Match List-I with List-II. Here [x] denotes the greatest integer function
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(f(x) = [x]\) | (I) | is continuous everywhere but not differentiable at \(x = -1\) |
| (B) | \(f(x)=|x-1|\) | (II) | is continuous everywhere except at all integral values |
| (C) | \(f(x) = e^{|x|}\) | (III) | is continuous everywhere but not differentiable at \(x = 1\) |
| (D) | \(f(x) = |x+1|\) | (IV) | is continuous everywhere but not differentiable at \(x = 0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
A modulus function \(|x - a|\) is continuous everywhere and has a corner (left and right derivatives −1 and 1) where the expression inside is zero, x = a.
(A) \(f(x) = [x]\)
This matches (II). [x] is constant between consecutive integers and jumps by 1 at every integer n:
\(\displaystyle \lim_{x \to n^-} [x] = n - 1,\quad [n] = n\)
So it is discontinuous at all integers and continuous elsewhere.
(B) \(f(x)=|x-1|\)
This matches (III). The expression inside is zero at x = 1, so the corner is at x = 1.
(C) \(f(x) = e^{|x|}\)
This matches (IV). It is continuous as a composition of continuous functions. At x = 0, with f(0) = 1:
\(\displaystyle \text{RHD} = \lim_{h \to 0^+} \frac{e^{h} - 1}{h} = 1\)
\(\displaystyle \text{LHD} = \lim_{h \to 0^-} \frac{e^{-h} - 1}{h} = -1\)
The one-sided derivatives differ, so f is not differentiable at x = 0. Elsewhere \(e^{|x|}\) is \(e^x\) or \(e^{-x}\), both differentiable.
(D) \(f(x) = |x+1|\)
This matches (I). The expression inside is zero at x = −1, so the corner is at x = −1.
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).