\(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\) If x neq 0 and x neq 1, and if x is replaced by 1/x everywhere in the expression...

1. Translate the problem requirements: We need to take the expression \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\) and replace every occurrence of x with \(\frac{1}{\mathrm{x}}\), then simplify to see which answer choice it matches
2. Perform the substitution systematically: Replace each x in the original expression with \(\frac{1}{\mathrm{x}}\) to get the new expression
3. Simplify the complex fraction: Work with the resulting expression \(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1}\) by finding common denominators and simplifying
4. Recognize the algebraic relationship: Compare the simplified result with the original expression to identify the pattern

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we need to do here. We start with the expression \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\). The problem asks us to replace every x with \(\frac{1}{\mathrm{x}}\) and see what we get.

Think of this like a simple substitution game - wherever we see the letter x, we're going to put \(\frac{1}{\mathrm{x}}\) in its place. So if our original expression is \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\), after substitution we'll have \(\left(\frac{\frac{1}{\mathrm{x}}+1}{\frac{1}{\mathrm{x}}-1}\right)^2\).

Process Skill: TRANSLATE - Converting the problem instruction into a clear mathematical operation

2. Perform the substitution systematically

Let's carefully replace each x with \(\frac{1}{\mathrm{x}}\) in our expression:

Original expression: \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\)
After substitution: \(\left(\frac{\frac{1}{\mathrm{x}}+1}{\frac{1}{\mathrm{x}}-1}\right)^2\)

Now we have \(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1}\), and we need to square this result.

Let's focus on simplifying the fraction \(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1}\) first, then we'll square it at the end.

3. Simplify the complex fraction

We need to work with \(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1}\). This looks complicated because we have fractions within fractions, but we can clean this up.

For the numerator \(\frac{1}{\mathrm{x}} + 1\):
To add these, we need a common denominator. Since \(1 = \frac{\mathrm{x}}{\mathrm{x}}\), we get:
\(\frac{1}{\mathrm{x}} + 1 = \frac{1}{\mathrm{x}} + \frac{\mathrm{x}}{\mathrm{x}} = \frac{1 + \mathrm{x}}{\mathrm{x}}\)

For the denominator \(\frac{1}{\mathrm{x}} - 1\):
Similarly, \(\frac{1}{\mathrm{x}} - 1 = \frac{1}{\mathrm{x}} - \frac{\mathrm{x}}{\mathrm{x}} = \frac{1 - \mathrm{x}}{\mathrm{x}}\)

So our fraction becomes:
\(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1} = \frac{\frac{1 + \mathrm{x}}{\mathrm{x}}}{\frac{1 - \mathrm{x}}{\mathrm{x}}}\)

When we divide fractions, we multiply by the reciprocal:
\(\frac{\frac{1 + \mathrm{x}}{\mathrm{x}}}{\frac{1 - \mathrm{x}}{\mathrm{x}}} = \frac{1 + \mathrm{x}}{\mathrm{x}} \times \frac{\mathrm{x}}{1 - \mathrm{x}} = \frac{1 + \mathrm{x}}{1 - \mathrm{x}}\)

Process Skill: SIMPLIFY - Breaking down complex fractions into manageable parts

4. Recognize the algebraic relationship

Now we need to square our result:
\(\left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right)^2\)

Let's rearrange this to see if it matches any of our answer choices. Notice that:

• \(1 + \mathrm{x}\) is the same as \(\mathrm{x} + 1\)
• \(1 - \mathrm{x}\) is the same as \(-(\mathrm{x} - 1)\)

So we have:
\(\left(\frac{\mathrm{x} + 1}{-(\mathrm{x} - 1)}\right)^2 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2 \times \left(\frac{1}{-1}\right)^2 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2 \times 1 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2\)

Wait, let me recalculate this more carefully:
\(\left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right)^2 = \left(\frac{\mathrm{x} + 1}{1 - \mathrm{x}}\right)^2\)

Since \(1 - \mathrm{x} = -(\mathrm{x} - 1)\), we get:
\(\left(\frac{\mathrm{x} + 1}{-(\mathrm{x} - 1)}\right)^2 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2 \times (-1)^2 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2 \times 1 = \left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right)^2\)

This is exactly our original expression!

Final Answer

After substituting x with \(\frac{1}{\mathrm{x}}\) in the original expression \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\), we get back to the same expression: \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\).

The answer is A.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the substitution instruction: Students may think they need to substitute only some instances of x with \(\frac{1}{\mathrm{x}}\), rather than replacing ALL occurrences of x. This leads to partial substitution and incorrect setup.

