Let $f$ be a continuous and differentiable function such that \[f\left( x \right)\] and \[f'\left( x \right)\] have opposite sign everywhere, then
(A) $f$ is increasing
(B) $f$ is decreasing
(C) $\left| f \right|$ is increasing and decreasing
(D) $\left| f \right|$ is decreasing

Hint: For the solution of the question, first find the nature of the given function, if the function is continuous and as well as differentiable and then find the inequality that the function holds because it tells that the function is increasing or decreasing.

Complete step by step solution:
As we know that the function is said to be continuous if it does not show any abrupt changes in values.
It is known that $\left| {f\left( x \right)} \right| = \left\{
  f\left( x \right),\;\;\;\;\;f\left( x \right) \geqslant 0 \\
   - f\left( x \right),\;\;\;f\left( x \right) < 0 \\
  \right.$.
It is given that the function is differentiable.
The differentiation of the function $\left| {f\left( x \right)} \right|$ with respect to $x$ is written as,
$\dfrac{d}{{dx}}\left| {f\left( x \right)} \right| = \left\{
  f'\left( x \right),\;\;\;\;\;f\left( x \right) \geqslant 0 \\
   - f'\left( x \right),\;\;\;f\left( x \right) < 0 \\
  \right.$
It is given that \[f\left( x \right)\] and \[f'\left( x \right)\] have opposite signs everywhere. So, when $f\left( x \right) \geqslant 0$, the $f'\left( x \right)$ will be negative, and when $f\left( x \right) < 0$, $f'\left( x \right)$ will be positive but negative sign with $f'\left( x \right)$ make it negative.
So, from the above explanation it can be concluded that at every point the function is less than zero ($f'\left( x \right) < 0$).
Therefore, the $f$ be a continuous and differentiable function such that \[f\left( x \right)\] and \[f'\left( x \right)\] have opposite sign everywhere, then $\left| f \right|$ is decreasing.

Hence, the correct option is D.

Note: As we know that the nature of the equation of the curve can be obtained by using the differentiation without drawing the curve. If \[f\left( x \right)\] is differentiable in the closed interval ${x_1},{x_2} \in \left[ {a,b} \right]$ such that ${x_1} < {x_2}$, there holds the inequality $f\left( {{x_1}} \right) \leqslant f\left( {{x_2}} \right)$, then the function is called increasing in this interval. If \[f\left( x \right)\] is differentiable in the closed interval ${x_1},{x_2} \in \left[ {a,b} \right]$ such that ${x_1} > {x_2}$, there holds the inequality $f\left( {{x_1}} \right) \geqslant f\left( {{x_2}} \right)$, then the function is called decreasing in this interval.