Question

If $\frac{a_0}{n+1}+\frac{a_1}{n}+\frac{a_2}{n-1}+\cdots +\frac{a_{n-1}}{2}+a_n=0$, then the equation $a_0{x}^n+a_1{x}^{n-1}+\cdots +a_{n-1}x+a_n=0$ has, in the interval $(0,1)$

• A. Exactly one root
• B. Atleast one root
• C. Atmost one root
• D. No root

Answer 1

The correct option is B Atleast one root

Consider the function
$f\left(x\right)=\frac{a_0{x}^{n+1}}{n+1}+\frac{a_1{x}^n}{n}+...\frac{a_{n-1}{x}^2}{2}+a_{n}x$
Then $f\left(0\right)=0$ ...(i)
And
$f\left(1\right)=\frac{a_0}{n+1}+\frac{a_1}{n}+...\frac{a_{n-1}}{2}+a_n$
$=0$ ...(given).
Hence
$f\left(0\right)=f\left(1\right)$.
Then applying Rolle's theorem, there exists atleast one 'c' $0<c<1$ such that
$f^{\prime }\left(c\right)=0$
Or
$[a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+...a_{n-1}x+a_n{]}_{x=c}=0$
Where $0<c<1$.