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23 Sep 2023, 05:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:57) correct
38%
(02:09)
wrong
based on 45
sessions
History
Date
Time
Result
Not Attempted Yet
23 Sep 2023, 07:53
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Bunuel
\(x^4-x^3+x^2-x > 0\)
\(x^3(x-1)+x(x-1) > 0\)
\((x-1)(x^3+x) > 0\)
\(x(x-1)(x^2+1) > 0\)
\(x^2 + 1\) is always positive, hence for the expression to hold true, \((x)(x-1) > 0\)
Case 1: \(x > 0\)
If \(x > 0, x - 1 > 0.\)
Therefore \(x > 1\)
Case 2: \(x < 0\)
If \(x < 0, x - 1 < 0.\)
Therefore \(x < 1\).
Combining the information above, either \(x > 1\) or \(x < 0\)
Statement 1
x > 1
We know from the analysis above that if x > 1, the expression holds true. Hence, we can conclude that the information provided in Statement 1 is sufficient to answer the question.
We can eliminate B, C, and E.
Statement 2
\(|x^3|>|x|\)
This statement tells us that on a number-line \(x\) lies in the highlighted region.
------------- -1 ------------- 0 ------------- 1 -------------
In both regions, x is either less than -1 (satisfies the condition \(x < 0\)) or x is greater than 1 (satisfies the condition \(x > 1\)). Hence, this statement is also sufficient to answer the question.
Option D
23 Sep 2023, 07:59
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Is \(x^4-x^3+x^2-x>0\)?
\(x(x^3 - x^2 + x - 1)>0\)
So either both x and \(x^3 - x^2 + x - 1\) are positive or both are negative for it to be true.
Only problem is when
0 < x < 1
as it leaves \(x^3 - x^2 + x - 1\) negative giving us a NO.
(1) \(x>1\)
x = 2
16 - 8 + 4 - 2 > 0
10 > 0
YES
x = 3/2
81/16 - 27/8 + 9/4 - 3/2 > 0
~5 - ~3.5 + 2.25 - 1.5 > 0
~ 2.25 > 0
SUFFICIENT.
(2) \(|x^3|>|x|\)
Taking clue from St. 1
x = 2 gives YES
x = - 2 gives YES too
Here x cannot be a fraction e.g. x = 1/2 or -1/2
Hence
SUFFICIENT.
Answer D.
\(x(x^3 - x^2 + x - 1)>0\)
So either both x and \(x^3 - x^2 + x - 1\) are positive or both are negative for it to be true.
Only problem is when
0 < x < 1
as it leaves \(x^3 - x^2 + x - 1\) negative giving us a NO.
(1) \(x>1\)
x = 2
16 - 8 + 4 - 2 > 0
10 > 0
YES
x = 3/2
81/16 - 27/8 + 9/4 - 3/2 > 0
~5 - ~3.5 + 2.25 - 1.5 > 0
~ 2.25 > 0
SUFFICIENT.
(2) \(|x^3|>|x|\)
Taking clue from St. 1
x = 2 gives YES
x = - 2 gives YES too
Here x cannot be a fraction e.g. x = 1/2 or -1/2
Hence
SUFFICIENT.
Answer D.
