Answer

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Hint: First and foremost break x, y in the possible intervals, then gets a relation between them and then find continuity and check differentiability.

Complete step-by-step answer:

In the question, we are given parametric equation of x and y,

$x=2t-\left| t-1 \right|$

$y=2{{t}^{2}}+t\left| t \right|$

Now let’s break or divide the values of x and y in intervals given below,

$\begin{align}

& x=2t-(1-t),t<0 \\

& x=3t-1,t<0 \\

& \\

& x=2t-(1-t),0\le t\le 1 \\

& x=3t-1,0\le t\le 1 \\

& \\

& x=2t-(t-1),t>1 \\

& x=t+1,t>1 \\

& \\

& y=2{{t}^{2}}-{{t}^{2}},t<0 \\

& y={{t}^{2}},t<0 \\

& \\

& y=2{{t}^{2}}+{{t}^{2}},0\le t\le 1 \\

& y=3{{t}^{2}},0\le t\le 1 \\

& \\

& y=2{{t}^{2}}+{{t}^{2}},t>1 \\

& y=3{{t}^{2}},t>1 \\

\end{align}$

So now considering the intervals we will find the relation of x and y in the above interval,

For $t<0$,

$\begin{align}

& x=3t-1 \\

& \Rightarrow t=\dfrac{x+1}{3} \\

& y={{t}^{2}}={{\left( \dfrac{x+1}{3} \right)}^{2}}=\dfrac{{{\left( x+1 \right)}^{2}}}{9} \\

\end{align}$

For $0\le t\le 1$,

$\begin{align}

& x=3t-1 \\

& \Rightarrow t=\dfrac{x+1}{3} \\

& y=3{{t}^{2}}=3{{\left( \dfrac{x+1}{3} \right)}^{2}}=\dfrac{{{\left( x+1 \right)}^{2}}}{3} \\

\end{align}$

For $t>1$,

$\begin{align}

& x=t+1 \\

& \Rightarrow t=x-1 \\

& y=3{{t}^{2}}=3{{\left( x-1 \right)}^{2}} \\

\end{align}$

Now summarizing the equation of y and x,

We get,

\[y=\dfrac{{{\left( x+1 \right)}^{2}}}{9}\] for $x<-1$ (or $t>0$).

\[y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}\] for $-1\le x\le 2$ (or $0\le t\le 1$).

$y=3{{\left( x-1 \right)}^{2}}$ for $x>2$ (or $t>1$).

Now consider the point at $x=-1$.

We will find left hand limit of y,

So,

$\begin{align}

& x=-{{1}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{9}=0 \\

\end{align}$

So the left hand limit is 0.

We will find right hand limit of y,

So,

$\begin{align}

& x=-{{1}^{+}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}=0 \\

\end{align}$

So the right hand limit is 0.

Now consider the point at $x=2$,

We will find Right hand limit of y,

So,

$\begin{align}

& x\to {{2}^{+}} \\

& y=3{{\left( x-1 \right)}^{2}}=3\times {{1}^{2}}=3 \\

\end{align}$

We will find left hand limit of y,

So,

$\begin{align}

& x\to {{2}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}=\dfrac{{{3}^{2}}}{3}=3 \\

\end{align}$

So in the cases of $x=-1$and $x=2$, both the left hand limit and right hand limit are the same.

Here y is continuous at -1 and 2.

At \[x=1\],

We will directly tell that it is continuous and differentiable as it is a polynomial function and unlike the other two points where we had to consider two polynomial functions.

As we already checked the continuity of $x=2$, we will check its left and right hand derivative.

So,

$\begin{align}

& x={{2}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3} \\

& y'=\dfrac{2{{\left( x+1 \right)}^{2}}}{3}=\dfrac{2\left( 2+1 \right)}{3}=2 \\

\end{align}$

Here the left hand derivative is 2.

So, at $x={{2}^{+}}$

$\begin{align}

& y=3{{\left( x-1 \right)}^{2}} \\

& y'=6\left( x-1 \right)=6(2-1)=6 \\

\end{align}$

Here the right hand derivative is 6.

Note: In these types of problems students always have problems finding left hand and right hand derivatives and limits. They make mistakes in left hand limit and right hand limit calculation.

