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Is k + 1/k < -1 ?

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Is k + 1/k < -1 ? [#permalink]

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New post 10 Oct 2017, 06:06
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Is k + 1/k < -1 ?

1) k^2 < 1
2) k^3 + k^2 + k < 0
[Reveal] Spoiler: OA

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Is k + 1/k < -1 ? [#permalink]

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New post 10 Oct 2017, 06:43
MathRevolution wrote:
Is k + 1/k < -1 ?

1) k^2 < 1
2) k^3 + k^2 + k < 0



Hi..
\(k+\frac{1}{k}<-1......(\frac{1+k+1}{k})<0.....\frac{(k^2+k+1)}{k}<0\)...
So both numerator and denominator should be of opposite sign..

If \(k<0, k^2+k+1>0\)....
If \(k>0, k^2+k+1<0\)....not possible

1) \(k^2<1\)
Insufficient

2) \(k^3+k^2+k<0.......
K(k^2+k+1)<0..\)
So \(k\) and \(k^2+k+1\) are of opposite sign.
This is same as the initial equation
Sufficient

B

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Re: Is k + 1/k < -1 ? [#permalink]

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New post 12 Oct 2017, 01:44
Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

k + 1/k < 1 is equivalent to k3 + k < -k2 or k3 + k2 + k < 0.
k(k2 + k + 1) < 0 implies k < 0, since k2 + k + 1 is always positive.
Thus, the simplified question is if k < 0.

Condition 1) k2< 1
It is equivalent to -1 < k < 1.
Thus,this is not sufficient.

Condition 2) k3 + k2 + k < 0
Thus, is equivalent to the question.
This is sufficient.

Therefore, the answer is B.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Then, D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.
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Re: Is k + 1/k < -1 ?   [#permalink] 12 Oct 2017, 01:44
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