Let ,$f:[-1,2] \rightarrow \mathbf{R}$ be given by $f(x)=2 x^2+x+\left[x^2\right]-[x]$, where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points, where $f$ is not continuous, is :
Solution
<p>$$\begin{aligned}
& f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\
& f(-1)=2+1+0=3 \\
& f\left(-1^{+}\right)=2+0+0=2 \\
& f\left(0^{-}\right)=0+1=1 \\
& f\left(0^{+}\right)=0+0+0=0 \\
& f\left(1^{+}\right)=2+1+0=3
\end{aligned}$$</p>
<p>$$\begin{aligned}
& f\left(1^{-}\right)=2+0+1=3 \\
& f\left(2^{-}\right)=8+3+1=12 \\
& f\left(2^{+}\right)=8+4+0=12
\end{aligned}$$</p>
<p>$\therefore$ discontinuous at $x=0, \sqrt{2}, \sqrt{3},-1$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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