Question

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

• A. 2
• B. 3
• C. 4
• D. -3

Answer 1

Answer: (B)

$g ( x )=\int\limits_{ x }^{\pi / 4}\left( f ^{\prime}( t ) \sec t +\tan t \sec tf ( t )\right) dt$
$g(x)=\int\limits_{x}^{\pi / 4} d(f(t) \cdot \sec t)=\left.f(t) \sec t\right|_{x} ^{\pi / 4}$
$g(x)=f\left(\frac{\pi}{4}\right) \sec \frac{\pi}{4}-f(x) \cdot \sec x$
$g(x)=2-f(x) \sec x=2-\left(\frac{f(x)}{\cos x}\right)$
$\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g(x)=2-\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}\left(\frac{f(x)}{\cos x}\right)$
Using L'Hopital Rule
$=2-\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} \frac{f^{\prime}(x)}{(-\sin x)}$
$=2+\frac{f^{\prime}\left(\frac{\pi}{2}\right)}{\sin \frac{\pi}{2}}=2+\frac{1}{1}=3$