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For all numbers x.... Help me please!

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For all numbers x.... Help me please!

by gmatpup » Fri Dec 23, 2011 6:39 pm
For all numbers x such that x does not equal 1, if g(x) is defined by g(x)= x^2+2 / x-1 , then (1/g(2))(1/2g(x)= ?

A. 6(x-1) / x^2+2

B. 6(x^2+2)/x-1

C. x^2+2 / 2(x-1)

D. x-1 / 6(x^2+2)

E. x^2+2 / 6(x-1)



Answer is D

I have no idea how to get the answer.. please advise! Thanks so much :)
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by Anurag@Gurome » Fri Dec 23, 2011 7:41 pm
gmatpup wrote:For all numbers x such that x does not equal 1, if g(x) is defined by g(x)= x^2+2 / x-1 , then (1/g(2))(1/2g(x)= ?

A. 6(x-1) / x^2+2
B. 6(x^2+2)/x-1
C. x^2+2 / 2(x-1)
D. x-1 / 6(x^2+2)
E. x^2+2 / 6(x-1)

Answer is D

I have no idea how to get the answer.. please advise! Thanks so much :)
g(x)= (x² + 2)/(x - 1)
g(2)= (2² + 2)/(2 - 1) = 6
(1/g(2))(1/2g(x)= [1/6] * [1/2 * (x² + 2)/(x - 1)] = (x - 1)/12(x² + 2), none of the options.
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by gmatpup » Fri Dec 23, 2011 7:56 pm
I made a huge error.. It should read:

(1/g(2))(1/g(x)) = ?
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by Anurag@Gurome » Fri Dec 23, 2011 7:57 pm
gmatpup wrote:I made a huge error.. It should read:

(1/g(2))(1/g(x)) = ?
g(x)= (x² + 2)/(x - 1)
g(2)= (2² + 2)/(2 - 1) = 6
(1/g(2))(1/g(x)= [1/6] * [(x² + 2)/(x - 1)] = (x - 1)/6(x² + 2)

The correct answer is D.
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