Reference answers and grading standards of test questions

Description:

1. When marking the test paper, please follow this grading standard, and the fill-in-the-blank questions only have 7 points and 0 points; Please grade other questions in strict accordance with the grading standards, without adding other intermediate grades.

2. If the candidate's problem-solving method is different from this method, as long as the thinking is reasonable and the steps are correct, the grading can be appropriately divided with reference to this grading standard, and at least four answers are one grade, without other intermediate grades.

Fill in the blanks (***8 small questions, 7 points for each small question, ***56 points)

1. If the function and, then.

answer

Analysis,

……

.

Therefore.

2. Given a straight line and a circle, the point is on the straight line, and, is two points on the circle, the middle, and passes through the center of the circle, then the abscissa range of the point is.

answer

Analytically speaking, the distance from the center of a circle to a straight line is obtained by the intersection of a straight line and a circle.

Solve.

3. On the coordinate plane, there are two regions, the sum of which is a variable region, which is determined by inequality. If the value range of is, the common area of and is a function.

answer

Analysis shows from the meaning of the problem

4. The value of the smallest positive integer that makes the inequality valid for all positive integers is.

answer

Analytical hypothesis. Obviously monotonically decreasing, then from the maximum value of, you can get.

5. Any two points on an ellipse, if, the minimum value of the product is.

answer

Analysis and design,

.

On the ellipse, there are

①

②

get

.

So when the minimum value is reached.

6. If the equation has only one real root, the value range of is.

Answer or

analyse

The only possibility is that

①

②

③

Right (3) is derived from the root formula.

, ④

Or ...

(i) When, it is decided by (3)

Therefore, both are negative roots.

Also known as ④

So the original equation has a solution.

(Ⅱ) If the original equation has a solution.

(iii) If it is timely, it will be obtained from (iii).

So they are all positive roots, irrelevant, give up.

To sum up, what you can get is what you want.

7. A digital table consisting of several rows of numbers. Starting from the second row, the number of each row is equal to the sum of the two numbers on its shoulder, and there is only one number in the last row. The first line is the one in which the positive integers in front are arranged from small to large, so the number in the last line is (can be expressed by exponent).

answer

The analysis is easy to understand:

(i) Table * * * has line 100;

(ii) Each line constitutes a arithmetic progression with a tolerance of

, , ,…,

(3) in order to seek.

Then let the first number be

……

Therefore.

8. Buses arrive at the station every day, but the arrival time is random, and the arrival time of the two is independent of each other. The legal provisions are as follows

arrival time

Probability; possibility

When passengers arrive at the station, the mathematical expectation of waiting time is (accurate to minutes).

Answer 27

The distribution table waits for the analysis of passengers.

Waiting time (minutes) 10 30 50 70 90

Probability; possibility

The mathematical expectation of waiting time is

Second, answer the question.

1. (The full score of this small question is 14) Let a straight line (in this case, an integer) intersect an ellipse at two different points and a hyperbola at two different points. Ask if there are any straight lines, so that the vector, if there are, indicates how many such straight lines there are. If it does not exist, please explain why.

Analysis is arranged by exclusion and simplification.

Set,, and then

①4 points.

By eliminating and simplifying

Set,, and then

(2) 8 points.

Because, so, at this time. By.

.

Therefore, it is still obtained from the above formula. When, by ① and ②. Because it is an integer, the values are,,,,. When, by ① and ②. Because it is an integer, there are 9 straight lines * * * that meet the conditions. .......................................................................................

2. (This subhead is 15) It is known to be a real number, the equation has two real roots, and the sequence satisfies,

(i) Find the general term formula of the sequence (represented by,,);

(ii) As the sum of the above.

Analysis method 1:

(i) According to Vieta's theorem, therefore,

,

arrange

Let's order then. So it is the geometric series of the common ratio.

The first item in this series is:

.

So, that's. So ...

(1) When,,, becomes. I'm sorting it out. So, the number sequence is transformed into a arithmetic progression, and the first term is. So ...

.

So the general formula for this sequence is

; Five points.

(2) When,

.

arrange

, .

So the first term of a geometric series whose series is common ratio is. So ...

Therefore, the general formula of this series is ......................................................................................................................................................................

(ii) If, then, at this time, from the result of step (i), the general term formula of the series is, so the sum of the first few terms of is.

Subtract the above two formulas to get the result.

Therefore, ......................................... 15.

Method 2:

(i) According to Vieta's theorem, therefore,

, .

