Last visit was: 05 Aug 2024, 12:30 It is currently 05 Aug 2024, 12:30
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) =

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94796
Own Kudos [?]: 646999 [17]
Given Kudos: 86860
Math Expert
Joined: 02 Sep 2009
Posts: 94796
Own Kudos [?]: 646999 [6]
Given Kudos: 86860
General Discussion
Manager
Joined: 03 Jul 2024
Posts: 77
Own Kudos [?]: 58 [0]
Given Kudos: 4
Manager
Joined: 11 Jun 2024
Posts: 77
Own Kudos [?]: 30 [0]
Given Kudos: 1
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
­for the function to work, the denominator can't be zero.

((x+9)^1/2 - (3-x)^1/2)^1/2 ≠ 0
(x+9)^1/2 - (3-x)^1/2 ≠ 0
(x+9)^1/2 ≠ (3-x)^1/2
x + 9 ≠ 3 - x
2x ≠ -6
x ≠ -3

Looking at the answer choices, D is the only that follow these
Manager
Joined: 12 Sep 2023
Posts: 82
Own Kudos [?]: 15 [0]
Given Kudos: 9
Location: Kenya
GMAT 1: 780 Q50 V48
GRE 1: Q167 V164
GPA: 3.7
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
Square of a negative number doesn't exist. In the upper equation x cannot be less than 2. In the left side equation on the lower side,x cannot be less than -9, on the right side equation on the lower side, x cannot be more than 3

Posted from my mobile device
Director
Joined: 16 Jul 2019
Posts: 659
Own Kudos [?]: 316 [1]
Given Kudos: 172
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
1
Kudos
1) |x|>=2 => x>=2 or x<=-2
2) x+9>=0 =>x>=-9
3) 3-x<=0 , x<=3
4) Denominator>0 => sqrt(x+9)-sqrt(3-x)>0
=> x+9>3-x => x>-3

Combining all 4 we get -3<x<=-2 or 2<=x<=3
we have vertical asymptote at x=-3 (hence exclusive)
B)
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 8023
Own Kudos [?]: 4264 [1]
Given Kudos: 243
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
1
Kudos
Bunuel wrote:
­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that:

A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$

B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$

D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more ­ ­real number will not be possible for function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ at values of x = -3, -9 option C,D,E can be eliminated option B covers complete range where we can get real number $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ OPTION B is correct Intern Joined: 03 Feb 2024 Posts: 10 Own Kudos [?]: 9 [1] Given Kudos: 249 Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] 1 Kudos Bunuel wrote: ­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that: A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$ B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$ D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­  This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

­IMO B

Making whatever is inside the root as non-negative

Numerator: |x|>=2 => x>=2, x<=-2

Denominator: \sqrt{ x+9 } -\sqrt{3 -x } > 0

x >-3

\sqrt{3 -x }3-x >=0
x<=3

­
Manager
Joined: 24 Dec 2023
Posts: 58
Own Kudos [?]: 46 [0]
Given Kudos: 42
Location: India
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
Bunuel wrote:
­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that:

