$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$
Where $a, b \in Z$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to ..........
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}=\left(\frac{8}{7}\right)^0=1$
$RHL$ at $\mathrm{x}=\frac{\pi}{2}$
$\lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}$
$=\mathrm{e}^{\left.\lim _{\left.\mathrm{x} \rightarrow \frac{\pi}{2} \right\rvert\, \cot x} \mathrm{~b}\left|\frac{\mathrm{b}}{\mathrm{a}}\right| \tan x \right\rvert\,}=\mathrm{e}^{\frac{\mathrm{b}}{\mathrm{a}}}$
$\Rightarrow 1=\mathrm{a}-8=\mathrm{e}^{\frac{\mathrm{b}}{\mathrm{a}}}$
$\Rightarrow \mathrm{a}=9, \mathrm{~b}=0$
$\Rightarrow a^2+b^2=81$
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$I.$ The range of $f$ is a closed interval.
$II.$ $f$ is continuous on $R$.
$III.$ $f$ is one-one on $R$