For all numbers x such that x != 1 (NOT EQUAL), if g(x) is defined by g(x) = (x^2 + 2) / (x - 1), then (1/g(2)) * (1/g(x)) =
Answer is (x-1)/6(x^2 + 2) according to GMat I just need some1 to help me understand how they got that.
I did this, (6/1) * (1 / ( (x^2 + 2)/(x - 1) ) = (6x + 6) / (x^2 +2) ,,,, no?
Functions junctions mess!
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Hi!moatik wrote:For all numbers x such that x != 1 (NOT EQUAL), if g(x) is defined by g(x) = (x^2 + 2) / (x - 1), then (1/g(2)) * (1/g(x)) =
Answer is (x-1)/6(x^2 + 2) according to GMat I just need some1 to help me understand how they got that.
I did this, (6/1) * (1 / ( (x^2 + 2)/(x - 1) ) = (6x + 6) / (x^2 +2) ,,,, no?
You made a couple of errors.
First, the first term is 1/g(2), not g(2). So, subbing in:
1/g(2) = 1/(6/1) = 1/6
and the second term is:
1/g(x) = 1/((x^2 + 2) / (x - 1)) = (x-1)/(x^2 + 2)
Taking the product of those two terms:
(1/6)*(x-1)/(x^2 + 2)
= (1)(x-1)/(6)(x^2 + 2)
= (x-1)/6(x^2 + 2)
which is what you posted as the accredited answer.
Hope that helps!
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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