Question

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

• A. 2
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

$F(x)=\int\limits_{-1}^x f(t) d t=\int\limits_1^x f(t) d t$
$G(x)=\int\limits_{-1}^x t|f(f(t))| d t=\int\limits_{-1}^x t|f(f(t))| d t$
$\displaystyle\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}$
$L^{\prime}$ hospitals $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)}{x|f(f(x))|}=\frac{1}{14}$
$\frac{\frac{1}{2}}{1\left|f\left(\frac{1}{2}\right)\right|}=\frac{1}{14}$
$f\left(\frac{1}{2}\right)=7$