Let f:RR be a differentiable function such that f(4)=√2, f(2)=0 and f'(2)=1, and let g(x)=…

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be a differentiable function such that $f\left(\dfrac\pi4ight)=\sqrt2$, $f\left(\dfrac\pi2ight)=0$ and $f'\left(\dfrac\pi2ight)=1$, and let $g(x)=\displaystyle\int_{\pi/4}^x\big(f'(t)\sec t+	an t\sec t\,f(t)\big)\,dt$ for $x\in\left(\dfrac\pi4,\dfrac\pi2ight)$. Then $\displaystyle\lim_{x	o(\pi/2)^-}g(x)$ is equal to:

Accepted Answer

Given : $f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$

$$
\begin{aligned}
&g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\
&=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x)
\end{aligned}
$$

Now, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$

$$
=\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right)
$$

$=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$

$=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$

$=3$

Solutions · 1