Question

If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x=0$, $a_1\ne 0$, $n\ge 2$, 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. smaller than $\alpha$
• B. greater than $\alpha$
• C. equal to $\alpha$
• D. greater than or equal to $\alpha$

Answer 1

Answer: (A)

smaller than $\alpha$

Let $f\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+...+a_{n}x$
Now for $x=0$
$f\left(0\right)=0$
And as $\alpha$ is one root
$f\left(\alpha \right)=0$
Hence $f^{\prime }\left(x\right)$ will have one root in $(0,\alpha )$
As $f^{\prime }\left(x\right)=n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+...+a_1=0$
Then root of $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+...+a_1=0$ is less than $\alpha$
Hence, option 'A' is correct.