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Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) --> \(x*|x|=x^2\)), negative when \(x<0\) (\(x<0\) --> \(x*|x|=-x^2\)) and equals to zero when \(x={0}\).

(1) x < 0. According to the above \(x*|x|<0<2^x\). Sufficient.

(2) x = -10. The same here \(x*|x|=-100<0<\frac{1}{2^{10}}\). Sufficient.

OA is D for the following reasons: When you first see a DS question, see if there is anyway to simplify the question stem In this case, since |x| is positive, we can divide both sides by |x| giving us a new question stem --> is x < 2?

S1: x<0, therefore x must be <2 = sufficient S2: x = -10 and -10 < 2 = sufficient

OA is D for the following reasons: When you first see a DS question, see if there is anyway to simplify the question stem In this case, since |x| is positive, we can divide both sides by |x| giving us a new question stem --> is x < 2?

S1: x<0, therefore x must be <2 = sufficient S2: x = -10 and -10 < 2 = sufficient

Let me know if this helps!

The OA is wrong here because of the following reasons:

(1) if x=-10 then -100<-20, on the other hand if x= -1, x<0 then the inequality changes from < to >, namely, -1 > -2 ; This statement is absolutely insufficient!

For x to be Negative LHS i.e. x|x| will always be NEGATIVE and 2^x will be positive for any value of x i.e. x|x|<2^x will always be true SUFFICIENT

Statement 1: x = -10 For x =-10 LHS i.e. x|x| will always be NEGATIVE (-100) and 2^x will be positive for given x (1/2^10) i.e. x|x|<2^x will always be true SUFFICIENT

Answer: option D
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Hi, we have an equation and the RHS 2^x will be positive irrespective of value of x and LHS xlxl will depend on the value of x.. 1) x is -ive .. so LHS is -ive and RHS is +ive.. suff 2) x=-10... again LHS is -ive and RHS is +ive.. suff ans D
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Re: If x is an integer, is x|x|<2^x ? [#permalink]

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24 Apr 2016, 15:22

This question can be solved as follows.

stmt1) it says that x /x/ < 2^x. and also we are told that x< 0. So if x is zero and the abs of x is always positive then we know that x/x/ will be negative. In addition to that, we know that 2^negative number will be positive because it will be in the form of 1/2^x, it will be less than 1 but it will be greater than a negative number. So stmt1 is SUFF.

stmt2) this is a repetition of stmt1 because the left side is negative and the right side is positive. SUFF.

Re: If x is an integer, is x|x|<2^x ? [#permalink]

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17 Apr 2017, 02:20

Bunuel wrote:

Walkabout wrote:

If x is an integer, is x|x|<2^x ?

(1) x < 0 (2) x= -10

If x is an integer, is x|x|<2^x ?

Notice that the RHS (right hand side) of the expression is always positive (\(2^x>0\)), but the LHS is positive when \(x>0\) (\(x>0\) --> \(x*|x|=x^2\)), negative when \(x<0\) (\(x<0\) --> \(x*|x|=-x^2\)) and equals to zero when \(x={0}\).

(1) x < 0. According to the above \(x*|x|<0<2^x\). Sufficient.

(2) x = -10. The same here \(x*|x|=-100<0<\frac{1}{2^{10}}\). Sufficient.

Answer: D.

just for the sake of time saving:

(2) x = -10 --> we could test it! done.... next question

gmatclubot

Re: If x is an integer, is x|x|<2^x ?
[#permalink]
17 Apr 2017, 02:20

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