ALEVEL IGCSE 数学试卷-1
2011 June igcse 英文数学试卷
This document consists of 19 printed pages and 1 blank page.IB11 06_0580_43/4RP© UCLES 2011[Turn over*8044643715*UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary EducationMATHEMATICS 0580/43Paper 4 (Extended) May/June 20112 hours 30 minutesCandidates answer on the Question Paper.Additional Materials: Electronic calculatorGeometrical instrumentsMathematical tables (optional)Tracing paper (optional)READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen.You may use a pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES.Answer all questions.If working is needed for any question it must be shown below that question. Electronic calculators should be used.If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For π use either your calculator value or 3.142.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 130.© UCLES 20110580/43/M/J/11For Examiner's Use1 Lucy works in a clothes shop.(a) In one week she earned $277.20.(i) She spent 81of this on food.Calculate how much she spent on food. Answer(a)(i) $ [1](ii) She paid 15% of the $277.20 in taxes. Calculate how much she paid in taxes. Answer(a)(ii) $ [2](iii) The $277.20 was 5% more than Lucy earned in the previous week. Calculate how much Lucy earned in the previous week. Answer(a)(iii) $ [3](b) The shop sells clothes for men, women and children.(i) In one day Lucy sold clothes with a total value of $2200 in the ratio men : women : children = 2 : 5 : 4. Calculate the value of the women’s clothes she sold. Answer(b)(i) $ [2](ii) The $2200 was 7344of the total value of the clothes sold in the shop on this day. Calculate the total value of the clothes sold in the shop on this day. Answer(b)(ii) $ [2]© UCLES 2011 0580/43/M/J/11[Turn overUsex(a) (i) Draw the reflection of shape X in the x -axis. Label the image Y . [2](ii) Draw the rotation of shape Y , 90° clockwise about (0, 0). Label the image Z . [2](iii) Describe fully the single transformation that maps shape Z onto shape X .Answer(a)(iii)[2](b) (i) Draw the enlargement of shape X , centre (0, 0), scale factor21. [2](ii) Find the matrix which represents an enlargement, centre (0, 0), scale factor 21.Answer(b)(ii)[2](c) (i) Draw the shear of shape X with the x -axis invariant and shear factor –1.[2](ii) Find the matrix which represents a shear with the x -axis invariant and shear factor –1.Answer(c)(ii)[2]© UCLES 20110580/43/M/J/11Use(x + 5) cm2x cmx cmNOT TO SCALEThe diagram shows a square of side (x + 5) cm and a rectangle which measures 2x cm by x cm. The area of the square is 1 cm 2 more than the area of the rectangle.(a) Show that x 2 – 10x – 24 = 0 . Answer(a) [3]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use(b) Find the value of x . Answer(b) x = [3](c) Calculate the acute angle between the diagonals of the rectangle. Answer(c) [3]© UCLES 2011 0580/43/M/J/11For Examiner's Use4NOT TO SCALEThe circle, centre O , passes through the points A , B and C . In the triangle ABC , AB = 8 cm, BC = 9 cm and CA = 6 cm. (a) Calculate angle BAC and show that it rounds to 78.6°, correct to 1 decimal place. Answer(a) [4](b) M is the midpoint of BC .(i) Find angle BOM . Answer(b)(i) Angle BOM = [1]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use(ii) Calculate the radius of the circle and show that it rounds to 4.59 cm, correct to 3 significantfigures.Answer(b)(ii) [3](c) Calculate the area of the triangle ABC as a percentage of the area of the circle. Answer(c) % [4]© UCLES 2011 0580/43/M/J/11ForExaminer's Use5 (a) Complete the table of values for the function f(x ), where f(x ) = x 2 + 21x , x ≠ 0 .xO 3 O 2.5 O 2 O 1.5 O 1 O 0.50.5 1 1.5 2 2.5 3 f(x ) 6.41 2.69 4.25 4.252.69 6.41[3](b) On the grid, draw the graph of y = f(x ) for O 3 Y x Y O 0.5 and 0.5 Y x Y 3 .[5]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use(c) (i) Write down the equation of the line of symmetry of the graph.Answer(c)(i)[1](ii) Draw the tangent to the graph of y = f(x ) where x = O 1.5. Use the tangent to estimate the gradient of the graph of y = f(x ) where x = O 1.5. Answer(c)(ii) [3](iii) Use your graph to solve the equation x 2 + 21x= 3.Answer(c)(iii) x = or x = or x = or x = [2](iv) Draw a suitable line on the grid and use your graphs to solve the equation x 2 + 21x = 2x .Answer(c)(iv) x =or x =[3]© UCLES 2011 0580/43/M/J/11For Examiner's Use6CumulativefrequencyMass (kilograms)mThe masses of 200 parcels are recorded. The results are shown in the cumulative frequency diagram above.