1. (Beijing Volume No.7) If positive numbers A, B, C and D satisfy a+b=cd=4, then
A.ab c+d, the equal sign holds, and the values of a, b, c and d are unique.
B.ab c+d, the equal sign holds, and the values of a, b, c and d are unique.
C.ab c+d, the equal sign holds, and the values of a, b, c and d are not unique.
D.ab c+d, the equal sign holds, and the values of a, b, c and d are not unique.
A: We can know from the inequality of mean value. Answer a.
Explain the conditions of equality of mean inequality and give specific values, so A, B, C and D have unique values.
2. (Hunan Volume No.2) The solution set of inequality is ()
A.B. C. D。
Answer: the original inequality can be summed up as option D.
3. ("Shandong" Volume 7) The negation of the proposition "arbitrariness" is ()
A. does not exist,
B. existence,
C. existence,
D. for any,
A: The negation of the full name proposition is an existential proposition. The answer is C.
Explain that the proposition is the content of the new curriculum standard, and it is not difficult to understand its connotation.
4. (Su JuanNo. 10) In the plane rectangular coordinate system, if the plane area is known, the area of the plane area is ().
A.B. C. D。
Solution: Let x+y=x, x-y=t, and we can get the plane area b = {(x, t) | s ≤ 1, s+t ≥ 0, s-t ≥ 0} from the meaning of the question. The answer is B.
5. (Question 6 of National Volume 2) The solution set of inequality: > 0 is
(A)( -2， 1) (B) ( 2，+∞)
(C) ( -2， 1)∩(2，+∞) (D) ( -∞，-2)∩( 1，+∞)
Answer: If the original inequality holds, B and D can be excluded, then if the original inequality still holds, A can be excluded, then you choose C. 。
It shows that in the selection branch of this question, the interval endpoint value only involves the root of the equation corresponding to the original inequality, so the main error lies in the missing solution in the process of solving inequality or intersection, so the result may be chosen as A or B. 。
6. (Question 9 of Tianjin Volume) If all the assumptions are positive numbers, and,,, then ()
A.B. C. D。
Answer:
So, there is a
7. (Chongqing Volume II) The negative proposition of the proposition "If, then" is ()
A. If, then or B. If, then
C. If or, then D. If or, then
Answer: A is the negative proposition of the known proposition, B is the inverse proposition, which is easier to know than C and D, and the answer is D.
8. (Fujian Volume No.7) The subtraction function is known, so the range of real numbers is ().
A.B. C. D。
A: Because f (x) is a decreasing function on R 。
So the solution is or, that is-1
9. (Hubei Volume No.21issue) It is known that m and n are positive integers.
(i) Prove by mathematical induction that when x >; At - 1，( 1+x)m≥ 1+MX；
(ii) For n≥6, it is known and verified that m= 1, 1, 2…, n;
(iii) Find all positive integers n that satisfy the equation 3n+4m+…+(n+2) m = (n+3) n. 。
When analyzing a problem with multiple questions, it is often necessary to apply the conclusion of the previous question to the later one.
The first question of this problem is not difficult, but the second question is quite tricky. We guess: can the conclusion of question (1) be used in question (2)? Question (iii) is more difficult. Can we use the conclusion of question (ii) again?
The practice of solving the problem proves that this conjecture is correct.
A: (1) Omission.
(2) Know the rules.
∴, that is (note: this is a prerequisite for using the first question (I))
According to (i),
But sometimes, there will still be.
(Note: The scaling method is used continuously here to prove the problem. )
(Ⅲ) When, directly calculate:
Obviously n=2 satisfies the condition:
When n=3, the left =33+43+53=2 16, the right =(3+3)3=2 16, ∴n=3 also meets the requirements.
When n=4, left = right =.
Note that the sum of two odd numbers must be odd, and the sum of any number of even numbers is even, so the left side is even and the right side is odd, so the two sides must be unequal, and ∴n=4 does not meet the condition.
