Question

Consider the function $y=f\left( x \right)$ defined parametrically by $x=2t-\left| t \right|,y={{t}^{2}}+t\left| t \right|,t\ in \mathbb{R}$. Then in the interval $-1\le x\le 1$, the number of points at which $f\left( x \right)$ is not differentiable is ____.

Answer 1

Hint: To find the point at which the function $f\left( x \right)$ is not differentiable, find all possible values of $t$ which satisfy the given conditions and then write the function by eliminating the variable $t$ and check the differentiability of the function in the possible domain.

We have a function $y=f\left( x \right)$ which is defined parametrically by $x=2t-\left|t \right|,y={{t}^{2}}+t\left| t \right|,t\ in \mathbb{R}$
We want to find the number of points at which the given function is not differentiable.
Hence, we will begin by writing the exact function by eliminating the variable $t$.
Consider the case when $t\ge 0$. Thus, we have $x=2t-\left| t \right|=2t-t=t$ and $y={{t}^{2}}+t\left| t \right|={{t}^{2}}+{{t}^{2}}=2{{t}^{2}}$.
We want the condition that $-1\le x\le 1$. Thus, we have $-1\le t\le 1$. But, we are only considering the values for $t\ge 0$.
Hence, we need that $0\le t\le 1$ such that $x=t,y=2{{t}^{2}}$.
By eliminating the variable $t$, we have $x=t,y=2{{t}^{2}}$ for $0\le t\le 1$.
Now, we will consider the case when $t\le 0$. Thus, we have $x=2t-\left| t \right|=2t+t=3t$ and $y={{t}^{2}}+t\left| t \right|={{t}^{2}}-{{t}^{2}}=0$.
We want the condition that $-1\le x\le 1$. Thus, we have $\dfrac{-1}{3}\le t\le \dfrac{1}{3}$. But, we are only considering the values for $t\le 0$.
Hence, we need that $\dfrac{-1}{3}\le t\le 0$ such that $x=3t,y=0$.
Thus, we have $y=0,x=3t$ for $\dfrac{-1}{3}\le t\le 0$.
Now, we will test the differentiability of the function $y=f\left( x \right)$.
We observe that for $\dfrac{-1}{3}\le t<0$, we have $y=0,x=3t$. Thus, the functions $x\left( t \right)$ and $y\left( t \right)$ are both polynomials.
We know that polynomials are always differentiable in the given range. Hence, there’s no point in the given range at which the functions are not differentiable.
Similarly, for range $0We know that polynomials are always differentiable in the given range. Hence, there’s no point in the given range at which the functions are not differentiable. Now, we need to check the differentiability around the point \[t=0$.
We know that a function $y=f\left( x \right)$ is differentiable around a point $x=a$ if $f'\left( {{a}^{-}} \right)=f'\left( {{a}^{+}} \right)=f'\left( a \right)$.
For$t=0$, we have $y=0,x=3t$ for $t<0$ and $x=t,y=2{{t}^{2}}$for$t>0$.
We know that differentiation of any function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
For $t<0$, we have $\dfrac{dy}{dt}=\dfrac{d\left( 0 \right)}{dt}=0,\dfrac{dx}{dt}=\dfrac{d\left( 3t \right)}{dt}=3$.
For $t<0$, we have $\dfrac{dy}{dt}=\dfrac{d\left( 2{{t}^{2}} \right)}{dt}=4t,\dfrac{dx}{dt}=\dfrac{d\left( t \right)}{dt}=1$.
As $t\to 0$, we have $\left( \dfrac{dy}{dt},\dfrac{dx}{dt} \right)=\left( 0,3 \right)$ for $t<0$.
As $t\to 0$, we have $\left( \dfrac{dy}{dt},\dfrac{dx}{dt} \right)=\left( 0,1 \right)$ for $t>0$.
Hence, we observe that ${{\left( \dfrac{dy}{dt},\dfrac{dx}{dt} \right)}_{t\to {{0}^{-}}}}\ne {{\left( \dfrac{dy}{dt},\dfrac{dx}{dt} \right)}_{t\to {{0}^{+}}}}$.
Thus, the function $y=f\left( x \right)$ is not differentiable at $t=0$.
Hence, the number of points of non-differentiability of the function is $1$.

Note: It’s very necessary to observe the domain of possible points. We can’t define the function beyond the possible domain.