Question

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

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

Answer 1

Answer: (A)

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