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 fixed points of $f$ equals

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

Answer 1

The correct option is B $1$

$f\left(x_{\circ }\right)=x_{\circ }$
Let $g\left(x\right)=f\left(x\right)-x$
Now, there exist atleast one real number $x_{\circ }$, for which $g\left(x_{\circ }\right)=0$
i.e., $f\left(x_{\circ }\right)-x_{\circ }=0,x_{\circ }\in R$
$\Rightarrow f\left(x_{\circ }\right)=x_{\circ }$
Let $x_1$ and $x_2$ be $2$ fixed points of $f$ i.e., $f\left(x_1\right)=x_1$ and $f\left(x_2\right)=x_2$
Let $x_1<x_2...\left(1\right)$
Since, $f$ is decreasing function.
$\therefore f\left(x_1\right)\ge f\left(x_2\right)$
$\Rightarrow x_1\ge x_2$ which contradict eqn(1).
Therefore, there is a only one fixed point.