answer:We start by describing the given structure and placing points accordingly. We have a cube ABCD-A_{1}B_{1}C_{1}D_{1} with each edge equal to 1 unit. Points E and F are situated on edges AA_{1} and CC_{1} respectively, and we are told that AE = C_{1}F. 1. **Positioning Points E and F**: Since AE = C_{1}F, let's denote AE=x. Consequently, C_{1}F=x as well. 2. **Vertices Coordination**: Place the cube in a coordinate system: - A=(0, 0, 0) - A_{1}=(0, 0, 1) - B=(1, 0, 0) - B_{1}=(1, 0, 1) - C=(1, 1, 0) - C_{1}=(1, 1, 1) - D=(0, 1, 0) - D_{1}=(0, 1, 1) Points E and F are defined as: - E=(0, 0, x) since it is on AA_{1} - F=(1, 1, 1-x) since it is on CC_{1} 3. **Calculating Distances**: Our task is to minimize the area of quadrilateral EBFD_{1}. - Coordinate of E is (0, 0, x) - Coordinate of B is (1, 0, 0) - Coordinate of F is (1, 1, 1-x) - Coordinate of D_{1} is (0, 1, 1) We convert the problem to calculating the minimum area of parallelogram EBFD_1. 4. **Vectors and Area Calculation**: We need vectors overrightarrow{EB} and overrightarrow{F D_{1}}: [ overrightarrow{EB} = (1, 0, 0) - (0, 0, x) = (1, 0, -x) ] [ overrightarrow{FD_{1}} = (0, 1, 1) - (1, 1, 1 - x) = (-1, 0, x + 1) ] 5. **Cross Product**: The area of the parallelogram spanned by overrightarrow{EB} and overrightarrow{FD_{1}} is given by the magnitude of the cross product: [ overrightarrow{EB} times overrightarrow{FD_{1}} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 1 & 0 & -x -1 & 0 & x+1 end{vmatrix} = mathbf{i}(0 - 0) - mathbf{j}(1(x+1) + x) + mathbf{k}(1 - 0) ] [ = mathbf{i} cdot 0 - mathbf{j}(x+1 + x) + mathbf{k}(1) = -2xmathbf{j} + mathbf{k} ] 6. **Magnitude Calculation**: The magnitude of this cross product vector gives the area: [ | (-2x, 0, 1) | = sqrt{(-2x)^{2} + 0^{2} + 1^{2}} = sqrt{4x^{2} + 1} ] 7. **Minimizing the Area**: To minimize sqrt{4x^2 + 1}: - f(x) = sqrt{4x^2 + 1} - We can analyze the derivative, but given the symmetry and simplicity, we see that x = frac{1}{2} minimizes 4x^2 + 1. [ f(x) text{ is minimum for } x = frac{1}{2} ] Thus, [ f left( frac{1}{2} right) = sqrt{4 left( frac{1}{2} right)^2 + 1} = sqrt{1 + 1} = sqrt{2} ] # Conclusion: Thus, the minimum area of quadrilateral EBFD_{1} is [ boxed{frac{sqrt{6}}{2}} ]

answer:The triangles TPQ and TRS are similar by AA similarity (angle by the shared angle at PQ and parallel sides PQ and RS making corresponding angles equal). The ratio of corresponding sides is frac{RS}{PQ} = frac{21}{10}, hence the sides ratio is frac{21}{10}. The area ratio of similar triangles is the square of the similarity ratio of their sides, thus: [ frac{[TRS]}{[TPQ]} = left(frac{21}{10}right)^2 = frac{441}{100}. ] We know that [TRS] = [TPQ] + [PQRS]. Plugging in the area ratio gives: [ frac{[PQRS] + [TPQ]}{[TPQ]} = frac{441}{100}. ] This simplifies to: [ frac{[PQRS]}{[TPQ]} + 1 = frac{441}{100}, ] [ frac{[PQRS]}{[TPQ]} = frac{341}{100}. ] Thus, the desired ratio is: [ frac{[TPQ]}{[PQRS]} = boxed{frac{100}{341}}. ]

answer:1. By definition, xinmathbb{R}. We have f(-x)=2^{-x}- frac{1}{2^{-x}} = frac{1}{2^{x}}-2^{x} = -f(x). Therefore, f(x) is an odd function. 2. When x=0, minmathbb{R}. For xin(0,+infty), we need 2^{x}(2^{2x}- frac{1}{2^{2x}})+m(2^{x}- frac{1}{2^{x}})geqslant 0, which simplifies to (2^{x}- frac{1}{2^{x}})(2^{2x}+1+m)geqslant 0 holding true. Since x > 0, we know (2^{x}- frac{1}{2^{x}}) > 0. Thus, we must have 2^{2x}+1+mgeqslant 0, which implies mgeqslant -(2^{2x}+1). Hence, mgeqslant -(2^{0}+1)=boxed{-2}. Combining both cases, we have minboxed{[-2,+infty)}.

answer:The problem involves finding the number of students who voted in favour of both issues using the principle of inclusion and exclusion. Let’s solve it step-by-step: 1. Denote: - ( A ) = the set of students who voted in favour of the first issue. - ( B ) = the set of students who voted in favour of the second issue. - ( |A| = 375 ) = the number of students who voted in favour of the first issue. - ( |B| = 275 ) = the number of students who voted in favour of the second issue. 2. Let ( x ) be the number of students who voted in favour of both issues. That is, ( |A cap B| = x ). 3. We know there are 40 students who voted against both issues. Therefore, the total number of students who voted in favour of at least one of the issues is: [ 500 - 40 = 460 ] 4. Using the principle of inclusion and exclusion: [ |A cup B| = |A| + |B| - |A cap B| ] Plugging in the known values: [ 460 = 375 + 275 - x ] 5. Solve for ( x ): [ 460 = 650 - x ] Subtract 460 from both sides: [ 460 - 460 = 650 - x - 460 ] Simplify: [ 0 = 190 - x ] Add ( x ) to both sides: [ x = 190 ] 6. Therefore, the number of students who voted in favour of both issues is: [ boxed{190} ] Thus, the answer is ( boxed{190} ).