Question

If $\underset{0}{\overset{x}{\int }}f(t)dt=x^2+\underset{x}{\overset{1}{\int }}{t}^{2}f(t)dt$, then f'(1/2) is

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

Answer 1

Answer: (B)

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