If $\int\limits_0^x f(t) d t=x+\int\limits_x^1 t f(t) d t$, then the value of $f(1)$, is
Answer & explanation
Correct answer: option 1
We have,
$\int\limits_0^x f(t) d t=x+\int\limits_x^1 t f(t) d t$
$\Rightarrow \frac{d}{d x}\left\{\int\limits_0^x f(t) d t\right\}=\frac{d}{d x}\left\{x+\int\limits_x^1 t f(t) d t\right\}$
$\Rightarrow f(x)=1+0-x f(x)$ [Using Leibnitz's rule]
$\Rightarrow f(x)=1-x f(x)$
$\Rightarrow f(x)=\frac{1}{x+1} \Rightarrow f(1)=\frac{1}{2}$