Let D be the set of real numbers d such that 2 - (d + 1)^(1/2) is

Question

Let D be the set of real numbers d such that \(2 - \sqrt{d+1}\) is a real number. Let R be the set of real numbers r such that r = 2 - \(\sqrt{d+1}\) for at least 1 value of d in D. The intersection of D and R is the set of all numbers v such that

A. -1 <= v<=2

B. -1 < v < 2

C. v >= -1

D. v <= 2

E. v is a real number.

Bunuel · Accepted Answer

Let D be the set of real numbers d such that \(2 - \sqrt{d+1}\) is a real number. Let R be the set of real numbers r such that r = 2 - \(\sqrt{d+1}\) for at least 1 value of d in D. The intersection of D and R is the set of all numbers v such that

A. -1 <= v<=2

B. -1 < v < 2

C. v >= -1

D. v <= 2

E. v is a real number.

A. -1 ≤ v ≤ 2
B. -1 < v < 2
C. v ≥ -1
D. v ≤ 2
E. v is a real number.

We are given that  D  is the set of real numbers  d  such that  2 - \sqrt{d+1}  is a real number:

For  2 - \sqrt{d+1}  to be a real number, the expression under the square root must be non-negative. Hence,  2 - \sqrt{d+1}  is a real number when  d \geq -1 . So, D consist of all number show in green below:

---------  (-1)----------------------

We are also told that  R  is the set of real numbers  r  such that  r = 2 - \sqrt{d+1}  for at least 1 value of  d  in  D . So, d in that expression is from D above ( d \geq -1 ), which makes  \sqrt{d+1}  defined. In this case, since the square root cannot give negative result, r would equal to 2 - (some non-negative value), which would make  r \leq 2  So, R consist of all number show in green below:

--------(-1)------------(2)  --------

The intersection of  D  and  R   consist of all number show in green below:

---------  (-1)------------(2)  --------

Therefore,  -1 ≤ v ≤ 2 .

Answer: A.

gmatophobia · Answer

Given  :

D  be the set of real numbers  d  such that  2 - \sqrt{d+1}  is a real number.

Inference : 
   d+1 \geq 0  → as the value under root has to be non-negative. Hence,  d \geq -1  
 Set D can have any value greater than or equal to -1 up to positive infinity i.e. D =

R  be the set of real numbers  r  such that  r  = 2 -  \sqrt{d+1}

Inference : 
   We are substracting a non negative value,  \sqrt{d+1} , from 2. Hence, the maximum value in set R equals to 2. 
 Set R will have any value less than or equal to 2 up to negative infinity i.e. R =

The intersection of  D  and  R  is the set of all numbers v

Range of v =

Lower Bound  = Higher of the lowest value of (Set D, Set R) = -1 
   Upper Bound  = Lower of the highest value of (Set D, Set R) = 2

Range of  v =    i.e  -1 \leq v \leq 2

Option A

KarishmaB · Answer

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

When is \(2 - \sqrt{d+1}\) a real number? The only thing that will make this number imaginary is if we have a negative under the sqaure root sign. We have (d+1) under the square root so when d = -1, we get 0 and when d < -1, we get negative.
Hence, as long as d >= -1, \(2 - \sqrt{d+1}\) will remain a real number.

Let R be the set of real numbers r such that \(r = 2 - \sqrt{d+1}\) for at least 1 value of d in D.

What are they talking about here? Here we have the same \(2 - \sqrt{d+1}\) that we were discussing in D. When we plug in various values of d (which range from - 1 to infinity), we will get the various values of r. So essentially set R is the range of D (so the question is discussing domain and range - terms that you don't really need to know but it helps if you can identify because everything becomes clear then)
When I plug in d = -1, I will get r = 2. When I plug in d = 0, I will get r = 1. When I plug in d = 8, I will get r = -1 etc. So r is taking values from 2 to -infinity.

What is the intersection of D and R? D >= -1 and R <= 2 so their common elements lie in the region -1 to 2 inclusive.

Answer (A)

DanTheGMATMan · Answer

This is just testing your recognition of some basic algebra rules regarding real numbers:
 https://www.youtube.com/watch?v=3oCA78PRPFY

ManifestDreamMBA · Answer

Since  2 - \sqrt{d+1}  is a real number
d+1>=0
d>=-1
This eliminates D and E, since overlap v cannot have anything < -1

Given  r = 2 - \sqrt{d+1} , max value of r is 2. eliminates C

If I substitute d = -1, I get r = 2, both -1 and 2 are inclusive, so A is the answer

GMATinsight · Answer

Things to note:    
- Real Number: A real number is one which can be represented on Number line 
- Non-Real Number: When we have negative value under the square root √(-ve) 
- Square root of any number is always greater than or equal to zero

2 - \sqrt{d+1}  is real i.e. (d+1) ≥ 0 i.e. d ≥ -1

r = 2 - \sqrt{d+1}  is real i.e. r ≤ 2

The intersection of r and d is values between -1 and 2 (both inclusive i.e. {-1, 0, 1, 2}

Answer: Option A

Kinshook · Answer

Let D be the set of real numbers d such that \(2 - \sqrt{d+1}\) is a real number. Let R be the set of real numbers r such that \(r = 2 - \sqrt{d+1}\) for at least 1 value of d in D. The intersection of D and R is the set of all numbers v such that

Let D be the set of real numbers d such that \(2 - \sqrt{d+1}\) is a real number.
d+1>=0
d>=-1

Let R be the set of real numbers r such that \(r = 2 - \sqrt{d+1}\) for at least 1 value of d in D.
Since d+1>=0
r<=2

The intersection of D and R is the set of all numbers v:-
v will satisfy conditions for both d & r
-1<=v<=2

A. -1 ≤ v ≤ 2
B. -1 < v < 2
C. v ≥ -1
D. v ≤ 2
E. v is a real number.

IMO A

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=83564 Harsha GMAT-Club-Forum-4c9pjoty.png

rak08 · Answer

how do we eliminate "E"

Bunuel · Answer

The reason we eliminate E ("v is a real number") is because the intersection is not  all  real numbers. The intersection is specifically bounded between -1 and 2. So E is too broad and doesn’t match the defined overlap of D and R.

Jxmes · Answer

just flagging that it's the entire continuum between -1 and 2 inclusive (i.e. all real numbers in that interval), not just the integers. just in case anyone reads this and is confused.

Bunuel , curious whether you think this question is tagged with the correct diff? does not seem to belong with other 805+ questions in my opinion.

Bunuel · Answer

The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question. So, this is an 805+ Level level question based on our statistics.

Jxmes · Answer

understood - thanks for clearing that up.

Let D be the set of real numbers d such that 2 - (d + 1)^(1/2) is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Let D be the set of real numbers d such that 2 - (d + 1)^(1/2) is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards