Let f : R $\to$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $f'\left( {{\pi \over 2}} \right) = 1$ and

let $g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}$ for $x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$. Then $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$ is equal to :

Given : $f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$ <br/><br/> $$ \begin{aligned} &g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\\\ &=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x) \end{aligned} $$ <br/><br/> Now, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$ <br/><br/> $$ =\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right) $$ <br/><br/> $=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$ <br/><br/> $=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$ <br/><br/> $=3$