Question

If the polynomial equation \[a_0 x^n + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_2 x^2 + a_1 x + a_0 = 0\] n positive integer, has two different real roots α and β, then between α and β, the equation \[n \ a_n x^{n - 1} + \left( n - 1 \right) a_{n - 1} x^{n - 2} + . . . + a_1 = 0 \text { has }\].

 

विकल्प

• exactly one root

• almost one root

• at least one root

• no root

Answer 1

 at least one root

We observe that, \[n a_n x^{n - 1} + \left( n - 1 \right) a_{n - 1} x^{n - 2} + . . . + a_1 = 0\] is the derivative of the polynomial \[a_n x^n + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_2 x^2 + a_1 x + a_0 = 0\]

Polynomial function is continuous every where in R and consequently derivative in R
Therefore, \[a_n x^n + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_2 x^2 + a_1 x + a_0\] is continuous on

\[\left[ \alpha, \beta \right]\] and derivative on \[\left( \alpha, \beta \right)\].

Hence, it satisfies the both the conditions of Rolle's theorem.
By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function \[f\left( x \right)\] , there exists at least one root of its derivative.

Hence, the equation \[n a_n x^{n - 1} + \left( n - 1 \right) a_{n - 1} x^{n - 2} + . . . + a_1 = 0\] will have at least one root between \[\alpha \text{ and } \beta\].