Question

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