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(C) Given equation is $f(x) + (x + \frac{1}{2}) f(1 - x) = 1$  $(i)$Replacing $x$ with $1 - x$ in $(i)$,we get:$f(1 - x) + (1 - x + \frac{1}{2}) f(x)

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$-2$

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(C) Given equation is $f(x) + (x + \frac{1}{2}) f(1 - x) = 1$  $(i)$Replacing $x$ with $1 - x$ in $(i)$,we get:$f(1 - x) + (1 - x + \frac{1}{2}) f(x) = 1$$f(1 - x) + (\frac{3}{2} - x) f(x) = 1$  $(ii)$From $(ii)$,$f(1 - x) = 1 - (\frac{3}{2} - x) f(x)$.Substitute this into $(i)$:$f(x) + (x + \frac{1}{2}) [1 - (\frac{3}{2} - x) f(x)] = 1$$f(x) + (x + \frac{1}{2}) - (x + \frac{1}{2})(\frac{3}{2} - x) f(x) = 1$$f(x) [1 - (\frac{3}{2}x - x^2 + \frac{3}{4} - \frac{1}{2}x)] = 1 - (x + \frac{1}{2})$$f(x) [1 - (-x^2 + x + \frac{3}{4})] = \frac{1}{2} - x$$f(x) [x^2 - x + \frac{1}{4}] = \frac{1}{2} - x$$f(x) (x - \frac{1}{2})^2 = - (x - \frac{1}{2})$For $x 
eq \frac{1}{2}$,$f(x) = \frac{-(x - \frac{1}{2})}{(x - \frac{1}{2})^2} = \frac{-1}{x - \frac{1}{2}} = \frac{2}{1 - 2x}$.Now,$f(0) = \frac{2}{1 - 0} = 2$ and $f(1) = \frac{2}{1 - 2} = -2$.Therefore,$2 f(0) + 3 f(1) = 2(2) + 3(-2) = 4 - 6 = -2$.

Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x) + (x + \frac{1}{2}) f(1 - x) = 1$,for all $x \in R$. Then $2 f(0) + 3 f(1)$ is equal to

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