Question

The function $f\left(x\right)=x-|x-x^2|,-1\le x\le 1$ is continuous on the interval

• A. $[-1,1]$
• B. $[-1,2]$
• C. $[-1,1]-\left\{0\right\}$
• D. $(-1,1)-\left\{0\right\}$

Answer 1

The correct option is A $[-1,1]$

Given,

$f\left(x\right)=x-|x-x^2|=x-|x(1-x)|$

$\therefore$ continuity is to be checked at $x=0$ and $x=1$
At $x=0$

$LHL=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}[-h-|-h(1+h\left)|\right]=0$

$RHL=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}[h-|h(1-h\left)|\right]=0$

Also, $f\left(0\right)=0$

Since $LHL=RHL=f\left(0\right)$

$\therefore f\left(x\right)$ is continuous at $x=0$
At $x=1$

$LHL=\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}\left[\right(1-h)-|(1-h\left)\right(1-1+h\left)|\right]=1$

$RHL=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}\left[\right(1+h)-|(1+h\left)\right(1-1-h\left)|\right]=1$

Also, $f\left(1\right)=1$

$\therefore f\left(x\right)$ is continuous at $x=1$

Hence $f\left(x\right)$ is continuous for all$x\in [-1,1]$