Question

If $f\left(x\right)=\frac{1}{x^2}{\int }_4^x(4t^2-2f^{\prime }(t\left)\right)dt$, then $f^{\prime }\left(4\right)$ equals

• A. $32$
• B. $\frac{32}{3}$
• C. $\frac{32}{9}$
• D. none of these

Answer 1

The correct option is C $\frac{32}{9}$

Given, $f\left(x\right)=\frac{1}{x^2}{\int }_4^x(4t^2-2f^{\prime }(t\left)\right)dt$
On differentiating once,$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2}(4x^2-2f^{\prime }(x\left)\right)-0)-\frac{2}{x^3}{\int }_4^x(4t^2-2f^{\prime }(t\left)\right)dt$
Given, $x=4$ on substituting.
$\Rightarrow f^{\prime }\left(4\right)=\frac{1}{4^2}(4\ast 4^2-2f^{\prime }(4\left)\right)-0)-\frac{2}{4^3}{\int }_4^4(4\ast 4^2-2f^{\prime }(4\left)\right)dt$
$\Rightarrow f^{\prime }\left(4\right)=\frac{1}{4^2}(64-2f^{\prime }(4\left)\right)-0$
$\Rightarrow 18f^{\prime }\left(4\right)=64$
$\Rightarrow f^{\prime }\left(4\right)=\frac{32}{9}$
Option C