Question

If the equation ${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .... + {a_1}x = 0$, ${a_1} \ne 0,\,n \geqslant 2$ has a positive root $x = \alpha $, then the equation $n{a_n}{x^{n - 1}} + (n - 1){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0$ has a positive root, which is
A. Greater than or equal to $a$
B. Equal to $\alpha $
C. Greater than $\alpha $
D. Smaller than $\alpha $

Answer 1

Hint: Write the first equation as $f(x)$ and if you differentiate $f(x)$ you will get the second equation in the question. Find two values a and b such that $f(a) = f(b)$. Check whether Rolle’s theorem is applicable and use it if it is applicable.

Formula used: $\dfrac{{d(a{x^n} + b{x^m})}}{{dx}} = na{x^{n - 1}} + mb{x^{m - 1}}$

Complete step-by-step solution:
Let $f(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .... + {a_1}x$
Rolle’s theorem states that if a function is continuous on $[a,b]$ and differentiable on $(a,b)$ and if $f(a) = f(b)$, then there exist a number $c \in (a,b)$ such that $f'(c) = 0$.
$f(0) = 0$
Since $\alpha $ is a root of $f(x)$, $f(\alpha ) = 0$.
$f(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .... + {a_1}x$ is continuous on $[0,\alpha ]$ and differentiable on $(0,\alpha )$ because it is a polynomial function.
Since $f(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + .... + {a_1}x$ is continuous on $[0,\alpha ]$ and differentiable on $(0,\alpha )$ and $f(0) = f(\alpha )$, there exists a number $c \in (0,\alpha )$ such that $f'(c) = 0$.
$f'(x) = n{a_n}{x^{n - 1}} + (n - 1){a_{n - 1}}{x^{n - 2}} + .... + {a_1}$
$f'(c) = 0$ means that $c$ is a root of $n{a_n}{x^{n - 1}} + (n - 1){a_{n - 1}}{x^{n - 2}} + .... + {a_1}$. The quadratic equation $n{a_n}{x^{n - 1}} + (n - 1){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0$ therefore has a root which is greater than $0$ and lesser than $\alpha $.
Therefore, the correct answer is D. Smaller than $\alpha $.

Note: The Rolle’s theorem cannot be applied if any of the following three conditions are not met: 1) The function needs to be continuous on the given interval (including the end points of the interval), 2) The function needs to be differentiable on the given interval (except the end points of the interval), 3) The value of the function needs to be the same at the end points of the interval being considered.