(a) Find(i) the median, Answer(a)(i) kg [1](ii) the lower quartile, Answer(a)(ii) kg [1](iii) the inter-quartile range, Answer(a)(iii) kg [1](iv) the number of parcels with a mass greater than 3.5 kg. Answer(a)(iv) [2]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use(b) (i) Use the information from the cumulative frequency diagram to complete the groupedfrequency table.Mass (m ) kg0 I m Y 44 I m Y 66 I m Y 77 I m Y 10Frequency 36 50[2](ii) Use the grouped frequency table to calculate an estimate of the mean. Answer(b)(ii) kg [4](iii) Complete the frequency density table and use it to complete the histogram.Mass (m ) kg 0 I m Y 4 4 I m Y 6 6 I m Y 7 7 I m Y 10Frequency density916.7FrequencydensityMass (kilograms)m[4]© UCLES 20110580/43/M/J/11ForExaminer's Use7 Katrina puts some plants in her garden.The probability that a plant will produce a flower is107. If there is a flower, it can only be red, yellow or orange.When there is a flower, the probability it is red is 32 and the probability it is yellow is 41.(a) Draw a tree diagram to show all this information. Label the diagram and write the probabilities on each branch. Answer(a) [5](b) A plant is chosen at random. Find the probability that it will not produce a yellow flower. Answer(b) [3](c) If Katrina puts 120 plants in her garden, how many orange flowers would she expect? Answer(c) [2]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use8A(a) Draw accurately the locus of points, inside the quadrilateral ABCD , which are 6 cm from thepoint D . [1](b) Using a straight edge and compasses only, construct(i) the perpendicular bisector of AB , [2](ii) the locus of points, inside the quadrilateral, which are equidistant from AB and from BC . [2](c) The point Q is equidistant from A and from B and equidistant from AB and from BC .(i) Label the point Q on the diagram. [1](ii) Measure the distance of Q from the line AB . Answer(c)(ii) cm [1](d) On the diagram, shade the region inside the quadrilateral which is• less than 6 cm from Dand• nearer to A than to Band• nearer to AB than to BC . [1]© UCLES 2011 0580/43/M/J/11For Examiner's Use9 f(x ) = 3x + 1 g(x ) = (x + 2)2(a) Find the values of(i) gf(2), Answer(a)(i)[2](ii) ff(0.5). Answer(a)(ii)[2](b) Find f –1(x ), the inverse of f(x ). Answer(b)[2](c) Find fg(x ). Give your answer in its simplest form. Answer(c)[2]© UCLES 2011 0580/43/M/J/11[Turn overFor Examiner's Use(d) Solve the equation x 2 + f(x ) = 0. Show all your working and give your answers correct to 2 decimal places. Answer(d) x = or x =[4]UseBABCD is a parallelogram.DC, M is the midpoint of BC and N is the midpoint of LM.pq.(i)Find the following in terms ofp and q, in their simplest form.(a)Answer(a)[1](b)Answer(a)[2](c)Answer(a)[2] (ii) N lies on the line AC.Answer(a)(ii) [1]© UCLES 2011 0580/43/M/J/11© UCLES 2011 0580/43/M/J/11[Turn overUseEH J2x°75°(x + 15)°NOT TO SCALEEFG is a triangle. HJ is parallel to FG . Angle FEG = 75°. Angle EFG = 2x ° and angle FGE = (x + 15)°.(i) Find the value of x . Answer(b)(i) x = [2](ii) Find angle HJG . Answer(b)(ii) Angle HJG = [1]© UCLES 2011 0580/43/M/J/11For Examiner's Use11 (a) (i) The first three positive integers 1, 2 and 3 have a sum of 6. Write down the sum of the first 4 positive integers. Answer(a)(i) [1](ii) The formula for the sum of the first n integers is21)(+n n . Show the formula is correct when n = 3. Answer(a)(ii) [1](iii) Find the sum of the first 120 positive integers. Answer(a)(iii) [1](iv) Find the sum of the integers121 + 122 + 123 + 124 + …………………………… + 199 + 200.Answer(a)(iv)[2](v) Find the sum of the even numbers 2 + 4 + 6 + …………………………+ 800.Answer(a)(v)[2]© UCLES 20110580/43/M/J/11For Examiner's Use(b) (i) Complete the following statements about the sums of cubes and the sums of integers.13 = 1 1 = 113 + 23 = 9 1 + 2 = 3 13 + 23 + 33 =1 +2 +3 =13 + 23 + 33 + 43 =1 +2 +3 +4 =[2](ii) The sum of the first 14 integers is 105. Find the sum of the first 14 cubes. Answer(b)(ii) [1](iii) Use the formula in part(a)(ii) to write down a formula for the sum of the first n cubes. Answer(b)(iii) [1](iv) Find the sum of the first 60 cubes. Answer(b)(iv) [1](v) Find n when the sum of the first n cubes is 278 784. Answer(b)(v) n = [2]BLANK PAGEPer mission to r epr oduce items wher e thir d-par ty owned mater ial pr otected by copyr ight is included has been sought and clear ed wher e possible. Ever y reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.© UCLES 2011 0580/43/M/J/11。
