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MathRevolution
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Is \(x^2-x>0?\)
\(x^2-x>0..........X(x-1)>0\)..
Two cases..
a) if \(x>0\), then \(x-1>0....X>1\)
b) if \(x<0....x-1<0....x<1\)
Hence question is asking
Is x<0 and x>1 or is it false that \(0\leq{X}\leq{1}\)
1) \(x>0\)
0<X<1....no
X>1....yes
Insufficient

2) \(x^3+x>0\)
\(x(x^2+1)>0\).
\(x^2+1>0\), so \(x>0\)..
Again \(0<x<1\)....no
\(x>1\)....yes
Insufficient

Combined..
Same cases remain
Insufficient

E

OA and solution is wrong
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MathRevolution
Is \(x^2-x>0?\)

1) \(x>0\)

2) \(x^3+x>0\)


To determine whether \(x^2 - x > 0\)

=> \(x * (x - 1) > 0\)

Statement 1

x > 0

x can be 9 => x(x-1) = 9*8 = 72 > 0

x can be 0.5 => x(x-1) = 0.5 * -0.5 = -0.25 < 0

Statement 1 is not sufficient

Statement 2

\(x^3 + x > 0\)

x can be 2 => \(x^3 + x > 0\) = 8 + 2 = 10 > 0

x can be -2 => \(x^3 + x > 0\) = -8 - 2 = -10 < 0

Statement 2 is not sufficient

Combining statements 1 and 2

x can be 2 => \(x^3 + x > 0\) = 8 + 2 = 10 > 0

x can be 0.5 => x(x-1) = 0.5 * -0.5 = -0.25 < 0

Statements 1 and 2 together are not sufficient

Hence option E

MathRevolution Can you please look into this ?
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The Q is basically asking to conclude that: x is NOT a proper fracton, positive or negative?

None of the two statements can confirm or dismiss this conclusively. Hence, Ans is E.

Posted from my mobile device
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I don't agree with the explanation.
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I think this is a high-quality question and I don't agree with the explanation. In case of first statement, if x lies between 0 and 1, lets say x=1/2, then x2-x<0, while if x>1, then x2-x>0, hence A is not sufficient.
The same goes for B if x lies in the range of 0 to 1
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The explaination is incorrect. The correct answer should be Both the conditions are insufficient for x>0 gives us no information whether x>1 or not.
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THE ANS SHOULD BE - E
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I think this is a high-quality question and I don't agree with the explanation. Hello! I believe the correct answer to this solution is E, not D.

A clear counterexample here is x = 1/2. This satisfies statement 1, but it does not satisfy x^2 - x > 0.

Alternatively, the question could be clarified such that x must be an integer.
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I think this the explanation isn't clear enough, please elaborate. if 0<x<1
then x^2-x <0, but x>0 holds and x^3+x >0 holds
if x>1
then x^2-x >0, but x>0 holds and x^3+x >0 holds
i am unable to understand the explanation, where am i going wrong
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I don't agree with the explanation. The answer posted here does not seem correct. If x = 0.5, the expression is not greater than 0 whereas if x = 10, the expression is greater than 0 so the answer should be E, shouldn't it?
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I think this is a high-quality question and I don't agree with the explanation. pls check. answer should be E because both statements given translate to x>0 but the question only has clear solutions if x>1 or x<0. Hence even both statements together arent sufficient
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I think this is a poor-quality question and I don't agree with the explanation. Hello, it seems there is an issue. Would like it if you elaborated on it:
We need either x larger than 1 or x smaller than 0

In condition 1, x is larger than zero, but it could be between 1 and 0 so it should be insufficient
similarly for the second condition.
Please let me know if I am wrong
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I think this is a high-quality question and I don't agree with the explanation. the solution is wrong.
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The solution is not correct for this problem. The correct answer would be E, proving this is as simple as substituting x with 2 which satisfies all equations and the condition in the posted question. Then, substitute x with 1 in both equations, which satisfies both conditions since the result is greater than 0 but substituting x for 1 in the original equation will yield 0 as a result and hence does not satify that condition.
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I think this is a high-quality question and I don't agree with the explanation. it's clearly e
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I think this is a poor-quality question and I don't agree with the explanation. as many have mentioned this solution is wrong and E should be the right choice just try substituting 2 and 1/2
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I think this is a high-quality question and I don't agree with the explanation. I do not agree with solution that x>0 is sufficient to judge x2-x>0.

1. for any value of x >0 and x<1, solution x2-x<0
2. for any value of x>1, solution x2-x>0
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