Question

The number of polynomial functions f of degree ≥ 1 satisfying f(x²) = (f (x)) ² = f (f(x)) is

• A. 0
• B. 1
• C. 2
• D. Infinitely many

Answer 1

The correct option is B

1

Let degree of f (x) be n.

deg of f(x²) = 2n

deg of ( f(x²)) = 2n

deg of f( f(x) = n²

n² = 2n iff n = 0 or 2. But n ≥ 1

Degree of f(x) = 2

Since f(x²) = ( f(x²)), if the leading coefficient is a then the coefficient of x⁴ in the LHS is a and in the RHS is a².

a = a² Þ a = 1 ( a = 0 not admissible )

f (x) = x² + bx + c

f (x²) = x⁴ + bx² + c = (f(x))² = (x² + bx + c)²

== x⁴ + 2bx² + b²x² + 2cx² + c² + 2bcx

Thus Þ 0 = 2b ; ∴ b = 0

b² + 2c = b Þ 2c = 0 Þ c = 0

∴ The only function is f (x) = x²