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(C) Given the function $f(x) = e^{-\left|\ln x\right|}$ for $x \in (0, \infty)$.We can rewrite the function by considering the definition of the absol

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(C) Given the function $f(x) = e^{-\left|\ln xight|}$ for $x \in (0, \infty)$.We can rewrite the function by considering the definition of the absolute value:$f(x) = \begin{cases} e^{-(-\ln x)} = e^{\ln x} = x, & 0 Now,let's check for continuity at $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} x = 1$Right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} \frac{1}{x} = 1$Value at $x = 1$: $f(1) = \frac{1}{1} = 1$Since the limits are equal,the function is continuous everywhere in its domain. Thus,$m = 0$.Now,let's check for differentiability at $x = 1$:Left-hand derivative: $f'(1^-) = \frac{d}{dx}(x) \big|_{x=1} = 1$Right-hand derivative: $f'(1^+) = \frac{d}{dx}(\frac{1}{x}) \big|_{x=1} = -\frac{1}{x^2} \big|_{x=1} = -1$Since $f'(1^-) 
eq f'(1^+)$,the function is not differentiable at $x = 1$. Thus,$n = 1$.Therefore,$m + n = 0 + 1 = 1$.

Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ are respectively the number of points at which $f$ is not continuous and $f$ is not differentiable,then $m+n$ is

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