When n=5, left = and right =.
Note that the fifth power of any integer is the same as its own unit number, so it is easy to judge that the left unit number is 5 and the right unit number is 2, which is still a Zuo Qi right pair, and ∴n=5 does not satisfy the condition.
So when n=2 or 3.
(Note: Extracurricular basic knowledge related to integer theory is also used in the college entrance examination of mathematics, and this trend is worth noting. )
When n≥6, assuming that it exists, there are:
But:
= .
According to formula (II), right formula
(1) contradicts (2), so when there is no positive integer satisfying the equation 3n+4m+…+(n+2)m=(n+3)n 。
(Note: Only two numbers, 2 and 3, meet the conditions, so we have guessed that qualified positive integers do not exist when n≥6. The strategy to prove the problem is to assume that it exists first, and then disprove this assumption by reducing to absurdity. )
To sum up, all positive integers suitable for this equation are only 2 and 3.
(8) Mathematical elites solve the problem of "conic curve"
1. (Hubei Volume 7, 2007) Hyperbolic C 1: (A > 0, b>0) is L, and the left focus and right focus are F 1 and F2 respectively; The directrix of parabola C2 is L and the focus is F2. The intersection of C 1 and C2 is m, which is equal to
A.- 1。
Solution: Let the centrifugal separation of hyperbola be E, as shown in the figure:
=
The answer is a.
It shows that MN is the intermediary of transformation and is skilled in using definitions.
2. (Question 9 of Hunan Volume) Let be the left focus and the right focus of the ellipse () respectively. If there is a point on its right alignment that allows the line segment centerline to pass through, the eccentricity range of the ellipse is ().
A.B. C. D。
Solution: the right directrix equation of an ellipse is the intersection rule of the middle vertical lines.
When, at least, that is, so choose D.
The answer is D.
It shows that we should make full use of the properties of conic section to find a breakthrough in solving problems.
3. (National Volume I, Question 4) It is known that the eccentricity of hyperbola is, the focus is, and the hyperbolic equation is ().
A.B. C. D。
Answer: c=4, e=2, then a=2. The focus is on the x axis. The answer is a.
explain
4. (Title 1 1 of National Volume I) The focus of the parabola is that the directrix is that the straight line with slope crosses the part of the parabola above the axis at this point, and the vertical foot is, then the area is ().
A.B. C. D。
Answer: |AK|=3-(- 1)=4,
.
The answer is C.
Explain that point A is the breakthrough point, as long as it is solved, it will be solved.
5. (Zhejiang Volume No.4) Sprinklers need to be installed on the square lawn with a side length of16m, so that the whole lawn can spray water. Assuming that the spraying range of each nozzle is a circular surface with a radius of 6 meters, the minimum number of nozzles installed is ().
A.B. C. D。
Answer: If there are at least two on each side and the symmetry is known, at least four should be installed.
And the answer is B.
6. (Zhejiang Volume 9) It is known that the left and right focuses of hyperbola are respectively a point on a quasi-straight line, then the eccentricity of hyperbola is ().
A.B. C. D。
A: Ⅷ
Set, and then solve,
Youyou
get
The answer is B.
Explain how to use vector solution to analyze geometry.
7. (Jiangsu Volume No.3) In the plane rectangular coordinate system, the center of the hyperbola is at the coordinate origin and the focus is on the axis. If the equation of an asymptote is, then its eccentricity is ().
A.B. C. D。
Solution: the slope of the asymptote.
The answer is a.
Explain the quirks.
8. (Title 1 1 in National Volume II) Let F 1 and F2 be the left and right focal points of hyperbola respectively. If there is a point A on the hyperbola, let ∠ F 1A2 = 90? , and |AF 1|=3|AF2|, then the eccentricity of hyperbola is
(A) (B) (C) (D)
Solution: Substitute |AF 1|=3|AF2| according to the problem to get the solution.
Which is the definition of hyperbola.
