Which of the following functions \(f(x)\) are differentiable at \(x = 0\)?
(A) \(|x|\)
(B) \(|x - 1|\)
(C) \(|\sin x|\)
(D) \(|\cos x|\)
(E) \(x^2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (D) and (E) only
A modulus function can fail to be differentiable only where the expression inside it changes sign. Near x = 0 the left-hand and right-hand derivatives decide each case.
(A) \(|x|\) — Incorrect. x changes sign at 0:
\(\displaystyle \text{RHD} = \lim_{h \to 0^+}\frac{|h|}{h} = 1\)
\(\displaystyle \text{LHD} = \lim_{h \to 0^-}\frac{|h|}{h} = -1\)
(B) \(|x - 1|\) — Correct. Near x = 0, \(x - 1 \lt 0\), so \(|x - 1| = 1 - x\), with derivative −1. The corner is at x = 1, not at x = 0.
(C) \(|\sin x|\) — Incorrect. \(\sin x\) changes sign at 0:
\(\displaystyle \text{RHD} = \lim_{h \to 0^+}\frac{\sin h}{h} = 1\)
\(\displaystyle \text{LHD} = \lim_{h \to 0^-}\frac{-\sin h}{h} = -1\)
(D) \(|\cos x|\) — Correct. Near x = 0, \(\cos x \gt 0\), so \(|\cos x| = \cos x\), with derivative \(-\sin 0 = 0\).
(E) \(x^2\) — Correct. A polynomial is differentiable everywhere; the derivative at 0 is \(2(0) = 0\).