1.r,; 2.3; 3. 1; 4.5; 5.;
6.2; 7 . y = 2x+3; 8. 1.5; 9.; 10.;
1 1. Necessary and sufficient; 12.- 1; 13.; 14.2.
Second, the solution: this big question is ***6 small questions, ***90 points. Write a written explanation, proof process or calculus steps when answering.
15. (The full score of this small question is 14)
The radius of the circumscribed circle of △ABC is 1, and the opposite sides of angles A, B and C are A, B and C respectively. Vector m =,
N= satisfies m//n/n.
The value range is (1);
(2) If the real number X satisfies abx=a+b, try to determine the value range of X. 。
Solution (1) Because m//n, so ... 2 points.
Because the radius of the circumscribed circle of triangle ABC is 1, it is obtained by sine theorem.
So ...
Because. Therefore, the triangle ABC is a right triangle with ................... 5 points.
, because,
So, therefore, seven points.
29 points.
Set, and then, ................. 1 1 min.
, because < 0, the function decreases monotonically on (1,].
So, so the range of real number x is ....................................................................................................................................................................
16. (The full score of this small question is 14)
In quadrilateral P-ABCD, quadrilateral ABCD is trapezoidal, with AD‖BC, ∠ ABC = 90, plane PAB⊥ plane ABCD, and plane PAD⊥ plane ABCD.
(1) verification: PA⊥ plane abcd;;
(2) If the plane PAB is PCD, Q: Can the straight line L be parallel to the plane ABCD?
Please explain the reason.
(1) proves that because ∠ ABC = 90, AD⊥AB.
Plane PAB⊥ plane ABCD, plane PAB plane ABCD=AB,
So AD⊥ airplane PAB, so advertising ⊥ PA .................................................................................. 3 points.
Similarly, you can get AB ⊥ PA ................................................................................................................................................................
Because AB and AD planes ABCD, and ABAD=C,
So PA⊥ plane ABCD ....................................................................................................................................................................
(2) The solution (method 1) is not parallel to .......................... 9 points.
Proof: suppose the straight line l‖ plane ABCD,
Because the L plane PCD, and the plane PCD plane ABCD=CD, so ‖ CD .............11min.
In the same way, you can get l‖AB, so the score of AB ‖ CD ..........................13.
This contradicts that AB and CD are the two waists of right-angled trapezoidal ABCD.
So the assumption is wrong, so the straight line L is not parallel to the plane ABCD .................................14 minute.
(Method 2) Because of AD‖BC in trapezoidal ABCD,
Therefore, the straight line AB intersects with the straight line CD, and let ABCD = T. .........................................................................................................................................................
T-plane PCD comes from TCD and CD-plane PCD.
Similarly, ............................................................................13 of PAB in T-plane.
That is, t is the common point of plane PCD and plane PAB, so PT is the intersection of plane PCD and plane PAB.
Therefore, the straight line is not parallel to the ABCD plane ............................................................ 14.
17. (The full score of this small question is 15)
Let a be a real number and the function is known.
(1) When a= 1, find the extreme value of the function.
(2) If equation =0 has three unequal real roots, find the range of a. 。
The solution (1) depends on the topic, so .......................... scored 2 points.
pass by
x 0 2
+ 0 - 0 +
↗ Maximum ↘ Minimum ↗
Five points.
Get the maximum value at and the minimum value at ................................. 7 points.
(2) Because ... Because ... So.
So the two roots of the equation are A- 1 and a+ 1.
Obviously, the maximum value of the function is obtained at x = a- 1, and the minimum value is ................................. 1 1 min at x=a+ 1.
Because equation =0 has three unequal real roots,
So get to know.
Therefore, the value range of a is .......................................................................................................................................................................
18. (The full score of this small question is 15)
As shown in the figure, the ellipse (a >;; B>0) are F 1 and F2, respectively, and m and n are two moving points on the right alignment line of the ellipse.
And ...
(1) Let C be a circle with a diameter of MN, and try to judge the positional relationship between the origin O and the circle C;
(2) Let the eccentricity of the ellipse be, and the minimum value of MN be, and find the elliptic equation.
Solution (1) Let the focal length of the ellipse be 2c (c >: 0),
Then the right directrix equation is x =, and f 1 (-c, 0), F2 (c, 0). ...............................................................................................................................
Let m,
Then =
................................., 4 points.
Because, therefore, that is.
So ∠MON is an acute angle.
