How can we solve this. Is x 0 2. x^3<0 OA: D

1) -2x>0 => 2x x if x^3 <0, then x has to be negative. D. This is what came to my mind, I am not expert Somebody, please explain. Thanks

Could you please confirm statement-2? chetanojha wrote:How can we solve this. Is x 0 2. x^3<0 OA: D

1. -2x > 0 - you have a negative number multiplied by x to get a positive number (since -2x is positive). This means that x must also be negative, since, if x were positive, then (negative)*(positive) <0. So 1 is sufficient. 2. Consider this: x^3 = (x^2)*x. x^2 will always be positive or equal to zero, so the sign of (x^2)*x (or x^3, as you prefer) will always be set by x. If x is positive, then (x^2)*x = (positive)(positive) = positive. If x is negative, then (x^2)*x = (positive)(negative) = negative. Since x^3 < 0, then x will "follow" the sign: x < 0. So 2 is sufficient as well. The point made with statement 2 can be expanded into a nice little rule: No matter the sign of x, x raised to an even power will always be positive or equal to zero (if x = 0). Odd powers of x will share the sign of x. If x is positive, then x raised to an odd power will also be positive. If x is negative, then x raised to an odd power will also be negative. Got it.

Thanks Guys. This question was from Quant Review. I got the concept now.

Odd powers of x will share the sign of x. If x is positive, then x raised to an odd power will also be positive. TRUE If x is negative, then x raised to an even power will also be negative. Not true. Probably u meant to say If x is negative, then x raised to an odd power will also be negative If x is negative or positive x raised to an even power will always be positive. Regards, CR