Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},ifx<\frac{\pi }{2}\\ a,ifx=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{(\pi -2x{)}^2},ifx>\frac{\pi }{2}\end{array}$. If $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, find $a$ and $b$.

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x},& \text{if}x<\frac{\pi }{2}\\ a,& \text{if}x=\frac{\pi }{2}\\ \frac{b(1-\sin x)}{{(\pi -2x)}^2},& \text{if}x>\frac{\pi }{2}\end{array}$

$\Rightarrow f\left(\frac{\pi }{2}\right)=a$

LHL at $x=\frac{\pi }{2}$, we have

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

$=\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}\frac{1-{\sin }^3(\frac{\pi }{2}-h)}{3{\cos }^2(\frac{\pi }{2}-h)}$

$=\frac{1}{3}\underset{h\to 0}{lim}\frac{1-{\cos }^{3}h}{{\sin }^{2}h}$

$=\frac{1}{3}\underset{h\to 0}{lim}\frac{(1-\cos h)(1+{\cos }^{2}h+\cos h)}{(1-\cos h)(1+\cos h)}$

$=\frac{1}{3}\times \frac{3}{2}=\frac{1}{2}$

RHL at $x=\frac{\pi }{2}$, we have

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

$=\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)$

$=\underset{h\to 0}{lim}\frac{b(1-\sin (\frac{\pi }{2}+h))}{{(\pi -2(\frac{\pi }{2}+h))}^2}$

$=\underset{h\to 0}{lim}\frac{b(1-\cos h)}{{(-2h)}^2}$

$=\underset{h\to 0}{lim}\frac{2b{\sin }^2(\frac{h}{2})}{16\left(\frac{h^2}{4}\right)}$

$=\frac{b}{8}\underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2$

$=\frac{b}{8}\times 1=\frac{b}{8}$

If $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, then

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

$\Rightarrow \frac{1}{2}=\frac{b}{8}=a$

$\frac{b}{8}=\frac{1}{2}\Rightarrow b=4$

$a=\frac{1}{2}$