Question

Let $f\left(x\right)=\frac{3x}{4}+1,f_2\left(x\right)=f\left(f\right(x\left)\right)$ and for $n\ge 2,f_{n+1}\left(x\right)=f\left(f_n\right(x\left)\right)$. If $\alpha =\underset{n\to \infty }{lim}{f}_n\left(x\right)$. Then

• A. $a$ has $9$ integral solutions for $|a-2|\le \alpha$
• B. $\underset{0}{\overset{2/3}{\int }}{f}_n\left(x\right)dx<0$
• C. the line $8y=\alpha$ intercepts a chord of length of $\sqrt{3}$ with $x^2+y^2=1$
• D. $\alpha$ is dependent on $x$

Answer 1

Answer: (A), (C)

The correct options are
A $a$ has $9$ integral solutions for $|a-2|\le \alpha$
C the line $8y=\alpha$ intercepts a chord of length of $\sqrt{3}$ with $x^2+y^2=1$

$f\left(x\right)=\frac{3x}{4}+1f_2\left(x\right)=f\left(f\right(x\left)\right)\Rightarrow f_2\left(x\right)=f(\frac{3x}{4}+1)\Rightarrow f_2\left(x\right)=\frac{3}{4}(\frac{3x}{4}+1)+1\Rightarrow f_2\left(x\right)={\left(\frac{3}{4}\right)}^{2}x+\frac{3}{4}+1$

$f_3\left(x\right)=f\left(f^2\right(x\left)\right)\Rightarrow f_3\left(x\right)=f({\left(\frac{3}{4}\right)}^{2}x+\frac{3}{4}+1)\Rightarrow f_3\left(x\right)=\frac{3}{4}({\left(\frac{3}{4}\right)}^{2}x+\frac{3}{4}+1)+1\Rightarrow f_3\left(x\right)={\left(\frac{3}{4}\right)}^{3}x+{\left(\frac{3}{4}\right)}^2+\frac{3}{4}+1$

Similarly we can write,
$\Rightarrow f_n\left(x\right)={\left(\frac{3}{4}\right)}^{n}x+{\left(\frac{3}{4}\right)}^{n-1}+\cdots +\frac{3}{4}+1\Rightarrow f_n\left(x\right)={\left(\frac{3}{4}\right)}^{n}x+\frac{1(1-{\left(\frac{3}{4}\right)}^n)}{(1-\frac{3}{4})}$

Now,
$\alpha =\underset{n\to \infty }{lim}{f}_n\left(x\right)\Rightarrow \alpha =\underset{n\to \infty }{lim}[{\left(\frac{3}{4}\right)}^{n}x+\frac{1(1-{\left(\frac{3}{4}\right)}^n)}{(1-\frac{3}{4})}]\Rightarrow \alpha =4$
So, $\alpha$ is independent of $x$.

$|a-2|\le \alpha \Rightarrow |a-2|\le 4\Rightarrow -4\le a-2\le 4\Rightarrow -2\le a\le 6$
So, $9$ integral solutions.

$8y=\alpha \Rightarrow y=\frac{1}{2}$


So, the length of the chord $=2\sqrt{1-{\left(\frac{1}{2}\right)}^2}=\sqrt{3}$

Now,
$\underset{0}{\overset{2/3}{\int }}{f}_n\left(x\right)dx=\underset{0}{\overset{2/3}{\int }}({\left(\frac{3}{4}\right)}^{n}x+\frac{1-{\left(\frac{3}{4}\right)}^n}{(1-\frac{3}{4})})dx={\left(\frac{3}{4}\right)}^n{\left[\frac{x^2}{2}\right]}_0^{2/3}+4(1-{\left(\frac{3}{4}\right)}^n)[x{]}_0^{2/3}$

$=\frac{8}{3}-\frac{22}{9}{\left(\frac{3}{4}\right)}^n$

Assuming
$\frac{8}{3}-\frac{22}{9}{\left(\frac{3}{4}\right)}^n<0\Rightarrow {\left(\frac{3}{4}\right)}^n>\frac{12}{11}$
which is not possible.
So, $\underset{0}{\overset{2/3}{\int }}{f}_n\left(x\right)dx>0$