Let `f(x) =ax^2+bx+c` where a,b,c are real numbers. Suppose `f(x)!=x` – InterviewSolution

1.	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» `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.

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