If `f(x)=x(x-1)` is a function from `[1/2,oo)to[-1/4,oo),` then `{x in r :f^(-1)(x)=f(x)}` is a. null set b. `{0,2}` c. `{2}` d. a set containing 3 elements

Options

A. null set
B. `{0,2}`
C. `{2}`
D. a set containing 3 elements

Answer

C

Explanation

`{x in R:f^(-1)(x)=f(x)}`
`={x in R:(f(x))=x}`
`f(f(x))=f(x(x-1))=[x(x-1)][x(x-1)-1]`
`=x(x-1)[x^(2)-x-1]`
Now, `f(f(x))=x`
`rArr" "x(x-1)(x^(2)-x-1)=x`
`rArr" "x(x^(3)-2x^(2))=0`
`rArr" "x=2 (because x in [(1)/(2),oo))`