Let `f(x)=3/4x+1,f^n(x)b ed efin e da sf^2(x)=f(f(x)),a n dforngeq2,f^(n+1)(x)=f(f^n(x))dotIflambda=(lim)_(nvecoo)f^n(x)` , then `lambda` is independent of `x` `lambda` is a linear polynomial in `x` the line `y=lambda` has slope `0.` the line `4y=lambda` touches the unit circle with centre at the origin.

Options

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 center at the origin.

Answer

A, C, D

Explanation

`f^(2)(x)=f((3)/(4)x+1)=(3)/(4)((3)/(4)x+1)+1=((3)/(4))^(2)x+(3)/(4)+1(1)`
` f^(3)(x)=f{f^(2)(x)}=(3)/(4){f^(2)(x)+1}`
`=(3)/(4){((3)/(4))^(2)x+(3)/(4)+1}+1`
`=((3)/(4))^(3)x+((3)/(4))^(2)+(3)/(4)+1`
` :. f^(n)(x)=((3)/(4))^(n)x+((3)/(4))^(n-1)+((3)/(4))^(n-2)+ ... +((3)/(4))+1`
`=((3)/(4))^(n)x+(1-((3)/(4))^(n))/(1-(3)/(4))`
` :. lambda=underset(n to oo)(lim) f^(n)(x)=0+4=4`