1. Given function

   $f(x)=\cos x-x+1, \qquad x\in \mathbb R$

We need to check the truth of:

• (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 $\left[\frac{\pi}{2},\pi\right]$.

2. Check monotonicity using derivative

Differentiate:

$f'(x)=-\sin x-1$

Now for $x\in[0,\pi]$, we know

$\sin x\in[0,1]$

So,

$f'(x)=-(\sin x+1)\le -1<0$

Thus $f'(x)$ is strictly negative for every $x\in[0,\pi]$.

Hence, $f(x)$ is strictly decreasing on the entire interval $[0,\pi]$.

So:

• it is decreasing on $\left[0,\frac{\pi}{2}\right]$,
• but it is not increasing on $\left[\frac{\pi}{2},\pi\right]$.

Therefore, (S2) is false.

3. Check number of zeros in $[0,\pi]$

Since $f(x)$ is continuous and strictly decreasing on $[0,\pi]$, it can have at most one zero there.

Now evaluate endpoints:

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

$f(\pi)=\cos \pi-\pi+1=-1-\pi+1=-\pi<0$

Since $f(0)>0$ and $f(\pi)<0$, by the Intermediate Value Theorem there is at least one root in $(0,\pi)$.

Because the function is strictly decreasing, there can be only one such root.

Therefore, (S1) is true.

4. Evaluate options

• (S1): True
• (S2): False

So the correct option is:

$\boxed{\text{B: Only (S1) is correct}}$

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.