JEE Main 2024 — Continuity and Differentiability Question with Solution
JEE Main 2024 (05 Apr Shift 2)
Question
Let be given by , where denotes the greatest integer less than or equal to . The number of points, where is not continuous, is :
Choose an option
Show full solutionCorrect option: D
Correct answer
D4
Step-by-step explanation
Doubtful points : at
at : Dis. at : $\left.\begin{array}{rl} \text {LHL } \Rightarrow f(x) & =8+2-1+3=12 \\ f(2) & =8+2-2+4=12 \end{array}\right\} \text { Cont. }$ at : $\left.\begin{array}{rl} \text {LHL } \Rightarrow & 0+0-(-1)+0=1 \\ & \mathrm{f}(0)=0 \end{array}\right\} \text { Dis. }$ at $\left.\begin{array}{rl} \text {LHL } \Rightarrow & 2+1-0+0=3 \\ & f(1)=3-1+1=3 \\ \text {RHL } \Rightarrow & 2+1-1+1=3 \end{array}\right\} \text { Cont. }$
at : Dis. at : $\left.\begin{array}{rl} \text {LHL } \Rightarrow f(x) & =8+2-1+3=12 \\ f(2) & =8+2-2+4=12 \end{array}\right\} \text { Cont. }$ at : $\left.\begin{array}{rl} \text {LHL } \Rightarrow & 0+0-(-1)+0=1 \\ & \mathrm{f}(0)=0 \end{array}\right\} \text { Dis. }$ at $\left.\begin{array}{rl} \text {LHL } \Rightarrow & 2+1-0+0=3 \\ & f(1)=3-1+1=3 \\ \text {RHL } \Rightarrow & 2+1-1+1=3 \end{array}\right\} \text { Cont. }$
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This is a previous-year question from JEE Main 2024, 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.