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# Root in $$(0,1)$$

Let $$a_0, a_1, \ldots, a_n$$ be real numbers which satisfy

$\frac{a_n} { n+1} + \frac{a_{n-1} } {n} + \ldots + \frac{a_2}{3} + \frac{a_1} { 2} + a_0 = 0 .$

Show that the equation

$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0$

has at least one solution in the interval $$(0,1)$$.

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

Note by Calvin Lin
3 years, 7 months ago

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Consider the polynomial function : $f(x)=\frac{a_n x^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n} +\dots +\frac{a_1x^2}{2}+a_0 x.$ Clearly $$f(0)=0$$, and $$f(1)=0$$ by the first given equality. By Rolle's theorem there is some $$x\in (0,1)$$, such that $$f'(x)=0$$ which can be written as : $a_n x^n +a_{n-1}x^{n-1}+\dots a_1 x+a_0=0.$

- 3 years, 7 months ago

Good work! I did the same, but with due respect "please avoid posting complete solutions, so that others can work on it"

- 3 years, 7 months ago

It appears to b a problem based on ROLLE's THEOREM........ On integrating the given equation we get a new equation whose roots are 0 &1. Now the new equation is a polynomial so it is continuos as well as differentiable in (0,1) . Now apply ROLLE'S THEOREM which states that between any 2 roots of a continuous and differentiable function there must b atleast one root of its differentiable function.

- 3 years, 7 months ago

JEE PROBLEM

- 2 years, 9 months ago

Rolle's theorem for interval (0,1). :)

- 3 years, 6 months ago