Let f(x) = begin cases 3x, & x < 0 \\ min\ 1 + x + [x], x + 2[x]\ , & 0 le x le 2 \\ 5, & x 2 end cases where [.] denotes greatest integer function. If α and β are the number of points, where f is not continuous and is not differentiable, respectively, then α + β equals .....…

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 28 January, Morning Shift — Question 24

Let $f(x) = \begin{cases} 3x, & x < 0 \\ \min\{1 + x + [x], x + 2[x]\}, & 0 \le x \le 2 \\ 5, & x > 2 \end{cases}$ where [.] denotes greatest integer function. If $\alpha$ and $\beta$ are the number of points, where $f$ is not continuous and is not differentiable, respectively, then $\alpha + \beta$ equals .....…

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

f(x) = \begin{cases} 3x, & x < 0 \\ \min\{1+x, x\}, & 0 \le x < 1 \\ \min\{2+x, x+2\}, & 1 \le x < 2 \\ 5, & x > 2 \end{cases}

f(x) = \begin{cases} 3x, & x < 0 \\ x, & 0 \le x < 1 \\ x+2, & 1 \le x < 2 \\ 5, & x > 2 \end{cases}

\text{Not continuous at } x \in \{1, 2\} \Rightarrow \alpha = 2

\text{Not diff. at } x \in \{0, 1, 2\} \Rightarrow \beta = 3

\alpha + \beta = 5

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Differentiability