Problem 1374

1. a) Since $f$ is differentiable, hence continuous, and $f^{\prime}(x)=\frac{-3}{(x-1)^{2}}<0 \Rightarrow f$ is strictly decreasing on $(1,+\infty) \Rightarrow \operatorname{Im} f=[f(4), f(2)]=[-1,1]$

$f(x \circ y)=f(x) f(y) \quad \forall x, y \in A$
b) $f$ is strictly monotonic on $A \Rightarrow f$ is injective on $A$.

It is shown that $f(x \circ y) \circ z=f(\circ(y \circ z)$, and since $f$ is injective $\Rightarrow$ the operation " $\circ$ " is associative. (1p)

We solve the equation $f(x \circ e)=f(x) \Rightarrow f(e)=1 \Rightarrow e=2 \in A$ is the neutral element.

To determine the invertible elements, we solve the equation $f\left(x \circ x^{\prime}\right)=f(e)$.

For $x \neq \frac{5}{2} \Rightarrow f(x) \neq 0$ and $f\left(x^{\prime}\right)=\frac{1}{f(x)} \in[-1,1] \Rightarrow x \in\{2,4\}$ which are the only invertible elements.

$$f\left(\prod_{k=1}^{n} \frac{6 k+5}{3 k+2}\right)=\prod_{k=1}^{n} f\left(\frac{6 k+5}{3 k+2}\right)=\prod_{k=1}^{n} \frac{k}{k+1}=\frac{1}{n+1}=f\left(\frac{5 n+6}{2 n+3}\right) \text {, and then }$$

$$\frac{11}{5} \circ \frac{17}{8} \circ \frac{23}{11} \circ \ldots \circ \frac{6 n+5}{3 n+2}=\frac{5 n+6}{2 n+3}$$