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 $\left[\frac{\pi}{2}, \pi\right]$.

<p>Let's analyze the function $f(x) = (\cos x) - x + 1$ over the interval $[0, \pi]$ and the statements provided.</p> <p>First, let's consider statement (S1):</p> <p>(S1) $f(x)=0$ for only one value of $x$ in $[0, \pi]$.</p> <p>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:</p> <p>$f(x) = \cos x - x + 1$</p> <p>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 $f(x)$:</p> <p>$f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1$</p> <p>The critical points occur when $f'(x) = 0$: <p>$- \sin x - 1 = 0 \Rightarrow \sin x = -1$.</p></p> <p>The equation $\sin x = -1$ does not hold for any $x$ in $[0, \pi]$. Note that: <ul> <li>For $x \in [0, \frac{\pi}{2}]$, $\sin x$ ranges from 0 to 1.</li><br> <li>For $x \in [\frac{\pi}{2}, \pi]$, $\sin x$ ranges from 1 to 0.</p></li> </ul> <p>Since $f'(x)$ is always negative (i.e., $f'(x) < 0$), $f(x)$ is a strictly decreasing function in $[0, \pi]$. Moreover, we evaluate:</p> <p>$f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2$</p> <p>$f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi$</p> <p>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).</p> <p>Now, let's consider statement (S2):</p> <p>(S2) $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$.</p> <p>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).</p> <p>Therefore, the correct option is:</p> <p><b>Option B</b>: Only (S1) is correct.</p>