As the polynomial, na_nx^n–1 + (n–1) a_n–1x^n–2 + … + a₁ = 0 is a derivative of the polynomial a₀xⁿ + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} +..............a₂x² + a₁x + a₀ = 0 --------- (i)

Putting x = 0 in equation (i),

f(0) = a₀ < 0, {Y – Intercept of the graph is negative}

On the other hand, an > 0 and ‘n’ is even, the leading term on xⁿ, is positive for only x.

For |x| to be large, the term anx_n will dominate, so

lim _x→-∞ f (x) = +∞

lim _x→+∞ f (x) = +∞

If lim _x→+∞ f (x) = +∞, there must exist

Same number α < 0, where f(α > 0)

f(0) = a₀ < 0,

α < β < 0, such that f(β) = 0

Also, there is some value 0 < α,

Where f(a) & so there exists,

0 < b < a with f(b) = 0

Additionally, the polynomial function (equation (i)) is continuous everywhere in R and consequently derivative in R.

a₀xⁿ + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} +..............a₂x² + a₁x + a₀ = 0 is continuous on α,β and derivative on α,β.

Thus, it satisfies both the conditions of Rolle' s Theorem.

As per the Rolle’s Theorem, between any two roots of a function f(x), there exists at least one root of its derivative.

Thus, the equation na_nx^n–1 + (n–1) a_n–1x^n–2 + … + a₁ = 0 will have at least one root between α & β.

Hence, Option (C) is the answer.