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M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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09 Jun 2012, 08:10
This post is edited to rectify the incorrect choice. If ab≠0 and a<b, which of the following must be negative? A. (a/b)− (b/a) B. (a−b)/(a+b) C. a^b−b^a D. a(b/(a−b)) E. (b−a)/b OE: Solution a<b means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.
To discard other options consider a=−1 and b=2.
Last edited by manulath on 09 Jun 2012, 08:37, edited 1 time in total.



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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09 Jun 2012, 08:16



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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09 Jun 2012, 08:28
manulath wrote: If ab≠0 and a<b, which of the following must be negative?
A. (a/b)− (b/b) B. (a−b)/(a+b) C. a^b−b^a D. a(b/(a−b)) E. (b−a)/b
With respect to the Choice A : (a/b)− (b/b) as ab≠0, what ever the value of b, b/b is always 1 choice A reduces to a/b1 as a<b ......... a/b will always be less than 1 scenario 1: a>0, b>0 clearly a/b <1 scenario 2: a<0, b<0 in a/b, the negative signs will cancel each other Hence once again a/b<1 scenario 3 and 4: a<0, b>0 or a>0,b<0 in a/b, one value is +ve and other is ve Hence a/b will be ve that is to say a/b<0 or to say a/b<1 As we see that in all the above cases a/b < 1 in choice A: (a/b)(b/b) will always be ve Where did I went wrong?Even if I put the values give in OE manulath wrote: OE: Solution a<b means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.
[highlight]To discard other options consider a=−1 and b=2[/highlight] A. (a/b)− (b/b) => (1/2)  (2/2) => 1/2  1 = 3/2 a negative value



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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09 Jun 2012, 08:29
manulath wrote: manulath wrote: If ab≠0 and a<b, which of the following must be negative?
A. (a/b)− (b/b) B. (a−b)/(a+b) C. a^b−b^a D. a(b/(a−b)) E. (b−a)/b
With respect to the Choice A : (a/b)− (b/b) as ab≠0, what ever the value of b, b/b is always 1 choice A reduces to a/b1 as a<b ......... a/b will always be less than 1 scenario 1: a>0, b>0 clearly a/b <1 scenario 2: a<0, b<0 in a/b, the negative signs will cancel each other Hence once again a/b<1 scenario 3 and 4: a<0, b>0 or a>0,b<0 in a/b, one value is +ve and other is ve Hence a/b will be ve that is to say a/b<0 or to say a/b<1 As we see that in all the above cases a/b < 1 in choice A: (a/b)(b/b) will always be ve Where did I went wrong?Even if I put the values give in OE manulath wrote: OE: Solution a<b means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.
[highlight]To discard other options consider a=−1 and b=2[/highlight] A. (a/b)− (b/b) => (1/2)  (2/2) => 1/2  1 = 3/2 a negative value There is a typo in option A. It should read a/bb/a.
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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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09 Jun 2012, 08:33
Bunuel wrote: A. \(\frac{a}{b}  \frac{b}{a}\)
Okay it seems that question I got was incorrect. I have send the screenshot to you as pm.



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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20 Sep 2012, 11:47
Hi Bunuel,
Could you please explain as to how did you know that a= 1 and b = 2 would make all the other answers invalid?was it just hit and trial or was there any specific reason for choosing these two values?
Thanks.



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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20 Sep 2012, 12:37



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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29 Apr 2013, 19:28
I thought x = \sqrt{x^2}? How do you arrive at a = a^2?



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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29 Apr 2013, 20:58
manulath wrote: This post is edited to rectify the incorrect choice. If ab≠0 and a<b, which of the following must be negative? A. (a/b)− (b/a) B. (a−b)/(a+b) C. a^b−b^a D. a(b/(a−b)) E. (b−a)/b OE: Solution a<b means that a2<b2 → a2−b2<0 → (a−b)(a+b)<0, so a−b and a+b have the opposite signs, which means that a−ba+b will always be negative.
To discard other options consider a=−1 and b=2. I've also done the way Bunuel explained ........................... as Given .......... \(a<b\) i.e., \(a^2<b^2\) Therefore, \(a^2b^2<0\) & therefore, \((ab)(a+b)<0\) & ultimately divide both sides with \((a+b)\) Hence, , I got ... \(\frac{ab}{a+b}\) < 0
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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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30 Apr 2013, 01:03



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Re: M14, 33 (If ab≠0 and a<b, which of the following must) [#permalink]
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30 Apr 2013, 15:42
Bunuel wrote: youngkacha wrote: I thought x = \sqrt{x^2}? How do you arrive at a = a^2? \(a^2<b^2\) is obtained by squaring \(a<b\) (we can safely square \(a<b\) since both sides of the inequality are nonnegative). Hope it's clear. Oh okay, I see now. This was a tough one for me. Thank you.




Re: M14, 33 (If ab≠0 and a<b, which of the following must)
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30 Apr 2013, 15:42







