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Question: Given that \(f'\left( x \right) > g'\left( x \right)\)for all x \[ \in \]R, and f (0) = g (0), then ...

Given that f′(x)>g′(x)f'\left( x \right) > g'\left( x \right)for all x ∈\inR, and f (0) = g (0), then f(x)<g(x)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

Explanation

Solution

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′(x)>g′(x)f'\left( x \right) > g'\left( x \right) andf(0)=g(0)f\left( 0 \right) = g\left( 0 \right) for all x ∈\in R
Let h(x)h\left( x \right) is a function whose value is f(x)−g(x)f\left( x \right)-g\left( x \right) for all x ∈\in R i.e. h(x)h\left( x \right) =f(x)−g(x)f\left( x \right)-g\left( x \right)for all x ∈\in R
So h′(x)h'\left( x \right) = f′(x)−g′(x)f'\left( x \right)-g'\left( x \right)
Since we know that f′(x)>g′(x)f'\left( x \right) > g'\left( x \right) for all x ∈\in R therefore
h′(x)h'\left( x \right) = f′(x)−g′(x)f'\left( x \right)-g'\left( x \right) > 0
Hence the h(x)h\left( x \right) is an increasing function
Now substituting the value of x = 0 we get
h(0)h\left( 0 \right) =f(0)−g(0)f\left( 0 \right)-g\left( 0 \right)
Since it is given that f(0)=g(0)f\left( 0 \right) = g\left( 0 \right)
Therefore h(0)h\left( 0 \right) = 0
For all x ∈\in (0,∞\infty )
Since we know that h(x)h\left( x \right) is an increasing function therefore
h(x)h\left( x \right) > h(0)h\left( 0 \right)
Substituting the value of h(x)h\left( x \right) in the above equation we get
f(x)−g(x)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, 0) Since h\left( x \right)isanincreasingfunctionthereforeis an increasing function therefore h\left( x \right)<<h\left( 0 \right)Substitutingthevalueof Substituting the value ofh\left( x \right)in the above equation we get $$f\left( x \right)-g\left( x \right)$$ < 0 \Rightarrow $$$f\left( x \right) < g\left( x \right)Thereforeforall Therefore for allf\left( x \right) < g\left( x \right)forallxfor all x \in (-$\infty $, 0) Since when x \in belongs to the interval (-$\infty $, 0) satisfies the given condition i.e.f\left( x \right) < g\left( x \right)$$ 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′(x)>0f'\left( x \right) > 0 in the given interval it is called an increasing function whereas when f′(x)<0f'\left( x \right) < 0 then the function is called a decreasing function in the given interval.