Question

If ${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt$, then the value of f(1) is [IIT 1998; AMU 2005]

• A. $\frac{1}{2}$
• B. 0
• C. 1
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}$

${\int }_0^{x}dt=x+{\int }_x^{1}tf\left(t\right)dt\Rightarrow {\int }_x^{1}tf\left(t\right)dt=x-{\int }_1^{x}tf\left(t\right)dt$
Differentiating w.r.t x, we get $f\left(x\right)=1+\{0-xf(x\left)\right\}$
$\Rightarrow f\left(x\right)=1-xf\left(x\right)\Rightarrow (1+x)f\left(x\right)=1\Rightarrow f\left(x\right)=\frac{1}{1+x}$
$\therefore f\left(1\right)=\frac{1}{1+1}=\frac{1}{2}$