Rolle's Theorem

Consider  $$g(x)=f(x)e^{-x}$$

Since

$$f$$is twice differentiable,

$$g$$ is continuous on $$[0,\pi]$$ and differentiable on $$(0,\pi).$$

Also,

$$g(0)=f(0)e^0=0$$ and $$g(\pi)=f(\pi)e^{-\pi}=0$$

Hence,

$$g(0)=g(\pi)$$

Therefore,

$$g(x)$$ satisfies Rolle’s theorem on $$[0,\pi].$$

So, statement ($$g(x)=f(x)e^{-x}$$ satisfies Rolle’s theorem on $$[0,\pi]$$) is correct.

Now,

$$g'(x)=e^{-x}(f'(x)-f(x))$$

By Rolle’s theorem, there exists $$c\in(0,\pi)$$ such that $$g'(c)=0$$

Therefore,

$$e^{-c}(f'(c)-f(c))=0$$

Since

$$e^{-c}\neq0,$$

$$f'(c)=f(c)$$

Hence, statement (There exists at least one $$c\in(0,\pi)$$ such that $$f'(c)=f(c)$$) is correct.

Given,

$$f''(x)-3f'(x)+2f(x)=e^x\cos x$$

At $$x=\frac\pi2,$$

$$e^{\pi/2}\cos\frac\pi2=0$$

Therefore,

$$f''\left(\frac\pi2\right)-3f'\left(\frac\pi2\right)+2f\left(\frac\pi2\right)=0$$

Hence,statement (There exists at least one $$c\in(0,\pi)$$ such that $$f''(c)-3f'(c)+2f(c)=0$$) is correct.

Therefore,all the given statements are correct