Question

The graph of the function $f\left(x\right)=x+\frac{1}{8}\sin \left(2\pi x\right),0\le x\le 1$ is shown below. Define $f_1\left(x\right)=f\left(x\right),f_{n+1}\left(x\right)=f\left(f_n\right(x\left)\right)$, for $n\ge 1$.
Which of the following statement are true?
I. There are infinitely many $xϵ[0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)=0$.
II. There are infinitely many $xϵ[0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)=\frac{1}{2}$.
III. There are infinitely many $xϵ[0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)=1$.
IV. There are infinitely many $xϵ[0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)$ does not exist.

• A. I and III only
• B. II only
• C. I, II, III only
• D. I, II, III and IV

Answer 1

The correct option is B II only

$f_1\left(x\right)=x+\frac{1}{8}\sin \left(2\pi x\right)$

$f_2\left(x\right)=x+\frac{1}{8}\sin 2\pi x+\frac{1}{8}\sin \left(2\pi \right(x+\frac{1}{8}\sin 2\pi x\left)\right)$

Similarly, $f_3\left(x\right)=x+\frac{1}{8}\left(\sin \right(2\pi x)+\sin (f_1\left(x\right))+\sin (f_2\left(x\right)\left)\right)$

$\therefore f_n\left(x\right)=x+\frac{1}{8}(\sin 2\pi x+\sin (f_1\left(x\right))+\dots \cdots +\sin (f_{n-1}\left(x\right)\left)\right)$

$f_n\left(x\right)=0$ at $x=0$ and $f_n\left(x\right)=1$ at $x=1$

Therefore I and III are not true

from the figure, $f_n\left(x\right)=\frac{1}{2}$ for many value of $x\in [0,1]$

statement II is true

for every $x,0\le f_1\left(x\right)\le 1$

therefore, for every value of $x,0\le f_n\left(x\right)\le 1$

${lim}_{n\to 0}{f}_n\left(x\right)$ always exists for $x\in [0,1]$

IV is also not true.