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If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0

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Bunuel
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If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 01:30
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If \(x≠0\), 3 or -3, is \(\frac{3}{x^2 - 3}> 0\)?


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

(2) \(\frac{-2}{x^2 - 2} < 0\)
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Bunuel
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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   07 Jul 2021, 01:15
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If \(x≠0\), 3 or -9, is \(\frac{3}{x^2 - 3}> 0\)?

For 3/(x^2 - 3) to be positive, the denominator there must be positive. Thus, this inequality holds true if x^2 - 3 > 0, so when x^2 > 3.

(1) \(\frac{1}{x^2 }> 0\). When x ≠ 0, then x^2 > 0 --> 1/x^2 > 0. So, this statement is absolutely useless. Not sufficient.

(2) \(\frac{-2}{x^2 - 2} < 0\). For this fraction to be negative the denominator must be positive, so this implies that x^2 > 2. Thus x^2 may be more than 3 as well as less than 3. Not sufficient.

(1)+(2) One useless statement + one insufficient statemented = not sufficient overall.

Answer: E.
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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 03:56
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From this equation (3)/(x^2 - 3) > 0 ?
We know that, for this equation to be positive, only the value of x^2 must be greater than 3

Numerator is positive! So the sign of equation whether positive or negative only depends on the denominator!

But, for what values the equation will be negative? Definitely x^2 < 3
Or we can say x < -\sqrt{3} && x > \sqrt{3}

Statement 1:
Here x can take any value except 0, 3, -9 This statement is insufficient.

Statement 2:
Numerator is negative.
So, for the equation to be true the value of x^2 must be greater than 2!
But still, x can take decimal values!
This statement too is insufficient.

Both statements together:
When statement 2, has narrower set of values. We can clearly say even combined both statements are insufficient!

Hence, answer is E

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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 04:25
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Bunuel wrote:
If \(x≠0\), 3 or -9, is \(\frac{3}{x^2 - 3}> 0\)?


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

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





Bunuel, it should be x≠0 \(x^2≠0\), 3 or -9 -3

\(\frac{3}{x^2 - 3}> 0\)......
As the numerator>0, so we are looking for : Is \(x^2-3>0.........x^2>3 \ \ or \ \ x>\sqrt{3}....x<-\sqrt{3}\)

(1) \(\frac{1}{x^2 }> 0\)
Will be true for all values of x.

(2) \(\frac{-2}{x^2 - 2} < 0\)
Since the numerator<0, \(x^2-2>0......x^2>2 \ \ or \ \ x>\sqrt{2}....x<-\sqrt{2}\)
x can be between \(\sqrt{2} \ \ and \ \ \sqrt{3}\).

Combined
Nothing more than what is given in statement 2.


E
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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 04:41
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Ask is

= \(\frac{3}{[x^2 - 3]}\)> 0 ;
= \(x^2-3 > 0\)

x= {-inf, -\(\sqrt{3}\)} or {\(\sqrt{3}\), inf}

Statement (1) \(\frac{1}{x^2 }> 0\)
True for all x ; Not sufficient

statement (2) \(\frac{-2}{x^2 - 2} < 0\)
\(x^2 > 2\)
True for x= x= {-inf, -\(\sqrt{2}\)} or {\(\sqrt{2}\), inf}

Not sufficient

Combined : No change in statement 2
E is the answer IMO

Bunuel wrote:
If \(x≠0\), 3 or -9, is \(\frac{3}{x^2 - 3}> 0\)?


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

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



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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 04:52
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Is 3/(x^2-3) > 0? => x^2 - 3 > 0? => x^2 > 3 ?
1) 1/x^2 > 0 => x != 0 (insufficient)
2) -2/(x^2-2) < 0 => x^2 - 2 > 0 => x^2 > 2 (insufficient)

Combine both statements, they are still insufficient. ans: E
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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 05:29
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If x≠ 0, 3 or -9, is 3 / (x^2−3) > 0


(1) 1 / x^2 > 0
x^2> 0
x > 0
Let x =1
3 / (x^2−3) = 3 / 1 -3 = -3/2

Let x =2
3 / (x^2−3) = 3 / 4-3 =3

Not sufficient.

(2) −2 / (x^2−2) < 0
2 / (x^2−2) > 0
2* (x^2−2) > 0
x^2 -2> 0
x^2 > 2

Let x^2 =2.5
3 / (x^2−3)
3 / (2.5 - 3) = -ve
Let x^2 =5
3 / 5 - 3 = 3/2 = +ve
Not sufficient.

Combining both statements
x> 0 and x^2 > 2 Still we are not able to get definite value ie +ve or -ve

IMO E
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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink] New post   06 Jul 2021, 07:39
Expert Reply
chetan2u wrote:
Bunuel wrote:
If \(x≠0\), 3 or -9, is \(\frac{3}{x^2 - 3}> 0\)?


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

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





Bunuel, it should be x≠0 \(x^2≠0\), 3 or -9 -3

\(\frac{3}{x^2 - 3}> 0\)......
As the numerator>0, so we are looking for : Is \(x^2-3>0.........x^2>3 \ \ or \ \ x>\sqrt{3}....x<-\sqrt{3}\)

(1) \(\frac{1}{x^2 }> 0\)
Will be true for all values of x.

(2) \(\frac{-2}{x^2 - 2} < 0\)
Since the numerator<0, \(x^2-2>0......x^2>2 \ \ or \ \ x>\sqrt{2}....x<-\sqrt{2}\)
x can be between \(\sqrt{2} \ \ and \ \ \sqrt{3}\).

Combined
Nothing more than what is given in statement 2.


E

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Re: If x0, 3 or -9, is 3/(x^2 - 3) > 0? (1) 1/x^2 > 0 (2) -2/(x^2 - 2) <0 [#permalink]
06 Jul 2021, 07:39
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