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Function gmatprep

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Function gmatprep

by vladmire » Sun Dec 07, 2008 4:19 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/g(x)) =

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

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

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


What basics do I need to know to figure this out?
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by pbanavara » Sun Dec 07, 2008 6:33 pm
I think the question is either written wrong

g(x) = (x^2+2)/(x-1)

Please use parenthesis while typing questions .. you need to know the fundamentals of algebra ... If you are absolutely new to Algebra.. it might not be a bad idea to pick up one of the many course materials offered by Kaplan, MGMAT and the like..

- pradeep
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Re: Function gmatprep

by canuckclint » Sun Dec 07, 2008 6:44 pm
vladmire 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/g(x)) =

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

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

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


What basics do I need to know to figure this out?
I would simplify and plug numbers. Solving algebraically each one will take too long!!

g(2) = 8 (just plug in x=2 for g(2))

1/g(2) * 1/g(x) = 1/8 * 1/g(x) = h(x)

lets see with x=3 (arbitrary number)

h(3) = 1/8 * 1/g(3) = 1/24

So one of the choices should equal 1/24 with x=3
if there is more than one you should pick another number besides x=3.

i`m not getting them to be equal..can someone help
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Gmatprep function

by vladmire » Sun Dec 07, 2008 6:53 pm
see attached sry
Attachments
equal.jpg
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by hwiya320 » Sun Dec 07, 2008 7:03 pm
that's much better, thanks for the attachment.

THe answer should be "D"

you solve one at a time.

when x = 2, then g(2) = ....

you plug in 2 for x, then you get 6.

g(x) is already given.

Now the question asks for 1/(g(2)) times 1(g(x))

you simply take the reciprocal of that and multiply them together.

rec. of 6 = 1/6 times rec of (x^2 + 2)/(x-1) = (x-1)/(x^2 + 2)
multiply together, you have (x-1) / 6(x^2 + 2)

answer, D

is that the OA?
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by vladmire » Sun Dec 07, 2008 7:16 pm
yes it is D
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by vladmire » Sun Dec 07, 2008 7:35 pm
rec. of 6 = 1/6 times rec of (x^2 + 2)/(x-1) = (x-1)/(x^2 + 2)
multiply together, you have (x-1) / 6(x^2 + 2)
I understand all the way up until getting (x-1)/6(x^2+2) why is the 6 under the division sign instead of 6(x-1)/(x^2+2)
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by hwiya320 » Sun Dec 07, 2008 7:59 pm
vladmire wrote:
rec. of 6 = 1/6 times rec of (x^2 + 2)/(x-1) = (x-1)/(x^2 + 2)
multiply together, you have (x-1) / 6(x^2 + 2)
I understand all the way up until getting (x-1)/6(x^2+2) why is the 6 under the division sign instead of 6(x-1)/(x^2+2)
if you solve the original equation with x=2, then you get the value of 6.

But that's not what the question is asking. question asks

what is 1 / (g(x)) times 1 / (g(2))

we know that g(2) is 6 because we solved the equation using x=2.

so 1 divided by g(2) is 1/6. does that clear things up?

i sometimes wished we had a drawing board... so much easier to write. :)
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by hwiya320 » Sun Dec 07, 2008 8:00 pm
when you solve these two equations, solve them separately so you won't get mixed up. you'll be able to see it clearly.
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by vivek.kapoor83 » Mon Dec 08, 2008 4:25 am
its pretty simple
put x =2 in g(x) & g(2) = 6
and dont do any thing for 1/g(x)

So, putting back it ll be (1/6)(x-1)/((x^2+2)
Hence D.
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