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Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2

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Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 00:38
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Is \(x^4 - x > 0\) ?

(1) \(x < 0\)
(2) \(x^4 > x^2\)


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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 01:15
3
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Is \(x^4−x>0\)?
Or, is \(x^4>x\)?

Note than \(x\) can be greater than \(x^4\) only in the range \(0≤x≤1\) so we can have an answer to the question if we know whether \(0≤x≤1\).

(1) \(x<0\)

We know that \(x\) is not in the range \(0≤x≤1\)

so yes, \(x^4>x\)

1 is sufficient


(2) \(x^4>x^2\)

If \(x^4>x^2\), \(x\) cannot be in the range \(-1≤x≤1\) which again means that \(x\) is not in the range \(0≤x≤1\)

so yes, \(x^4>x\)

2 is sufficient

Answer is (D)
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 02:17
1
Kudos
IMO D
x^4 - x
= x(x-1)(x^2 + x + 1)
As complex numbers are out of scope of gmat, we only need to check for x(x-1)>0
ie is x^2 > x?

1) will always satisfy the above
2) as both sides are positive, we can just square root them

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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 02:18
1
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Is x^4 - x > 0? Yes ?
Or x^4 -x <= 0? No?

(1) x< 0
(-ve)^4 - (-ve) = +ve + +ve = +ve Yes! (Sufficient)

(2) x^4 > x^2
Say x= -2 —> (-2)^4 > (-2)^2
Now (-2)^4 - (-2) >0 yes

Say x= 2 —> (2)^4 > (2)^2
Now (2)^4 - (2)> 0 yes
(Sufficient)

Hit that D

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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 02:27
1
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Is \(x^4 - x > 0\) ?

(1) \(x < 0\)
It means x = (-) negative. You can also consider example such as X = -1, -2
\(x^4\) will be positive
\( x \) will be negative but \( -x \) will be positive
Hence, \(x^4 - x > 0\) will be always positive

A D / B C E

(2) \(x^4 > x^2\)
It means X is either negative or positive but greater than 1. So, the value of X will increase with increase in its power.
Sufficient.

'D' is the winner.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 02:29
Explanation:

x^4 - x > 0
x*(x^3 - 1^3) > 0
x*(x-1)*(x^2+x+1) > 0

1. x<0
Put any value less than x
let x = -1, Result will be +ve.
Sufficient.

2. x^4>x^2
x^4 - x^2 >0
x^2 * (x^2-1)>0
x^2*(x+1)*(x-1)>0

Sufficient.

IMO-D
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 02:42
1
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x^4−x>0? ---> x(x-1)(x^2+x+1)>0
The real question (or condition) is whether x<0 or x>1

(1) x<0
Yes, this satisfies the condition above.
SUFFICIENT

(2) x^4>x^2 ---> x^2(x-1)(x+1)>0
So, x<-1 or x>1. This condition also satisfies the condition above.
SUFFICIENT

FINAL ANSWER IS (D)
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 03:01
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Ans: D

By using both the equation a & b, we can reach the solution.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 03:11
given
x*(x^3-1)>0
#1
x<0
test with -ve integer and fractions we get always yes ; sufficient
#2
x^4>x^2
always true for integer values whether +/-ve
and also yes for x^4−x> 0
sufficient
IMO D
Is x^4−x> 0?

(1) x<0
(2) x4>x2
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Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   Updated on: 30 Jan 2020, 04:11
Quote:
Is x^4−x>0 ?

(1) x<0
(2) x^4>x^2


\(x^4−x>0…x^3(x-1)>0\)
\(case[1]:If…x>0…then…x-1>0…x>1\)
\(case[2]:If…x<0…then…x-1<0…x<1\)

(1) x<0 sufic

x<0, which is less than x<1 (case[2]).

(2) x^4>x^2 sufic

\(x^4-x^2>0…(x^2-x)(x^2+x)>0…[x(x-1)][(x^2+1)]>0\)
\([(x)(x-1)][x(x+1)]>0…x^2(x-1)(x+1)>0\)
\(If…x>0…[pos](x-1)[pos]>0…then…x-1>0…x>1\)
\(If…x<0…[pos][neg](x+1)>0…then…x+1<0…x<-1\)

Ans (D)

Originally posted by exc4libur on 29 Jan 2020, 03:48.
Last edited by exc4libur on 30 Jan 2020, 04:11, edited 1 time in total.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 04:49
1
Kudos
We are to determine if x^4-x>0.
x^4-x>0
x(x^3-1)>0
range of values x that satisfy the inequality above is: x<0 and x>1.
So basically we are to determine if x falls within the range given above.

Statement 1: x<0
Sufficient since we know that the range of values that satisfy x^4-x>0 includes x<0.

