JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Continuity and Differentiability

The two expressions inside the maximum meet where $2+3x^2=6x$, that is $3x^2-6x+2=0$, giving $x=\frac{3\pm \sqrt{3}}{3}$. Between these roots $6x$ is the larger and outside them $2+3x^2$ is larger, so $\max \{6x,2+3x^2\}$ has a corner at each of $x=\frac{3-\sqrt{3}}{3}$ and $x=\frac{3+\sqrt{3}}{3}$, where the two branches meet with different slopes. In the second term $\cos \left|x^2-\frac{1}{4}\right|=\cos \left(x^2-\frac{1}{4}\right)$ because cosine is even, so that factor is differentiable everywhere; the factor $|x-1|$ has a corner at $x=1$, and since $\cos \left(1-\frac{1}{4}\right)=\cos \frac{3}{4}\ne 0$, the product $|x-1|\cos \left|x^2-\frac{1}{4}\right|$ is non-differentiable at $x=1$. The three abscissae $\frac{3-\sqrt{3}}{3},1,\frac{3+\sqrt{3}}{3}$ are distinct and all lie in $(-\pi ,\pi )$. The number of points of non-differentiability is $3$.