# Using Rolle’s Theorem, the equation ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$ has at least one root between 0 and 1, if

$

{\text{A}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{\text{n}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{{\text{n - 1}}}} + ...... + {{\text{a}}_{{\text{n - 1}}}} = 0 \\

{\text{B}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{{\text{n - 1}}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{{\text{n - 2}}}} + ...... + {{\text{a}}_{{\text{n - 2}}}} = 0 \\

{\text{C}}{\text{. n}}{{\text{a}}_0} + \left( {{\text{n - 1}}} \right){{\text{a}}_1} + ...... + {{\text{a}}_{{\text{n - 1}}}} = 0 \\

{\text{D}}{\text{. }}\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}}{\text{ + }}\dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ...... + {{\text{a}}_{\text{n}}} = 0 \\

$

Answer

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Hint – Observing the equation given in the question we consider a polynomial function, and check its properties. Then we check if our polynomial function holds the conditions of Rolle’s Theorem to determine the answer.

Complete step-by-step answer:

Given data, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.

Consider the function f defined by

f(x) = ${{\text{a}}_0}\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}} + {{\text{a}}_{\text{n}}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{\text{n}}} + ....... + {{\text{a}}_{{\text{n - 1}}}}\dfrac{{{{\text{x}}^2}}}{2} + {{\text{a}}_{\text{n}}}{\text{x}}$

Since f(x) is a polynomial, it is continuous and differentiable for all x.

f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1).

Also f(0) = 0.

And let us say,

f(1) = $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$

i.e. f(0) = f(1)

Thus, all three conditions of Rolle’s Theorem are satisfied. Hence there is at least one value of x in the open interval (0, 1) where ${\text{f'}}$(x) = 0.

i.e. ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.

Hence, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$ has one root between 0 and 1 if $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$.

Option D is the correct answer.

Note – In order to solve this type of questions the key is to assume a polynomial function and verify if it holds all the required conditions.

The Three Conditions of Rolle’s Theorem, for a function f(x) are,

(a and b are the first and last of the values x takes)

f is continuous on the closed interval [a, b], f is differentiable on the open interval (a, b) and f(a) = f(b).

Complete step-by-step answer:

Given data, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.

Consider the function f defined by

f(x) = ${{\text{a}}_0}\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}} + {{\text{a}}_{\text{n}}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{\text{n}}} + ....... + {{\text{a}}_{{\text{n - 1}}}}\dfrac{{{{\text{x}}^2}}}{2} + {{\text{a}}_{\text{n}}}{\text{x}}$

Since f(x) is a polynomial, it is continuous and differentiable for all x.

f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1).

Also f(0) = 0.

And let us say,

f(1) = $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$

i.e. f(0) = f(1)

Thus, all three conditions of Rolle’s Theorem are satisfied. Hence there is at least one value of x in the open interval (0, 1) where ${\text{f'}}$(x) = 0.

i.e. ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$.

Hence, ${{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n - 1}}}} + ..... + {{\text{a}}_{\text{n}}} = 0$ has one root between 0 and 1 if $\dfrac{{{{\text{a}}_0}}}{{{\text{n + 1}}}} + \dfrac{{{{\text{a}}_1}}}{{\text{n}}} + ..... + \dfrac{{{{\text{a}}_{{\text{n - 1}}}}}}{{\text{2}}} + {{\text{a}}_{\text{n}}} = 0$.

Option D is the correct answer.

Note – In order to solve this type of questions the key is to assume a polynomial function and verify if it holds all the required conditions.

The Three Conditions of Rolle’s Theorem, for a function f(x) are,

(a and b are the first and last of the values x takes)

f is continuous on the closed interval [a, b], f is differentiable on the open interval (a, b) and f(a) = f(b).

Last updated date: 22nd Sep 2023

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