Question

If the function $f(x)=\{\begin{array}{ll}x+a\sqrt{2}\sin x& 0\le x<\frac{\pi }{4}\\ 2x\cot x+b& \frac{\pi }{4}\le x\le \frac{\pi }{2}\\ a\cos 2x-b\sin x& \frac{\pi }{4}<x\le \pi \end{array}$ 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 $x=\frac{\pi }{2}$
$\therefore \underset{x\to {\left(\frac{\pi }{4}\right)}^{-}}{\lim }f(x)=\underset{x\to {\left(\frac{\pi }{4}\right)}^{+}}{\lim }f(x)$
$\Rightarrow \frac{\pi }{4}+a=\frac{\pi }{2}+b\dots (1)$
and $\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }f(x)=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{\lim }f(x)$
$\Rightarrow 0+b=-a-b\dots (2)$
By solving (1) and (2), we get,
$a=\frac{\pi }{6},b=-\frac{\pi }{12}$