Complete step-by-step answer:

In the question, we are given parametric equation of x and y,

$x=2t-\left| t-1 \right|$

$y=2{{t}^{2}}+t\left| t \right|$

Now let’s break or divide the values of x and y in intervals given below,

$\begin{align}

& x=2t-(1-t),t<0 \\

& x=3t-1,t<0 \\

& \\

& x=2t-(1-t),0\le t\le 1 \\

& x=3t-1,0\le t\le 1 \\

& \\

& x=2t-(t-1),t>1 \\

& x=t+1,t>1 \\

& \\

& y=2{{t}^{2}}-{{t}^{2}},t<0 \\

& y={{t}^{2}},t<0 \\

& \\

& y=2{{t}^{2}}+{{t}^{2}},0\le t\le 1 \\

& y=3{{t}^{2}},0\le t\le 1 \\

& \\

& y=2{{t}^{2}}+{{t}^{2}},t>1 \\

& y=3{{t}^{2}},t>1 \\

\end{align}$

So now considering the intervals we will find the relation of x and y in the above interval,

For $t<0$,

$\begin{align}

& x=3t-1 \\

& \Rightarrow t=\dfrac{x+1}{3} \\

& y={{t}^{2}}={{\left( \dfrac{x+1}{3} \right)}^{2}}=\dfrac{{{\left( x+1 \right)}^{2}}}{9} \\

\end{align}$

For $0\le t\le 1$,

$\begin{align}

& x=3t-1 \\

& \Rightarrow t=\dfrac{x+1}{3} \\

& y=3{{t}^{2}}=3{{\left( \dfrac{x+1}{3} \right)}^{2}}=\dfrac{{{\left( x+1 \right)}^{2}}}{3} \\

\end{align}$

For $t>1$,

$\begin{align}

& x=t+1 \\

& \Rightarrow t=x-1 \\

& y=3{{t}^{2}}=3{{\left( x-1 \right)}^{2}} \\

\end{align}$

Now summarizing the equation of y and x,

We get,

\[y=\dfrac{{{\left( x+1 \right)}^{2}}}{9}\] for $x<-1$ (or $t>0$).

\[y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}\] for $-1\le x\le 2$ (or $0\le t\le 1$).

$y=3{{\left( x-1 \right)}^{2}}$ for $x>2$ (or $t>1$).

Now consider the point at $x=-1$.

We will find left hand limit of y,

So,

$\begin{align}

& x=-{{1}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{9}=0 \\

\end{align}$

So the left hand limit is 0.

We will find right hand limit of y,

So,

$\begin{align}

& x=-{{1}^{+}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}=0 \\

\end{align}$

So the right hand limit is 0.

Now consider the point at $x=2$,

We will find Right hand limit of y,

So,

$\begin{align}

& x\to {{2}^{+}} \\

& y=3{{\left( x-1 \right)}^{2}}=3\times {{1}^{2}}=3 \\

\end{align}$

We will find left hand limit of y,

So,

$\begin{align}

& x\to {{2}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3}=\dfrac{{{3}^{2}}}{3}=3 \\

\end{align}$

So in the cases of $x=-1$and $x=2$, both the left hand limit and right hand limit are the same.

Here y is continuous at -1 and 2.

At \[x=1\],

We will directly tell that it is continuous and differentiable as it is a polynomial function and unlike the other two points where we had to consider two polynomial functions.

As we already checked the continuity of $x=2$, we will check its left and right hand derivative.

So,

$\begin{align}

& x={{2}^{-}} \\

& y=\dfrac{{{\left( x+1 \right)}^{2}}}{3} \\

& y'=\dfrac{2{{\left( x+1 \right)}^{2}}}{3}=\dfrac{2\left( 2+1 \right)}{3}=2 \\

\end{align}$

Here the left hand derivative is 2.

So, at $x={{2}^{+}}$

$\begin{align}

& y=3{{\left( x-1 \right)}^{2}} \\

& y'=6\left( x-1 \right)=6(2-1)=6 \\

\end{align}$

Here the right hand derivative is 6.

Note: In these types of problems students always have problems finding left hand and right hand derivatives and limits. They make mistakes in left hand limit and right hand limit calculation.

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