dzyubam wrote:
We should probably revise the difficulty level of the question.
I like your approach. +1. Can't see anything wrong with it.
dpgxxx wrote:
Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.
S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient
S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.
Answer: B
Ok. I belive there is something which is wrong.
S1) As per comments above X>-1 but X =-2 satisfies
-1/2 > -1 True
We cannot multiply variables in inequality without knowing their signs.
1/X > -1 will be solved as
Case 1) X>0 => -X<1 or X > -1 but X>0
so X>0
Case 2) X<0 => Multiply by X and change sign -X>1 or X < -1
S2) 1/X^5 > 1/X^3
1/(X*X^4) > 1/(X*X^2)
(X^2 and X^4 are positive)
X^2/X > X^4/X
or X > X^3
X (X^2-1) < 0
0<X<1 or X<-1
Therefore Ans B
Question deserves to be at this level.
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Please +1 KUDO if my post helps. Thank you.
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