The function f(x) = frac{|x - 1|}{x^2} is mon | vedclass

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(D) Given $f(x) = \frac{|x - 1|}{x^2}$.Case $1$: If $x < 1$ and $x 
eq 0$,then $f(x) = \frac{-(x - 1)}{x^2} = \frac{1 - x}{x^2} = x^{-2} - x^{-1}$.$f

Vedclass · Accepted Answer

$(0, 1) \cup (2, \infty)$

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(D) Given $f(x) = \frac{|x - 1|}{x^2}$.Case $1$: If $x $f'(x) = -2x^{-3} + x^{-2} = \frac{-2 + x}{x^3} = \frac{x - 2}{x^3}$.For $f(x)$ to be decreasing,$f'(x) This holds for $x \in (0, 1)$ and $x \in (2, \infty)$ is not applicable here as we assumed $x Thus,for $x \in (0, 1)$,$f'(x) Case $2$: If $x > 1$,then $f(x) = \frac{x - 1}{x^2} = x^{-1} - x^{-2}$.$f'(x) = -x^{-2} + 2x^{-3} = \frac{-x + 2}{x^3} = \frac{-(x - 2)}{x^3}$.For $f(x)$ to be decreasing,$f'(x)  0$.This holds for $x \in (2, \infty)$.Combining both cases,the function is decreasing in $(0, 1) \cup (2, \infty)$.

The function $f(x) = \frac{|x - 1|}{x^2}$ is monotonically decreasing in-

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