Let f: R → R be a differentiable function such that f((π)/(4)) = 2...

Given,

$g\left(x\right) = {∫}_{x}^{\dfrac{\pi}{4}}\left({f}^{'}\left(t\right)\sec t + \tan t \sec tf\left(t\right)\right)dt$

$⇒ g\left(x\right) = {∫}_{x}^{\dfrac{\pi}{4}}d\left(f\left(t\right) \cdot \sec t\right)$⇒ g\left(x\right) = {\left[f\left(t\right)\sec t\right]}_{x}^{\dfrac{\pi}{4}}$</p><p>$g\left(x\right) = f\left(\dfrac{\pi}{4}\right)sec\dfrac{\pi}{4} - f\left(x\right) \cdot \sec x$</p><p>$g\left(x\right) = 2 - f\left(x\right)\sec x = 2 - \left(\dfrac{f\left(x\right)}{\cos x}\right)$</p><p>Now taking limit both side, we get</p><p>$\lim_{x \to {\left(\dfrac{\pi}{2}\right)}^{-}}g\left(x\right) = 2 - \lim_{x \to {\left(\dfrac{\pi}{2}\right)}^{-}}\left(\dfrac{f\left(x\right)}{\cos x}\right)$</p><p>Using L'Hospital Rule</p><p>$= 2 - \lim_{x \to {\left(\dfrac{\pi}{2}\right)}^{-}}\dfrac{{f}^{'}\left(x\right)}{\left(- \sin x\right)}$</p><p>$= 2 + \dfrac{{f}^{'}\left(\dfrac{\pi}{2}\right)}{sin\dfrac{\pi}{2}} = 2 + \dfrac{1}{1} = 3$