## SUBJECT III

On the set of real numbers, the composition law is defined as $\quad x * y=x y+a x+b y+c, a, b, c \in R$.

a) Determine $a, b$ and $c$ such that the law "*" is stable on $(2,+\infty)$ and has a neutral element.

b) Determine $a, b$ and $c$ such that $((2,+\infty), *)$ is a group.

c) Determine the differentiable morphisms $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*},\right), f(x)=a x+b$.
## SUBJECT III

On the set of real numbers, the composition law is defined as $\quad x * y=x y+a x+b y+c, a, b, c \in R$.

a) Determine $a, b$ and $c$ such that the law "*" is stable on $(2,+\infty)$ and has a neutral element.

b) Determine $a, b$ and $c$ such that $((2,+\infty), *)$ is a group.

c) Determine the differentiable morphisms $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*}, \cdot\right), f(x)=a x+b$.

Selected by Prof. Haiduc Sorina

Let the law be $x * y=x y+a x+b y+c, a, b, c \in R$.

If this law admits a neutral element $e \in[2 ;+\infty)$, then we have $x^{*} e=e^{*} x=x$, for any $x \in[2 ;+\infty)$.

From $x^{*} e=x \Rightarrow e(x+b)+(a-1)\left(x+\frac{c}{a-1}\right)=0, a \neq 1$

It is observed that for the existence of the neutral element, we must have $b=\frac{c}{a-1}, a \neq 1$.

Similarly, from $x^{*} e=x \Rightarrow a=\frac{c}{b-1}, b \neq 1$.

From here, we conclude that $a=b$ and $c=a^{2}-a$ $1 p$

Our law now looks like this: $x * y=x y+a x+a y+a^{2}-a$, where $a \in R \backslash\{1\}$

From $H=(2,+\infty)$ being a stable part of $R$ with respect to the law *, it follows that $\forall x, y \in H \Rightarrow x^{*} y \in H$, i.e., from $x>2, y>2$ we get $(x-2)(y-2)>0$ or $x * y-(a+2) x-(a+2) y-a^{2}+a+6>2, \forall x, y \in H$

From here, $(a+2)(x+y)+a^{2}-a-6>0, \forall x, y \in H$

It follows that $a^{2}+3 a+2 \geq 0$, i.e., $a \in(-\infty ;-2] \cup[-1 ;+\infty)$

But $e=1-a \in(2,+\infty) \Rightarrow a2, \forall x>2$ which is satisfied if and only if $a=-2$.

Therefore, $((2,+\infty), *)$ is a group if and only if $a=b=-2, c=6$ $1 \mathrm{p}$

c) From b) it follows that $x * y=x y-2 x-2 y+6, \forall x, y>2$.

We seek morphisms of the form $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*}, \cdot\right), f(x)=a x+b$

Since $\quad e=3>2$ is the neutral element in the group $((2,+\infty), *)$ and $e^{\prime}=1$ is the neutral element in the group $\left(R_{+}^{*}, \cdot\right)$, it follows that $f(3)=1$, i.e., $b=1-3 a$. It follows that $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*} \cdot\right), f(x)=a x+1-3 a$. $1 \mathrm{p}$

From the definition of the morphism, $f\left(x^{*} y\right)=f(x) \cdot f(y), \forall x, y>2$, we get $a=0$ or $a=1$.

For $a=0$, we obtain the morphism $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*},\right), f(x)=1$, and for $a=1$, we obtain the morphism $f:((2,+\infty), *) \rightarrow\left(R_{+}^{*}, \cdot\right), f(x)=x-2$.

Clearly, both morphisms are differentiable. $1 \mathrm{p}$