answer: Solution:# Part (1): Monotonic Intervals of f(x)Given f(x)=-xln(-x), we first determine its domain, which is (-infty, 0) because the logarithm function requires its argument to be positive, and -x is positive when x is negative.To find the monotonic intervals, we calculate the derivative of f(x):[f'(x) = frac{d}{dx}(-xln(-x)) = -ln(-x) - 1]This is because the derivative of -xln(-x) with respect to x is the derivative of -x times ln(-x) plus -x times the derivative of ln(-x), which simplifies to -ln(-x)-1.To determine where f(x) is increasing or decreasing, we set f'(x) > 0 and f'(x) 0: -ln(-x) - 1 > 0 Rightarrow -ln(-x) > 1 Rightarrow ln(-x) frac{1}{e} Rightarrow x 0 Rightarrow ln(-x) -frac{1}{e}. Thus, f(x) is monotonically increasing on (-frac{1}{e}, 0).- For f'(x) 0 because both e^{x} and ln(-x) are positive. - For -1 0 and h'(-frac{1}{e}) = e^{-frac{1}{e}} - 1 < 0, there exists x_0 in (-1, -frac{1}{e}) such that h'(x_0) = 0. This implies h(x) has a maximum at x_0, and since h'(x_0) = 0, we find e^{x_0} = -ln(-x_0), leading to x_0 = -e^{x_0} and ln(-x_0) = x_0. Thus, h(x_0) = x_0^2 - 2x_0.Since x_0 in (-1, -frac{1}{e}), it follows that h(x_0) < h(-1) = 3, and h(x) leq h(x_0). Therefore, we conclude:[boxed{g(x)-f(x) < 3}]

answer:To solve this problem, we should visualize the motion of the square as it rotates around one of its sides. Since the side around which the square is rotating functions as the axis of rotation, the resulting geometric body will be a cylinder.Given that a square with an area of 1 implies that each side of the square has a length of 1, we can then deduce the dimensions of the cylinder. The side of the square becomes the height of the cylinder (h = 1), and the rotation of one side around the axis will trace a circle with a circumference equal to the square's perimeter, which in this case is 4 times the length of the side. But since the square rotates about one side, the circle's circumference will be equal to the length of one side itself (C=1).Therefore, the radius of the circle (base of the cylinder) can be found knowing the circumference of the circle relates to its radius by the formula C = 2pi r. Solving for r, we get r = frac{C}{2pi} = frac{1}{2pi}.However, we notice here that the result for the radius stated above is not necessary, as we are directly given the perimeter (circumference for the cylinder's base) by the length of the rotating square's side. Hence, the lateral surface area of a cylinder is given by the formula A_{text{lateral}} = h cdot C = h cdot 2pi r. In our case, h = 1 and C = 1 (since it's the length of the square's side and so the circumference of cylinder's base), making the calculation straightforward:A_{text{lateral}} = 1 cdot 1 = 2pi cdot frac{1}{2pi}Simplifying the above equation, we find:A_{text{lateral}} = boxed{2pi}

answer:To solve for |overrightarrow{BN}|, we start by expressing vectors overrightarrow{BM} and overrightarrow{BN} in terms of the given information:1. Let overrightarrow{AN} = lambda overrightarrow{AD}. This means we are expressing the vector overrightarrow{AN} as a scalar multiple of overrightarrow{AD}, where lambda is the scalar.2. We know that overrightarrow{BM} = overrightarrow{BC} + overrightarrow{CM} and since M is the midpoint of CD, overrightarrow{CM} = frac{1}{2}overrightarrow{CD}. However, since overrightarrow{CD} and overrightarrow{BA} are parallel and of the same length, we can write overrightarrow{CM} = frac{1}{2}overrightarrow{BA}. Therefore, overrightarrow{BM} = overrightarrow{BC} + frac{1}{2}overrightarrow{BA}.3. Similarly, overrightarrow{BN} = overrightarrow{BA} + overrightarrow{AN} = overrightarrow{BA} + lambda overrightarrow{AD}. Since overrightarrow{AD} and overrightarrow{BC} are also parallel and of the same length, we can replace overrightarrow{AD} with overrightarrow{BC}. Thus, overrightarrow{BN} = overrightarrow{BA} + lambda overrightarrow{BC}.4. Given that the side length of square ABCD is 4, we know that overrightarrow{BA} cdot overrightarrow{BC} = 0 because they are perpendicular.5. The dot product overrightarrow{BM} cdot overrightarrow{BN} = (overrightarrow{BC} + frac{1}{2}overrightarrow{BA}) cdot (overrightarrow{BA} + lambda overrightarrow{BC}) simplifies to 16lambda + 8 because overrightarrow{BA} cdot overrightarrow{BC} = 0 and the lengths of overrightarrow{BA} and overrightarrow{BC} are both 4. Therefore, we have 16lambda + 8 = 20.6. Solving for lambda, we get 16lambda = 12, which means lambda = frac{3}{4}.7. To find |overrightarrow{BN}|, we use the Pythagorean theorem in the context of the vector's components. Since overrightarrow{BN} = overrightarrow{BA} + lambda overrightarrow{BC} and both overrightarrow{BA} and overrightarrow{BC} have a length of 4, the length of overrightarrow{BN} is sqrt{16 + (3/4)^2 cdot 16} = sqrt{16 + 9} = 5.Therefore, the magnitude of vector overrightarrow{BN} is boxed{5}, which corresponds to choice boxed{C}.

