answer:(I) From the given function, we have the following conditions:begin{cases} 2 - t > 0 t - 1 geq 0 end{cases}Solving these inequalities, we obtain 1 leq t < 2. Thus, the domain is D = [1, 2).(II) Rewrite the function g(x) as follows:g(x) = x^{2} + 2mx - m^{2} = (x + m)^{2} - 2m^{2}The axis of symmetry for this quadratic function is x = -m. Now we consider three cases based on the value of m:1. If -m geq 2 (i.e., m leq -2), then g(x) is decreasing on the interval [1, 2), and thus, it does not have a minimum value.2. If 1 < -m < 2 (i.e., -2 < m < -1), then g(x) is decreasing on [1, -m) and increasing on (-m, 2]. In this case, the minimum value is g(-m) = -2m^{2} neq 2, so there is no such m.3. If -m leq 1 (i.e., m geq -1), then g(x) is increasing on [1, 2). In this case, g(x) achieves its minimum value at x = 1:g(1) = 1 + 2m - m^{2} = 2Solving the equation above, we find that m = 1.In summary, the value of m is boxed{m = 1}.

answer:7. 2 .Let the complex number corresponding to the center of the square be m. Then, after translating the origin of the complex plane to m, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation (z-m)^{4}=n (where n is a complex number).From z^{4}+a z^{3}+b z^{2}+c z+d=(z-m)^{4}-n, comparing coefficients we know that m=-frac{a}{4} is a rational number.Furthermore, from -4 m^{3}=c, we know that m is an integer.Then, from d=m^{4}-n, we know that n is an integer.Thus, the roots of the equation (z-m)^{4}=n arez_{k}=m+sqrt[4]{|n|}left(cos frac{k pi}{2}+mathrm{i} sin frac{k pi}{2}right) .Therefore, the length of the diagonal of the square is 2 sqrt[4]{|n|}, and its area is2 sqrt{|n|} geqslant 2 text {. }

answer:(1) Since point F(-1,0) is one of the foci of the ellipse,we have c = 1,and given that b^{2} = 3,we obtain a^{2} = b^{2} + c^{2} = 4,thus the equation of the ellipse is frac{x^{2}}{4} + frac{y^{2}}{3} = 1.(2) When the slope of line l does not exist, the line equation is x=-1,in this case, Dleft(-1, frac{3}{2}right), Cleft(-1, -frac{3}{2}right), and the areas of Delta ABD and Delta ABC are equal,thus, |S_{1} - S_{2}| = 0.When the slope of line l exists,let the line equation be y = k(x+1), (k neq 0),and denote C(x_{1}, y_{1}), D(x_{2}, y_{2}). Obviously, y_{1} and y_{2} have opposite signs.From begin{cases} frac{x^{2}}{4} + frac{y^{2}}{3} = 1 y = k(x+1) end{cases},we get (3 + 4k^{2})x^{2} + 8k^{2}x + 4k^{2} - 12 = 0. Clearly, Delta > 0, the equation has real roots,and x_{1} + x_{2} = frac{8k^{2}}{3 + 4k^{2}}, x_{1}x_{2} = frac{4k^{2} - 12}{3 + 4k^{2}}.Thus, |S_{1} - S_{2}| = 2||y_{1}| - |y_{2}|| = 2|y_{1} + y_{2}|,= 2|k(x_{1} + 1) + k(x_{2} + 1)|,= 2|k(x_{1} + x_{2}) + 2k|,= frac{12|k|}{3 + 4k^{2}}.Since k neq 0, we havefrac{12|k|}{3 + 4k^{2}} = frac{12}{frac{3}{|k|} + 4|k|} leqslant frac{12}{2sqrt{frac{3}{|k|} cdot 4|k|}} = sqrt{3},and the equality holds if and only if k = pm frac{sqrt{3}}{2}.Therefore, the maximum value of |S_{1} - S_{2}| is boxed{sqrt{3}}.

answer:# Answer: 864.Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).To ensure divisibility by three, we proceed as follows. We will choose three digits arbitrarily (this can be done in 6 cdot 6 cdot 6 ways), and the fourth digit will be chosen so that the sum of all the digits of the number is divisible by 3. Since among the given digits, there are two digits that are divisible by 3 (0 and 9), two digits that give a remainder of 1 when divided by 3 (4 and 7), and two digits that give a remainder of 2 when divided by 3 (2 and 5), this choice can be made in two ways.Applying the rule of product, we get that in total 2 cdot 6 cdot 6 cdot 6 cdot 2=864 ways.

answer:To solve this problem, we start by setting overrightarrow{b}=(1,0) and overrightarrow{a}=(x,y) in the Cartesian coordinate system. This leads to the following vectors:- overrightarrow{a}+overrightarrow{b}=(x+1,y)- overrightarrow{a}+2overrightarrow{b}=(x+2,y)Given that the angle between overrightarrow{a}+overrightarrow{b} and overrightarrow{a}+2overrightarrow{b} is 30^{circ}, we can use the cosine of this angle to set up an equation:[cos30^{circ} = frac{(overrightarrow{a}+overrightarrow{b})cdot(overrightarrow{a}+2overrightarrow{b})}{|overrightarrow{a}+overrightarrow{b}||overrightarrow{a}+2overrightarrow{b}|} = frac{(x+1)(x+2)+y^{2}}{sqrt{(x+1)^{2}+y^{2}}cdotsqrt{(x+2)^{2}+y^{2}}} = frac{sqrt{3}}{2}]Simplifying this equation, we obtain:[4[(x+1)(x+2)+y^{2}]^{2}=3[(x+1)^{2}+y^{2}][(x+2)^{2}+y^{2}]]Further simplification leads to:[[(x+1)(x+2)+y^{2}]^{2}=3y^{2}]This simplifies to:[(x+1)(x+2)+y^{2}=sqrt{3}|y|]Considering the symmetry about the x-axis, we focus on ygeqslant 0 and rewrite the equation as:[(x+frac{3}{2})^{2}+(y-frac{sqrt{3}}{2})^{2}=1 quad (ygeqslant 0)]This represents a circle centered at C(-frac{3}{2},frac{sqrt{3}}{2}) with radius 1. The distance from C to the origin O is:[|OC|=sqrt{(frac{3}{2})^{2}+(frac{sqrt{3}}{2})^{2}}=sqrt{3}]The line OC is described by:[y=-frac{sqrt{3}}{3}x]By substituting y from the line equation into the circle equation, we get:[frac{4}{3}x^{2}+4x+2=0]Solving this quadratic equation, we find that it has two solutions, indicating two intersection points with the circle. Thus, the minimum and maximum values of |overrightarrow{a}| are:- |overrightarrow{a}|_{min}=|OC|-1=sqrt{3}-1- |overrightarrow{a}|_{max}=|OC|+1=sqrt{3}+1Therefore, the range of values for |overrightarrow{a}| is [sqrt{3}-1, sqrt{3}+1]. Hence, the correct answer is boxed{B}.

answer:According to the product rule (uv)'=u'v+uv', we have y'=x'cos x + x(cos x)'=cos x - xsin x Therefore, the correct option is boxed{A}. By using the product rule (uv)'=u'v+uv' and the derivative formula cos x'=-sin x, we can find the derivative of the function. When calculating the derivative of a function, first select the appropriate derivative operation rule and derivative formula based on the form of the function. This is a basic question.