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x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!

I think you misread the second statement: it's x^2 - x^3 > 2 not (2) x^2 - x^3 > 0.
_________________

Bunuel, can you explain Stat 2...I did half way ..\(x^2\) ( 1 - x ) > 2 and taken \(x^2\) > 0 as it is square root and tried with 1-x >2...and struck here.

(x^2 - 2x + 2) has no real roots so it doesn't cut the x axis. It is positive for all real values of x. So for (x+1)(x^2 - 2x + 2) to be negative, (x+1) must be negative. So x + 1 < 0 x < -1 If x < -1, x is greater than x^3.

Sufficient

Answer (B)

Note: Know the relation between x, x^2 and x^3 on the 4 sections of the number line: x > 1, 0 < x< 1, -1 < x < 0 and x < -1
_________________

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_________________

Statement 2 : x^2 - x^3 > 2 x^2> 2+x^3 For x>1, x^3 > x^2 So, x^3 +2 > x^2. This is not the case For 0<x<1, x^2 <1 So, x^2 <2 +x^3 This is also not the case. For -1<x<0, x^2<1 , -1<x^3<0 So, x^2 <2 +x^3 This is also not the case. For x = -1 , x^2=2+x^3 This is also not the case. For x<-1, x^2>2+x^3 This is the case.

Now if x<-1, Clearly x>x^3 Hence Answer B _________________

I didn't quite understand how you factored the cubic polynomial x^3 - x^2 + 2 Would appreciate it if you could walk me through the steps. Thanks in advance!

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