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

Question

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

Bunuel · Accepted Answer

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.

harshtandel838 · Answer

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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chetan2u · Answer

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

sumitkrocks · Answer

Ask is

=  \frac{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

kantapong · Answer

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

asishron29181 · Answer

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

Bunuel · Answer

____________________
Than you! Fixed.

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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