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(D) Given, $f(x) f^{\prime}(x) < 0$. This implies that $f(x)$ and $f^{\prime}(x)$ have opposite signs for all $x \in R$. Consider the function $g(x) =

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$|f(x)|$ must be a decreasing function

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(D) Given, $f(x) f^{\prime}(x) Consider the function $g(x) = |f(x)|$. Then $g(x)^2 = |f(x)|^2 = f(x)^2$. Differentiating both sides with respect to $x$, we get $2g(x) g^{\prime}(x) = 2f(x) f^{\prime}(x)$. Thus, $g^{\prime}(x) = \frac{f(x) f^{\prime}(x)}{g(x)} = \frac{f(x) f^{\prime}(x)}{|f(x)|}$. Since $f(x) f^{\prime}(x)  0$ for all $x$ where $f(x) 
eq 0$, it follows that $g^{\prime}(x) Therefore, $|f(x)|$ is a decreasing function.

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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