Question

Using Rolle's theorem, the equation $a_0{x}^n+a_1{x}^{n-1}+....+a_n=0$ has atleast one root between $0$ and $1$, if

• A. $\frac{a_0}{n}+\frac{a_1}{n-1}+....+a_{n-1}=0$
• B. $\frac{a_0}{n-1}+\frac{a_1}{n-2}+....+a_{n-2}=0$
• C. $n{a}_0+(n-1)a_1+....+a_{n-1}=0$
• D. $\frac{a_0}{n+1}+\frac{a_1}{n}+....+a_n=0$

Answer 1

The correct option is D $\frac{a_0}{n+1}+\frac{a_1}{n}+....+a_n=0$

Consider the function $f$ defined by
$f\left(x\right)=a_0\frac{x^{n+1}}{n+1}+a_n\frac{x^n}{n}+....+a_{n-1}\frac{x^2}{2}+a_{n}x$
Since, $f\left(x\right)$ is a polynomial, so it is continuous and differentiable for all $x$.
$f\left(x\right)$ is continuous in the closed interval $[0,1]$ and differentiable in the open interval $(0,1)$.
Also, $f\left(0\right)=0$
and
$f\left(1\right)=\frac{a_0}{n+1}+\frac{a_1}{n}+...+\frac{a_{n-1}}{2}+a_n=0\left[say\right]$
i.e. $f\left(0\right)=f\left(1\right)$
Thus, all the three conditions of Rolle's theorem are satisfied. Hence, there is atleast one value of $x$ in the open interval $(0,1)$ where $f^{\prime }\left(x\right)=0$
i.e. $a_0{x}^n+a_1{x}^{n-1}+....+a_n=0$

Hence, $a_0{x}^n+a_1{x}^{n-1}+....+a_n=0$ has one root between $0$ and $1$ if $\frac{a_0}{n+1}+\frac{a_1}{n}+...+\frac{a_{n-1}}{2}+a_n=0$.