Reference answers and grading standards of mathematics test papers
1. Multiple choice questions: (There are 6 questions in this big question, 4 points for each question, out of 24 points).
1.c； 2.a； 3.b； 4.d； 5.b； 6.C。
Fill in the blanks: (4 points for each question, out of 48 points)
7.； 8.； 9.； 10 . x = 2； 1 1. decrease; 12.；
13. 1350; 14.4; 15.； 16. 17; 17.； 18. 1 or 7.
Third, answer: (This big question is ***7 questions, out of 78 points)
19. (The full mark of this question is 10)
Solution: From ① ....................................................... (2 points)
From ② ............................................. (2 points)
Solution ............................................ (2 points)
Therefore, the solution set of the original inequality group is ................................................. (2 points).
If the solution set of the inequality group is expressed on the number axis, you will get 2 points correctly, without removing the end point deduction 1 point.
20. (The full mark of this question is 10)
Solution: Multiply both sides by the simplest common denominator, and you get
................................. (2 points)
Upon completion, you will get a .................................................. (3 points).
Solution, ....................................... (2 points)
It is proved to be the root of the original equation, so it is discarded; Is the root of the original equation ........... (2 points).
Therefore, the root of the original equation is x = 4 .................................... (1min).
2 1. (This title is ***2, item (1) is 4 points, item (2) is 6 points, and the full score is 10).
Solution: (1) Let the resolution function between y and x be (k ≠ 0).
According to the meaning of the question, get ............................... (2 points)
Solution ............................... (1)
Therefore, the resolution function sought is ...................................... (1 min).
(2) Let the sales price of this day be X yuan .......................................................... (1 min).
According to the meaning of the question, get ................................. (2 points)
Upon completion, you will receive .......................................... (1 minute).
Solution, .................................. (1 min)
∵50 & gt； 38, ∴x = 50 is irrelevant, it has been discarded.
A: The sales price of this day should be 33 yuan. ....................................................................................................................................................................
22. (This question ***2 is a minor issue, the first question (1) is 5 points, the second question (2) is 5 points, and the full score is 10).
Proof: (1)∵PC // OB, PD // OA,
∴ Quadrilateral OCPD is a parallelogram, and ∠ECP =∠O, ∠ FDP = ∠ O...( 1 min)
∴PC = OD, PD = OC, ∠ ECP = ∠ FDP ....................... (1min)
∴∠pec ∵pe⊥oa,pf⊥ob =∠pdf = 90。
∴△ PCE ∽△ PDF ....................................... (1min)
That is ∴ ................................ (1min).
∴ ..................................... (1min)
(2) When the point p is on the bisector of ∠AOB∞, the quadrilateral CODP is a rhombic ................................................................................................................. (1 min).
When point P is on the bisector of ∠AOB, PE = PF. It is obtained from PE⊥OA,PF⊥OB.
Then, from △PCE∽△PDF, PC = PD ............................................................. (2 points).
∵ Quadrilateral OCPD is a parallelogram, and∴ Quadrilateral OCPD is a diamond .............................. (1 min).
When point P is not on the bisector of ∠AOB, PE ≠ PF. You can get it, that is, PC ≠ PD. Available.
∴ When the point P is not on the bisector of ∞∠AOB, the quadrilateral OCPD is not a diamond ... (1 min)
23 (This topic is ***2, the first item (1) is 5 points, and the second item (7 points, full score 12 points).
Solution: (1) Add ad.
Ab = ac = 8, D is the midpoint of the side length, BC ∴ AD BC ⊥ ................................... (1min).
In Rt△ABD, ∴ BD = CD = 5 ................. (1).
