If $f(x)$ is a continuous and differentiable function such that the equation $\int\limits_0^x {\sin t.f\left( {x - t} \right)dt} = f\left( x \right) - \sin x$, holds for any real number x, then
(A) $f'\left( 2 \right)=3$
(B) $f'\left( 3 \right)=1$
(C) $f\left( 2 \right)=2$
(D) $f'\left( 1 \right)=0$
My approach is as follow
$\int\limits_0^x {\sin t.f\left( {x - t} \right)dt} = f\left( x \right) - \sin x$
At x=0, $\int\limits_0^0 {\sin t.f\left( {x - t} \right)dt} = f\left( 0 \right) - \sin 0 \Rightarrow f\left( 0 \right) = 0$
$\sin x.f\left( {x - x} \right)\frac{d}{{dx}}x - \sin 0.f\left( {x - 0} \right)\frac{d}{{dx}}0 = f'\left( x \right) - \cos x$
$ \Rightarrow f'\left( x \right) - \cos x = 0 \Rightarrow \frac{{dy}}{{dx}} = \cos x \Rightarrow \int {dy} = \int {\cos xdx} + C \Rightarrow y = \sin x + C$
None of the answer is matching