Let $f:\mathrm{R}\to \mathrm{R}$ be a differentiable function such that $\mathrm{f}\left(\frac{\mathrm{\pi }}{4}\right)=\sqrt{2}, \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=0\text{ and }{\mathrm{f}}^{\mathrm{\prime }}\left(\frac{\mathrm{\pi }}{2}\right)=1$ and let$\mathrm{g}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^{\mathrm{\pi }/4} \left({\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{t}\right)\sec \mathrm{t}+\tan \sec \mathrm{tf}\left(\mathrm{t}\right)\right)\mathrm{dt}$ for $\mathrm{x}\in \left[\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)$. Then $\underset{\mathrm{x}\to {\left(\frac{\mathrm{\pi }}{2}\right)}^{-}}{lim} \mathrm{g}\left(\mathrm{x}\right)$  is equal to :

$\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^{\mathrm{\pi }/4} \left({\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{t}\right)\sec \mathrm{t}+\tan \mathrm{tsec}\mathrm{tf}\left(\mathrm{t}\right)\right)\mathrm{dt}\\ \mathrm{g}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^{\mathrm{\pi }/4} \mathrm{d}\left(\mathrm{f}\right(\mathrm{t})\cdot \sec \mathrm{t})={\left.\mathrm{f}\left(\mathrm{t}\right)\sec \mathrm{t}\right|}_{\mathrm{x}}^{\mathrm{\pi }/4}\\ \\ \mathrm{g}\left(\mathrm{x}\right)=2-\mathrm{f}\left(\mathrm{x}\right)\sec \mathrm{x}=2-\left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\cos \mathrm{x}}\right)\\ \underset{\mathrm{x}\to {\left(\frac{\mathrm{\pi }}{2}\right)}^{-}}{\lim } \mathrm{g}\left(\mathrm{x}\right)=2-\underset{\mathrm{x}\to {\left(\frac{\mathrm{\pi }}{2}\right)}^{-}}{lim} \left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\cos \mathrm{x}}\right)\end{array}$

$\begin{array}{l}=2-\underset{\mathrm{x}\to {\left(\frac{\mathrm{\pi }}{2}\right)}^{-}}{lim} \frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)}{(-\sin \mathrm{x})} \left(by L-Hospital rule\right)\\ =2+\frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\frac{\mathrm{\pi }}{2}\right)}{\sin \frac{\mathrm{\pi }}{2}}=2+\frac{1}{1}=3\end{array}$