JEE Main 2026 Apr 4 Shift 1, Mathematics Q25: Differentiability Of A Function

Examine the two summands separately.

Max piece. Set $6x=2+3x^2\Rightarrow 3x^2-6x+2=0\Rightarrow x=\frac{3\pm \sqrt{3}}{3}$. Both roots lie in $(-\pi ,\pi )$ and at each the two graphs cross transversally, so $\max \{6x,2+3x^2\}$ has a corner there. That gives $2$ non-differentiable points.

$|x-1|\cos \left|x^2-\frac{1}{4}\right|$ piece. The outer absolute value $|x-1|$ contributes a potential corner only at $x=1$. At $x=1$, $\cos \left|1-\frac{1}{4}\right|=\cos \frac{3}{4}\ne 0$. Since the cosine factor is non-zero at $x=1$, the kink of $|x-1|$ survives and the product is non-differentiable at $x=1$.

The inner $|x^2-\frac{1}{4}|$ has kinks at $x=\pm \frac{1}{2}$, but composing with the smooth, even function $\cos$ removes them (since $\cos |u|=\cos u$ is smooth).

Total non-differentiable points: $2+1=3$.