Question

The function $y=f\left(x\right)$ is defined by $x=2t-|t|,y=t^2+t|t|,t\in R$ in the interval $x\in [-1,1]$ then $f\left(x\right)$ is

• A. continuous every where
• B. not continuous at $x=0$
• C. continuous but not differentiable at $x=0$
• D. a constant function

Answer 1

The correct option is A continuous every where

When $t\ge 0,x=2t-t=t$
$y=t^2+t^2=2t^2$
$\Rightarrow y=2x^2,x\ge 0$
When $t<0$
$x=2t-(-t)=3t,y=t^2+t(-t)=0$
$\Rightarrow y=0,x<0$
So, we can write
$f\left(x\right)=\left\{\begin{array}{cc}2x^2& 0\le x\le 1\\ 0& -1\le x<0\end{array}\right\}$
Clearly, $f\left(x\right)$ is continuous and differentiable everywhere.