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For all values of x where x > 2, which of the following is equivalent

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For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 23 Nov 2016, 06:51
00:00
A
B
C
D
E

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Question Stats:

87% (01:28) correct 13% (01:54) wrong based on 88 sessions

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Re: For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 23 Nov 2016, 06:59
Bunuel wrote:
For all values of x where x > 2, which of the following is equivalent to \(\frac{x!+(x−1)!}{(x+2)!}\)?

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



IMO B
\(\frac{x!+(x−1)!}{(x+2)!}\)can be simplified as \(\frac{x* (x-1)! + (x-1)!}{(x+2)(x+1)x(x-1)!}\)
cancelling common terms we are left with \(\frac{1}{x(x+2)}\)
which is B
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Re: For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 23 Nov 2016, 07:58
I say choice (B).

I plugged x=3 into the answer choices and x!+(x−1)!/(x+2)! simplified to 1/15.

Answer choice (B) yields 1/9+6=1/15.
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Re: For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 23 Nov 2016, 09:17
Bunuel wrote:
For all values of x where x > 2, which of the following is equivalent to \(\frac{x!+(x−1)!}{(x+2)!}\)?

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


= \(\frac{x!+(x−1)!}{(x+2)!}\)

= \(\frac{x( x - 1) !+(x−1)!}{(x+2)( x + 1) x ( x - 1)!}\)

= \(\frac{( x - 1) ! ( x + 1)}{(x+2)( x + 1) x ( x - 1)!}\)

= \(\frac{1}{(x+2) x}\)

= \(\frac{1}{(x^2+2x)}\)

Hence, answer will be (B)


PS : I will ordinarily employ gracie90 's plug in approach for solving this question quickly
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Re: For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 08 Oct 2018, 13:32
Bunuel wrote:
For all values of x where x > 2, which of the following is equivalent to \(\frac{x!+(x−1)!}{(x+2)!}\)?

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



when we have variables in question stem and in answer choices , it's better to assign values for the variables.

let x be 3

\(\frac{x!+(x−1)!}{(x+2)!}\)

= 3! + 2! / 5!

= 8 / 120

= 1 /15


Put x=3 in option B . the result is same.



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Re: For all values of x where x > 2, which of the following is equivalent  [#permalink]

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New post 10 Oct 2018, 18:45
Bunuel wrote:
For all values of x where x > 2, which of the following is equivalent to \(\frac{x!+(x−1)!}{(x+2)!}\)?

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


[x! + (x - 1)!]/(x + 2)!

In the numerator, we factor out the common (x - 1)! from both terms. We also expand the factorial in the denominator, obtaining:

[(x - 1)!(x + 1)]/[(x + 2)(x + 1)(x)(x - 1)!]

1/[(x + 2)(x)]

1/(x^2 + 2x)

Answer: B
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Re: For all values of x where x > 2, which of the following is equivalent   [#permalink] 10 Oct 2018, 18:45
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