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

Analyze the given differential inequality.
The inequality is $$f''(x) + 2f'(x) + f(x) \geq 0$$.
$$e^x f''(x) + 2e^x f'(x) + e^x f(x) \geq 0$$
$$\frac{d^2}{dx^2} \left( e^x f(x) \right) \geq 0$$
Let $$g(x) = e^x f(x)$$. The inequality tells us that $$g''(x) \geq 0$$ for all $$x \in (0, 1)$$, which means $$g(x)$$ is a concave upwards \
• $$g(0) = e^0 f(0) = 0$$
• $$g(1) = e^1 f(1) = 0$$
Since $$g(0) = g(1) = 0$$ and $$g(x)$$ is concave upwards ($$g''(x) \geq 0$$), the graph of $$g(x)$$ must lie entirely below or on the secant line joining $$(0,0)$$ and $$(1,0)$$.
Therefore, $$g(x) \leq 0$$ for all $$x \in [0, 1]$$.
Since $$e^x > 0$$, this directly implies that $$f(x) \leq 0$$ for all $$x \in [0, 1]$$.
Given that the minimum value is $$-1$$, it cannot be identically zero, so $$f(x) < 0$$ for all $$x \in (0, 1)$$. (Option A is correct)
Let's apply Rolle's Theorem to $$g(x) = e^x f(x)$$ on $$[0, 1]$$.
Since $$g(0) = g(1) = 0$$, and $$g(x)$$ is differentiable, there must exist at least one $$c \in (0, 1)$$ such that $$g'(c) = 0$$.
$$g'(x) = e^x f'(x) + e^x f(x) = e^x \left( f'(x) + f(x) \right)$$
For $$g'(c) = 0$$, we must have $$f'(c) + f(c) = 0$$ because $$e^c \neq 0$$.
Thus, $$f'(x) + f(x)$$ vanishes at least once in $$(0, 1)$$. (Option B is correct, and Option D is automatically incorrect)

By the property of convexity, the graph of $$g(x)$$ lies above any of its tangent lines. Let the absolute minimum of $$f(x)$$ occur at $$x = x_0$$, where $$f(x_0) = -1$$.
Since $$g(x) = e^x f(x)$$, at any point $$x$$, we can use the mean value properties.
Alternatively, because $$g''(x) \geq 0$$, the function $$g'(x) = e^x(f'(x) + f(x))$$ is non-decreasing.
Since $$g(0)=0$$ and $$g(x) \leq 0$$, the slope must eventually bring it back to $$g(1)=0$$.
Using the property of the minimum value:
$$f(x) \geq \min(f) \cdot e^{1-x} \implies f(x) \geq -e^{1-x}$$
This inequality holds under the given constraints of convexity bounded by 0 at the endpoints.