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# GMAT Prep DS

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Intern
Joined: 18 Apr 2006
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07 Aug 2006, 01:21
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SVP
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07 Aug 2006, 03:21
B

sqr(x2) = -x when x<0
sqr(x2) = x when x>0

Hence we have to prove that x-3 is negative. this can be true when x is less than 3.

1) x!= 3. X can be less than 3 also greater than 3.
Not suff

2) -x|x| > 0
This can be true only when x < 0

Hence x-3 is <0 Hence sqr{(x-3)^2} = -(x-3) = 3-x
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07 Aug 2006, 10:13
I have a doubt here..
Stmt 2 )
for x< 0 lets say -2

then sqrt(-2-3)^2) != 3- -2

because sqrt(25) != 5
because it can be + 5 or -5.
Please let me know what u think . where am i going wrong ?
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07 Aug 2006, 20:53
techieGuy wrote:
I have a doubt here..
Stmt 2 )
for x< 0 lets say -2

then sqrt(-2-3)^2) != 3- -2

because sqrt(25) != 5
because it can be + 5 or -5.
Please let me know what u think . where am i going wrong ?

It cannot be +5. Put +5 in place of x and you will get a different answer.
Manager
Joined: 26 Jun 2006
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07 Aug 2006, 21:04
Agree with (B)

The function sqtr(x) is defined for non-negative X's can take non-negative values...

sqrt(25) = 5 (can not be -5)

similar, sqrt((-5)^2) = 5 (can not be -5)
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11 Aug 2006, 20:40
sqrt((x-3)^2) = |x-3|
|x-3| = x-3 for x>=3
|x-3| = 3-x for x< 3

(1) x!= 3
x could be < 3 or >=3 Insuff
BCE
(2) -x|x| > 0
|x| is always +ve
Hence -x>0
i.e x<0
Hence |x-3| = 3-x

Suff
Hence
B
Heman
11 Aug 2006, 20:40
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