Is |x| x^3 (2) x^2 x^3

Question

Is |x| < 1

(1) x^5 > x^3
(2) x^2 > x^3

Bunuel · Answer

Is |x| < 1 Is |x| < 1 ? --> is -1 x^3. Reduce by x^2: x^3>x --> (x+1)x(x-1)>0 --> -11 . Not sufficient. (2) x^2 > x^3. Reduce by x^2: 1>x ( x eq{0} ). Not sufficient. (1)+(2) Intersection of the ranges is -1MSP824621h8d31791668agg000062b1e415iai4e4i2.gif Sufficient. Answer: C.

rango · Answer

How do you do it (x+1)x(x-1)>0 --> -1 1 or x>-1 or x>0 hw that -1 :shock:

KarishmaB · Answer

One statement says "x must be either greater than 1 OR between -1 and 0"
Another says "x must be less than 1"

So what values can you give x such that both statements are satisfied? You cannot have x greater than 1 since second statement says x must be less than 1. So x must be between -1 and 0. This satisfies both the statements.
-1 < x < 0

rango · Answer

Thanks , Sorry to say BUT Do not got the point. :shock:

Bunuel · Answer

I guess you ask about the first statement: (1) x^5 > x^3. Reduce by x^2: x^3>x --> x^3-x>0 --> x(x^2-1)>0 --> x(x-1)(x+1)>0 --> (x+1)x(x-1)>0 . The "roots" are -1, 0, and 1 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-1 ; -11 . Test some extreme value: for example if x is very large number then all multiples ((x + 1), x, and (x - 1)) will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive. Now the trick: as in the 4th range the expression is positive then in 3rd it'll be negative, in 2nd positive and finally in 1st it'll be negative again: - + - + . So, the ranges when the expression is positive are: -11 (4th range). Theory on Inequalities: x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope this helps.

PiyushK · Answer

For second option I tried to solve it with handy trick.
x^2(x-1)<0

critical points on number line --(+)---0--(-)--1---(+)-- as equation is < 0 trick suggest to pick range between 0-1, this trick mostly worked in all cases even in option 1. But in option 2 it is not working.

You can refer trick on link : inequalities-trick-91482.html#p804990

I am not sure where I am making mistake with this trick.

Bunuel · Answer

The squared term, x^2, should be omitted because it's always positive and thus doesn't affect the sign of the left side.

PiyushK · Answer

Ohh Thanks Bunuel I understood and can relate it to this trick.

jlgdr · Answer

Ahhh now I get it, that's awesome thanks. Still I don't quite know what's wrong with including x^2 in the key points, I know it will give you a different valid range but I would like to understando the logic behind it.

Cheers
J

KarishmaB · Answer

First understand why we have transition points at all - at the transition point, the sign of one factor changes from positive to negative (starting from right to left). That reverses the sign of the entire inequality. Therefore, we plot the number line with the transition points and make that wave to give different signs to adjoining regions. Detailed explanation of transition points: https://youtu.be/PWsUOe77__E Now what happens when you have x^2? Does x^2 change sign? No, it is ALWAYS positive (or 0). It never becomes negative and hence never reverses the sign of the inequality.

CrackverbalGMAT · Answer

Statement I is insufficient:

x = 2 (NO), x = -0.5 (YES)

Statement II is insufficient:

x = 0.5 (NO), x = -3 (YES)

Combining is sufficient:
X has to be a negative fraction as x cannot be a positive number or negative number less than -1.

Hence answer is C

bumpbot · Answer

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Is |x| < 1 (1) x^5 > x^3 (2) x^2 > x^3

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