Question

Let \(f\) be a continuous real valued function, defined by, \(f(x) = \sin x\), for all \(x \in [-\frac{\pi}{2}, \frac{\pi}{2}]\). Then which of the following does not hold.

• A. \(f'\) is continuous on \((-\frac{\pi}{2}, \frac{\pi}{2})\)
• B. \(f'\) is bounded on \((-\frac{\pi}{2}, \frac{\pi}{2})\)
• C. \(f'(x) = 0\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\)
• D. \(f'(x) = 1\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\)

Hint:

When analyzing properties of a function on an interval, pay close attention to whether the interval is open `()` or closed `[]`. Endpoints are included in closed intervals but excluded from open intervals, which can be critical for questions about attaining maximums, minimums, or specific values like zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given the function \(f(x) = \sin x\) on the closed interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). We need to analyze its derivative, \(f'(x)\), on the open interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) and determine which of the given statements is false.
Step 2: Detailed Explanation:
First, we find the derivative of the function \(f(x)\):
\(f(x) = \sin x\)
\(f'(x) = \cos x\)
Now, we evaluate each statement for \(f'(x) = \cos x\) on the open interval \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\).
(A) \(f'\) is continuous on \((-\frac{\pi}{2}, \frac{\pi}{2})\): The function \(f'(x) = \cos x\) is continuous for all real numbers. Thus, it is continuous on this interval. This statement is TRUE.
(B) \(f'\) is bounded on \((-\frac{\pi}{2}, \frac{\pi}{2})\): On the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), the value of \(\cos x\) is strictly greater than 0 and less than or equal to 1. The range is \((0, 1]\). Since all values are between 0 and 1, the function is bounded. This statement is TRUE.
(C) \(f'(x) = 0\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\): We need to check if the equation \(\cos x = 0\) has a solution within the open interval. The solutions to \(\cos x = 0\) are \(x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \dots\). None of these values lie *inside* the open interval \((-\frac{\pi}{2}, \frac{\pi}{2})\). This statement is FALSE.
(D) \(f'(x) = 1\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\): We need to check if the equation \(\cos x = 1\) has a solution within the open interval. The solution is \(x = 0\), and \(0\) is clearly within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\). This statement is TRUE.
Step 3: Final Answer:
The question asks which statement does not hold (i.e., is false). Based on our analysis, statement (C) is the one that is false.