A differentiable function satisfies $f\left(x\right)={\int }_o^x\left\{f\right(t)\cos t-\cos (t-x\left)\right\}dt.$ which is of the following hold good?

$f\left(x\right)={\int }_0^x\left\{f\right(t)\cos t-\cos (t-x\left)\right\}dx$

$={\int }_0^{x}f\left(t\right)\cos tdt-{\int }_0^x\cos (-t)dt\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx\right]$

$f\left(x\right)={\int }_0^{x}f\left(t\right)\cos tdt-\sin x$

Differentiating both the sides, we get

$f^{\text{'}}\left(x\right)=f\left(x\right)\cos x-\cos x$

Let             $f\left(x\right)=y;f^{\text{'}}\left(x\right)=\frac{dy}{dx}$

$\frac{dy}{dx}-y\cos x=-\cos x$

$IF=e^{-\int \cos xdx}=e^{-\sin x}$

Therefore, $y.e^{-\sin x}=-\int e^{-\sin x}\cos xdx;$

$y.e^{-\sin x}=c+e^{-\sin x};y=C{e}^{\sin x}+1$

If  x=0;                       y=0                   (from the given relation)

$\Rightarrow c=-1$

Therefore $f\left(x\right)=1-e^{\sin x}$

Now, minimum value =1-e

Maximum value     $=1-e^{-1}$

$f^{\text{'}}\left(x\right)=-e^{\sin x}\cos x$

Therefore,           $f^{\text{'}}\left(o\right)=-1$

$f^{\text{'}\text{'}}\left(x\right)=-[{\cos }^{2}x.e^{\sin x}-e^{\sin x}.\sin x]$

$f^{\text{'}\text{'}}\left(\frac{\pi }{2}\right)=e$

Hence, (a),(b)and (c) are the correct answer s.