Is x^(k+1) > x^k?

C)


i) NS,

since X could be positive or negative, and since (k+1) could be odd or even.

ii) NS

Since X could be a fraction



i) + ii)

Sufficient since i) makes sure X it is not a fraction <1, and, ii) makes sure (k+1) is even.

Bunuel, I am not understanding why the answer is C.
Here is my reasoning:
1.
X can be positive or negative, x is not equal to zero. X can be any number except 0 and greater than 1 and less than -1
Insufficient

2. K+1 is even so K is odd.
If X is 0 , x^(k+1) = x^k=0
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
Also, there is condition for fractions
Insufficient

1+2
If X is +ve,x^(k+1) > x^k
If X is negative, x^(k+1) > x^k
and for fractions the statement changes.

Ans: E
I do not know why we are not considering fractions

Not clear how you concluded the red part there.

This following post shows why the answer is C: answerhttps://gmatclub.com/forum/is-x-k ... l#p1394406

Simplifying the prompt we get,

\(x^k*(x - 1) > 0\)

Hence , we have 3 cases
i) x > 0, k - odd/even, then the given Expression is > 0
ii) x < 0, k - odd, then the given Expression is > 0
iii)x < 0, k - even, then the given Expression is < 0


Statement 1 - |x| > 1, hence x < -1 or x > 1

However, No information about k

Statement 1 alone is Not Sufficient.


Statement 2: (k + 1) = 6p, hence k = 6p - 1 = (even) - 1.
k = odd

However, No information about x

Statement 2 alone is Not Sufficient.


Combining both, we get x > 1 or x < -1 & k = odd

We can see from the 3 cases above, the given conditions satisfy cases i) & ii).

Hence the given expression > 0

Combining is Sufficient.


Hence Answer is C.



Thanks,
GyM