Looking for Guangdong 20 12 Mathematics Examination Classification Analysis Compilation (12 Special Topic)

20 12 Guangdong province senior high school entrance examination mathematics examination paper

First, multiple-choice questions (***5 small questions)

The absolute value of 1.(20 1 1 Henan) -5 is ().

A.﹣5﹣

Test center: absolute value.

Solution: According to the fact that the absolute value of a negative number is equal to its reciprocal, we get |-5 | = 5. So choose a.

2.(20 12 Guangdong) radius of the earth is about 6,400,000 meters, which is expressed by scientific notation as ().

A.b . 6.4× 106 c . 64× 105d . 640× 104

Test center: scientific notation-indicating larger numbers.

Solution: Solution: 6400000 = 6.4× 106.

So choose B.

3. The pattern of (20 1 2 Guangdong) data 8, 8, 6, 5, 6,16 is ().

A. 1 B. 5 C. 6 D. 8

Test sites: most.

Answer: Solution: 6 appears the most times, so the pattern is 6.

So choose C.

4.(20 12 Guangdong) The front view of the geometry shown in the figure is ().

A.B. C. D。

Test center: simple assembly three views.

Solution: From the front, the main view of this diagram consists of three columns, and the number of small squares from left to right are: 1, 3, 1.

Therefore, choose: B.

5. (Yue, 20 12) Given that the lengths of two sides of a triangle are 4 and 10 respectively, the length of the third side of this triangle may be ().

A.5 B. 6 C. 1 1 D. 16

Test site: triangular trilateral relationship.

Solution: Let the length of the third side of this triangle be x, then 10-4 < x < 10+4, that is, 6 < x < 14, and only1of the four options meets the requirements.

So choose C.

2. Fill in the blanks (***5 small questions)

6.(20 12 Guangdong) Decomposition factor: 2x2- 10x = 2x (x-5).

Test center: factorization-common factor method.

Solution: Solution: Original formula = 2x (x-5).

So the answer is 2x (x-5).

7. The solution set of (2012 Guangdong) inequality 3x-9 > 0 is x > 3.

Test center: Solve linear inequality of one variable.

Solution: solution: shift term, 3x > 9,

If the coefficient is 1, x > 3.

So the answer is: x > 3.

8.(20 12 Guangdong) As shown in the figure, A, B and C are three points on ⊙O, and ∠ ABC = 25, then ∠AOC is 50.

Test site: fillet theorem.

Answer: solution: ∵ central angle ∠AOC and circumferential angle ∠ABC,

∴∠AOC=2∠ABC, and ∠ ABC = 25,

∠ AOC = 50。

So the answer is: 50.

9.(20 12 Guangdong) If both x and y are real numbers and satisfy | x | 3 |+= 0, then the value of () 20 12 is 1.

Test center: non-negative nature: arithmetic square root; The nature of non-negative number: absolute value.

A: A: According to the meaning of the question,

Solution:.

Then () 2012 = () 2012 =1.

So the answer is: 1.

10.(20 12 Guangdong) As shown in the figure, in? In ABCD, AD=2, AB=4, ∠ A = 30, with point A as the center and the length of AD as the radius, draw the arc intersection point AB at point E to connect CE, then the area of the shadow part is 3 ∠ π (π is retained as a result).

Test center: fan-shaped area calculation; Properties of parallelogram.

Solution: Through point D, do DF⊥AB at point F.

AD = 2，AB=4，∠A=30，

∴DF=AD？ sin30 = 1,EB=AB﹣AE=2，

∴ area of shaded part:

4× 1﹣ ﹣2× 1÷2

=4﹣ π﹣ 1

=3﹣ π.

So the answer is: 3-π.

Three. Answer the question (*** 12)

1 1.(20 12 Guangdong) calculation:-2sin45-(1+) 0+2-1.

Test center: the operation of real numbers; Zero exponential power; Negative integer exponential power; Trigonometric function value of special angle.

Solution: Solution: Original formula =-2×- 1+

=﹣ .

12.(20 12 Guangdong) Simplify first and then evaluate: (x+3)(x﹣3)﹣x(x﹣2), where x = 4.

Test center: mixed operation of algebraic expressions-simplified evaluation.

Solution: Original formula = x2-9-x2+2x.

=2x﹣9，

When x=4, the original formula = 2× 4-9 =- 1.

13.(20 12 Guangdong) Solution equation:.

Test site: solving binary linear equations.

Solution: Solution: ①+②, 4x=20,

The solution is x=5,

Substitute x=5 into ①, 5-y = 4,

The solution is y= 1,

Therefore, the solution of the inequality group is:

14.(20 12 Guangdong) As shown in the figure, in △ABC, AB=AC, ∠ ABC = 72.