2. Confusing the original expression structure: The expression \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\) can be misread due to unclear parentheses. Students might interpret it as \(\mathrm{x} + \left(\frac{1}{\mathrm{x}-1}\right)^2\) or other variations, leading to wrong substitution targets.

3. Overlooking the constraint implications: Students may not recognize that the constraints \(\mathrm{x} \neq 0\) and \(\mathrm{x} \neq 1\) are crucial for ensuring the substitution \(\frac{1}{\mathrm{x}}\) is valid and the denominators don't become zero.

Errors while executing the approach

1. Complex fraction simplification errors: When dealing with \(\frac{\frac{1}{\mathrm{x}} + 1}{\frac{1}{\mathrm{x}} - 1}\), students often make mistakes in finding common denominators or incorrectly combine the fractions in numerator and denominator, leading to wrong intermediate expressions.

2. Sign errors during algebraic manipulation: Students frequently make errors when handling \(1-\mathrm{x}\) versus \(\mathrm{x}-1\), not recognizing that \(1-\mathrm{x} = -(\mathrm{x}-1)\). This leads to incorrect signs in the final expression and missing the equivalence to the original.

3. Premature squaring: Some students attempt to square the expression before fully simplifying the fraction, making the algebra unnecessarily complex and increasing chances of computational errors.

Errors while selecting the answer

1. Failing to recognize algebraic equivalence: Even after correct computation, students may not recognize that their result \(\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)^2\) is identical to the original expression, leading them to select a different answer choice that looks more 'transformed'.

2. Getting distracted by negative signs: Students who made sign errors during execution might be drawn to answer choice E with the negative sign, thinking their work shows the expression becomes negative after substitution.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a strategic value for x

Let's choose \(\mathrm{x} = 2\). This satisfies our constraints (\(\mathrm{x} \neq 0\) and \(\mathrm{x} \neq 1\)) and will give us clean arithmetic.

Step 2: Calculate the original expression with x = 2

Original expression: \(\frac{(\mathrm{x}+1)^2}{(\mathrm{x}-1)^2}\)

\(= \frac{(2+1)^2}{(2-1)^2}\)

\(= \frac{3^2}{1^2}\)

\(= \frac{9}{1} = 9\)

Step 3: Calculate the new expression after substitution

When we replace x with \(\frac{1}{\mathrm{x}}\), we get \(\mathrm{x} = \frac{1}{2}\)

New expression: \(\frac{\left(\frac{1}{2}+1\right)^2}{\left(\frac{1}{2}-1\right)^2}\)

\(= \frac{\left(\frac{1}{2} + \frac{2}{2}\right)^2}{\left(\frac{1}{2} - \frac{2}{2}\right)^2}\)

\(= \frac{\left(\frac{3}{2}\right)^2}{\left(-\frac{1}{2}\right)^2}\)

\(= \frac{\frac{9}{4}}{\frac{1}{4}}\)

\(= \frac{9}{4} \times \frac{4}{1} = 9\)

Step 4: Test the answer choices with x = 2

1. \(\frac{(\mathrm{x}+1)^2}{(\mathrm{x}-1)^2} = 9\) ✓
2. \(\frac{(\mathrm{x}-1)^2}{(\mathrm{x}+1)^2} = \frac{1^2}{3^2} = \frac{1}{9}\) ✗
3. \(\frac{\mathrm{x}^2+1}{1-\mathrm{x}^2} = \frac{5}{1-4} = \frac{5}{-3} = -\frac{5}{3}\) ✗
4. \(\frac{\mathrm{x}^2-1}{\mathrm{x}^2+1} = \frac{3}{5}\) ✗
5. \(-\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^2 = -\left(\frac{1}{3}\right)^2 = -\frac{1}{9}\) ✗

Step 5: Verify with another value

Let's verify with \(\mathrm{x} = 3\):

Original: \(\frac{(3+1)^2}{(3-1)^2} = \frac{16}{4} = 4\)

After substitution (\(\mathrm{x} = \frac{1}{3}\)): \(\frac{\left(\frac{1}{3}+1\right)^2}{\left(\frac{1}{3}-1\right)^2} = \frac{\left(\frac{4}{3}\right)^2}{\left(-\frac{2}{3}\right)^2} = \frac{\frac{16}{9}}{\frac{4}{9}} = 4\)

Answer choice A with \(\mathrm{x} = 3\): \(\frac{(3+1)^2}{(3-1)^2} = 4\) ✓

Conclusion: The answer is A.