官方SAT2数学真题解析 Subject Tests Answer Explanations Math

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SAT 2 数学2 college board官网全部原题加答案讲解 SAT Subject Test Math Level 2

SAT Subject Test Practice - Results Summary Mathematics Level 21Your answer Omitted!What is the distance in space between the points with coordinates and ?(A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is D.The distance between the points with coordinates and is given by the distance formula: .Therefore, the distance between the points with coordinates and is:,which simplifies to .2Your answer Omitted!If , what value does approach as gets infinitely larger?(A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is E.One way to determine the value that approaches as gets infinitely larger is to rewrite the definition of the function to use only negative powers of and then reason about the behavior of negative powers of as gets infinitely larger. Since the question is only concerned with what happens to as gets infinitely larger, one can assume that is positive. For , theexpression is equivalent to the expression . As gets infinitely larger, the expression approaches the value , so as gets infinitely larger, the expression approaches the value . Thus, as gets infinitely larger, approaches .Alternatively, one can use a graphing calculator to estimate the height of the horizontal asymptote for the function . Graph the function on an interval with “large”, say, from to .By examining the graph, the all seem very close to . Graph the function again, from, say, to .The vary even less from . In fact, to the scale of the coordinate plane shown, the graph of the function is nearly indistinguishable from the asymptotic line . This suggests that as gets infinitely larger, approaches , that is, .Note: The algebraic method is preferable, as it provides a proof that guarantees that the value approaches is . Although the graphical method worked in this case, it does not provide a complete justification; for example, the graphical method does not ensure that the graph resembles a horizontal line for “very large”such as .3Your answer Omitted!If is a factor of , then(A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is A.By the Factor Theorem, is a factor of only when is a root ofthat is, , which simplifies to . Therefore, .Alternatively, one can perform the division of by and then find a value for so that the remainder of the division is .Since the remainder is , the value of must satisfy . Therefore, .4Your answer Omitted!Alison deposits into a new savings account that earns percent interest compounded annually. If Alison makes no additional deposits or withdrawals, how many years will it take for the amount in the account to double?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumAfter year, the amount in the account is equal to . After years, the amount isequal to , and so on. After years, the amount is equal to . You needto find the value of for which . There are several ways to solve this equation. You can use logarithms to solve the equation as follows.Since , it will take more than years for the amount in the account to double. Thus, you need to round up to .Another way to find is to use your graphing calculator to graph and . From the answer choices, you know you need to set the viewing window with values from to about and values extending just beyond . The of the point of intersection is approximately . Thus you need to round up to .5Your answer Omitted!In the figure above, when is subtracted from , what is the length of the resultant vector?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe resultant of can be determined by . The length of the resultant is:6Your answer Omitted!In the -plane, what is the area of a triangle whose vertices are , , and ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt is helpful to draw a sketch of the triangle:The length of the base of the triangle is and the height of the triangle is . Therefore, the area of the triangle is . The correct answer is B.7Your answer Omitted!A right circular cylinder has radius and height . If and are two points on its surface, what is the maximum possible straight-line distance between and ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe maximum possible distance occurs when and are on the circumference of opposite bases: You can use the Pythagorean Theorem:The correct answer is (B).8Your answer Omitted!Note: Figure not drawn to scale.In the figure above, and the measure of is . What is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThere are several ways to solve this problem. One way is to use the law of sines. Since ,the measure of is and the measure of is . Thus, and . (Make sure your calculator is in degree mode.)You can also use the law of cosines:Since is isosceles, you can draw the altitude to the triangle.9Your answer Omitted!The function is defined by for .What is the difference between the maximum and minimum values of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt is necessary to use your graphing calculator for this question. First graph the function. It is helpful to resize the viewing window so the -values go fromto . On this interval the maximum value of is and the minimum value of is. The difference between these two values is , which rounds to .10Your answer Omitted!Suppose the graph of is translated units left and unit up. If the resulting graph represents , what is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt may be helpful to draw a graph of and .The equation for is . Therefore,. The correct answer is B.11Your answer Omitted!A sequence is recursively defined by , for . If and , what is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe values for and are given. is equal to . is equal to. is equal to . is equal to.If your graphing calculator has a sequence mode, you can define the sequence recursively and findthe value of . Let , since the first term is . Define . Let , since we have to define the first two terms and . Then examining a graph or table, you can find .12Your answer Omitted!The diameter and height of a right circular cylinder are equal. If the volume of the cylinder is , what is the height of the cylinder?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is A.To determine the height of the cylinder, first express the diameter of the cylinder in terms of theheight, and then express the height in terms of the volume of the cylinder.The volume of a right circular cylinder is given by , where is the radius of the circular base of the cylinder and is the height of the cylinder. Since the diameter and height are equal, . Thus . Substitute the expression for in the volume formula to eliminate :. Solving for gives . Since the volume of the cylinder is , theheight of the cylinder is .13Your answer Omitted!If ,then(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is E.One way to determine the value of is to apply the sine of difference of two angles identity: . Since and , the identity gives . Therefore, .Another way to determine the value of is to apply the supplementary angle trigonometric identity for the sine: . Therefore, .14Your answer Omitted!A line has parametric equations and , where is the parameter. The slope of the line is(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is B.One way to determine the slope of the line is to compute two points on the line and then use the slope formula. For example, letting gives the point on the line, and letting gives the point on the line. Therefore, the slope of the line is equal to .Alternatively, one can express in terms of . Since and , it follows that . Therefore, the slope of the line is .15Your answer Omitted!What is the range of the function defined by ?(A) All real numbers(B) All real numbers except(C) All real numbers except(D) All real numbers except(E) All real numbers between andExplanationDifficulty: MediumThe correct answer is D.The range of the function defined by is the set of such thatfor some .One way to determine the range of the function defined by is to solve the equation for and then determine which correspond to at least one . To solve for , first subtract from both sides to get and then take the reciprocal of both sides to get . The equation shows that for anyother than , there is an such that , and that there is no such for . Therefore, the range of the function defined by is all real numbers except .Alternatively, one can reason about the possible values of the term . The expression can take on any value except , so the expression can take on any value except . Therefore, the range of the function defined by is all real numbers except .16Your answer Omitted!The table above shows the number of digital cameras that were sold during a three-day sale. The prices of models , , and were , , and , respectively. Which of the following matrix representations gives the total income, in dollars, received from the sale of the cameras for each of the three days?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is C.A correct matrix representation must have exactly three entries, each of which represents the total income, in dollars, for one of the three days. The total income for Day is given by the arithmetic expression , which is the single entry of the matrix product; in the same way, the total income for Day is given by, the single entry of ; and the total income for Day is given by , the single entry of. Therefore, the matrix representationgives the total income, in dollars, received from the sale of the cameras for each of the three days. Although it is not necessary to compute the matrix product in order to answer the question correctly, equals .17Your answer Omitted!The right circular cone above is sliced horizontally forming two pieces, each of which has the sameheight. What is the ratio of the volume of the smaller piece to the volume of the larger piece?(A)(B)(C)(D)(E)ExplanationDifficulty: HardIt is helpful to label the figure.The top piece is a cone whose height is one-half the height of the original cone . Using the properties of similar right triangles, you should realize the radii of these two cones must be in the same ratio. So if the top cone has radius , the original cone has radius .The volume of the top piece is equal to . The volume of the bottom piece is equal to the volume of the original cone minus the volume of the top piece.The ratio of the volume of the smaller piece to the volume of the larger piece is .18Your answer Omitted!In the figure above, is a regular pentagon with side of length . What is the -coordinate of ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe sum of the measures of the interior angles of a regular pentagon is equal to . Each interior angle has a measure of . Using supplementary angles, has a measure of . You can use right triangle trigonometry to find the -coordinate of point .Since , is about . Since the length of each side of the pentagon is , the -coordinate of point is . Putting the information together tells us that the -coordinate of point is . The correct answer is (B).19Your answer Omitted!For a class test, the mean score was , the median score was , and the standard deviation of the scores was . The teacher decided to add points to each score due to a grading error. Which of the following statements must be true for the new scores?I. The new mean score is .II. The new median score is .III. The new standard deviation of the scores is .(A) None(B) only(C) only(D) and only(E) , , andExplanationDifficulty: HardFor this type of question you need to evaluate each statement separately. Statement is true. If you add to each number in a data set, the mean will also increase by . Statement is also true. The relative position of each score will remain the same. Thus, the new median score will be equal to more than the old median score. Statement is false. Since each new score is more than the old score, the spread of the scores and the position of the scores relative to the mean remain the same. Thus, the standard deviation of the new scores is the same as the standard deviation of the old scores.20Your answer Omitted!A game has two spinners. For the first spinner, the probability of landing on blue is . Independently, for the second spinner, the probability of landing on blue is What is the probability that the first spinner lands on blue and the second spinner does not land on blue?(A)(B)(C)(D)(E)ExplanationDifficulty: HardSince the two events are independent, the probability that the first spinner lands on blue and the second spinner does not land on blue is the product of the two probabilities. The first probability is given. Since the probability that the second spinner lands on blue is the probability that thesecond spinner does not land on blue is Therefore, . The correct answer is (E).21Your answer Omitted!In January the world’s population was billion. Assuming a growth rate of percent per year, the world’s population, in billions, for years after can be modeled by theequation . According to the model, the population growth from January to January was(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is C.According to the model, the world’s population in January was and in January was . Therefore, according to the model, the population growth from January to January , in billions, was , or equivalently,.22Your answer Omitted!What is the measure of one of the larger angles of a parallelogram in the that has vertices with coordinates , , and ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is C.First, note that the angle of the parallelogram with vertex is one of the two larger angles of the parallelogram: Looking at the graph of the parallelogram in the makes this apparent. Alternatively, the sides of the angle of the parallelogram with vertex are a horizontal line segment with endpoints and and a line segment of positive slope with endpoints and that intersects the horizontal line segment at its left endpoint , so the angle must measure more than Since the sum of the measures of the four angles of aparallelogram equals , the angle with vertex must be one of the larger angles.One way to determine the measure of the angle of the parallelogram with vertex is to apply the Law of Cosines to the triangle with vertices , , and . The length of the two sides of the angle with vertex are and; the length of the side opposite the angle is . Let represent the angle with vertex and apply the Law of Cosines: , so. Therefore, the measure of one of the larger angles of the parallelogram is .Another way to determine the measure of the angle of the parallelogram with vertex is to consider the triangle , , and . The measure of the angle of this triangle with vertex is less than the measure of the angle of the parallelogram with vertex . The angle of the triangle has opposite side of length and adjacent side of length , so the measure of this angle is . Therefore, the measure of the angle of the parallelogram withvertex is .Yet another way to determine the measure of the angle of the parallelogram with vertex is to use trigonometric relationships to find the measure of one of the smaller angles, and then use the fact that each pair of a larger and smaller angle is a pair of supplementary angles. Consider the angle of the parallelogram with vertex ; this angle coincides with the angle at vertex of the right triangle with vertices at , , and , with opposite side of lengthand adjacent side of length , so the measure of this angle is . This angle, together with the angle of the parallelogram with vertex , form a pair of interior angles on the same side of a transversal that intersects parallel lines, so the sum of the measures of the pair of angles equals . Therefore, the measure of the angle of the parallelogram with vertex is.23Your answer Omitted!For some real number , the first three terms of an arithmetic sequence are, and . What is the numerical value of the fourth term?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is E.To determine the numerical value of the fourth term, first determine the value of and then apply the common difference.Since , and are the first three terms of an arithmetic sequence, it must be true that, that is, Solving for gives . Substituting for in the expressions of the three first terms of the sequence, one sees that they are , , and , respectively. The common difference between consecutive terms for this arithmetic sequence is , and therefore, the fourth term is .24Your answer Omitted!In a group of people, percent have brown eyes. Two people are to be selected at random from the group. What is the probability that neither person selected will have brown eyes?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is A.One way to determine the probability that neither person selected will have brown eyes is to count both the number of ways to choose two people at random from the people who do not have brown eyes and the number of ways to choose two people at random from all people, and then compute the ratio of those two numbers.Since percent of the people have brown eyes, there are people with brown eyes, and people who do not have brown eyes. The number of ways of choosing two people, neither of whom has brown eyes, is : there are ways to choose a first person and ways to choose a second person, but there are ways in which that same pair of people could be chosen. Similarly, the number of ways of choosing two people at random from the people is . Therefore, the probability that neither of the two people selected has brown eyes is.Another way to determine the probability that neither person selected will have brown eyes is to multiply the probability of choosing one of the people who does not have brown eyes at random from the people times the probability of choosing one of the people who does not have brown eyes at random from the remaining people after one of the people who does not have brown eyes has been chosen.Since percent of the people have brown eyes, the probability of choosing one of the people who does not have brown eyes at random from the people is . If one of the people who does not have brown eyes has been chosen, there remain people who do not have brown eyes out of a total of people; the probability of choosing one of the people who does not have brown eyes at random from the people is . Therefore, if two people are to be selected from the group at random, the probability that neither person selected will have brown eyes is .25Your answer Omitted!If , what is ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is E.One way to determine the value of is to solve the equation for . Since , start with the equation , and cube both sides to get. Isolate to get , and apply the cube root to both sides of the equation to get .Another way to determine the value of is to find a formula for and then evaluate at Let and solve for : cubing both sides gives , so , and. Therefore, , and .26Your answer Omitted!Which of the following equations best models the data in the table above?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is D.One way to determine which of the equations best models the data in the table is to use a calculator that has a statistics mode to compute an exponential regression for the data.The specific steps to be followed depend on the model of calculator, but can be summarized as follows: Enter the statistics mode, edit the list of ordered pairs to include only the four points givenin the table and perform an exponential regression. The coefficients are, approximately, for the constant and for the base, which indicates that the exponential equation is the result of performing the exponential regression. If the calculator reports a correlation, it should be a number that is very close , to which indicates that the data very closely matches the exponential equation. Therefore, of the given models, best fits the data.Alternatively, without using a calculator that has a statistics mode, one can reason about the data given in the table.The data indicates that as increases, increases; thus, options A and B cannot be candidates for such a relationship. Evaluating options C, D and E at shows that option D is the one that gives a value of that is closest to In the same way, evaluating options C, D and E at each of the other given data points shows that option D is a better model for that one data point than either option C or option E. Therefore, is the best of the given models for the data.27Your answer Omitted!The linear regression model above is based on an analysis of nutritional data from 14 varieties of cereal bars to relate the percent of calories from fat to the percent of calories from carbohydrates . Based on this model, which of the following statements must be true?I. There is a positive correlation between and .II. When percent of calories are from fat, the predicted percent of calories from carbohydrates is approximately .III. The slope indicates that as increases by , decreases by .(A) II only(B) I and II only(C) I and III only(D) II and III only(E) I, II, and IIIExplanationDifficulty: HardThe correct answer is D.Statement I is false: Since , high values of are associated with low values of which indicates that there is a negative correlation between and .Statement II is true: When percent of calories are from fat, and the predicted percent of calories from carbohydrates is .Statement III is true: Since the slope of the regression line is , as increases by , increases by ; that is, decreases by .28Your answer Omitted!The number of hours of daylight, , in Hartsville can be modeled by , where is the number of days after March . The day with the greatest number of hours of daylight has how many more daylight hours than May ? (March and May have days each. April and June have days each.)(A) hr(B) hr(C) hr(D) hr(E) hrExplanationDifficulty: HardThe correct answer is A.To determine how many more daylight hours the day with the greatest number of hours of daylight has than May , find the maximum number of daylight hours possible for any day and then subtract from that the number of daylight hours for May .To find the greatest number of daylight hours possible for any day, notice that the expressionis maximized when , which corresponds to , so. However, for this problem, must be a whole number, as it represents a count of days after March . From the shape of the graph of the sine function, either or corresponds to the day with the greatest number of hours of daylight, and since, the expression is maximized when days after March . (It is not required to find the day on which the greatest number of hours of daylight occurs, but it is days after March ,that is, June .)Since May is days after March , the number of hours of daylight for May is .Therefore, the day with the greatest number of hours of daylight hasmore daylight hours than May .。

