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(B) The region is bounded by $y = \frac{1}{x}$,$y = \frac{5x-1}{4}$,and $y = \frac{17-4x}{4}$.First,find the intersection points:For $y = \frac{1}{x}$

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$\frac{33}{8} - \log_e 4$

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(B) The region is bounded by $y = \frac{1}{x}$,$y = \frac{5x-1}{4}$,and $y = \frac{17-4x}{4}$.First,find the intersection points:For $y = \frac{1}{x}$ and $y = \frac{5x-1}{4}$,$4 = 5x^2 - x \implies 5x^2 - x - 4 = 0 \implies (5x+4)(x-1) = 0$. Since $x > 0$,$x = 1$,so $y = 1$. Point is $(1, 1)$.For $y = \frac{1}{x}$ and $y = \frac{17-4x}{4}$,$4 = 17x - 4x^2 \implies 4x^2 - 17x + 4 = 0 \implies (4x-1)(x-4) = 0$. Points are $(\frac{1}{4}, 4)$ and $(4, \frac{1}{4})$.For $y = \frac{5x-1}{4}$ and $y = \frac{17-4x}{4}$,$5x-1 = 17-4x \implies 9x = 18 \implies x = 2$,so $y = \frac{9}{4}$. Point is $(2, \frac{9}{4})$.The area is given by $\int_{1/4}^{1} (\frac{17-4x}{4} - \frac{5x-1}{4}) dx + \int_{1}^{2} (\frac{17-4x}{4} - \frac{1}{x}) dx$.Area $= \int_{1/4}^{1} (\frac{18-9x}{4}) dx + \int_{1}^{2} (\frac{17-4x}{4} - \frac{1}{x}) dx$.Area $= [\frac{18x}{4} - \frac{9x^2}{8}]_{1/4}^{1} + [\frac{17x}{4} - \frac{x^2}{2} - \log_e x]_{1}^{2}$.Area $= (\frac{9}{2} - \frac{9}{8}) - (\frac{9}{8} - \frac{9}{128}) + ((\frac{17}{2} - 2 - \log_e 2) - (\frac{17}{4} - \frac{1}{2} - 0))$.Area $= \frac{27}{8} - \frac{117}{128} + \frac{13}{2} - \log_e 2 - \frac{15}{4} = \frac{33}{8} - \log_e 4$.

Let $R$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0\}$ is

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