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

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

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22 Mar 2010, 21:59
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If x is an integer, is x|x| < 2^x?

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

[Reveal] Spoiler:
I can understand the second part:
-10|-10| < 2^-10 --> -10 * 10 < 1/2 ^ 10
|-10| --> reduced to 10 as its numeric.. is my reasoning correct?
B is sufficient

For (1) .. however i am not able to decipher anything..
-x|-x| < 2^-x --> -x * -x < 1/2 ^x
|-x| --> reduced to -x as x < 0 .. is my reasoning correct?
[Reveal] Spoiler: OA

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23 Mar 2010, 02:42
Thanks Kp.

But if x<0 so we get |-x| => -x
Am i missing something?

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23 Mar 2010, 06:33
2
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Expert's post
rohitgoel15 wrote:
Sorry to open a new thread for an already existing question. was not satisfied with the answers.
another-absolute-value-question-41274.html

If x is an integer, is x|x| < 2^x?
(1) x<0
(2) x=-10

I can understand the second part:
-10|-10| < 2^-10 --> -10 * 10 < 1/2 ^ 10
|-10| --> reduced to 10 as its numeric.. is my reasoning correct?
B is sufficient

For (1) .. however i am not able to decipher anything..
-x|-x| < 2^-x --> -x * -x < 1/2 ^x
|-x| --> reduced to -x as x < 0 .. is my reasoning correct?

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

Question: is $$x|x| < 2^x$$? Notice that the right hand side (RHS), $$2^x$$, is always positive for any value of $$x$$.

(1) $$x<0$$ --> $$LHS=x*|x|=negative*positive=negative$$ --> $$(LHS=negative)<(RHS=positive)$$. Sufficient.

(2) $$x=-10$$ --> LHS is negative --> $$(LHS=negative)<(RHS=positive)$$. Sufficient.

rohitgoel15 wrote:
But if x<0 so we get |-x| => -x
Am i missing something?

For $$x<0$$, $$|x|=-x$$ yes. So for (1) $$LHS=x*(-x)$$, $$x$$ is negative, $$-x$$ is positive. So $$LHS=x*(-x)=negative*positive=negative$$.

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

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22 Feb 2014, 12:25
Bumping for review and further discussion.
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Re: If x is an integer, is x|x| < 2^x? [#permalink]

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14 Apr 2014, 03:30
If x is an integer, is x|x| < 2^x?

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

Sol.

(1) Pick two numbers for x
a. x = -2 =>
LHS = -2.|-2| = -4
RHS = 2^-2 i.e. positive
=> LHS<RHS
b.LHS = -1/3.|-1/3| = -9
RHS = 2^-1/3 i.e. positive
=> LHS<RHS
therefore, (1) is sufficient to answer

(2) case covered in statement (1) a
hence, sufficient

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

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07 Mar 2016, 20:57
rohitgoel15 wrote:
If x is an integer, is x|x| < 2^x?

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

[Reveal] Spoiler:
I can understand the second part:
-10|-10| < 2^-10 --> -10 * 10 < 1/2 ^ 10
|-10| --> reduced to 10 as its numeric.. is my reasoning correct?
B is sufficient

For (1) .. however i am not able to decipher anything..
-x|-x| < 2^-x --> -x * -x < 1/2 ^x
|-x| --> reduced to -x as x < 0 .. is my reasoning correct?

here =>
see in first case x<0 => |x|=-x
=> x*-x<2^x
=> -x^2<2^x

statement 2 i would say you should not even check as the specific value will yield a result
hence sufficient.

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

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13 Mar 2016, 06:15
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A quick takeaway from this question is that whenever we encounter a negative sign in front of a square => its a negative value
and offcourse the second big takeaway => if the base if positive => THE value is positive irrespective of the exponent being positive or negative
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If x is an integer, is x|x| < 2^x? [#permalink]

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18 Jul 2016, 02:07
rohitgoel15 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?

(1) x < 0
Means x is negative , the product on LHS will be negative
therefore RHS will become $$2^{-x}$$
$$2^{-x}$$ is equal to $$\frac{1}{2^{x}}$$

$$\frac{1}{2^{x}}$$ (which is a positive decimal/positive fraction) will always be GREATER than x (which is a negative integer)

SUFFICIENT

(2) x = -10
We just proved it, through statement 1
SUFFICIENT

EACH ALONE IS SUFFICIENT

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Last edited by LogicGuru1 on 18 Jul 2016, 23:54, edited 2 times in total.

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

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18 Jul 2016, 23:45
hrs always +ve in this case.

so D

coz both statements say LHS is negative

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

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30 Jun 2017, 02:19
I feel like this question doesn't really touch upon a more tricky inequality when x is positive (both statements indicate x as negative, so it's enough to answer the question).

Notice how things change when x > 0

x x*|x| 2^x
1 1 2 Y
2 4 4 N
3 9 8 N
4 16 16 N
5 25 32 Y
6 36 64 Y
7 49 128 Y
8 64 256 Y... so on

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Re: If x is an integer, is x|x| < 2^x?   [#permalink] 30 Jun 2017, 02:19
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