answer:Given the equation bcos A - acos B = a, we can manipulate it as follows:begin{align*}bcos A - acos B &= a Rightarrow bcos A &= a(cos B + 1).end{align*}By applying the Law of Sines, frac{a}{sin A} = frac{b}{sin B}, we can rewrite the equation as:begin{align*}frac{b}{sin B}cos A &= frac{a}{sin A}(cos B + 1) Rightarrow sin Bcos A &= sin Acos B + sin A Rightarrow sin(B-A) &= sin A.end{align*}This implies that either B-A = A or B-A+A = pi. Solving these gives B = 2A or B = pi (which we discard since triangle ABC is acute).Given B = 2A, we find C = pi - A - B = pi - 3A. Since triangle ABC is acute, we have:begin{align*}0 < A < frac{pi}{2}, 0 < 2A < frac{pi}{2}, 0 < pi - 3A < frac{pi}{2}.end{align*}Solving these inequalities gives frac{pi}{6} < A < frac{pi}{4}.Now, we evaluate sqrt{3}sin B + 2sin^2 A:begin{align*}sqrt{3}sin B + 2sin^2 A &= sqrt{3}sin 2A + 2sin^2 A &= sqrt{3}sin 2A + 1 - cos 2A &= 2sinleft(2A - frac{pi}{6}right) + 1.end{align*}Given frac{pi}{6} < A < frac{pi}{4}, we have:begin{align*}frac{pi}{6} < 2A - frac{pi}{6} < frac{pi}{3}.end{align*}This implies:begin{align*}frac{1}{2} < sinleft(2A - frac{pi}{6}right) < frac{sqrt{3}}{2}.end{align*}Therefore, we find:begin{align*}2 < 2sinleft(2A - frac{pi}{6}right) + 1 < sqrt{3} + 1.end{align*}Thus, the range of sqrt{3}sin B + 2sin^2 A is (2, sqrt{3} + 1).Therefore, the answer is: boxed{B}.

answer:To solve the polynomial x^{2}+x^{10}=a_{0}+a_{1}(x+1)+a_{2}(x+1)^{2}+ldots+a_{9}(x+1)^{9}+a_{10}(x+1)^{10}, on the right side of the equation, only the last term a_{10}(x+1)^{10} in its expanded form contains x^{10}, and the coefficient of x^{10} is a_{10}. On the left side, the coefficient of x^{10} is 1, therefore a_{10}=1, The coefficient of x^{9} on the right side comes from the last two terms, and the coefficient of x^{9} is a_{9}+binom{10}{9}cdot a_{10}. On the left side, the coefficient of x^{9} is 0, therefore a_{9}+10=0, therefore a_{9}=-10, Therefore, the correct choice is: boxed{A}. By utilizing the coefficients from the binomial theorem, first find the coefficient of x^{10}, then from a_{9}+binom{10}{9}cdot a_{10}, the coefficient of x^{9} can be determined, thus obtaining the answer. This question mainly tests the application of the binomial theorem, where understanding the coefficients of each term is key.

answer:AnalysisThis question examines the geometry and operations of complex numbers. Based on the point corresponding to the complex number being in the second quadrant of the complex plane, we can set up inequalities to solve it.SolutionGiven: (1−i)(a−i)= a-1-(1+a)i corresponds to a point in the second quadrant of the complex plane,Therefore, begin{cases}a-1 0end{cases},Solving these, we find a < -1,Hence, the correct choice is boxed{text{B}}.