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28 Jul 2009, 10:15
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If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?
1) x < 1
2) n is even.
1) x < 1
2) n is even.
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28 Jul 2009, 10:28
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Quote:
is x^n > x^(n+1)?
1) x < 1
x can be between 0 and 1 or x can be negative
If x is b/w 0 and 1, then x^n > x^(n+1)
if x is negative, x^n > x^(n+1) - here it depends on n
not suff
2) n is even
not suff because no information on x
Together
If x is b/w 0 and 1 and n is even then x^n > x^(n+1)
ex 0.1 -->x^2=0.01
0.1 -->x^2+1=x^3=0.001
0.01>0.001
if x is negative and n is even then x^n > x^(n+1)
ex x=-2-->x^2=4
-2 -->x^2+1=x^3=-8
Hence C
28 Jul 2009, 11:41
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[quote="amitgovin"]If n is a positive integer and x does not equal zero, is x^n > x^(n+1)?
1) x 0
ie: is??
x^n(1 - x)>0 ie is both x^n and (1-x) have the same sign?? only possibloe if we have 2 conditions
1>x , n is even....................
obviously each alone is not suff
both are
C
1) x 0
ie: is??
x^n(1 - x)>0 ie is both x^n and (1-x) have the same sign?? only possibloe if we have 2 conditions
1>x , n is even....................
obviously each alone is not suff
both are
C
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
