roots - exponents - absolutes

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roots - exponents - absolutes

by Brent@GMATPrepNow » Sun Dec 28, 2008 3:01 pm
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by cramya » Sun Dec 28, 2008 3:12 pm
I would go with C)

I am asuming = sign is missing in the diagram

sqrt(|x|) ^ 2 is |x| which is x if x is positive or 0

Stmt II

x^0 = 1

x can be positive or negative

INSUFF

Toegther x has to be psoitive since 0^0 is not equal to 1
Last edited by cramya on Sun Dec 28, 2008 3:16 pm, edited 1 time in total.

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by parallel_chase » Sun Dec 28, 2008 3:15 pm
Statement I

(sqrtlxl)^2 = x

if x = -1

lxl = 1

(sqrt 1)^2 = 1 but this is not equal to x, which is -1

Hence only way this could be possible when x = positive.

Sufficient

Statement II

x^0 = 1

irrespective of any value of x, any value when raised to the power of 0 will always result in 1, Therefore x could be positive or negative.

Insufficient.

Hence A.
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by cramya » Sun Dec 28, 2008 3:17 pm
PC I thought of A too initially then decided C) for reasons posted above.

Let me know what u think

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by parallel_chase » Sun Dec 28, 2008 3:23 pm
cramya wrote:PC I thought of A too initially then decided C) for reasons posted above.

Let me know what u think
Nice work cramya, missed that 0, I hate to make silly mistakes.

Answer should be C.

I also have a doubt in statement II

-2^0 = -1

2^0 = 1

are these correct or is it I am right in my logic to assume that any number positive or negative, when raised to the power of 0 is 1.

thanks.
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by Brent@GMATPrepNow » Sun Dec 28, 2008 3:23 pm
(1) x can be positive or zero --> insuff
(2) x can be any number EXCEPT zero (0^0 is undefined)

(1) and (2) combine to tell us that x must be positive.
Answer: C
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by cramya » Sun Dec 28, 2008 3:45 pm
I also have a doubt in statement II

-2^0 = -1

2^0 = 1

are these correct or is it I am right in my logic to assume that any number positive or negative, when raised to the power of 0 is 1

PC its always 1.

-2^0 = 1

2^0 = 1

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by parallel_chase » Sun Dec 28, 2008 3:47 pm
cramya wrote:
I also have a doubt in statement II

-2^0 = -1

2^0 = 1

are these correct or is it I am right in my logic to assume that any number positive or negative, when raised to the power of 0 is 1

PC its always 1.

-2^0 = 1

2^0 = 1
Thanks Buddy.
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by vivek.kapoor83 » Mon Dec 29, 2008 10:36 am
pls tell me ...cramya...howcm u arrvied it on...pls elaborate...
i face too much prob in absolute values.

pls elaborate ur sol.

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by vivek.kapoor83 » Mon Dec 29, 2008 10:42 am
cramya wrote:
sqrt(|x|) ^ 2 is |x| which is x if x is positive or 0



This is creating probs. as u said , sqrt(|x|) ^ 2 is |x|
now it boils down to |x| =x
now if x is +ve it holds
if x is -ve, then LHS = x and RHS = -x , not equal....


why and hw u hv taken 0 in consideration is my ques.

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by cramya » Mon Dec 29, 2008 2:31 pm
why and hw u hv taken 0 in consideration is my ques
Hi Vivek,

Basic but important piece of information:

The absolute value(never negative) or magnitude of a real number x is denoted
by |x| and is defined by

|x| = x if x ≥ 0 ( x is greater than or equal to 0)
|x| = &#8722;x if a < 0


stmt I tells us it either a 0 or a positive number. Hence INSUFF

Together it cant be 0 since 0^0 is undefined so has to be a positive number


Hope this helps!