Solve this

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

We have

$\left(\mathrm{LHL}\right.$ at $\left.x=\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)$

$=\lim _{h \rightarrow 0}\left(\frac{1-\sin ^{3}\left(\frac{x}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{1-\cos ^{3} h}{3 \sin ^{2} h}\right)$

$=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{(1-\cos h)\left(1+\cos ^{2} h+\cos h\right)}{(1-\cos h)(1+\cos h)}\right)$

$=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{\left(1+\cos ^{2} h+\cos h\right)}{(1+\cos h)}\right)$

$=\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2}$

$\left(\mathrm{RHL}\right.$ at $\left.x=\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)$

$=\lim _{h \rightarrow 0}\left(\frac{b\left[1-\sin \left(\frac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\frac{\pi}{2}+h\right)\right]^{2}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{b(1-\cos h)}{[-2 h]^{2}}\right)$

$=\lim _{h \rightarrow 0}\left(\frac{2 b \sin ^{2} \frac{h}{2}}{4 h^{2}}\right)$

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

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

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

$=\frac{b}{8}$

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

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

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

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

$\Rightarrow a=\frac{1}{2}$ and $b=4$