answer:The regions enclosed by the curve xy = 1 and the lines y = x and y = 2 are bounded by: 1. The hyperbola xy = 1, which can be rewritten as y = frac{1}{x} when x neq 0. 2. The line y = x. 3. The line y = 2. To find the enclosed area, we need to determine where these lines intersect. The intersection of y = frac{1}{x} and y = x occurs when frac{1}{x} = x; solving for x gives x^2 = 1 or x = 1 (since x is positive in the region of interest). The intersection of y = frac{1}{x} and y = 2 occurs when frac{1}{x} = 2; solving for x gives x = frac{1}{2}. We can now set up the integral to find the area between x = frac{1}{2} and x = 1 for the curve y = frac{1}{x}, with the line y = x being the lower bound and the line y = 2 the upper bound for x values between frac{1}{2} and 1. The region we're interested in is beneath the line y = 2 and above the curve y = frac{1}{x}, so we can express the area as an integral: text{Area} = int_{frac{1}{2}}^{1} (2 - frac{1}{x}) , dx Evaluating the integral: begin{align*} text{Area} & = left[ 2x - ln|x| right]_{frac{1}{2}}^{1} & = left[ 2(1) - ln|1| right] - left[ 2(frac{1}{2}) - ln|frac{1}{2}| right] & = (2 - 0) - (1 + ln 2) & = 2 - 1 - ln 2 & = 1 - ln 2 end{align*} Therefore, the area of the region is: boxed{1 - ln 2}

answer:To clarify why the given proposition is a false statement, consider the geometric interpretations of x, y, and z: - If all of x, y, z are lines (Option A), and if x perp y and y parallel z, it would follow that x perp z, making the statement true, which does not align with our initial condition that the proposition is false. - If all of x, y, z are planes (Option B), and if x perp y and y parallel z, then by the nature of planes being infinitely extended flat surfaces, x would also be perpendicular to z (since a plane perpendicular to another would be perpendicular to anything parallel to that other plane), again making the proposition true. - If x, z are lines, and y is a plane (Option C), and if x perp y and y parallel z, then x would be perpendicular to line z as well (as line z would lie in the plane y, and x is perpendicular to every line in that plane), which would mean the proposition is true. From the above reasoning, it's clear that Options A, B, and C all lead to the proposition being true, contrary to our requirement that it be false. Therefore, the only remaining option is: - If x, y are planes, and z is a line (Option D), when x perp y and y parallel z, it does not necessarily imply that x perp z. In this case, line z could be skew to plane x, meaning they do not intersect and are not perpendicular, which would make the original proposition false. This is the scenario where the proposition "If x perp y, y parallel z, then x perp z" does not hold. Thus, the correct answer is boxed{text{D: x, y are planes, z is a line}}.

answer:Since overrightarrow{a} parallel overrightarrow{b}, we have 4-n-2m=0, which simplifies to n+2m=4. Given m>0 and n>0, we can rewrite frac{1}{m} + frac{8}{n} as frac{1}{4}(n+2m)(frac{1}{m} + frac{8}{n}) = frac{1}{4}(10 + frac{n}{m} + frac{16m}{n}) geq frac{1}{4}(10 + 2sqrt{frac{n}{m} times frac{16m}{n}}) = frac{9}{2}. Equality holds if and only if n=4m=frac{8}{3}. Therefore, the minimum value of frac{1}{m} + frac{8}{n} is boxed{frac{9}{2}}. This conclusion is reached by utilizing the property that overrightarrow{a} parallel overrightarrow{b} implies n+2m=4, and then applying the "multiply by 1 method" and the properties of basic inequalities. This problem tests the properties of the "multiply by 1 method" and basic inequalities, the theorem of collinear vectors, and examines reasoning and computational skills. It is considered a medium-level question.

answer:To find the range of a for which frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} < 0 holds for all x_1, x_2 in mathbb{R} with x_1 neq x_2, we analyze the function f(x) piecewise and apply the given condition. 1. **For x < 1:** The function is f(x) = (2a-1)x + 4a. For f(x) to be monotonically decreasing, the coefficient of x must be negative, i.e., [2a - 1 < 0 implies a < frac{1}{2}.] 2. **For x geq 1:** The function is f(x) = frac{a}{x}. For f(x) to be monotonically decreasing, a must be positive (since x is positive and the function decreases as x increases), i.e., [a > 0.] 3. **At x = 1:** We must ensure the function is continuous and monotonically decreasing across x = 1. This requires that the value of f(x) just before x = 1 is greater than or equal to the value at x = 1. Thus, [(2a-1) cdot 1 + 4a geq frac{a}{1} implies 6a - 1 geq a implies 5a geq 1 implies a geq frac{1}{5}.] Combining these conditions, we find that a must satisfy: [a > 0, quad a < frac{1}{2}, quad text{and} quad a geq frac{1}{5}.] Therefore, the range of a is the intersection of these conditions, which is: [boxed{left[frac{1}{5}, frac{1}{2}right)}.]