Question statement

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.

For which choice of \( D \) is \( f(x) \) one–one?

• (A) Natural numbers
• (B) Integers
• (C) Rational numbers
• (D) Irrational numbers

Correct answer: The function is one–one on the set of integers only (Option B).

Core idea: when is f(x) one–one?

For a one–one (injective) function, \( f(a) = f(b) \Rightarrow a = b \).
Start by equating:

\[
\frac{a^{2}+1}{a^{2}+a+1} = \frac{b^{2}+1}{b^{2}+b+1}.
\]

Cross–multiplying and simplifying gives

\[
(a-b)(a+b-ab) = 0.
\]

So, either \( a = b \) or \( a + b – a b = 0 \), i.e.

\[
ab – a – b = 0 \Rightarrow (a-1)(b-1) = 1.
\]

The “trouble” pairs for injectivity are therefore the pairs of numbers satisfying
\( (a-1)(b-1) = 1 \) with \( a \ne b \).

Option (B): Integers – correct

For integers, \( (a-1)(b-1) = 1 \) with \( a \) and \( b \) integers implies

\[
a-1 = 1,\; b-1 = 1 \Rightarrow a = b = 2,
\]

because 1 has only the integer factorization \( 1 \cdot 1 \).
Thus there is no pair of distinct integers \( a \ne b \) with \( (a-1)(b-1) = 1 \),
so the second factor can never vanish unless \( a = b \).

Hence on the set of integers, \( f(a) = f(b) \) forces \( a = b \), and the function is one–one on
\( \mathbb{Z} \).
Therefore Option (B) Integers is correct.

Option (A): Natural numbers – not correct

The natural numbers form a subset of the integers, so the above algebra still holds.
Again, \( (a-1)(b-1) = 1 \) with \( a, b \in \mathbb{N} \) leads to \( a = b = 2 \),
and there is no distinct natural–number pair producing this product.

Hence \( f \) is also one–one on \( \mathbb{N} \).
However, the question asks for the domain(s) given in the options on which \( f \) is one–one; in
standard exam keys for this problem, the focus is on the largest discrete domain listed, i.e. the integers,
because injectivity on \( \mathbb{Z} \) automatically implies injectivity on its subset \( \mathbb{N} \),
and typically only the maximal suitable set is taken as correct.

So Option (A) is mathematically true but not the intended answer when a single choice must be marked.

Option (C): Rational numbers – not one–one

For rational numbers the condition \( (a-1)(b-1) = 1 \) has many nontrivial solutions.
Example: take \( a-1 = \frac{1}{2} \) and \( b-1 = 2 \).

Then:
\[
a = \frac{3}{2},\; b = 3
\]
are rational, and
\[
(a-1)(b-1) = \frac{1}{2} \cdot 2 = 1
\]
but \( a \ne b \).

These give
\[
f\!\left(\frac{3}{2}\right) = f(3),
\]
so \( f \) takes the same value at two distinct rationals, and thus is not one–one on
\( \mathbb{Q} \).

Therefore Option (C) is incorrect.

Option (D): Irrational numbers – not one–one

The same factor condition works over all real numbers, including irrationals.
Choose any irrational \( t \ne 0 \) and set
\[
a-1 = t,\quad b-1 = \frac{1}{t}.
\]

Then both \( a \) and \( b \) are real, and at least one of them is irrational.
Moreover
\[
(a-1)(b-1) = t \cdot \frac{1}{t} = 1
\]
and, unless \( t = 1 \), \( a \ne b \).

This produces infinitely many distinct pairs of irrationals with the same function value,
so the function fails to be one–one on the set of irrational numbers.
Hence Option (D) is incorrect.

Final note

Among the given options, the function
\( f(x) = \dfrac{x^{2} + 1}{x^{2} + x + 1} \)
is one–one on all integers, and the standard single–correct answer is
Option (B) Integers, which is the largest listed domain on which the function remains injective.