Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 4 April, Evening Shift — Question 46

Let $f(x) = \begin{cases} e^{x-1}, & x<0 \\ x^2-5x+6, & x\ge 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$ . If the number of points where g is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha+\beta$ is equal to

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Graph of $\mathrm{f}(\mathrm{x})$

Now graph of $\mathrm{f}(|\mathrm{x}|)$

$\mathrm{f}(|\mathrm{x}|)$ is continuous function and it is non diff. at $\mathrm{x}=0$ Graph of $|\mathrm{f}(\mathrm{x})|$

$|\mathrm{f}(\mathrm{x})|$ is discontinuous at $\mathrm{x}=0$ $|\mathrm{f}(\mathrm{x})|$ is non-diff. at $\mathrm{x}=0,2,3$ $\mathrm{g}(\mathrm{x})=\mathrm{f}(|\mathrm{x}|)+|\mathrm{f}(\mathrm{x})|$ $\mathrm{g}(\mathrm{x})$ will be discontinuous at $\mathrm{x}=0$ $\mathrm{g}(\mathrm{x})$ will be non diff at $\mathrm{x}=0,2,3$ $\alpha=1, \beta=3$ $\alpha+\beta=1+3=4$

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Differentiability