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 $f_1\left(x\right)=f\left(x\right),f_{n+1}\left(x\right)=f\left(f_n\right(x\left)\right),\text{for}n\ge 1$



Which of the following statements are true ?

I. There are infinitely many $x\in [0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)=0$
II. There are infinitely many $x\in [0,1]$ for which $\underset{n\to \infty }{lim}=\frac{1}{2}$
III. There are infinitely many $x\in [0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)=1$
IV. There are infinitely many $x\in [0,1]$ for which $\underset{n\to \infty }{lim}{f}_n\left(x\right)$ does not exist

• A. $I\text{and}II$ only
• B. $II$ only
• C. $I,II,III$ only
• D. $I,II,III\text{and}IV$ only

Answer 1

Answer: (B)

$II$ only

Given function,

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

$\underset{n\to \infty }{lim}f\left(x\right)={lim}_{n\to \infty }x+\frac{1}{8}\sin \left(2n\pi \right)=\infty \Rightarrow f_n\left(x\right)$ does not exist

Now for $x\in (0,\frac{1}{2})$
$f\left(x\right)>x$
So when $n\to \infty$ then $f_n\left(x\right)\to \frac{1}{2}$

Similarly for $x\in (\frac{1}{2},1)$
$f\left(x\right)<x$
So when $n\to \infty$ then $f_n\left(x\right)\to \frac{1}{2}$

Hence $f_n\left(x\right)\to \frac{1}{2}$ as $n\to \infty$

Hence, option (D) is correct.