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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

1

This post received KUDOS

Expert's post

If \(ab\neq{0}\) and \(|a|<|b|\), which of the following must be negative?

A. \(\frac{a}{b} - \frac{b}{a}\)

B. \(\frac{a-b}{a+b}\)

C. \(a^b-b^a\)

D. \(\frac{ab}{a-b}\)

E. \(\frac{b-a}{b}\)

\(|a|<|b|\) means that \(a^2<b^2\) --> \(a^2-b^2<0\) --> \((a-b)(a+b)<0\), so \(a-b\) and \(a+b\) have the opposite signs, which means that \(\frac{a-b}{a+b}\) will always be negative.

Answer: B.

To discard other options consider \(a=-1\) and \(b=2\). _________________

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/b-1

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

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

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09 Jun 2012, 08:29

Expert's post

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/b-1

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/b-b/a. _________________

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?

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

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20 Sep 2012, 12:37

Expert's post

negative wrote:

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.

With a certain algebraic manipulations with the expression in the stem we can get that B is the correct answer. So, the example which discard other options are just to illustrate that they are wrong, and yes you can find the proper numbers for that by trial and error. _________________

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, \(a^2-b^2<0\) & therefore, \((a-b)(a+b)<0\) & ultimately divide both sides with \((a+b)\) Hence, , I got ... \(\frac{a-b}{a+b}\) < 0 _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.