The answer is B.
Explain that this question will not be wrong except for the wrong part of the question. For example, if the condition |AF 1|=3|AF2| is wrongly selected as |AF 1|=2|AF2|, it may be wrongly selected as A, etc.
9. (Question 12 of National Volume II) Let f be the focus of the parabola y2=4x, and A, B and C are the three points on the parabola. If =0, then | FA |+| FB | | FC | =
9 (B) 6 (C) 4 (D) 3
Answer: To find |FA|+|FB|+|FC|, according to the definition of parabola, we only need the sum of the abscissas of the three points A, B and C, and let the coordinates of the three points A, B and C on the parabola y2=4x be,, respectively.
Because the focal coordinate of parabola y2=4x is,
, and by =0,
Then |FA|+|FB|+|FC|=, then B.
The answer is B.
Explain that if the focus coordinate of parabola is wrongly calculated as (this kind of error is easy to appear), it is wrongly selected as A; If the sum of the abscissas of a vector is wrongly calculated as, it is wrongly selected as d.
10. (Tianjin Volume No.4) Let the eccentricity of hyperbola be, and one of its directrix coincides with that of parabola, then the equation of this hyperbola is ().
A.B.
C.D.
Answer:
The answer is D.
It means that eccentricity is connected with A and C. If you find it, B will know.
1 1. (Liaoning VolumeNo. 1 1) Let P be a point on another curve, and F 1 and F2 are the two focuses of this hyperbola. If | PF 1 |: | PF2 | = 3: 2, then △ PF66.
A. BC 12 years
Solution: According to hyperbolic definition |PF 1|-|PF2|=2. And | PF 1 |: | PF2 | = 3: 2, the solution is |PF 1|=6, |PF2|=4.
According to hyperbolic equation, c2= 13. ∴|F 1F2|=2c=. And ∴| pf1| 2+| pf2 | 2 = | f1F2 | 2, ∴ PF65438+.
∴ .
The answer is B.
Explain that this topic examines the definition, properties and basic operation ability of hyperbola.
12. (Fujian Volume No.6) The equation of a circle with the right focus of the hyperbola as the center and tangent to its asymptote is ().
A.B.
C.D.
Solution: We know that the coordinate of the center of the circle should be (5,0), excluding C and D, and because the distance from point (5,0) to the asymptote is 4, we can verify that item A is correct.
The answer is a.
Explain that this question examines the basic operation of hyperbola and the related knowledge of straight lines and circles.
(3) Mathematical elites solve the "sequence" problem
1. (Guangdong Volume No.5) If the sum of the first n terms of the sequence {0} is known and the k term satisfies 5 < < 8, then k=
(A)9 (B)8 (C)7 (D)6
Answer: The series B is arithmetic progression, which consists of five.
2. (Tianjin Volume 8) Let arithmetic progression's tolerance not be 0. If it is equal to the middle term of, then ().
A.2 B.4 C.6 D.8
Answer: from the meaning of the question, an=(n+8)d, a,
∴(k+8)2d2=9d(2k+8)d.∴k=4.
The answer is B.
3. (Question 6 of Hubei Volume) If the series {an} satisfies N*, it is called {an} as an "equal square ratio series".
Answer: The sequence {an} is an equal square ratio sequence; B: The sequence {an} is a geometric series.
A.a is a sufficient condition for B, but not a necessary condition.
B.a is a necessary but not sufficient condition for B.
C.a is a necessary and sufficient condition for b.
D.A. is neither a sufficient condition nor a necessary condition for B.
Answer: So this series {an} is not a geometric series; If {an} is geometric progression, then the series {an} is an equal square ratio series.
The answer is B.
Explain that 1, 2, 4, 8,-16, -32, ... are equidistant series, but not geometric series.
4. (Hubei Volume No.8) It is known that the sum of the first n terms of two arithmetic progression {An} And {Bn} is an and Bn, respectively, so the number of positive integer n is
A.2 B.3 C.4 D.5
Solution: Use the mean value theorem.
