Let function # be defined as follows:

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Let function # be defined as follows:

by AAPL » Wed Nov 01, 2017 8:24 am
Let function f be defined as follows:
$$f\left(x\right)=\frac{x!+\left(x+1\right)!}{\left(x+2\right)!}$$
What is the value of f(6)?

$$A.\ \frac{1}{10}$$
$$B.\ \frac{1}{9}$$
$$C.\ \frac{1}{8}$$
$$D.\ \frac{1}{7}$$
$$E.\ \frac{1}{6}$$

The OA is D.

I need help to solve this PS question. Please, can any expert assist me with it? Thanks.
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Answer

by EconomistGMATTutor » Wed Nov 01, 2017 9:04 am
Hello AAPL.

It is simple. You just have to eval f on 6 and the simplify. I will show you:

$$f\left(6\right)=\frac{6!+\left(6+1\right)!}{\left(6+2\right)!}=\frac{6!+7!}{8!}.$$

Now, we can write 7! and 8! as follows: $$7!=7\cdot6!\ \ \ \ \ \ and\ \ \ \ 8!=8\cdot7\cdot6!$$

Now, let's replace this on f(6). We will get

$$f\left(6\right)=\frac{6!+7\cdot6!}{8\cdot7\cdot6!}=\frac{6!\cdot\left(1+7\right)}{6!\cdot\left(8\cdot7\right)}=\frac{8}{8\cdot7}=\frac{1}{7}.$$

So, the correct answer is D.

I hope this help you.

Regards.
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by Scott@TargetTestPrep » Fri Nov 01, 2019 7:06 pm
AAPL wrote:Let function f be defined as follows:
$$f\left(x\right)=\frac{x!+\left(x+1\right)!}{\left(x+2\right)!}$$
What is the value of f(6)?

$$A.\ \frac{1}{10}$$
$$B.\ \frac{1}{9}$$
$$C.\ \frac{1}{8}$$
$$D.\ \frac{1}{7}$$
$$E.\ \frac{1}{6}$$

The OA is D.

I need help to solve this PS question. Please, can any expert assist me with it? Thanks.
f(6) = (6! + 7!)/8!

= 6!(1 + 7)/(6! x 7 x 8)

= 8/(7 x 8)

= 1/7

Answer: D

Scott Woodbury-Stewart
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[email protected]

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