Answer:
Option A
Explanation:
We have,
$\int f(x) \cos x dx= \frac{1}{2} (f(x))^{2}+C$
On differentiating w.r.t 'x' , we get
$f(x) \cos x=\frac{1}{2} \times 2 f(x).f'(x)$
$\Rightarrow$ $f(x) \cos x= f(x).f'(x)$
$\Rightarrow$ $f'(x)=\cos x$
$\Rightarrow$ $f'(0)= \cos 0$
$\Rightarrow$ $f'(0)=1$