Question

Let f(x) be a polynomial function of degree n satisfying the condition f(x) + f(1/x) = f(x)f (1/x) ∀x ∈ R − {0}. Then, prove that f(x) = 1 ± xⁿ. 

Answer 1

Let us consider that f(x) = a₀ + a₁x + ⋅⋅⋅ + a_nxⁿ. Thus,

Multiplying on both sides with xⁿ, we get

Equating the coefficients of x²ⁿ, x^{2n − 1}, ... , x^{n + 1} on both sides, we get

a_n = a₀a_n ⇒ a₀ = 1    [since an ≠ 0]

That is,

a_{n − 1} = a_{n − 1}a₀ + a_na₁ ⇒ a_na₁ = 0 ⇒ a₁ = 0

Similarly, we get 

a₂ = a₃ = ... = a_{n − 1} = 0

Now, equate the coefficient of xⁿ on both sides, we get

Hence, f(x) = 1 ± xⁿ.