Key Concepts and Formulas

• Monotonicity of a Function: A function $f(x)$ is increasing if $f'(x) > 0$ and decreasing if $f'(x) < 0$.
• Intermediate Value Theorem (IVT): If $f(x)$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.
• Derivative of Trigonometric Functions: $\frac{d}{dx}(\cos x) = -\sin x$.

Step-by-Step Solution

Step 1: Analyze the function $f(x) = \cos x - x + 1$.

We are given the function $f(x) = \cos x - x + 1$ and we need to analyze its behavior in the interval $[0, \pi]$.

Step 2: Analyze Statement (S1): $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.

Step 2.1: Calculate the first derivative of $f(x)$.

To determine the monotonicity of $f(x)$, we find its first derivative: $f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1$

Step 2.2: Determine the sign of $f'(x)$ in the interval $[0, \pi]$.

Since $0 \le x \le \pi$, we have $0 \le \sin x \le 1$. Therefore, $-\sin x$ will be between $-1$ and $0$, inclusive. Thus, $-\sin x - 1$ will be between $-2$ and $-1$, inclusive. This means $f'(x) = -\sin x - 1 \le -1 < 0$ for all $x \in [0, \pi]$.

Step 2.3: Conclude the monotonicity of $f(x)$.

Since $f'(x) < 0$ for all $x \in [0, \pi]$, the function $f(x)$ is strictly decreasing in the interval $[0, \pi]$.

Step 2.4: Evaluate $f(0)$ and $f(\pi)$.

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

Step 2.5: Apply the Intermediate Value Theorem (IVT).

Since $f(0) = 2 > 0$ and $f(\pi) = -\pi < 0$, and $f(x)$ is continuous on $[0, \pi]$, by the IVT, there exists at least one $c \in (0, \pi)$ such that $f(c) = 0$.

Step 2.6: Determine the number of roots.

Since $f(x)$ is strictly decreasing on $[0, \pi]$, it can have at most one root in this interval. Combining this with the conclusion from IVT, $f(x)$ has exactly one root in $[0, \pi]$. Therefore, statement (S1) is correct.

Step 3: Analyze Statement (S2): $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$.

Step 3.1: Analyze the sign of $f'(x)$ in $\left[0, \frac{\pi}{2}\right]$.

For $x \in \left[0, \frac{\pi}{2}\right]$, $0 \le \sin x \le 1$. Therefore, $f'(x) = -\sin x - 1 \le -1 < 0$. This means $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$.

Step 3.2: Analyze the sign of $f'(x)$ in $\left[\frac{\pi}{2}, \pi\right]$.

For $x \in \left[\frac{\pi}{2}, \pi\right]$, $0 \le \sin x \le 1$. Therefore, $f'(x) = -\sin x - 1 \le -1 < 0$. This means $f(x)$ is decreasing in $\left[\frac{\pi}{2}, \pi\right]$.

Step 3.3: Conclude about statement (S2).

Since $f(x)$ is decreasing in both $\left[0, \frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2}, \pi\right]$, statement (S2) is incorrect because it states that $f(x)$ is increasing in $\left[\frac{\pi}{2}, \pi\right]$.

Step 4: Final Conclusion.

Statement (S1) is correct, and statement (S2) is incorrect.

Common Mistakes & Tips

• Remember to check the sign of the derivative carefully to determine monotonicity.
• The Intermediate Value Theorem only guarantees the existence of a root, not its uniqueness. Monotonicity helps determine uniqueness.
• Don't assume a function is increasing or decreasing without proof. Always analyze the derivative.

Summary

We analyzed the function $f(x) = \cos x - x + 1$ by finding its derivative and evaluating its sign in the given intervals. We determined that $f(x)$ is strictly decreasing in $[0, \pi]$ and thus has only one root in that interval, making statement (S1) correct. Statement (S2) is incorrect as $f(x)$ is decreasing in both $\left[0, \frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2}, \pi\right]$. Therefore, only statement (S1) is correct.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).