JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 7)
Let f : R $$\to$$ R be a differentiable function such that $$f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$$ and $$f'\left( {{\pi \over 2}} \right) = 1$$ and
let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :
let $$g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $$ for $$x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$$. Then $$\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$$ is equal to :
2
3
4
$$-$$3
Explanation
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, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$
$$ =\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right) $$
$=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$
$=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$
$$ =3 $$
$$ \begin{aligned} &g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\\\ &=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x) \end{aligned} $$
Now, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$
$$ =\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right) $$
$=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$
$=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$
$$ =3 $$
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