Question

Let $f:R\to R$ 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 $g(x)=\underset{x}{\overset{\pi /4}{\int }}\left(f^{\prime }(t)\sec t+\tan t\sec tf(t)\right)dt$ for $x\in \left[\frac{\pi }{4},\frac{\pi }{2}\right)$. Then $\underset{x\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }g(x)$ is equal to

• A. 2
• B. 3
• C. 4
• D. -3

Answer 1

Answer: (B)

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