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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 25

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 ________.

Answer and worked solution

Official answer: 3

The max. 6x=2+3x26x=2+3x^2 at x=1±13x=1\pm\frac1{\sqrt3}, both in (−π,π)(-\pi,\pi). The max has a corner at each.

The second term. cos⁡∣x2−14∣=cos⁡(x2−14)\cos\left|x^2-\frac14\right|=\cos\left(x^2-\frac14\right) is smooth. ∣x−1∣|x-1| has a corner at x=1x=1, where cos⁡34≠0\cos\frac34\ne0, so ff is not differentiable there. (Near x=1x=1 the max is the smooth 6x6x.)

Total 3 points.

Answer: 3

Topics:Differentiability
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