The values of $a$ and $b$, so that the function $\mathrm{f}(x)= \begin{cases}x+\mathrm{a} \sqrt{2} \sin x & …
The values of $a$ and $b$, so that the function
$\mathrm{f}(x)= \begin{cases}x+\mathrm{a} \sqrt{2} \sin x & , 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+\mathrm{b} & , \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ \mathrm{a} \cos 2 x-\mathrm{b} \sin x & , \frac{\pi}{2} \lt x \leq \pi\end{cases}$
is continuous for $0 \leq x \leq \pi$, are respectively given by
$+\frac{\pi}{12},-\frac{\pi}{6}$
$-\frac{\pi}{6},-\frac{\pi}{12}$
$\frac{\pi}{6}, \frac{\pi}{12}$
$\frac{\pi}{6},-\frac{\pi}{12}$
Solution
As the given function is continuous at $x=\frac{\pi}{4}$ and
$\frac{\pi}{2}$, we get
$\lim _{x \rightarrow \frac{\pi^{-}}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}} \mathrm{f}(x)$
$\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{4}}(x+a \sqrt{2} \sin x)=\lim _{x \rightarrow \frac{\pi}{4}}(2 x \cot x+b) \\
& \therefore \quad \frac{\pi}{4}+a=\frac{2 \pi}{4}+b \\
& \therefore \quad a-b=\frac{\pi}{4}...(i)
\end{aligned}$
Also, $\lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x)$
$\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{2}}(2 x \cot x+b)=\lim _{x \rightarrow \frac{\pi}{2}}(a \cos 2 x-b \sin x) \\
& \therefore \quad 0+b=-a-b \\
& \therefore \quad a+2 b=0...(ii)
\end{aligned}$
Solving equations (i) and (ii), we get $a=\frac{\pi}{6}$ and $b=\frac{-\pi}{12}$