Question

Consider the function $f\left(x\right)=x-|x-x^2|,-1\le x\le 2.$ The points of discontinuities of $f\left(x\right)$ for $x\in [-1,2]$ are:

• A. $x=0,1$
• B. $x=1,2$
• C. $x=0,\frac{1}{2},1$
• D. None of these

Answer 1

The correct option is D None of these

$\begin{array}{c}f\left(x\right)=x-\left|x-x^2\right|x\in \left[-1,2\right]\\ x-\left(x^2-x\right)-1\le x<0\\ =2x-x^2\\ x-\left(x-x^2\right)0\le x<1\\ =x^2\\ x-\left(x^2-x\right)1\le x\le 2\\ =2x-x^2\\ clearlyf:\left(x\right)iscontinuousatatallpoint.\\ Hence,\\ optionDiscorrectanswer.\end{array}$