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Seeking the Mathematics Examination Paper of Jiangsu College Entrance Examination in 2008 (with the answer)
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In 2008, the national unified entrance examination for ordinary colleges and universities (Jiangsu volume)

mathematics

This paper is divided into two parts: the first volume (fill in the blanks) and the second volume (answer questions). Candidates should answer the questions on the answer sheet, which is invalid. After the exam, they should return the test paper and the answer sheet together.

Precautions:

1. Before answering questions, candidates should fill in their names and admission ticket numbers on the answer sheet, and carefully check the bar code.

Admission ticket number and name, and stick the bar code in the designated position.

2. Use 2B's multiple-choice answers

Fill it in with a pencil, wipe it clean with an eraser if necessary, and then choose other answer labels; Nonselection

The answer to the question is written with 0.5mm black gel pen (signature pen) or carbon pen, with neat font and clear handwriting.

3. Please answer each question in the answer area (black wireframe) according to the question number. The answer written outside the answer area is invalid.

4. Keep the card surface clean and don't fold or damage it.

5. Candidates should answer the questions according to the requirements of the questions when choosing the exam questions, and use 2B pencil to blacken the labels corresponding to the selected questions on the answer sheet.

Reference formula:

Standard deviation of sample data,,,.

Where is the sample average.

Cylinder volume formula

Where is the bottom area and where is the height.

Fill in the blanks: this big question is * *1small question, with 5 points for each small question and 70 points for * *.

The minimum positive period of 1. Yes, where = ▲.

This small question examines the periodic formula of trigonometric functions.

10

2. The probability that a dice is thrown twice in a row and the sum of points is 4 ▲.

This little question examines classical probability. There are * * 6× 6 basic events, and the sum of (1, 3), (2, 2), (3 1)* * 3 points is 4, so

3. Expressed as, then = ▲.

This problem examines the division of complex numbers. ∫∴= 0, = 1, so

1

4.A=, then the number of elements of A Z ▲.

This small question examines the operation of sets and the solution of a quadratic inequality. It is concluded that ∵ δ < 0 and ∴ set A is 0, so the element of A Z does not exist.

5. The included angle is, then ▲.

This little problem examines the linear operation of vectors.

= , 7

seven

6. In the plane rectangular coordinate system, let d be the area formed by points whose absolute values of abscissa and ordinate are not more than 2, and e be the area formed by points whose distance to the origin is not more than 1. If you throw a point into D at random, the probability of falling into E will bring.

This little question examines classical probability. As shown in the figure, area D represents the interior (including the boundary) of a square with a side length of 4, and area E represents the unit circle and its interior.

7. Themes of algorithms and statistics

8. If the straight line is the tangent of the curve, then the real number b = ▲.

This topic examines the geometric meaning of the derivative and the solution of the tangent, thus obtaining the tangent point (2, ln2), and substituting it into the linear equation, so that b = LN2- 1.

ln2- 1

9 In the plane rectangular coordinate system, let the vertices of the triangle ABC be A(0, a), B(b, 0), C (c, 0) and P(0, p) on the line segment AO (different from the endpoint), let AC, AB, C and P all be non-zero real numbers, and the straight lines BP and CP intersect AC and AB at points E and CP respectively.

( ▲ ) .

This small question examines the solution of the equation of a straight line. Sketch can be guessed by symmetry. In fact, the equation of the straight line can be obtained by subtracting these two formulas from the intercept formula. Obviously, the intersection F of straight line AB and CP satisfies this equation, and the origin O also satisfies this equation, so it is an equation for finding straight line of.

10. Arrange all positive integers into a triangular number array:

1

2 3

4 5 6

7 8 9 10

. . . . . . .

According to the above arrangement, the third number from left to right in the nth row (n ≥3) is ▲.

This small question examines inductive reasoning and sum formula of arithmetic sequence. The n- 1 line * * has a positive integer of 1+2+…+(n- 1), that is, one, so the third number in the n-th line is +3 of all positive integers, that is.

