# 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 \\

$

Last updated date: 25th Mar 2023

•

Total views: 305.4k

•

Views today: 8.82k

Answer

Verified

305.4k+ views

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).

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?