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