数学二真题详细答案解析

数学二真题详细答案解析数学是一门抽象而又具有广泛应用的学科，它在现代社会中扮演着重要的角色。

无论是在科学研究，金融交易，还是在日常生活中，我们都会遇到各种与数学有关的问题。

因此，深入理解数学的原理和拓展能力对我们来说至关重要。

下面，我将对数学二真题进行详细解答，以帮助大家更好地掌握数学知识。

题目一：计算题某公司去年的年利润为100万元，今年的年利润比去年增长了20%。

今年的年利润是多少万元？解析：首先，我们需要知道“增长20%”意味着增加了原来的20%，即原数加上原数的1/5。

所以，今年的年利润为100万元 + 100万元/5 = 120万元。

题目二：代数题若函数f(x) = x^2 + 3x - 4, 求f(2)的值。

解析：将x=2代入函数f(x)的表达式中，可以得到f(2) = 2^2 + 3*2 - 4 = 4 + 6 - 4 = 6。

题目三：几何题已知正方形ABCD的边长为a，点E是AD边上一点，连接CE，并延长交BC于点F。

若BE=3，CF=5，求正方形ABCD的边长a。

解析：利用类似三角形的性质，我们可以发现三角形BEC与三角形CFD相似，因此可以得到BE/BC = CE/CF。

根据已知条件，可以得到3/（3+a）= （a+3）/5。

经过化简计算，可以得到a^2 - 11a + 12 = 0，这是一个一元二次方程。

求解方程，可以得到a = 1 或 a = 12。

但由于正方形的边长必须为正数，因此a = 12。

通过以上几道题目，我们可以看到数学在解决实际问题中所起到的重要作用。

通过运用数学的方法和原理，我们可以更好地理解和应对各种问题。

这样的能力不仅可以帮助我们解决日常生活中的困难，也对我们在学术和职业生涯中有所帮助。

在学习数学的过程中，我们还要注意培养抽象思维和逻辑推理能力。

数学不仅仅是一门死记硬背的学科，更是培养思维能力的工具。

在解答问题时，我们需要运用各种数学原理和方法，进行思维的灵活转换。

2023年全国统一高考数学试卷以及答案解析(全国2卷)

2023年全国统一高考数学试卷以及答案
解析(全国2卷)
简介
本文档为2023年全国统一高考数学试卷及答案解析提供了全
国2卷的详细内容。

试卷由相关教育机构编写，并经过严格审核确
保质量。

以下是试卷和答案解析的概要。

试卷内容
试卷分为多个部分，涵盖了数学的各个领域和知识点。

主要的
考查内容包括但不限于：代数、几何、概率与统计、函数与解析几
何等。

试卷设置了不同难度的题目，旨在全面考查学生的数学能力
和应试能力。

答案解析
答案解析部分为每个试题提供了详细的解题方法和步骤。

通过
阅读答案解析，学生能够理解每道题目的解题思路和方法。

答案解
析还包括常见错误的解释和注意事项，帮助学生避免犯同样的错误。

注意事项
1. 本文档提供的试卷及答案解析仅供研究和参考，不可作为学生高考成绩的依据。

2. 学生在参考本文档时应保持独立思考，不应完全依赖答案解析提供的答案。

3. 文档中提供的内容经过审核，但仍有可能存在错误或遗漏，敬请谅解。

结束语
希望本文档能为广大学生提供有价值的研究参考。

祝愿各位同学在2023年全国统一高考中取得优异的成绩！
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SAT II Subject Test Math (Level 2)历年真题及其详解(代数和函数)【圣才出品】

代数和函数真题精选及详解◆代数和函数真题精选及详解（一）1. If 3a=8b and 15b=4c, then a/c= _____.(A) 4/5(B) 8/15(C) 32/45(D) 15/18(E) 11/9【答案】C【解析】Divide the equations: (3a)/(4c)=(8b)/(15b).Multiply both sides by 4/3：a/c=(8/15)×(4/3)=32/45.2. If x2=16, y2=25, and(x-4)(y+5)≠0, then x3+y3=?(A) 125(B) 61(C) 52(D) 20(E) 9【答案】B【解析】If x2=16, then x=4 or -4, and if y2=25, then y=5 or -5. But if (x-4)(y+5)≠0,then x≠4 and y≠-5. Therefore, x3 +y3 =(-4) 3 +53 =61.3. If f(x)=3x-5, and if the domain of x consists of all real numbers defined by the inequality -3<x<5, then the range of f(x) contains all of the following members except：(A) 0(B) 5(C) 10(D) -5(E)【答案】C【解析】f(-3)=3(-3)-5=-14. f(5)=3(5)-5=10. So -14<f(x)<10, or R={x|-14<x<10}.4. If g is a linear function, and if g(3)=4 and g(5)=2, then g(-3)= _____.(A) -36(B) -15(C) 81(D) 10(E) 12【答案】D【解析】The linear function g includes the two functional pairs(3, 4)and(5, 2), so its slope is m=(2-4)/(5-3)=1, then the equation of the line is y-2=-1(x-5), org(x)=y=-x+7. Hence, g(-3)=-(-3)+7=10.5. Let the function g be defined by g = {(-2, 4), (-1, 1), (0, 6), (1, -1), (2, 8)}. Which statement is true?(A) g(1) = g(-1)(B) g(2)+g(-2) = 0(C) g(-2) = (1/2)g(2)(D) g(0) = g(1)+g(-1)(E) g(g(-1)) = g(g(1))【答案】C【解析】For each ordered pair included in the set of ordered pairs that defines function g, the first member is x and the second member is g(x). For example, g(-2) = 4, since (-2,4) belongs to the function g. Check the statement in each choice until you find the one that is true:~ Choice (A): g(1) = -1 and g(-1) = 1. Hence, g(1) ≠ g(-1).~ Choice (B): g(2) = 8 and g(-2) = 4. Hence, g(2)+g(-2) ≠ 0.~ Choice (C):g(-2) = 4 and 1/2g(2) = 1/2(8)= 4. Hence, g(-2) = 1/2g(2). This is the correct choice.6. If the function k is defined by k(h) = (h + 1)2, then k(x-2) = _____.(A) x2-x(B) x2-2x(C) x 2-2x + 1(D) x 2 + 2x- 1(E) x 2-1【答案】C 【解析】If k(h) = (h+1)2, then k(x-2) = ((x-2)+1)2= (x-1)2= (x-1)(x-1) =x 2 -2x+1.7. If 22155x x -=, then _____. I. x=0II. x=1III. x = -1(A) I only(B) II only(C) I and II(D) I and III(E) II and III【答案】B【解析】Since 2211555x x --==, x 2-2x = -1. Then, x 2-2x +1 = 0, so (x - 1)(x - 1)= 0, which means x = 1 is the only root. Hence, only Roman numerical choice II is correct.8. Rosa has 35 coins that are worth a total of $2. 50. If the coins are all nickels and dimes, how many more nickels than dimes does she have?(A) 25(B) 20(C) 15(D) 10 (E) 5 【答案】E 【解析】Let x the number of nickels, then 35-x is the number of dimes. 5x+10(35-x)=250. x=20.9.If the broken curve above is the graph of the function ()f x x =, then the solid curve could be the graph of the function _____.(A) ()4g x x =+(B) ()4g x x =-(C) ()4g x x =-(D) ()4g x x =+ (E) ()4g x x =【答案】D 【解析】The solid graph represents the vertical shift of the broken graph 4 units vertically up so that the y-coordinate of each point on the solid graph is 4 more than the y-coordinate of its corresponding point on the broken graph. Hence, if the broken graph is the graph of ()f x x =, then the solid graph must be the graph of ()4g x x =+.10. The weekly profit for a magazine company is given by the function P(x ) = 3/2x -300, where x represents the number of magazine sold. What is the weekly profit if 400 magazines are sold?(A) $100。

SAT2数学level2经典考题讲解

解析：这道题目最本质的考点是：syntheticdivision(综合除法)。

但是对于考生而言，首先要理解的关键数学术语是zero，零点。

什么是零点呢?零点是使得方程的因变量(dependent variable, 即f(x))等于0 (即：图像与x轴相交)的x的值。

以下给出这个方程的图像，从中我们要找出蓝色的曲线和X轴(横轴)相交的三个点的x坐标值。

就本题而言，既然题目说-1是一个零点，那么就将x=-1带入方程，会发现f(x)确实等于0.那么，另外的使方程等于0的x的值应该怎么求得呢?方法1：画图求解，(考场上实际操作可能性为0)。

如上图所示，三个让方程与横轴相交的点的x值分别为-3.5，-1，3。

所以，-1给出来的情况下，另外两个点应该选择C。

但是，考场上根本不可能有时间和工具供我们画图，因此这个方法不是实际操作可行的方法，不予考虑。

方法2：代入法既然是选择题，就可以用带入法求解。

带入法求解的时候，要遵循以下两个原则：(1)带入常见的数值来检验是否正确;(2)在常见数值中带入易于计算的数值来看是否正确。

按照这两个原则，在5个选项中首先带入1来运算，因为好计算，还因为出现了3次。

当x=1时候，f(x)=-36，显然不是0，因此有1的ADE都不是答案。

下一步，在剩下的BC两个选项中，要检验的是-3，因为两项都有3，说明3一定是答案，而-3和-7/2中，-3更方便于计算，于是决定用-3来检验。

当x=-3时候，f(x)=12，显然不是0，因此答案不是B而是C.由此可见，代入法是一个smart strategy，是带有一点小聪明的策略。

在代入法的过程中耗时的长短体现了考生的判断和逻辑分析。

但是，即使可以凭借这种方法选择答案，这也不。

2023年全国卷Ⅱ数学试题及其参考答案

2023年全国卷Ⅱ数学试题及其参考答案第一题试题请计算下列数列的前n项和：\[a_n = 3n + 2\]参考答案数列$\{a_n\}$的前n项和可表示为：\[S_n = \sum_{k=1}^{n} a_k = \frac{n}{2}(a_1 + a_n)\]代入数列$\{a_n\}$的表达式，得：\[S_n = \frac{n}{2}(2a_1 + (n-1)d)\]其中，$a_1$为数列的首项，$d$为数列的公差。

根据题目中给定的数列，首项$a_1 = 5$，公差$d = 3$。

将这些值代入上述公式，可得数列$\{a_n\}$的前n项和的表达式为：\[S_n = \frac{n}{2}(5 + 3(n-1))\]第二题试题已知正整数$n$满足条件：$\log_{n}{2} + \log_{2}{n} = 2$，求$n$的值。

参考答案根据题目中给定的条件，我们可以转化为指数方程：\[\left(\frac{1}{n}\right)^2 + 2^{\log_{2}{n}} = 2\]化简可得：\[\frac{1}{n^2} + n = 2\]将该方程转化为二次方程：\[n^3 - 2n^2 + n - 2 = 0\]通过试算，我们发现$n = 2$是该方程的一个解。

通过带入韦达定理可知该方程有且仅有一个正整数解。

所以，$n = 2$。

第三题试题已知函数$f(x) = x^3 + x^2$，求函数$g(x) = f^{-1}(x)$的表达式。

参考答案对于函数$f(x) = x^3 + x^2$，我们需要求其反函数$f^{-1}(x)$的表达式。

首先，我们令$y = f(x)$，得：$y = x^3 + x^2$将该方程转化为关于$x$的二次方程：$x^3 + x^2 - y = 0$通过试算，我们发现$x = -1$是该方程的一个根。

普通高等学校招生全国统一考试(课标Ⅱ卷)理数答案解析(正式版)(解析版).docx

一、选择题：本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要求的。

1．已知集合21,01,2A =--｛，，｝，{}(1)(20B x x x =-+<,则A B =I （）A ．{}1,0A =-B ．{}0,1C ．{}1,0,1-D ．{}0,1,2 【答案】A考点：集合的运算．2．若a 为实数且(2)(2)4ai a i i +-=-，则a =（） A ．1- B ．0 C ．1 D ．2 【答案】B 【解析】试题分析：由已知得24(4)4a a i i +-=-，所以240,44a a =-=-，解得0a =，故选B ．考点：复数的运算．3．根据下面给出的2004年至2013年我国二氧化硫排放量（单位：万吨）柱形图。