A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$

B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$

D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more ­ ­Expression under sq root cannot be negative; and denominator cannot be zero. These two conditions we need to find the values for x. |x|>2 => x>=2; x<=-2 Denominator not equal to zero => sqrt(x+9) should not be equal to sqrt(3-x) => x should be equal to -3 x>=-9; x<=3 Combining all -9=<X<-3; -3<X<=-2; 2<=x<=3 IMO D Manager Joined: 30 May 2024 Posts: 54 Own Kudos [?]: 41 [0] Given Kudos: 1 Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] ­ The domain of the function f(x)=|x|−2‾‾‾‾‾√x+9‾‾‾‾√−3−x‾‾‾‾√‾‾‾‾‾‾‾‾‾‾‾‾‾√𝑓(𝑥)=|𝑥|−2𝑥+9−3−𝑥 is the set of real numbers x such that: A. −3<x≤−2−3<𝑥≤−2 or 2≤x<32≤𝑥<3 B. −3<x≤−2−3<𝑥≤−2 or 2≤x≤32≤𝑥≤3 C. −3≤x≤−2−3≤𝑥≤−2 or 2≤x≤32≤𝑥≤3 D. −9≤𝑥<−3 or −3<𝑥≤−2 or 2≤𝑥≤3 E. −9≤x≤−2 or 2≤x≤3 Explanation: sq rt of (x+9)>= 0 x + 9 >= 0 x >= -9 sq rt of (3-x)>= 0 3-x >=0 3>=x sq rt of (|x|-2) >= 0 |x|-2 >= 0 |x|>= 2 x >= 2 -x >= 2 -2 >= x sq rt of (x+9) cannot be equal to sq rt of (3-x) x+9 = 3-x 2x = -6 x = -3 x cannot be -3 Therefore we get -9 ≤𝑥<−3 or −3<𝑥≤−2 or 2≤𝑥≤3 Answer is D Manager Joined: 03 May 2024 Posts: 76 Own Kudos [?]: 51 [0] Given Kudos: 1 Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] ­Ans C Numerator needs to be non negative ∣x∣−2≥0 =>∣x∣≥2 This gives two intervals: x≤−2 or x≥2 roots are non negative : x+9≥0=>x≥−9 3−x≥0 => x≤3 So the interval is -9<= x <=3 For real no , Denomenator cannot be zero x+9≥3−x\sqrt{x + 9} Squaring both sides: Squaring both sides: x+9≥3−x Simplifying: 2x≥−6 ⟹ x≥−3 −3≤ x≤3 −3≤x≤−2 or 2≤x≤3 Manager Joined: 24 Jan 2024 Posts: 92 Own Kudos [?]: 67 [1] Given Kudos: 140 Location: Israel Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] 1 Kudos After the opening day of the week, We're back to business. I feel the luck is going to shift to our Green team as we get closer to the end of the Champions. Let's get started with our explanation for this topic: ­Glance - the Question&Answer Choices: Question: We have here an Algebra question. Answer Choices: Seperated to 2 parts. We can eliminate half of question after knowing one side. Rephrase - Reading and Understanding the question: [?]: What X cannot be here? First, let's break into part all our domains here: Solve: a) let's start with |x|-2. we can deconstruct it to: |x| < and euqal to 2 X <= -2 or 2 <= X b) x+9 >= 0 -9 <= X c) 3-x >= 0 X <= 3 d) Roots of (x+9) - Roots of (3-x) >= 0 we can see that the domain here is -3 <= X if we put -5 we can see that: Roots of (-5+9) - Roots of (-3+5) => Roots of (4) - Roots of (2) = Negative. It is not in our domain. only when we have a number bigger that -3 (EXCLUDE) the result is -3 < X because the denominator cannot be 0. so let's unite all information we have until now: X <= -2 or 2 <= X -9 <= X X <= 3 -3 <X Let's draw a line and put it all together: ________-9______________________-3__________-2____________________2__________3_______ *------------------------------------------------------------------------------------------------------> <-----------------* *-----------------> -------------------------------------------------------------------> <-------------------* We can see now clearly that our answer is: -3 < X <= -2 or 2 <= X <= 3 Answer Choice B THE END I hope you liked the explanation, I have tried my best here. Let me know if you have any questions about this question or my explanation. ­ GMAT Club Legend Joined: 03 Jun 2019 Posts: 5326 Own Kudos [?]: 4276 [1] Given Kudos: 161 Location: India GMAT 1: 690 Q50 V34 WE:Engineering (Transportation) Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] 1 Kudos ­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that: $$|x| - 2 \geq 0$$ $$x \geq 2$$ or (1) $$x \leq -2$$ (2) $$x + 9 \geq 0$$ $$​​​​​​​x \geq - 9$$​​​​​​​ (3) $$3-x \geq 0$$ $$​​​​​​​x \leq 3$$​​​​​​​ (4) $$\sqrt{x + 9}- \sqrt{3 - x}​​​​ > 0​​​$$ $$\sqrt{x + 9} > \sqrt{3 - x}​​​​​$$ x + 9 > 3 - x 2x > -6 x > -3 (5) Combing (1) - (5), we get $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ IMO B Manager Joined: 03 Jun 2024 Posts: 74 Own Kudos [?]: 65 [0] Given Kudos: 9 Location: India Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] From the question |x| >= 2 -> implies x >= 2 or x <= -2 x+9 >= 0 -> implies x >= -9 3-x >= 0 -> implies x <= 3 And sqroot(x+9) >= sqroot(3-x) -> implies x >= -3 Combining all these, we have x belongs to [-3, -2] U [2, 3] Hence answer C. Manager Joined: 11 Aug 2018 Posts: 128 Own Kudos [?]: 130 [1] Given Kudos: 33 Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] 1 Kudos Bunuel wrote: ­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that: A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$ B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$ D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­  This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

