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QJEEM24Mac87e: JEE Main 2024 Mathematics - Class 12 JEEM24Mac87e Solution: Let , be given by , where denotes the greatest integer less…

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 :

Multiple Choice Options

  • A 4
  • B 3
  • C 5
  • D 6

Step-by-Step Verified Solution

amp;f(x)=2x2+x+[x2]−[x]=2x2+[x2]+{x}amp;f(−1)=2+1+0=3amp;f(−1+)=2+0+0=2amp;f(0−)=0+1=1amp;f(0+)=0+0+0=0amp;f(1+)=2+1+0=3\begin{aligned}& f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\& f(-1)=2+1+0=3 \\& f\left(-1^{+}\right)=2+0+0=2 \\& f\left(0^{-}\right)=0+1=1 \\& f\left(0^{+}\right)=0+0+0=0 \\& f\left(1^{+}\right)=2+1+0=3\end{aligned}

amp;f(1−)=2+0+1=3amp;f(2−)=8+3+1=12amp;f(2+)=8+4+0=12\begin{aligned}& f\left(1^{-}\right)=2+0+1=3 \\& f\left(2^{-}\right)=8+3+1=12 \\& f\left(2^{+}\right)=8+4+0=12\end{aligned}

∴\therefore discontinuous at x=0,2,3,−1x=0, \sqrt{2}, \sqrt{3},-1

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