0606_m19_qp_12(IGCSE 数学考试2019年3月真题
Identities Formulae for ∆ABC
2. TRIGONOMETRY
sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A
a sin A
=
b sin B
=
c sin C
a2 = b2 + c2 – 2bc cos A
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80.
∆
=
1 2
bc
sin
A
© UCLES 2019
0606/12/F/M/19
3
1 (a) Given that
பைடு நூலகம்
= { x : 1 1 x 1 20},
A = {multiples of 3},
B = {multiples of 4},
find
(i)
n (A),
[1]
(ii)
n (A + B) .
[1]
*6715046212*
Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education
Igcse 数学 历年真题
4400/4HEdexcel IGCSEMathematicsPaper 4HHigher TierFriday 11 June 2010 – AfternoonTime: 2 hoursMaterials required for examination Items included with question papersRuler graduated in centimetres and Nilmillimetres, protractor, compasses,pen, HB pencil, eraser, calculator.Tracing paper may be used.Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions. Write your answers in the spaces provided in this question paper.You must NOT write on the formulae page. Anything you write on the formulae page will gain NO credit.If you need more space to complete your answer to any question, use additional answer sheets.Information for CandidatesThe marks for individual questions and the parts of questions are shown in round brackets: . (2).There are 22 questions in this question paper. The total mark for this paperis 100.You may use a calculator.Advice to CandidatesWrite your answers neatly and in good English.This publication may be reproduced only in accordance with Edexcel Limited copyright policy.©2010 Edexcel Limited.Printer’s Log. No. N36905AIGCSE MATHEMATICS 4400 FORMULA SHEET – HIGHER TIERAnswer ALL TWENTY TWO questions.Write your answers in the spaces provided.You must write down all stages in your working.1. Solve 6 y – 9 = 3 y + 7y = ................................(Total 3 marks) 2. The diagram shows two towns, A and B, on a map.(a) By measurement, find the bearing of B from A.....................................(2)C is another town.The bearing of C from A is 050.(b) Find the bearing of A from C.....................................(2)(Total 4 marks)3. A spinner can land on red or blue or yellow.The spinner is biased.The probability that it will land on red isThe probability that it will land on blue isImad spins the spinner once.(a) Work out the probability that it will land on yellow......................................(2)Janet spins the spinner 30 times.(b)Work out an estimate for the number of times the spinner will land on blue......................................(2)(Total 4 marks)4. Rosetta drives 85 kilometres in 1 hour 15 minutes.(a) Work out her average speed in kilometres per hour...................................... km/h(2)Rosetta drives a total distance of 136 kilometres.(b) Work out 85 as a percentage of 136................................. %(2)Sometimes Rosetta travels by train to save money.The cost of her journey by car is £12The cost of her journey by train is 15% less than the cost of her journey by car.(c)Work out the cost of Rosetta’s journ ey by train.£ ...................................(3)(Total 7 marks)5.Calculate the value of x.Give your answer correct to 3 significant figures.x = ................................(Total 3 marks)6. A = {2, 3, 4, 5}B = {4, 5, 6, 7}(a)(i) List the members of A B......................................(ii) How many members are in A B?.....................................(2)ℰ = {3, 4, 5, 6, 7}P = {3, 4, 5}Two other sets, Q and R, each contain exactly three members.P Q = {3, 4}P R = {3, 4}Set Q is not the same as set R.(b)(i) Write down the members of a possible set Q......................................(ii) Write down the members of a possible set R......................................(2)(Total 4 marks)7. Rectangular tiles have width (x + 1) cm and height (5x – 2) cm.Some of these tiles are used to form a large rectangle.The large rectangle is 7 tiles wide and 3 tiles high.The perimeter of the large rectangle is 68 cm.(a) Write down an equation in x...............................................................................................................(3)(b) Solve this equation to find the value of x.x = ................................