answer:31. First, classify these 100 numbers based on their remainders when divided by 3.Bybegin{array}{l}1 leqslant 3 k leqslant 100 Rightarrow frac{1}{3} leqslant k leqslant 33 frac{1}{3}, 1 leqslant 3 k+1 leqslant 100 Rightarrow 0 leqslant k leqslant 33 .end{array}Thus, there are 33 numbers of the form 3 k, 34 numbers of the form 3 k+1, and 33 numbers of the form 3 k+2. These are denoted as A type, B type, and C type, respectively. The numbers that meet the criteria can be divided into 4 categories:(1) 3 A type; (2) 3 B type; (3) 3 C type; (4) 1 A type, 1 B type, 1 C type.Therefore, the required probability isP=frac{C_{33}^{3}+C_{34}^{3}+C_{33}^{3}+C_{33}^{1} cdot C_{34}^{1} cdot C_{33}^{1}}{C_{100}^{3}}=frac{817}{2450} .

answer:Solution: A. Inferring the properties of a spatial tetrahedron from the properties of a plane triangle is analogical reasoning, B. All metals can conduct electricity, uranium is a metal, therefore uranium can conduct electricity, is deductive reasoning, C. There are 12 classes participating in military training in the first year of high school, with 51 students in class 1, 53 students in class 2, and 52 students in class 3, from this it is inferred that each class has more than 50 students, is inductive reasoning, D. In the sequence ({a_n}), (a_1=2), (a_n=2a_{n-1}+1(ngeqslant 2)), from this the general formula of ({a_n}) is induced, is inductive reasoning. Therefore, the answer is: boxed{B} By understanding the definitions of inductive reasoning, analogical reasoning, and deductive reasoning, one can make the judgment accordingly. This question mainly examines the judgment of the truth of propositions, involving the judgment of inductive reasoning, analogical reasoning, and deductive reasoning. Understanding the corresponding definitions is key to solving this question. It is quite basic.

answer:Solution: frac{2+i}{i}=1-2i, thus a=1, b=-2; then frac{b}{a}=-2, therefore, the correct choice is boxed{A}. By multiplying both the numerator and the denominator of the complex number by the conjugate of the denominator, the complex number is simplified into the form of a+bi (a,b in mathbb{R}). This question tests the basic concept of complex numbers and the operations of multiplication and division in algebraic form of complex numbers, and it is a basic question.

answer:(1) When a=1, b=-2, we have f(x) = x^2 - x - 3 = x Leftrightarrow x^2 - 2x - 3 = 0 Leftrightarrow (x-3)(x+1) = 0 Leftrightarrow x = 3 or x = -1,Therefore, the fixed points of f(x) are x = 3 or x = -1.So, the fixed points of f(x) are boxed{x = 3 text{ or } x = -1}.(2) For any real number b, f(x) always has two distinct fixed pointsLeftrightarrow For any real number b, ax^2 + (b+1)x + b - 1 = x i.e., ax^2 + bx + b - 1 = 0 always has two distinct real rootsLeftrightarrow For any real number b, Delta = b^2 - 4a(b-1) > 0 always holdsLeftrightarrow For any real number b, b^2 - 4ab + 4a > 0 always holdsLeftrightarrow Delta' = (4a)^2 - 4 times 4a < 0Leftrightarrow a^2 - a < 0Leftrightarrow 0 < a < 1.Therefore, the range of values for a is boxed{0 < a < 1}.

answer:## Solutionx_{t}^{prime}=left(ln frac{1-t}{1+t}right)^{prime}=frac{1+t}{1-t} cdot frac{-1 cdot(1+t)-(1-t) cdot 1}{(1+t)^{2}}==frac{-1-t-1+t}{1-t^{2}}=-frac{2}{1-t^{2}}y_{t}^{prime}=left(sqrt{1-t^{2}}right)^{prime}=frac{1}{2 sqrt{1-t^{2}}} cdot(-2 t)=-frac{t}{sqrt{1-t^{2}}}We obtain:y_{x}^{prime}=frac{y_{t}^{prime}}{x_{t}^{prime}}=left(-frac{t}{sqrt{1-t^{2}}}right) /left(-frac{2}{1-t^{2}}right)=frac{t cdotleft(1-t^{2}right)}{2 sqrt{1-t^{2}}}==frac{t cdot sqrt{1-t^{2}}}{2}## Problem Kuznetsov Differentiation 16-16

answer:8. frac{5 sqrt{74}}{37}Analysis: It is known that frac{1}{z}=frac{bar{z}}{|z|^{2}}, so the arguments of frac{1}{z} and z on the complex plane are opposite numbers. Suppose z=r e^{i theta}, then frac{1}{z}=frac{e^{-i theta}}{r}. The area of the parallelogram can be expressed as: |z| cdotleft|frac{1}{z}right| cdot sin 2 theta. Given that sin 2 theta=frac{35}{37}, then left|z+frac{1}{z}right|^{2}=left(r e^{i theta}+frac{e^{-i theta}}{r}right) cdotleft(r e^{-i theta}+frac{e^{i theta}}{r}right)=r^{2}+e^{i cdot 2 theta}+e^{-i cdot 2 theta}+frac{1}{r^{2}}=r^{2}+frac{1}{r^{2}}+2 cos 2 theta geq frac{50}{37}. Therefore, left|z+frac{1}{z}right| geq frac{5 sqrt{74}}{37}.