How many polynomials of degree >=1 satisfy f(x2)=[f(x)]2=f(f(x)
a) more than 2
b) 2
c) 0
d) 1

• let f(x) be a polynomial of degree n>=1
  then comparing the degrees of f(x^2) , (f(x))^2 and f(f(x)) we have 2n=2n=n^2
  since n not = 0 we have n=2
  so the polynomial f(x) be ax^2+bx+c where a not = 0
  then putting this in f(x^2)=(f(x))^2=f(f(x)) and expanding gives
  ax^4+bx^2+c = a^2x^4+2abx^3+(2ac+b^2)x^2+2bcx+c^2=a^3x...
  comparing the co-efficients of f(x^2) , (f(x))^2 and f(f(x)) respectively
  we have
  a=a^2=a^3
  0=2ab=2a^2b
  b=2ac+b^2=ab^2+2a^2c+ab
  0=2bc=2abc+b^2
  c=c^2=bc+c
  solving these we get a=1 , b=0, and c=0
  therefore the polynomial is f(x)=x^2 this is the only polynomial satisfying the given conditions
  
• 10 years agoHelpfull: Yes(1) No(4)
• ans will be only one ans: 1
  
• 10 years agoHelpfull: Yes(0) No(0)