answer:1. **Introduction and Setup**: Given n geq 3, and we have a sequence of positive real numbers x_2, x_3, ldots, x_n such that their product equals 1: [ x_2 times x_3 times cdots times x_n = 1. ] We aim to demonstrate the following inequality: [ left(1 + x_2right)^2 times left(1 + x_3right)^3 times cdots times left(1 + x_nright)^n > n^n. ] 2. **Aiming for a Lower Bound**: Let j be an integer such that 2 leq j leq n. We need to find a lower bound for left(1 + x_jright)^j. We start by considering: [ 1 + x_j = frac{1}{j-1} + frac{1}{j-1} + cdots + frac{1}{j-1} + x_j. ] Using the AM-GM inequality on the j terms (with j-1 terms being frac{1}{j-1} and one term being x_j), we can write: [ 1 + x_j = frac{1}{j-1} + cdots + frac{1}{j-1} + x_j geq j left(frac{x_j^{1/j} cdot left(frac{1}{j-1}right)^{(j-1)/j}}right). ] 3. **Refinement Using AM-GM**: Simplifying the earlier bound: [ 1 + x_j geq j left(frac{x_j}{(j-1)^{left(j-1right)/j}}right), ] we raise both sides to the j-th power: [ left(1 + x_jright)^j geq left(j frac{x_j^{1/j}(j-1)^{-(j-1)/j}}right)^j = x_j frac{j^j}{(j-1)^{j-1}}. ] Therefore: [ left(1 + x_jright)^j geq x_j frac{j^j}{(j-1)^{j-1}}. ] 4. **Product of Terms**: Next, we consider the product of all such terms from j=2 to j=n: [ left(1 + x_2right)^2 times left(1 + x_3right)^3 times cdots times left(1 + x_nright)^n geq x_2 cdot x_3 cdot cdots cdot x_n times frac{2^2}{1^1} times frac{3^3}{2^2} times cdots times frac{n^n}{(n-1)^{n-1}}. ] 5. **Simplification and Final Product**: Since x_2 cdot x_3 cdot cdots cdot x_n = 1, we get: [ left(1 + x_2right)^2 times left(1 + x_3right)^3 times cdots times left(1 + x_nright)^n geq frac{2^2}{1^1} times frac{3^3}{2^2} times cdots times frac{n^n}{(n-1)^{n-1}}. ] Noting that the product of the fractions on the right-hand side results in n^n, we have: [ left(1 + x_2right)^2 times left(1 + x_3right)^3 times cdots times left(1 + x_nright)^n geq n^n. ] 6. **Ensuring Strict Inequality**: To ensure the inequality is strict, we examine the equality case in the AM-GM inequality. AM-GM equality holds if all terms are equal. However, if each x_j = frac{1}{j-1}, their product is less than 1. This contradicts the initial condition x_2 times x_3 times cdots times x_n = 1. Hence, we cannot have equality. # Conclusion: [ boxed{left(1 + x_2right)^2 times left(1 + x_3right)^3 times cdots times left(1 + x_nright)^n > n^n} ]

answer:Firstly, evaluate at x=0, yielding: [ 2 sin theta > 0 implies sin theta > 0. ] This implies theta in (0, pi). Next, evaluate at x=1, giving: [ cos theta > 0. ] This implies theta in (0, frac{pi}{2}). For the general case, consider: [ x^2 cos theta - x(1 - x) + 2(1 - x)^2 sin theta = x^2 cos theta - x(1-x) + 2(1-x)^2 sin theta. ] Rewriting this, we have: [ x^2 cos theta - x(1-x) + 2(1-x)^2 sin theta = (x sqrt{cos theta} - sqrt{2}(1-x) sqrt{sin theta})^2 + x(1-x)(2 sqrt{2} sqrt{cos theta sin theta} - 1). ] Setting x sqrt{cos theta} = sqrt{2}(1-x) sqrt{sin theta}, we solve for x: [ x = frac{sqrt{2} sqrt{sin theta}}{sqrt{cos theta} + sqrt{2} sqrt{sin theta}}. ] This x must satisfy 0 leq x leq 1. The remaining expression simplifies to: [ x(1-x)(2 sqrt{2} sqrt{cos theta sin theta} - 1) > 0. ] This gives: [ 2 sqrt{2} sqrt{cos theta sin theta} > 1 implies 8 cos theta sin theta > 1 implies sin 2theta > frac{1}{8}. ] Finding the range of theta for which sin 2theta > frac{1}{8}: [ frac{pi}{12} < theta < frac{11pi}{12}. ] Combining all conditions: [ 0 < theta < frac{pi}{2}. ] Thus, the solution is theta in boxed{left( 0, frac{pi}{2} right)}.

answer:First, apply the square of a binomial formula: [ (47+21)^2 = 47^2 + 21^2 + 2 cdot 47 cdot 21 ] So, [ (47+21)^2 - (47^2 + 21^2) = 2 cdot 47 cdot 21 ] Calculating 2 cdot 47 cdot 21 gives: [ 2 cdot 47 cdot 21 = 1974 ] Now, subtract 7 times 47 from 1974: [ 1974 - (7 cdot 47) = 1974 - 329 = 1645 ] Thus, the value of the expression (47 + 21)^2 - (47^2 + 21^2) - 7 times 47 is boxed{1645}.

answer:Let the equation of the circle be x^2+(y-b)^2=b^2, By combining it with x^2=4y, we get y^2+(4-2b)y=0. Therefore, 4-2b=0, Thus, b=2, Therefore, the maximum value of the circumference of circle C is 2pi times 2 = 4pi. Hence, the correct option is: boxed{C}. **Analysis:** By setting the equation of the circle as x^2+(y-b)^2=b^2 and combining it with x^2=4y, we obtain y^2+(4-2b)y=0. Using 4-2b=0 to solve for b, we can find the maximum value of the circumference of circle C.