Let α, β and γ be three positive real numbers. Let f(x)=α x⁵+β x³+γ x, x ∈ R and g: R → R be such that g(f(x))=x for all x ∈ R. If a₁, a₂, a₃, …, aₙ be in arithmetic progression with mean zero, then the value of f(g(1/n Σᵢ₌₁ⁿ f(aᵢ))) is equal to :

Solution

Maths sir · 4 steps

An AP with mean zero is symmetric about 0, and f is odd, so the f-values cancel in pairs.

1. g undoes f. f'(x) = 5α x⁴ + 3β x² + γ > 0, so f is strictly increasing. Being an odd-degree polynomial, it also takes every real value. So f is a bijection, g = f⁻¹, and f(g(y)) = y for every real y.
2. f is odd. Every power in f is odd, so f(-x) = -f(x).
3. Pair the terms. In an AP, terms equally far from the two ends add to twice the mean, so aᵢ + aₙ₊₁₋ᵢ = 0. That gives aₙ₊₁₋ᵢ = -aᵢ, so f(aᵢ) + f(aₙ₊₁₋ᵢ) = 0. For odd n, the middle term is 0 and f(0) = 0.

   1/nΣᵢ₌₁ⁿ f(aᵢ) = 0

4. So. The expression is f(g(0)) = 0, which is option (a).

The idea to keep. Spot the symmetry first. An odd function summed over points placed symmetrically about 0 always gives 0. And f(g(y)) = y holds once g is the true inverse of a bijective f.

Answer from the NTA final answer key. Solution by PrepWiser.