First, multiple choice questions
1. There are 6 classes in a grade, and 3 Chinese teachers are sent to teach, and each teacher teaches 2 classes, so the number of different teaching methods is ().
A.C26C24C22 B.A26A24A22
C.C26C24C22C33 D.A26C24C22A33
[Answer] A.
2. Take five different letters from the word "equation" and line them up. There are () in different arrangements with "go" (where "go" is linked together and the order remains unchanged).
A. 120 species B.480 species
About 720 species and about 840 species
[answer] b
[Analysis] Choose the back row first, and choose three of the six letters except inflections. There are C36 permutations, and then take qu as a whole (equivalent to an element) and make a full permutation with the selected three letters. There are A44 permutations, and C36A44=480 permutations based on the principle of step multiplication and counting.
3. Choose three kinds of four different seeds numbered 1, 2, 3 and 4, and try planting on three different plots, one on each plot, among which 1 seed must be tried, so the different method of trial planting is ().
A.24 species B. 18 species
C. 12 species and D.96 species.
[answer] b
[resolution] first choose the back C23A33 = 18, so choose B. 。
4. Take the six numbers 0, 1, 2, 3, 4 and 5, take three different numbers at a time, and put the largest number in the hundredth place to arrange three digits. Such three digits are ().
120
From 360 to 720 A.D.
[Answer] A.
[Resolution] There are C36 ways to select three different numbers, and then put the largest number in the hundredth place, and the other two different numbers in the tenth place and the unit place, so there are A22 permutation methods, so * * has C36A22 = 40 three digits.
5.(20 10 Hunan Province, 7) In a certain information transmission process, a message is represented by the arrangement of four numbers (numbers are allowed to be repeated), and different arrangements represent different messages. If only 0 and 1 are used, then the number of messages is the same as (?
A. 10 B. 1 1
c 12d 15
[answer] b
[Resolution] Information with the same number 0 1 10 has at most two corresponding positions, including three categories:
Category I: There are only two numbers in the corresponding position, and the information is 0 1 10, and C24 = 6 (pieces).
Category II: There is only one number corresponding to information 0 1 10, with C 14 = 4 (pieces).
Category III: Numbers without corresponding information 0 1 10, with C04 = 1 (pieces).
Information 0 1 10 corresponds to at most two pieces of information with the same number, of which 6+4+1=1(pieces).
6. During the opening of Fortune Global Forum in Beijing, 0/4 volunteers from a university participated in the reception. If there are 4 people in each class every morning, noon and evening, and each person is worth at most one class every day, then the number of people in different classes on the first day of school is ().
A.c 4 14c 4 12c 48 B . c 12 14c 4 12c 48
C.c 12 14c 4 12c 48 a33 D . c 12 14c 4 12c 48 a33
[answer] b
[Resolution] Solution 1: According to the meaning of the question, the number of different shifts is: c 414c410c46 =14×13×12×166. 10×9×8×74! 6×52! = c 12 14c 4 12c 48。
So choose B.
Option 2: You can also choose 12 people first and then arrange them in the following order: c1214c412c48c44, that is, choose B. 。
7. (In 2009, Hunan University of Science and Technology) selected three college graduates from 10 to serve as village head assistants, so at least 1 of A and B were selected, and the number of people with different selection methods that C was not selected was ().
A.85 B.56
C.49 D.28
[answer] c
"Analysis" examines the constrained combination problem.
(1) There are two ways to choose 1 from Party A and Party B. There are C27 ways to choose 2 from 7 people except Party A, Party B and Party C. According to the principle of step-by-step multiplication and counting, * * * has 2c27 = 42 ways.
(2) Select Party A and Party B, and then select 1 person from 7 other people except Party C. 。
According to the principle of classified counting, * * has 42+7 = 49 different selection methods.
8. The tetrahedron with the vertex of the regular triangular prism as the vertex has ().
A.6 B. 12
C. 18 D.30
[answer] b
[resolution] C46-3 = 12, so choose B. 。
9. (Liaoning Province, May, 2009) Choose 3 doctors from 5 male doctors and 4 female doctors to form a medical team, which requires both male and female doctors, so different team building schemes are * * * ().
A.B. 80 kinds
C. 100 species and D. 140 species.
[Answer] A.
[Analysis] Examine the knowledge about permutation and combination.
Solution: It can be divided into two categories: 2 male doctors, 1 female doctors or 1 male doctors and 2 female doctors.
* * * C25C 14+C 15C24 = 70, ∴ choose A. 。
10. Let set Ⅰ = {1,2,3,4,5}. Choose two nonempty subsets A and B of Ⅰ. If the minimum number in B is greater than the maximum number in A, there are different selection methods * * * ().
