Question

Let $f\left(x\right)=\frac{3}{4}x+1,$ and $f^n\left(x\right)$ be defined as $f^2\left(x\right)=f\left(f\right(x\left)\right)$ and for $n\ge 2,f^{n+1}\left(x\right)=f\left(f^n\right(x\left)\right).$ If $\lambda =\underset{n\to \infty }{lim}{f}^n\left(x\right),$ then

• A. $\lambda$ is independent of $x$
• B. $\lambda$ is a linear polynomial in $x$
• C. the line $y=\lambda$ has slope $0$
• D. the line $4y=\lambda$ touches the unit circle with centre at the origin

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\lambda$ is independent of $x$
C the line $y=\lambda$ has slope $0$
D the line $4y=\lambda$ touches the unit circle with centre at the origin

$f\left(x\right)=\frac{3}{4}x+1$
$f^2\left(x\right)=f(\frac{3}{4}x+1)=\frac{3}{4}(\frac{3}{4}x+1)+1$
$={\left(\frac{3}{4}\right)}^{2}x+\frac{3}{4}+1\cdots \left(1\right)$
$f^3\left(x\right)=f\left(f^2\right(x\left)\right)=\frac{3}{4}\left(f^2\right(x\left)\right)+1$
$=\frac{3}{4}[{\left(\frac{3}{4}\right)}^{2}x+\frac{3}{4}+1]+1$
$={\left(\frac{3}{4}\right)}^{3}x+{\left(\frac{3}{4}\right)}^2+\frac{3}{4}+1$
$⋮$
$⋮$
$\therefore f^n\left(x\right)={\left(\frac{3}{4}\right)}^{n}x+{\left(\frac{3}{4}\right)}^{n-1}+{\left(\frac{3}{4}\right)}^{n-2}+\cdots +\left(\frac{3}{4}\right)+1$
$={\left(\frac{3}{4}\right)}^{n}x+\frac{1-{\left(\frac{3}{4}\right)}^n}{1-\frac{3}{4}}$
$\therefore \lambda =\underset{n\to \infty }{lim}{f}^n\left(x\right)=0+4=4$