Let's analyze the function $$f(x) = (\cos x) - x + 1$$ over the interval $$[0, \pi]$$ and the statements provided.

First, let's consider statement (S1):

(S1) $$f(x)=0$$ for only one value of $$x$$ in $$[0, \pi]$$.

To examine this statement, we need to explore the zeros of the function $$f(x)$$ within the given interval. Let's define and analyze the function:

$$f(x) = \cos x - x + 1$$

We seek to determine if $$f(x) = 0$$ has only one solution in the interval $$[0, \pi]$$. To do this, we can use the Intermediate Value Theorem and the behavior of the function's derivative. First, compute the derivative of $$f(x)$$:

$$f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1$$

The critical points occur when $$f'(x) = 0$$:

$$- \sin x - 1 = 0 \Rightarrow \sin x = -1$$.

The equation $$\sin x = -1$$ does not hold for any $$x$$ in $$[0, \pi]$$. Note that:

• For $$x \in [0, \frac{\pi}{2}]$$, $$\sin x$$ ranges from 0 to 1.


• For $$x \in [\frac{\pi}{2}, \pi]$$, $$\sin x$$ ranges from 1 to 0.

Since $$f'(x)$$ is always negative (i.e., $$f'(x) < 0$$), $$f(x)$$ is a strictly decreasing function in $$[0, \pi]$$. Moreover, we evaluate:

$$f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2$$

$$f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi$$

Given the continuous and strictly decreasing nature of $$f(x)$$ in $$[0, \pi]$$, by the Intermediate Value Theorem, there is exactly one value of $$x$$ in the interval $$[0, \pi]$$ where $$f(x) = 0$$, confirming (S1).

Now, let's consider statement (S2):

(S2) $$f(x)$$ is decreasing in $$\left[0, \frac{\pi}{2}\right]$$ and increasing in $$\left[\frac{\pi}{2}, \pi\right]$$.

We already analyzed that $$f'(x)$$ shows that $$f'(x) = -\sin x - 1$$ is always less than zero in $$[0, \pi]$$. No interval exists where the derivative is positive. This means that $$f(x)$$ is strictly decreasing throughout the entire interval of $$[0, \pi]$$, invalidating (S2).

Therefore, the correct option is:

Option B: Only (S1) is correct.