JEE Main 2026 — Continuity and Differentiability Question with Solution

Let $f(x)=g(x)+h(x)$, where
$g(x)=\max \{6x,2+3x^2\}$
$h(x)=|x-1|\left|\cos \left|x^2-\frac{1}{4}\right|\right|$

We find the points of non-differentiability of each part in $x\in (-\pi ,\pi )$.

Non-differentiability of $g(x)$:
$g(x)=\max \{6x,2+3x^2\}$ is non-differentiable where the two curves intersect with different slopes.

$6x=2+3x^2\Rightarrow 3x^2-6x+2=0$
$x=\frac{6\pm \sqrt{36-24}}{6}=\frac{6\pm 2\sqrt{3}}{6}=1\pm \frac{1}{\sqrt{3}}$

Both values lie in $(-\pi ,\pi )$. At these points, the derivatives of the two inner functions are $6$ and $6x$; since $x\ne 1$, the slopes differ.
So $g(x)$ contributes $2$ points.

Non-differentiability of $h(x)$:
Since $\cos |y|=\cos y$:
$h(x)=|x-1|\left|\cos \left(x^2-\frac{1}{4}\right)\right|$

An expression of the form $|u(x)|$ is non-differentiable at simple zeros of $u(x)$.

Case A: $x-1=0\Rightarrow x=1$
At $x=1$: $\cos \left(1-\frac{1}{4}\right)=\cos \frac{3}{4}\ne 0$, so $x=1$ is a point of non-differentiability.
Contribution: $1$ point.

Case B: $\cos \left(x^2-\frac{1}{4}\right)=0$
$x^2-\frac{1}{4}=(2n+1)\frac{\pi }{2}\Rightarrow x^2=\frac{1}{4}+(2n+1)\frac{\pi }{2}$

Domain restriction: $x\in (-\pi ,\pi )\Rightarrow x^2\in [0,{\pi }^2)$, and ${\pi }^2\approx 9.87$.

$n=-1$: $x^2=0.25-\frac{\pi }{2}<0$ (rejected)
$n=0$: $x^2\approx 1.82<9.87$ (gives $2$ points)
$n=1$: $x^2\approx 4.96<9.87$ (gives $2$ points)
$n=2$: $x^2\approx 8.10<9.87$ (gives $2$ points)
$n=3$: $x^2\approx 11.25>9.87$ (rejected)

Contribution: $2+2+2=6$ points.

Checking for overlap:
The points $x=1\pm \frac{1}{\sqrt{3}}$ are not equal to $1$, and for these, $x^2-\frac{1}{4}=\frac{13}{12}\pm \frac{2}{\sqrt{3}}$, which are not odd multiples of $\frac{\pi }{2}$. So the non-differentiable points of $g(x)$ and $h(x)$ are disjoint.

Total number of points of non-differentiability:
$2+1+6=9$

Hence, the answer is $9$.