answer:1. Given the functions ( f(x) = sin x ) and ( g(x) = sqrt{pi^{2} - x^{2}} ), with the domain for both functions being ( [-pi, pi] ). 2. To find the area enclosed by the curves of the given functions ( y = sin x ) and ( y = sqrt{pi^{2} - x^{2}} ): - Note that ( y = sqrt{pi^{2} - x^{2}} ) represents the upper semicircle of the circle ( x^2 + y^2 = pi^2 ). - The area of a full circle with radius (pi) is: [ text{Area}_{text{circle}} = pi times (pi)^2 = pi^3 ] - Hence, the area of the semicircle (upper half) is: [ text{Area}_{text{semicircle}} = frac{1}{2} pi^3 ] 3. To ensure we correctly calculate the area between the curve ( sqrt{pi^{2} - x^{2}} ) and ( sin x ), we can use integration: Construct the integral to find the area under (y = sqrt{pi^2 - x^2}) from (-pi) to (pi): [ A_1 = int_{-pi}^{pi} sqrt{pi^2 - x^2} , dx ] And the area under (y = sin x) from (-pi) to (pi): [ A_2 = int_{-pi}^{pi} sin x , dx ] 4. The area of the upper semicircle (since (y = sqrt{pi^2 - x^2})) can be visualized as: Since it's the upper semicircle, from (-pi) to (pi): [ A_1 = frac{1}{2} pi^3 ] 5. The integral of (y = sin x) over (-pi) to (pi) balances out due to symmetry: [ A_2 = int_{-pi}^{pi} sin x , dx = 0 ] 6. Hence, the net area is actually equivalent to the semicircle area, which is half the area of the whole circle: [ text{Total Area} = frac{1}{2} pi^3 ] # Conclusion: (boxed{frac{pi^3}{2}}).

answer:First, consider lfloor 3x rfloor is an integer; then lfloor lfloor 3x rfloor - 1/3 rfloor = lfloor 3x rfloor - 1 since subtracting 1/3 doesn't cross an integer boundary unless lfloor 3x rfloor is at a .33 repeating limit, which does not apply in general subtraction of 1/3. On the other side of the equation lfloor x + 3 rfloor = lfloor x rfloor + 3 because adding a whole number doesn't change the fraction part of x. Hence, our new equation becomes, [ lfloor 3x rfloor - 1 = lfloor x rfloor + 3. ] Set n = lfloor x rfloor, then n leq x < n+1. Consider how lfloor 3x rfloor is formulated: - If x < n + frac{1}{3}, then 3n leq 3x < 3n + 1, so lfloor 3x rfloor = 3n. Substituting gives: [ 3n - 1 = n + 3, ] leading to n = 2. - If x geq n + frac{1}{3}, consider split cases: - If x < n + frac{2}{3}, then 3n + 1 leq 3x < 3n + 2, so lfloor 3x rfloor = 3n + 1. Substituting gives: [ 3n = n + 3, ] leading to n = 1.5, which is not possible as n is an integer. - If x geq n + frac{2}{3}, then 3n + 2 leq 3x < 3n + 3, so lfloor 3x rfloor = 3n + 2. Substituting gives: [ 3n + 1 = n + 3, ] leading to n = 1. Both solutions gives: Continuing with n = 2, the solution is x in left[ frac{7}{3}, 3 right). For n = 1, x in left[ frac{5}{3}, 2 right). The combined set of solutions is boxed{left[ frac{5}{3}, 3 right)}.

answer:Let alpha = a + bi and gamma = c + di, where a, b, c, and d are real numbers. Then begin{align*} f(1) &= (3 - 2i) + alpha + gamma = (a + c + 3) + (b + d - 2)i, f(-i) &= (3 - 2i)(-1) + alpha (-i) + gamma = (b - c - 2) + (-a + d - 3)i. end{align*} Since f(1) and f(-i) are both real, b + d - 2 = 0 and -a + d - 3 = 0, so a = d - 3 and b = 2 - d. Then begin{align*} |alpha| + |gamma| &= sqrt{a^2 + b^2} + sqrt{c^2 + d^2} &= sqrt{(d - 3)^2 + (2 - d)^2} + sqrt{c^2 + d^2} &= sqrt{2d^2 - 10d + 13} + sqrt{c^2 + d^2} &ge sqrt{13} - 5 + |d|. end{align*} Equality occurs when a = -3, b = 2, c = 0, and d = 0. Therefore, the minimum value is boxed{sqrt{13}}.

answer:According to the principle that the negation of a universal proposition is an existential proposition, the negation of the proposition "For any x in mathbb{R}, it holds that x^{2} geqslant 0" is "There exists an x_{0} in mathbb{R}, such that x_{0}^{2} < 0". Therefore, the correct option is boxed{text{A}}. The key to solving the problem is to master that the negation of a universal proposition "For any x in M, p(x)" is the existential proposition "There exists an x_{0} in M, neg p(x)".