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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 : \[a_n x^n +a_{n-1}x^{n-1}+\dots a_1 x+a_0=0.\]

Haroun Meghaichi - 3 years, 7 months ago

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

Piyal De - 3 years, 7 months ago

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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.

Abhishek Pal - 3 years, 7 months ago

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JEE PROBLEM

U Z - 2 years, 9 months ago

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

Vishal Sharma - 3 years, 6 months ago

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