For the function $$f(x) = (\cos x) - x + 1, x \in \mathbb{R}$$, between the following two statements (S1) $$f(x) = 0$$ for only one value of $$x$$ in $$[0, \pi]$$. (S2) $$f(x)$$ is decreasing in $$\left[0, \frac{\pi}{2}\right]$$ and increasing in $$\left[\frac{\pi}{2}, \pi\right]$$.

Given ( $$f(x)=\cos x-x+1$$) on$$([0,\pi])$$

(S1): (f(x)=0) has only one solution in ([0,\pi])

Check endpoints:
$$f(0)=1-0+1=2>0,\quad f(\pi)=\cos\pi-\pi+1=-1-\pi+1=-\pi<0$$

Since (f) is continuous, there is at least one root.

Now derivative:
$$f'(x)=-\sin x-1$$

$$Since(\sin x\ge0)on([0,\pi]),$$
$$f'(x)=-(\sin x+1)<0$$

So (f) is strictly decreasing, hence it can cross zero only once.

(S1) is true 

(S2): decreasing in ([0,\pi/2]) and increasing in ([\pi/2,\pi])

But we already have:
$$f'(x)<0\quad\text{for all }x\in[0,\pi]$$

So (f) is decreasing everywhere — never increasing.

 (S2) is false

Only (S1) is correct