(3)(Total 6 marks)8. Show that 121 141 = 1519. The depth of water in a reservoir increases from 14 m to m.Work out the percentage increase.................................. %(Total 3 marks)10. Quadrilaterals ABCD and PQRS are similar.AB corresponds to PQ.BC corresponds to QR.CD corresponds to RS.Find the value of(a) xx = ...............................(2)(b) yy = ...............................(1)(Total 3 marks)11. Simplify fully6x + 43x.....................................(Total 3 marks)12.(a)Find the equation of the line L......................................(3)(b) Find the three inequalites that define the unshaded region shown in the diagram below................................................................................................................(3)(Total 6 marks)13. (a) Solve x 2– 8x + 12 = 0.....................................(3)(b) Solve the simultaneous equationsy = 2x4x – 5y = 9x = ................................y = ................................(3)(Total 6 marks)14.The area of the triangle is cm2.The angle x° is acute.Find the value of x.Give your answer correct to 1 decimal place.x = ................................(Total 3 marks)15. The unfinished histogram shows information about the heights, h metres, ofsome trees.(a) Calculate an estimate for the number of trees with heights in theinterval < h ≤ 10.....................................(3)(b) There are 75 trees with heights in the interval 10 < h ≤ 13Use this information to complete the histogram.(2)(Total 5 marks)16. A bag contains 3 white discs and 1 black disc.John takes at random 2 discs from the bag without replacement.(a) Complete the probability tree diagram.First disc Second disc(3)(b)Find the probability that both discs are white......................................(2)All the discs are now replaced in the bag.Pradeep takes at random 3 discs from the bag without replacement.(c)Find the probability that the disc left in the bag is white......................................(3)(Total 8 marks)17. The diagram shows a sector of a circle, radius 45 cm, with angle 84°.Calculate the area of the sector.Give your answer correct to 3 significant figures.............................. cm2(Total 3 marks) 18.Calculate the length of AC.Give your answer correct to 3 significant figures................................ cm(Total 3 marks)19. A cone has slant height 4 cm and base radius r cm.The total surface area of the cone is 433π cm 2.Calculate the value of r .r = ................................(Total 4 marks)20. f(x) = (x – 1)2(a) Find f(8).....................................(1)The domain of f is all values of x where x ≥ 7(a)Find the range of f......................................(2)xg(x) =x1(c) Solve the equation g(x) =.....................................(2)(d) (i) Express the inverse function g –1 in the form g –1(x) = .......g –1(x) = ...................................(ii) Hence write down gg(x) in terms of x.gg(x) = ....................................(6)(Total 11 marks)21.In the diagram OA= a and OC= c.(a) Find CA in terms of a and c......................................(1)The point B is such that AB=1c.2(b) Give the mathematical name for the quadrilateral OABC......................................(1)The point P is such that OP= a + k c, where k ≥ 0(c) State the two conditions relating to a + k c that must be true for OAPCto be a rhombus.(2)(Total 4 marks)22. (a) Work out × 102+ × 104Give your answer in standard form......................................(2)a × 102 +b × 104 =c × 104(b) Express c in terms of a and b.c = ................................(2)(Total 4 marks)TOTAL FOR PAPER = 100 MARKSEND。
国际学校高中数学期末试卷(美高几何)
Multiple Choice:(18’)1、If there are no dancers that aren't slim and no singers that aren't dancers, then which statements are always true?( ).A.There is not one slim person that isn't a dancerB.All singers are slimC.Anybody slim is also a singerD.None of the above2、In the equation (x-5)2+(y-2)2=16, the center of the circle is answer choices( ).A.(5,2)B.(-5, -2)C.(-5, -3)D.(5,3)3、In the following diagram,which of the following is not an example of an inscribed angle of circle O( ).A.NST∠ D.∠ C.SNT∠ B.MNS∠NSMQ3 Q4 Q54、What can you NOT conclude from the diagram at the right? ( ).A. c=dB. a=bC. c2 + e2=b2D. e=d5、If m∠KLM=20° and measure of arc MP=30 ° ,what is m∠KNP?