Question

If $f$ is a real-valued differentiable function such that $f\left(x\right)f^{\prime }\left(x\right)<0$ for all real $x$, then

• A. $f\left(x\right)$ must be a increasing function
• B. $f\left(x\right)$ must be a decreasing function
• C. $|f\left(x\right)|$ must be a increasing function
• D. $|f\left(x\right)|$ must be a decreasing function

Answer 1

The correct option is D $|f\left(x\right)|$ must be a decreasing function

Given, $f\left(x\right)f^{\prime }\left(x\right)<0$
$\Rightarrow f\left(x\right)$ and $f^{\prime }\left(x\right)$ must be of opposite sign.

(i) Let $f\left(x\right)=e^{-x}$
$\therefore f^{\prime }\left(x\right)-e^{-x}$
$\Rightarrow f\left(x\right)>0$ and $f^{\prime }\left(x\right)<0,\forall x\in R$

(ii) Let $f\left(x\right)=-e^{-x}$
$\therefore f^{\prime }\left(x\right)=e^{-x}$
$\Rightarrow f\left(x\right)<0$ and $f^{\prime }\left(x\right)>0,\forall x\in R$

But $|f\left(x\right)|=|\pm e^{-x}|=e^{-x}$ in both cases

$\therefore \frac{d}{dx}|f\left(x\right)|-e^{-x}<0$ in both case, $\forall \cdot x\in R$

$\Rightarrow |f\left(x\right)|$ must be a decreasing function.