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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Continuity and Differentiability

Let f:[-1,2] arrow R be given by f(x)=2 x^2+x+[x^2]-[x] , where [t]...

Let f:[−1,2]→Rf:[-1,2] \rightarrow \mathbf{R} be given by f(x)=2x2+x+[x2]−[x]f(x)=2 x^2+x+\left[x^2\right]-[x], where [t][t] denotes the greatest integer less than or equal to tt. The number of points, where ff is not continuous, is :

Options

  1. A5
  2. B6
  3. C3
  4. D4

Correct answer

D. 4

Step-by-step solution

Doubtful points : −1,0,1,2,3,2-1,0,1, \sqrt{2}, \sqrt{3}, 2 at x=2,3x=\sqrt{2}, \sqrt{3} (f(x)=(2x2+x−[x])+[x2]=Discount↓↓Cont.Cont.\begin{array}{cc} f(x)=\left(2 x^2+x-[x]\right)+\left[x^2\right]= \text{Discount} \\ \downarrow \qquad\qquad\quad \downarrow \\ \text{Cont.} \qquad\qquad \text{Cont.} \end{array})
at x=−1x=-1 :  RHL ⇒f(x)=(2−1−(−1))+0=2f(−1)=2−1−(−1)+1=3\begin{array}{rl}\text{ RHL } \Rightarrow & f(x)=(2-1-(-1))+0=2 \\ & f(-1)=2-1-(-1)+1=3\end{array} Dis. at x=2x=2 : LHL ⇒f(x)=8+2−1+3=12f(2)=8+2−2+4=12\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 x=0\mathrm{x}=0 : LHL ⇒0+0−(−1)+0=1f(0)=0\begin{array}{rl} \text{LHL } \Rightarrow & 0+0-(-1)+0=1 \\ & \mathrm{f}(0)=0 \end{array}\right\} \text{ Dis. } at \mathrm{x}=1 \left.LHL ⇒2+1−0+0=3f(1)=3−1+1=3RHL ⇒2+1−1+1=3\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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