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: Problem Overview:
We analyze \(f(x) = \sin x\) on \([-\frac{\pi}{2}, \frac{\pi}{2}]\). We examine \(f'(x)\) on \((-\frac{\pi}{2}, \frac{\pi}{2})\) to find the false statement.
Step 2: Analysis:
First, find the derivative of \(f(x)\):
\(f(x) = \sin x\)
\(f'(x) = \cos x\)
Evaluate each statement for \(f'(x) = \cos x\) on \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\).
(A) \(f'\) is continuous on \((-\frac{\pi}{2}, \frac{\pi}{2})\): \(f'(x) = \cos x\) is continuous everywhere, including this interval. TRUE.
(B) \(f'\) is bounded on \((-\frac{\pi}{2}, \frac{\pi}{2})\): On \((-\frac{\pi}{2}, \frac{\pi}{2})\), \(0<\cos x \le 1\). The function is bounded. TRUE.
(C) \(f'(x) = 0\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\): Solve \(\cos x = 0\). Solutions are \(x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \dots\). None are within \((-\frac{\pi}{2}, \frac{\pi}{2})\). FALSE.
(D) \(f'(x) = 1\) for some \(x \in (-\frac{\pi}{2}, \frac{\pi}{2})\): Solve \(\cos x = 1\). Solution is \(x = 0\), which is in \((-\frac{\pi}{2}, \frac{\pi}{2})\). TRUE.
Step 3: Conclusion:
The false statement is (C).