(1) Use a ruler and compasses as the bisector BD of ∠ABC, and intersect with AC at point D (drawing traces are left, and writing method is not required);

(2) The number of times to find ∠BDC after making ∠ABC's bisector BD in (1).

Test center: drawing-basic drawing; Properties of isosceles triangle.

Solution: Solution: (1)① Draw an arc with point B as the center and any length as the radius, and AB and BC intersect at point E and point F respectively;

② Draw a circle with the center at point E and point F and the radius greater than EF. Connect BG angle AC to point D compared with point G. 。

(2)∵ In △ABC，AB=AC，∠ ABC = 72，

∴∠a= 180 ﹣2∠abc= 180 ﹣ 144 = 36，

∵AD is the bisector of∝∠ ∝∠ABC,

∴∠ABD= ∠ABC= ×72 =36，

∫∠BDC is the outer corner of △ABD,

∴∠BDC=∠A+∠ABD=36 +36 =72。

15. (Guangdong, 20 12) It is known that in a quadrilateral ABCD, ABCD, diagonal AC and BD intersect at point O, and Bo = do.

It is proved that the quadrilateral ABCD is a parallelogram.

Test center: the judgment of parallelogram; Congruent triangles's judgment and nature.

Answer: Proof: ∵AB∨CD,

∴∠ABO=∠CDO，

At △ A wave and △CDO,

∵ ,

∴△ABO≌△CDO，

∴AB=CD，

The quadrilateral ABCD is a parallelogram.

16. (Guangdong, 20 12) According to media reports, the total number of Chinese citizens traveling abroad in 2009 was about 50 million, and the total number of citizens traveling abroad in 20 1 1 year was about 72 million. If 20 10, 20 165438,

(1) Find the average annual growth rate of the total number of Chinese citizens traveling abroad in these two years;

(2) If the average annual growth rate in 20 12 remains unchanged, how many tens of thousands of China citizens will travel abroad in 20 12?

Test center: the application of a quadratic equation. www。 xkb 1。 com

Answer: Answer: (1) Suppose that the average annual growth rate of the total number of Chinese citizens traveling abroad in these two years is X.

5000( 1+x)2 =7200。

The solution is x 1 =0.2=20%, x2 =﹣2.2 (irrelevant, omitted).

A: In the past two years, the average annual growth rate of the total number of China citizens traveling abroad has been 20%.

(2) If the average annual growth rate of 20 12 remains unchanged,

In 20 12, the total number of China citizens traveling abroad was 7,200 (1+x) = 7,200×120% = 86.4 million.

A: It is predicted that the total number of China citizens traveling abroad in 20 12 will be about 86.4 million.

17. (Guangdong, 20 12) As shown in the figure, the straight line y= 2x-6 and the inverse proportional function y= the image where point A (4,2) intersects with the X axis at point B. 。

(1) Find the value of k and the coordinates of point B;

(2) Is there a point C on the X axis that makes AC=AB? If it exists, find the coordinates of point C; If it does not exist, please explain why.

Test center: inverse proportional function synthesis problem.

Solution: Solution: (1) Substitute (4,2) into the inverse proportional function y= to obtain

k=8，

Substituting y=0 into y = 2x ~ 6, you can get

x=3，

Therefore, k = 8；; The coordinate of point B is (3,0);

(2) If it exists and the coordinate of point C is (a, 0), then

AB = AC，

∴ = ,

That is, (4-a) 2+4 = 5,

The solution is a=5 or a=3 (this point coincides with B and is discarded).

So the coordinate of point C is (5,0).

18. (Guangdong, 20 12) As shown in the figure, the slope AC of the hill is tanα=, and the elevation angle of the top A is 26.6 at D, which is 200 meters away from the foot of the mountain. Find the height AB of the hill (the result is an integer: reference data: sin26.6 =0.45, COS 26.6).

Test center: solve the application of right triangle-elevation angle and depression angle; Application of solving the problem of right triangle-inclination angle.

Solution: in the right triangle ABC, =tanα=, xkb 1.co m.

∴BC=

In the right triangle ADB,

= Tan 26.6 =0.50

That is BD=2AB.

∵BD﹣BC=CD=200

∴2AB﹣ AB=200

Solution: AB=300 meters,

The height of this mountain is 300 meters.

19.(20 12 Guangdong) Observe the following equation:

Equation1:a1= =× (1-);

Equation 2: A2 = =× (﹣); ;

The third equation: A3 = =× (﹣); ;

Equation 4: A4 = =× (﹣); ;

…

Please answer the following questions:

(1) List the fifth equation according to the above law: a5 = =

(2) The nth equation is expressed by an algebraic expression containing n: an= = (n is a positive integer);

(3) Find the value of a1+A2+A3+A4+…+A100.

