Seeking answers to the true questions (multiple-choice questions) in the history of college entrance examination (2 volumes of 2065438+2004 national new curriculum standards)

Where are you a candidate?

This is the answer in Chinese.

First, multiple choice questions

( 1)c

(2)b

(3)b

⑷a⑸b⑹d

(7)b

(8)d

(9)c

( 10)a

( 1 1)a

( 12)d

Second, fill in the blanks

( 13)-6 ( 14) ( 15) ( 16)

Third, answer questions.

(18) solution:

(1) Because

Derived from cosine theorem

So bd

+ad

abdominal muscle

, so bd

Police again

Bottom abcd, bd

pitch diameter of thread

So bd

flat gasket

therefore

dad

Bottom fuze (abbreviation of base detonating)

(2) As shown in the figure, D is the coordinate origin, the length of ad is the unit length, and the ray da is

The positive semi-axis of the shaft establishes the spatial rectangular coordinate system d-

, then

Let the normal vector of the plane pab be n=(x, y, z), then

that is

Therefore, we hope that n=

Let the normal vector of the plane pbc be m, then

M=(0，- 1，

)

Therefore, the cosine of dihedral angle a-pb-c is

(19) solution

(1) According to the test results, the product quality yield of Formula A is

Therefore, the estimated ratio of high-quality products produced by Formula A is 0.3.

According to the test results, the frequency of producing high-quality products by Formula B is as follows

Therefore, the estimated ratio of high-quality products produced by Formula B is 0.42.

(2) The quality index value of 100 product produced by Formula B is within this range.

The frequencies are 0.04, 0.054 and 0.42, respectively, so

p(x=-2)=0.04，p(x=2)=0.54，

p(x=4)=0.42，

In other words, the distribution list of x is

The mathematical expectation of x ex=-2×0.04+2×0.54+4×0.42=2.68.

(20)

solve

(i) Let m(x, y) be given by known b (x, -3) and a (0,-1).

therefore

=(-x，- 1-y)，

=(0，-3-y)，

=(x，-2)。

From the meaning of the question (

=0,

That is, (-x, -4-2y)? (x，-2)=0。

So the equation of curve c is y=

-2.

(i) Establishment

, by

Yes, when?

When,

. but

, so

while

When,

, available

When x

( 1,+

), h (x) < 0, available.

h(x)>0

So when x>0 and x

In 1, f(x)- (

)>0, that is, f(x) >

(ii) Set 0 < k< 1. Because when x

( 1,

)、(k- 1)(x

+ 1)+2x & gt； 0, so

(x)>0 and

H( 1)=0, so when x

( 1,

)，h(x)>； 0, available

h(x)& lt； 0, which contradicts the topic.

(3) Let k

1. At this moment

(x)>0, and h( 1)=0, so when x.

( 1,+

)，h(x)>； 0, available

h(x)& lt； 0, which contradicts the topic.

On the whole, the value range of k is (-

,0]

(22)

Solution:

(I) connect de, in delta ade and delta △acb according to the meaning of the question,

ad×ab=mn=ae×ac，

that is

And ∠dae=∠cab, so △ade∽△acb.

Therefore ∠ade=∠acb.

So c, b, d and e are four * * * cycles.

(ⅱ)m=4，

When n=6, equation x

-14x+mn=0, two of them are X ..

=2，x

= 12.

therefore

ad=2，ab= 12。

Take the midpoint G of ce and the midpoint F of DB, cross G and F respectively as the vertical lines of ac and ab, and the two vertical lines intersect at point H to connect dh. Because C, B, D and E are * * * circles, the center of the circle where C, B, D and E are located is H and the radius is dh.

Because ∠ a = 90.

, so GH∨ab,

HF∑AC。

hf=ag=5，df=

( 12-2)=5.

Therefore, the radius of the circle where C, B, D and E are located is 5.

Questioner's question

20 1 1-06- 13

1 1:07

Well, thank you. What about the two volumes? Report to the authorities

My supplement

20 1 1-06- 13