Limit Continuity And Differentiability Ques 59
- Let $f: R \rightarrow R$ be a differentiable function such that
$f(0)=0, f\left(\frac{\pi}{2}\right)=3$ and $f^{\prime}(0)=1$.
If $g(x)=\int_x^{\frac{\pi}{2}}\left[f^{\prime}(t) \operatorname{cosec} t-\cot t \operatorname{cosec} t f(t)\right] d t$
for $x \in\left(0, \frac{\pi}{2}\right]$, then $\lim _{x \rightarrow 0} g(x)=$
(2017 Adv.)
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Answer:
Correct Answer: 59.$(2)$
Solution: $g(x)=\int_x^{\frac{\pi}{2}}\left(f^{\prime}(t) \operatorname{cosec} t-\cot t \operatorname{cosec} t f(t)\right) d t$
$\therefore \quad g(x)=f\left(\frac{\pi}{2}\right) \operatorname{cosec} \frac{\pi}{2}-f(x) \operatorname{cosec} x$
$\Rightarrow \quad g(x)=3-\frac{f(x)}{\sin x}$
$ \begin{aligned} \lim _{x \rightarrow 0} g(x) & =\lim _{x \rightarrow 0}\left(\frac{3 \sin x-f(x)}{\sin x}\right) \\ & =\lim _{x \rightarrow 0} \frac{3 \cos x-f^{\prime}(x)}{\cos x} \\ & =\frac{3-1}{1}=2 \end{aligned} $
BETA
