p(1) = 1
Let p(x) = x2 + bx = c
then, p(p(x)) = (x2 + bx + c)² + b(x2 + bx + c) + c
Sum of roots of p(p(x)) is given by = - (coefficient of x3/constant term)
= -2b/c2+bc+c = f(a,b) (say)
For critical points,
∂f/∂c = 0 and ∂f/∂b = 0
= 2b(2c+b+1)/c2+bc+c = 0 and - {( c2+bc+c)²- 2b (c)/ ( c2+bc+c)²}= 0
= b(2c+b+1) =0
= b= 0 or b= - (1+2c) = -(2c(c+1))/c2+bc+c = 0
If c=0 , b= -1 => c= 0 or c=-1
If c = -1 ,b = 1
thus we have the following possibility for (b,c)
(0,0) , (0,-1) , (-1,0) ,(1,-1).
At (0,0) f(0,0) = not defined
(0,-1) f(0,-1) = 0
(1,-1) f(1,-1) = 2
(-1,0) is not possible.
p(x) = (x2 +x-1)² + (x2 +x-1) -1
= p(1) = (1+1-1)² + (1+1-1) -1
= 1
Hence, p(1) = 1
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