Statement 2: x^4>x^2
x^4>x^2
x^4-x^2>0
x^2(x-1)(x+1)>0
Range of values of x that satisfy the above inequality are: x<-1 and x>1
But x<-1 is a subset of x<0 while x>1 is a range of x values that satisfies x^4-x>0. Hence all values of x that satisfy x^4>x^2 always yield a value of x^4-x > 0.
Statement 2 is also sufficient.

The answer is D.

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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 06:40
1
Kudos
\(x^4 − x > 0\)
--> \(x(x^3 - 1) > 0\)
--> \(x < 0\) or \(x^3 > 1\)
--> \(x < 0\) or \(x > 1\)
E.g: \((-1)^4 > -1\) or \(2^4 > 2\)

(1) \(x < 0\) --> Sufficient

(2) \(x^4 > x^2\)
--> \(x^4 - x^2 > 0\)
--> \(x^2(x^2 - 1) > 0\)
Since \(x^2\) can NEVER be negative, \(x^2 - 1 > 0\) ALWAYS
--> \(x^2 - 1 > 0\)
--> \((x - 1)(x + 1) > 0\)
--> \(x < -1\) or \(x > 1\)
This solution also falls within the solution range of \(x < 0\) or \(x > 1\) --> Sufficient

Option D
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 09:12
1
Kudos
Is \(x^{4}−x>0\)?

(Statement1) \(x<0\)
\(x^{4}−x= x(x^{3}-1)>0\)
\(x\) and \(x^{3}-1\) are negative. --> (-)*(-)= (+)
Sufficient .

(Statement2): \(x^{4}>x^{2}\)
\(x^{4} -x^{2}= x^{2} (x^{2}-1)= x^{2} (x-1)(x+1) >0\)
--> \(x<-1 \) and \(x >1\)

\(x(x^{3}-1)>0\)
\(x(x-1)(x^{2}+x+1) >0\)
\(x<0\) and \(x>1\)
Always YES
Sufficient

The answer is D.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 10:22
1
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Quote:
Is \(x^4 - x > 0 \)
(1) \(x<0\)
(2) \(x^4 > x^2\)


\(x^4 - x = x*(x^3-1)\)

(1)\( x<0 => x^3 - 1 <0 => x*(x^3-1) > 0 \)
=> \(x^4 - x > 0\)
=> SUFF

(2) \(x^4 > x^2 => x^4 - x^2 >0 => x^2*(x^2-1) > 0\)
\( => x^2 - 1 >0 \) (Because \(x^2 \geq{0}\))
=> \((x+1)*(x-1) >0 =>\) \(x>1\) or \(x<-1\)
Case \(x<-1 => x^4 - x > 0\) (reference (1) is suff above)
Case \(x>1 => x^2 > 0\) and \(x^2-1 >0 \)
\(=> x^4 - x > 0\)
=> (2) Suff

=> Choice D
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Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   Updated on: 30 Jan 2020, 00:54
1
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Is \(x^4 − x > 0\) ?
\(x(x^3 - 1) > 0\) So,
Either both x > 0 and \(x^3 - 1 > 0\) i.e. x > 1
OR
both x < 0 and \(x^3 - 1 < 0\) i.e. x < 0

(1) x < 0
For x = -1,
\(x^4 - x > 0 = (-1)^4 - (-1) = 2 > 0 \)

SUFFICIENT.

(2) \(x^4 > x^2\)
\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1) > 0\)
i.e. \( x^2 > 1 \)
Hence x > 1 or x < -1

For x = 2, \(x^4 − x > 0 \)
\(2^4 − 2 > 0\)
14 > 0

For x = -2, \((-2)^4 − (-2) > 0\)
18 > 0

SUFFICIENT.

Answer D.

Originally posted by unraveled on 29 Jan 2020, 11:26.
Last edited by unraveled on 30 Jan 2020, 00:54, edited 1 time in total.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 11:46
x^(4 )− x > 0 ?
Or, x^(4 )>x, this is true for (I) all negative integers and negative fractions
(II) all positive integers.
Statement (1) sufficient.
Statement (2) is true for all negative and positive integers but not for real fractions.
Answer:A
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 18:19
True.

X^4>X^2 iff x>1 or x<-1
Since it's given that x<0
Then it must be the casé that x<-1
Therefore x^4-x is a positive minus a negative, therefore result is positive.
x^4-x>0

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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   29 Jan 2020, 19:48
1
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The question asks whether x^4 - x > 0 , or x^4 > x. This can be true when x is an integer, or when x is a negative fraction.

1) x < 0. sufficient as x^4 will always be greater than x in this case.

2) x^4 > x^2, sufficient as it means x will be an integer. As we know when x is a fraction (no matter whether it's a positive or negative), its value decreases as power increases.

D is the answer.
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Re: Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2 [#permalink] New post   12 Jun 2023, 12:25
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