So the origin O is outside the circle C, and .................................................... is 7 points.
(2) Because the eccentricity of the ellipse is 0, a=2c, so …
So M, and ........................, 9 points.
Mn2 =(y 1-y2)2 = y 12+y22-2y 1 y2 ......................................................................................................................
If and only if y 1 =-y2 = or y2 =-y 1 =, take the "=" sign.
So (MN)min= 2c=2 = 2, so c= 1, so a = 2, b =,
So the elliptic equation is 15 point.
19. (Full score for this small question 16)
The following series is called "Sendram Screen" and marked as S. Its characteristic is that each row and column is arithmetic progression, and the number of row I and column J is marked as Aij.
1 4 7 10 13 …
4 8 12 16 20 …
7 12 17 22 27 …
10 16 22 28 34 …
13 20 27 34 4 1 …
… … … …
(1) proves that there is a constant, which is always a composite number for any positive integer I and J;
(2) setting? The numbers 1, 8, 17, 28, 4 1, … on the main diagonal of S form a series. It is proved that there are no positive integers k and m, which makes it a geometric series.
(3) Are there positive integers p and r for the sequence in (2)? And make it arithmetic progression. If it exists, write a solution (no need to write reasoning process); If it does not exist, please explain why.
(1) It is proved that the sequence {A 1j}(j= 1, 2, ...) constitutes a arithmetic progression with a tolerance of 3, A 1J = 1+(j-65438
Sequence {a2j} (j = 1, 2, ...) is a arithmetic progression with a tolerance of 4.
Therefore, a2j = 4+(j-1) × 4 = 4 j. ........................................................................................................................................
So A2J-A 1J = 4J-(3J-2) = J+2,
Therefore, the sequence {aiji} (I = 1, 2, ...) is a arithmetic progression with the first term of 3j-2 and the tolerance of j+2.
Therefore, aiji = 3j-2+(I-1) × (j+2) = ij+2i+2j-4 = (I+3) (j+2) 8. ........................................................................................
Therefore, Aij+8 = (I+3) (J+2) is a composite number.
So when = 8, for any positive integer I and J, it always adds up to 6 points.
(2) Prove that (reduction to absurdity) has k and m, making it a geometric series.
That's seven points.
∫bn = Ann =(n+2)2-4
∴
OK,
That is 10.
And ∵, and k, m∈N, ∴k≥2, m≥3,
∴, which contradicts∈ z, so there are no positive integers k and m, which becomes a geometric series. ...........................................................................................................................................
(3) Assuming that the solution has a satisfactory condition, then
That is 14 points.
Might as well arrive on time
Therefore, there is a score of ..................................... 16 for the arithmetic series.
(Note: The array in question (3) is not unique, for example, it can be. )
20. (Full score for this small question 16)
F(a), f(b) and f(c) are also three sides of any triangle. As long as its three sides A, B and C are within the definition of function f(x), then f(x) is called "triangle preserving function".
(1) Judge whether the following functions are "triangle preserving functions" and prove your conclusion:
①f(x)=; ② g(x)=sinx (x∈(0,π))。
(2) If the function h (x) = lnx (x ∈ [m, +∞)) is a triangle preserving function, find the minimum value of m. 。
(1) the answer F (x) = is a triangle preserving function, and G (x) = sinx (x ∈ (0, π)) is not a triangle preserving function.
It is proved that ① f (x) = is a triangle preserving function.
For three sides of any triangle A, B and C, then a+b > c+a>b+c > a, c+a>b,
f(a)=,f(b)=,f(c)=。
Because (+) 2 = A+2+B > C+2 > () 2, so+>
It can also be proved that:+>,+>.
Therefore, f(a), f(b) and f(c) are also the lengths of three sides of a triangle, so f (x) = is a triangle preserving function. .....................................................................................................................
(2) g (x) = sinx (x ∈ (0, π)) is not a triangle preserving function. Obviously, these three numbers can be used as one.
The length of three sides of a triangle, and SIN = 1, SIN =, cannot be used as the length of three sides of a triangle.
Therefore, g (x) = sinx (x ∈ (0, π)) is not a triangle preserving function .......................................... 8 points.
(2) The minimum value of solution M is 2. ..................................................................................................................................................................
(i) First, it is proved that the function h (x) = lnx (x ∈ [m, +∞)) is a triangle preserving function when M≥2.
For any triangle whose three sides are A, B, c ∈ [m, +∞] and a+b > c+a>b+c > a+b>c+a > b,
Then h (a) = lna, h (b) = lnb, and h (c) = lnc.