∠∠EDC =∠b+∠BED =∠EDF+∠CDF，∠EDF =∠B，
∴∠ Bed =∠ CDF .............................................................. (1min)
∫ab = ac,∴∠b =∠c
∴△ BDE ∽△ CFD ..................................... (1min)
BE = 4，...........................................................( 1)
(2)∫△bde∽△CFD, ∴ ........................... (1min)
Bd = cd, ∴ .............................................. (1min)
∠EDF =∠B，∴△ BDE ∽△ DFE。 ∴∠ bed =∠ def ............................................ (1min)
∫ef//BC, ∴∠ BDE = ∠ def ..................................................................... (1min)
∴∠ BDE = ∠ bed. ∴ Be = BD = 5 ............................................................................ (1min)
Therefore, from AB = 8, AE = 3. ..
∵EF // BC，∴............................................( 1)
∫ BC = 10, ∴ ............................... (1min)
24. (This question ***2 is a minor issue, with the score of (1) being 5, the score of (2) being 7 and the full score being 12).
Solution: (1)∫ The image of quadratic function passes through point m (1, 0),
∴ .................................... (1min)
∴ m =-3 .......................................... (1min)
The analytical formula of the function is .................................................................................................................................. (1 min).
In addition, the coordinate of vertex ∴ is (2,1) ....................... (2 points).
(2) The symmetry axis of the quadratic function image obtained from (1) is the straight line x = 2, ∴ d (2 2,0) .................. (1min).
Judging from the meaning of the question, A (0), B(0, B), c (2 2,4+B) ................................ (2 points).
∵ Symmetry axis straight line x = 2 is parallel to the y axis,
∴△ AOB ∽△ ADC ........................................ (1min)
That is ∴ ................................. (1min).
Solution, ......................................... (2 points)
It is verified that,, are all the values of m that meet the conditions.
25. (This title is ***3, item (1) is 4 points, item (2) is 5 points, item (3) is 5 points, and the full score is 14).
(1) Proof: Cut the line segment AH on the AB side, make AH = PC, and connect pH. 。
From the square of ABCD, we get ∠ B = ∠B =∠BCD =∠D = 90, AB = BC = AD...( 1).
∠∠APF = 90° ,∴∠apf =∠b
∠∠APC =∠b+∠BAP =∠APF+∠FPC，
∴∠ PAH =∠ FPC ........................................... (1min)
∠∠BCD =∠DCE = 90, CF ∠DCE, ∴∠ FCE = 45.
∴∠PCF = 135。
And ∵AB = BC, AH = PC, ∴BH = BP, that is ∠ BPH = ∠ BHP = 45.
∴∠∠ AHP = 135, that is ∠ AHP = ∠ PCF .................................................................... (1min).
In △AHP and △PCF, ∠PAH =∠FPC, AH = PC, ∠AHP =∠PCF,
∴△ Analytic Hierarchy Process△ ?△PCF. ∴ AP = PF .............................. (1min)
(2) solution: circumscribe the positional relationship between ⊙P and ⊙ g.
Extend CB to point m, make BM = DG, and connect AM.
By AB = AD, ∠ abm = ∠ d = 90, BM = DG,
Get △ adg △ abm, namely AG = AM, ∠ mab = ∠ gad ........................... (1min).
ap = fp，∠APF = 90° ,∴∠paf = 45°。
∠∠bad = 90, ∴∠BAP +∠DAG = 45, that is ∠ Map = ∠ PAG = 45. (1 min).
Then, AM = AG, ∠MAP =∠PAG, AP = AP,
Get △ APM△ APG. ∴ PM = PG。
That is Pb+DG = PG ................................................. (2 points).
∴⊙P and ⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙ ...............................................................................
(3) solution: from PG // CF, ........................................... (∠ GPC = ∠ FCE = 45) was obtained.
Therefore, from < BCD = 90, < GPC = < PGC = 45.
∴ PC = GC。 DG = BP .................................................. (1min)
Let BP = x, DG = X. From AB = 2, PC = GC = 2–X.
∫p b+ DG = pg,∴pg = 2 x
At Rt△PGC, ∠ PCG = 90, ............. (1) is obtained.
I see. Get its .............................. (1 min)
∴ (if applicable), pg//cf ...............................................................................................................................................................
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