Question

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

• A. Both (S1) and (S2) are correct
• B. Only (S1) is correct
• C. Both (S1) and (S2) are incorrect
• D. Only (S2) is correct

Answer 1

The Correct Option is B

The function is:
\[ f(x) = \cos x - x + 1. \]
The derivative is:
\[ f'(x) = -\sin x - 1. \]
Since \( -\sin x - 1 < 0 \) for all \( x \in \mathbb{R} \), \( f(x) \) is strictly decreasing in \([0, \pi]\).
At \( x = 0 \),
\[ f(0) = \cos(0) - 0 + 1 = 2. \]
At \( x = \pi \),
\[ f(\pi) = \cos(\pi) - \pi + 1 = -\pi < 0. \]
By the intermediate value theorem, \( f(x) = 0 \) has exactly one root in \([0, \pi]\). Thus, (S1) is correct.
(S2) is incorrect because \( f(x) \) is strictly decreasing in \([0, \pi]\).
Final Answer: Only (S1) is correct.