- Home
- Standard 12
- Mathematics
The graph of the function $f(x)=x+\frac{1}{8} \sin (2 \pi x), 0 \leq x \leq 1$ is shown below. Define $f_1(x)=f(x), f_{n+1}(x)=f\left(f_n(x)\right)$, for $n \geq 1$.
Which of the following statements are true?
$I.$ There are infinitely many $x \in[0,1]$ for which $\lim _{n \rightarrow \infty} f_n(x)=0$
$II.$ There are infinitely many $x \in[0,1]$ for which $\lim _{n \rightarrow \infty} f_n(x)=\frac{1}{2}$
$III.$ There are infinitely many $x \in[0,1]$ for which $\lim _{n \rightarrow \infty} f_n(x)=1$
$IV.$ There are infinitely many $x \in[0,1]$ for which $\lim _{n \rightarrow \infty} f_n(x)$ does not exist.
$I$ and $III$ only
$II$ only
$I,II,III$ only
$I, II, III$ and $IV$
Solution
(b)
$f_1(x)=x+\frac{1}{8} \sin (2 \pi x)$
$f_2(x)=x+\frac{1}{8} \sin 2 \pi x+\frac{1}{8} \sin \left(2 \pi\left(x+\frac{1}{8} \sin 2 \pi x\right)\right)$
Similarly, $f _3( x )= x +\frac{1}{8}\left(\sin (2 \pi x)+\sin \left( f _1( x )\right)+\sin \left( f _2( x )\right)\right)$
$\therefore f_n(x)=x+\frac{1}{8}\left(\sin 2 \pi x+\sin \left(f_1(x)\right)+\ldots \cdots+\sin \left(f_{n-1}(x)\right)\right)$
$f_n(x)=0$ at $x=0$ and $f_n(x)=1$ at $x=1$
Therefore $I$ and $III$ are not true
from the figure, $f _{ n }( x )=\frac{1}{2}$ for many value of $x \in[0,1]$
statement $II$ is true
for every $x , \quad 0 \leq f _1( x ) \leq 1$
therefore, for every value of $x, \quad 0 \leq f_n(x) \leq 1$
$\lim _{n \rightarrow 0} f_n(x)$ always exists for $x \in[0,1]$
$IV$ is also not true.
