20 10 mathematical answers of model science in Haidian No.1 middle school

Haidian district third grade second semester mid-term exercise.

Mathematics (literature)

Reference answer and grading standard 20 10.4

Note: Reasonable answers may be scored as appropriate, but shall not exceed the original score.

The first volume (multiple choice questions ***40 points)

First, multiple-choice questions (this big question is ***8 small questions, each with 5 points and ***40 points)

Title number

1 2 3 4 5 6 7 8

answer

A C B C D A B A

Second coupon (non-multiple choice question * * 1 10 score)

2. Fill in the blanks (this big question is ***6 small questions, with 5 points for each small question, there are two empty small questions, 3 points for the first empty question, 2 points for the second empty question, and 30 points for * * *).

9.4 10. 1 1.6 12.30 13. 14.,

15. (The full score of this small question is 13)

Solution: (i) As can be seen from the figure, a =1.............1min.

So .................... scored two points.

So .................. scored three points.

Again, and

So ..................... scored five points.

So ................. scored six points.

(II) by deleting item (i),

So =

............................, eight.

Nine points

................ 10 point

Because, therefore,

Therefore,

When, get the maximum .............. 13.

16. (The full score of this small question is 13)

Solution: (i) Let "A gets the coupon" as the event A................ 1.

Because it is assumed that the possibility of the pointer stopping at any position is equal and the areas of the three parts given in the question are equal,

So the pointer stops in 20 yuan, 10 yuan, and the probability in 0 yuan area is ................................................................................................................................................

Customer A gets a coupon indicating that the pointer stops in 20 yuan or 10 yuan area.

According to the probability of mutually exclusive events, there are: ...................................................................................................................................................................

Therefore, the probability that customer A will get a coupon with a face value greater than that of 0 yuan is.

(2) Let "the amount of coupons obtained by Party B is not less than that of 20 yuan" as event B.

Because customer B rotates the turntable twice, it is assumed that the coupon amount obtained by customer B for the first time is RMB.

If the coupon amount obtained for the second time is RMB, the basic event space can be expressed as:

Nine points

In other words, there are 9 basic events, and the probability of each basic event is ..........................................................................................................................................................

And the coupon amount obtained by B is not lower than that of 20 yuan, that is,

Therefore, event B contains six basic events, and the score is 1 1.

Therefore, the probability that B gets the par value not lower than that of 20 yuan is .................... 13.

Answer: The probability that A gets a coupon with a denomination greater than 0 yuan is, and the probability that B gets a coupon with a denomination not less than 20 yuan is.

17. (The full score of this small question is 14)

Prove: (i) AB=BC because ABCD is a diamond.

So AB=BC=AC, .............. 1 min.

And m is the midpoint of BC, so it is 2 o'clock.

And plane ABCD, plane ABCD, so ...

Again, so the plane ................. 5 points.

(2) Because ...

And the bottom, so

Therefore, the volume of the triangular pyramid is 8 points.

.........................., 9 points.

(iii) Existence .........................................................................................................................................................................

Take the midpoint e of PD and connect ne, EC and AE.

....................... 1 1 min because n and e are the midpoint of PA and PD, respectively.

Also in diamond ABCD,

In other words, MCEN is a parallelogram with ............................... 12 points.

So,,,

Plane, plane

So the plane, .................. 13 points.

That is, there is a point e on PD, so that the plane,

At this moment, .............. 14.

18. (The full score of this small question is 14)

Solution: (1) Because,

So the point is ............... 1 min on the image of the function at the same time.

Because, ..................... scored 3 points.

Five points.

From what is known, we get, so, that is, ...................................... scored 6 points.

(2) Because

So .................... scored eight points.

When,

Because, therefore, it applies to constants,

Therefore, it is monotonically increasing, and the infinite value is 10 point;

When,

Order, get (give up) .....................................................................................................................................................................

So when, the changes are as follows:

0 +

minimum value

................. 13 o'clock

So when, get the minimum, and

................ 14 o'clock

To sum up, when there is no value on the function;

When is, the function takes the minimum value at.

19. (The full score of this small question is 13)

Solution: (1) Let the equation of ellipse C be, which can be obtained from the meaning of the question.

Again, so .................... scored 2 points.

Because ellipse C passes through (1,), it is substituted into the elliptic equation as follows.