以下结论不正确的是( )A ．逐年比较，2008年减少二氧化硫排放量的效果最显著B ．2007年我国治理二氧化硫排放显现C ．2006年以来我国二氧化硫年排放量呈减少趋势D ．2006年以来我国二氧化硫年排放量与年份正相关【答案】D 【解析】试题分析：由柱形图得，从2006年以来，我国二氧化硫排放量呈下降趋势，故年排放量与年份负相关，故选D ．2004年 2005年 2006年 2007年 2008年 2009年 2010年 2011年 2012年 2013年考点：正、负相关．4．等比数列｛a n ｝满足a 1=3，135a a a ++ =21，则357a a a ++= ( )A ．21B ．42C ．63D ．84 【答案】B考点：等比数列通项公式和性质．5．设函数211log (2),1,()2,1,x x x f x x -+-<⎧=⎨≥⎩,2(2)(log 12)f f -+=( )A ．3B ．6C ．9D ．12 【答案】C 【解析】试题分析：由已知得2(2)1log 43f -=+=，又2log 121>，所以22log 121log 62(log 12)226f -===，故2(2)(log 12)9f f -+=，故选C ．考点：分段函数．6．一个正方体被一个平面截去一部分后，剩余部分的三视图如右图，则截去部分体积与剩余部分体积的比值为( ) A ．81 B ．71 C ．61 D ．51【答案】D 【解析】试题分析：由三视图得，在正方体1111ABCD A B C D -中，截去四面体111A A B D -，如图所示，，设正方体棱长为a ，则11133111326A A B D V a a -=⨯=，故剩余几何体体积为3331566a a a -=，所以截去部分体积与剩余部分体积的比值为51，故选D ．考点：三视图．A1A7．过三点(1,3)A ，(4,2)B ，(1,7)C -的圆交y 轴于M ，N 两点，则||MN =( ) A ．26 B ．8 C ．46 D ．10 【答案】C【解析】由已知得321143AB k -==--，27341CB k +==--，所以1AB CB k k =-，所以AB CB ⊥，即ABC ∆为直角三角形，其外接圆圆心为(1,2)-，半径为5，所以外接圆方程为22(1)(2)25x y -++=，令0x =，得2y =±，所以MN =C ．考点：圆的方程．8．右边程序框图的算法思路源于我国古代数学名著《九章算术》中的“更相减损术”．执行该程序框图，若输入,a b 分别为14,18，则输出的a =( )【答案】B 【解析】试题分析：程序在执行过程中，a ，b 的值依次为14a =，18b =；4b =；10a =；6a =；2a =；2b =，此时2a b ==程序结束，输出a 的值为2，故选B ．考点：程序框图．9．已知A,B 是球O 的球面上两点，∠AOB=90,C 为该球面上的动点，若三棱锥O-ABC 体积的最大值为36，则球O 的表面积为( )A ．36π B.64π C.144π D.256π【答案】C 【解析】试题分析：如图所示，当点C 位于垂直于面AOB 的直径端点时，三棱锥O ABC -的体积最大，设球O 的半径为R ，此时2311136326O ABC C AOB V V R R R --==⨯⨯==，故6R =，则球O 的表面积为24144S R ππ==，故选C ．考点：外接球表面积和椎体的体积．10．如图，长方形ABCD 的边2AB =，1BC =，O 是AB 的中点，点P 沿着边BC ，CD 与DA 运动，记BOP x ∠=．将动P 到A 、B 两点距离之和表示为x 的函数()f x ，则()y f x =的图像大致为（）【答案】B 【解析】DPCBOAx考点：函数的图象和性质．11．已知A ，B 为双曲线E 的左，右顶点，点M 在E 上，∆ABM 为等腰三角形，且顶角为120°，则E 的离心率为（）A 5．2 C 32 【答案】D 【解析】试题分析：设双曲线方程为22221(0,0)x y a b a b-=>>，如图所示，AB BM =，0120ABM ∠=，过点M作MN x ⊥轴，垂足为N ，在Rt BMN ∆中，BN a =，3MN a =，故点M 的坐标为(23)M a a ，代入双曲线方程得2222a b a c ==-，即222c a =，所以2e =D ．考点：双曲线的标准方程和简单几何性质．12．设函数'()f x 是奇函数()()f x x R ∈的导函数，(1)0f -=，当0x >时，'()()0xf x f x -<，则使得()0f x >成立的x 的取值范围是（）A ．(,1)(0,1)-∞-UB ．(1,0)(1,)-+∞UC ．(,1)(1,0)-∞--UD ．(0,1)(1,)+∞U【答案】A 【解析】试题分析：记函数()()f x g x x=，则''2()()()xf x f x g x x -=，因为当0x >时，'()()0xf x f x -<，故当0x >时，'()0g x <，所以()g x 在(0,)+∞单调递减；又因为函数()()f x x R ∈是奇函数，故函数()g x 是偶函数，所以()g x 在(,0)-∞单调递减，且(1)(1)0g g -==．当01x <<时，()0g x >，则()0f x >；当1x <-时，()0g x <，则()0f x >，综上所述，使得()0f x >成立的x 的取值范围是(,1)(0,1)-∞-U ，故选A ．考点：导数的应用、函数的图象与性质．第II 卷（非选择题，共90分）本卷包括必考题和选考题两部分。

