answer:When x in [-infty, 0], f(x) < 0 implies x in (-2, 0]. Since the function is even, it is symmetric about the y-axis. Therefore, the solution set for f(x) < 0 is (-2, 2), Hence, the correct choice is boxed{text{D}}.

answer:To prove that if (0 < alpha + beta leqslant pi), then [ (sin alpha - sin beta)(cos alpha - cos beta) leqslant 0. ] we will proceed as follows: 1. Start by using the trigonometric identity for the difference of sines: [ sin alpha - sin beta = 2 cos left( frac{alpha + beta}{2} right) sin left( frac{alpha - beta}{2} right). ] 2. Similarly, for the difference of cosines: [ cos alpha - cos beta = -2 sin left( frac{alpha + beta}{2} right) sin left( frac{alpha - beta}{2} right). ] 3. Multiply these two expressions: [ (sin alpha - sin beta)(cos alpha - cos beta) = left[ 2 cos left( frac{alpha + beta}{2} right) sin left( frac{alpha - beta}{2} right) right] left[ -2 sin left( frac{alpha + beta}{2} right) sin left( frac{alpha - beta}{2} right) right]. ] 4. Simplify the expression: [ (sin alpha - sin β)(cos α - cos β) = -4 cos left( frac{alpha + beta}{2} right) sin^2 left( frac{alpha - beta}{2} right) sin left( frac{alpha + beta}{2} right). ] 5. Notice that (sin^2 left( frac{alpha - beta}{2} right) geq 0) since it is the square of a sine function, and that (cos left( frac{alpha + beta}{2} right) sin left( frac{alpha + beta}{2} right) geq 0) for (0 < alpha + beta leq pi). 6. Therefore, the product: [ -cos left( frac{alpha + beta}{2} right) sin left( frac{alpha + beta}{2} right) sin^2 left( frac{alpha - beta}{2} right) ] is non-positive. 7. This verifies that: [ (sin alpha - sin beta)(cos alpha - cos beta) leq 0. ] # Conclusion: Therefore, [ (sin alpha - sin beta)(cos alpha - cos beta) leqslant 0. ] (blacksquare)

answer:1. **Convert the logarithmic equation to exponential form**: Given log_b 1024 = n, rewrite this as b^n = 1024. 2. **Factorize 1024**: Realizing that 1024 = 2^{10}, express b^n = 2^{10}. 3. **Determine valid forms of b**: Since b^n = 2^{10}, b must be a power of 2, say b = 2^k where k divides 10. The divisors of 10 are 1, 2, 5, and 10. 4. **List possible values of b**: - If k = 1, then b = 2^1 = 2. - If k = 2, then b = 2^2 = 4. - If k = 5, then b = 2^5 = 32. - If k = 10, then b = 2^{10} = 1024. 5. **Count possible values for b**: We have 2, 4, 32, and 1024. 6. **Conclusion**: There are 4 positive integers b for which log_b 1024 is a positive integer, resulting in text{4}. The final answer is boxed{(E) 4}

answer:In the given code language, we have the following codes: - 'item' is coded as 'pencil' - 'pencil' is coded as 'mirror' - 'mirror' is coded as 'board' We are looking for the useful item to write on paper, which is coded as '2'. In normal circumstances, the useful item to write on paper would be a 'pencil'. However, in this code language, 'pencil' is coded as 'mirror'. Therefore, the original item that is coded as boxed{'2'} must be 'mirror', which in the code language actually means 'pencil'.