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QJEEM25Maa05c: JEE Main 2025 Mathematics - Class 12 JEEM25Maa05c Solution: Let where [.] denotes greatest integer function. If and are the …

Let f(x)={3x,amp;xlt;0min⁡{1+x+[x],x+2[x]},amp;0≤x≤25,amp;xgt;2\mathrm{f}(x)=\left\{\begin{array}{lc}3 x, & x<0 \\ \min \{1+x+[x], x+2[x]\}, & 0 \leq x \leq 2 \\ 5, & x>2\end{array}\right.

where [.] denotes greatest integer function. If α\alpha and β\beta are the number of points, where ff is not continuous and is not differentiable, respectively, then α+β\alpha+\beta equals _______ .

Step-by-Step Verified Solution

amp;f(x)={3xamp;;amp;xlt;0min⁡{1+x,x}amp;;amp;0≤xlt;1min⁡{2+x,x+2}amp;;amp;1≤xlt;25amp;;amp;xgt;2amp;f(x)={3xamp;;amp;xlt;0xamp;;amp;0≤xlt;1x+2amp;;amp;1≤xlt;25amp;;amp;xgt;2\begin{aligned}& f(x)=\left\{\begin{array}{ccc}3 x & ; & x<0 \\\min \{1+x, x\} & ; & 0 \leq x<1 \\\min \{2+x, x+2\} & ; & 1 \leq x<2 \\5 & ; & x>2\end{array}\right. \\& f(x)=\left\{\begin{array}{ccl}3 \mathrm{x} & ; & x<0 \\\mathrm{x} & ; & 0 \leq x<1 \\\mathrm{x}+2 & ; & 1 \leq x<2 \\5 & ; & x>2\end{array}\right.\end{aligned}

JEE Main 2025 (Online) 28th January Morning Shift Mathematics - Limits, Continuity and Differentiability Question 1 English Explanation

Not continuous at x∈{1,2}⇒α=2\mathrm{x} \in\{1,2\} \Rightarrow \alpha=2

Not diff. at x∈{0,1,2}⇒β=3\mathrm{x} \in\{0,1,2\} \Rightarrow \beta=3

α+β=5\alpha+\beta=5

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