数学二真题及答案解析

数学二真题及答案解析在高考数学中，数学二是一项非常重要的科目。

它包括了数学的许多重要概念和方法，如函数、微积分、概率与统计等。

因此，对于考生来说，掌握数学二的知识和技巧是非常关键的。

本文将对数学二的真题及其答案进行解析和讨论，希望能够帮助考生更好地准备和应对数学二的考试。

一、函数与图像在数学二的考试中，函数与图像是一个重要的考点。

考生需要掌握函数的定义和性质，了解不同类型函数的特点和变化规律。

同时，他们还需要能够通过函数的图像进行解题，如确定函数的极值、最值等。

下面，我们来看一道典型的函数与图像题目。

题目：已知函数f(x) = x^3 - 3x^2 + 3x - 1，求函数f(x)的最小值。

解析：要求函数的最小值，我们可以使用求导方法来解决这个问题。

首先，对函数f(x)求导得到f'(x) = 3x^2 - 6x + 3。

然后，令f'(x) = 0，即可求出极值点。

解方程3x^2 - 6x + 3 = 0得到x = 1，将x = 1代入原函数f(x)中，得到f(1) = -1。

因此，函数f(x)的最小值为-1。

通过这道题目，我们可以看出数学二考试中，函数与图像的题目既需要运用基本的概念和性质，又需要运用一定的计算和推理能力。

二、微积分微积分是数学二考试中的重要部分。

它涉及到函数的导数、积分、微分方程等内容。

掌握微积分的原理和应用是考生取得高分的关键。

下面，我们来看一个与微积分相关的题目。

题目：已知函数f(x) = sin(x)，求函数f(x)在区间[0,π]上的定积分值。

解析：要求函数在一个区间上的定积分值，我们可以先求出函数积分的原函数，然后根据定积分的性质进行计算。

对于函数f(x) = sin(x)，它的原函数是F(x) = -cos(x)。

因此，函数f(x)在区间[0,π]上的定积分值为F(π) - F(0) = -cos(π) - (-cos(0)) = -(-1) - (-1) = 2。

全国2卷真题及答案解析

全国2卷真题及答案解析一、数学部分第一题: 计算题该题是关于三角函数的计算题。

根据给定的角度和三角函数的运算规则，我们可以轻松地计算出答案。

第二题: 证明题该题是一道证明题，要求使用等式推导的方法证明一个数学定理。

我们可以先把已知条件列出来，然后运用一系列等式推导，最终得出结论。

第三题: 应用题该题是一个应用题，涉及到数学在实际问题中的应用。

我们需要根据题目中给定的条件和要求，运用数学知识进行分析和计算，得出最终答案。

二、语文部分第一题: 阅读理解该题是一篇阅读理解题，需要阅读一篇文章，并回答一系列问题。

我们可以先通读文章，理解文章的主旨和要点，然后再根据问题找出相关的句子和段落，进行逐个解答。

第二题: 短文改错该题是一个短文改错题，要求根据给定的短文，找出并改正其中的错误。

我们需要仔细阅读短文，理解其内容和语法结构，然后逐个找出错误，并进行修改。

第三题: 作文题该题是一个作文题，要求根据提供的素材和要求，写一篇文章。

我们可以先分析提供的素材，理清要求和主题，然后构思文章的结构和内容，进行逐步写作。

三、英语部分第一题: 完形填空该题是一篇完形填空题，需要填入适当的词语，使得短文的意思通顺连贯。

我们可以先通读短文，理解上下文的意思，然后根据语法和语义的要求，选择适当的词语填空。

第二题: 阅读理解该题是一篇阅读理解题，需要阅读一篇文章，并回答一系列问题。

我们可以先快速浏览文章，了解其大意和结构，然后再细读文章，找出与问题相关的句子和段落，进行逐个解答。

第三题: 书面表达该题是一个书面表达题，要求根据给定的要求和提示，写一篇文章。

我们可以先理清要求和提示的意思，然后构思文章的结构和内容，进行逐步写作。

四、物理部分第一题: 计算题该题是一道物理计算题，要求根据给定的物理量和公式，计算出答案。

我们可以先列出已知条件和要求，然后根据物理公式和运算规则，进行逐步计算，得出最终结果。

第二题: 解答题该题是一个物理解答题，要求回答一系列问题，解释物理现象和原理。

2021年普通高等学校招生全国统一考试(全国新课标Ⅱ卷)数学试题(理科)解析版

新课标II 卷数学试卷（理科）第I 卷一、选择题（本大题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一个选项是符合题目要求的）1. 设集合M ={0,1,2}，20{|32}N x x x =-≤+，则M N ⋂=( )A. {1}B. {2}C. {0,1}D. {1,2} 【答案解析】D解析：把0，1，2代人2203x x +≥-验证，只有1，2满足不等式，故选D. 考点：考查集合与一元二次不等式的知识，简单题.2. 设复数12,z z 在复平面内的对应点关于虚轴对称，12z i =+，则12z z =（）A. -5B.5C.-4+iD.-4-i 【答案解析】A. 解析：12i z =+ 与2z 关于虚轴对称，∴2z =-2+i∴12(2i)(2i)5z z =+-+=- ，故选A. 考点：考查复数的基本知识，简单题. 3. 设向量,a b 满足10a b +=，6a b -=，则a b •=（）A. 1B.2C. 3D.5 【答案解析】A. 解析：||10,6|4=41=+=-=∴+⋅+⋅+∴⋅∴⋅=-=2222a b a b a 2a b b a 2a b b a b a b 故选A.考点：考查平面向量的数量积，中等题. 4. 钝角三角形ABC 的面积是12，AB =1，BC =AC =（）A.5 【答案解析】B. 解析：∵△ABC 面积为12，1,AB BC ==∴111sin 45,13522B B B ⋅=⇒=⇒=︒︒当B=45°时，222cos 452122212112BC A AC AB BC C AB ⋅︒=+-⋅⋅⋅=⇒=-=+此时，AC=AB=1,故A=90°，这与△ABC 为钝角三角形矛盾.当B=135°时，222cos135212212525AC AB B BC C A AC B =+-⋅︒=++⋅⋅⋅=⇒= 故选B.考点：考查正余弦定理的应用，中等题.A. 0.8B.0.75C. 0.6D.0.45 【答案解析】A.解析：设第i 天空气优良记着事件i A ，则1(A )0.75,(A A )0.6(i 1,2,)i i i P P +=== ,∴第1天空气优良，第2天空气也优良这个事件的概率为12211()0.60.8((|).75)0A A P A A P P A ===，故选A.考点：考查条件概率的概率，简单题.6. 如图，网格纸上正方形小格子的边长为1(表示1cm)，图中粗线画出的是某零件的三视图，该零件由一个底面半径为3cm ，高为6cm 的圆柱体毛胚切削而得到，则切削掉部分的体积与原来毛胚体积的比值为（）A.1727 B. 59 C. 1027 D.13【答案解析】C.解析：毛胚的体积23654V ππ⋅⋅==制成品的体积 221322434V πππ⋅⋅+⋅⋅==∴切削掉的体积与毛胚体积之比为：13454101127V V ππ-=-=，故选C. 考点：考查三视图于空间几何体的体积，中等题.7. 执行右图的程序框图，如果输入的x ，t 均为2，则输出的S =( )A.4B.5C.6D.7 【答案解析】D.解析：第1次循环M=2,S=5,k=1第2次循环，M=2,S=7,k=2第3次循环k=3>2,故输出S=7考点：考查算法的基本知识，简单题.8. 设曲线y=ax-ln(x+1)在点(0,0)处的切线方程为y=2x，则a=（）A.0B.1C.2D.3【答案解析】D.解析：0ln(121)1,1|xy ax xy axy a==-+∴'=-'=∴-+=故a=3，选D.考点：考查导数的几何应用，中等题.9. 设x，y满足约束条件3103507x yxx yy≤⎧⎪-+≤⎨⎪--≥-⎩+，则z=2x-y的最大值为( )A.10B.8C.3D.2【答案解析】B.解析：作图即可.考点：考查二元一次不等式组的应用，中等题.10. 设F为抛物线23C y x=：的焦点，过F且倾斜角为30°的直线交C于A，B两点，O为坐标原点，则△OAB 的面积为（） A.33 B. 93 C. 6332 D.94【答案解析】D 解析：∵23y x =∴抛物线C 的焦点的坐标为：()3,04F 所以直线AB 的方程为：330an )t (4y x ︒-=故233()43x y y x ⎧==-⎪⎨⎪⎩从而2122161689012x x x x -+=+=⇒ ∴弦长12||=3122x x AB ++= 又∵O 点到直线43:430AB x y --= 的距离2238(4=3)4d =+ ∴13129428OAB S ⋅⋅==,故选D. 考点：综合考查抛物线的知识，弦长计算与分析直线和圆锥曲线位置关系的能力，难度为困难题.11. 直三棱柱111ABC A B C -中，∠BCA=90°，M ，N 分别是11A B ，11AC 的中点，1BC CA CC ==，则BM 与AN 的夹角的余弦值为( ) A.110 B. 25C. 30D.2【答案解析】C.解析：设AC =2，12BC CA CC ∴===(2,0,0),(1,0,2),(0,2,0),(1,1,2)A N B M ∴ (1,1,2),(1,2,0)BM AN ∴=-=-30cos ,||||065013AN BM AN BM AN BM ⋅∴<>===-=-⋅⋅ 故选C. 考点：考查空间夹角问题.中等题.12. 设函数)n(f x xmπ=，若存在f (x )的极值点0x 满足22200[()]x f x m +<，则m 的取值范围是（）A.,6()(6,)∞-⋃+∞-B. ,4()(4,)∞-⋃+∞-C. ,2()(2,)∞-⋃+∞-D.,1()(1,)∞-⋃+∞- 【答案解析】C.解析：()()cosxxf x f m mx mππ=⇒'=令()0cos0()2xxx k k Z mf mππππ'=⇒=⇒=+∈(21)2k m x +=∴ ，即f (x )的极值点0(2)1()2mx k k Z =+∈ ∵存在f (x )的极值点0x ，满足22200()f x x m +<∴ 2220(2)m 31sin []2x k m m π+<+ 又∵222202sin )(21)m sin s s in in (222()1k k x k m m ππππππ===++⋅=+∴存在k Z ∈，使得221[]2(2)m 3k m ++<∴存在k Z ∈,使得223(2)141k m +<-∴223(21)13[1]|m |24414max k m +<-=⇒->= ，故选C. 考点：考查导数与极值，三角函数，不等式的知识，为困难题.第II 卷二、填空题（本大题共4小题，每小题5分，共20分）13. 10()x a +的展开式中，7x 的系数为15，则a = .（用数字填写答案）【答案解析】12解析：1010110(0(),1,,10)r r rr T C x r x a a -+==+展开式的通项为∴10()x a +展开式中7x 的系数为31031125C a a ⇒== 考点：考查二项展开式的通项公式，简单题.14. 函数()sin(2)2sin cos()x x f x ϕϕϕ=+-+的最大值为 .【答案解析】1．解析：()(2)2()cos()sin 2sin cos()cos()si ()sin()cos sin()cos n sin .f x sin x sin cos x x f x x x x x x ϕϕϕϕϕϕϕϕϕϕϕϕϕ=+-+∴++-++=-+==+故填写1.考点：本题考查和差角公式，为中等题.15. 已知偶函数f (x )在[0,+∞)上单调递减，f (2)=0,若f (x -1)>0,则x 的取值范围是 . 【答案解析】(-1,3).解析：作出函数f (x )的示意图，如图所示因为(1213)201x x f x ⇒-<-<⇒--<<> 考点：本题考查函数的单调性与奇偶性.简单题.16. 设点0(,1)M x ，若在园22:1O x y +=上存在点N ，使得∠OMN =45°，则0x 的取值范围是 . 【答案解析】[-1,1]解析：设N 点的坐标为,s (cos )in θθ （1）当00,1x ≠± 时 ∵0(,1)M x 点的坐标为 ∴OM ,MN 的斜率分别为：001s n c s ,i o 1OM MN k x k x θθ-==- ∵45OMN ∠=︒ ∴1tan 45()1MN OMMN OM MN OM MN OMk k k k k k k k -︒=±⇒=-++±即000011sin 1()11sin cos cos ()x x x x θθθθ--±-=--+⋅*取正号时，化简（*）式得：2000(1)sin 11()cos x x x θθ+-=++ 取负号化简（*）式得：2000(1)sin 1(1)cos x x x θθ++=+-∴2220000(1)(1)sin()1x x x θϕ++-+=+ ∴222400000(1)(1)11||1x x x x x +-≥+⇒≤⇒≤+故0||<1x 且00x ≠（2）当00x =时，取(1,0)N ,此时满足题设. （3）当01x =±时，取(0,1)N ，此时也满足题设. 综上所述，011x -≤≤考点：考查应用斜率与倾斜角的概念，直线方程，园的方程,分析问题的能力.困难题. 三、解答题（本大题共8小题） 17. (12分)已知数列{}n a 满足111,31n n a a a +==+.(I)证明{12}n a +是等比数列，并求{}n a 的通项公式； (II)证明2111132n a a a +++<. 【答案解析】解析：(I)∵131n n a a +=+11331111)223(22n n n n a a a a ++∴⇒+=+++=+ 1112132a a =+⇒=∴{12}n a +是首项为32 ，公比为3的等比数列∴1*131333,2222n n n n n a a n N --⋅+==∈=⇒ (II)由(I)知，*13,2n n a n N -=∈，故 121213111111231(13)n n a a a +++=++-+-- 12110331112()3333n n --+-≤+-+12111()11131331(1()).133323213nn n --=++++==⋅-<- 考点：考查等比数列的通项公式,求和公式，考查放缩法证明不等式的技巧.中等题. 18. (12分)如图，四棱锥P-ABCD 中，底面ABCD 为矩形，D A BC P A ⊥平面，E 为PD 中点.(I)证明:PB ||平面AEC ;(II)设二面角D-AE-C 为60°，AP =1，3AD =，求三棱锥E-ACD 的体积.【答案解析】解析：(I)连接EF ,因为四边形ABCD 是矩形，故F 为AC 中点，又因为E 为PD 中点，故EF 是△PBD 的中位线，从而||EF PB ,故||.PB AEC 面(II)建立坐标系如图所示.因为1,3AP AD == ，E 为PD 中点，∴313,(0,0,1),(0,)0(0,)2,,P D E ，设||CD a = ，则(,0,0),(,3,0)B a C a∴31(0,,),(,3,0)22AE AC a == ∵PA ABCD ⊥面，ABCD 平面是矩形∴(,0,0)PAD AB AB a ⊥⇒=面是平面ADE 的法向量设平面AEC 的法向量为=(,,)n x y z ，则3102302n AE y z n AC ax y ⎧⋅=+=⎪⎨⎪⋅=+=⎩令3y = ，得,33x a z =-=- ，故3(,3,3)n a=-- ∵二面角D AE C --的大小为60° ∴22||||991239cos60||n AB n AB aa a⋅︒===⋅+⋅++解得32a =∵三棱锥E ACD -的高为111||1222PA =⋅= ∴11111313(||||)(||)(3)3223222E ACD AD CD P V A -⋅⋅⋅⋅=⋅⋅⋅⋅==考点：考查空间线面关系，椎体的体积计算和向量法解决立体几何问题的技能，中等题.19. (12分)某地区2007年至2013年农村居民家庭人均纯收入y (单位：千元)的数据如下表：(I)求y 关于t 的线性回归方程；(II)利用(I)中的回归方程，分析2007年至2013年该地区农村居民家庭人均纯收入的变化情况，并预测该地区2015年农村居民人均纯收入. 附：回归直线的斜率和截距的最小二乘估计公式分别为：2()()()ˆniiini it y t t y bt =---∑∑，ˆˆab y t =-. 【答案解析】解析：（I ）7117221ˆ0.5.28ˆˆ 4.30.542()()14(.)3iii i i t t y bt t y b y at ==∑====∑=-=-⋅--=-∑∑∴回归方程为：0.5 2.3y t =+(II)由于ˆ0.50b=> ，故y 与t 是正线性相关的，因此从2007年到2013年农村居民的人均纯收入是逐年上升的.当9t =时，9 2.3 6.80.5y ⋅+==，即2015年农村居民的人均纯收入预测将达到6.8千元.考点：考查线性回归方程，线性相关的概念的应用.难度中等. 20. (12分)设12,F F 分别是椭圆22221(0):x y C a a b b+=>>的左右焦点，M 是C 上一点且2MF 与x轴垂直，直线1MF 与C 的另一个交点是N . (I)若直线MN 的斜率为34，求C 的离心率; (II)若直线MN 在y 轴上的截距为2，且1|MN |5||F N =，求a ，b .【答案解析】解析：（I ）∵2MF x ⊥轴（不妨设M 在x 轴的上方）∴M 的坐标满足方程组222221(,)x b M c a a y bx c⎧⎪⇒⎨⎪⎩=+= ∵MN 的斜率为34∴2234322b a ac cb =⇒= ∵222222()3a c a a c c b =-⇒-=又∵222(1)32320ce e e e e a⇒+-⇒-=== ∴椭圆离心率为12e = .(II)∵MN 在y 轴上的截距为2，O 为12,F F 的中点 ∴M 的坐标为(c ,4)(不妨设M 在x 轴的上方)由（I ）得24b a= (*)∵1||5||MN NF = ∴11||4||MF NF =作1NF x ⊥轴于T ,由于112~TF F N MF ,故有24,4M N Ny cy c x =--=- ∴321,14N M N y y c x =-=-=- ，即,3()12c N --把N 点的坐标代人椭圆方程得：2221419c a b+=∴2222222)111(9(9544**)4a b b a b a b +=⇒-=- 把（*）与（**）联立得：772a b ==⎧⎪⎨⎪⎩考点：考查椭圆的几何性质以及直线与椭圆的位置关系，难题.21. (12分)已知函数()e 2x xf x e x -=--.(I)讨论f(x)的单调性;(II)设()(2)4()g x f x bf x =-，当0x >时，()0g x >，求b 的最大值;(III)已知1.4142 1.4143<<，估计ln2的近似值(精确到0.001).【答案解析】解析：(I)∵22()x e x f e x ---=∴2()2x x x f e e -'=≥+=-∴()f x 在R 上递增.（II ）()2x x e e f x x --=-224(2)x x x x f e e ---∴=22()(2)4()()4b(4)2x x x x g x f x bf x e x e x e e --∴=-=-----22()()4b()0((0,)4)2x x x x g x e e e x x e x ----∴=-->∀+∞-∈（注意这里用分离变量法处里恒成立无法进行下去!）∵22()0,(2)22420x x x x f e x e f x e e --'=≥'=++-≥-()(0)0,(2)(0)0f x f f x f ∴>=>=又∵22()2244(2)x x x x g e b e e x e --+--+-'=2()24(81(()))x x x x g x e e e b e b --+-++∴-'=2[()2(1[])2])(x x x x e b e e e --⋅=-+-+-令()02(11)x x x g e b x e e b -+=-=-'=⇒⇒（1）当1b ≤时2(10()0)x x e e b g x -≥⇒'+-≥-∴()(0)0g x g >= 成立,故1b ≤成立.(2)当12b <≤时),2(201b <-≤ 而2x x e e -+≥ ，此时22(1)x x b e e -≥≥-+∴()0()(0)0x x g g g '≥⇒>= 成立，故12b <≤（3）当2b >时2221((1)12)b b b b b b =-+>--⇒->2222(2)4422(2)2(2)2b b b b b b b b b b b -⇒--=-+<--<-=-∴ (2)1ln(1(2))001b b b b b b -<⇒-----<<又∵(2)01(2)12b b b b b b ->⇒-+-+>-∴(l 21)0)n(b b b --+>若x R ∈时， 2(1))0(0x x g e b x e -+-->⇔>'ln(1(2))x b b b ⇔<--- 或ln(1(2))b x b b >-+-2(1))0(0x x g e b x e -+--<⇔<'ln(1(2ln(1)))(2)b b b x b b b ⇔---<<--+∴区间(ln(1(2)),ln()())21b b b b b b --+---是()g x 的减区间∵(0)0g =∴ (ln(1)0(2)b b g b --+<即在区间ln(1)02))(,(b b b -+-上()0g x < 这与()g x 在区间(0,)+∞上大于0矛盾，故(2,)b ∈+∞/综上所述，(,2]b ∈-∞ ，故2max b = .(III)由（I ）得，当0x >时，()(0)0f x f >= ，由(II)得，当2b ≤时，(2)4()00()0f x bf x x g x >⇒->⇒> ，从而(2)8()0f x f x ->（注：由于题目给出了2 的近似值，该怎么取x 的值代人是显然的）令1n 2l x = ，则1111221111(2)8()48(2)x x x x f x f x x x e e e e -->-->--⇒∴1232ln 28(2ln 2)ln 2282212-->--⇒>- 由（II ）的证明过程（3）知道，当b >2时，在区间ln(1]02))[,(b b b -+-上()0g x <. 若32ln(1(2))224ln 101b b b b ≤≤-+-⇒+≤≤+，令3214b =+，由()0g x <，得2222221324(21)(()4ln 2l 24n )0ln ln ln ln e g x e e e --=-----+< ∴34ln 2282(23)21822+<++= ∴ 8ln 2212822318<<-+ 下面进行误差估计：∵1.41422 1.4143<<∴23 1.414230.6928121212888.316-⋅->== ∴18219.412828430.6934+<= ∴符合精度要求的值为ln 20.693(0.001)≈精确到 .考点：本题考查利用导数研究函数性质的能力，考查分类讨论的能力及误差估计的思想，思路背景为常规思路，构建函数()g x 的图像即可，难度压轴题.22. (10分)选修4-1：几何证明选讲如图，P 是O 外一点，PA 是切线，A 为切点，割线PBC 与O 相交于点B ，C ，PC =2PA ，D 为PC 中点，AD 的延长线交O 于点E ，证明：(I) BE = EC(II) 22DE B AD P ⋅=【答案解析】解析：(I)连接OA ,OD 交BC 于F ,设PAD α∠=，因PA 是O 的切线，则90-EAO OEA α∠=∠=︒∵2,2PC PA PC PD ==∴P A D P PD A ⇒=是等腰三角形∴ PDA EDF α∠=∠=∵(90)90EDF OEA αα∠+∠=+︒-=︒∴OE BC ⊥故OE 平分弧BC ,从而BE = EC.(II)∵2,2PC PC PA D PB P ⋅==∴22PA PB PD ⋅=由（I ）知PD PA =∴222PA PA PB PB PA ⋅⇒==∴()()DE BD DC BD PA PD PB PA A PA D PA PB ⋅=⋅=⋅=-⋅=-⋅2()PA PB PC PA PB PC PA PA PB PB ⋅=⋅-⋅=⋅-=-()PC PD PB DC PB PA PB ⋅-=⋅=⋅=把2PA PB =代人上式，得222PA PB B P PB P B ⋅=⋅=∴22DE B AD P ⋅=考点：考查与园有关的角的知识和圆幂定理的应用.难度中等.23. (10分)选修4-4：坐标系与参数方程在直角坐标系xOy 中，以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，半圆C 的极坐标方程为 2cos ,[0,]2πρθθ=∈. (I)求C 的参数方程(II)设点D 在C 上，C 在D 处的切线与直线2:3l y x =+垂直，根据(I)中你得到的参数方程，确定D 的坐标.【答案解析】解析：（I ）∵极坐标方程为2cos ,[0,]2πρθθ=∈∴22cos ρρθ= ∴对应的普通方程为：220()02x y x y =≥+- ，即22(01)1()x y y -+=≥∴对应的参数方程为[0,]sin 1cos ,x y ϕϕπϕ⎧∈=+⎨=⎩(II)设半圆的圆心为A ，则A (1,0),又由（I ）知，可以设D 点坐标为(1cos n ),si ϕϕ+ ∴直线DA 的斜率tan k ϕ=∵切线与直线32y x =+垂直∴tan 3=3([0,])πϕϕϕπ⇒=∈∴3,sin 231cos 2ϕϕ==+ 即D 点坐标为3(3,22) 考点：本题考查园的极坐标方程参数方程以及参数方程的简单应用，难度中等题.24. (10分)选修4-5：不等式选讲设函数()||||()10a f x x x a a =++->.(I)证明：()2;f x ≥(II)若(3)5f <，求a 的取值范围.【答案解析】解析：（I ）∵()||||()10a f x x x a a =++-> ∴1111,2x ,(12),a aa a x f x aa a x a x x aa ⎧⎪⎪⎪+-≤≤⎨-+-<-=⎪⎪-+>⎪⎩∴()f x 在递增(,)a +∞，在递减(-1)a ∞，-，在[]1,a a -上为常数∴()f x的最小值为()(11)2f a f a a a ≥-=+==∴()2f x ≥（II ）（1）当3a ≥时，1(3)5f a a +<=∴2510a a a ⇒<<-+<∴523a ≤<（2）当03a <<时，2(3)61510f a a a a <⇒-+-->=∴a <或a >3a <<综上所述15(22a +∈考点：考查带有绝对值的不等式的应用能力，考查函数与不等式的关系，中等题.。

