answer:1. We can simplify f(x) as follows: begin{align*} f(x) &= cos^4x - 2sin xcos x - sin^4x &= (cos^2x + sin^2x)(cos^2x - sin^2x) - sin 2x &= 1 cdot (cos^2x - sin^2x) - sin 2x &= cos 2x - sin 2x &= sqrt{2}cosleft(2x + frac{pi}{4}right). end{align*} Consequently, since the cosine function has a period of 2pi, the minimal positive period T for f(x), considering the coefficient 2 in 2x, is: [ T = frac{2pi}{2} = pi. ] boxed{T = pi} 2. For the monotonicity intervals, observe that f(x) in its trigonometric form will be increasing whenever the cosine function is increasing. We can determine these intervals given the restriction that - pi leq 2x + frac{pi}{4} leq 2kpi, where k in mathbb{Z} (integers). Translating these bounds to the x-domain gives us: [ -frac{5pi}{8} + kpi leq x leq frac{3pi}{8} + kpi, ; k in mathbb{Z}. ] Therefore, the intervals where f(x) is monotonically increasing are given by: [ boxed{left[-frac{5pi}{8} + kpi, -frac{pi}{8} + kpiright], ; k in mathbb{Z}.} ] 3. To find the range of f(x) when x in left[0, frac{pi}{2}right], we can analyze the function's behavior on the period including this interval. Given the previous step, we know that f(x) is monotonically decreasing on left[0, frac{3pi}{8}right] and monotonically increasing on left[frac{3pi}{8}, frac{pi}{2}right]. Therefore, the minimum value of f(x) on this interval will be at the boundary of the aforementioned monotonically increasing interval, which is fleft(frac{3pi}{8}right); thus, it is: [ min f(x) = -sqrt{2}. ] The maximum value will be either at x = 0 or x = frac{pi}{2}. Calculating both yields: [ f(0) = sqrt{2}cosleft(0 + frac{pi}{4}right) = 1, ] [ fleft(frac{pi}{2}right) = sqrt{2}cosleft(pi + frac{pi}{4}right) = -1. ] As such, the range of f(x) for x in left[0, frac{pi}{2}right] is boxed{left[-sqrt{2}, 1right]}.

answer:Since sin x leq 1, the maximum value of y = 1 + sin x is 2. Therefore, the answer is boxed{2}.

answer:- We firstly observe that since n^2 - 41n = n(n-41), at least one factor must be even (since one number between n and n-41 is even), implying n^2 - 41n is even. Thus, n^2 - 41n + 408 is also even. Therefore, the prime p must be 2. - We solve for n in the equation n^2 - 41n + 408 = 2, or equivalently n^2 - 41n + 406 = 0. - The product of the solutions to the quadratic ax^2 + bx + c = 0 is c/a. Here, a = 1, b = -41, and c = 406. The product of the roots is 406. - The sum of the roots is given by -b/a = 41. Since the sum and the product of the roots are both positive integers, both solutions are positive integers. Conclusion: The product of all positive integral values of n that satisfy the equation is boxed{406}.

answer:Let's start by figuring out the ages of each person step by step. 1. Beckett is 12 years old. 2. Olaf is 3 years older than Beckett, so Olaf is 12 + 3 = 15 years old. 3. Shannen is 2 years younger than Olaf, so Shannen is 15 - 2 = 13 years old. 4. Jack is five more than twice as old as Shannen, so Jack is 2 * 13 + 5 = 26 + 5 = 31 years old. Now, let's add up the ages of all four people: Beckett (12) + Olaf (15) + Shannen (13) + Jack (31) = 12 + 15 + 13 + 31 = 71 The sum of the ages of all four people is boxed{71} years.