f(x) =

x² + 1
x² + x + 1

where D ⊆ 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

Domain on Which f(x) = (x² + 1)/(x² + x + 1) Is One-to-One: Detailed Solution

Correct Options: (A) Natural numbers and (B) Integers

Understanding What a One-to-One Function Means

The problem asks us to identify the domains on which the function
f(x) = (x² + 1)/(x² + x + 1)
is one-to-one. A one-to-one function, also known as an injective function, is a function in which two different inputs cannot produce the same output.

Mathematically, a function f is one-to-one on a domain D if:

f(x) = f(y) ⇒ x = y

Therefore, the most reliable way to solve this question is to assume that two elements x and y from the chosen domain produce the same function value. We then determine the exact condition under which this equality is possible.

The answer depends strongly on the selected domain. The same formula may be one-to-one on natural numbers or integers but fail to be one-to-one on rational or irrational numbers. This makes the analysis of the domain the central idea of the problem.

Checking Whether the Function Is Defined for All Real Numbers

Before testing the one-to-one property, it is useful to examine the denominator:

x² + x + 1

Completing the square gives:

x² + x + 1
=
(x + 1/2)² + 3/4

Since a square is always non-negative,

(x + 1/2)² + 3/4 > 0

for every real value of x. Therefore, the denominator never becomes zero, and the function is defined for all real numbers. Hence, natural numbers, integers, rational numbers, and irrational numbers can all be considered as possible restricted domains.

Applying the One-to-One Function Test

To determine when the function gives equal outputs, suppose:

f(x) = f(y)

Using the definition of the function, we obtain:

x² + 1
x² + x + 1

 =

y² + 1
y² + y + 1

Since both denominators are non-zero, we can cross-multiply:

(x² + 1)(y² + y + 1)
=
(y² + 1)(x² + x + 1)

Expanding both sides gives:

x²y² + x²y + x² + y² + y + 1

=
x²y² + xy² + y² + x² + x + 1

The common terms on both sides cancel, leaving:

x²y + y = xy² + x

Bringing all terms to one side,

x²y − xy² + y − x = 0

Factoring the expression gives:

xy(x − y) − (x − y) = 0

(x − y)(xy − 1) = 0

Therefore, if f(x) = f(y), one of the following two conditions must hold:

x = y

or

xy = 1

The first condition, x = y, is perfectly consistent with a one-to-one function. The second condition is the important one. If a domain contains two distinct numbers x and y whose product is 1, then those two different inputs produce the same output, and the function is not one-to-one on that domain.

The Reciprocal Property of the Function

The condition xy = 1 means:

y = 1/x

Therefore, the function has an important symmetry:

f(x) = f(1/x)

whenever x is non-zero. This means that a domain will fail the one-to-one test if it contains a number and its distinct reciprocal.

For example, if a domain contains both 2 and 1/2, then:

f(2) = f(1/2)

even though 2 ≠ 1/2. Therefore, the entire problem reduces to examining whether each proposed domain contains distinct reciprocal pairs.

Detailed Explanation of Every Option

Option (A): Natural Numbers

Consider the domain of natural numbers:

ℕ = {1, 2, 3, 4, …}

For the function to fail the one-to-one test, there must be two distinct natural numbers x and y satisfying:

xy = 1

Among positive natural numbers, the only possible solution is:

x = 1 and y = 1

However, these are not two distinct inputs because x = y. If x = 2, its reciprocal 1/2 is not a natural number. Similarly, the reciprocals 1/3, 1/4, and so on are not natural numbers.

Therefore, no two distinct natural numbers can satisfy xy = 1. Hence, f(x) = f(y) implies x = y on the natural number domain.

Therefore, the function is one-to-one on the natural numbers.

Thus, Option (A) is correct.

Option (B): Integers

Now consider the domain of integers:

ℤ = {…, −3, −2, −1, 0, 1, 2, 3, …}

Again, we need to determine whether two distinct integers can satisfy:

xy = 1

The only integer solutions are:

x = 1, y = 1

and

x = −1, y = −1

In both cases, x = y. For example, the reciprocal of 2 is 1/2, which is not an integer, while the reciprocal of −2 is −1/2, which is also not an integer.

Therefore, the integer domain does not contain any pair of distinct elements whose product is 1. Consequently, equal function values can occur only when the inputs themselves are equal.

Therefore, the function is one-to-one on the integers.

Thus, Option (B) is correct.

Option (C): Rational Numbers

The rational numbers contain many distinct reciprocal pairs. For example:

x = 2 and y = 1/2

Both 2 and 1/2 are rational numbers, they are distinct, and their product is:

2 × 1/2 = 1

Therefore, according to the condition derived earlier:

f(2) = f(1/2)

even though:

2 ≠ 1/2

We can verify this directly:

f(2) = (2² + 1)/(2² + 2 + 1) = 5/7

and

f(1/2) = 5/7

Thus, two different rational inputs produce the same output. Therefore, the function is not one-to-one on the rational numbers.

Hence, Option (C) is incorrect.

Option (D): Irrational Numbers

The irrational numbers also contain distinct reciprocal pairs. Consider:

x = √2

Its reciprocal is:

y = 1/√2

Both √2 and 1/√2 are irrational numbers, and they are clearly distinct. Their product is:

√2 × 1/√2 = 1

Therefore:

f(√2) = f(1/√2)

while:

√2 ≠ 1/√2

This provides a direct counterexample to the one-to-one property on the domain of irrational numbers. Hence, the function is not one-to-one on the irrational numbers.

Therefore, Option (D) is incorrect.

Why the Domain Changes the One-to-One Property

An important concept demonstrated by this problem is that the one-to-one nature of a function depends not only on its formula but also on its domain. The formula satisfies f(x) = f(1/x), so any domain containing a distinct number and its reciprocal will cause the function to fail the injectivity test.

The natural numbers and integers avoid such distinct reciprocal pairs. In these domains, the only elements whose reciprocals remain in the same domain are 1 and, in the case of integers, −1. Both are equal to their own reciprocals, so they do not create two different inputs with the same output.

In contrast, rational numbers contain pairs such as 2 and 1/2, while irrational numbers contain pairs such as √2 and 1/√2. Therefore, the function fails to be one-to-one on both of these larger domains.

Final Answer

Starting from f(x) = f(y), we obtain:

(x − y)(xy − 1) = 0

Thus, equal function values occur when either x = y or xy = 1. The natural numbers and integers contain no distinct pair of elements whose product is 1. Therefore, the function is one-to-one on both of these domains.

However, rational numbers contain distinct reciprocal pairs such as 2 and 1/2, while irrational numbers contain distinct reciprocal pairs such as √2 and 1/√2. Hence, the function is not one-to-one on either of these domains.

Correct Options: (A) Natural numbers and (B) Integers