SAT 2 数学2 college board官网全部原题加答案讲解 SAT Subject Test Math Level 2

SAT Subject Test Practice - Results SummaryMathematics Level 21Your answer Omitted!What is the distance in space between the points with coordinates and ? (A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is D.The distance between the points with coordinates and is given by the distance formula: .Therefore, the distance between the points with coordinates and is:,which simplifies to .2Your answer Omitted!If , what value does approach as gets infinitely larger?(A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is E.One way to determine the value that approaches as gets infinitely larger is to rewrite the definition of the function to use only negative powers of and then reason about the behavior of negative powers of as gets infinitely larger. Since the question is only concerned with what happens to as gets infinitely larger, one can assume that is positive. For , the expression is equivalent to the expression . As gets infinitelylarger, the expression approaches the value , so as gets infinitely larger, the expressionapproaches the value . Thus, as gets infinitely larger, approaches .Alternatively, one can use a graphing calculator to estimate the height of the horizontal asymptote for the function . Graph the function on an interval with “large”, say, from to .By examining the graph, the all seem very close to . Graph the function again, from, say, to .The vary even less from . In fact, to the scale of the coordinate plane shown, the graph of the function is nearly indistinguishable from the asymptotic line . This suggests that as gets infinitely larger, approaches , that is, .Note: The algebraic method is preferable, as it provides a proof that guarantees that the value approaches is . Although the graphical method worked in this case, it does not provide a complete justification; for example, the graphical method does not ensure that the graph resembles a horizontal line for “very large” such as .3Your answer Omitted!If is a factor of , then(A)(B)(C)(D)(E)ExplanationDifficulty: EasyThe correct answer is A.By the Factor Theorem, is a factor of only when is a root ofthat is, , which simplifies to . Therefore, .Alternatively, one can perform the division of by and then find a value for so that the remainder of the division is .Since the remainder is , the value of must satisfy . Therefore, .4Your answer Omitted!Alison deposits into a new savings account that earns percent interest compounded annually. If Alison makes no additional deposits or withdrawals, how many years will it take for the amount in the account to double?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumAfter year, the amount in the account is equal to . After years, the amount isequal to , and so on. After years, the amount is equal to . You needto find the value of for which . There are several ways to solve this equation. You can use logarithms to solve the equation as follows.Since , it will take more than years for the amount in the account to double. Thus, you need to round up to .Another way to find is to use your graphing calculator to graph and . From the answer choices, you know you need to set the viewing window with values from to about and values extending just beyond . The of the point of intersection is approximately . Thus you need to round up to .5Your answer Omitted!In the figure above, when is subtracted from , what is the length of the resultant vector?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe resultant of can be determined by. The length of the resultant is:6Your answer Omitted!In the -plane, what is the area of a triangle whose vertices are , , and ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt is helpful to draw a sketch of the triangle:The length of the base of the triangle is and the height of the triangle is . Therefore, the area of the triangle is . The correct answer is B.7Your answer Omitted!A right circular cylinder has radius and height . If and are two points on its surface, what is the maximum possible straight-line distance between and ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe maximum possible distance occurs when and are on the circumference of opposite bases:You can use the Pythagorean Theorem:The correct answer is (B).8Your answer Omitted!Note: Figure not drawn to scale.In the figure above, and the measure of is . What is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThere are several ways to solve this problem. One way is to use the law of sines. Since ,the measure of is and the measure of is . Thus, and . (Make sure your calculator is in degree mode.)You can also use the law of cosines:Since is isosceles, you can draw the altitude to the triangle.9Your answer Omitted!The function is defined by for .What is the difference between the maximum and minimum values of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt is necessary to use your graphing calculator for this question. First graph the function. It is helpful to resize the viewing window so the -values go fromto . On this interval the maximum value of is and the minimum value of is. The difference between these two values is , which rounds to .10Your answer Omitted!Suppose the graph of is translated units left and unit up. If the resulting graph represents , what is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumIt may be helpful to draw a graph of and .The equation for is . Therefore,. The correct answer is B.11Your answer Omitted!A sequence is recursively defined by , for . If and , what is the value of ?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe values for and are given. is equal to . is equal to. is equal to . is equal to.If your graphing calculator has a sequence mode, you can define the sequence recursively and find the value of . Let , since the first term is . Define . Let , since we have to define the first two terms and . Then examining a graph or table, you can find .12Your answer Omitted!The diameter and height of a right circular cylinder are equal. If the volume of the cylinder is , what is the height of the cylinder?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is A.To determine the height of the cylinder, first express the diameter of the cylinder in terms of the height, and then express the height in terms of the volume of the cylinder.The volume of a right circular cylinder is given by , where is the radius of the circular base of the cylinder and is the height of the cylinder. Since the diameter and height are equal, . Thus . Substitute the expression for in the volume formula to eliminate :. Solving for gives . Since the volume of the cylinder is , theheight of the cylinder is .13Your answer Omitted!If , then(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is E.One way to determine the value of is to apply the sine of difference of two angles identity: . Since and , the identity gives . Therefore, .Another way to determine the value of is to apply the supplementary angle trigonometric identity for the sine: . Therefore, .14Your answer Omitted!A line has parametric equations and , where is the parameter. The slope of the line is(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is B.One way to determine the slope of the line is to compute two points on the line and then use the slope formula. For example, letting gives the point on the line, and letting gives the point on the line. Therefore, the slope of the line is equal to .Alternatively, one can express in terms of . Since and , it follows that . Therefore, the slope of the line is .15Your answer Omitted!What is the range of the function defined by ?(A) All real numbers(B) All real numbers except(C) All real numbers except(D) All real numbers except(E) All real numbers between andExplanationDifficulty: MediumThe correct answer is D.The range of the function defined by is the set of such thatfor some .One way to determine the range of the function defined by is to solve the equationfor and then determine which correspond to at least one . To solve for , first subtract from both sides to get and then take the reciprocal of both sides to get . The equation shows that for anyother than , there is an such that , and that there is no such for . Therefore, the range of the function defined by is all real numbers except .Alternatively, one can reason about the possible values of the term . The expression can take on any value except , so the expression can take on any value except . Therefore, the range of the function defined by is all real numbers except .16Your answer Omitted!The table above shows the number of digital cameras that were sold during a three-day sale. The prices of models , , and were , , and , respectively. Which of the following matrix representations gives the total income, in dollars, received from the sale of the cameras for each of the three days?(A)(B)(C)(D)(E)ExplanationDifficulty: MediumThe correct answer is C.A correct matrix representation must have exactly three entries, each of which represents the total income, in dollars, for one of the three days. The total income for Day is given by the arithmetic expression , which is the single entry of the matrix product; in the same way, the total income for Day is given by, the single entry of ; and the total income for Day is given by , the single entry of. Therefore, the matrix representationgives the total income, in dollars, received from the sale of the cameras for each of the three days. Although it is not necessary to compute the matrix product in order to answer the question correctly, equals .17Your answer Omitted!The right circular cone above is sliced horizontally forming two pieces, each of which has the same height. What is the ratio of the volume of the smaller piece to the volume of the larger piece?(A)(B)(C)(D)(E)ExplanationDifficulty: HardIt is helpful to label the figure.The top piece is a cone whose height is one-half the height of the original cone . Using the properties of similar right triangles, you should realize the radii of these two cones must be in the same ratio. So if the top cone has radius , the original cone has radius .The volume of the top piece is equal to . The volume of the bottom piece is equal to the volume of the original cone minus the volume of the top piece.The ratio of the volume of the smaller piece to the volume of the larger piece is .18Your answer Omitted!In the figure above, is a regular pentagon with side of length . What is the -coordinate of ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe sum of the measures of the interior angles of a regular pentagon is equal to . Each interior angle has a measure of . Using supplementary angles, has a measure of . You can use right triangle trigonometry to find the -coordinate of point .Since , is about . Since the length of each side of the pentagon is , the -coordinate of point is . Putting the information together tells us that the -coordinate of point is . The correct answer is (B).19Your answer Omitted!For a class test, the mean score was , the median score was , and the standard deviation of the scores was . The teacher decided to add points to each score due to a grading error. Which of the following statements must be true for the new scores?I.The new mean score is .II.The new median score is .III.The new standard deviation of the scores is .(A) None(B) only(C) only(D) and only(E) , , andExplanationDifficulty: HardFor this type of question you need to evaluate each statement separately. Statement is true. If you add to each number in a data set, the mean will also increase by . Statement is also true. The relative position of each score will remain the same. Thus, the new median score will be equal tomore than the old median score. Statement is false. Since each new score is more than the old score, the spread of the scores and the position of the scores relative to the mean remain the same. Thus, the standard deviation of the new scores is the same as the standard deviation of the old scores.20Your answer Omitted!A game has two spinners. For the first spinner, the probability of landing on blue is . Independently, for the second spinner, the probability of landing on blue is What is the probability that the first spinner lands on blue and the second spinner does not land on blue?(A)(B)(C)(D)(E)ExplanationDifficulty: HardSince the two events are independent, the probability that the first spinner lands on blue and the second spinner does not land on blue is the product of the two probabilities. The first probability is given. Since the probability that the second spinner lands on blue is the probability that thesecond spinner does not land on blue is Therefore, . The correct answer is (E).21Your answer Omitted!In January the world’s population was billion. Assuming a growth rate of percent per year, the world’s population, in billions, for years after can be modeled by theequation . According to the model, the population growth from January to January was(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is C.According to the model, the world’s population in January was and in January was . Therefore, according to the model, the population growth from Januaryto January , in billions, was , or equivalently,.22Your answer Omitted!What is the measure of one of the larger angles of a parallelogram in the that has vertices with coordinates , , and ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is C.First, note that the angle of the parallelogram with vertex is one of the two larger angles of the parallelogram: Looking at the graph of the parallelogram in the makes this apparent. Alternatively, the sides of the angle of the parallelogram with vertex are a horizontal line segment with endpoints and and a line segment of positive slope with endpoints and that intersects the horizontal line segment at its left endpoint , so the angle must measure more than Since the sum of the measures of the four angles of aparallelogram equals , the angle with vertex must be one of the larger angles.One way to determine the measure of the angle of the parallelogram with vertex is to apply the Law of Cosines to the triangle with vertices , , and . The length of the twosides of the angle with vertex are and; the length of the side opposite the angle is . Let represent the angle with vertex and apply the Law of Cosines: , so. Therefore, the measure of one of thelarger angles of the parallelogram is .Another way to determine the measure of the angle of the parallelogram with vertex is to consider the triangle , , and . The measure of the angle of this triangle with vertex is less than the measure of the angle of the parallelogram with vertex . The angleof the triangle has opposite side of length and adjacent side of length , so the measure of this angle is . Therefore, the measure of the angle of the parallelogram with vertex is .Yet another way to determine the measure of the angle of the parallelogram with vertex is to use trigonometric relationships to find the measure of one of the smaller angles, and then use the fact that each pair of a larger and smaller angle is a pair of supplementary angles. Consider the angle of the parallelogram with vertex ; this angle coincides with the angle at vertex of the right triangle with vertices at , , and , with opposite side of lengthand adjacent side of length , so the measure of this angle is . This angle, together with the angle of the parallelogram with vertex , form a pair of interior angles on the same side of a transversal that intersects parallel lines, so the sum of the measures of the pair of angles equals . Therefore, the measure of the angle of the parallelogram with vertex is.23Your answer Omitted!For some real number , the first three terms of an arithmetic sequence are, and . What is the numerical value of the fourth term?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is E.To determine the numerical value of the fourth term, first determine the value of and then apply the common difference.Since , and are the first three terms of an arithmetic sequence, it must be true that, that is, Solving for gives . Substituting for in the expressions of the three first terms of the sequence, one sees that they are , , and , respectively. The common difference between consecutive terms for this arithmetic sequence is , and therefore, the fourth term is .24Your answer Omitted!In a group of people, percent have brown eyes. Two people are to be selected at random from the group. What is the probability that neither person selected will have brown eyes?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is A.One way to determine the probability that neither person selected will have brown eyes is to count both the number of ways to choose two people at random from the people who do not have brown eyes and the number of ways to choose two people at random from all people, and then compute the ratio of those two numbers.Since percent of the people have brown eyes, there are people with brown eyes, and people who do not have brown eyes. The number of ways of choosing twopeople, neither of whom has brown eyes, is : there are ways to choose a first person andways to choose a second person, but there are ways in which that same pair of people could bechosen. Similarly, the number of ways of choosing two people at random from the people is. Therefore, the probability that neither of the two people selected has brown eyes is.Another way to determine the probability that neither person selected will have brown eyes is to multiply the probability of choosing one of the people who does not have brown eyes at random from the people times the probability of choosing one of the people who does not have brown eyes at random from the remaining people after one of the people who does not have brown eyes has been chosen.Since percent of the people have brown eyes, the probability of choosing one of the people who does not have brown eyes at random from the people is . If one of the people who does not have brown eyes has been chosen, there remain people who do not have brown eyes out of a total of people; the probability of choosing one of the people who does not have brown eyes at random from the people is . Therefore, if two people are to be selected from the group at random, the probability that neither person selected will have brown eyes is.25Your answer Omitted!If , what is ?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is E.One way to determine the value of is to solve the equation for . Since , start with the equation , and cube both sides to get. Isolate to get , and apply the cube root to both sides of the equation to get .Another way to determine the value of is to find a formula for and then evaluate at Let and solve for : cubing both sides gives , so , and. Therefore, , and .26Your answer Omitted!Which of the following equations best models the data in the table above?(A)(B)(C)(D)(E)ExplanationDifficulty: HardThe correct answer is D.One way to determine which of the equations best models the data in the table is to use a calculator that has a statistics mode to compute an exponential regression for the data.The specific steps to be followed depend on the model of calculator, but can be summarized as follows: Enter the statistics mode, edit the list of ordered pairs to include only the four points given in the table and perform an exponential regression. The coefficients are, approximately, for the constant and for the base, which indicates that the exponential equation is the result of performing the exponential regression. If the calculator reports a correlation, it should be a number that is very close , to which indicates that the data very closely matches the exponential equation. Therefore, of the given models, best fits the data.Alternatively, without using a calculator that has a statistics mode, one can reason about the data given in the table.The data indicates that as increases, increases; thus, options A and B cannot be candidates for such a relationship. Evaluating options C, D and E at shows that option D is the one that gives a value of that is closest to In the same way, evaluating options C, D and E at each of the other given data points shows that option D is a better model for that one data point than either option C or option E. Therefore, is the best of the given models for the data.27Your answer Omitted!The linear regression model above is based on an analysis of nutritional data from 14 varieties of cereal bars to relate the percent of calories from fat to the percent of calories from carbohydrates . Based on this model, which of the following statements must be true?I.There is a positive correlation between and .II.When percent of calories are from fat, the predicted percent of calories from carbohydrates is approximately .III.The slope indicates that as increases by , decreases by .(A) II only(B) I and II only(C) I and III only(D) II and III only(E) I, II, and IIIExplanationDifficulty: HardThe correct answer is D.Statement I is false: Since , high values of are associated with low values of which indicates that there is a negative correlation between and .Statement II is true: When percent of calories are from fat, and the predicted percent of calories from carbohydrates is .Statement III is true: Since the slope of the regression line is , as increases by ,increases by ; that is, decreases by .28Your answer Omitted!The number of hours of daylight, , in Hartsville can be modeled by , where is the number of days after March . The day with the greatest number of hours of daylight hashow many more daylight hours than May ? (March and May have days each. April and June have days each.)(A) hr(B) hr(C) hr(D) hr(E) hrExplanationDifficulty: HardThe correct answer is A.To determine how many more daylight hours the day with the greatest number of hours of daylight has than May , find the maximum number of daylight hours possible for any day and then subtract from that the number of daylight hours for May .To find the greatest number of daylight hours possible for any day, notice that the expression is maximized when , which corresponds to , so. However, for this problem, must be a whole number, as it represents a count of days after March . From the shape of the graph of the sine function, either orcorresponds to the day with the greatest number of hours of daylight, and since, the expression is maximized when days after March . (It is not required to find the day on which the greatest number of hours of daylight occurs, but it is days after March ,that is, June .)Since May is days after March , the number of hours of daylight for Mayis .Therefore, the day with the greatest number of hours of daylight hasmore daylight hours than May .。

