Question

Let $f:R\to R$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\frac{1}{2}$. Suppose that $F(x)=\underset{-1}{\overset{x}{\int }}f(t)dt$ for all $x\in [-1,2]$ and $G(x)=\underset{-1}{\overset{x}{\int }}t|f(f(t))|$ dt for all $x\in [-1,2]$. If $\underset{x\to 1}{\lim }\frac{F(x)}{G(x)}=\frac{1}{14}$, then the value of $f\left(\frac{1}{2}\right)$ is.

• A. 2
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

$F(x)=\underset{-1}{\overset{x}{\int }}f(t)dt=\underset{1}{\overset{x}{\int }}f(t)dt$
$G(x)=\underset{-1}{\overset{x}{\int }}t|f(f(t))|dt=\underset{-1}{\overset{x}{\int }}t|f(f(t))|dt$
$\underset{x\to 1}{\lim }\frac{F(x)}{G(x)}$
$L^{\prime }$ hospitals $\underset{x\to 1}{\lim }\frac{f(x)}{x|f(f(x))|}=\frac{1}{14}$
$\frac{\frac{1}{2}}{1|f\left(\frac{1}{2}\right)|}=\frac{1}{14}$
$f\left(\frac{1}{2}\right)=7$