Question

The number of points where the function $f\left(x\right)=\{\begin{array}{ccc}|2x^2-3x-7|,& \text{if}& x\le -1\\ [4x^2-1],& \text{if}& -1<x<1\\ |x+1|+|x-2|& \text{if},& x\ge 1\end{array},$ where $\left[t\right]$ denotes the greatest integer $\le t$, is discontinuous is

• A. 7
• B. 07
• C. 7.00
• D. 7.0

Answer 1

Answer: (A), (B), (C), (D)

Given, $f\left(x\right)=\{\begin{array}{cc}|2x^2-3x-7|,& \text{if}x\le -1\\ [4x^2-1],& \text{if}-1<x<1\\ |x+1|+|x-2|,& \text{if}x\ge 1\end{array}$

Since $f(-1^{-})=2=f(-1^{+})=f(-1)$
$\therefore f\left(x\right)$ is continuous at $x=-1$

and $f\left(1^{-}\right)=2$ and $f\left(1^{+}\right)=3=f\left(1\right)$
$\therefore f\left(x\right)$ is discontinuous at $x=1$

and also whenever $4x^2-1=0,1$ or $2$
$\Rightarrow x=\pm \frac{1}{2},\pm \frac{1}{\sqrt{2}}$ and $\pm \frac{\sqrt{3}}{2}$
So, there are total $7$ points of discontinuity.