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 $\displaystyle\lim_{n\to\infty} f _{n}\left(x\right)=0$
II. There are infinitely many $x \in\left[0, 1\right]$ for which $\displaystyle\lim_{n\to\infty} f_{n}(x)=\frac{1}{2}$
III. There are infinitely many $x \in\left[0, 1\right]$ for which $\displaystyle\lim_{n\to\infty} f_{n}(x)=1$
IV. There are infinitly many $x \in\left[0, 1\right]$ for which $\displaystyle\lim_{n\to\infty} f_{n}(x)$ 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,
$f (x)=x+\frac{1}{8} \sin (2\pi\,x), x \in [0,1]$

$\displaystyle\lim_{n\to\infty} f _{n}\left(x\right)=(f (f (f (\ldots\infty {\text{times }} \left(x\right)))$
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, $f_{n}\rightarrow\frac{1}{2} n \rightarrow\infty$
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, $f_{n} \rightarrow \frac{1}{2}$ as $n \rightarrow\infty$
$\therefore $ Only II is true