If f(x) is a polynomial function satisfying f(x) f ( 1 x )=f(x)+f ( 1 x ) and f(4)=257 , then f(3)=

Step-by-step solution

Let f(x)=a₀ x^n+a₁ x^ n-1 +a₂ x^ n-2 + +a_ n-1 x+a_n Then, f(x) f ( 1 x )=f(x)+f ( 1 x ) aligned (a₀ x^n+a₁ x^ n-1 + +a_n ) & ( a₀ x^n + a₁ x^ n-1 + +a_n ) aligned On comparing the coefficient of x^n , we have a₀ a_n=a₀ a_n=1 Comparing the coefficient of x^ n-1 , we have aligned & a₀ a_ n-1 +a_n a₁=a₁ & a₀ a_ n-1 +a₁=a₁ & [ as a_n=1 ] & a₀ a_ n-1 =0 & a_ n-1 =0 [ as a₀ 0 ] & aligned Similarly, a_ n-1 =a_ n-2 = =a₁=0 and a₀= 1 aligned & f(x)=1 x^n & f(4)=1 4^n=257 4^n=256 & 4^n=256 4=4 & f(x)=1+x^4 & aligned So, f(3