Let f be any function continuous on [a, b] an | vedclass

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(C) Given that $f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) < 0$ for all $x \in (a, b)$,the function $f$ is strictly increasing and concave downwards

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$\frac{c-a}{b-c}$

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(C) Given that $f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) Let $m_1$ be the slope of the secant line passing through $(a, f(a))$ and $(c, f(c))$,and $m_2$ be the slope of the secant line passing through $(c, f(c))$ and $(b, f(b))$.$m_1 = \frac{f(c)-f(a)}{c-a}$ and $m_2 = \frac{f(b)-f(c)}{b-c}$.Since the function is concave downwards,the slope of the secant line decreases as we move to the right. Therefore,$m_1 > m_2$.Substituting the expressions for $m_1$ and $m_2$,we get:$\frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$.Rearranging the terms to isolate the required ratio:$\frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$.

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$. If for all $x \in (a, b)$,$f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) < 0$,then for any $c \in (a, b)$,$\frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than

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