Question

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

• A. $32$
• B. $\frac{32}{3}$
• C. $\frac{32}{9}$
• D. $35$

Answer 1

Answer: (C)

$\frac{32}{9}$

$F\left(x\right)=\frac{1}{x^2}{\int }_4^x(4t^2-F^{{}^{\prime }}\left(t\right))dt$
$F^{{}^{\prime }}\left(x\right)=-\frac{2}{x^3}{\int }_4^x(4t^2-{2F}^{{}^{\prime }}\left(t\right))dt+\frac{1}{x^2}\frac{d}{dx}{\int }_4^x(4t^2-{2F}^{{}^{\prime }}\left(t\right))dt$
$=-\frac{2}{x^3}{\int }_4^x(4t^2-{2F}^{{}^{\prime }}\left(t\right))dt+\frac{1}{x^2}\left((4x^2-{2F}^{{}^{\prime }}\left(x\right))\right)$
Put $x=4$,
$F^{{}^{\prime }}\left(4\right)=-\frac{2}{64}{\int }_4^4(4t^2-{2F}^{{}^{\prime }}\left(t\right))dt+\frac{1}{16}(64-{2F}^{{}^{\prime }}\left(4\right))=4-\frac{1}{8}{F}^{{}^{\prime }}\left(4\right)$
$\Rightarrow \frac{9}{8}{F}^{{}^{\prime }}\left(4\right)=4\Rightarrow F^{{}^{\prime }}\left(4\right)=\frac{32}{9}$