Let f(x) = min ({x}, {e^{-x}}) for x in [0, 1 | vedclass

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(C) The function is $f(x) = \min (\{x\}, \{e^{-x}\})$.$1$. Discontinuity: The fractional part function $\{x\}$ is discontinuous at all integers $x = 1

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$28$

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(C) The function is $f(x) = \min (\{x\}, \{e^{-x}\})$.$1$. Discontinuity: The fractional part function $\{x\}$ is discontinuous at all integers $x = 1, 2, \dots, 9$. At $x=10$,the interval is closed,so we consider the limit from the left. Thus,there are $9$ points of discontinuity $(x=1, 2, \dots, 9)$. So,$C = 9$.$2$. Non-differentiability: The function is non-differentiable at:- Points of discontinuity: $9$ points.- Points where $\{x\} = \{e^{-x}\}$: In each interval $(n, n+1)$,$\{x\} = x-n$ and $\{e^{-x}\} = e^{-x} - \lfloor e^{-x} \rfloor$. Since $e^{-x}$ is strictly decreasing,$\{e^{-x}\}$ is also piecewise defined. For $x \in [0, 10]$,$\{e^{-x}\}$ has discontinuities at $x = -\ln(k)$ for $k \in \mathbb{Z}$.- Sharp corners: Within each interval $(n, n+1)$,the function switches between $\{x\}$ and $\{e^{-x}\}$. There are $10$ such intervals,and in each,there is exactly one intersection point where the graph has a sharp corner. Thus,there are $10$ such points.- Total non-differentiable points $D = C + 10 = 9 + 10 = 19$.Therefore,$C + D = 9 + 19 = 28$.

List $I$	List $II$
$A. 3x^4 - 2x^3 - 6x^2 + 6x + 1$	$(I)$ has minimum value at $x = 4$
$B. x + \frac{1}{x}, \forall x < 0$	$(II)$ has maximum value at $x = -1$
$C. x^4(7 - x)^3$	$(III)$ has maximum value at $x = 4$
$D. x^4 + (8 - x)^4$	$(IV)$ is decreasing in $[2, \infty)$
	$(V)$ is increasing in $[2, \infty)$

Let $f(x) = \min (\{x\}, \{e^{-x}\})$ for $x \in [0, 10]$. If $C$ and $D$ are the number of points where $f(x)$ is discontinuous and non-differentiable respectively,then $(C + D)$ is equal to (where $\{.\}$ denotes the fractional part function).

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