1 1. given, the minimum value of ▲.

This little question examines the application of binary basic inequalities.

Take "=" if and only if = 3.

three

12. In the plane rectangular coordinate system, if the focal length of the ellipse 1( 0) is 2, and the circles with O as the center and radius are perpendicular to each other, the eccentricity = ▲.

Let the tangents PA and PB be perpendicular to each other and the radius OA be perpendicular to PA, then △OAP is an isosceles right triangle, so the solution is obtained.

13. If AB = 2 AB=2 and AC= BC, what is the maximum value?

This small question examines the triangle area formula, cosine theorem and function thought. Let BC =, then AC =,

According to the area formula =, according to the cosine theorem

, substituted into the above formula.

=

Accord to that solution of triangle trilateral relation,

Therefore, when the maximum value is obtained.

14. If it is always ≥0, then = ▲.

This topic examines the comprehensive application of monotonicity of functions. If x = 0, no matter what value is taken, ≥0 is obviously true; When x > 0, that is ≥0, it can be changed to,

If, then, it increases monotonously in the interval and decreases monotonously in the interval, so it is ≥ 4;

When x < 0, that is ≥0, it can be changed to,

It increases monotonically in the interval, so ≤4, in short = 4.

four

Second, problem solving: the idea of solving problems should be clearly written, explaining the process or calculus steps.

15. As shown in the figure, in the plane rectangular coordinate system, two acute angles are made with the axis as the starting edge, and the terminal edge intersects with the unit circle at two points A and B, and the abscissas of A and B are known respectively.

(i) Find the value of tan ();

(ii) the value of.

This little question examines the definition of trigonometric function, the tangent of the sum of two angles and the tangent formula of two angles.

Conditional, because it is an acute angle, so =

therefore

(1) Tan () =

(2) So,

∫ is acute, ∴, ∴ =

16. in tetrahedral AB, BD, CB= CD, AD⊥BD, e and f are the midpoint of AB and BD respectively.

Verification: (i) straight line EF ‖ surface ACD

(ii) surface EFC⊥ surface alkaline catalytic decomposition.

This small problem examines the determination of the positional relationship between spatial straight lines and planes, and between planes.

(I)e and f are the midpoint of AB and BD respectively,

∴EF is the center line of∴ ef ‖ ad △ Abd,

∫ef plane ACD, AD plane ACD, ∴ straight line EF‖ plane ACD.

(2) ∵ AD⊥BD ,EF‖AD,∴ EF⊥BD.

Cb = cd, and f is the midpoint of ∴ cf ⊥ bd.bd.

And EF CF=F, ∴BD⊥ faces EFC. ∵ BD BCD, ∴ EFC⊥ BCD.

17. There are three factories in a certain place, which are located at the vertices A and B of rectangular ABCD and the midpoint P of CD. It is known that AB=20km.

CB = 10km。 In order to treat the sewage from three factories, a sewage treatment plant was built on the rectangular ABCD area (including the boundary), and the points A, B and A were equidistant, and sewage pipes ao, bo and op were laid, with a total length of km.

(1) Write functional relationships according to the following requirements:

(1) Let ∠BAO= (rad) and express it as a functional relationship;

② Let OP (km) be expressed as a function of x 。

(ii) Please select a functional relationship in (i) to determine the location of the sewage treatment plant, so as to minimize the total length of the three sewage pipes.

This small problem mainly investigates the application of the maximum value of the function.

(Ⅰ) ① It can be known from the conditions that PQ is divided vertically into AB, if ∠BAO= (rad), then, therefore.

And op = 10- 10ta,

So,

The functional relationship is as follows

② If OP= (km) and OQ = 10-, then OA =OB=

The functional relationship is as follows

(2) Select the functional model ①,

Let 0 get sin, because, so =,

When,, is a subtraction function; When, is increasing function, so when =,. At this time, the point P is located on the middle vertical line of the line segment AB, far from the AB side.

