Question

Let f be a polynomial function such that f(3x) = f' (x) , f" (x), for all x e r. Then

Answer 1

Answer:

3x³/2

Step-by-step explanation:

Let f(x) be nth degree polynomial,

then f(x) is of the form

f(x)= a₀xⁿ+a₁xⁿ⁻¹+a₂xⁿ⁻²+a₃xⁿ⁻³+...,a₀≠0

f'(x) = na₀xⁿ⁻¹+(n-1)a₁xⁿ⁻²+(n-2)a₂xⁿ⁻³+(n-3)a₃xⁿ⁻⁴+...

f''(x) =n(n-1)a₀xⁿ⁻²+(n-1)(n-2)a₁xⁿ⁻³+(n-2)(n-3)a₂xⁿ⁻⁴+(n-3)(n-4)a₃xⁿ⁻⁵+...

f(3x) =a₀(3x)ⁿ+a₁(3x)ⁿ⁻¹+a₂(3x)ⁿ⁻²+a₃(3x)ⁿ⁻³+...

Given f(3x) = f'(x)f''(x)

=> a₀(3x)ⁿ+a₁(3x)ⁿ⁻¹+a₂(3x)ⁿ⁻²+a₃(3x)ⁿ⁻³+..

=(na₀xⁿ⁻¹+(n-1)a₁xⁿ⁻²+(n-2)a₂xⁿ⁻³+(n-3)a₃xⁿ⁻⁴+...)*(n(n-1)a₀xⁿ⁻²+(n-1)(n-2)a₁xⁿ⁻³+(n-2)(n-3)a₂xⁿ⁻⁴+(n-3)(n-4)a₃xⁿ⁻⁵+...).

We can observe that L.H.S is an expression with degree 3, where as R.H.S

is an expression with degree (n-1)+(n-2)

=2n-3

Since both should be equal expressions on either sides, we get

2n-3 = n,

=> n = 3.

Hence expression is of form a₀x³ + a₁x² + a₂x + a₃

f(3x) = 27a₀x³ +9 a₁x² + 3a₂x + a₃

f'(x) = 3a₀x² + 2a₁x + a₂

f"(x) = 6a₀x + 2a₁

Using f(3x) = f'(x)f''(x), we get

27a₀x³ +9 a₁x² + 3a₂x + a₃

=(3a₀x² + 2a₁x + a₂ )(6a₀x + 2a₁)

=18a₀²x³ + (6a₀a₁ + 12a₁a₀)x² + (4a₁²+6a₀a₂)x + 2a₁a₂

Now comparing L.H.S and R.H.S wew get

27a₀ = 18a₀²

=> a₀ = 3/2

6a₀a₁ + 12a₁a₀ = 9a₁

Substituting the value of a₀,we get

=>a₁ = 0

4a₁²+6a₀a₂ = 3a₂

=>a₂ = 0

Comparing the constants, we get a₃ = 0,

hence f(x) = 3/2x³.