Question

The value of $a$ and $b$ respectively so that the function
$f\left(x\right)=\{\begin{array}{cc}x+a\sqrt{2}\sin x,& 0\le x<\frac{\pi }{4}\\ 2x\cot x+b,& \frac{\pi }{4}\le x\le \frac{\pi }{2}\\ a\cos 2x-b\sin x,& \frac{\pi }{2}<x\le \pi \end{array}$
is continuous for $x\in [0,\pi ]$ is

• A. $\frac{\pi }{6},\frac{\pi }{12}$
• B. $\frac{\pi }{6},\frac{-\pi }{12}$
• C. $\frac{-\pi }{4},\frac{\pi }{12}$
• D. $\frac{\pi }{4},\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{6},\frac{-\pi }{12}$

$f\left(x\right)=\{\begin{array}{cc}x+a\sqrt{2}\sin x,& 0\le x<\frac{\pi }{4}\\ 2x\cot x+b,& \frac{\pi }{4}\le x\le \frac{\pi }{2}\\ a\cos 2x-b\sin x,& \frac{\pi }{2}<x\le \pi \end{array}$

At $x=\frac{\pi }{4},$
$LHL=\underset{x\to \pi /4^{-}}{lim}f\left(x\right)$
$=\underset{x\to \pi /4^{-}}{lim}(x+a\sqrt{2}\sin x)=\frac{\pi }{4}+a$

$RHL=\underset{x\to \pi /4^{+}}{lim}f\left(x\right)$
$=\underset{x\to \pi /4^{+}}{lim}(2x\cot x+b)=\frac{\pi }{2}+b$
$f\left(\frac{\pi }{4}\right)=2\times \frac{\pi }{4}\cot \left(\frac{\pi }{4}\right)+b=\frac{\pi }{2}+b$

For continuity,
$LHL=RHL=f\left(\frac{\pi }{4}\right)$
$\Rightarrow \frac{\pi }{4}+a=\frac{\pi }{2}+b$
$\Rightarrow a-b=\frac{\pi }{4}...\left(1\right)$

Now, at $x=\frac{\pi }{2}$
$LHL=\underset{x\to \pi /2^{-}}{lim}f\left(x\right)$
$=\underset{x\to \pi /2^{-}}{lim}(2x\cot x+b)=0+b$
$RHL=\underset{x\to \pi /2^{+}}{lim}(a\cos 2x-b\sin x)=-a-b$
$f\left(\frac{\pi }{2}\right)=0+b$

So, for continuity,
$f\left(\frac{\pi }{2}\right)=LHL=RHL$
$\Rightarrow a+2b=0...\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$b=-\frac{\pi }{12},a=\frac{\pi }{6}$