Quadratic-1

f(f(x))=x is same as f(x)=f(inverse)(x) in some bijective domain (The real domain has to be divided into 2 in the quadratic case so that each is a bijective one). Since f(x) != x for any real x and f(inverse) is just the reflection of 'f' about y=x, the equation has no real solution.

As We Can Notice That f(x) = x does not have a real solution .

Let c and d denote the imaginary roots of the above function

so for f(f(x)) = x to have a solution the input in function f should be the roots of the above equation which are c and d .

so for f(f(x)) = x to hold f(x) = c or d then only it is possible that given condition holds.

But as a ,b,c are real numbers and we have been given that x is real so it is not possible for the function with real coefficients and for real x

to attain imaginary values .

therefore the number of real solution are 0 .

If $f(x) - x > 0$

$\Rightarrow f(f(x)) - f(x) > 0 \Rightarrow f(f(x)) - x > f(x) - x > 0$

Similarly if $f(x) - x < 0$

$\Rightarrow f(f(x)) - f(x) < 0 \Rightarrow f(f(x)) - x < f(x) - x < 0$

Both the cases lead to no solution.

Hence $\boxed{0}$ real values.

Let t = f(x). We will have system f(t) = x; f(x) = t => x = t or a(x + t) + b + 1 = 0..

In case x = t => f(x) = x, this function has no root

In case a(x + t) + b + 1 = 0

=> a^2x^2 + a(b + 1)x + ac + b + 1 = 0

delta = a^2 (b + 1)^2 - 4a^2 (ac + b + 1)

=a^2 [ b^2 + 2b + 1 - 4ac - 4b - 4]

= a^2[(b - 1)^2 - 4ac - 4]

As f(x) = x don't have root => (b-1)^2 - 4ac < 0 => delta < 0 => This case doesn't have root either.