answer:Since the sequence {a_n} is an arithmetic sequence, and a_3 + a_{11} = 50, also a_4 = 13, we have 2a_1 + 12d = 50, a_1 + 3d = 13, solving these equations, we get a_1 = 1, d = 4, therefore a_2 = 5, thus, the correct choice is boxed{C}.

answer:Solution. Let 2^{x}=a and 3^{x}=b. The equation transforms into1+a^{2}+b^{2}-a-b-a b=0which is equivalent to the equation(1-a)^{2}+(a-b)^{2}+(b-1)^{2}=0From the last equation, it follows that a=b=1, from which we get that x=0.

answer:In an isosceles triangle, the two base angles are equal. Given that one of the base angles is 72^{circ}, we can find the vertex angle by first understanding that the sum of all angles in any triangle is 180^{circ}. Therefore, we can calculate the vertex angle as follows:- Sum of all angles in a triangle = 180^{circ}- Sum of the two base angles = 72^{circ} + 72^{circ}- Vertex angle = Total sum of angles - Sum of base anglesSubstituting the values, we get:[ text{Vertex angle} = 180^{circ} - (72^{circ} + 72^{circ}) ][ text{Vertex angle} = 180^{circ} - 144^{circ} ][ text{Vertex angle} = 36^{circ} ]Therefore, the vertex angle of the isosceles triangle is boxed{36^{circ}}.

answer:Given that in triangle ABC, overrightarrow {AB} = (2, 3), overrightarrow {AC} = (3, 4),Therefore, overrightarrow {BC} = overrightarrow {AC} - overrightarrow {AB} = (1, 1).Thus, overrightarrow {AB} cdot overrightarrow {BC} = 1 times 2 + 1 times 3 = 5.Hence, the answer is boxed{5}.To solve this problem, we first find overrightarrow {BC}, and then use the coordinate operations of vectors to calculate the dot product. This question tests the calculation of the dot product of vectors and the coordinate operations of vectors, focusing on computational skills.

answer:Reference answer: 21Exam point: Average number problem

answer:(1) From the given information, we have 2a = 2sqrt{3}b, which implies a = sqrt{3}b, c = sqrt{a^2 - b^2} = sqrt{a^2 - frac{1}{3}a^2} = frac{sqrt{6}}{3}a, thus, the eccentricity e = frac{c}{a} = frac{sqrt{6}}{3}; (2) Proof: Given that c = sqrt{2}, from (1), we have a = sqrt{3}, b = 1, the equation of the ellipse becomes frac{x^2}{3} + y^2 = 1, when the slope of line l does not exist, let the equation of line l be x = m, substituting into the ellipse equation, we get y = pm sqrt{1 - frac{m^2}{3}}, since OP perp OQ, we have m^2 - (1 - frac{m^2}{3}) = 0, solving this gives m = pm frac{sqrt{3}}{2}, the distance from the center (0,0) to the line x = m is frac{sqrt{3}}{2}, thus, line l is tangent to the circle x^2 + y^2 = frac{3}{4}; when the slope of the line exists, let l: y = kx + t, substituting into the ellipse equation x^2 + 3y^2 = 3, we get (1 + 3k^2)x^2 + 6ktx + 3t^2 - 3 = 0, let P(x_1,y_1), Q(x_2,y_2), we have x_1 + x_2 = -frac{6kt}{1 + 3k^2}, x_1x_2 = frac{3t^2 - 3}{1 + 3k^2}, y_1y_2 = (kx_1 + t)(kx_2 + t) = k^2x_1x_2 + kt(x_1 + x_2) + t^2, since OP perp OQ, we have x_1x_2 + y_1y_2 = 0, which is (1 + k^2)x_1x_2 + kt(x_1 + x_2) + t^2 = 0, thus, (1 + k^2) cdot frac{3t^2 - 3}{1 + 3k^2} + kt(-frac{6kt}{1 + 3k^2}) + t^2 = 0, simplifying gives 4t^2 = 3 + 3k^2, the distance from the center (0,0) to the line y = kx + t is d = frac{|t|}{sqrt{1 + k^2}} = frac{|t|}{sqrt{frac{4}{3}t^2}} = frac{sqrt{3}}{2}, which is the radius. Therefore, line l is always tangent to the circle x^2 + y^2 = frac{3}{4}.Thus, the eccentricity of the ellipse E is boxed{frac{sqrt{6}}{3}}, and it is proven that the line l is always tangent to the circle x^2 + y^2 = frac{3}{4}.