2023年新高考全国Ⅱ卷数学试题(附答案解析)

绝密★启用前2023年普通高等学校招生全国统一考试(新高考全国Ⅱ卷)数学本试卷共4页，22小题，满分150分。

考试用时120分钟。

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一、选择题:本大题共8小题, 每小题5分, 共40分. 在每小题给出的四个选项中, 只有一项是符合题目要求的。

1.在复平面内, 1+3i3-i对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限【答案】A【解析】1+3i3-i=6+8i,故对应的点在第一象限，选A。

2.设集合A={0,-a},B={1,a-2,2a-2}, 若A⊆B, 则a=()A.2B.1C.23D.-1【答案】B【解析】若a-2=0,则a=2,此时A=0,-2},B=1,0,2},不满足题意;若2a-2=0,则a =1,此时A={0,-1},B={1,-1,0},满足题意。

选B。

3.某学校为了解学生参加体育运动的情况, 用比例分配的分层随机抽样方法作抽样调查, 拟从初中部和高中部两层共抽取60名学生, 已知该校初中部和高中部分别有400名和200名学生, 则不同的抽样结果共有()A.C45400⋅C15200种 B.C20400⋅C40200种 C.C30400⋅C30200种 D.C40400⋅C20200种【答案】D【解析】根据按比例分配的分层抽样可知初中部抽40人，高中部抽20人，选D。

精品解析：2022年全国新高考II卷数学试题(解析版)