Test center: regular type: types of numbers.

Answer: Solution: According to observation, the answer is:

( 1) ; ;

(2) ; ;

(3)a 1+A2+A3+A4+…+a 100

= ×( 1﹣ )+ ×( ﹣ )+ ×( ﹣ )+ ×( ﹣ )+…+ ×

= ( 1﹣ + ﹣ + ﹣ + ﹣ +…+ ﹣ )

= ( 1﹣ )

= ×

= .

20. (Guangdong, 20 12) has three cards, with the numbers ﹣2, ﹣ 1 and 1 written on the front respectively, which are exactly the same on the back. After washing three cards with their backs up, randomly draw one card, take the positive number as the value of X, put it back in the card, and then wash it evenly from the top.

(1) Use tree diagram or list method to represent (x, y) all possible results;

(2) Find the probability that the score+meaningful (x, y) appears;

(3) Simplify the score+find the probability of (x, y) that makes the value of the score an integer.

Test sites: list method and tree diagram method; Conditions for meaningful scores; Simplified calculation of fractions.

Solution: Solution: (1) Use a tree diagram to represent (x, y). All possible outcomes are as follows:

(2) ∫(x, y) that makes the score+meaningful is (﹣ 1, ﹣2), (﹣ 1, ﹣2), (﹣2, ﹣.

The probability that ∴ score+meaningful (x, y) appears is,

(3)∵ + =

(x, y) that makes the score an integer is (﹣2, ﹣2), (﹣ 1, ﹣ 1), (﹣ 1, ﹣/.

∴(x, y) The probability of making the value of a fraction an integer is.

2 1. (Guangdong, 20 12) As shown in the figure, in rectangular paper ABCD, AB=6 and BC = 8. Fold △BCD along diagonal BD, so that point C falls at point C ′, and BC ′ intersects with AD at point G; E and f are points on C ′ d and BD, respectively. The line segment EF and AD intersect at H point, and the △FDE is folded along EF, so that the D point falls on D ′, and the D ′ point coincides with the A point.

(1) verification: △ abg △ c 'dg;

(2) Find the value of tan∠ABG;

(3) Find the length of EF.

Test center: folding transformation (folding problem); Congruent triangles's judgment and nature; The nature of rectangle; Solve right triangle.

Answer: (1) Proof: ∫△BDC' is folded from △BDC.

∴∠C=∠BAG=90，c′d = ab = CD，∠agb =∠dgc′，

∴∠ABG=∠ADE，

in:△ABG?△C ' DG，

∵ ,

∴△abg≌△c′dg；

(2) Solution: ∵ From (1), we can know △ abg △ c ′ dg.

∴GD=GB，

∴AG+GB=AD, let AG=x, then GB = 8 x,

At Rt△ABG,

∵AB2+AG2=BG2, that is, 62+x2 = (8-x) 2, and x=,

∴tan∠abg= = =；

(3) Solution: ∫△AEF is folded from △DEF.

∴EF vertical division advertisement,

∴HD= AD =4 years,

∴tan∠ABG=tan∠ADE=，

∴EH=HD× =4× =，

∵EF vertical division AD, AB⊥AD

∴HF is the center line of △ABD,

∴HF= AB= ×6=3，

∴EF=EH+HF= +3=。

22.(20 12 Guangdong) As shown in the figure, the parabola y = x2-x-9 intersects with the X axis at points A and B, and intersects with the Y axis at point C, connecting BC and AC.

(1) Find the length of AB and OC;

(2) Starting from point A, point E moves to point B along the X-axis (point E does not coincide with points A and B), and the point E is a straight line L parallel to BC, which intersects with point D, let the length of AE be m and the area of ADE be s, find the functional relationship between s and m, and write the value range of independent variable m;

(3) Under the condition of (2), connect CE to find the maximum value of △CDE area; At this time, find the area of the circle tangent to BC with point E as the center (the result keeps π).

Test center: Quadratic function synthesis problem.

Solution: Solution: (1) Known: parabola y = x2-x-9;

When x=0 and y=﹣9, then: C(0, ﹣ 9);

When y=0, x2-x-9 = 0, then: x1=-3, x2﹣ x﹣9=0 6, then: a (-3,0), b (6 6,0);

∴AB=9,OC=9.

②∫ED∨BC，

∴△AED∽△ABC，

∴ =( )2, that is: =( )2, so: s = m2 (0 < m < 9).

(3)S△AEC= AE？ OC= m，S△AED = S = m2；

Then: s △ EDC = s △ AEC-s △ AED =-m2+m =-(m-) 2+;

∴△ Maximum area of ∴△CDE, where AE=m=, BE = ABAE =.

If e is EF⊥BC in F, then Rt△BEF∽Rt△BCO, we get: