If the equation $a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x=0$, $a_{1} \neq 0$, $n \geq 2$, has a positive root $x=\alpha$, then the equation $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ has a positive root, which is

(C) greater than or equal to $\alpha$

JEE Main · 2005 · 08 May 2005 · Offline · Application of Derivatives

If the equation

a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x = 0

a_{1} \neq = 0

n \geq 2

, has a positive root

x = α

, then the equation

n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \dots + a_{1} = 0

has a positive root, which is

Answer: (B)

Let

f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x

. It has no constant term, so

f (0) = 0

, and by hypothesis

f (α) = 0

for some

α > 0

. The second equation given is precisely

f^{'} (x) = 0

. Since

f

is a polynomial, hence differentiable, and

f (0) = f (α) = 0

, Rolle's theorem guarantees a point

c

(0, α)

with

f^{'} (c) = 0

. This root is positive and lies strictly between

0

and

α

, so it is smaller than

α

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option