Question

A function $f : D \to \mathbb{R}$ is defined as $f(x) = \dfrac{x^2 + 1}{x^2 + x + 1}$ where $D \subseteq \mathbb{R}$ is the domain. The domain(s) on which the function $f(x)$ is one-to-one is/are

• A. Natural numbers
• B. Integers
• C. Rational numbers
• D. Irrational numbers

Hint:

To test one-to-one: find $f'(x)$ sign. If sign doesn’t change in interval, function is monotonic → one-to-one.

Answer 1

The Correct Option is A, B

Step 1: Analyze the function. 
\[ f(x) = \frac{x^2 + 1}{x^2 + x + 1} \] The denominator never becomes zero for real $x$. 
Step 2: Check monotonicity. 
Compute derivative: \[ f'(x) = \frac{(2x)(x^2 + x + 1) - (x^2 + 1)(2x + 1)}{(x^2 + x + 1)^2} = \frac{x(x-1)}{(x^2 + x + 1)^2} \] $f'(x)>0$ for $x>1$ and $f'(x)<0$ for $0<x<1$. Thus, the function is decreasing for $x<1$ and increasing for $x>1$. 
Step 3: Identify one-to-one intervals. 
The function is one-to-one on $(-\infty, 0]$ and $[1, \infty)$. For natural numbers ($x = 1, 2, 3, ...$), it is strictly increasing → one-to-one. 
Step 4: Conclusion. 
Hence, $f(x)$ is one-to-one on natural numbers.