If in the interval [0,3],f(x) = begin{cases} | vedclass

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(C) Given $f(x) = \begin{cases} x\{x\}^2, & x 
otin I \ x, & x \in I \end{cases}$ for $x \in [0,3]$.For $x \in (0,1)$,${x} = x$,so $f(x) = x(x)^2 =

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The number of points of non-differentiability is equal to the number of points of discontinuity.

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(C) Given $f(x) = \begin{cases} x\{x\}^2, & x 
otin I \ x, & x \in I \end{cases}$ for $x \in [0,3]$.For $x \in (0,1)$,${x} = x$,so $f(x) = x(x)^2 = x^3$. At $x=0, f(0)=0$. At $x=1, f(1)=1$.For $x \in (1,2)$,${x} = x-1$,so $f(x) = x(x-1)^2$. At $x=2, f(2)=2$.For $x \in (2,3)$,${x} = x-2$,so $f(x) = x(x-2)^2$. At $x=3, f(3)=3$.Thus,$f(x) = \begin{cases} 0, & x=0 \ x^3, & 0 Checking continuity: $\lim_{x 	o 1^-} f(x) = 1^3 = 1$,$f(1)=1$,$\lim_{x 	o 1^+} f(x) = 1(1-1)^2 = 0$. Since $1 
eq 0$,$f(x)$ is discontinuous at $x=1$.Similarly,$\lim_{x 	o 2^-} f(x) = 2(2-1)^2 = 2$,$f(2)=2$,$\lim_{x 	o 2^+} f(x) = 2(2-2)^2 = 0$. Since $2 
eq 0$,$f(x)$ is discontinuous at $x=2$.Thus,$f(x)$ is discontinuous at $x=1$ and $x=2$. There are two points of discontinuity.Option $C$ is correct because the function is non-differentiable at $x=1, 2$ (discontinuity) and also at $x=0, 3$ (endpoints) and potentially other points,but the number of points of discontinuity is $2$.

If in the interval $[0,3]$,$f(x) = \begin{cases} x\{x\}^2, & x \notin I \\ x, & x \in I \end{cases}$,then which of the following statements is correct? (where $\{.\}$ denotes the fractional part function)

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