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19) Inequality DS

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Source: — Data Sufficiency |
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by liferocks » Sat May 01, 2010 7:02 am
we have to prove x^5+x^3>2x^4---1
now if X <=0 relation 1 will not hold true.Hence condition 2 is not sufficient

from 1 we get (x-1)>0
or (x-1)^2>0
or (x^2-2x+1)>0
or (x^2+1)>2x
or x^3(x^2+1)>2x^4..since x is positive multiplying both sides by x^3
or x^5-x^4>x^4-x^3...sufficient

Ans option A
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by boazkhan » Sat May 01, 2010 2:37 pm
since 1st statement is X>1 -- The question then can be reduced to X>1 (since X is positive we can divide) --- which is statement 1.

Hence A.

Is that the correct approach?
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by tanviet » Sun May 02, 2010 5:23 am
correct approach

x^4(x-1)>x^3(x-1)

x^4(x-1)-x^3(x-1)>0

(x-1)(x^4-x^3)

x^3(x-1)^2>0.......correct
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by tryingtocrack » Sun May 02, 2010 7:13 am
Is X^5-X^4>X^4-X^3?
1. X>1
2. X is integer

My take on this q

x^4(x-1) > x ^3(x-1)
or simply say x^4>x^3 ...rephrase it as is X +ve ?

Statement A - x >1 x + ve

Statement B x can be +ve or -ve

Answer A

Please let me know if this make sense
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