Question

If ${\int }_0^{x}f\left(t\right)dt=x^2+{\int }_x^1{t}^{2}f\left(t\right)dt$, then $f^{\prime }\left(1/2\right)$ is:

• A. $\frac{6}{25}$
• B. $\frac{24}{25}$
• C. $\frac{18}{25}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{24}{25}$

${\int }_0^{x}f\left(t\right)dt=x^2+{\int }_x^1{t}^{2}f\left(t\right)dt$ $f^{\prime }\left(\frac{1}{2}\right)=?$
Differentiate w.r.t. '$x$'
$f\left(x\right)=2x+0-x^{2}f\left(x\right)$
$f\left(x\right)=\frac{2x}{1+x^2}\Rightarrow f^{\prime }\left(x\right)=\frac{(1+x^2)2-2x\left(2x\right)}{(1+x^2{)}^2}$
$f^{\prime }\left(x\right)=\frac{2x^2-4x^2+2}{(1+x^2{)}^2}$
$f^{\prime }\left(\frac{1}{2}\right)=\frac{2-2\left(\frac{1}{4}\right)}{{(1+\frac{1}{4})}^2}=\frac{\left(\frac{3}{2}\right)}{\frac{25}{16}}=\frac{48}{50}=\frac{24}{25}$