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# A function F(x, y) = (x^y)^1/x for positive integers x and y

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A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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29 May 2018, 02:19
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Difficulty:

95% (hard)

Question Stats:

24% (02:12) correct 76% (02:10) wrong based on 70 sessions

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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) $$F(b^2,b^2)>F(a,a)^2$$
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A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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29 May 2018, 09:21
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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 Insufficient

Statement 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. Sufficient

Option $$B$$
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Re: A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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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.
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Re: A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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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 mernaka123

This 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.
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Re: A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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26 Jun 2018, 22:41

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.
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A function F(x, y) = (x^y)^1/x for positive integers x and y  [#permalink]

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26 Jun 2018, 23:15
mernaka123 wrote:

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 mernaka123

What 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
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