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# If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS

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Director
Joined: 08 Jun 2013
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Location: India
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If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS  [#permalink]

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11 Aug 2018, 08:59
00:00

Difficulty:

(N/A)

Question Stats:

71% (01:14) correct 29% (00:35) wrong based on 7 sessions

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If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ?

(1) -2/(x2 - 2) < 0

(2) 1/x2 > 0

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Director
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Joined: 01 Oct 2017
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WE: Supply Chain Management (Energy and Utilities)
If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS  [#permalink]

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11 Aug 2018, 09:50
Harshgmat wrote:
If x≠0, 3 or -9, is $$3/(x^2 - 3) > 0$$?

(1) $$\frac{-2}{(x^2 - 2)} < 0$$

(2) $$\frac{1}{x^2}> 0$$

Re-phrasing question stem:-
$$\frac{3}{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}>0$$
Or, $$x<-\sqrt{3}$$, $$\:x>\sqrt{3}$$
Or, $$x:\left(-\infty \:,\:-\sqrt{3}\right)\cup \left(\sqrt{3},\:\infty \:\right)$$

St1:- $$\frac{-2}{(x^2 - 2)} < 0$$
Or, $$-\frac{2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}<0$$
Or, $$x:\left(-\infty \:,\:-\sqrt{2}\right)\cup \left(\sqrt{2},\:\infty \:\right)$$
Insufficient.

St2:- $$\frac{1}{x^2} > 0$$
$$x:\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)$$
Insufficient.

Combined, the question stem is still inconsistent in the domain of x.
Insufficient.

Ans. (E)

N.B:- The cut-off points are to be excluded from the intervals.
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Re: If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS  [#permalink]

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11 Aug 2018, 09:57
Harshgmat wrote:
If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ?

(1) -2/(x2 - 2) < 0

(2) 1/x2 > 0

Harshgmat

This question is already discussed over here. Please search the forum before creating new questions.

https://gmatclub.com/forum/if-x-0-3-or- ... 75521.html

Thanks.
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Re: If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS &nbs [#permalink] 11 Aug 2018, 09:57
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# If x≠0, 3 or -9, is 3/(x2 - 3) > 0 ? - Algebra - DS

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