The two roots of the characteristic equation are.

(1) when, general terms by also.

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

(2), generally refers to. From, get.

Therefore, solve

.................................... 10.

(2) Same as method 1.

3. (Full score of this small question 15) Find the maximum and minimum values of the function.

The domain of analytic function is. because

When the equal sign holds, the minimum value is ....................................................................................................................................................................

It is also derived from Cauchy inequality.

Therefore, .......................................... 10.

The conditions of Cauchy inequality are obtained and solved. Therefore, the equal sign was established at that time. Therefore, the maximum value is .........................................................................................................................................

In 2009, the national senior high school mathematics entrance examination was added.

Reference answers and grading standards for test questions (Volume A)

Description:

1. When marking papers, please grade them in strict accordance with the grading standards.

2. If the candidate's answer method is different from this answer, as long as the thinking is reasonable and the steps are correct, you can divide the grades appropriately with reference to this grading standard. 10 is divided into one grade, and no other intermediate grades are added.

Fill in the blanks (***4 small questions, each with 50 points, ***200 points)

9. As shown in the figure, they are the midpoint of the arc on the circumscribed circle () of the acute triangle. The intersection points intersect at a point, which is the heart of a triangle, connecting and extending the intersection points.

(1) Verification:

(2) Take a point (,,) on the arc, excluding this point, and record its heart respectively.

Proof:,,, four * * circles.

Analysis 1 is cheap. Because,,, * * is a circle and an isosceles trapezoid. Therefore,.

Even, and pay, because

,

So ... In the same way.

.

therefore

, .

Therefore, a quadrilateral is a parallelogram. So, (same bottom, same height).

There is also a four-point * * * circle, therefore, by the triangle area formula.

So ...

Because,

So, in the same way, you get it.

Therefore, as proved by (1),

.

Also because

,

have

.

Therefore, this way

.

So, four o'clock * * *.

10. Proof of inequality:

, ,2,…

Analytic proof: First, prove an inequality:

⑴ , .

In fact, make

, .

Right,

, .

therefore

, .

Accepted (1)

⑵ .

So, order,

Therefore.

because

.

therefore

.

1 1. Set two given positive integers. It is proved that there are infinite positive integers, which makes sum coprime.

Analytic proof 1: For any positive integer, let. We prove it.

Let it be any prime factor, as long as it is proved.

If so, then by

.

And, and, know and. Therefore.

Proof 2: For any positive integer, we prove that.

Let it be any prime factor, as long as it is proved.

If so, then by

.

Is that the above formula is inseparable, so.

If set, but ... so by

And, and, know and. Therefore.

12. In a table composed of non-negative numbers

The numbers of each row in are different from each other, and the sum of the three numbers of each column in the first six columns is a table composed of the first three columns in which 1,,,,, and are all greater than.

The following properties are satisfied: For any column (0, 2, …, 9) in the numerical table, there exists one.

manufacture

⑶ .

Verification:

(i) Minimum values 0, 2 and 3 must come from different columns of the table.

(ii) There is only one row in the table, and 2 and 3 constitute the table.

Still natural.

Analysis (i) Assuming that the minimum values of 0, 2 and 3 are not taken from different columns of the table, one column contains no values. Let's assume 0, 2, 3. Because any two elements in the same row in the table are not equal, 0, 2 and 3. On the other hand, because of the nature of the table, if you take it from 3, there is a contradiction.

(2) Starting from the principle of painting.

, ,

At least two values in are in the same column. You might as well set

, .

From the previous conclusion, we know that the first column of the table must contain a certain number, so it can only be. Similarly, the second column must also contain a number. Let's assume that it is a number on the diagonal in the table.

Remember, make a collection.

.

Obviously, and 1, 2. Because,,, so.

Therefore, the existence makes it obvious, 2, 3.

The following proof table

Have nature.

As can be seen from the above selection method, this shows that

, .

Starting from the satisfactory nature, we deduce from (3), which proves that there is some certainty that 2,3 is arbitrary. If not, then, 3 and. This is the biggest contradiction with. Therefore, the table satisfies nature.

Uniqueness of the next certificate. There is a table to do.

For nature, without losing generality, we assume that

⑷

.

Because, there is (me), there is. And from (i), we can know: or, or.

If true, the table has attributes, and then

,

⑸ ,

.

If the table satisfies the attribute, then at least one formula makes, by and (4) and (6), so there can only be. Similarly, by satisfying the nature and deductibility, therefore.