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)

# 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 $f\left(x\right)={a}_{0}{x}^{n}+{a}_{1}{x}^{n-1}+{a}_{2}{x}^{n-2}+\dots +{a}_{n}$ has in $\left(0,1\right)$

1. A

at least one zero

2. B

at most one zero

3. C

only 3 zeros

4. D

only 2 zeros

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### Solution:

Consider the function

$\varphi \left(x\right)={a}_{0}\frac{{x}^{n+1}}{n+1}+{a}_{1}\frac{{x}^{n}}{n}+{a}_{2}\frac{{x}^{n-1}}{n-1}+\dots +{a}_{n-1}\frac{{x}^{2}}{2}+{a}_{n}x.$

Clearly $\varphi \left(x\right),$ being a polynomial, is continuous on [0, 1] and differentiable on (0, 1)

Also, $\varphi \left(0\right)=0$

and, $\varphi \left(1\right)=\frac{{a}_{0}}{n+1}+\frac{{a}_{1}}{n}+\frac{{a}_{2}}{n-1}+\dots +{a}_{n}=0$          [Given]

Thus $\varphi \left(x\right)$ satisfies conditions of Rolle's theorem on [0, 1]

Consequently, there exists $c\in \left(0,1\right)$ such that ${\varphi }^{\mathrm{\prime }}\left(c\right)=0$ i.e.

$c\in \left(0,1\right)$ is a zero of ${\varphi }^{\mathrm{\prime }}\left(x\right)={a}_{0}{x}^{n}+{a}_{1}{x}^{n-1}+\dots +{a}_{n}=f\left(x\right).$

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