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If $f$ is a real-valued differentiable function such that $f(x) f^{\prime}(x) < 0$ for all real $x,$ then
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The correct answer is:
$|f(x)|$ must be a decreasing function
Given, $f(x) f^{\prime}(x) < 0$ $\Rightarrow f(x)$ and $f^{\prime}(x)$ must be of opposite sign.
(i) Let $f(x)=e^{-x}$
$\therefore \quad f^{\prime}(x)=-e^{-x}$
$\Rightarrow f(x)>0$ and $f^{\prime}(x) < 0, \forall x \in R$
(ii) Let $f(x)=-e^{-x}$
$\therefore \quad f^{\prime}(x)=e^{-x}$
$\Rightarrow f(x) < 0$ and $f^{\prime}(x)>0, \forall x \in R$
But $|f(x)|=\left|\pm e^{-x}\right|=e^{-x}$ in both cases
$\therefore \frac{d}{d x}|f(x)|=-e^{-x} < 0$ in both case, $\forall x \in R$
$\Rightarrow|f(x)|$ must be a decreasing function
(i) Let $f(x)=e^{-x}$
$\therefore \quad f^{\prime}(x)=-e^{-x}$
$\Rightarrow f(x)>0$ and $f^{\prime}(x) < 0, \forall x \in R$
(ii) Let $f(x)=-e^{-x}$
$\therefore \quad f^{\prime}(x)=e^{-x}$
$\Rightarrow f(x) < 0$ and $f^{\prime}(x)>0, \forall x \in R$
But $|f(x)|=\left|\pm e^{-x}\right|=e^{-x}$ in both cases
$\therefore \frac{d}{d x}|f(x)|=-e^{-x} < 0$ in both case, $\forall x \in R$
$\Rightarrow|f(x)|$ must be a decreasing function
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