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question:In triangle ABC, a, b, c are the lengths of the sides opposite to angle A, angle B, angle C respectively. It is known that sqrt{2}sin A= sqrt{3cos A}. (1) If a^{2}-c^{2}=b^{2}-mbc, find the value of the real number m; (2) If a= sqrt{3}, find the maximum area of triangle ABC.
answer:Solution: (1) Squaring both sides of sqrt{2}sin A= sqrt{3cos A}, we get: 2sin^{2}A=3cos A, which simplifies to (2cos A-1)(cos A+2)=0, Solving this, we find cos A= frac{1}{2}, And a^{2}-c^{2}=b^{2}-mbc can be transformed into frac{b^{2}+c^{2}-a^{2}}{2bc}= frac{m}{2}, That is, cos A= frac{m}{2}= frac{1}{2}, thus m=1. (2) From (1), we know cos A= frac{1}{2}, hence sin A= frac{sqrt{3}}{2}. Also, frac{b^{2}+c^{2}-a^{2}}{2bc}= frac{1}{2}, Therefore, bc=b^{2}+c^{2}-a^{2}geqslant 2bc-a^{2}, which implies bcleqslant a^{2}. Thus, the area of triangle ABC is S_{triangle ABC}= frac{bc}{2}sin Aleqslant frac{a^{2}}{2}cdot frac{sqrt{3}}{2}= frac{3sqrt{3}}{4}. So, the answers are: (1) m=1, so boxed{m=1}. (2) The maximum area of triangle ABC is boxed{frac{3sqrt{3}}{4}}.
question:For a given positive integer n(ge 2), find maximum positive integer A such that a polynomial with integer coefficients P(x) with degree n that satisfies the condition below exists.bullet P(1), P(2), cdots P(A) is a multiple of Abullet P(0) = 0 and the coefficient of the first term of P(x) is 1, In other words, P(x) has the following form:P(x) = c_nx^n + c_{n-1}x^{n-1}+dots+c_2x^2+x (c_n neq 0)
answer:1. Claim: The maximum positive integer ( A ) such that a polynomial ( P(x) ) with integer coefficients and degree ( n ) satisfies the given conditions is ( n! ).2. Constructing a Polynomial: Consider the polynomial ( P(x) = x(x+1)(x+2)cdots(x+n-1) ). This polynomial has the following properties: - It is a monic polynomial of degree ( n ). - The coefficient of the linear term is 1. - ( P(0) = 0 ).3. Verifying the Conditions: - For ( P(i) ) where ( i = 1, 2, ldots, n ): [ P(i) = i(i+1)(i+2)cdots(i+n-1) ] This can be rewritten using the factorial notation: [ P(i) = frac{(i+n-1)!}{(i-1)!} ] Since ( (i+n-1)! ) is divisible by ( n! ), it follows that ( P(i) ) is a multiple of ( n! ) for ( i = 1, 2, ldots, n ).4. Upper Bound for ( A ): - We need to show that ( A leq n! ). Consider the ( n )-th finite difference of ( P(x) ), denoted by ( Delta^n P(x) ). - For a polynomial ( P(x) ) of degree ( n ), the ( n )-th finite difference ( Delta^n P(x) ) is a constant and equals ( n! ). - By the given condition, ( P(z) ) is a multiple of ( A ) for all positive integers ( z ). Therefore, ( A ) must divide ( Delta^n P(0) ), which is ( n! ).5. Conclusion: Since ( A ) must divide ( n! ) and we have constructed a polynomial where ( P(i) ) is a multiple of ( n! ) for ( i = 1, 2, ldots, n ), the maximum value of ( A ) is ( n! ).(blacksquare)The final answer is ( boxed{ n! } )
question:Each of the 2N = 2004 real numbers x_1, x_2, ldots , x_{2004} equals either sqrt 2 -1 or sqrt 2 +1. Can the sum sum_{k=1}^N x_{2k-1}x_2k take the value 2004? Which integral values can this sum take?
