Question

If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x=0$ has a positive root $x=\alpha$, then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1=0$ has a positive root, which is

• A. Smaller than $\alpha$
• B. Greater than $\alpha$
• C. Equal to $\alpha$
• D. Greater than or equal to $\alpha$

Answer 1

The correct option is A Smaller than $\alpha$

Considering
$f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+...a_2{x}^2+a_{1}x$
Hence, $f\left(0\right)=0$.
And it is also given that $f\left(\alpha \right)=0$.
Hence, $f\left(0\right)=f\left(\alpha \right)=0$.
Then by Rolle's theorem there exists '$c$' where $0<c<\alpha$, such that
$f^{\prime }\left(c\right)=0$
$[n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+....2a_{2}x+a_1{]}_{x=c}=0$.
Hence, $0<c<\alpha$.