A.50 species B. 49 species
C.D. 47 species
[answer] b
The analysis mainly examines the basic knowledge of set, arrangement and combination, and examines the thinking method of classified discussion.
Because the largest element in set A is smaller than the smallest element in set B, the elements in A are taken from 1, 2,3,4, and the elements in B are taken from 2,3,4,5. Because a and b are not empty, there must be at least one element.
1 When a = {1}, there are 24- 1 = 15 schemes to choose from.
When a = {2}, there are 23- 1 = 7 schemes to choose from.
When a = {3}, there are 22- 1 = 3 options for choosing B.
When a = {4}, there are 2 1- 1 = 1 schemes to choose from.
Therefore, when A is a single element set, B has 15+7+3+ 1 = 26 kinds.
When 2 A is a set of two elements,
The largest element in A is 2, and there are 1, and there are 23- 1 = 7 schemes for choosing B. 。
The largest element in A is 3, with C 12, and the scheme of choosing B is 22- 1 = 3. Therefore, * * * has 2× 3 = 6.
The largest element in A is 4, and there are three kinds of C 13. The scheme of choosing B is 2 1- 1 = 1 species, so there are 3× 1 = 3 species * *.
Therefore, when there are two elements in A, there are 7+6+3 = 16 kinds of * *.
When 3 A is a set of three elements,
The largest element in A is 3, and there are 1, and there are 22- 1 = 3 schemes for choosing B. 。
The largest element in A is 4, there are C23 = 3 kinds, and there are 1 kind of' scheme of choosing B'.
∴ * * There are 3× 1 = 3 kinds.
When a is three elements, there are 3+3 = 6 kinds of * *.
When 4 A is four elements, it can only be A = {1, 2, 3, 4}, so B can only be {5}, and there is only one kind.
∴ * * There are 26+ 16+6+ 1 = 49 species.
Second, fill in the blanks
1 1. A middle school in Beijing will send nine computers of the same model to three hope primary schools in the western region, and each primary school will get at least two computers. * * * There are _ _ _ different delivery methods.
[Answer] 10
[Analysis] First, each school should get one set, and then divide the remaining six sets into three parts and solve them by plug-in. There are C25 = 10 kinds of * *.
12.3 people have 7 seats in a row, and two people are not allowed to be adjacent. The total number of all different permutations is _ _ _ _ _.
[Answer] 60
[Resolution] For any sitting method, you can see that four spaces are 0 and three people are 1, 2, 3, so all the species numbers of different sitting methods can be regarded as a code of four zeros and 1, 2, 3, which requires that 1, 2, 3 cannot be adjacent, so four zeros make up five.
There are 35 = 60 different arrangements.
13.(09 Hainan Ningxia Li15) Six of the seven volunteers were arranged to participate in community public welfare activities on Saturday and Sunday. If three people are arranged every day, there are _ _ _ _ _ _ different arrangement schemes (answer with numbers).
[Answer] 140
[Analysis] This question mainly examines the knowledge of permutation and combination.
If three people are arranged every day, the different arrangements are as follows
C37C34 = 140 species.
14.20 10 During the Shanghai World Expo, five volunteers were assigned to venues in three different countries to participate in the reception work, and the number of schemes for assigning at least one volunteer to each venue was _ _ _ _ _.
[Answer] 150
[Resolution] There are C35+C25C232 species in the * * group, and then there are A33 species, so * * * has (C35+C25C232) A33 = 150 schemes.
Third, answer questions.
15. Solve the equation cx2+3x+216 = c5x+516.
[resolution] because cx2+3x+216 = c5x+516, x2+3x+2 = 5x+5 or (x2+3x+2)+(5x+5) = 16, that is, x2-2x-.
16. ∠ There are five points different from O ON the edge of mon, and four points different from O on the edge of ON. How many triangles can you get with this 10 point (including O point) as the vertex?
[Analysis] SolutiON 1: (Direct method) Consider several cases: in a triangle with O as the vertex, the other two vertices must be on OM and ON respectively, so there is c15c14; A triangle with O as its vertex, with two vertices on OM and one vertex on C25C 14. There are 24 vertices in C15c. Because this is a classification problem, using the principle of classification addition counting, * * has c15c14+c25c15c24 = 5× 4+10× 4.
SolutiON 2: (indirect method) does not consider the problem of * * * line points, and the combination number of any three points in 10 is C3 10, but any three of the six points on om (including O points) cannot get triangles, and any three of the five points on ON (including O points) cannot get triangles, so *. That is, c 310-c36-c35 =10× 9× 81× 2× 3-6× 5× 4/kloc-0 /× 2× 3-5× 41× 2 =/.
Solution 3: You can also think of point O as a point on the edge of OM. Firstly, take two points from six points on om (including point O) and one point from four points on ON (excluding point O) to get C26C 14 triangle, and then take one point from five points on OM (excluding point O) and four points on ON (excluding point O).