2022年普通高等学校招生全国统一考试（新高考全国Ⅱ卷）数学一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知集合{}{}1,1,2,4,11A B x x =-=-≤，则A B = （）A.{1,2}-B.{1,2}C.{1,4}D.{1,4}-【答案】B 【解析】【分析】求出集合B 后可求A B .【详解】{}|02B x x =≤≤，故{}1,2A B = ，故选：B.2.(22i)(12i)+-=（）A.24i-+ B.24i-- C.62i+ D.62i-【答案】D 【解析】【分析】利用复数的乘法可求()()22i 12i -.【详解】()()22i 12i 244i 2i 62i +-=+-+=-，故选：D.3.图1是中国古代建筑中的举架结构，,,,AA BB CC DD ''''是桁，相邻桁的水平距离称为步，垂直距离称为举，图2是某古代建筑屋顶截面的示意图．其中1111,,,DD CC BB AA 是举，1111,,,OD DC CB BA 是相等的步，相邻桁的举步之比分别为11111231111,0.5,,DD CC BB AAk k k OD DC CB BA ====．已知123,,k k k 成公差为0.1的等差数列，且直线OA 的斜率为0.725，则3k =（）A.0.75B.0.8C.0.85D.0.9【答案】D 【解析】【分析】设11111OD DC CB BA ====，则可得关于3k 的方程，求出其解后可得正确的选项.【详解】设11111OD DC CB BA ====，则111213,,CC k BB k AA k ===，依题意，有31320.2,0.1k k k k -=-=，且111111110.725DD CC BB AA OD DC CB BA +++=+++，所以30.530.30.7254k +-=，故30.9k =，故选：D4.已知向量(3,4),(1,0),t ===+ a b c a b ，若,,<>=<>a cbc ，则t =（）A.6- B.5- C.5D.6【答案】C 【解析】【分析】利用向量的运算和向量的夹角的余弦公式的坐标形式化简即可求得【详解】解：()3,4c t =+ ,cos ,cos ,a c b c =,即931635t tc c+++=,解得5t =,故选：C5.有甲、乙、丙、丁、戊5名同学站成一排参加文艺汇演，若甲不站在两端，丙和丁相邻，则不同排列方式共有（）A.12种B.24种C.36种D.48种【答案】B【解析】【分析】利用捆绑法处理丙丁，用插空法安排甲，利用排列组合与计数原理即可得解【详解】因为丙丁要在一起，先把丙丁捆绑，看做一个元素，连同乙，戊看成三个元素排列,有3!种排列方式；为使甲不在两端，必须且只需甲在此三个元素的中间两个位置任选一个位置插入，有2种插空方式；注意到丙丁两人的顺序可交换，有2种排列方式，故安排这5名同学共有：3!2224⨯⨯=种不同的排列方式，故选：B6.若sin()cos()sin 4παβαβαβ⎛⎫+++=+ ⎪⎝⎭，则（）A.()tan 1αβ-=B.()tan 1αβ+=C.()tan 1αβ-=-D.()tan 1αβ+=-【答案】C 【解析】【分析】由两角和差的正余弦公式化简，结合同角三角函数的商数关系即可得解.【详解】由已知得：()sin cos cos sin cos cos sin sin 2cos sin sin αβαβαβαβααβ++-=-,即：sin cos cos sin cos cos sin sin 0αβαβαβαβ-++=，即：()()sin cos 0αβαβ-+-=,所以()tan 1αβ-=-,故选：C7.已知正三棱台的高为1，上、下底面边长分别为为（）A.100πB.128πC.144πD.192π【答案】A 【解析】【分析】根据题意可求出正三棱台上下底面所在圆面的半径12,r r ，再根据球心距，圆面半径，以及球的半径之间的关系，即可解出球的半径，从而得出球的表面积．【详解】设正三棱台上下底面所在圆面的半径12,r r ，所以1233432,2sin 60sin 60r r ==，即123,4r r ==，设球心到上下底面的距离分别为12,d d ，球的半径为R ，所以1d =2d =121d d -=或121d d +=1=1+=，解得225R =符合题意，所以球的表面积为24π100πS R ==．故选：A ．8.已知函数()f x 的定义域为R ，且()()()(),(1)1f x y f x y f x f y f ++-==，则221()k f k ==∑（）A.3-B.2- C.0D.1【答案】A 【解析】【分析】根据题意赋值即可知函数()f x 的一个周期为6，求出函数一个周期中的()()()1,2,,6f f f 的值，即可解出．【详解】因为()()()()f x y f x y f x f y ++-=，令1,0x y ==可得，()()()2110f f f =，所以()02f =，令0x =可得，()()()2f y f y f y +-=，即()()f y f y =-，所以函数()f x 为偶函数，令1y =得，()()()()()111f x f x f x f f x ++-==，即有()()()21f x f x f x ++=+，从而可知()()21f x f x +=--，()()14f x f x -=--，故()()24f x f x +=-，即()()6f x f x =+，所以函数()f x 的一个周期为6．因为()()()210121f f f =-=-=-，()()()321112f f f =-=--=-，()()()4221f f f =-==-，()()()5111f f f =-==，()()602f f ==，所以一个周期内的()()()1260f f f +++= ．由于22除以6余4，所以()()()()()221123411213k f k f f f f ==+++=---=-∑．故选：A ．二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分.9.已知函数()sin(2)(0π)f x x ϕϕ=+<<的图像关于点2π,03⎛⎫⎪⎝⎭中心对称，则（）A.()f x 在区间5π0,12⎛⎫⎪⎝⎭单调递减B.()f x 在区间π11π,1212⎛⎫- ⎪⎝⎭有两个极值点C.直线7π6x =是曲线()y f x =的对称轴D.直线2y x =-是曲线()y f x =的切线【答案】AD 【解析】【分析】根据三角函数的性质逐个判断各选项，即可解出．【详解】由题意得：2π4πsin 033f ϕ⎛⎫⎛⎫=+=⎪⎪⎝⎭⎝⎭，所以4ππ3k ϕ+=，k ∈Z ，即4ππ,3k k ϕ=-+∈Z ，又0πϕ<<，所以2k=时，2π3ϕ=，故2π()sin 23f x x ⎛⎫=+ ⎪⎝⎭．对A ，当5π0,12x ⎛⎫∈ ⎪⎝⎭时，2π3π2,32x ⎛⎫+ ⎪⎝⎭，由正弦函数sin y u =图象知()y f x =在5π0,12⎛⎫ ⎪⎝⎭上是单调递减；对B ，当π11π,1212x ⎛⎫∈- ⎪⎝⎭时，2ππ5π2,322x ⎛⎫+∈ ⎪⎝⎭，由正弦函数sin y u =图象知()y f x =只有1个极值点，由2π3π232x +=，解得5π12x =，即5π12x =为函数的唯一极值点；对C ，当7π6x =时，2π23π3x +=，7π()06f =，直线7π6x =不是对称轴；对D ，由2π2cos 213y x ⎛⎫'=+=- ⎪⎝⎭得：2π1cos 232x ⎛⎫+=- ⎪⎝⎭，解得2π2π22π33x k +=+或2π4π22π,33x k k +=+∈Z ,从而得：πx k =或ππ,3x k k =+∈Z ,所以函数()y f x =在点0,2⎛⎫ ⎪ ⎪⎝⎭处的切线斜率为02π2cos 13x k y =='==-，切线方程为：3(0)2y x -=--即2y x =-．故选：AD ．10.已知O 为坐标原点，过抛物线2:2(0)C y px p =>焦点F 的直线与C 交于A ，B 两点，其中A 在第一象限，点(,0)M p ，若||||AF AM =，则（）A.直线AB的斜率为 B.||||OB OF =C.||4||AB OF > D.180OAM OBM ∠+∠<︒【答案】ACD 【解析】【分析】由AF AM =及抛物线方程求得3(,)42p A ，再由斜率公式即可判断A 选项；表示出直线AB 的方程，联立抛物线求得6(,)33p B -，即可求出OB 判断B 选项；由抛物线的定义求出2512p AB =即可判断C 选项；由0OA OB ⋅< ，0MB ⋅<求得AOB ∠，AMB ∠为钝角即可判断D 选项.【详解】对于A ，易得(,0)2p F ，由AF AM =可得点A 在FM 的垂直平分线上，则A 点横坐标为3224ppp +=，代入抛物线可得2233242p y p p =⋅=，则3(,42p A ，则直线AB的斜率为62342p p =-，A 正确；对于B，由斜率为AB的方程为2p x y =+，联立抛物线方程得220y py p -=，设11(,)B x y ，则16626p y p +=，则163y =-，代入抛物线得2123p x ⎛⎫-=⋅ ⎪ ⎪⎝⎭，解得13p x =，则6(,33p B -，则32pOB OF =≠=，B 错误；对于C ，由抛物线定义知：325244312p p p AB p p OF =++=>=，C 正确；对于D ，23663663(,)(,0423343234p p p p p OA OB ⎛⎫⋅=⋅-=⋅+⋅-=-< ⎪ ⎪⎝⎭ ，则AOB ∠为钝角，又2225(,)(,)0423343236p p p p p MA MB ⎛⎫⎛⎫⋅=-⋅--=-⋅-+⋅-=-< ⎪ ⎪ ⎪⎝⎭⎝⎭，则AMB ∠为钝角，又360AOB AMB OAM OBM ∠+∠+∠+∠= ，则180OAM OBM ∠+∠< ，D 正确.故选：ACD.11.如图，四边形ABCD 为正方形，ED ⊥平面ABCD ，,2FB ED AB ED FB ==∥，记三棱锥E ACD -，F ABC -，F ACE -的体积分别为123,,V V V ，则（）A.322V V =B.31V V =C.312V V V =+ D.3123V V =【答案】CD 【解析】【分析】直接由体积公式计算12,V V ，连接BD 交AC 于点M ，连接,EM FM ，由3A EFM C EFM V V V --=+计算出3V ，依次判断选项即可.【详解】设22AB ED FB a ===，因为ED ⊥平面ABCD ，FB ED ，则()2311114223323ACD V ED S a a a =⋅⋅=⋅⋅⋅= ，()232111223323ABC V FB S a a a =⋅⋅=⋅⋅⋅= ，连接BD 交AC 于点M ，连接,EM FM ，易得BD AC ⊥，又ED ⊥平面ABCD ，AC ⊂平面ABCD ，则ED AC ⊥，又ED BD D = ，,ED BD ⊂平面BDEF ，则AC ⊥平面BDEF ，又12BM DM BD ===，过F 作FG DE ⊥于G ，易得四边形BDGF 为矩形，则,FG BD EG a ===，则EM ===，3EF a =，222EM FM EF +=，则EM FM ⊥，213222EFM S EM FM a =⋅= ，AC =，则33123A EFM C EFM EFM V V V AC S a --=+=⋅= ，则3123V V =，323V V =，312V V V =+，故A 、B 错误；C 、D 正确.故选：CD.12.若x ，y 满足221+-=x y xy ，则（）A.1x y +≤B.2x y +≥-C.222x y +≤D.221x y +≥【答案】BC 【解析】【分析】根据基本不等式或者取特值即可判断各选项的真假．【详解】因为22222a b a b ab ++⎛⎫≤≤⎪⎝⎭（,a b ÎR ），由221+-=x y xy 可变形为，()221332x y x y xy +⎛⎫+-=≤ ⎪⎝⎭，解得22x y -≤+≤，当且仅当1x y ==-时，2x y +=-，当且仅当1x y ==时，2x y +=，所以A 错误，B 正确；由221+-=x y xy 可变形为()222212x y x y xy ++-=≤，解得222x y +≤，当且仅当1x y ==±时取等号，所以C 正确；因为221+-=x y xy 变形可得223124y x y ⎛⎫-+= ⎪⎝⎭，设3cos ,sin 22y x y θθ-==，所以cos ,x y θθθ==，因此2222511cos sin cos 12cos 2333x y θθθθ=θ-θ+=+++42π2sin 2,23363θ⎛⎫⎡⎤=+-∈ ⎪⎢⎥⎝⎭⎣⎦，所以当33,33x y ==-时满足等式，但是221x y +≥不成立，所以D 错误．故选：BC ．三、填空题：本题共4小题，每小题5分，共20分.13.已知随机变量X 服从正态分布()22,N σ，且(2 2.5)0.36P X <≤=，则( 2.5)P X >=____________．【答案】0.14##750．【解析】【分析】根据正态分布曲线的性质即可解出．【详解】因为()22,X N σ，所以()()220.5P X P X <=>=，因此()()()2.522 2.50.50.360.14P X P X P X >=>-<≤=-=．故答案为：0.14．14.曲线ln ||y x =过坐标原点的两条切线的方程为____________，____________．【答案】①.1ey x =②.1ey x =-【解析】【分析】分0x >和0x <两种情况，当0x >时设切点为()00,ln x x ，求出函数的导函数，即可求出切线的斜率，从而表示出切线方程，再根据切线过坐标原点求出0x ，即可求出切线方程，当0x <时同理可得；【详解】解：因为ln y x =，当0x >时ln y x =，设切点为()00,ln x x ，由1y x'=，所以001|x x y x ='=，所以切线方程为()0001ln y x x x x -=-，又切线过坐标原点，所以()0001ln x x x -=-，解得0e x =，所以切线方程为()11e e y x -=-，即1ey x =；当0x <时()ln y x =-，设切点为()()11,ln x x -，由1y x '=，所以111|x x y x ='=，所以切线方程为()()1111ln y x x x x --=-，又切线过坐标原点，所以()()1111ln x x x --=-，解得1e x =-，所以切线方程为()11e ey x -=+-，即1ey x =-；故答案为：1ey x =；1e y x=-15.设点(2,3),(0,)A B a -，若直线AB 关于y a =对称的直线与圆22(3)(2)1x y +++=有公共点，则a 的取值范围是________．【答案】13,32⎡⎤⎢⎥⎣⎦【解析】【分析】首先求出点A 关于y a =对称点A '的坐标，即可得到直线l 的方程，根据圆心到直线的距离小于等于半径得到不等式，解得即可；【详解】解：()2,3A -关于y a =对称的点的坐标为()2,23A a '--，()0,B a 在直线y a =上，所以A B '所在直线即为直线l ，所以直线l 为32a y x a -=+-，即()3220a x y a -+-=；圆()()22:321C x y +++=，圆心()3,2C --，半径1r =，依题意圆心到直线l 的距离1d =≤，即()()2225532a a -≤-+，解得1332a ≤≤，即13,32a ⎡⎤∈⎢⎥⎣⎦；故答案为：13,32⎡⎤⎢⎥⎣⎦16.已知直线l 与椭圆22163x y +=在第一象限交于A ，B 两点，l 与x轴，y 轴分别交于M ，N 两点，且||||,||MANB MN ==l 的方程为___________．【答案】0x +-=【解析】【分析】令AB 的中点为E ，设()11,A x y ，()22,B x y ，利用点差法得到12OE AB k k ⋅=-，设直线:AB y kx m =+，0k <，0m >，求出M 、N 的坐标，再根据MN 求出k 、m ，即可得解；【详解】解：令AB 的中点为E ，因为MA NB =，所以ME NE =，设()11,A x y ，()22,B x y ，则2211163x y +=，2222631x y +=，所以2222121206633x x y y -+-=，即)()()()12121212063x x x x y y y y -++-+=所以()()()()1212121212y y y y x x x x +-=--+，即12OE AB k k ⋅=-，设直线:AB y kx m =+，0k <，0m >，令0x =得y m =，令0y =得m x k =-，即,0m M k ⎛⎫- ⎪⎝⎭，()0,N m ，所以,22m m E k ⎛⎫- ⎪⎝⎭，即1222mk m k⨯=--，解得22k =-或2k =（舍去），又MN =，即MN =，解得2m =或2m =-（舍去），所以直线2:22AB y x =-+，即0x +-=；故答案为：0x +-=四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．17.已知{}n a 为等差数列，{}n b 是公比为2的等比数列，且223344a b a b b a -=-=-．（1）证明：11a b =；（2）求集合{}1,1500k m k b a a m =+≤≤中元素个数．【答案】（1）证明见解析；（2）9．【解析】【分析】（1）设数列{}n a 的公差为d ，根据题意列出方程组即可证出；（2）根据题意化简可得22k m -=，即可解出．【小问1详解】设数列{}n a 的公差为d ，所以，()11111111224283a d b a d b a d b b a d +-=+-⎧⎨+-=-+⎩，即可解得，112db a ==，所以原命题得证．【小问2详解】由（1）知，112d b a ==，所以()1111121k k m b a a b a m d a -=+⇔⨯=+-+，即122k m -=，亦即[]221,500k m -=∈，解得210k ≤≤，所以满足等式的解2,3,4,,10k = ，故集合{}1|,1500k m k b a a m =+≤≤中的元素个数为10219-+=．18.记ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，分别以a ，b ，c 为边长的三个正三角形的面积依次为123,,S S S，已知1231,sin 23S S S B -+==．（1）求ABC 的面积；（2）若sin sin 3A C =，求b ．【答案】（1）8（2）12【解析】【分析】（1）先表示出123,,S S S ，再由12332S S S -+=求得2222a c b +-=，结合余弦定理及平方关系求得ac ，再由面积公式求解即可；（2）由正弦定理得22sin sin sin b acB A C=，即可求解.【小问1详解】由题意得222212313333,,22444S a a S b S c =⋅⋅===，则2221234442S S S a b c -+=-+=，即2222a c b +-=，由余弦定理得222cos 2a c b B ac +-=，整理得cos 1ac B =，则cos 0B >，又1sin 3B =，则22cos 3B ==，132cos 4ac B ==，则12sin 28ABC S ac B == ；【小问2详解】由正弦定理得：sin sin sin b a c B A C ==，则223294sin sin sin sin sin 423b ac ac B A C A C =⋅==，则3sin 2b B =，31sin 22b B ==.19.在某地区进行流行病学调查，随机调查了100位某种疾病患者的年龄，得到如下的样本数据的频率分布直方图：（1）估计该地区这种疾病患者的平均年龄（同一组中的数据用该组区间的中点值为代表）；（2）估计该地区一位这种疾病患者的年龄位于区间[20,70)的概率；（3）已知该地区这种疾病的患病率为0.1%，该地区年龄位于区间[40,50)的人口占该地区总人口的16%.从该地区中任选一人，若此人的年龄位于区间[40,50)，求此人患这种疾病的概率．（以样本数据中患者的年龄位于各区间的频率作为患者的年龄位于该区间的概率，精确到0.0001）.【答案】（1）47.9岁；（2）0.89；（3）0.0014．【解析】【分析】（1）根据平均值等于各矩形的面积乘以对应区间的中点值的和即可求出；（2）设A ={一人患这种疾病的年龄在区间[20,70)}，根据对立事件的概率公式()1(P A P A =-即可解出；（3）根据条件概率公式即可求出．【小问1详解】平均年龄(50.001150.002250.012350.017450.023x =⨯+⨯+⨯+⨯+⨯550.020650.017750.006850.002)1047.9+⨯+⨯+⨯+⨯⨯=（岁）．【小问2详解】设A ={一人患这种疾病的年龄在区间[20,70)}，所以()1()1(0.0010.0020.0060.002)1010.110.89P A P A =-=-+++⨯=-=．【小问3详解】设{B =任选一人年龄位于区间}[40,50)，{C =任选一人患这种疾病}，则由条件概率公式可得()0.1%0.023100.0010.23(|)0.00143750.0014()16%0.16P BC P C B P B ⨯⨯⨯====≈．20.如图，PO 是三棱锥P ABC -的高，PA PB =，AB AC ⊥，E 是PB 的中点．（1）证明：//OE 平面PAC ；（2）若30ABO CBO ∠=∠=︒，3PO =，5PA =，求二面角C AE B --的正弦值．【答案】（1）证明见解析（2）1113【解析】【分析】（1）连接BO 并延长交AC 于点D ，连接OA 、PD ，根据三角形全等得到OA OB =，再根据直角三角形的性质得到AO DO =，即可得到O 为BD 的中点从而得到//OE PD ，即可得证；（2）过点A 作//Az OP 角函数的基本关系计算可得；【小问1详解】证明：连接BO 并延长交AC 于点D ，连接OA 、PD ，因为PO 是三棱锥P ABC -的高，所以PO ⊥平面ABC ，,AO BO ⊂平面ABC ，所以PO AO ⊥、PO BO ⊥，又PA PB =，所以POA POB ≅△△，即OA OB =，所以OAB OBA ∠=∠，又AB AC ⊥，即90BAC ∠=︒，所以90OAB OAD ∠+∠=︒，90OBA ODA ∠+∠=︒，所以ODA OAD∠=∠所以AO DO =，即AO DO OB ==，所以O 为BD 的中点，又E 为PB 的中点，所以//OE PD ，又OE ⊄平面PAC ，PD ⊂平面PAC ，所以//OE 平面PAC【小问2详解】解：过点A 作//Az OP ，如图建立平面直角坐标系，因为3PO =，5AP =，所以4OA ==，又30OBA OBC ∠=∠=︒，所以28BD OA ==，则4=AD，AB =，所以12AC =，所以()2,0O，()B，()2,3P ，()0,12,0C，所以32E ⎛⎫ ⎪⎝⎭，则32AE ⎛⎫= ⎪⎝⎭，()AB =，()0,12,0AC = ，设平面AEB 的法向量为(),,n x y z =，则3020n AE y z nAB ⎧⋅=++=⎪⎨⎪⋅==⎩，令2z =，则3y =-，0x =，所以()0,3,2n =-；设平面AEC 的法向量为(),,m a b c =，则302120m AE b c m AC b ⎧⋅=++=⎪⎨⎪⋅==⎩，令a =6c =-，0b =，所以)6m =-；所以43cos ,13n m n m n m⋅==-设二面角C AE B --为θ，由图可知二面角C AE B --为钝二面角，所以43cos 13θ=-，所以11sin 13θ==故二面角C AE B --的正弦值为1113；21.已知双曲线2222:1(0,0)x y C a b a b-=>>的右焦点为(2,0)F，渐近线方程为y =．（1）求C 的方程；（2）过F 的直线与C 的两条渐近线分别交于A ，B 两点，点()()1122,,,P x y Q x y 在C 上，且1210,0x x y >>>．过P且斜率为的直线与过QM .从下面①②③中选取两个作为条件，证明另外一个成立：①M 在AB 上；②PQ AB ∥；③||||MA MB =．注：若选择不同的组合分别解答，则按第一个解答计分.【答案】（1）2213y x -=（2）见解析【解析】【分析】（1）利用焦点坐标求得c 的值，利用渐近线方程求得,a b 的关系，进而利用,,a b c 的平方关系求得,a b 的值，得到双曲线的方程；（2）先分析得到直线AB 的斜率存在且不为零，设直线AB 的斜率为k ,M (x 0,y 0),由③|AM |=|BM |等价分析得到200283k x ky k +=-；由直线PM 和QM 的斜率得到直线方程，结合双曲线的方程，两点间距离公式得到直线PQ 的斜率03x m y =，由②//PQ AB 等价转化为003ky x =，由①M 在直线AB 上等价于()2002ky k x =-，然后选择两个作为已知条件一个作为结论，进行证明即可.【小问1详解】右焦点为(2,0)F ，∴2c =,∵渐近线方程为y =，∴ba=∴b =，∴222244c a b a =+==，∴1a =，∴b =．∴C 的方程为：2213y x -=；【小问2详解】由已知得直线PQ 的斜率存在且不为零，直线AB 的斜率不为零，若选由①②推③或选由②③推①：由②成立可知直线AB 的斜率存在且不为零；若选①③推②，则M 为线段AB 的中点，假若直线AB 的斜率不存在，则由双曲线的对称性可知M 在x 轴上，即为焦点F ,此时由对称性可知P 、Q 关于x 轴对称，与从而12x x =，已知不符；总之，直线AB 的斜率存在且不为零.设直线AB 的斜率为k ,直线AB 方程为()2y k x =-,则条件①M 在AB 上，等价于()()2000022y k x ky k x =-⇔=-；两渐近线的方程合并为2230x y -=,联立消去y 并化简整理得：()22223440k x k x k --+=设()()3334,,,A x y B x y ,线段中点为(),N N N x y ,则()2342226,2233N N N x x k kx y k x k k +===-=--,设()00,M x y ,则条件③AM BM =等价于()()()()222203030404x x y y x x y y -+-=-+-,移项并利用平方差公式整理得：()()()()3403434034220x x x x x y y y y y ⎡⎤⎡⎤--++--+=⎣⎦⎣⎦，()()3403403434220y y x x x y y y x x -⎡⎤⎡⎤-++-+=⎣⎦⎣⎦-,即()000N N x x k y y -+-=,即200283k x ky k +=-；由题意知直线PM的斜率为,直线QM∴由))10102020,y y x x y y x x -=--=-,∴)121202y y x x x -=+-,所以直线PQ的斜率)1201212122x x x y y m x x x x +--==---,直线)00:PM y x x y =-+,即00y y =,代入双曲线的方程22330x y --=,即)3yy +-=中，得：()()00003y y ⎡⎤+-+=⎣⎦,解得P的横坐标：100x y ⎛⎫=+⎪⎪⎭,同理：200x y ⎛⎫=-⎪⎪⎭，∴0012012002222000033,2,33y x x x y x x x x y x y x ⎛⎫-=++-=--⎪--⎭∴03x m y =,∴条件②//PQ AB 等价于003m k ky x =⇔=，综上所述：条件①M 在AB 上，等价于()2002ky kx =-；条件②//PQ AB 等价于003ky x =；条件③AM BM =等价于200283k x ky k +=-；选①②推③:由①②解得：2200002228,433k k x x ky x k k =∴+==--,∴③成立；选①③推②：由①③解得：20223k x k =-，20263k ky k =-，∴003ky x =，∴②成立；选②③推①：由②③解得：20223k x k =-，20263k ky k =-，∴02623x k -=-，∴()2002ky kx =-，∴①成立.22.已知函数()e e ax x f x x =-．（1）当1a =时，讨论()f x 的单调性；（2）当0x >时，()1f x <-，求a 的取值范围；（3）设n *∈Nln(1)n ++>+ ．【答案】（1）()f x 的减区间为(),0-∞，增区间为()0,+∞.（2）12a ≤（3）见解析【解析】【分析】（1）求出()f x ¢，讨论其符号后可得()f x 的单调性.（2）设()e e 1axxh x x =-+，求出()x ''，先讨论12a >时题设中的不等式不成立，再就102a <≤结合放缩法讨论()h x '符号，最后就0a ≤结合放缩法讨论()h x 的范围后可得参数的取值范围.（3）由（2）可得12ln t tt<-对任意的1t >恒成立，从而可得()ln 1ln n n +-<*n N ∈恒成立，结合裂项相消法可证题设中的不等式.【小问1详解】当1a =时，()()1e xf x x =-，则()e xf x x '=，当0x <时，()0f x ¢<，当0x >时，()0f x ¢>，故()f x 的减区间为(),0-∞，增区间为()0,+∞.【小问2详解】设()e e 1axxh x x =-+，则()00h =，又()()1e e ax x h x ax '=+-，设()()1e e ax xg x ax =+-，则()()22e e ax x g x a a x '=+-，若12a >，则()0210g a '=->，因为()g x '为连续不间断函数，故存在()00,x ∈+∞，使得()00,x x ∀∈，总有()0g x ¢>，故()g x 在()00,x 为增函数，故()()00g x g >=，故()h x 在()00,x 为增函数，故()()01h x h >=-，与题设矛盾.若102a <≤，则()()()ln 11e e e e ax ax ax x x h x ax ++'=+-=-，下证：对任意0x >，总有()ln 1x x +<成立，证明：设()()ln 1S x x x =+-，故()11011x S x x x-'=-=<++，故()S x 在()0,+∞上为减函数，故()()00S x S <=即()ln 1x x +<成立.由上述不等式有()ln 12e e e e e e 0ax ax x ax ax x ax x +++-<-=-≤，故()0h x '≤总成立，即()h x 在()0,+∞上为减函数，所以()()01h x h <=-.当0a ≤时，有()e e e 1100ax x ax h x ax '=-+<-+=，所以()h x 在()0,+∞上为减函数，所以()()01h x h <=-.综上，12a ≤.【小问3详解】取12a =，则0x ∀>，总有12e e 10x x x -+<成立，令12e x t =，则21,e ,2ln x t t x t >==，故22ln 1t t t <-即12ln t t t<-对任意的1t >恒成立.所以对任意的*n N ∈，有2ln <整理得到：()ln1lnn n+-<，()ln2ln1ln3ln2ln1lnn n+>-+-+++-()ln1n=+，故不等式成立.【点睛】思路点睛：函数参数的不等式的恒成立问题，应该利用导数讨论函数的单调性，注意结合端点处导数的符号合理分类讨论，导数背景下数列不等式的证明，应根据已有的函数不等式合理构建数列不等式.。

