Answer:
Option A
Explanation:
Here, limx→1F(x)G(x)=114
⇒ limx→1F′(x)G′(x)=114
[ Using L' Hospital's rule ]...........(i)
As F(x)=∫x−1f(t)dt
⇒ F'(x)= f(x) .........(ii)
and G(x)=∫x−1t|f{f(t)}|dt
⇒ G′(x)=x|f{f(x)}| ...........(iii)
∴ \lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\lim_{x \rightarrow 1}\frac{F'(x)}{G'(x)}=\lim_{x \rightarrow 1}\frac{f(x)}{x|f\left\{f(x)\right\}|}
= \frac{f(1)}{1|f\left\{f(1)\right\}|}=\frac{1/2}{|f(1/2)|} ........(iv)
Given, \lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}
\therefore \frac{\frac{1}{2}}{|f\left(\frac{1}{2}\right)|}=\frac{1}{14}\Rightarrow |f\left(\frac{1}{2}\right)|=7