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# 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 ____.

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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, thereâ€™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, thereâ€™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: Itâ€™s very necessary to observe the domain of possible points. We canâ€™t define the function beyond the possible domain.
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