
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
514.5k+ views
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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