2023新高考全国II卷数学试题及详细答案（解析）

2023新高考全国II卷数学试题及详细答案（解析）2023新高考全国II卷数学试题及详细答案（解析）2023新高考全国II卷数学试题及详细答案是什么?高考部分科目已经结束，许多人都对高考真题十分好奇，那么今年高考真题都有哪些呢?下面是小编为大家搜集整理的关于2023新高考全国II卷数学试题及详细答案，供大家参考，快来一起看看吧!2023新高考全国II卷数学试题及详细答案2023高考数学时间分配原则对于高考数学基础比较薄弱的同学，重在保简易题。

鉴于高考数学客观题部分主要是对基础知识点的考察，可以稍稍放慢速度，把时间控制在50-60分钟，力求做到准确细致，尽量保证70分的基础分不丢分。

之后的三道简易高考数学解答题每题平均花10-15分钟完成。

至于后三道高考数学大题，建议先阅读完题目，根据题意把可以联想到的常考知识点写出来，例如涉及函数单调性、切线斜率的可对函数求导，圆锥曲线的设出标准方程、数列里求出首项等等。

如果没有其它的思路，不要耽误太多时间，把剩下的时间倒回去检查前面的题目。

高考数学题要认真仔细对于一道具体的习题，解题时最重要的环节是审题。

审题的第一步是读题，这是获取信息量和思考的过程。

所以，在高考数学实际解题时，应特别注意，审题要认真、仔细。

对于高考数学题，第一重要的是数学知识点的掌握，第二是对答题技巧的掌握，考生在答高考数学题的时候，一定不要把所有时间都浪费在一道题上，否则会影响整张数学试卷的作答。

2023高考数学选择题题型及分布规律1.集合交并补运算2.充分必要条件，命题真假3.复数四则运算4.三视图恢复与，体积表面积内外截球计算5.算法循环结构6.概率，排列组合计算，积分计算6.函数奇偶周期对称抽象函数与导函数(及结论)7.分段函数8空间几何平行垂直夹角体积计算9线性规划10三角函数求值11解三角形相关夹角面积周长12向量共线垂直乘积夹角模长最值及向量有关三角形计算等13.数列通项，某一项，求和，最值14.复杂图形辨别及导数相关图形辨别15.函数比较大小，非常规(指数，对数，三角，抽象)不等式求解及恒成立，参数范围求解。

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Mathematics Level 1 and Level 2
SAT Subject Tests™
Answer Explanations to Practice Questions
from Getting Ready for the SAT Subject Tests
Visit the SAT® student website to get more practice and study tips for the Subject Tests in Mathematics Level 1 and Level 2.
pairs (10, 255.90) and (20, 261.80). The line has slope
261.80 − 255.90 20 − 10
=
0.59 and y-intercept 250, which is
the cost, in dollars, for the band to produce 0 CDs after
Getting Ready for the SAT Subject Tests: Answer Explanations to Practice Questions
1
SAT® Subject Tests in Mathematics Level 1 and Level 2
4. Difficulty: 3
Alternatively, one can invert the operations that were done to n. Apply the inverse of each operation, in the reverse order: First cube −0.5 to get −0.125, and then decrease this value by 8 to find that n = −0.125 − 8 = −8.125.
purchasing the equipment. The equation of this line is
y = 0.59n + 250, where n is the number of CDs used.
Alternatively, note that the total cost to produce 10 CDs
5. Difficulty: 3
Choice (C) is correct.
One way to determine all values of x for which 4 − x2 ≥ x − 2 is first to rewrite the inequality in an equivalent form that compares a factored expression to 0 and then reason about the arithmetic sign of the product of the factors. The inequality 4 − x2 ≥ x − 2 is equivalent to 4 − x2 − x + 2 ≥ 0, which in turn is equivalent to −x2 − x + 6 ≥ 0. Since −x2 − x + 6 factors as (−1)(x2 + x − 6) = (−1)(x + 3)(x − 2), the original inequality is equivalent to (−1)(x + 3)(x − 2) ≥ 0 or (x + 3)(x − 2) ≤ 0. To solve (x + 3)(x − 2) ≤ 0, notice that this inequality is satisfied by a value of x precisely when either x = −3, x = 2 or the product of the factors (x + 3) and (x − 2) is negative; this last condition is true for −3 < x < 2, as shown in the table below:
y°
x° z°
E
A
D
One way to determine x in terms of y and z is to find the measure of each of the angles of ∆EDC in terms of x, y and z and then apply the triangle sum theorem. Since ∠CED is supplementary to ∠BEC and the measure of ∠BEC is given to be x°, ,the measure of ∠CED is (180 − x)°. Since
( ) produce n CDs is
250 + 5.90
n 10
= 250 + 0.59n dollars.
2. Difficulty: 2
Choiceoint of intersection of lines AC and BD.
B
C
Alternatively, one can apply the interior angle sum theorem to pentagon ABECD. Since line AD is a transversal to the parallel lines AB and CD, it follows that ∠BAD and ∠ADC are supplementary; that is, the sum of the measures of these two angles is 180°. The measure of ∠BEC, interior to polygon ABECD, is (360 − x)°. The measure of ∠EBA is given to be y°, and the measure of ∠DCE is given to be z°. Therefore, the sum of the measures of the interior angles of pentagon ABECD is 180 + (360 − x) + y + z. The interior angle sum theorem applied to pentagon ABECD gives the equation 180 + (360 − x) + y + z = 540, which can be solved for x to arrive at x = y + z.
One way to determine the correct expression is to find
the slope-intercept form of the equation of the line that
passes through the two points given by the ordered
line AC is a transversal to the parallel lines AB and CD, the alternate interior angles ∠BDC and ∠DBA are of equal measure. Thus, the measure of ∠BDC is y, which is also the measure of ∠EDC. The measure of ∠DCE is given to be z°. Therefore, the sum of the angle measures of ∆EDC, in degrees, is (180 − x) + y + z. The triangle sum theorem applied to ∆EDC gives the equation (180 − x) + y + z = 180, which can be solved for x to arrive at x = y + z.
Choice (A) is correct.
To determine the value of b, apply the fact that two complex numbers are equal if and only if the real and pure imaginary parts are equal. Since (a + b) + 5i = 9 + ai, this gives the two equations a + b = 9 and 5i = ai;, that is, a + b = 9 and 5 = a. Therefore, 5 + b = 9, and b = 4.
The cost of the equipment is $250, and each package of 10 blank CDs costs $5.90. The total cost for the band to produce 10 CDs, that is, 1 package, is 250 + 5.90 = 255.90 dollars, and the total cost to produce 20 CDs, that is, 2 packages, is 250 + 5.90(2) = 250 + 11.80 = 261.80 dollars.
3. Difficulty: 2
Choice (C) is correct.