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

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

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New post 05 Dec 2017, 07:16
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.


what is wrong in my approach :

x |x| < 2^x
x *sqrt(x^2) < 2^x
square on both sides,
x^2 * x^2 < 2^2x
x^4 < 2^2x

given 1 stmt, x as -ve, always x^4 > 2^2x, whereas I know i am making some mistake.
are we not allowed to take square on both sides?

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

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New post 05 Dec 2017, 07:24
Avinash_R1 wrote:
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.


what is wrong in my approach :

x |x| < 2^x
x *sqrt(x^2) < 2^x
square on both sides,
x^2 * x^2 < 2^2x
x^4 < 2^2x

given 1 stmt, x as -ve, always x^4 > 2^2x, whereas I know i am making some mistake.
are we not allowed to take square on both sides?


We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). Here x|x| is negative if x is negative, so we cannot square.


RAISING INEQUALITIES TO EVEN/ODD POWER

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Adding, subtracting, squaring etc.: Manipulating Inequalities.

9. Inequalities



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

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New post 13 Dec 2017, 20:48
Hi All,

We're told that X is an integer. We're asked if X|X| < 2^X. This is a YES/NO question. We can answer it with a bit of Number Property knowledge.

1) X < 0

With Fact 1, we know that X is NEGATIVE. By definition, that means...
X|X| = (Neg)|Neg| = Negative
2^(Negative) = Positive
Thus, X|X| will ALWAYS be less than 2^X and the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

2) X = -10

With the value of X, we can absolutely answer the question (we would just need to plug in that value:
Is (-10)|-10| < 2^(-10)?
The answer to the question IS yes, but we don't have to actually do that work. There would be just one answer to the question, so it doesn't really matter what that one answer is.
Fact 2 is SUFFICIENT

Final Answer:
[Reveal] Spoiler:
D


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

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New post 05 Jan 2018, 02:00
adkikani wrote:
Bunuel mikemcgarry IanStewart shashankism Engr2012

I know this is an OG Q, but is it not strange that st 2 is an integral part of st 1.
During actual exam I do not think to need to evaluate S2 separately since x=-10
is always going to be x<0. Let me know if my understanding is correct


Hi Adkikani,

I agree! This thinking can save us some time :-)

Aiena.

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

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New post 05 Jan 2018, 05:14
adkikani wrote:
Bunuel mikemcgarry IanStewart shashankism niks18

I know this is an OG Q, but is it not strange that st 2 is an integral part of st 1.
During actual exam I do not think to need to evaluate S2 separately since x=-10
is always going to be x<0. Let me know if my understanding is correct


Hi adkikani

Here RHS will be always positive irrespective of the value of \(x\) and LHS can be positive or negative.

So the question becomes very simple if we know the value of \(x\). Incidentally both the statements clearly mention the same thing. Hence this is one of the easiest question and the stats also confirms it.

You are right in assuming that if x<0 hold true then x=-10 will be true here.

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Re: If x is an integer, is x|x|<2^x ?   [#permalink] 05 Jan 2018, 05:14

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