MCQ
If $f(x)$ is continuous in $[0, \pi]$, where
$f(x)=\left\{\begin{array}{ll}x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2}, \text { then } \\a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi\end{array}\right.$
  • A
    $a=\frac{\pi}{6}, b=\frac{\pi}{12}$
  • B
    $a=-\frac{\pi}{6}, b=\frac{\pi}{12}$
  • $a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
  • D
    $a=-\frac{\pi}{6}, b=-\frac{\pi}{12}$

Answer

Correct option: C.
$a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
(C)
Sincc $f(x)$ is continuous in $[0, \pi]$.
$\therefore $ it is continuous at $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2}$.
$\therefore \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}}(x+ a \sqrt{2} \sin x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}}(2 x \cot x+ b )$
$\Rightarrow \frac{\pi}{4}+ a \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=2\left(\frac{\pi}{4}\right)(1)+ b$
$\Rightarrow \frac{\pi}{4}+ a =\frac{\pi}{2}+ b$
$\Rightarrow a-b=\frac{\pi}{4}$ $\quad\ldots(i)$
Also, $\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(2 x \cot x+ b )=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}}( a \cos 2 x- b \sin x)$
$\Rightarrow 2\left(\frac{\pi}{2}\right)(0)+ b = a (-1)- b (1)$
$\Rightarrow b =- a - b$
$\Rightarrow a+2 b=0$ $\quad\ldots(ii)$
From (i) and (ii), we get
$a=\frac{\pi}{6}$ and $b=\frac{-\pi}{12}$

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