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Calculus
Continuity and Differentiability, Absolute Value Functions
jee_main_2026_april_4_shift_2
Grade 12

Question:

Let f(x) = |x^3 - 6x^2 + 11x - 6|. The number of points in the interval (0, 4) where f(x) is non-differentiable is:
A. 0
B. 1
C. 2
D. 3

Step-by-Step Solution

Key Concept: Non-differentiability occurs at roots of the polynomial inside absolute value.
Step 1: Factor x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3). Step 2: The absolute value function is non-differentiable where the expression inside changes sign: at x = 1, 2, 3. Step 3: All three roots lie in (0, 4). Hence number of points = 3.
Correct Answer: D
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