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question:## Task B-2.3.Given is the equation x^{2}-p x+q=0, where p and q are positive real numbers. If the difference between the solutions of the equation is 1, and the sum of the solutions is 2, calculate p and q.
answer:## First Solution.According to Vieta's formulas, x_{1}+x_{2}=p, x_{1} cdot x_{2}=q, and according to the conditions of the problem,begin{array}{r}x_{1}+x_{2}=2 left|x_{1}-x_{2}right|=1end{array}from which it follows that p=2.begin{aligned}left|x_{1}-x_{2}right| & =sqrt{left(x_{1}-x_{2}right)^{2}}=sqrt{left(x_{1}+x_{2}right)^{2}-4 x_{1} cdot x_{2}} 1 & =sqrt{4-4 q} q & =frac{3}{4}end{aligned}## Note:The difference in solutions can also be obtained by direct substitution into the formula for the solutions of a quadratic equation:begin{aligned}& left|x_{1}-x_{2}right|=left|frac{-b+sqrt{D}}{2}-frac{-b-sqrt{D}}{2}right|=|sqrt{D}| quad Rightarrow quad D=1 & 4-4 q=1 & q=frac{3}{4}end{aligned}Note that the absolute value is not used even if it is not applied.
question:3、Given f(x)=x^{2}-2 x+3, g(x)=k x-1, then “ |k| leq 2 ” is “ f(x) geq g(x) holds for all x in R ” ( )A、Sufficient but not necessary conditionB、Necessary but not sufficient conditionC、Sufficient and necessary conditionD、Neither sufficient nor necessary condition
answer:3、A The translation retains the original text's line breaks and formatting.
question:Use the proof by contradiction to prove the proposition: "If a > b > 0, then |a| > |b|". The assumption should be ( )A: |a| = |b|B: |a| |b| and |a| = |b|
answer:Solution: Since the negation of the conclusion |a| > |b| is |a| leq |b|, when proving a proposition using proof by contradiction, we must first assume that the negation of the conclusion holds, thus, we should assume: |a| leq |b|, from which a contradiction is derived. Therefore, the correct choice is: C. The negation of the conclusion |a| > |b| is |a| leq |b|, from which the conclusion is drawn. This question mainly examines the proof of mathematical propositions using proof by contradiction, negating the conclusion to be proved, obtaining the opposite of the conclusion to be proved, thereby obtaining the desired result, and is considered a basic question.The final answer is boxed{C}.
question:5. Find the number of such sequences: of length n, each term is 0, 1 or 2, and 0 is neither the preceding term of 2 nor the following term of 2.
answer:5. Let in the said sequence, there are a_{n} ending with 0, b_{n} ending with 1, and c_{n} ending with 2, then a_{1}=b_{1}= c_{1}=1, and the recurrence relations are a_{n+1}=a_{n}+b_{n}, b_{n+1}=a_{n}+b_{n}+c_{n}, c_{n+1}=b_{n}+c_{n}. Thus, a_{n+1}+c_{n+1}+sqrt{2} b_{n+1}=1+ sqrt{2}left(a_{n}+c_{n}+sqrt{2} b_{n}right), hence a_{n}+c_{n}+sqrt{2} b_{n}=(1+sqrt{2})^{n-1}left(a_{1}+c_{1}+sqrt{2} b_{1}right)=(1+sqrt{2})^{n-1}(2+sqrt{2})=sqrt{2}(1+sqrt{2})^{n}, taking the conjugate gives a_{n}+c_{n}-sqrt{2} b_{n}=-sqrt{2}(1-sqrt{2})^{n}. Therefore, a_{n}+c_{n}=frac{sqrt{2}}{2}left[(1+sqrt{2})^{n}-(1-sqrt{2})^{n}right], b_{n}=frac{1}{2}left[(1+sqrt{2})^{n}right. left.+(1-sqrt{2})^{n}right]. Hence, a_{n}+b_{n}+c_{n}=frac{1}{2}left[(1+sqrt{2})^{n+1}+(1-sqrt{2})^{n+1}right]. This is the desired result.
question:3. In triangle A B C, angle C=90^{circ}, angle A and angle B are bisected and intersect at point P, and P E perp A B at point E. If B C=2, A C=3,then A E cdot E B= qquad
answer:3. 3Solution: Draw P D perp A C at D, P F perp B C at F, then:begin{aligned}C D & =C F & =frac{1}{2}(A C+B C-A B) & =frac{1}{2}(5-sqrt{13}), A D & =A C-C D & =frac{1}{2}(sqrt{13}+1), B F & =B C-C F=frac{1}{2}(sqrt{13}-1), therefore quad & A E cdot E B=A D cdot B F=3 .end{aligned}
question:Compare the sizes of a=2^{0.6} and b=0.6^{2} (use , or =).
answer:a=2^{0.6}>1, b=0.6^{2}=0.36b.Hence, the answer is boxed{a>b}.