Question

{"given":"We have a polynomial equation $f(x) = a_n x^n + a_{n-1} x^{n-1} + \\ldots + a_1 x = 0$ with $a_1 \\neq 0$ and $n \\geq 2$. This equation has a positive root $x = \\alpha$. We need to find the relationship between $\\alpha$ and the positive root of the derivative $f'(x) = na_n x^{n-1} + (n-1)a_{n-1} x^{n-2} + \\ldots + a_1 = 0$.","key_observation":"Since $f(x) = x(a_n x^{n-1} + a_{n-1} x^{n-2} + \\ldots + a_1)$, we have $f(0) = 0$ and $f(\\alpha) = 0$. The function $f(x)$ is continuous and differentiable on $[0, \\alpha]$. By Rolle's theorem, there exists at least one point $k \\in (0, \\alpha)$ such that $f'(k) = 0$. This means the derivative has a root in the open interval $(0, \\alpha)$, which is necessarily smaller than $\\alpha$.","option_analysis":[{"label":"(A)","text":"greater than $\\alpha$","verdict":"incorrect","explanation":"This contradicts Rolle's theorem. Since $f(0) = 0$ and $f(\\alpha) = 0$, the derivative must have a root between these points, not beyond $\\alpha$."},{"label":"(B)","text":"smaller than $\\alpha$","verdict":"correct","explanation":"By Rolle's theorem, since $f(0) = 0$ and $f(\\alpha) = 0$, there exists $k \\in (0, \\alpha)$ such that $f'(k) = 0$. Therefore, the positive root of $f'(x) = 0$ is smaller than $\\alpha$."},{"label":"(C)","text":"greater than or equal to $\\alpha$","verdict":"incorrect","explanation":"This is incorrect because Rolle's theorem guarantees the existence of a root strictly between 0 and $\\alpha$, not at or beyond $\\alpha$."},{"label":"(D)","text":"equal to $\\alpha$","verdict":"incorrect","explanation":"The root cannot be equal to $\\alpha$ because Rolle's theorem places it strictly in the open interval $(0, \\alpha)$, excluding the endpoints."}],"answer":"(B)","formula_steps":[]}