If f(x)`=(x-1)/(x+1),f^(2)(x)=f(f(x)),……..,……..f^(k+1)(x)=f(f^(k)(x))`,k=1,2,3,……and `g(x)=f^(1998)(x)` then `int_(1//e)^(1) g(x)dx` is equal to

Options

A. 0
B. 1
C. -1
D. e

Answer

C

Explanation

We have,
f(x)`=(x-1)/(x+1)`
`rArr f^(2)(x)=f(f(x))=f((x-1)/(x+1))==((x-1)/(x+1)-1)/((x-1)/(x+1)+1)=-(1)/(x)`
`rArr f^(4) (x)=f^(2)(f^(2)f(x))=f^(2)(-(1)/(x))=(-1)/(-(1)/(x))=x`
`:. G(x)=f^(1998)(x)=f^(2)"of"^(1996)(x)`
`rArr g(x)=f^(2)(f^(1996)(x))`
`g(x)=f^(2)(x)[:'f^(1996)(x)=(f^(4)"of"^(4)"of"^(4)"o"......"of"^(4))(x)=x]`
`rArr g(x)=-(1)/(x)`
`:. overset(1)underset(1//e)int g(x)dx=overset(1)underset(1//e)int-(1)/(x)dx=-[log_(e )x]_(1//e)^(1)`
`rArr overset(1)underset(1//e)int g(x)dx=-[log_(e )1-log_(e )(1//e)]=-[0+1]=-1`