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(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.We know that $e^{x}$ is differentiable for all

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$R-\{1\}$

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(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.We know that $e^{x}$ is differentiable for all $x \in R$.The function $g(x) = |x-1|$ is continuous everywhere but is not differentiable at $x = 1$ because the left-hand derivative and right-hand derivative at $x = 1$ are not equal.Specifically,at $x = 1$,the left-hand derivative is $-e^{1} = -e$ and the right-hand derivative is $e^{1} = e$.Since the product of a non-differentiable function and a non-zero differentiable function at a point is non-differentiable at that point,$f(x)$ is not differentiable at $x = 1$.Therefore,the set of points where $f(x)$ is differentiable is $R - \{1\}$.

The set of points where the function $f(x)=|x-1| e^{x}$ is differentiable,is

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