# The function $f(x)=\left\{\begin{array}{1 1}x+a\sqrt 2\sin x&0\leq x < \large\frac{\pi}{4}\\2x\cot x+b&\large\frac{\pi}{4}\normalsize \leq x < \frac{\pi}{2}\\a\cos 2x-b\sin x&\large\frac{\pi}{2}\normalsize < x \leq \pi\end{array}\right.$ is continuous for $0 \leq x \leq \pi$ then a,b are

$\begin{array}{1 1}(a)\;\large\frac{\pi}{6},\frac{\pi}{12}&(b)\;\large\frac{\pi}{3},\frac{\pi}{6}\\(c)\;\large\frac{\pi}{6},-\frac{\pi}{12}&(d)\;\text{None of these}\end{array}$

We apply the test of continuity at $x=\large\frac{\pi}{4}$ and $x=\large\frac{\pi}{2}$ to get the values of a and b.
At $x=\large\frac{\pi}{4}$ LHL =RHL =value
$\Rightarrow \large\frac{\pi}{4}$$+a\sqrt 2\times \large\frac{1}{\sqrt 2} \Rightarrow 2.\large\frac{\pi}{4}$$.(1)+b$
$\Rightarrow a-b=\large\frac{\pi}{4}$------(1)
At $x=\large\frac{\pi}{2}$$,2\big(\large\frac{\pi}{2}\big)$$(0)+b=a(-1)-b(1)$
$\Rightarrow 2b+a=0$------(2)
Solving (1) and (2) we get
$a=\large\frac{\pi}{6}$
$b=-\large\frac{\pi}{12}$
Hence (c) is the correct answer.