For ` x in R , x ne0, 1, ` let `f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x)),n=0,1,2…..` Then the value of `f_(100)(3)+f_(1)((2)/(3))+f_(2)((3)/(2))` is equal to

A. `(4)/(3)`

B. `(1)/(3)`

C. `(5)/(3)`

D. `(8)/(3)`

A. `(4)/(3)`

B. `(1)/(3)`

C. `(5)/(3)`

D. `(8)/(3)`

Correct Answer – C

We have , `f_(0)(x)=(1)/(1-x) and f_(n+1)(x)=f_(0)(f_(n)(x))`.

`:. f_(1)(x)=f_(0)(f_(0)(x))=f_(0)((1)/(1-x))=(1)/(1-(1)/(1-x))=(x-1)/(x)`

`and , f_(2)(x)=f_(0)(f_(1)(x))=f_(0)((x-1)/(x))=(1)/(1-(x-1)/(x))=x`

Thus, ` f_(3)(x)=f_(0)(f_(2)(x))=f_(0)(x)`

`f_(4)(x)=f_(0)(f_(3)(x))=f_(0)(f_(0)(x))=f_(1)(x)`

`f_(5)(x)=f_(0)(f_(4)(x))=f_(0)(f_(1)(x))=f_(2)(x)=x`

`f_(6)(x)=f_(0)(f_(5)(x))=f_(0)(f_(2)(x))=f_(0)(x)` and so on.

In general `f_(3n)(x)=f_(0)(x)`

` f_(3n_+1)(x)=f_(1)(x) and f_(3n+2)(x)=f_(2)(x)=x ` for ` n=1,2,`

`:. f_(100)(x)=f_(1)(x)=(x-1)/(x)`.

Hence, `f_(100)(x)+f_(1)((2)/(3))+f_(2)((3)/(2))=f_(1)(3)+f_(1)((2)/(3))+f_(2)((3)/(2))`

`=(30-1)/(3)+((2)/(3)-1))/((2)/(3))+(3)/(2)=(2)/(3)-(1)/(2)+(3)/(2)=(5)/(3)`