( ).A.25°B.50°C.35°D.70°6、What is the negation of the statement ''The coat is blue''? ( ).A. the coat is greenB. the coat is sometimes blueC. the coat is not blueD. it is not true that the coat is not blue7、Which of the following is equal to cos35°( ).A. Sin35°B. Cos55°C. Sin55°D. cos145°8、Look at this series: 80, 10, 70, 15, 60, … What number should come next?( ).A. 20B. 25C. 30D. 509、What is the contrapositive of the proposition ''If a polygon has three sides then it is a triangle''?( ).A.''If a polygon is a triangle, then it has less than three sides.''B.''If a polygon is not a triangle, then it does not have three sides.''C.''If a polygon is not a triangle, then it has more than three sides.''D.''If a polygon is a triangle, then it does not have three sides.''Fill in the blank with sometimes, always, or never. (7’)1)Two tangents to the same circle from the same point are_______ congruent to each other.2)If two inscribed angles are congruent, then their intercepted arcs are________ congruent.3)If a line bisects an arc, then it_______bisects the chord of the arc.4)The measure of a central angle is______equal to the measure of its intercepted arc.5)If a radius bisects a chord of a circle, then it____bisects the minor arc of the chord.6)If two arcs are congruent, an inscribed angle of one arc is______congruent to an inscribed angle of the other arc.7)If two chords of a circle are not congruent then the shorter chord is_______ closer to the center of the circle.Find the value of x in each figure below.(6’)The circle C has equation(6’)0195242022=+--+y x y xThe centre of C is at the point M1)Find the coordinate of the point M and the radius of the circle C.2)N is the point with coordinate (25,32), find the length of the line MN.Use the method above to solve the following:(12’)()⎪⎭⎫ ⎝⎛-43sin 1π ()⎪⎭⎫ ⎝⎛629cos 2π ()()︒420tan 3()⎪⎭⎫ ⎝⎛49sec 4π— ()()︒510csc 5 ()⎪⎭⎫ ⎝⎛319cot 6πFor each the following statement ，do each of the following(8’)：“All Park View students will graduate.”Conditional: _______________________;Symbol________________;Converse: ________________________;Symbol________________;Inverse: __________________________;Symbol________________;Contrapositive: ____________________;Symbol________________.In circle O, diameter AB and ED intersect at center O;chord BD and tangent CB , Secant CDA,if the measure of arc EA equals 80°,the measure of arc AD equals 100°,find (5’)：1)m∠EOB=_____________; 2)m∠BAD=_____________; 3)m∠C=_____________;4)m∠CBD=_____________; 5)m∠EDB=_____________.Find the m∠DGF to the nearest whole degree.(4’)：Extra Questions:(10’)1、The8×18rectangle ABCD is cut into two congruent hexagons, as shown, in sucha way that the two hexagons can be repositioned without overlap to form a square. What is Y?2、Square ABCD has side length s, a circle centered at E has radius r,and r and s are both rational. The circle passes through D, and D lies on BE. Point F lies on the circle, on the same side of BE as A. Segment AF is tangent to the circle, and 259+=AF . What is sr ?3、All of the triangles in the diagram below are similar to isosceles triangle ABC, in which AB=AC. Each of the 7 smallest triangles has area 1, and triangle ABC has area40. What is the area of trapezoid DBCE?。
Igcse-数学-历年真题-2
4400/4HEdexcel IGCSEMathematicsPaper 4HHigher TierFriday 11 June 2010 – AfternoonTime: 2 hoursMaterials required for examination Items included with question papersRuler graduated in centimetres and Nilmillimetres, protractor, compasses,pen, HB pencil, eraser, calculator.Tracing paper may be used.Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials and signature.Check that you have the correct question paper.Answer ALL the questions. Write your answers in the spaces provided in this question paper.You must NOT write on the formulae page. Anything you write on the formulae page will gain NO credit.If you need more space to complete your answer to any question, use additional answer sheets.Information for CandidatesThe marks for individual questions and the parts of questions are shown in round brackets: e.g. (2).There are 22 questions in this question paper. The total mark for this paperis 100.You may use a calculator.Advice to CandidatesWrite your answers neatly and in good English.This publication may be reproduced only in accordance with Edexcel Limited copyright policy.©2010 Edexcel Limited.Printer’s Log. No. N36905AIGCSE MATHEMATICS 4400 FORMULA SHEET – HIGHER TIERAnswer ALL TWENTY TWO questions.Write your answers in the spaces provided.You must write down all stages in your working.1. Solve 6 y – 9 = 3 y + 7y = ................................