Question

Let $f:R\to R$ be a continuous decreasing function. A point $x_0\in R$ is said to be a fixed point of $f$ if $f\left(x_0\right)=x_0.$
The number of distinct $3$-tuples $(x,y,z)$ satisfying the system
$x=f\left(y\right),y=f\left(z\right),z=f\left(x\right)$
equals

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Answer 1

The correct option is B $1$

$x=f\left(y\right)y=f\left(z\right)z=f\left(x\right)$

$x=f\left(f\right(z\left)\right)$
$=f\left(f\right(f\left(x\right)\left)\right)$
$=(f\circ f\circ f)\left(x\right)...\left(1\right)$
Similarly,
$y=(f\circ f\circ f)\left(y\right)...\left(2\right)$
$z=(f\circ f\circ f)\left(z\right)...\left(3\right)$

Now, let $x,y\in D_f=R$ and $x<y$
$\Rightarrow f\left(x\right)\ge f\left(y\right)$
$\Rightarrow f\left(f\right(x\left)\right)\le f\left(f\right(y\left)\right)$
​​​​​​​$\Rightarrow f\left(f\right(f\left(x\right)\left)\right)\ge f\left(f\right(f\left(y\right)\left)\right)$
​​​​​​​​​​​​​​​​​​​​​$\Rightarrow (f\circ f\circ f)\left(x\right)\ge (f\circ f\circ f)\left(y\right)$
​​​​​​​​​​​​​​​​​​​​​$\Rightarrow f\circ f\circ f$ is decreasing function on $R$.
​​​Therefore, it has only one fixed point.

Since, $f\circ f\circ f$ has only one fixed point.
$\therefore x=y=z$

Hence, there is only one distinct $3$-tuples $(x,y,z)$ that satisfying the system of equations.