Practicing Success
Let a function $y=f(x)$ be defined as $x=2 t-|t|, y=t^2+t|t|$, where $t \in R$. Then, $f(x)$ is |
continuous and differentiable in $[-1,1]$ continuous but not differentiable in $[-1,1]$ continuous in $[-1,1]$ and differentiable in $(-1,1)$ only none of these |
continuous and differentiable in $[-1,1]$ |
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 $[-1,1]$. |