Question

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

Answer 1

Given: f is continuous on  $$[0,\pi ]$$ .

∴ f is continuous at x =   $$\frac{\pi }{4}$$ and  $$\frac{\pi }{2}$$

At x =  $$\frac{\pi }{4}$$, we have

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

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

At x =  $$\frac{\pi }{2}$$ , we have

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

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

Since f is continuous at x = $$\frac{\pi }{4}$$and x = $$\frac{\pi }{2}$$  we get 

$$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)\text{ and }\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)$$

$$\Rightarrow -b-a=b\text{ and }\frac{\pi }{4}+a=\frac{\pi }{2}+b$$
$$\Rightarrow b=\frac{-a}{2}...\left(1\right)\text{ and }\frac{-\pi }{4}=b-a...\left(2\right)$$
$$\Rightarrow \frac{-\pi }{4}=\frac{-3a}{2}\left[\text{ Substituting the value of b in eq .}\left(2\right)\right]$$
$$\Rightarrow a=\frac{\pi }{6}$$
$$\Rightarrow b=\frac{-\pi }{12}[\text{ From eq }.\left(1\right)]$$