(Total 3 marks) 2. The diagram shows two towns, A and B, on a map.(a) By measurement, find the bearing of B from A.....................................︒(2)C is another town.The bearing of C from A is 050︒.(b) Find the bearing of A from C.....................................︒(2) (Total 4 marks)3. A spinner can land on red or blue or yellow.The spinner is biased.The probability that it will land on red is 0.5The probability that it will land on blue is 0.2Imad spins the spinner once.(a) Work out the probability that it will land on yellow......................................(2)Janet spins the spinner 30 times.(b)Work out an estimate for the number of times the spinner will land on blue......................................(2)(Total 4 marks)4. Rosetta drives 85 kilometres in 1 hour 15 minutes.(a) Work out her average speed in kilometres per hour...................................... km/h(2)Rosetta drives a total distance of 136 kilometres.(b) Work out 85 as a percentage of 136................................. %(2)Sometimes Rosetta travels by train to save money.The cost of her journey by car is £12The cost of her journey by train is 15% less than the cost of her journey by car.(c)Work out the cost of Rose tta’s journey by train.£ ...................................(3)(Total 7 marks)5.Calculate the value of x.Give your answer correct to 3 significant figures.x = ................................(Total 3 marks)6. A = {2, 3, 4, 5}B = {4, 5, 6, 7}(a)(i) List the members of A ⋂B......................................(ii) How many members are in A ⋃B?.....................................(2)ℰ = {3, 4, 5, 6, 7}P = {3, 4, 5}Two other sets, Q and R, each contain exactly three members.P ⋂Q = {3, 4}P ⋂R = {3, 4}Set Q is not the same as set R.(b)(i) Write down the members of a possible set Q......................................(ii) Write down the members of a possible set R......................................(2)(Total 4 marks)7. Rectangular tiles have width (x + 1) cm and height (5x – 2) cm.Some of these tiles are used to form a large rectangle.The large rectangle is 7 tiles wide and 3 tiles high.The perimeter of the large rectangle is 68 cm.(a) Write down an equation in x...............................................................................................................(3)(b) Solve this equation to find the value of x.x = ................................(3)(Total 6 marks)8. Show that 121 141 = 1519. The depth of water in a reservoir increases from 14 m to 15.75 m.Work out the percentage increase.................................. %(Total 3 marks) 10. Quadrilaterals ABCD and PQRS are similar.AB corresponds to PQ.BC corresponds to QR.CD corresponds to RS.Find the value of(a) xx = ...............................(2)(b) yy = ...............................(1)(Total 3 marks)11. Simplify fully6x + 43x.....................................(Total 3 marks)12.(a)Find the equation of the line L......................................(3)(b) Find the three inequalites that define the unshaded region shown in the diagram below................................................................................................................(3)(Total 6 marks)13. (a) Solve x 2– 8x + 12 = 0.....................................(3)(b) Solve the simultaneous equationsy = 2x4x – 5y = 9x = ................................y = ................................(3)(Total 6 marks)14.The area of the triangle is 6.75 cm2.The angle x° is acute.Find the value of x.Give your answer correct to 1 decimal place.x = ................................(Total 3 marks)15. The unfinished histogram shows information about the heights, h metres, ofsome trees.(a) Calculate an estimate for the number of trees with heights in theinterval 4.5 < h ≤ 10.....................................(3)(b) There are 75 trees with heights in the interval 10 < h ≤ 13Use this information to complete the histogram.(2)(Total 5 marks)16. A bag contains 3 white discs and 1 black disc.John takes at random 2 discs from the bag without replacement.(a) Complete the probability tree diagram.First disc Second disc(3)(b)Find the probability that both discs are white......................................(2)All the discs are now replaced in the bag.Pradeep takes at random 3 discs from the bag without replacement.