JEE Main 2025 — Continuity and Differentiability Question with Solution
JEE Main 2025 (4 Apr Shift 1)
Question
Let and be the number of points at which the function , is not differentiable and not continuous, respectively. Then is equal to ________ .
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Show full solutionCorrect answer: 3
Correct answer
3
Step-by-step explanation
$f(x)=\left\{\begin{array}{cc}
x, & x \lt -1 \\ x^{21}, & -1 \leq x \lt 0 \\ x, & 0 \leq x \lt 1 \\ x^{21}, & x \geq 1
\end{array}\right.f(x)\begin{aligned}
& \therefore \mathrm{n}=0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}
1, & \mathrm{x} \lt -1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x} \lt 0 \\ 1, & 0 \lt \mathrm{x} \lt 1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1
\end{array}\right.
\end{aligned}\therefore \mathrm{f}(\mathrm{x})\mathrm{x}=-1,0,1\begin{aligned}
& \therefore \mathrm{m}=3 \\ & \mathrm{~m}+\mathrm{n}=3
\end{aligned}$
x, & x \lt -1 \\ x^{21}, & -1 \leq x \lt 0 \\ x, & 0 \leq x \lt 1 \\ x^{21}, & x \geq 1
\end{array}\right.f(x)\begin{aligned}
& \therefore \mathrm{n}=0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}
1, & \mathrm{x} \lt -1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x} \lt 0 \\ 1, & 0 \lt \mathrm{x} \lt 1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1
\end{array}\right.
\end{aligned}\therefore \mathrm{f}(\mathrm{x})\mathrm{x}=-1,0,1\begin{aligned}
& \therefore \mathrm{m}=3 \\ & \mathrm{~m}+\mathrm{n}=3
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Continuity and Differentiability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.