If k is a non zero integer, is k > 0? (1) |k - 4| = |k| + 4 (2) k>k^3

Statement 1:
The absolute values give us the regions we should check. There are three cases, k > 4, 0 < k < 4, and k < 0. If k > 4, the equation is k - 4 = k + 4 which has no solution. If 0 < k < 4, we get 4 - k = k + 4 and k = 0, which isn't within the region but plugging in k = 0 tells us it's a viable solution. Therefore there are no positive k solutions, k cannot be positive. Sufficient.

Statement 2:
Whenever we are comparing powers of a single variable, we should check the 4 regions: k < -1, -1 < k < 0, 0 < k < 1, and k > 1. We can check the regions by picking a point in the region and if the point satisfies the inequality, the entire region is a solution set. For example, k = 0.5 satisfies this inequality so 0 < k < 1 is a solution set. By plugging in k = -2, we can see the same is true for k < -1. So we can have positive k or negative k, insufficient.

Ans: A

Hi chetan2u ,

Although i did arrive at D , I could not algebraically solve the first equation as the equation got very messy and time consuming to solve. So i substitutd numbers to reach the correct answer.

Could you please show us the correct approach to move with the first stsatement ?



Regards

ShankSouljaBoi If k is a non zero integer, is \(k > 0\)?

(1) \(|k - 4| = |k| + 4\)
Now both sides are positive, so you can square two sides
\(k^2-8k+16=k^2+8|k|+16......|k|=-k\). This will be true ONLY when k is negative.
OR
Logically if you see what is happening to two sides.
The right side is adding two quantities but the left will add only when k and -4 have same sign, so k has to be 0 or <0
Suff

(2) \(k > k^3\)
\(k(1-k^2)>0\)..
If k>0, 1-k^2>0....k^2<1...-1<k<1 but no integer value exists as k is non zero.
When k<0, 1-k^2<0....k^2>1...k<-1
Suff

D

how did u solve k^2>1 and k^2<1 ?

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