First slide

Mean value theorems

Question

Using rolle’s theorem equation $a_{0} x^{n} + a_{1} x^{n - 1} + .. . . . . . . . . . . . + a_{n} = 0$ has at least one root between $0$ and $1$ , if :

Moderate

A

$\frac{a_{0}}{n} - \frac{a_{1}}{n - 1} + ... + a_{n - 1} = 0$
B

$\frac{a_{0}}{n - 1} + \frac{a_{1}}{n - 2} + ... + a_{n - 2} = 0$
C

$n a_{0} + (n - 1) a_{1} + ... + a_{n - 1} = 0$
D

$\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + ... + a_{n} = 0$

Solution

Consider the function, $ϕ (x) = \int (a_{0} x^{n} + a_{1} x^{n - 1} + ... + a_{n}) d x$
$= \frac{a_{0} x^{n + 1}}{n + 1} + \frac{a_{1} x^{n}}{n} + ... + a_{n} x + A$ …(1)

Where $A$ is constant.

$ϕ (x)$ being algebraic polynomial is continuous in the closed interval $[0, 1]$ and differentiable in the open interval $] 0, 1 [$

If $f (0) = f (1)$ , then we have $\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + ... + a_{n} = 0$ …(2)

Thus $ϕ (x)$ is given in (1) by satisfying all conditions of Rolle’s theorem, so there exists atleast one $x = c$ s.t. $0 < c < 1$ , where $ϕ' (c) = 0$

i.e., atleast one root of $ϕ' (x) = 0$

i.e., $a_{0} x^{n} + a_{1} x^{n - 1} + ... + a_{n} = 0$ lie between 0 and 1 if (2) is satisfied

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