time-consuming inequality Q

Question

what is a quick way to solve this besides picking various numbers and testing.

If -1 > y > 0, then which of this must be true
1. y^2 > 1/(y+2)
2. y^2 1/(y^3 - y)
4. None

Thanks.

AK · Answer

-1 > y > 0 is not correct.
Please verify the question.

oops · Answer

sorry, just checked back after a while - you are quite right - i made a typo - the correct question is : If -1 < y < 0, then which of this must be true 1. y^2 > 1/(y+2) 2. y^2 < 1/(y+3) 3. y^2 > 1/(y^3 - y) 4. None Plz provide an algebraic soln. Thanks.

ywilfred · Answer

(1) can be written as:
y+2 > 1/y^2
y > (1/y^2)-2 --> out because 1/y^2 is going to be always > 2, and y is always going to be a fraction.

(2) can be written as:
y+3 < 1/y^2
y <1> Always true. Same reasoning above.

(3) can be written as:
y^3-y > 1/y^2
y^3-(1/y^2) > y --> Is not true for negative y.

Ans B

LM · Answer

ANswer should be 2. There is not other way...just put the value of Y= -1/2. Only 2 will satisfy it.

Y does not lie between -1 and 0.

gmatiscoming · Answer

even though you had a typo my brain somehow knew what you were trying to say. strange.

anyway my strategy was to recognize from the statement right away that y was some sort of negative fraction.

second part of my strategy was to ignore choice 4 (at least until you have exhausted other options) b/c I never trust those types of answers

choice 1:

well i had to plug in here

-1/2 squared = 1/4 1/(-1/2+2) = 1/(3/2) =2/3

1/4<2/3 so choice 1 is false.

choice 2:

1/4<2/5 true.

choice three:
i just ignored.

probably not the best strategy!! lol

-1/2 squared = 1/4 1/(-1/2+2) = 1/(3/2) =2/3 1/4<2/3 so choice 1 is false. choice 2: 1/4<2/5 true. choice three: i just ignored. probably not the best strategy!! lol

ronron · Answer

I go for 3 as well

Left hand side is always positive, right hand side is always negative

Caas · Answer

3 it is 

y^2 - positive
(y^3 - y) - negative

vidyasagar · Answer

Applying the reverse method: assuming 1), 2) or 3) to be true and finding value of 'y', whether it comes in the range:   as y^2 is +ve. For this inequality to be true, we must have y+3 > 0] i.e. y > -3 ((no-no as it does specify whether  

'3)  .For this inequality to be true, we must have either
case1:   and  > 0 i.e.   and   
 THIS IMPLIES y < -1 (again no-no)

Is the ans. D ? OR I AM MAKING SOME SERIOUS MISTAKE SOMEWHERE. KINDLY POINT OUT.

terp26 · Answer

I am getting that none satisfy the equation.

Can anyone verify an OA?

maratikus · Answer

none satisfy the equation.

1) y^2 > 1/(y+2), y = -1/2, 1/4 > 1/(-1/2+2)=2/3 - incorrect
2) y^2 < 1/(y+3), y = -1 (very close to -1, because -1 is not included) 1 < 1/2 - incorrect
3) y^2 > 1/(y^3-y)= 1/(y*(y^2-1)), y = -1/2, 1/4 > 1/((-1/2)*(1/4-1))=2*4/3 - incorrect

mtgmat · Answer

Yes none satisfies the inequalities. Ans should be D.
No better way to explain than what maratikus did.

JuliaS · Answer

Can you please explain why you picked -1 to plug into answer #2, if the condition states that y is greater then -1? I've tried plugging in -1/2 and -2/3 and both satisfy the answer. What am I doing wrong here?

mtgmat · Answer

pl recheck. -2/3 does not saisfy the inequality at answer#2.

'-1' is better to pick instead of say '-0.99' for ease of calculation. however one has to take precaution that the curve {y^2 > 1/(y^3-y)} is continuos near y =-1, which is indeed the case here.

time-consuming inequality Q

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