Analyzing Function Properties

The problem asks us to analyze two statements about the given function \({\rm{f}}\left( {\rm{a}} \right) = \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}}\). We need to determine which of these statements is/are correct by evaluating the function at different inputs.

The given function is: \({\rm{f}}\left( {\rm{a}} \right) = \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}}\)

Evaluating Statement 1: \({\rm{f}}\left( {2{\rm{a}}} \right) = {\rm{f}}\left( {\rm{a}} \right) + 1\)

Let's evaluate both sides of the equation in Statement 1.

First, let's find \({\rm{f}}\left( {2{\rm{a}}} \right)\). We replace \({\rm{a}}\) with \(2{\rm{a}}\) in the function definition:

$${\rm{f}}\left( {2{\rm{a}}} \right) = \frac{{\left( {2{\rm{a}}} \right) - 1}}{{\left( {2{\rm{a}}} \right) + 1}} = \frac{{2{\rm{a}} - 1}}{{2{\rm{a}} + 1}}$$

Next, let's find \({\rm{f}}\left( {\rm{a}} \right) + 1\). We take the original function definition and add 1:

$${\rm{f}}\left( {\rm{a}} \right) + 1 = \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}} + 1$$

To add 1, we find a common denominator, which is \({\rm{a}} + 1\):

$${\rm{f}}\left( {\rm{a}} \right) + 1 = \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}} + \frac{{{\rm{a}} + 1}}{{{\rm{a}} + 1}} = \frac{{\left( {{\rm{a}} - 1} \right) + \left( {{\rm{a}} + 1} \right)}}{{{\rm{a}} + 1}} = \frac{{{\rm{a}} - 1 + {\rm{a}} + 1}}{{{\rm{a}} + 1}} = \frac{{2{\rm{a}}}}{{{\rm{a}} + 1}}$$

Now we compare \({\rm{f}}\left( {2{\rm{a}}} \right)\) and \({\rm{f}}\left( {\rm{a}} \right) + 1\):

• \({\rm{f}}\left( {2{\rm{a}}} \right) = \frac{{2{\rm{a}} - 1}}{{2{\rm{a}} + 1}}\)
• \({\rm{f}}\left( {\rm{a}} \right) + 1 = \frac{{2{\rm{a}}}}{{{\rm{a}} + 1}}\)

These two expressions are generally not equal. For example, if \({\rm{a}} = 2\):

• \({\rm{f}}\left( {4} \right) = \frac{{4 - 1}}{{4 + 1}} = \frac{3}{5}\)
• \({\rm{f}}\left( {2} \right) + 1 = \frac{{2 - 1}}{{2 + 1}} + 1 = \frac{1}{3} + 1 = \frac{1}{3} + \frac{3}{3} = \frac{4}{3}\)

Since \(\frac{3}{5} \neq \frac{4}{3}\), Statement 1 is incorrect.

Evaluating Statement 2: \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = - {\rm{f}}\left( {\rm{a}} \right)\)

Let's evaluate both sides of the equation in Statement 2.

First, let's find \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right)\). We replace \({\rm{a}}\) with \(\frac{1}{{\rm{a}}}\) in the function definition:

$${\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = \frac{{\left( {\frac{1}{{\rm{a}}}} \right) - 1}}{{\left( {\frac{1}{{\rm{a}}}} \right) + 1}}$$

To simplify this complex fraction, we can multiply the numerator and denominator by \({\rm{a}}\) (assuming \({\rm{a}} \neq 0\)):

$${\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = \frac{{\left( {\frac{1}{{\rm{a}}}} - 1 \right) \times {\rm{a}}}}{{\left( {\frac{1}{{\rm{a}}}} + 1 \right) \times {\rm{a}}}} = \frac{{1 - {\rm{a}}}}{{1 + {\rm{a}}}}$$

Next, let's find \(- {\rm{f}}\left( {\rm{a}} \right)\). We take the original function definition and multiply by -1:

$$- {\rm{f}}\left( {\rm{a}} \right) = - \left( \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}} \right) = \frac{{ - \left( {{\rm{a}} - 1} \right)}}{{{\rm{a}} + 1}} = \frac{{ - {\rm{a}} + 1}}{{{\rm{a}} + 1}} = \frac{{1 - {\rm{a}}}}{{{\rm{a}} + 1}}$$

Now we compare \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right)\) and \(- {\rm{f}}\left( {\rm{a}} \right)\):

• \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = \frac{{1 - {\rm{a}}}}{{1 + {\rm{a}}}}\)
• \(- {\rm{f}}\left( {\rm{a}} \right) = \frac{{1 - {\rm{a}}}}{{{\rm{a}} + 1}}\)

Since \(1 + {\rm{a}}\) is the same as \({\rm{a}} + 1\), the two expressions are equal. Therefore, Statement 2 is correct.

Conclusion

Based on our analysis:

• Statement 1: \({\rm{f}}\left( {2{\rm{a}}} \right) = {\rm{f}}\left( {\rm{a}} \right) + 1\) is incorrect.
• Statement 2: \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = - {\rm{f}}\left( {\rm{a}} \right)\) is correct.

Thus, only Statement 2 is correct.

The correct option is the one stating that only 2 is correct.

Revision Table: Key Concepts

Additional Information: Function Properties & Transformations

This problem explores properties of a specific rational function. Checking if \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = - {\rm{f}}\left( {\rm{a}} \right)\) is a test for a specific type of symmetry or transformation property. Functions can exhibit various properties based on how inputs and outputs relate:

• Even Functions: A function \({\rm{f}}\left( {\rm{x}} \right)\) is even if \({\rm{f}}\left( { - {\rm{x}}} \right) = {\rm{f}}\left( {\rm{x}} \right)\) for all \({\rm{x}}\) in its domain. Even functions are symmetric about the y-axis.
• Odd Functions: A function \({\rm{f}}\left( {\rm{x}} \right)\) is odd if \({\rm{f}}\left( { - {\rm{x}}} \right) = - {\rm{f}}\left( {\rm{x}} \right)\) for all \({\rm{x}}\) in its domain. Odd functions are symmetric about the origin.
• Transformations: Evaluating \({\rm{f}}\left( {c{\rm{x}}} \right)\), \({\rm{f}}\left( {\rm{x}} + c \right)\), \({\rm{f}}\left( {\rm{x}} \right) + c\), \(c{\rm{f}}\left( {\rm{x}} \right)\), etc., represents transformations of the function's graph (stretching/compressing, shifting, reflections).

The second statement \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = - {\rm{f}}\left( {\rm{a}} \right)\) tests a different kind of property involving the reciprocal of the input. This function \({\rm{f}}\left( {\rm{a}} \right) = \frac{{{\rm{a}} - 1}}{{{\rm{a}} + 1}}\) is sometimes called a Mobius transformation or fractional linear transformation. These functions have interesting properties under composition and inversion.