Let’s test -9 from Option D and E, if we do the denominator becomes negative inside square root which is not possible .

Next check -3 option C, we get 0 in denominator which makes function invalid and hence it is out

Between option A and B, difference is x=3. So at this point numerator is 1 and denominator becomes sq root ( sq root 12))
even though sq root of (3-x) is 0
So x=3 is a valid case

Posted from my mobile device
Intern
Joined: 02 Nov 2020
Posts: 4
Own Kudos [?]: 1 [0]
Given Kudos: 3
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
To determine the domain, we need to ensure that all expressions under the square roots are non-negative and that the denominator is not zero.
After which the eqations are compared and combined to determine result

Posted from my mobile device
Manager
Joined: 27 Nov 2023
Posts: 61
Own Kudos [?]: 26 [1]
Given Kudos: 4
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
1
Kudos
We need to find the domain separately for the numerator and denominator.
Numerator - x>=2 and x<=-2
Denominator - x>-3

combining and substituing the values so that the underroot is negative we get B as the domain
Manager
Joined: 13 Oct 2023
Posts: 98
Own Kudos [?]: 13 [1]
Given Kudos: 78
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
1
Kudos
Bunuel wrote:
­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that:

A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$

B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$

D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$

E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­

 This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more ­ ­We have to make sure denominator should not be 0 or even values underroot can never be negative since we dont have complex numbers, so X cant be equal to -3 or -9 and best option seems to be 1st or 2 nd but then we can also include 3 for the value of the function will be accurate. Hence option B Manager Joined: 27 May 2024 Posts: 72 Own Kudos [?]: 53 [1] Given Kudos: 0 Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink] 1 Kudos Bunuel wrote: ­The domain of the function $$f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}$$ is the set of real numbers x such that: A. $$-3 < x ≤ -2$$ or $$2 ≤ x < 3$$ B. $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ C. $$-3 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$ D. $$-9 ≤ x < -3$$ or $$-3 < x ≤ -2$$ or $$2 ≤ x ≤ 3$$ E. $$-9 ≤ x ≤ -2$$ or $$2 ≤ x ≤ 3$$­  This question was provided by GMAT Club for the GMAT Club Olympics Competition Win over$30,000 in prizes such as Courses, Tests, Private Tutoring, and more

­

­√|x|−2 : |x| - 2 >= 0 => |x| >= 2 => x <= -2 or x >= 2

√x+9 : x + 9 >= 0 => x >= -9

√3-x : 3 - x >= 0 => x <= 3

x <= -2 or x >= 2 ...(i)
x >= -9 ...(ii)
x >= 3 ...(iii)

Combining all 3 inequalities we get...

-9 <= x <= -2 or 2 <= x <= 3

But if we put x <= -3 then √x+9 - √3-x <= 0 which is not defined.

So, -3 < x <= -2 or 2 <= x <= 3

Answer: B. -3 < x <= -2 or 2 <= x <= 3
Manager
Joined: 17 Mar 2019
Posts: 120
Own Kudos [?]: 81 [0]
Given Kudos: 37
Location: India
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
­It's between D and E.
Both represent the same domain except that D says x can not take -3.
If we test the given equation with x=-3, the denominator becomes 0;
Thus, D over E.
D is the correct option.
Re: GMAT Club Olympics 2024 (Day 7): ­The domain of the function f(x) = [#permalink]
1   2   3   4   5
Moderator:
Math Expert
94796 posts