Question

If \[f'\left( x \right) > 0\] and \[g'\left( x \right) < 0\], \[x \in R\], then which of the following is true?
A. \[f\left( {g\left( x \right)} \right) > f\left( {g\left( {x + 1} \right)} \right)\]
B. \[f\left( {g\left( x \right)} \right) < f\left( {g\left( {x + 1} \right)} \right)\]
C. \[g\left( {f\left( x \right)} \right) \not > g\left( {f\left( {x + 1} \right)} \right)\]
D. \[g\left( {f\left( x \right)} \right) > g\left( {f\left( {x - 1} \right)} \right)\]

Answer 1

Hint: First we will find whether the functions \[f\left( x \right)\] and \[g\left( x \right)\] are increasing or decreasing. Then we will check each composite function by using the concept of increasing function and decreasing function.

Formula Used: If the derivative of the first order of a function is greater than zero then the function is increasing.
If the derivative of the first order of a function is less than zero then the function is decreasing.

Complete step by step solution: Given that, \[f'\left( x \right) > 0\] and \[g'\left( x \right) < 0\],\[x \in R\]
Since the first order derivative of \[f\left( x \right)\] is greater than zero. So, the function is an increasing function.
That is \[f\left( x \right) < f\left( {x + 1} \right)\]. …….(i)
Since the first order derivative of \[g\left( x \right)\] is greater than zero. So, the function is an increasing function.
That is \[g\left( x \right) > g\left( {x + 1} \right)\]. …..(ii)

Now putting \[x = g\left( x \right)\] and \[x + 1 = g\left( {x + 1} \right)\] in the inequality (i)
Since \[g\left( x \right) > g\left( {x + 1} \right)\] and \[f\left( x \right)\] is an increasing function.
[If the function is an increasing function, then the inequality sign of the composite function will be same as the inequality of inputs.]
Therefore \[f\left( {g\left( x \right)} \right) > f\left( {g\left( {x + 1} \right)} \right)\]
Thus, option A is correct and option B is incorrect.

Now put \[x = f\left( x \right)\] and \[x + 1 = f\left( {x + 1} \right)\] in the inequality (ii)
Since \[f\left( x \right) < f\left( {x + 1} \right)\] and \[g\left( x \right)\] is a decreasing function.
[If the function is an increasing function, then the inequality sign of the composite function will be reverse of the inequality of inputs.]
\[g\left( {f\left( x \right)} \right) > g\left( {f\left( {x + 1} \right)} \right)\]
Thus, option C is incorrect.

We know that \[x > x - 1\] for all \[x \in R\].
Since \[f\left( x \right)\] is an increasing function.
Thus \[f\left( x \right) > f\left( {x - 1} \right)\]
Now put \[x = f\left( x \right)\] and \[x + 1 = f\left( {x - 1} \right)\] in the inequality (ii)
Since \[f\left( x \right) > f\left( {x - 1} \right)\] and \[g\left( x \right)\] is a decreasing function.
\[g\left( {f\left( x \right)} \right) < g\left( {f\left( {x - 1} \right)} \right)\]
Thus, option D is incorrect

Option ‘A’ is correct

Note: Students often confused the inequality sign of composite function.
Remember following the points:
(1) If the function is an increasing function, then the inequality sign of the composite function will be same as the inequality of inputs.
(2) If the function is an increasing function, then the inequality sign of the composite function will be the reverse of the inequality of inputs.