Question

The function $f\left(x\right)=x-|x-x^2|$ is

• A. continuous at $x=1$
• B. discontinuous at $x=0$
• C. not defined at $x=1$
• D. not defined at $x=0$

Answer 1

The correct option is A continuous at $x=1$

$f\left(x\right)=x-|x(1-x)|$
$=\{\begin{array}{cc}x+x(1-x)& x\ge 1\\ x-x(1-x)& 0\le x<1\\ x+x(1-x)& x<0\end{array}$
So, doubtfull points are $x=0,1$
at $x=1,f\left(x\right)=1=f\left(1^{+}\right)$ and
$\underset{x\to 1^{-}}{lim}f\left(x\right)=1$,
at $x=0,f\left(x\right)=0=f\left(0^{+}\right)$ and
$\underset{x\to 0^{-}}{lim}f\left(x\right)=0$
Hence, $f\left(x\right)$ is continuous at $x\in R$