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

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Manager
Joined: 07 Nov 2009
Posts: 242
If x is an integer, is x|x| < 2^x?  [#permalink]

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22 Mar 2010, 20:59
2
00:00

Difficulty:

5% (low)

Question Stats:

84% (01:09) correct 16% (01:26) wrong based on 407 sessions

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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?
Manager
Joined: 07 Nov 2009
Posts: 242

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23 Mar 2010, 01:42
Thanks Kp.

But if x<0 so we get |-x| => -x
Am i missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 52385

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23 Mar 2010, 05:33
3
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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Posts: 52385
Re: If x is an integer, is x|x| < 2^x?  [#permalink]

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22 Feb 2014, 11:25
Bumping for review and further discussion.
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Joined: 14 Jul 2013
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Re: If x is an integer, is x|x| < 2^x?  [#permalink]

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14 Apr 2014, 02:30
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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Posts: 2626
Schools: Boston U '20 (M)
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If x is an integer, is x|x| < 2^x?  [#permalink]

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07 Mar 2016, 19:57
rohitgoel15 wrote:
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?

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, 05:15
1
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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Updated on: 18 Jul 2016, 22:54
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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Originally posted by LogicGuru1 on 18 Jul 2016, 01:07.
Last edited by LogicGuru1 on 18 Jul 2016, 22:54, 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, 22:45
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, 01:19
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]

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10 Mar 2018, 22:46
rohitgoel15 wrote:
If x is an integer, is x|x| < 2^x?

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

Question: Is x|x| < 2^x?

Given x = Integer

St 1: x < 0

|x| = -x , when x < 0

means we have, x * -x = 2^-x

or -(x)^2 = 1/(2)^x

As LHS is negative and RHS always be positive, we can say

x|x| < 2^x Sufficient.

St 2: x = -10

means -10|-10| < 2^-10

-100 < 2^-10

Sufficient

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

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15 Jul 2018, 04:33
stonecold wrote:
rohitgoel15 wrote:
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?

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

I always gets confused many times in absolute value questions.

|x| = -x when x < 0, now my question is;

You wrote x * |x| = x * -x, but why not

-x * -x.

Does it not affect the other "x" which is not inside the modulus sign?

It may sound silly, but I lose many points because of this confusion.

.
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Re: If x is an integer, is x|x| < 2^x? &nbs [#permalink] 15 Jul 2018, 04:33
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