Question

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

Answer 1

Given: f is continuous on $\left[0,\mathrm{\pi }\right]$.

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


At x = $\frac{\mathrm{\pi }}{4}$, we have

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

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


At x = $\frac{\mathrm{\pi }}{2}$, we have

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

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

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

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

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