answer:1. **Expand with binomial theorem**: [ (n+i)^6 = sum_{k=0}^6 binom{6}{k} n^{6-k} i^k. ] Each term simplifies as follows using i^2 = -1, hence: i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = (-1). [ (n+i)^6 = n^6 - 15n^4 + 15n^2 - 1 + (6n^5i - 20n^3i + 6ni). ] 2. **Separate real and imaginary parts**: The real part is n^6 - 15n^4 + 15n^2 - 1, and the imaginary part is 6n^5i - 20n^3i + 6ni. The imaginary part must sum to zero: [ 6n^5i - 20n^3i + 6ni = i(6n^5 - 20n^3 + 6n) = 0. ] Factoring, [ 6n(n^4 - (10/3)n^2 + 1) = 0. ] Solving the quadratic n^4 - (10/3)n^2 + 1 = 0 is impractical within integers, suggesting n=0 is the only solution. 3. **Conclusion**: Substituting n=0 yields: [ (0+i)^6 = i^6 = -1. ] That confirms for n = 0, (n+i)^6 is an integer. Thus, there is 1 integer n such that (n+i)^6 is an integer. The final answer is boxed{text{(B)} 1}

answer:To solve the problem, we will follow these steps: 1. Analyze the functions (cos x) and (tan x) to determine their concavity on the given intervals. 2. Apply Jensen's Inequality, which states that the value of a convex function at the arithmetic mean of a set of points is less than or equal to the arithmetic mean of the value of the function at those points, and similarly for concave functions. Step 1: Analyze (cos x) On the interval ( left[ -frac{pi}{2}, frac{pi}{2} right] ), consider the second derivative of (cos x): [ (cos x)'' = -cos x. ] For (-frac{pi}{2} leq x leq frac{pi}{2}), (cos x geq 0). Thus, [ -cos x leq 0. ] This implies that ( cos x ) is a concave function on (left[ -frac{pi}{2}, frac{pi}{2} right]). According to Jensen's Inequality for concave functions: [ cosleft(frac{x_1 + x_2 + cdots + x_n}{n}right) geq frac{cos x_1 + cos x_2 + cdots + cos x_n}{n}. ] Therefore, this inequality can be written as: [ frac{cos x_1 + cos x_2 + cdots + cos x_n}{n} leq cos left( frac{x_1 + x_2 + cdots + x_n}{n} right). ] Step 2: Analyze (tan x) On the interval ( left[ 0, frac{pi}{2} right) ), consider the second derivative of (tan x): [ (tan x)'' = frac{2 sin x}{cos^3 x}. ] For (0 leq x < frac{pi}{2}), both (sin x) and (cos x) are positive, and thus (frac{2 sin x}{cos^3 x} > 0). This implies that (tan x) is a convex function on (left[ 0, frac{pi}{2} right)). According to Jensen's Inequality for convex functions: [ tanleft(frac{x_1 + x_2 + cdots + x_n}{n}right) leq frac{tan x_1 + tan x_2 + cdots + tan x_n}{n}. ] Therefore, this inequality is written as: [ frac{tan x_1 + tan x_2 + cdots + tan x_n}{n} geq tan left( frac{x_1 + x_2 + cdots + x_n}{n} right). ] # Conclusion: Both parts of the problem have been solved using the properties of concave and convex functions together with Jensen's Inequality. boxed{frac{cos x_1 + cos x_2 + cdots + cos x_n}{n} leq cos left( frac{x_1 + x_2 + cdots + x_n}{n} right)} and boxed{frac{tan x_1 + tan x_2 + cdots + tan x_n}{n} geq tan left( frac{x_1 + x_2 + cdots + x_n}{n} right)}

answer:Firstly, to demonstrate that the condition is not sufficient, consider the function y = |x|. This function achieves a minimum value at x=0 with f(0) = 0, but f'(x) is not defined at x=0. Hence, f'(0) = 0 does not hold, showing that the condition is not sufficient. Secondly, to show that the condition is not necessary, let's take the function f(x) = x^3. Its derivative is f'(x) = 3x^2. Even though f'(0) = 0 at x=0, this point is not an extremum for f(x). Therefore, the condition is not necessary either. From these two counterexamples, we conclude that f'(x) = 0 is neither a sufficient nor a necessary condition for f(x) to have an extremum at x=a. Thus, the correct answer is [boxed{D: Neither sufficient nor necessary}]

answer:1. **Calculate the number of discs needed**: Given that each disc can hold up to 70 minutes and the total reading time is 480 minutes, the number of discs required is calculated as: [ frac{480}{70} approx 6.857 ] Since we cannot use a fraction of a disc, we round up to the nearest whole number, resulting in 7 discs. 2. **Determine the minutes of reading per disc**: To find out how many minutes of reading each of these 7 discs would contain, divide the total reading time by the number of discs: [ frac{480}{7} approx 68.571 ] Thus, each disc will contain approximately 68.57 minutes of reading. Conclusion: Each disc will contain 68.57 minutes of reading. The final answer is boxed{68.6} (Choice C).