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.

Let f(x)=(3)/(4)x+1,f^(n)(x) be defined as f^(2)(x)=f(f(x)), and for n>=2,f^(n+1)(x)=f(f^(n)(x)). If lambda=(lim)_(n rarr oo)f^(n)(x), then (a)lambda is independent of x( b) lambda is a linear polynomial in 4y=lambda touches the unit circle 0. (d)the line 4y=lambda touches the unit circle with centre at the origin.

If f(x)=(a-x^n)^(1/n) then find f(f(x))

If f(x)=x+(1)/(x) , such that f^3 (x)=f(x^(3))+lambdaf((1)/(x)) , then lambda=

Let f(x)=ax+b with a,b in R , f_(1)(x)=f(x) and f_(n+1)(x)=f(f_(n)(x)),n in N , If f_(7)(x)=128x+381 , Then a+b is

If f(x)=(2018x-2019)/(x+lambda) and f(f(x))=x then lambda=

f_(n)(x)=e^(f_(n-1)(x))" for all "n in N and f_(0)(x)=x," then "(d)/(dx){f_(n)(x)} is

Let f_(1)(x)=e^(x),f_(2)(x)=e^(f_(1)(x)),......,f_(n+1)(x)=e^(f_(n)(x)) for all n>=1. Then for any fixed n,(d)/(dx)f_(n)(x) is

Let f: N to N be defined by f(x) = x-(1)^(x) AA x in N , Then f is