Because a≥2, b≥2 and a+b > c, (a-1) (b-1) ≥1,ab ≥ a+b > c, so lnab>lnc,
That is, LNA+LNB > LNC. Similarly, LNB+LNC > lna+lnb>lnc+LNA > LNB can also be proved.
So lna, lnb and lnc are three sides of a triangle.
Therefore, the function h (x) = lnx (x ∈ [m, +∞], M≥2) is a triangle preserving function. ......................................................................................................................
Secondly, it is proved that when 0
When 0
Because 0 < m < 2 and m+m = 2m > M2, m, m and M2 are three sides of a triangle.
And lnM+lnM = 2lnM = lnM2, so lnM, lnm and lnM2 can't be three sides of a triangle.
So h (x) = lnx is not a triangle preserving function.
Therefore, when m < 2, h (x) = lnx (x ∈ [m, +∞)) is not a trigonometric preserving function.
To sum up, the minimum value of m is 2 ....................................................................................................................................................................
Additional question part
2 1. (multiple choice questions) This big question includes four questions: A, B, C and D * * *. Please choose two questions from these four questions. 10 for each question, and ***20. Please fill in the question mark accurately on the answer sheet. Write a written description, proof process or calculation steps when answering.
Elective course 4- 1: Selected lectures on geometric proof
As shown in the figure, PA intercepts ⊙O at the point, and D is the midpoint of the point, which leads to it.
The secant crosses o at two o'clock.
Prove that because it is tangent to the circle,
So, ..................................... scored 2 points.
Because d is the midpoint of PA, DP=DA,
So dp2 = db DC, that is to say, ...................................................................... scored 5 points.
Because, so ∽, ....................................... 8 points.
Therefore, ................... 10.
Elective 4-2: Matrix and Transformation
It is known that under the transformation of a second-order matrix m, points become points and points become points. Find the matrix m.
Solution, 2 points.
By,, .................................. 5 points.
Get .......................... 8 points.
Therefore, ........................ 10 points.
Elective 4-4: Coordinate System and Parameter Equation
In the polar coordinate system, it is known that the center coordinate of circle C is C (2,) and the radius is R=, and the polar coordinate equation of circle C is found.
Solution 1: Let P(ρ, θ) be any point on the circle, then PC = r = ..........................................................................................................................................
According to the cosine theorem, ρ 2+22-2× 2× ρ cos (θ-) = 5. ...................................................................................................................................
Simplify and get ρ 2-4 ρ cos (θ-)+ 1 = 0, which is the equation of circle C. .......................................................................................................................
Solution 2: Change the center of the circle C (2,) into a rectangular coordinate of (1,) with radius R=, ............................................................................................................................
So the equation of circle C is (x- 1) 2+(y-) 2 = 5. ...................................................................................................................................
C is converted into polar coordinate equation, and (ρ cos θ- 1) 2+(ρ cos θ-) 2 = 5 ... get 6 points.
Simplify and get ρ 2-4 ρ cos (θ-)+ 1 = 0, which is the equation of circle C. .......................................................................................................................
Elective Course 4-5: Special Lecture on Inequality
Understand and verify:
Prove that because ...
So, so, 10.
22. You must do the problem, and this small question is 10. Write a written explanation, proof process or calculus steps when answering.
The probability of throwing three commemorative coins A, B and C is shown in the following table.
commemorative coin
Probability a
Throw these three commemorative coins at the same time, and set the number of heads up.
(1) distribution table and mathematical expectation;
(2) In the probability (I = 0, 1, 2, 3), if the value of A is the largest, find the range of A. 。
The solution (1) is a probability that the front face is up and the back face is up. The possible values are 0, 1, 2, 3.
,
,
, ... 4 points
So the distribution list is
The mathematical expectation is
....................., five points.
(2) ,
,
.
To sum up, the value range of a is ............... 10.
23. You must do the problem, and this small question is 10. Write a written explanation, proof process or calculus steps when answering.
Known. Prove by mathematical induction.
It is proved that (1) when n=2 and left and right =, the inequality holds.
.............................., two points.
(2) Suppose that when n=k (), the inequality holds, that is, ... 4 points.
Because of this, ................ scored six points.
When n=k+ 1,
.
That is, when n=k+ 1, the inequality also holds ........... 9 points.
Combining (1) and (2), we can see that the inequality always holds.
.................... 10 point