Question

Given that $f'\left( x \right) > g'\left( x \right)$for all x \[ \in \]R, and f (0) = g (0), then $f\left( x \right) < g\left( x \right)$ for all x belonging to

(A). (0, $\infty $)
(B). (-$\infty $, 0)
(C). R
(D). None of these

Answer 1

Hint: In this question remember the concept of functions, increasing function and decreasing functions and also remember to take h (x) as a function which is equal to f (x) – g (x) for all x \[ \in \] R, use this function to check which one of the options satisfies this function.

Complete step-by-step answer:
According to the given information we have $f'\left( x \right) > g'\left( x \right)$ and\[f\left( 0 \right) = g\left( 0 \right)\] for all x \[ \in \] R
Let $h\left( x \right)$ is a function whose value is \[f\left( x \right)-g\left( x \right)\] for all x \[ \in \] R i.e. $h\left( x \right)$ =\[f\left( x \right)-g\left( x \right)\]for all x \[ \in \] R
So $h'\left( x \right)$ = \[f'\left( x \right)-g'\left( x \right)\]
Since we know that $f'\left( x \right) > g'\left( x \right)$ for all x \[ \in \] R therefore
$h'\left( x \right)$ = \[f'\left( x \right)-g'\left( x \right)\] > 0
Hence the $h\left( x \right)$ is an increasing function
Now substituting the value of x = 0 we get
$h\left( 0 \right)$ =\[f\left( 0 \right)-g\left( 0 \right)\]
Since it is given that \[f\left( 0 \right) = g\left( 0 \right)\]
Therefore $h\left( 0 \right)$ = 0
For all x \[ \in \] (0,$\infty $)
Since we know that $h\left( x \right)$ is an increasing function therefore
$h\left( x \right)$ > $h\left( 0 \right)$
Substituting the value of $h\left( x \right)$ in the above equation we get
\[f\left( x \right)-g\left( x \right)\] > 0
$ \Rightarrow $\[f\left( x \right) > g\left( x \right)\]
Therefore \[f\left( x \right) > g\left( x \right)\] for all x \[ \in \] (0,$\infty $)
Since when x \[ \in \] to the interval (0,$\infty $) doesn’t satisfies the given condition i.e. \[f\left( x \right) > g\left( x \right)\] for the given function
Now for all x \[ \in \] (-$\infty $, 0)
Since $h\left( x \right)$ is an increasing function therefore
$h\left( x \right)$ < $h\left( 0 \right)$
Substituting the value of $h\left( x \right)$ in the above equation we get
\[f\left( x \right)-g\left( x \right)\] < 0
$ \Rightarrow $\[f\left( x \right) < g\left( x \right)\]
Therefore for all \[f\left( x \right) < g\left( x \right)\] for all x \[ \in \] (-$\infty $, 0)
Since when x \[ \in \] belongs to the interval (-$\infty $, 0) satisfies the given condition i.e. \[f\left( x \right) < g\left( x \right)\] for the given function
Hence option B is the correct option.

Note: In the above solution we came across the term function which can be explained as the relation of a set of outputs with its inputs such that each input of the function is related to one output, the representation of function is given as suppose we have a function from a to b then it will be represented as f: a $ \to $ b. When $f'\left( x \right) > 0$ in the given interval it is called an increasing function whereas when $f'\left( x \right) < 0$ then the function is called a decreasing function in the given interval.