So it is a positive integer if and only if n= 1, 2,3,5, 1 1.
The answer is D.
5. (Question 4 of Liaoning Volume) Let the sum of the first n items of arithmetic progression {an} be Sn, if S3=9 and S6=36, then a7+a8+a9= ().
A.63 B.45 C.36 D.27
Analysis of 1: let the first term of arithmetic progression be a 1 and the tolerance be d,
rule
∴a7+a8+a9=3a8=3(a 1+7d)=3×( 1+7×2)=45.
Analysis 2: From the essence of arithmetic progression:
S′3 = S6-S3 = 36-9 = 27，d′= S′3-S3 = 27-9 = 18。
∴s″3=s3+2d′=9+2× 18=45.
The answer is B.
6. (Fujian Volume No.2) The sum of the first few items in the series is, if, equal to ()
65438 AD+0 BC
A: by, by,
The answer is B.
7. (National Volume INo. 15) The sum of the first few terms of geometric series is known as the common ratio of arithmetic progression.
Solution 1: substitute S2 = (1+q) S 1 and S3 = (1+q+Q2) S 1 into 4.
Note that q≠0, the common ratio q=
Solution 2: Set by the problem
Simplified a2=3a3, so the common ratio q=
Solution 3: S2-S 1+3S3 is obtained from 4S2 = S 1 = 3 (S3-S2), that is, a2=3a3, so the common ratio q=
8. (National Volume I, Question 22) Among the known series, …
(i) General formula for searching;
(2) If,,,
Prove:
Solution: (1) Solution 1: Set by the topic:
.
Therefore, the sequence is a geometric series whose first term is the common ratio.
,
That is, the general formula is.
Solution 2: Setup
arrange
Known by
Comparison coefficient.
∴ .
That is, the order
∴，(n∈N+)
(2) Solution 1: Proved by mathematical induction.
(a) when, because, so.
The conclusion is valid.
(ii) Suppose that when the conclusion is established, that is,
That is.
When,
Say it again,
therefore
.
In other words, when the conclusion holds.
According to (1) and (2).
Solution 2: Pass
therefore
manufacture
have
∵
∴ series is a geometric series, the first term is 1+, and the common ratio is (3+ )2.
∴ ,
Say it again,
To prove this,
Prove this.
Total income
(9) Mathematical elites solve the problem of "solid geometry"
1. (Hubei Volume No.4, 2007) There are two straight lines M and N outside the plane α. If the projections of m and n on plane α are M' and N' respectively, the following four propositions are given:
①m'⊥n' m⊥n； ②m⊥n m'⊥n'
③m' and n' intersect, and M and N intersect or overlap; ④M' and N' are parallel, and M and N are parallel or coincident.
The number of incorrect propositions is
A. 1
Analysis D takes the classroom space as a cuboid model, M' and N' as the root lines of the basement wall, and M and N are selected on the wall, which is easy to know.
M'⊥n' is a necessary or insufficient condition for m⊥n, so ① ② is a false proposition. M' and n' intersect or are parallel, and M and n can be different planes; So ③ ④ is also a false proposition.
Explain that the abstract line (surface) relationship is concrete. That is to find a spatial model, the rectangular classroom is a "cost-free" model. When necessary, candidates can also use the ruler and triangle in their hands as a "graphic combination".
2. (Beijing Volume 3, 2007) A sufficient condition for plane α ‖ plane β is
A. there is a straight line a, a‖α, a‖β.
B. there is a straight line a, a‖β.
C there are two parallel straight lines a, b, a, A ‖ β, B ‖α.
D has two straight lines A, B, A, a‖β, b‖α.
Taking the ceiling and a wall of the examination room as α and β, we can find that different straight lines A and B satisfy items A, B and C, thus excluding the first three items.
It shows that the classroom itself is a good cuboid model, which is simple and clear when judging the relationship between lines.
