Naming a calculus question is hard than solving it!

The given equation can be distributed over the subtraction sign. So the equation becomes:- $f(x) = \int_{0}^{x} f(t)\cos (t)dt - \int_{0}^{x}\cos (t-x)dt$ Solve the second integral :- $\int_{0}^{x}\cos(t-x)dt$ Put $t-x = k$ (Note that in the integral x is constant and hence will be treated as such. ) $\implies dt=dk$ Limits will change as when $t=0, k=-x$

when $t=x,k=0$ .

So the integral becomes $\int_{-x}^{0}\cos kdk$ $=-\int_{0}^{-x}\cos kdk$ Now in the original equation $f(x) = \int_{0}^{x} f(t)\cos(t)dt + \int_{0}^{-x} \cos(k) dk$ Differentiating the equation wrt x, we get $f'(x) = f(x) cos(x) + cos(-x)(-1)$ $\dfrac{f'(x)}{f(x)-1}=cos(x)$ Integrating we get $ln(|f(x)-1|) = sin(x)+c$ But the equation passes through (0,0). (This can be easily seen in the given equation.).

Keeping (0,0) in the equation we find that:-

We have to use the negative sign of the mod sign and c = 0. So we can write $f(x) = 1-e^{sin(x)}$ Differentiating we will get:- $f'(x) = -cos(x) e^{sin(x)}$ Put x= 0 $f'(0) = -(1)(1)$ $f'(0)= - 1$ Hence the option $f'(0) = 1$ is wrong.