Question

Let `f(x) =ax^2+bx+c` where a,b,c are real numbers. Suppose `f(x)!=x` for any real number x. Then the number of solutions for f(f(x))=x in real numbers x is

Answer 1

`f(x)=ax^2+bx+c`
`g(x)=x`
`f(f(x))=x`
x=d is solution
`f(f(d))=d`
`f(d)=e`
`y=f(x)`
When`!=`e
(d,p)(,p,d)
When d=p
MI with respect to t=x
which is not possible
`f(d)=d->f(x)!=x`
option c is correct.