Question

The graph of the function $f(x)=x+\frac{1}{8}\sin (2\pi x),0\le x\le 1$ is shown below. Define $f_1(x)=f(x),f_{n+1}=f(f_n(x))$, for $n\ge 1$

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

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

Answer 1

Answer: (B)

We have,
$f(x)=x+\frac{1}{8}\sin (2\pi x),x\in [0,1]$

$\underset{n\to \infty }{\lim }{f}_n\left(x\right)=(f(f(f(\dots \infty \text{times }\left(x\right)))$
Now, for $x_1\in \left(0,\frac{1}{2}\right)$
$f\left(x_1\right)>x_1$ as $f\left(x\right)$ is concave-downward
Thus, $f_n\to \frac{1}{2}n\to \infty$
Similarly, for $x_1\in \left(\frac{1}{2},1\right)$
$f\left(x_1\right)<x_1$ as $f\left(x\right)$ is concave upward.
Thus, $f_n\to \frac{1}{2}$ as $n\to \infty$
$\therefore$ Only II is true