Question

Let $y=f\left( x \right)$ be a function defined parametrically by $x=2t-\left| t-1 \right|$and $y=2{{t}^{2}}+t\left| t \right|$ then $f$ is
A. continuous at x=-1
B. continuous at x=2
C. differentiable at x=1
D. not differentiable at x=2

Answer 1

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.