Question

Let $f\left(x\right)=\frac{x+1}{x-1}$ for all $x\ne 1$.

Let
$f^1\left(x\right)=f\left(x\right),f^2\left(x\right)=f\left(f\right(x\left)\right)$ and generally
$f^n\left(x\right)=f\left(f^{n-1}\right(x\left)\right)$ for $n>1$
Let $P=f^1\left(2\right)f^2\left(3\right)f^3\left(4\right)f^4\left(5\right)$
Which of the following is a multiple of P ?

• A. $125$
• B. $375$
• C. $250$
• D. $147$

Answer 1

Answer: (B)

$375$

$f^1\left(x\right)=\frac{x+1}{x-1}$

$f^2\left(x\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}$

$=\frac{2x}{2}=x$

Now, $f^3\left(x\right)=f\left(f^2\right(x\left)\right)=f\left(x\right)=f^1\left(x\right)$

This will repeat so,

$f^3\left(x\right)=f^1\left(x\right)$ $f^4\left(x\right)=f^2\left(x\right)$

We have,

$f^1\left(2\right)=3,$ $f^2\left(3\right)=3,$ $f^3\left(4\right)=\frac{5}{3},$ $f^4\left(5\right)=5$

So, $P=75$ and $375$ is the multiple of $75$.

Hence, the answer is $375$.