answer:1. Given (2N = 2004), we have (N = 1002). Each (x_i) is either (sqrt{2} - 1) or (sqrt{2} + 1).2. Let (x = sqrt{2} + 1). Then, we have: [ x + frac{1}{x} = 2sqrt{2} ] Squaring both sides: [ x^2 + 2 + frac{1}{x^2} = 8 implies x^2 + frac{1}{x^2} = 6 ]3. Since (x_i in {sqrt{2} - 1, sqrt{2} + 1}), we have: [ x_i x_{i+1} in {x^2, frac{1}{x^2}, 1} ]4. We need to find the sum: [ sum_{k=1}^{N} x_{2k-1} x_{2k} ] Let (lambda) be the number of pairs ((x_{2k-1}, x_{2k})) such that (x_{2k-1} x_{2k} = x^2), and (mu) be the number of pairs such that (x_{2k-1} x_{2k} = frac{1}{x^2}). The remaining pairs will have (x_{2k-1} x_{2k} = 1).5. Therefore, we can write: [ sum_{k=1}^{N} x_{2k-1} x_{2k} = lambda x^2 + mu frac{1}{x^2} + (N - lambda - mu) ]6. Given that the sum is 2004, we have: [ lambda x^2 + mu frac{1}{x^2} + (N - lambda - mu) = 2004 ]7. Substituting (x^2 = 6) and (frac{1}{x^2} = frac{1}{6}), we get: [ 6lambda + frac{mu}{6} + (1002 - lambda - mu) = 2004 ]8. Simplifying the equation: [ 6lambda + frac{mu}{6} + 1002 - lambda - mu = 2004 ] [ 5lambda + 1002 + frac{mu}{6} - mu = 2004 ] [ 5lambda + 1002 - frac{5mu}{6} = 2004 ] [ 5lambda - frac{5mu}{6} = 1002 ] [ 30lambda - 5mu = 6012 ] [ 6lambda - mu = 1202.4 ]9. Since (lambda) and (mu) must be integers, the above equation does not hold. Therefore, the sum cannot be 2004.10. Now, we need to find the possible integral values of the sum. We know: [ 6lambda + frac{mu}{6} + (1002 - lambda - mu) ] Simplifying: [ 5lambda + 1002 - frac{5mu}{6} ]11. Since (lambda) and (mu) are integers, the possible values of (lambda) range from 0 to 501 (since (N/2 = 501)).12. Therefore, the possible values of the sum are: [ 4lambda + 1002 ] where (lambda in {0, 1, 2, ldots, 501}).13. Thus, the possible integral values of the sum are: [ {1002, 1006, 1010, ldots, 3006} ]The final answer is ( boxed{ {1002, 1006, 1010, ldots, 3006} } ).
question:Define the length of the intervals (a,b), [a,b), (a,b], [a,b] to be d=b-a. The length of the union of multiple intervals is the sum of the lengths of each interval, for example, the length d of (1,2) cup [3,5) is d=(2-1)+(5-3)=3. Let [x] denote the largest integer not greater than x and let {x}=x-[x], where x in mathbb{R}. Suppose f(x)=[x]{x} and g(x)=x-1. When 0 leq x leq k, the length of the solution interval for the inequality f(x) < g(x) is 5, find the value of k.A: 6B: 7C: 8D: 9
answer:f(x)=[x]{x}=[x](x-[x])=[x]x-[x]^2 and g(x)=x-1.The inequality f(x)1.So, the solution set emptyset.For x in [1,2), [x]=1, the inequality does not hold since it simplifies to 0>0.So, the solution set emptyset.For x in [2,3), [x]=2, and since [x]-1>0, the inequality simplifies to x<[x]+1=3.Therefore, the solution interval for f(x) < g(x) when x in [2,3) is [2,3) which has a length d=3-2=1.Similarly, for x in [3,4) the solution interval for f(x) < g(x) has length d=4-2=2.Since the length of the solution interval for f(x) < g(x) is 5, we have k-2=5.Thus, k=7.So, the correct choice is boxed{B}.
question:1. The total number of students in the school must be divisible by 7 and 4, which means it must be divisible by 28. Since the number is no more than 40, there were 28 students in total. The number of prize winners is (1 / 7 + 1 / 4 + 1 / 4) cdot 28 = 18 people. Therefore, 10 students did not receive any medals.
answer:Answer: x=10.Answer to the variant: 4-2: x=18.
question:If sin (α- dfrac {7π}{4})= dfrac {1}{2}, then cos ( dfrac {π}{4}-α)= _____ .
answer:Given: sin (α- dfrac {7π}{4})= dfrac {1}{2}Hence, cos ( dfrac {π}{4}-α)=cos [(α- dfrac {7π}{4})+ dfrac {3π}{2}]=sin (α- dfrac {7π}{4})= dfrac {1}{2}.Therefore, the answer is: boxed{dfrac {1}{2}}.The solution can be obtained by simplifying and calculating using the given information and the trigonometric identities. This problem primarily tests the application of trigonometric identities in simplifying and finding the value of trigonometric functions, and it requires a transformative thinking. It is a basic problem.