1、C 2、A 3、A 4、D 5、C 6、B
7、D 8、A 9。 d 10 . d 1 1 . b 12 . b
Fill-in-the-blank question: This big question has four small questions, each with 5 points and ***20 points. Fill in the answers on the lines in the questions.
13、 9 14、2 15、3/8 16、 1/6
.
Third, the solution: this big question is ***6 small questions, with ***70 points. The solution should be written in proof process or calculus steps.
17. Analysis: (i) Derived from sine theorem
a=
acosB-bcosA=( )c
=
=
=
Set by theme
The solution is tanAcotB=4.
(II) tanA=4tanB is known from (i), so a and b are acute angles, so tanb >；; 0
Tan ()
=
≤ ,
And when tanB=, the above formula is equal sign, then the maximum value of tan(A-B) is
18. Solution:
(1) For AO⊥BC, the vertical foot is O and connected with OD. From the title, it can be seen that the bottom BCDE of AO⊥ BC and O are the midpoint of BC.
You know, Rt△OCD∽Rt△CDE,
So ∠ODC=∠CED, so CE⊥OD,
According to the theorem of three vertical lines, AD⊥CE
(2) Judging from the meaning of the question, BE⊥BC is like this, BE⊥ one side ABC, and Abe is like this, ABE⊥ one side ABC.
For CF⊥AB, the vertical foot is F, which is connected with FE, so the CF⊥ plane is ABE.
Therefore, ∠CEF is the angle formed by CE and plane ABE, ∠ cef = 45.
From CE=，CF=
And BC=2, so ∠ ABC = 60, so △ABC is an equilateral triangle.
For CG⊥AD, the vertical foot is G, which is connected with GE.
According to (I), CE⊥AD and CE∩CG=C,
Therefore, AD⊥ plane CGE, AD⊥GE and ∠CGE are the plane angles of dihedral angle c-ad-e.
CG=
GE=
cos∠CGE=
So the dihedral angle C-AD-E is arccos ()
Solution 2:
(i) If AO⊥BC and vertical foot are O, then the bottom BCDE of AO⊥, O is the midpoint of BC, O is the coordinate origin, and ray OC is the positive direction of X axis, and a rectangular coordinate system O-xyz as shown in the figure is established.
Let a (0,0, t) be given by the following known conditions.
C( 1，0，0)，D( 1，，0)，E(- 1，，0)，
So, go to AD⊥CE.
(II) CF⊥AB, whose vertical leg is F, connected with FE,
Let F(x, 0, z) be =(x- 1, 0, z),
Therefore, CF⊥BE and AB∩BE=B, so CF⊥ plane ABE,
∠CEF is the angle formed by CE and plane ABE ∠ CEF = 45.
From CE=，CF=
And CB=2, so ∠ FBC = 60, △ABC is an equilateral triangle, so a (0 0,0,).
Let CG⊥AD, the vertical foot is G, and connected with GE. In Rt△ACD, |AG|= |AD| is obtained.
Therefore, G[]
and
So the included angle is equal to the plane angle of dihedral angle c-ad-e.
By cos( )=
Know that dihedral angle C-AD-E is arccos ()
(19) solution:
(1) f? (x)=3x2+2ax+ 1, and the discriminant δ = 4 (A2-3).
(I) if a> or a 0, f(x) is increasing function;
Including f? (x)& lt； 0, f(x) is a decreasing function;
On the f? (x)>0, and f(x) is increasing function.
(ii) If
(iii) if a=, then f? () =0, and all x≦ have f? (x)>0, so when a=, f(x) is the increasing function on R. ..
(ii) Starting from (i), only when a >; Or a < F(x) is a subtraction function.
Therefore ≤ ①
And ≥ ②
When |a| > is, a≥2 is obtained from ① and ②.
So the range of a is [2, +∞).
(20) Solution:
Notes A 1 and A2 indicate that 1 and 2 tests are carried out according to Scheme A, respectively.
B 1 and B2 respectively indicate that it needs to be tested for 2 times and 3 times according to Scheme B,
A means that the number of tests required by Scheme A is not less than that required by Scheme B. A2 and B2 are independent.
(Ⅰ)
, , 。
P( )=P(A 1+A2？ B2)
=P(A 1)+P(A2？ B2)
=P(A 1)+P(A2)？ P(B2)
=
=
So P(A)= 1-P( )= =0.72.
(ii) The possible values of ξ are 2,3.
P(B 1)=，P(B2)=，P(ξ=2)=P(B 1)=，P(ξ=3)=P(B2)=，
So Eξ= (times).
(2 1) solution:
(i) Let the hyperbolic equation be (a >；; 0, b>0), and the right focus is f (c, 0) (c >; 0)，c2=a2+b2。
Let l 1: bx-ay = 0, and L2: bx+ay = 0.
Then,
.
Because 2+ 2= 2 and
=2 - ,
So 2+ 2=(2-)2,
So tan∠AOB=
And the direction is the same, so ∠AOF= ∠AOB,
therefore
The solution is Tan ∠AOF=, or Tan ∠AOF=-2 (discard).
Therefore.
So the eccentricity of hyperbola e= =
(ii) From a=2b, the hyperbolic equation can be simplified as
x2-4y2=4b2 ①
From the slope of l 1, c= b, the equation of straight line AB is
y=-2(x- b) ②
Substitute ② into ① and simplify to obtain
15x2-32 bx+84b2=0
Let the coordinates of the two intersections of AB and hyperbola be (x 1, y 1) and (x2, y2) respectively, then
x 1+x2=，x 1？ x2= ③
Length of hyperbola section AB.
l= ④
Substitute ③ into ④ and simplify to get l=. Since l=4 is known, b=3 and a=6.
So the hyperbolic equation is
22. Solution:
(1) When 0
f′(x)= 1-lnx- 1 =-lnx & gt； 0
So the function f(x) is the increasing function in the interval (0, 1),
(2) When 0
It is also known from (i) that f(x) is continuous at x= 1
When 0
Therefore, when 0
Proved by mathematical induction as follows: 0
(1) From 0
(ii) If n=k, inequality ② holds, that is, 0.
Then 0
Therefore, when n=k+ 1, inequality ② also holds.
Combining (i) and (ii) to prove
(3) According to (2), {an} increases item by item, so if there is a positive integer m≤k, so that am≥b, then AK+1>; am≥b
Otherwise, if I
AML Nam≤a 1 lnam & lt； a 1 lnb & lt； 0 ③
ak+ 1=ak-aklnak
= AK- 1-AK- 1 lnak- 1-aklnak
……
=a 1- amlnam
From ③, we know AML Nam.
So AK+1>; a 1+k|a 1lnb|
≥a 1+(b-a 1)
=b