2} \\ a \cos 2x-b \sin x & \frac{\pi }{4} < x\leq \pi \end{cases}$ is continuous in $\left[0 , \pi \right]$ , then the values of $a$ and $b$ respectively are

• A. $\frac{\pi }{6},-\frac{\pi }{12}$
• B. $-\frac{\pi }{6},\frac{\pi }{4}$
• C. $-\frac{\pi }{3},\frac{\pi }{12}$
• D. None of these

Answer 1

Answer: (A)

$\because f(x)$ is continuous in $[0, \pi]$ so it is continuous at $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2}$
$\therefore \displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}} f(x)=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}} f(x)$
$\Rightarrow \frac{\pi}{4}+a=\frac{\pi}{2}+b \ldots(1)$
and $\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f(x)=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f(x)$
$\Rightarrow 0+b=-a-b \ldots(2)$
By solving (1) and (2), we get,
$a=\frac{\pi}{6}, b=-\frac{\pi}{12}$