Continuity & DifferentiabilityhardPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Non-Differentiable Points of a Max-plus-Modulus Function: 3 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The number of points, at which the function f(x)=max⁡{6x, 2+3x2}+∣x−1∣cos⁡∣x2−14∣f(x)=\max\{6x,\ 2+3x^2\}+|x-1|\cos\left|x^2-\dfrac14\right|, x∈(−π,π)x\in(-\pi,\pi), is not differentiable, is
Solution
Answer: 3 (± 0.01)
Step 1: The two pieces of the max are equal when 2+3x2=6x⇒3x2−6x+2=0⇒x=3±232+3x^2=6x\Rightarrow3x^2-6x+2=0\Rightarrow x=\dfrac{3\pm\sqrt2}{3}. At these 22 points the max function has corners. Step 2: The term ∣x−1∣cos⁡∣x2−14∣|x-1|\cos\left|x^2-\dfrac14\right| has a corner at x=1x=1 (from ∣x−1∣|x-1|, since the cosine factor is non-zero there). The modulus ∣x2−14∣\left|x^2-\tfrac14\right| does not create corners because cos⁡∣u∣=cos⁡u\cos|u|=\cos u is smooth at u=0u=0. Step 3: The corner points are x=3−23, 3+23x=\dfrac{3-\sqrt2}{3},\ \dfrac{3+\sqrt2}{3} (from the max) and x=1x=1 (from ∣x−1∣|x-1|) — all distinct. Step 4: Total number of non-differentiable points =2+1=3=2+1=3. Correct answer: 3
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