answer:(This question is worth 12 points) Solution: (1) Since alpha in left( frac{pi}{2}, pi right) and sinalpha= frac{3}{5}, then cosalpha= -sqrt{1-sin^2alpha}= -frac{4}{5}. (2 points) Therefore, sinleft( frac{pi}{4}+alpha right) = sin frac{pi}{4}cosalpha + cos frac{pi}{4}sinalpha = frac{sqrt{2}}{2}(cosalpha + sinalpha) = -frac{sqrt{10}}{10} = -frac{sqrt{2}}{10}. (6 points) (2) Since sin2alpha=2sinalphacosalpha= -frac{24}{25}, and cos2alpha=cos^2alpha-sin^2alpha= frac{7}{25}, (8 points) Therefore, cosleft( frac{pi}{6}-2alpha right) = cos frac{pi}{6}cos2alpha + sin frac{pi}{6}sin2alpha = frac{sqrt{3}}{2} times frac{7}{25} + frac{1}{2} times left(-frac{24}{25}right) = frac{7sqrt{3}-24}{50}. (12 points) Thus, the final answers are: (1) boxed{-frac{sqrt{2}}{10}} (2) boxed{frac{7sqrt{3}-24}{50}}

answer:Given the ellipse C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 where a > b > 0 and its foci F_{1}(-1,0) and F_{2}(1,0), we are also given the parabola E: y^2 = 4x which shares the same foci with the ellipse C. We are to find the length of the major axis of the ellipse C when the line PF_{1}, tangent to the parabola at point P, intersects the ellipse in the first quadrant. 1. The equation of the line PF_{1} can be represented as y = k(x+1), given that it passes through F_{1}(-1,0). 2. Substituting y from y = k(x+1) into the parabola equation y^2 = 4x, we get k^2(x+1)^2 = 4x. 3. Expanding and rearranging gives us k^2x^2 + 2k^2x + k^2 - 4x = 0, which simplifies to k^2x^2 + (2k^2 - 4)x + k^2 = 0. 4. For this quadratic equation in x to have a unique solution (since the line is tangent to the parabola), the discriminant must be zero. Thus, 4(k^2 - 2)^2 - 4k^4 = 0. 5. Solving the equation for k, we find k = 1 or k = -1. Since we are considering the intersection in the first quadrant and the slope of the tangent line must be positive, we discard k = -1 and keep k = 1. 6. Substituting k = 1 back into the equation of the tangent line, we get y = x + 1. Substituting y in terms of x into the parabola equation, we solve for x and find x = 1. 7. With x = 1, substituting back into the parabola equation gives y^2 = 4(1), so y = 2 (considering the first quadrant). 8. Therefore, point P is at (1,2). The distance |PF_{2}| is the distance from P to F_{2}(1,0), which is 2 units. 9. The distance |PF_{1}| is the distance from P to F_{1}(-1,0), calculated as 2sqrt{2} units. 10. The sum of the distances from P to both foci, |PF_{2}| + |PF_{1}| = 2 + 2sqrt{2}, equals the length of the major axis 2a. 11. Therefore, the length of the major axis of the ellipse C is 2 + 2sqrt{2}. Hence, the correct option is boxed{text{B: } 2sqrt{2}+2}.

answer:**Step 1: Verify solution (1, 2, 3, 4)** 1. **Check the first equation:** [ x + y + z + w = 1 + 2 + 3 + 4 = 10 ] This satisfies the first equation. 2. **Check the second equation:** [ x^2 + y^2 + z^2 + w^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 ] This satisfies the second equation. 3. **Check the third equation:** [ x^3 + y^3 + z^3 + w^3 = 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100 ] This satisfies the third equation. 4. **Check the fourth equation:** [ xyzw = 1 cdot 2 cdot 3 cdot 4 = 24 ] This satisfies the fourth equation. Since (1, 2, 3, 4) satisfies all four equations, it is a valid solution. **Step 2: Consider all permutations of (1, 2, 3, 4)** The system of equations is symmetric with respect to (x, y, z, w). Therefore, any permutation of (1, 2, 3, 4) will also satisfy all the equations. **Permutations of (1, 2, 3, 4):** [ (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1) ] Each permutation is a distinct solution which does not introduce any new values, but rather rearranges the existing ones. **Conclusion:** All solutions to the system are given by the 24 permutations of the solution (1, 2, 3, 4). [ boxed{(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)} ]

answer:(1) Since the solution set of the inequality is {x | x < -3 text{ or } x > -1}, we can deduce that k < 0. Also, -3 and -1 are the roots of the quadratic equation kx^2 - 2x + 3k = 0. Therefore, we have the system of equations: begin{cases} (-3) cdot (-1) = 3 (-3) + (-1) = frac{2}{k} end{cases} Solving this system, we find that k = boxed{-frac{1}{2}}. (2) Because the solution set of the inequality kx^2 - 2x + 3k < 0 is the empty set, if k = 0, then the inequality simplifies to -2x < 0, which implies x > 0; this is not consistent with having an empty solution set. If k neq 0, we consider the discriminant of the quadratic equation: Delta = b^2 - 4ac = (2)^2 - 4 cdot k cdot 3k = 4 - 12k^2 geq 0 In order for the solution set to be empty, the inequality must never be satisfied, hence k must be greater than 0, and the discriminant must be less than 0: begin{cases} k > 0 Delta < 0 end{cases} Solving this, we get 0 < k leq frac{sqrt{3}}{3}. Thus, the range of real numbers k is boxed{(0, frac{sqrt{3}}{3}]}.