The function f(x) = |x| + |x − 1| is:

Correct option is A

Given:

f(x) = |x| + |x − 1|

Concept used:
A function involving absolute values is continuous everywhere because |x| is continuous for all x.

Solution:
Let’s check piecewise:

For x < 0:
f(x) = −x + (−x + 1) = 1 − 2x

For 0 ≤ x < 1:
f(x) = x + (−x + 1) = 1

For x ≥ 1:
f(x) = x + (x − 1) = 2x − 1

Check continuity at x = 0:
Left-hand limit (x → 0⁻): f(x) = 1 − 2x → 1
Right-hand limit (x → 0⁺): f(x) = 1
f(0) = |0| + |−1| = 1
All equal => continuous at x = 0

Check continuity at x = 1:
Left-hand limit (x → 1⁻): f(x) = 1
Right-hand limit (x → 1⁺): f(x) = 2x − 1 = 1
f(1) = |1| + |0| = 1
All equal => continuous at x = 1

Correct answer is (a) continuous at x = 0 as well as at x = 1