Question

If $\int\limits^{x}_0 f(t) dt = x^2 + \int \limits^1_x t^2 f(t) dt$, 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^{x}_{0} f\left(t\right)dt =x^{2} +\int^{1}_{x} t^{2} f\left(t\right)dt ,\, \, f'\left(\frac{1}{2}\right) =? $
Differentiate w.r.t. 'x'
$f\left(x\right)=2x+0-x^{2}f\left(x\right) $
$f\left(x\right)= \frac{2x}{1+x^{2}} \Rightarrow f'\left(x\right)= \frac{\left(1+x^{2}\right)2-2x\left(2x\right)}{\left(1+x^{2}\right)^{2}}$
$ f'\left(x\right)=\frac{2x^{2}-4x^{2}+2}{\left(1+x^{2}\right)^{2}} $
$ f'\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} $