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Seek the answer of science mathematics in the national college entrance examination in 2008 1 (online, etc. )
First, multiple choice questions

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