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

To solve this problem, we need to analyze the function \( f(x) = \cos x - x + 1 \) over the interval \([0, \pi]\). We will examine each statement (S1) and (S2) individually and determine their correctness.

Step 1: Analyzing Statement (S1)

The statement (S1) claims that \( f(x) = 0 \) for only one value of \( x \) in the interval \([0, \pi]\).

To examine this, let's first consider the endpoints of the interval:

• At \( x = 0 \): \(f(0) = \cos(0) - 0 + 1 = 1 + 1 = 2\).
• At \( x = \pi \): \(f(\pi) = \cos(\pi) - \pi + 1 = -1 - \pi + 1 = -\pi\).

Since \( f(0) = 2 \) and \( f(\pi) = -\pi \), the function changes sign over the interval \([0, \pi]\). Therefore, by the Intermediate Value Theorem, there must be at least one solution \( x \) in \([0, \pi]\) where \( f(x) = 0 \).

Next, let's find \( f'(x) \) to determine if there could be more than one root:

• \(f'(x) = -\sin x - 1\).
• Since \( -1 \leq \sin x \leq 1 \), we have \( f'(x) = -\sin x - 1 \leq 0 - 1 = -1\).

This implies that \( f(x) \) is a strictly decreasing function over the interval \([0, \pi]\).

As \( f(x) \) is strictly decreasing and continuous, it can have at most one root in \([0, \pi]\). Thus, (S1) is correct.

Step 2: Analyzing Statement (S2)

The statement (S2) claims that \( f(x) \) is decreasing in \([0, \pi/2]\) and increasing in \([\pi/2, \pi]\).

As we previously found, \( f'(x) = -\sin x - 1 \leq -1 \) for all \( x \in [0, \pi]\). This means \( f(x) \) is strictly decreasing throughout the entire interval \([0, \pi]\).

Therefore, statement (S2) is incorrect since there is no subinterval of \([0, \pi]\) where \( f(x) \) is increasing.

Conclusion

Based on the analysis above, we conclude that:

• (S1) is correct because \( f(x) \) has exactly one root in the interval \([0, \pi]\).
• (S2) is incorrect as the function is constantly decreasing in that interval.

Thus, the correct answer is: Only (S1) is correct.

Answer 2

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.