Let \(f(x) = |\sin x|\), \(x \in \mathbf{R}\). Consider the following statements:
(A) f is continuous at every real number x.
(B) f is differentiable at \(x = 0\).
(C) f is differentiable at \(x = \displaystyle\frac{\pi}{2}\).
(D) \(f'\left(\displaystyle\frac{3\pi}{2}\right) = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
f is the modulus of the continuous function sin x, so it is continuous everywhere.
Near x = 0, f(x) = sin x for x > 0 and f(x) = −sin x for x < 0. The one-sided derivatives at 0 are
\(Rf'(0) = \cos 0 = 1\), \(Lf'(0) = -\cos 0 = -1\)
Near \(\frac{\pi}{2}\), sin x > 0, so f(x) = sin x and \(f'(x) = \cos x\).
Near \(\frac{3\pi}{2}\), sin x < 0, so f(x) = −sin x and \(f'(x) = -\cos x\).
So f is continuous everywhere, has a corner at 0, and is differentiable at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) with derivative 0 at both.
Checking the statements
(A) f is continuous at every real number x. — Correct. It is the modulus of the continuous function sin x.
(B) f is differentiable at \(x = 0\). — Incorrect. The left derivative is −1 and the right derivative is 1.
(C) f is differentiable at \(x = \frac{\pi}{2}\). — Correct. There f(x) = sin x, with derivative \(\cos\frac{\pi}{2} = 0\).
(D) \(f'\left(\frac{3\pi}{2}\right) = 0\) — Correct. There f(x) = −sin x, so \(f'\left(\frac{3\pi}{2}\right) = -\cos\frac{3\pi}{2} = 0\).