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Let $\mathrm{f: \mathbf{R} \rightarrow \mathbf{R}}$ be a differentiable function such that $\mathrm{f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0\: and \: \: f^{\prime}\left(\frac{\pi}{2}\right)=1}$ and let $\mathrm{g(x)=\int_{x}^{\pi / 4}\left(f^{\prime}(\mathrm{t})\right.sect \left.+\tan t \operatorname{sect} f(\mathrm{t})\right) dt \: \: for \: \: x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right).}$ Then $\lim_{x\rightarrow\left ( \frac{\pi }{2} \right )^{-} }g(x)$ is equal to:
Option: 1
$2$

Option: 2
$3$

Option: 3
$4$

Option: 4
$-3$

Answers (1)

best_answer

$\mathrm{\text { Since } \frac{d}{d x}(f(x) \sec x)=f^{\prime}(x) \sec x+f(x) \sec x \tan x}$

$\mathrm{\therefore g(x) =\int_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \cdot \sec t \cdot f(t)\right) d t} \\$

$\mathrm{=\left.f(t) \cdot \sec (t)\right|_{x} ^{\pi / 4}}$

$\mathrm{=f\left(\frac{\pi}{4}\right) \cdot \sec \left(\frac{\pi}{4}\right)-f(x) \sec (x)} \\$

$\mathrm{=\sqrt{2} \cdot \sqrt{2}-f(x) \cdot \sec x} \\$

$\mathrm{=2-f(x) \cdot \sec x}$

$\mathrm{\text { Now } \lim _{x \rightarrow \frac{\pi}{2}^{-}} g(x)=2-\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x) \cdot \sec x}$

$\mathrm{=2-\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{f(x)}{\cos x}}$

$\mathrm{Using\: L'H \: Rule }$

$\mathrm{=2-\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{f^{\prime}(x)}{-\sin x}}$

$\mathrm{=2-\frac{f^{\prime}(\pi / 2)}{-1}} \\$

$\mathrm{=2+1 }\\$

$\mathrm{=3}$

Hence the correct answer is option 2

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