3. (Question 8 of Hunan Volume 2007) The eight vertices of a cube with a side length of 1 are all on the surface of the ball, which are the midpoints of the sides, so the line segment of the straight line cut by the ball is ().
A.B. C. D。
By analyzing the radius of the circular surface obtained by the D-plane truncated sphere,
The line segment cut by the ball O is the diameter of the circular surface, so D is selected.
Explain the relevant knowledge points: the combination of balls
(1) combination of sphere and cuboid:
The diameter of the circumscribed ball of a cuboid is the diagonal length of the cuboid.
(2) Combination of sphere and cube:
The diameter of the inscribed sphere of the cube is the side length of the cube, the diameter of the inscribed sphere of the cube is the diagonal length of the cube, and the diameter of the circumscribed sphere of the cube is the diagonal length of the cube.
(3) The combination of sphere and regular tetrahedron:
The radii of the inscribed sphere and circumscribed sphere of a regular tetrahedron with side length are.
As shown in the figure, in a regular quadrangular prism, the cosine of the angle formed by straight lines in different planes is ().
A.B. C. D。
If d is connected with CD 1, ∠AD 1C is the angle formed by a non-planar straight line A 1B and AD 1.
Let AB= 1,.
It shows that finding the angle formed by straight lines on different planes is the key to the problem.
5. (Zhejiang Volume 6, 2007) If it is any point outside two straight lines with different planes, then ()
There is one and only one straight line parallel to the two straight lines at the intersection of A.
B only one straight line is perpendicular to all intersections.
The intersection point has one and only one straight line and all intersections.
D. There is only one straight line and different planes at the intersection.
Analysis b for option a, if there is a straight line n parallel to l and m at the intersection p, then l‖m is inconsistent with l and m; For B, a straight line passing through point P and perpendicular to L and M is a straight line passing through point P and parallel to the common perpendicular of L and M; For option c, p may not have a straight line intersection with l and m; For d, there may be countless straight lines passing through points p and l and m.
Explain the spatial line-line relationship and find the spatial model.
6. (Shandong Volume III, 2007) Of the three views of the following geometry, only two are the same ()
A.①② B.①③ C.①④ D.②④
All three views of analytic D cube are the same; The two views of the cone are the same; The three triangular terraces are different; The two views of the regular pyramid are the same.
Explain the exertion of spatial imagination.
7. (Jiangsu Volume 4, 2007) Two straight lines and two planes are known. Give the following four propositions:
① , ;
② , , ;
③ , ;
④ , , .
The serial number of the correct proposition is ()
A.①、③ B.②、④ C.①、④ D.②、③
Analysis C For ②, there are two kinds of positional relationships of straight lines in two parallel planes: parallel or different planes; For ③, one of the parallel lines is parallel to the plane, and the other may be parallel to or in the plane.
8. (Question 7, National Volume II, 2007) It is known that the side length of the regular triangular prism ABC-a1b1is equal to the side length of the bottom surface, then the sine of the angle formed by Ab 1A 1 is equal to.
(A) (B) (C) (D)
The key to analyze the angle between A and the side ACC 1 is to find the projection of straight line AB 1 on ACC 1A 1 plane, that is, to find the projection of B 1 on this plane. According to the properties of regular prism and the theorem that the plane is perpendicular to the plane, it is easy to know B 1.
This is what you want. Depending on the theme, you can calculate the sine of the angle as:
So choose a.
Explain that if the relationship between the angle and the side in the right triangle is confused, it is easy to choose B; If correct
The concept of the included angle between a straight line and a plane is unclear, and it is easy to choose the wrong C or D.
9. (Question 6 of Tianjin Volume in 2007) Let it be two straight lines and two planes. Among the following four propositions, the correct proposition is ().
