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moatik: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sat Mar 26, 2011 11:47 am

Functions junctions mess!

by moatik » Tue Apr 12, 2011 6:46 pm

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?

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Source: Beat The GMAT — Problem Solving | Tue Apr 12, 2011 6:46 pm

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Tue Apr 12, 2011 7:29 pm

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?

Hi!

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!