Author 
Message 
Manager
Joined: 26 Feb 2018
Posts: 53
Location: India
WE: Web Development (Computer Software)

A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
29 May 2018, 02:19
Question Stats:
24% (02:12) correct 76% (02:10) wrong based on 70 sessions
HideShow timer Statistics
A function \(F(x, y) =(x^y)^1^/^x\) for positive integers x and y. Is \(F(a, b) > a\), where a and b are positive integers? (1) \(F(a,ab)≥F(a,a^2)\) (2) \(F(b^2,b^2)>F(a,a)^2\)
Official Answer and Stats are available only to registered users. Register/ Login.



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1213
Location: India
GPA: 3.82

A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
29 May 2018, 09:21
PrayasT wrote: A function \(F(x, y) =(x^y)^1^/^x\) for positive integers x and y. Is \(F(a, b) > a\), where a and b are positive integers?
(1) \(F(a,ab)≥F(a,a^2)\)
(2) \(F(b^2,b^2)>F(a,a)^2\) Is \(F(a, b) > a\) \(=>a^\frac{b}{a}>a^1\) => Is \(\frac{b}{a}>1=>\) Is \(b>a\) ? Statement 1: implies \(a^\frac{ab}{a}≥a^\frac{(a^2)}{a}\) \(=>a^b≥a^a=>b≥a\). Here \(b\) can be equal to \(a\) or greater than \(a\). Hence InsufficientStatement 2: implies \((b^2)^{\frac{b^2}{b^2}}>(a^{\frac{a}{a}})^2\) \(=>b^2>a^2=>b>a\), because \(a\) & \(b\) are positive. SufficientOption \(B\)



Intern
Joined: 23 Jun 2018
Posts: 2

Re: A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
26 Jun 2018, 21:17
Counterexample: A=1, B=2
This fulfills both conditions 1 and 2:
1. F(1,2) = 1 ≥ F(1,1) = 1
2. F(4,4) = 4 > F(1,1)^2 = 1
Yet does not meet the inequality in the prompt:
F(1,2) = 1 is NOT greater than 1.
Choosing A ≥ 2 would make the answer B, but because we made no such restriction, the answer is E.



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1213
Location: India
GPA: 3.82

Re: A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
26 Jun 2018, 22:00
mernaka123 wrote: Counterexample: A=1, B=2
This fulfills both conditions 1 and 2:
1. F(1,2) = 1 ≥ F(1,1) = 1
2. F(4,4) = 4 > F(1,1)^2 = 1
Yet does not meet the inequality in the prompt:
F(1,2) = 1 is NOT greater than 1.
Choosing A ≥ 2 would make the answer B, but because we made no such restriction, the answer is E. Hi mernaka123This is a "Is" type question, where you need a definite "Yes" or a definite "No". if you are getting sometime "yes" and sometime "no" then you can reject that option. As you have mentioned that Statement 1 is true only for certain condition and not for other, so you can reject Statement 1 but for Statement 2 are you getting both a "Yes" or a "No"? if not then it will be your answer.



Intern
Joined: 23 Jun 2018
Posts: 2

Re: A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
26 Jun 2018, 22:41
Here are your definites:
1 is definitely a positive integer and fits the prompt.
1 to the power of anything is definitely equal to 1.
A=1, B=2 definitely gives F(A,B) = A and disproves your setup. The question is NOT asking if B > A. Your simplification assumes A cannot be 1, which was never implied by the prompt.
You definitely only need one counterexample to disprove a rule.
Since there is at least one set of positive integers that fits both conditions 1 and 2 and makes the statement TRUE, and at least one set of positive integers that fits both conditions 1 and 2 and makes the statement FALSE, the answer is definitely E.
Oh and I definitely got a Q51.



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1213
Location: India
GPA: 3.82

A function F(x, y) = (x^y)^1/x for positive integers x and y
[#permalink]
Show Tags
26 Jun 2018, 23:15
mernaka123 wrote: Here are your definites:
1 is definitely a positive integer and fits the prompt.
1 to the power of anything is definitely equal to 1.
A=1, B=2 definitely gives F(A,B) = A and disproves your setup. The question is NOT asking if B > A. Your simplification assumes A cannot be 1, which was never implied by the prompt.
You definitely only need one counterexample to disprove a rule.
Since there is at least one set of positive integers that fits both conditions 1 and 2 and makes the statement TRUE, and at least one set of positive integers that fits both conditions 1 and 2 and makes the statement FALSE, the answer is definitely E.
Oh and I definitely got a Q51. Hi mernaka123What you are trying to do is to refute Statement 2 which is not possible because it is a FACT for the question. As per statement 2 we have \(a≠b\), rather \(b>a\) so \(a=b=1\) is not a possible substitution. You need to arrive at the solution using this statement (if possible). Given the definition of the Function and Statement 2, can you prove that F(a,b)=a or a=b ?




A function F(x, y) = (x^y)^1/x for positive integers x and y &nbs
[#permalink]
26 Jun 2018, 23:15






