Consider the following statements:Statement I | vedclass

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(D) For Statement $I$: Let $F(x) = a_0 x + \frac{a_1 x^2}{2} + \frac{a_2 x^3}{3} + \ldots + \frac{a_n x^{n+1}}{n+1}$.$F(x)$ is a polynomial, so it is

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Both $(I)$ and $(II)$ are true

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(D) For Statement $I$: Let $F(x) = a_0 x + \frac{a_1 x^2}{2} + \frac{a_2 x^3}{3} + \ldots + \frac{a_n x^{n+1}}{n+1}$.$F(x)$ is a polynomial, so it is continuous on $[0,1]$ and differentiable on $(0,1)$.$F(0) = 0$ and $F(1) = a_0 + \frac{a_1}{2} + \ldots + \frac{a_n}{n+1} = 0$ (given).By Rolle's Theorem, there exists $c \in (0,1)$ such that $F^{\prime}(c) = 0$.Since $F^{\prime}(x) = a_0 + a_1 x + \ldots + a_n x^n = P(x)$, we have $P(c) = 0$. Thus, Statement $I$ is true.For Statement $II$: Let $g(x) = \frac{f(x)}{x}$. Since $f$ is continuous on $[a, b]$ and $a>0$, $g(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$.Given $g(a) = \frac{f(a)}{a} = \frac{f(b)}{b} = g(b)$.By Rolle's Theorem, there exists $c \in (a, b)$ such that $g^{\prime}(c) = 0$.$g^{\prime}(x) = \frac{x f^{\prime}(x) - f(x)}{x^2}$.Setting $g^{\prime}(c) = 0$ implies $c f^{\prime}(c) - f(c) = 0$, or $c f^{\prime}(c) = f(c)$. Thus, Statement $II$ is true.

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