Question

If $F\left(x\right)=\frac{1}{x^2}\underset{4}{\overset{x}{\int }}[4t^2-2F^{\prime }(t\left)\right]dt,$ then $\frac{9F^{\prime }\left(4\right)}{4}$ is

Answer 1

$F\left(x\right)=\frac{1}{x^2}\underset{4}{\overset{x}{\int }}[4t^2-2F^{\prime }(t\left)\right]dt$
$\Rightarrow F^{\prime }\left(x\right)=\frac{1}{x^2}[4x^2-2F^{\prime }(x\left)\right]+\frac{-2}{x^3}\underset{4}{\overset{x}{\int }}[4t^2-2F^{\prime }(t\left)\right]dt\Rightarrow F^{\prime }\left(4\right)=\frac{1}{16}[64-2F^{\prime }(4\left)\right]-\frac{1}{32}\underset{4}{\overset{4}{\int }}[4t^2-2F^{\prime }(t\left)\right]dt$
$\Rightarrow (1+\frac{1}{8})F^{\prime }\left(4\right)=4-0\Rightarrow F^{\prime }\left(4\right)=\frac{32}{9}$
$\therefore \frac{9F^{\prime }\left(4\right)}{4}=8$