A. If it is equal to the angle of formation, then
B.if,, then
C. If so, then
D. If,, then
In analytic D A, A and B can be parallel, intersecting and not in the same plane;
In B, A and B can be parallel, intersect and not in the same plane;
A and b in c can be parallel to the intersection of α and β at the same time;
A and b in d can be regarded as normal vectors of α and β.
Explain that a corner of the classroom can also be used as a model, and then choose different wall lines as counterexamples.
10. (Chongqing Volume 3, 2007) If three planes intersect in pairs and the three intersecting lines are parallel to each other, the three planes divide the space into ().
A. part b, part c and part d.
The analysis of C replaces lines with points and surfaces with lines. The schematic diagram can be drawn as follows:
The explanatory diagram is intuitive and does not require reasoning.
11.(Liaoning Volume No.7, 2007) If m and n are two different straight lines and α, β and γ are three different planes, the correct proposition in the following propositions is ().
A. if m, then B. if ∩ =m, ∩ =n, m‖n, then ‖.
C. If m, m‖, then D. If, then
In analytic C A, there are various possibilities for the positional relationship between straight line M and plane α. In B, planes α and β may also intersect; In c ∫m‖, if the intersection plane α of plane γ is in m', then m∫m'. And ∵ m, ∴m'. According to the judgment theorem of vertical plane; In d, planes β and γ can also intersect or be parallel.
Explain that this question examines the positional relationship between straight lines, straight lines and planes, and planes and planes.
12. (Fujian Volume No.8, 2007) is called two different straight lines and two different planes, and the correct one of the following propositions is ().
A.
B.
C.
D.
Analysis D For A, when M and N are two parallel straight lines, A is wrong. For b, m and n may be straight lines out of plane, and for c, straight line n may be in plane α.
It shows that this problem mainly investigates the position relationship between line and surface in space.
13. (Fujian Volume10,2007) If the vertices are in a regular quadrangular prism on the same sphere, the spherical distance between two points is ().
A.B. C. D。
Analysis b is shown in the figure below.
Let the radius of the ball be r, then there is, connecting AC, connecting AC' and A'C intersect at point O, then O is the center of the circumscribed ball.
In △AOC, AO=OC= 1 and AC=, so ∠AOC=.
So the spherical distance between point a and point c is.
Explain that this question examines the knowledge of the assembly.
13 (question 16, Volume I, 2007) The three vertices of an isosceles right triangle are on the three sides of a regular triangular prism. If the side length of the bottom surface of a regular triangular prism is known to be 2, the length of the hypotenuse of the triangle is.
Solution: The isosceles right triangle is ABC, A is the right vertex, and the section parallel to the bottom of A is α.
If b and c are on the same side of α (figure 1), it is easy to prove that ∠ABC is an acute angle, which is irrelevant;
If B and C are on opposite sides of α (Figure 2), then the intersection point B is the section BPQ parallel to the bottom, and it is easy to prove that CP=2AQ according to "isosceles". Taking the midpoint G of BC, the midpoint H of BP, and even AG, GH and HQ, we can prove that AGHQ is rectangular, so BC=2AG=2HQ=.
The key to this solution is to "guess" the chart and do mental arithmetic. Of course, let AQ = x and CP = 2x in Figure 2, and it is also very simple to solve by Pythagorean theorem.
Figure 1 Figure 2
Only graphically, there seems to be no essential difference between Figure 1 and Figure 2. This is because the author did not specify which plane is α, so it seems that B and C are on the same side of plane α. If so, classification is unnecessary.
The next solution to this problem is:
Analytically extend MN and CB to P and even AP.
No. 1, it can be proved that m is the midpoint of PN. Let MD‖BC and D span CC 1. Obviously, it is △ AMB △ MND. So DN=BM=CD, that is, BM= CN is the midpoint of △PNC, and ∴M is the midpoint of PN.
Secondly, it can be inferred from the middle vertical that AM is PN△APN is an isosceles right triangle.
The following is obtained by BA=BP=2, ABP = 120 in △ABP, thus the edge.