17. A football match *** 12 teams participate in three stages.
(1) Group Match: Divided into Group A and Group B by drawing lots. Six teams in each group will play a single round robin match, and the top two teams will be selected by points and net remaining balls;
(2) Semi-final: the first place in Group A and the second place in Group B, and the first place in Group B and the second place in Group A will play a home and away cross-elimination match (one for each team) to decide the outcome;
(3) Final: The two winning teams will participate in a final to decide the outcome.
How many games does the whole schedule need?
[resolution] (1) group match, each group of 6 teams plays a round, that is, any two teams of 6 teams play once, and the required number of games is the combination number of any two elements in 6 elements, so the group match * * * plays 2c26 = 30 (field).
(2) In the semi-final, the first place in Group A and the second place in Group B (or the first place in Group B and the second place in Group A) will play one game at home and away, and the required number of games is the arrangement number of two elements in the two elements, so the semi-final * * * will play 2A22 = 4 (games).
(3) Only 1 game is needed for the final.
So the whole race needs 30+4+ 1 = 35 (games).
18. There are 9 different extracurricular books distributed to students A, B and C. How many kinds of books are there in the following situations?
(1) A has 4 copies, B has 3 copies and C has 2 copies;
(2) 4 copies for one person, 3 copies for one person and 2 copies for one person;
(3) Party A, Party B and Party C each hold 3 copies.
[Resolution] The following main information can be obtained from the topic:
① Distribute 9 different extracurricular books to students A, B and C;
The conditions of the three questions in the title are different.
To solve this problem, first judge whether it is related to order, and then use relevant knowledge to solve it.
[Resolution] (1) is completed in three steps:
Step 1: Choose 4 books from 9 different books for A, and there are C49 methods.
Step 2: Choose three books from the remaining five books for B, and there are C35 methods.
Step 3: There are C22 ways to give the remaining books to C,
∴ * * * has different classifications: C49c35c22 = 1260 (species).
(2) It is completed in two steps:
The first step: divide 4 books, 3 books and 2 books into three groups, and there are 22 methods in C49C35C22;
Step 2: Distribute three groups of books to A, B and C, with 33 methods.
* * * There are C49C35C22A33=7560 (species).
(3) Solve by the same method as (1),
Get c39c36c33 = 1680 (species).
Senior two math test questions and answers 2
First, multiple choice questions
1. Given an+ 1=an-3, the sequence {an} is ().
A. Increasing sequence B. Decreasing sequence
C. Constant sequence D. Swing sequence
Analysis: ∫an+ 1-an =-30. From the definition of descending order, we can see that option B is correct, so we choose B.
Answer: b
2. Let an =1n+1n+2+1n+3+12n+1(nn *), then ().
An+1 an
C.an+ 1
Analysis: an+1-an = (1n+2+1n+3+12n+0+12n+2+12n+3)-(.
nN *，an+ 1-an0。 So, C.
Answer: c
3. The general formula of 1, 0,1,0 is ()
a . 2n- 1 b . 1+- 1 N2
c . 1- 1 N2 d . n+- 1 N2
Analysis: Solution 1: Substitution verification method.
Solution 2: Each term can be converted into 1+ 12, 1- 12, 1+02, 1- kloc-0/2, and even terms are/kloc-0.
Answer: c
4. If the known sequence {an} satisfies a 1=0, an+1= an-33an+1(nn *), then a20 is equal to ().
A.0 B.-3
C.3 D.32
Analysis: From a2=-3, a3=3, a4=0 and a5=-3, we know that the minimum positive period of this series is 3, a20=a36+2=a2=-3, so we choose B. 。
Answer: b
5. Given the general term an {an} = N2 N2+ 1, then 0.98 ().
A is the term of this series, and n=6.
B. not an item in this series.
C is the term of this series, and n=7.
D is the term of this series, and n=7.
Analysis: from n2n2+ 1=0.98, we get 0.98n2+0.98=n2, n2=49.n=7(n=-7 omitted), so we choose C. 。
Answer: c
6. If the general formula of the series {an} is an=7(34)2n-2-3(34)n- 1, then () of the series {an}
A The longest period is a5 and the shortest period is a6.
B The maximum duration is a6 and the minimum duration is a7.
The maximum term of c is a 1 and the minimum term is a6.
D The maximum duration is a7 and the minimum duration is a6.
Analysis: let t=(34)n- 1, nN+, then t(0, 1], and (34)2n-2=[(34)n- 1]2=t2.
So an=7t2-3t=7(t-3 14)2-928.
