Question

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

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

Answer 1

The correct option is D

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

Finding the value:

Given ,

$f\left(x\right)f\text{'}\left(x\right)<0$

$f\left(x\right)$ and $f\text{'}\left(x\right)$ must be of the opposite sign.

Let $f\left(x\right)=e^{-x}$

$f\text{'}\left(x\right)=-e^{-x}$

$f\left(x\right)>0$ and $f\text{'}\left(x\right)<0$

Let $f\left(x\right)=-e^{-x}$

$f\text{'}\left(x\right)=e^{-x}$

$f\left(x\right)<0$ and $f\text{'}\left(x\right)>0,\forall x\in R$

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

$d\left[\left|f\left(x\right)\right|\right]/dx-e^{-x}<0$ in both cases, $\forall x\in R$

Hence , option D is correct