IIT-JEE 2004Question 1
Using Rolle's theorem, prove that there is at least one root in $\left(45^{1/100},\,46\right)$ of the polynomial $P(x)=51x^{101}-2323x^{100}-45x+1035$.
Differentiation and its Application
Answerproved
Solution
Concept. Take an antiderivative $Q$ of $P$. If $Q$ has equal values at $45^{1/100}$ and $46$, Rolle gives a root of $Q'=P$ between them. Formulas used. (1) Rolle: $Q(\alpha)=Q(\beta)\Rightarrow Q'(c)=0$ for some $c\in(\alpha,\beta)$Step 1 - Antiderivative. Since $\frac{51}{102}=\frac12$ and $\frac{2323}{101}=23$: $$Q(x)=\frac{x^{102}}{2}-23x^{101}-\frac{45x^2}{2}+1035x.$$ Step 2 - Factor. $$Q(x)=\tfrac12(x-46)\left(x^{101}-45x\right)=\tfrac12\,x(x-46)\left(x^{100}-45\right).$$ (Expanding gives back Step 1.) Step 3 - Equal values. $Q(46)=0$ and $Q\left(45^{1/100}\right)=0$. Step 4 - Apply (1). $Q$ is a polynomial, so it is continuous and differentiable. Hence $P=Q'$ vanishes at some $c\in\left(45^{1/100},46\right)$. Hence proved.