Question

Let $f\left(x\right)=x|x-x^2|,x\in [-1,1]$. Then the number of points at which f(x) is discontinuous is

• A. 1
• B. 2
• C. 0
• D. None of these

Answer 1

The correct option is C 0

Given $f\left(x\right)=x|x-x^2|$.................[$xϵ[-1,1]$]

Case1: $x-x^2\ge 0$

$⟹x(1-x)\ge 0$

Hence $xϵ[0,1]$

Case2:$x-x^2<0$

$⟹x(1-x)<0$

Hence $xϵ[-1,0)$

So,

$x^2-x^3$......for $0\le x\le 1$

$f\left(x\right)=\{$

$x^3-x^2$......for$-1\le x<0$

Since,

${lim}_{x\to 0^{-}}={lim}_{x\to 0^{+}}$.

So, function is continuous at $x=0$ So function is continuous in all points..