Question

For \( x > -\dfrac{1}{2} \), let \( f_1(x) = \dfrac{2x}{1+2x} \), \( f_2(x) = \log_e(1 + 2x) \) and \( f_3(x) = 2x \). Then which one of the following is TRUE?
 

• A. \( f_3(x) < f_2(x) < f_1(x) \) for \( 0 < x < \frac{\sqrt{3}}{2} \)
• B. \( f_1(x) < f_3(x) < f_2(x) \) for \( x > 0 \)
• C. \( f_1(x) + f_2(x) < \frac{f_3(x)}{2} \) for \( x > \frac{\sqrt{3}}{2} \)
• D. \( f_2(x) < f_1(x) < f_3(x) \) for \( x > 0 \)

Hint:

For functions involving rational, logarithmic, and linear terms, compare their growth rates to determine their relative order for different values of \( x \).

Answer 1

The Correct Option is D

Step 1: Comparing the functions.
We are given three functions and need to analyze their order for different values of \( x \). We start by evaluating the functions for different intervals of \( x \).

Step 2: Behavior of the functions.
- For \( f_1(x) = \frac{2x}{1 + 2x} \), this is a rational function and is increasing for \( x > 0 \).
- For \( f_2(x) = \log_e(1 + 2x) \), this is a logarithmic function and is also increasing for \( x > 0 \).
- For \( f_3(x) = 2x \), this is a linear function and increases linearly with \( x \).

Step 3: Analyzing the inequalities.
Comparing \( f_1(x) \), \( f_2(x) \), and \( f_3(x) \) for \( x > 0 \), we find that \( f_2(x) \) is smaller than \( f_1(x) \), and \( f_1(x) \) is smaller than \( f_3(x) \), making option (D) the correct choice.

Step 4: Conclusion.
The correct answer is (D) \( f_2(x) < f_1(x) < f_3(x) \) for \( x > 0 \).