If x is an integer, is x|x| < 2^x?
1) x < 0
2) x=-10
OA: D
I chose E. Can someone explain why my reasoning below is incorrect?
1) Insufficient:
Picked x = - 1
-1*|x| < 1/2
|x| > - 1/2
|x| = +/- 1
a. |x| = -1 => -1 < -1/2 => 1 > 1/2 => true
b. |x| = 1 => 1 < -1/2 => false
2) Insufficient
a. |x| = 10 => -10*10 = < 2^10 => 100 > 2^10 => true
b. |x| = -10 => 100 < 2^-10 => false
x|x|<2^x?
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- santhoshsram
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If you pick x = -1, then lets look at LHS
LHS is -1*|-1| = -1*(+1) = -1 => Negative
RHS is 2^x. For any x positive or negative or zero, 2^x is always positive
So LHS (negative) < RHS (positive)
Sufficient.
In your explanation,
Picked x = - 1
-1*|x| < 1/2
|x| > - 1/2
|x| = +/- 1 ===>>> This is the issue (I think).
As long you don't know the value of x (+ or -) then |x| = +/-(x). But when you know the value of x, you don't need the +/- anymore, have a deterministic answer for the value of |x|. Whatever x is, |x| is always positive, so if x = -1, then |x| = 1 and not +/-1
Hope that helps.
LHS is -1*|-1| = -1*(+1) = -1 => Negative
RHS is 2^x. For any x positive or negative or zero, 2^x is always positive
So LHS (negative) < RHS (positive)
Sufficient.
In your explanation,
Picked x = - 1
-1*|x| < 1/2
|x| > - 1/2
|x| = +/- 1 ===>>> This is the issue (I think).
As long you don't know the value of x (+ or -) then |x| = +/-(x). But when you know the value of x, you don't need the +/- anymore, have a deterministic answer for the value of |x|. Whatever x is, |x| is always positive, so if x = -1, then |x| = 1 and not +/-1
Hope that helps.
-- Santhosh S
- neerajkumar1_1
- Master | Next Rank: 500 Posts
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- Joined: Wed Apr 07, 2010 9:00 am
- Thanked: 24 times
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try to simplify your every data question...stoy4o wrote:If x is an integer, is x|x| < 2^x?
1) x < 0
2) x=-10
OA: D
I chose E. Can someone explain why my reasoning below is incorrect?
1) Insufficient:
Picked x = - 1
-1*|x| < 1/2
|x| > - 1/2
|x| = +/- 1
a. |x| = -1 => -1 < -1/2 => 1 > 1/2 => true
b. |x| = 1 => 1 < -1/2 => false
2) Insufficient
a. |x| = 10 => -10*10 = < 2^10 => 100 > 2^10 => true
b. |x| = -10 => 100 < 2^-10 => false
the equation looks daunting at first.. but what is it really testing u on..
Look at equation closely
x|X|<2^x
now RHS will always be +ve ( for e.g 2^2 = 4 and 2^(-2)=1/4 ...)
Now Lets look LHS.
|x| is always +ve
we are left with x
for the equation to be true x will have to be negative.. so that u can get -ve < +ve
Hence u r checking for whether x is +ve or -ve.
Both statements tell u x is negative.
Hence IMO: D
Hope it helps..
- neerajkumar1_1
- Master | Next Rank: 500 Posts
- Posts: 270
- Joined: Wed Apr 07, 2010 9:00 am
- Thanked: 24 times
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try to simplify your every data question...stoy4o wrote:If x is an integer, is x|x| < 2^x?
1) x < 0
2) x=-10
OA: D
I chose E. Can someone explain why my reasoning below is incorrect?
1) Insufficient:
Picked x = - 1
-1*|x| < 1/2
|x| > - 1/2
|x| = +/- 1
a. |x| = -1 => -1 < -1/2 => 1 > 1/2 => true
b. |x| = 1 => 1 < -1/2 => false
2) Insufficient
a. |x| = 10 => -10*10 = < 2^10 => 100 > 2^10 => true
b. |x| = -10 => 100 < 2^-10 => false
the equation looks daunting at first.. but what is it really testing u on..
Look at equation closely
x|X|<2^x
now RHS will always be +ve ( for e.g 2^2 = 4 and 2^(-2)=1/4 ...)
Now Lets look LHS.
|x| is always +ve
we are left with x
for the equation to be true x will have to be negative.. so that u can get -ve < +ve
Hence u r checking for whether x is +ve or -ve.
Both statements tell u x is negative.
Hence IMO: D
Hope it helps..