JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q22

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MathematicsInteger+4 / -1Question 22 of 75

Let $f(x)=\begin{cases}e^{x-1}, & x<0\\ x^2-5x+6, & x\ge0\end{cases}$ 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 ________.

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Answer and worked solution

Official answer: 4

Continuity. $f$ jumps at 0 ( $e^{-1}$ vs 6), so $|f(x)|$ does too, while $f(|x|)$ is continuous. So $g$ is discontinuous only at 0: $\alpha=1$ .

Differentiability. $g$ is not differentiable at 0 (discontinuous) and at 2 and 3, the corners of $|x^2-5x+6|$ : $\beta=3$ .

Answer: 4

Topics:Differentiability Continuity

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