Limit Continuity And Differentiability Ques 31
- Let $f: R \rightarrow R$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\frac{1}{2}$.
Suppose that $F(x)=\int_{-1}^x f(t) d t$
for all $x \in[-1,2]$ and $G(x)=\int_{-1}^x t|f\{f(t)\}| d t \quad$
for all $\quad x \in[-1,2]$. If $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$, then the value of $f\left(\frac{1}{2}\right)$ is
(2015 Adv.)
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Answer:
Correct Answer: 31.$(7)$
Solution: Here, $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$
$\Rightarrow \quad \lim _{x \rightarrow 1} \frac{F^{\prime}(x)}{G^{\prime}(x)}=\frac{1}{14} \quad$ [using L’Hospital’s rule]
As $\quad F(x)=\int_{-1}^x f(t) d t \Rightarrow F^v(x)=f(x)$
and
$\Rightarrow \quad G^{\prime}(x)=x|f\{f(x)\}|$
$\therefore \quad \lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\lim _{x \rightarrow 1} \frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim _{x \rightarrow 1} \frac{f(x)}{x \mid f\{f(x)\}}$
$ =\quad \frac{f(1)}{1|f\{f(1)\}|}=\frac{1 / 2}{|f(1 / 2)|} $
Given, $\quad \lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$
$\therefore \quad \frac{\frac{1}{2}}{f\left(\frac{1}{2}\right)}=\frac{1}{14} \Rightarrow\left|f\left(\frac{1}{2}\right)\right|=7$
BETA
