Here, f(x) = |x - 3| f(x) = x - 3 when x - 3 ≥ 0 ⇒ x ≥ 3 And f(x) = 3 - x when x - 3 < 0 ⇒ x < 3 For, x = 0, f(x) = 3 - x Function is not changing at x = 0 (∵ function is linear) So, f(x) is continuous at x = 0 Now, x = 0, f(x) = 3 - x ⇒ f'(x) = -1 f'(0−) = -1 and f'(0+) = -1 So, f(x) is differentiable at x = 0 Hence, option (3) is correct.