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If $\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + \dots + \frac{a_{n - 1}}{2} + a_{n} = 0 .$ Then the function $f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n}$ has in $(0, 1)$

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detailed solution

Correct option is A

Consider the function

$ϕ (x) = a_{0} \frac{x^{n + 1}}{n + 1} + a_{1} \frac{x^{n}}{n} + a_{2} \frac{x^{n - 1}}{n - 1} + \dots + a_{n - 1} \frac{x^{2}}{2} + a_{n} x .$

Clearly $ϕ (x),$ being a polynomial, is continuous on [0, 1] and differentiable on (0, 1)

Also, $ϕ (0) = 0$

and, $ϕ (1) = \frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + \dots + a_{n} = 0$ [Given]

$∴ ϕ (0) = ϕ (1)$

Thus $ϕ (x)$ satisfies conditions of Rolle's theorem on [0, 1]

Consequently, there exists $c \in (0, 1)$ such that $ϕ^{'} (c) = 0$ i.e.

$c \in (0, 1)$ is a zero of $ϕ^{'} (x) = a_{0} x^{n} + a_{1} x^{n - 1} + \dots + a_{n} = f (x) .$

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If a0n+1+a1n+a2n−1+…+an−12+an=0. Then the function f(x)=a0xn+a1xn−1+a2xn−2+…+an has in (0,1)