Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 5 April, Shift 2 — Question 1

Let f:[−1,2]→Rf:[-1,2] \rightarrow \mathrm{R} be given by f(x)=2x2+x+[x2]−[x]f(\mathrm{x})=2 \mathrm{x}^{2}+\mathrm{x}+\left[\mathrm{x}^{2}\right]-[\mathrm{x}], where [t][\mathrm{t}] denotes the greatest integer less than or

equal to tt. The number of points, where ff is not continuous, is :

  1. Option A:

    6

  2. Option B:

    3

  3. Option C:

    4

    Correct
  4. Option D:

    5

Answer: C

Step-by-step solution

Doubtful points :

−1,0,1,2,3,2-1,0,1, \sqrt{2}, \sqrt{3}, 2  at x=2,3f(x)=(2x2+x↓ Cont. −[x])+[x2] Cont. = Discount \begin{aligned} & \text { at } \mathrm{x}=\sqrt{2}, \sqrt{3} & \left.\mathrm{f}(\mathrm{x})=\underset{\underset{\text { Cont. }}{\downarrow}}{\left(2 \mathrm{x}^{2}+\mathrm{x}\right.}-[\mathrm{x}]\right)+\underset{\text { Cont. }}{\left[\mathrm{x}^{2}\right]}=\text { Discount } \end{aligned}

at x=−1x=-1 :

\left. \begin{array}{*{35}{r}}\text{RHL}\Rightarrow & f\left( \text{x} \right)=\left( 2-1-\left( -1 \right) \right)+0=2 \\{} & \text{f}\left( -1 \right)=2-1-\left( -1 \right)+1=3 \\\end{array} \right\}\text{ }\!\!~\!\!\text{ Dis}\text{. }\!\!~\!\!\text{ }

at x=2\text{x}=2 :

\left. \begin{array}{*{35}{r}}\text{LHL}\Rightarrow & \text{f}\left( \text{x} \right)=8+2-1+3=12 \\{} & \text{f}\left( 2 \right)=8+2-2+4=12 \\\end{array} \right\}\text{ }\!\!~\!\!\text{ Cont}\text{. }\!\!~\!\!\text{ }

at x=0\text{x}=0 :

\left. \begin{array}{*{35}{r}}\text{ }\!\!~\!\!\text{ LHL }\!\!~\!\!\text{ }\Rightarrow & 0+0-\left( -1 \right)+0=1 \\{} & \text{f}\left( 0 \right)=0 \\\end{array} \right\}\text{ }\!\!~\!\!\text{ Dis}\text{. }\!\!~\!\!\text{ }

at x=1\text{x}=1

\left. \begin{array}{*{35}{r}}\text{LHL}\Rightarrow & 2+1-0+0=3 \\{} & \text{f}\left( 1 \right)=3-1+1=3 \\\text{RHL}\Rightarrow & 2+1-1+1=3 \\\end{array} \right\}\text{ }\!\!~\!\!\text{ Cont}\text{. }\!\!~\!\!\text{ }

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Continuity