Question

Find the values of $a$ and $b$ so that the function $f\left(x\right)$ defined by
$f\left(x\right)=\{\begin{array}{c}x+a\sqrt{2}\sin x,if0\le x<\frac{\pi }{4}\\ 2x\cot x+b,if\frac{\pi }{4}<\frac{\pi }{2}\\ a\cos 2x-b\sin x,if\frac{\pi }{2}\le x\le \pi \end{array}$ becomes continuous on $[0,\pi ]$.

Answer 1

Given,

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

$f\left(x\right)$ is continuous on $[0,\pi ]$.

$\Rightarrow f\left(x\right)$ is continuous at $x=\frac{\pi }{4}$ and $x=\frac{\pi }{2}$.

Now, at $x=\frac{\pi }{4}$, we have

$\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{4}-h)=\underset{h\to 0}{lim}[(\frac{\pi }{4}-h)+a\sqrt{2}\sin (\frac{\pi }{4}-h)]=\frac{\pi }{4}+a\sqrt{2}\sin \frac{\pi }{4}=a+\frac{\pi }{4}$

$\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{4}+h)=\underset{h\to 0}{lim}2(\frac{\pi }{4}+h)\cot (\frac{\pi }{4}+h)+b=\frac{\pi }{2}\cot \frac{\pi }{4}+b=\frac{\pi }{2}+b$

Since $f\left(x\right)$ is continuous at $x=\frac{\pi }{4}$, we have

$\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)$

$\Rightarrow a+\frac{\pi }{4}+\frac{\pi }{2}+b$

$\Rightarrow a-b=\frac{\pi }{4}.....\left(1\right)$

$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)=\underset{h\to 0}{lim}2(\frac{\pi }{2}-h)\cot (\frac{\pi }{2}-h)+b=b$

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)=\underset{h\to 0}{lim}a\cos \left(2(\frac{\pi }{2}+h)\right)-b\sin (\frac{\pi }{2}+h)=-(a+b)$

Since $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, we have

$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)$

$\Rightarrow b=-a-b$

$\Rightarrow b=\frac{-a}{2}.....\left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get

$a=\frac{\pi }{6}$ and $b=-\frac{\pi }{12}$