JEE Main 2026 — Mathematics Calculus

Let $f(x) = g(x) + h(x)$f(x)=g(x)+h(x), where

$$g(x) = \max\{6x,\ 2 + 3x^2\}$$g(x)=max{6x, 2+3x2}

$$h(x) = |x - 1|\left|\cos\left|x^2 - \dfrac{1}{4}\right|\right|$$h(x)=∣x−1∣​cos​x2−41​​​

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

Non-differentiability of $g(x)$g(x):

$g(x) = \max\{6x,\ 2 + 3x^2\}$g(x)=max{6x, 2+3x2} is non-differentiable where the two curves intersect with different slopes.

$$6x = 2 + 3x^2 \Rightarrow 3x^2 - 6x + 2 = 0$$6x=2+3x2⇒3x2−6x+2=0

$$x = \dfrac{6 \pm \sqrt{36 - 24}}{6} = \dfrac{6 \pm 2\sqrt{3}}{6} = 1 \pm \dfrac{1}{\sqrt{3}}$$x=66±36−24​​=66±23​​=1±3​1​

Both values lie in $(-\pi, \pi)$(−π,π). At these points, the derivatives of the two inner functions are $6$6 and $6x$6x; since $x \neq 1$x=1, the slopes differ.

So $g(x)$g(x) contributes $2$2 points.

Non-differentiability of $h(x)$h(x):

Since $\cos|y| = \cos y$cos∣y∣=cosy:

$$h(x) = |x - 1|\left|\cos\left(x^2 - \dfrac{1}{4}\right)\right|$$h(x)=∣x−1∣​cos(x2−41​)​

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

Case A: $x - 1 = 0 \Rightarrow x = 1$x−1=0⇒x=1

At $x = 1$x=1: $\cos\left(1 - \dfrac{1}{4}\right) = \cos\dfrac{3}{4} \neq 0$cos(1−41​)=cos43​=0, so $x = 1$x=1 is a point of non-differentiability.

Contribution: $1$1 point.

Case B: $\cos\left(x^2 - \dfrac{1}{4}\right) = 0$cos(x2−41​)=0

$$x^2 - \dfrac{1}{4} = (2n + 1)\dfrac{\pi}{2} \Rightarrow x^2 = \dfrac{1}{4} + (2n + 1)\dfrac{\pi}{2}$$x2−41​=(2n+1)2π​⇒x2=41​+(2n+1)2π​

Domain restriction: $x \in (-\pi, \pi) \Rightarrow x^2 \in [0, \pi^2)$x∈(−π,π)⇒x2∈[0,π2), and $\pi^2 \approx 9.87$π2≈9.87.

$n = -1$n=−1: $x^2 = 0.25 - \dfrac{\pi}{2} < 0$x2=0.25−2π​<0 (rejected)

$n = 0$n=0: $x^2 \approx 1.82 < 9.87$x2≈1.82<9.87 (gives $2$2 points)

$n = 1$n=1: $x^2 \approx 4.96 < 9.87$x2≈4.96<9.87 (gives $2$2 points)

$n = 2$n=2: $x^2 \approx 8.10 < 9.87$x2≈8.10<9.87 (gives $2$2 points)

$n = 3$n=3: $x^2 \approx 11.25 > 9.87$x2≈11.25>9.87 (rejected)

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

Checking for overlap:

The points $x = 1 \pm \dfrac{1}{\sqrt{3}}$x=1±3​1​ are not equal to $1$1, and for these, $x^2 - \dfrac{1}{4} = \dfrac{13}{12} \pm \dfrac{2}{\sqrt{3}}$x2−41​=1213​±3​2​, which are not odd multiples of $\dfrac{\pi}{2}$2π​. So the non-differentiable points of $g(x)$g(x) and $h(x)$h(x) are disjoint.

Total number of points of non-differentiability:

$$2 + 1 + 6 = 9$$2+1+6=9

Hence, the answer is $9$9.