It's kilometers away.

18. Let the image of the quadratic function in the plane rectangular coordinate system have three intersections with the two coordinate axes, and the circle passing through these three intersections is called C. Find:

(i) the range of the actual number b;

(ii) the equation for finding circle c;

(iii) Does circle C pass through a fixed point (its coordinates have nothing to do with B)? Please prove your conclusion.

This small question mainly examines the image and properties of quadratic function and the solution of circular equation.

(i) Let = 0, and the intersection of parabola and axis is (0, b);

Therefore, from the problem that means b≠0, δ > 0, we can get b < 1, b ≠ 0.

(Ⅱ) Let the general equation of a circle be

Let = 0 and = 0 be the same equation, so d = 2 and f =.

Let = 0 get = 0. This equation has a root of b, and if it is substituted, E =-b- 1 is obtained.

So the equation of circle c is.

(iii) Circle C must pass through the fixed points (0, 1) and (-2, 1).

The proof is as follows: (0, 1) is substituted into the equation of circle C, and the left side = 0+1+2× 0-(b+1)+b = 0 and the right side = 0.

Therefore, circle C must pass through the fixed point (0, 1).

It can also be proved that circle C must pass through a fixed point (-2, 1).

19. (I) Let arithmetic progression () be non-zero with tolerance. If an item is deleted from the series, the series is a geometric series (in the original order):

(1) When n =4, the numerical value; (2) All possible values;

(2) Prove that for a given positive integer n(n≥4), there exists a arithmetic progression whose term and tolerance are not zero, and any three of them (in the original order) cannot form a geometric progression.

This paper mainly investigates the comprehensive application of arithmetic progression and geometric progression.

(Ⅰ) ① When n = 4, the first item or the last item cannot be deleted, otherwise three consecutive items in arithmetic progression will become geometric progression, and d = 0 is deduced.

If it is deleted, there will be

Simplification = 0, because ≠0, so = 4;

If it is deleted, it is = 1.

In summary = 1 or -4.

② When n = 5, the first item or the last item cannot be deleted.

If deleted, there is =, that is = 6;

If deleted, then =, that is.

Simplified to 3 = 0, because d≠0, it cannot be deleted;

If deleted, there is =, that is = 2.

When n≥6, there is no such arithmetic progression. In fact, in the sequence,,,,

Because the first item or the last item cannot be deleted, if it is deleted, there must be =, which is inconsistent with d≠0; Similarly, if you delete

Go also has =, which is contradictory to d≠0; If you delete any of,,, you must have.

=, which is in contradiction with d≠0.

All in all, n.

(2) Omission

20. If,, is a constant,

and

(i) Find the necessary and sufficient conditions of all real numbers (expressed by);

(ii) is set to two real numbers, and if

It is proved that the sum of the lengths of monotonically increasing intervals on intervals is (the length of closed intervals is defined as).

This small question examines the comprehensive application of necessary and sufficient conditions, exponential function, absolute value function and inequality.

(i) Ongoing establishment

(*)

because

Therefore, it only needs (*) to keep.

To sum up, the necessary and sufficient conditions for the establishment of all real numbers are:

(ii)1If, the image is symmetrical about the straight line. Because, so the interval is symmetrical about a straight line.

Because the subtraction interval is 0 and the increase interval is 0, the sum of the lengths of monotone increase intervals is 0.

2 if.

(1) When? ,

When, because, therefore,

Therefore =

When, because, so.

Therefore =

Because, so, so that is.

When, when, then, then,

When, so =

,, so =

The sum of the lengths of monotonically increasing intervals on an interval

=

(2) When,

When, because, therefore,

Therefore =

When, because, so.

Therefore =

Because, so, so.

When, when, then, then,

When, so =

,, so =

The sum of the lengths of monotonically increasing intervals on an interval

=

To sum up, the sum of the lengths of monotonically increasing intervals on the interval is