Let D be the set of real numbers d such that 2 - sqrt(d+1) is a real number. Let R...

1. Translate the problem requirements: Set D contains all real numbers d where \(2 - \sqrt{\mathrm{d}+1}\) is real (meaning \(\sqrt{\mathrm{d}+1}\) must be real). Set R contains all possible output values \(\mathrm{r} = 2 - \sqrt{\mathrm{d}+1}\) when d is in D. We need to find which numbers appear in both D and R.
2. Determine the domain D: Find all values of d for which \(\sqrt{\mathrm{d}+1}\) is real by identifying when the expression under the square root is non-negative.
3. Determine the range R: Find all possible output values of \(\mathrm{r} = 2 - \sqrt{\mathrm{d}+1}\) as d varies over the domain D, considering the behavior of the square root function.
4. Find the intersection D ∩ R: Identify which values appear in both sets by finding the overlap between the domain and range.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have a mathematical expression \(2 - \sqrt{\mathrm{d}+1}\), and we need to understand when this expression makes sense and what values it can produce.

Set D contains all the input values d where the expression \(2 - \sqrt{\mathrm{d}+1}\) is a real number. This means we need the square root \(\sqrt{\mathrm{d}+1}\) to be a real number, which happens when what's inside the square root is not negative.

Set R contains all the possible output values that the expression \(2 - \sqrt{\mathrm{d}+1}\) can produce when we use valid inputs from set D.

The intersection \(\mathrm{D} ∩ \mathrm{R}\) means we want to find numbers that can serve as both valid inputs AND appear as possible outputs.

Process Skill: TRANSLATE - Converting the set notation and function language into concrete mathematical requirements

2. Determine the domain D

For \(2 - \sqrt{\mathrm{d}+1}\) to be a real number, we need \(\sqrt{\mathrm{d}+1}\) to be real. A square root is real when the expression under the radical is greater than or equal to zero.

So we need: \(\mathrm{d} + 1 \geq 0\)
Solving this: \(\mathrm{d} \geq -1\)

Therefore, \(\mathrm{D} = \{\mathrm{d} | \mathrm{d} \geq -1\}\), which means D contains all real numbers greater than or equal to -1.

3. Determine the range R

Now we need to find all possible values of \(\mathrm{r} = 2 - \sqrt{\mathrm{d}+1}\) when \(\mathrm{d} \geq -1\).

Let's think about this step by step:

• When \(\mathrm{d} = -1\) (the smallest value in D): \(\sqrt{\mathrm{d}+1} = \sqrt{0} = 0\), so \(\mathrm{r} = 2 - 0 = 2\)
• As d increases from -1, \((\mathrm{d}+1)\) increases from 0, so \(\sqrt{\mathrm{d}+1}\) increases from 0
• As \(\sqrt{\mathrm{d}+1}\) gets larger, the expression \(2 - \sqrt{\mathrm{d}+1}\) gets smaller
• Since \(\sqrt{\mathrm{d}+1}\) can become arbitrarily large as d increases, \(2 - \sqrt{\mathrm{d}+1}\) can become arbitrarily small (approaching negative infinity)

Therefore, the range R contains all values r such that \(\mathrm{r} \leq 2\).
So \(\mathrm{R} = \{\mathrm{r} | \mathrm{r} \leq 2\}\)

Process Skill: INFER - Recognizing that as the square root increases without bound, the function value decreases without bound

4. Find the intersection D ∩ R

Now we need to find which numbers appear in both sets:

• \(\mathrm{D} = \{\mathrm{d} | \mathrm{d} \geq -1\}\) (all numbers greater than or equal to -1)
• \(\mathrm{R} = \{\mathrm{r} | \mathrm{r} \leq 2\}\) (all numbers less than or equal to 2)

The intersection \(\mathrm{D} ∩ \mathrm{R}\) consists of numbers that satisfy both conditions:

• Must be \(\geq -1\) (to be in D)
• Must be \(\leq 2\) (to be in R)

Therefore: \(\mathrm{D} ∩ \mathrm{R} = \{\mathrm{v} | -1 \leq \mathrm{v} \leq 2\}\)

4. Final Answer

The intersection of D and R is the set of all numbers v such that \(-1 \leq \mathrm{v} \leq 2\).

This matches answer choice A: "\(-1 \leq \mathrm{v} \leq 2\)"

Common Faltering Points

Errors while devising the approach

Students often confuse the intersection with other operations or think it means "values that make the function undefined" rather than "values that can serve as both valid inputs AND possible outputs of the function."

Some students may overthink this phrase and assume it requires special conditions or multiple solutions, when it simply means the function produces that output value for some valid input.

Students may mix up which set represents valid inputs (domain D) versus possible outputs (range R), leading them to set up the wrong inequalities or constraints.

Errors while executing the approach

When finding the range R, students may forget to check the boundary behavior carefully. They might miss that when \(\mathrm{d} = -1\), we get \(\mathrm{r} = 2\), or incorrectly conclude that r can exceed 2.

As d increases, \(\sqrt{\mathrm{d}+1}\) increases, so \(2 - \sqrt{\mathrm{d}+1}\) decreases. Students may incorrectly conclude the function is increasing rather than decreasing, leading to wrong range bounds.

Students may incorrectly allow negative values under the square root or forget that \(\sqrt{\mathrm{d}+1} \geq 0\), which is crucial for determining that \(\mathrm{r} \leq 2\).

Errors while selecting the answer

Even with correct calculations, students may select answer choice B ("\(-1 < \mathrm{v} < 2\)") using strict inequalities instead of the correct choice A ("\(-1 \leq \mathrm{v} \leq 2\)") with inclusive boundaries.

Students who correctly find \(\mathrm{D} = \{\mathrm{d} | \mathrm{d} \geq -1\}\) and \(\mathrm{R} = \{\mathrm{r} | \mathrm{r} \leq 2\}\) may mistakenly select answer choice C ("\(\mathrm{v} \geq -1\)") or choice D ("\(\mathrm{v} \leq 2\)") instead of recognizing that the intersection requires BOTH conditions.