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Is x^5 > x^4? (1) x^3 > x (2) 1/x < x

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Mck2023
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Re: Is x^5 > x^4? (1) x^3 > x (2) 1/x < x [#permalink] New post   07 Mar 2021, 20:18
HI Bunuel
Can you please help me understand this:

You reduced by x^4. i.e. x^4(x-1)>0, I guess!!
But I was not confident to do this because I had seen somewhere in 'finding the range of value of question', using wavy line method, where reducing directly like this got me wrong answer!! Can you please contrast between these two scenarios? Thank You!
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Bunuel
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Re: Is x^5 > x^4? (1) x^3 > x (2) 1/x < x [#permalink] New post   08 Mar 2021, 00:18
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Mck2023 wrote:
HI Bunuel
Can you please help me understand this:

You reduced by x^4. i.e. x^4(x-1)>0, I guess!!
But I was not confident to do this because I had seen somewhere in 'finding the range of value of question', using wavy line method, where reducing directly like this got me wrong answer!! Can you please contrast between these two scenarios? Thank You!


We cannot divide an inequality by the variable if we don't know its sign. Here though x^4 cannot be negative, so we can divide and write x > 1.
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kavitaverma
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Re: Is x^5 > x^4? (1) x^3 > x (2) 1/x < x [#permalink] New post   28 Apr 2023, 07:09
Bunuel wrote:
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.



For option 2, \(-1 < x < 0\) or \(x > 1\).

isn't this range sufficient for the question to answer? IMO x^5 > x^4 for both the ranges. Or am I missing something here?

Bunuel @sayantanck please shed some light
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Re: Is x^5 > x^4? (1) x^3 > x (2) 1/x < x [#permalink] New post   28 Apr 2023, 07:42
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kavitaverma wrote:
Bunuel wrote:
Is x^5 > x^4?

Is \(x^5 > x^4\)? --> reduce by x^4: is \(x > 1\)?

(1) x^3 > −x --> \(x(x^2 + 1) > 0\). Since \(x^2 + 1\) is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.

(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): \(x < x^3\) --> \(x(x^2 -1) > 0\) --> \((x + 1)x(x - 1) > 0\) --> \(-1 < x < 0\) or \(x > 1\). Not sufficient.

(1)+(2) Intersection of ranges from (1) and (2) is \(x > 1\). Sufficient.

Answer: C.



For option 2, \(-1 < x < 0\) or \(x > 1\).

isn't this range sufficient for the question to answer? IMO x^5 > x^4 for both the ranges. Or am I missing something here?

Bunuel @sayantanck please shed some light


The inequality x^5 > x^4 doesn't hold true when -1 < x < 0. For instance, when x = -1/2, x^5 becomes negative while x^4 is positive, making x^5 actually smaller than x^4, not larger.

I hope this helps.
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Re: Is x^5 > x^4? (1) x^3 > x (2) 1/x < x [#permalink]
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