JEE Main 2026 — Continuity and Differentiability Question with Solution

We are given the function:

$f(x)=\{\begin{array}{ll}{e}^{x-1},& x<0\\ x^2-5x+6,& x\ge 0\end{array}$

We need to analyze the continuity and differentiability of $g(x)=f(|x|)+|f(x)|$.

For $x<0$, $|x|=-x>0$. Thus, $f(|x|)=(-x{)}^2-5(-x)+6=x^2+5x+6$.
Also, for $x<0$, $f(x)=e^{x-1}>0$, so $|f(x)|=e^{x-1}$.
Therefore, for $x<0$, $g(x)=x^2+5x+6+e^{x-1}$.

For $x\ge 0$, $|x|=x$. Thus, $f(|x|)=f(x)=x^2-5x+6$.
Therefore, for $x\ge 0$, $g(x)=x^2-5x+6+|x^2-5x+6|$.

Let us check the continuity of $g(x)$ at $x=0$:
${\lim }_{x\to 0^{-}}g(x)={\lim }_{x\to 0^{-}}(x^2+5x+6+e^{x-1})=6+\frac{1}{e}$
${\lim }_{x\to 0^{+}}g(x)={\lim }_{x\to 0^{+}}(x^2-5x+6+|x^2-5x+6|)=6+6=12$
$g(0)=12$

Since ${\lim }_{x\to 0^{-}}g(x)\ne {\lim }_{x\to 0^{+}}g(x)$, $g(x)$ is discontinuous at $x=0$. For all other $x$, $g(x)$ is a sum of continuous functions and is therefore continuous.
Thus, the number of points of discontinuity is $\alpha =1$.

Now, let us check the differentiability of $g(x)$.
Since $g(x)$ is discontinuous at $x=0$, it is not differentiable at $x=0$.

For $x<0$, $g(x)=x^2+5x+6+e^{x-1}$, which is differentiable everywhere in its domain.

For $x>0$, we can rewrite $g(x)$ by analyzing the sign of $x^2-5x+6=(x-2)(x-3)$:
$g(x)=\{\begin{array}{ll}2(x^2-5x+6),& x\in (0,2]\cup [3,\infty )\\ 0,& x\in (2,3)\end{array}$

Differentiating $g(x)$ for $x>0,x\ne 2,3$:
$g^{\prime }(x)=\{\begin{array}{ll}4x-10,& x\in (0,2)\cup (3,\infty )\\ 0,& x\in (2,3)\end{array}$

Checking differentiability at $x=2$:
$g^{\prime }(2^{-})=4(2)-10=-2$
$g^{\prime }(2^{+})=0$
Since $g^{\prime }(2^{-})\ne g^{\prime }(2^{+})$, $g(x)$ is not differentiable at $x=2$.

Checking differentiability at $x=3$:
$g^{\prime }(3^{-})=0$
$g^{\prime }(3^{+})=4(3)-10=2$
Since $g^{\prime }(3^{-})\ne g^{\prime }(3^{+})$, $g(x)$ is not differentiable at $x=3$.

Thus, $g(x)$ is not differentiable at exactly three points: $x=0,2,3$.
So, the number of points of non-differentiability is $\beta =3$.

Finally, $\alpha +\beta =1+3=4$.

Answer: $4$