Let a function y=f(x) be defined as x=2 t-|t|, y=t^2+t|t|, where t ∈…

As $t \in R$. So, the following cases arise:

CASE I: When $t \geq 0$

In this case, we have

$x =2 t-t=t$ and $y=t^2+t^2=2 t^2$

$\Rightarrow y =2 x^2, x \geq 0$

CASE II: When $t<0$

In this case, we have

$x =2 t+t=3 t$ and  $y=t^2-t^2=0$

$\Rightarrow y =0$  for all $x<0$

Thus, we have

$y=f(x)=\left\{\begin{aligned} 2 x^2, & x \geq 0 \\ 0, & x<0 \end{aligned}\right.$

Clearly, f(x) is everywhere continuous and differentiable.

Hence, f(x) is continuous and differentiable on $[-1,1]$.