The function f(t)=7t2-3t is a decreasing function at (0,3 14) and a increasing function at [3 14, 1], so a 1 is the largest term, so c 。
Answer: c
7. If the sum of the first n terms of the series {an} is Sn=32an-3, the general term formula of the series is ().
A.an=23n- 1 B.an=32n
C.an=3n+3 D.an=23n
Analysis:
①-② anan- 1=3。
∫a 1 = s 1 = 32a 1-3，
A 1=6，an=23n。 So I chose D.
Answer: d
8. In the sequence {an}, an=(- 1)n+ 1(4n-3), and the sum of the first n items is Sn, then S22-S 1 1 is equal to ().
A.-85 B.85
C.-65 D.65
Analysis: s22 =1-5+9-13+17-21+-85 =-44,
s 1 1 = 1-5+9- 13 ++ 33-37+4 1 = 2 1，
S22-S 1 1=-65。
Or s22-s11= a12+a13+a22 = a12+(a13+a14)+(.
Answer: c
9. In the sequence {an}, a 1= 1, a2=5, an+2=an+ 1-an, then a2007 is equal to ().
A.-4 B.-5
C.4 D.5
Analysis: The first calculation is 1, 5,4,-1,-5,4, 1, 5,4, and the discovery period is 6, so a2007=a3=4. So, C.
Answer: c
10. In the sequence {an}, an = (23) n-1[(23) n-1-1], then the following statement is correct ().
A the maximum duration is a 1 and the minimum duration is a3.
The maximum term of b is a 1, and the minimum term does not exist.
The maximum term of c does not exist, and the minimum term is a3.
The maximum term of d is a 1 and the minimum term is a4.
Analysis: let t=(23)n- 1, then t = 1, 23, (23)2, when t(0, 1), an=t(t- 1), an = t (.
So the maximum term is a 1=0.
When n=3, t=(23)n- 1=49, a3 =-2081;
When n=4, t=(23)n- 1=827, A4 =-152729;
Ni a3
A: A.
Second, fill in the blanks
1 1. The general formula of known sequence {an} an=
Then its first eight items are _ _ _ _ _.
Analysis: substitute n= 1, 2, 3 and 8 into the general formula in turn to get it.
Answer: 1, 3, 13, 7, 15,1,17, 15.
12. The general term formula of a given series {an} is an=-2n2+29n+3, and the largest term in {an} is _ _ _ _ _ _ _.
Analysis: an=-2(n-294)2+8658. When n=7, an is the largest.
Answer: 7
13. If the sum of the first n items of the series {an} is Sn=log3(n+ 1), then a5 is equal to _ _ _ _ _ _ _ _ _.
Analysis: A5 = S5-S4 = log3 (5+1)-log3 (4+1) = log365.
Answer: log365
14. Give the following formula:
①an=sinn
②an=0, n is even,-1n, n is odd;
③an =(- 1)n+ 1. 1+- 1n+ 12；
④an = 12(- 1)n+ 1[ 1-(- 1)n]。
Where: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Parse: It can be obtained by enumeration.
Answer: ①
Third, answer questions.
15. Find the general formula of the sequence 1, 1, 2,2,3,3.
Analysis: This series is 1+ 12, 2+02, 3+ 12, 4+02, 5+ 12, 6+02. According to the molecular law, the former constitutes a positive natural number series, and the latter constitutes/kloc.
an=n+ 1 - 1n22，
That is, an =14 [2n+1-(-1) n] (nn *).
It can also be expressed as the following parts.
16. The general formula of known sequence {an} an = (-1) n12n+1. Look for a3, a 10, a2n- 1.
Analysis: replace n in the general formula with 3, 10, 2n- 1 respectively, and get
a3 =(- 1)3 123+ 1 =- 17，
a 10 =(- 1) 10 10+ 1 = 12 1，
a2n- 1 =(- 1)2n- 1 122n- 1+ 1 =- 14n- 1。
17. In the sequence {an}, it is known that the general formula of a 1=3, a7= 15, {an} is a linear function about the number of terms n. 。
(1) Find the general term formula of this series;
(2) Take out all even terms in this series, form a new series {bn} according to the original series, and find the general term formula of the series {bn}.
Analysis: (1) According to the meaning of the question, the general formula can be set to an=pn+q,
P+q=3，7p+q= 15。 The solution is p=2 and q= 1.
The general formula of {an} is an=2n+ 1.
(2) According to the meaning BN = A2n = 2 (2n)+1= 4n+1,
The general formula of {bn} is bn=4n+ 1.
18. Given an=9nn+ 1 10n(nN*), does the sequence have the largest term? If yes, find out the biggest item, if not, explain the reason.
Analysis: ∫ an+1-an = (910) (n+1) (n+2)-(910) n (n+1) = (.
When n7, an+ 1-an
When n=8, an+1-an = 0;
At n9, an+ 1-an0.
a 1
Therefore, the term of sequence {an} is the largest, a8=a9=99 108.