Let f: [-1, 2] → R be given by f(x) = 2x^2 + x + [x^2] - [x], where [t] denotes the greatest integer

Question

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

(1) 6

(2) 3

(3) 4

(4) 5

Answer 1

Correct option is (3) 4

Doubtful points: \(-1,0,1, \sqrt{2}, \sqrt{3}, 2\)

at \(\mathrm{x}=\sqrt{2}, \sqrt{3}\)

\(\mathrm{f}(\mathrm{x})=(\underset{\substack{\;\;\;\downarrow \\ \;\;\;\text { Cont. }}}{\mathrm{2x^2 +x}}-[x])+\underset{\substack{\downarrow \\ \text { Cont. }}}{[x^2]}=\text{Discount}\)

at \(\mathrm{x}=-1\):

\(\left.\begin{array}{rl}\text {RHL } \Rightarrow \mathrm{f}(\mathrm{x})=(2-1-(-1))+0=2 \\ \mathrm{f}(-1)=2-1-(-1)+1=3\end{array}\right\} \text{Dis.}\)

at \(\mathrm{x}=2\):

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

at \(\mathrm{x}=0\):

\(\left.\begin{array}{rl} \text {LHL } \Rightarrow & 0+0-(-1)+0=1 \\ & f(0)=0 \end{array}\right\} \text { Dis. }\)

at \(\mathrm{x}=1\):

\(\left.\begin{array}{rl} \mathrm{LHL} \Rightarrow & 2+1-0+0=3 \\ & \mathrm{f}(1)=3-1+1=3 \\ \mathrm{RHL} \Rightarrow & 2+1-1+1=3 \end{array}\right\} \text { Cont. }\)