JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Continuity and Differentiability

Here $f(x)=e^{x-1}$ for $x<0$ and $f(x)=(x-2)(x-3)$ for $x\ge 0$. Since $|x|\ge 0$ always, $f(|x|)=|x{|}^2-5|x|+6=x^2-5|x|+6$, which is continuous everywhere and differentiable except at $x=0$, where the $-5|x|$ term produces a corner. For $|f(x)|$: at $x=0$, ${\lim }_{x\to 0^{-}}f(x)=e^{-1}\ne 6=f(0)$, so $f$ and hence $|f(x)|$ has a jump at $x=0$; on $x\ge 0$, $|(x-2)(x-3)|$ has corners at $x=2$ and $x=3$ where the product changes sign. Adding, $g(x)=f(|x|)+|f(x)|$ is discontinuous only at $x=0$, so $\alpha =1$. It fails to be differentiable at $x=0$ (jump), $x=2$ and $x=3$, so $\beta =3$. Therefore $\alpha +\beta =4$.