If c = frac{1}{2} and f(x) = 2x - x^2,then th | vedclass

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(C) For the function $f(x) = 2x - x^2$,the derivative is $f^{\prime}(x) = 2 - 2x$.According to $LMVT$,there exists a point $c \in (a, b)$ such that $f

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$(0, 1)$

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(C) For the function $f(x) = 2x - x^2$,the derivative is $f^{\prime}(x) = 2 - 2x$.According to $LMVT$,there exists a point $c \in (a, b)$ such that $f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}$.Given $c = \frac{1}{2}$,we have $f^{\prime}(\frac{1}{2}) = 2 - 2(\frac{1}{2}) = 2 - 1 = 1$.Now,calculate the slope of the secant line: $\frac{f(b) - f(a)}{b - a} = \frac{(2b - b^2) - (2a - a^2)}{b - a} = \frac{2(b - a) - (b^2 - a^2)}{b - a} = \frac{2(b - a) - (b - a)(b + a)}{b - a} = 2 - (a + b)$.Equating the two: $1 = 2 - (a + b)$,which implies $a + b = 1$.For $LMVT$ to be applicable on $(a, b)$,the function must be continuous on $[a, b]$ and differentiable on $(a, b)$. Since $f(x)$ is a polynomial,it is continuous and differentiable everywhere.Checking the options,for $(0, 1)$,$a=0$ and $b=1$,so $a+b = 1$. This satisfies the condition $c = \frac{1}{2} \in (0, 1)$.

If $c = \frac{1}{2}$ and $f(x) = 2x - x^2$,then the interval $(a, b)$ of $x$ in which the Lagrange's Mean Value Theorem $(LMVT)$ is applicable for the function $f(x)$ is:

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