If xy = 1, what is the value of \(\frac{2^{(\mathrm{x}+\mathrm{y})^2}}{2^{(\mathrm{x}-\mathrm{y})^2}}\)?

1. Translate the problem requirements: We need to find the value of a fraction with exponential expressions in both numerator and denominator, where the exponents are \((x+y)^2\) and \((x-y)^2\) respectively, and we're given that \(xy = 1\).
2. Apply exponent rules to simplify the fraction: Use the property that when dividing powers with the same base, we subtract exponents to get \(2^{[(x+y)^2 - (x-y)^2]}\).
3. Expand and simplify the exponent expression: Expand both squared terms and combine like terms to find the simplified exponent.
4. Use the given constraint to evaluate: Apply \(xy = 1\) to determine the final numerical value.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in plain English. We have a fraction where both the top and bottom are powers of 2. The top has 2 raised to the power of \((x+y)^2\), and the bottom has 2 raised to the power of \((x-y)^2\). We need to find the actual numerical value of this entire expression, and we're told that \(xy = 1\).

In mathematical terms, we want to find the value of: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\)

Process Skill: TRANSLATE - Converting the complex exponential expression into a clear mathematical goal

2. Apply exponent rules to simplify the fraction

When we have the same base (in this case, 2) in both the numerator and denominator, we can use a fundamental rule: dividing powers means subtracting the exponents.

Think of it this way: if you have \(2^3 \div 2^1\), that's the same as \(2^{3-1} = 2^2\). The same principle applies here, even though our exponents are more complex.

So our fraction becomes: \(2^{[(x+y)^2 - (x-y)^2]}\)

Now we need to figure out what \((x+y)^2 - (x-y)^2\) equals.

3. Expand and simplify the exponent expression

Let's expand each squared term step by step using the basic algebra pattern \((a+b)^2 = a^2 + 2ab + b^2\):

\((x+y)^2 = x^2 + 2xy + y^2\)
\((x-y)^2 = x^2 - 2xy + y^2\)

Now we subtract the second from the first:
\((x+y)^2 - (x-y)^2 = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)\)

When we distribute the negative sign and combine like terms:
\(= x^2 + 2xy + y^2 - x^2 + 2xy - y^2\)
\(= (x^2 - x^2) + (y^2 - y^2) + (2xy + 2xy)\)
\(= 0 + 0 + 4xy\)
\(= 4xy\)

So our expression becomes: \(2^{4xy}\)

Process Skill: MANIPULATE - Using algebraic expansion and combining like terms to simplify

4. Use the given constraint to evaluate

Now we can use the fact that \(xy = 1\). Since we found that our exponent is \(4xy\), we can substitute:

\(2^{4xy} = 2^{4 \times 1} = 2^4\)

Calculating \(2^4\):
\(2^4 = 2 \times 2 \times 2 \times 2 = 16\)

Process Skill: APPLY CONSTRAINTS - Using the given condition xy = 1 to reach the final numerical answer

Final Answer

The value of \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}\) is 16.

This matches answer choice D, confirming our solution is correct.

Common Faltering Points

Errors while devising the approach

• Misunderstanding the constraint xy = 1: Students may think they need to find specific values of x and y (like x = 1, y = 1) rather than recognizing that the product \(xy = 1\) is sufficient to solve the problem. This leads them down unnecessary paths of trying to solve for individual variables.
• Not recognizing the exponent rule application: Students might attempt to evaluate \(2^{(x+y)^2}\) and \(2^{(x-y)^2}\) separately instead of recognizing they can use the quotient rule for exponents (\(a^m \div a^n = a^{(m-n)}\)) to simplify the fraction first.
• Overlooking algebraic expansion strategy: Students may not realize they need to expand \((x+y)^2 - (x-y)^2\) algebraically and might try to substitute values or use other complex approaches unnecessarily.

Errors while executing the approach

• Algebraic expansion errors: When expanding \((x+y)^2\) and \((x-y)^2\), students commonly make sign errors, especially with \((x-y)^2 = x^2 - 2xy + y^2\). They might write it as \(x^2 + 2xy + y^2\) or forget the middle term entirely.
• Combining like terms incorrectly: When subtracting \((x+y)^2 - (x-y)^2\), students often make errors distributing the negative sign or combining terms. They might get \(2xy\) instead of \(4xy\), or make other arithmetic mistakes in the simplification.
• Substitution timing errors: Students might try to substitute \(xy = 1\) too early in the process, before properly simplifying the algebraic expression, leading to confusion and incorrect calculations.

Errors while selecting the answer

• Calculation errors with powers of 2: Even after correctly arriving at \(2^4\), students might miscalculate this basic power. Common errors include \(2^4 = 8\) (confusing it with \(2^3\)) or \(2^4 = 32\) (confusing it with \(2^5\)).

Alternate Solutions

Smart Numbers Approach

Step 1: Choose specific values that satisfy xy = 1

Let's select \(x = 2\) and \(y = 1/2\), since \(2 \times (1/2) = 1\). These values are chosen because:

• They satisfy our constraint \(xy = 1\)
• They give us nice, workable numbers for \(x + y\) and \(x - y\)
• The calculations will be straightforward

Step 2: Calculate the sum and difference

• \(x + y = 2 + 1/2 = 5/2\)
• \(x - y = 2 - 1/2 = 3/2\)

Step 3: Calculate the squared terms

• \((x + y)^2 = (5/2)^2 = 25/4\)
• \((x - y)^2 = (3/2)^2 = 9/4\)

Step 4: Substitute into the original expression

\(\frac{2^{(x+y)^2}}{2^{(x-y)^2}} = \frac{2^{25/4}}{2^{9/4}}\)

Step 5: Apply exponent rules

\(\frac{2^{25/4}}{2^{9/4}} = 2^{25/4 - 9/4} = 2^{16/4} = 2^4 = 16\)

Verification with different smart numbers:

Let's verify with \(x = 1/2\) and \(y = 2\):

• \(x + y = 1/2 + 2 = 5/2\)
• \(x - y = 1/2 - 2 = -3/2\)
• \((x + y)^2 = 25/4\)
• \((x - y)^2 = (-3/2)^2 = 9/4\)

This gives us the same result: \(2^{25/4 - 9/4} = 2^4 = 16\)

The answer is D: 16.