(c)Find the probability that the disc left in the bag is white......................................(3)(Total 8 marks)17. The diagram s hows a sector of a circle, radius 45 cm, with angle 84°.Calculate the area of the sector.Give your answer correct to 3 significant figures.............................. cm2(Total 3 marks) 18.Calculate the length of AC.Give your answer correct to 3 significant figures................................ cm(Total 3 marks)19. A cone has slant height 4 cm and base radius r cm.The total surface area of the cone is 433π cm 2.Calculate the value of r .r = ................................(Total 4 marks)20. f(x) = (x – 1)2(a) Find f(8).....................................(1)The domain of f is all values of x where x ≥ 7(a)Find the range of f......................................(2)xg(x) =x1(c) Solve the equation g(x) = 1.2.....................................(2)(d) (i) Express the inverse function g –1 in the form g –1(x) = .......g –1(x) = ...................................(ii) Hence write down gg(x) in terms of x.gg(x) = ....................................(6)(Total 11 marks)21.In the diagram = a and = c.(a) Find CA in terms of a and c......................................(1)The point B is such that AB=1c.2(b) Give the mathematical name for the quadrilateral OABC......................................(1)The point P is such that = a + k c, where k ≥ 0(c) State the two conditions relating to a + k c that must be true for OAPCto be a rhombus.(2)(Total 4 marks)22. (a) Work out 5.2 × 102+ 2.3 × 104Give your answer in standard form......................................(2)a × 102 +b × 104 =c × 104(b) Express c in terms of a and b.c = ................................(2)(Total 4 marks)TOTAL FOR PAPER = 100 MARKS END。
IGCSE math 数学试卷0580_s17_qp_43
This document consists of 19 printed pages and 1 blank page.DC (NH/JG) 130218/2© UCLES 2017[Turn over*0731247115*MATHEMATICS 0580/43Paper 4 (Extended) May/June 20172 hours 30 minutesCandidates answer on the Question Paper.Additional Materials: Electronic calculator Geometrical instrumentsTracing paper (optional).READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use an HB pencil for any diagrams or graphs.Do not use staples, paper clips, glue or correction fluid.DO NOT WRITE IN ANY BARCODES.Answer all questions.If working is needed for any question it must be shown below that question.Electronic calculators should be used.If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.For π, use either your calculator value or 3.142.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.The total of the marks for this paper is 130.Cambridge International ExaminationsCambridge International General Certificate of Secondary Education1 (a) In 2016, a company sold 9600 cars, correct to the nearest hundred.(i) Write down the lower bound for the number of cars sold. (1)(ii) The average profit on each car sold was $2430, correct to the nearest $10.Calculate the lower bound for the total profit.Write down the exact answer.$ (2)(iii) Write your answer to part (a)(ii) correct to 4 significant figures.$ (1)(iv) Write your answer to part (a)(iii) in standard form.$ (1)(b) In April, the number of cars sold was 546.This was an increase of 5% on the number of cars sold in March.Calculate the number of cars sold in March. (3)© UCLES 20170580/43/M/J/17(c) The price of a new car grows exponentially by 3% per year.A new car has a price of $3000 in 2013.Find the price of a new car 4 years later.$ (2)© UCLES 2017[Turn over0580/43/M/J/170580/43/M/J/17© UCLES 20172 (a)y °x °z °DABPQSRCE NOT TO SCALE24°38°PQ is parallel to RS .ABC and ADE are straight lines. Find the values of x , y and z .x = ..................................................y = ..................................................z = ..................................................[3] (b)ABDCNOT TOSCALE42°The points A , B , C and D lie on the circumference of the circle. AB = AD , AC = BC and angle ABD = 42°. Find angle CAB .Angle CAB = (3)(c)NOT TOSCALEThe points P, Q, R and S lie on the circumference of the circle, centre O.Angle QOS =146°.Find angle QRS.QRS = . (2)Angle© UCLES 2017[Turn over0580/43/M/J/170580/43/M/J/17© UCLES 20173The table shows some values for 24y x x 32=+.x –2.2–2–1.5–1–0.500.50.8y–1.940.753.58(a) Complete the table.[4](b) Draw the graph of 24y x x 32=+ for 2.20.8x G G - .[4] (c) Find the number of solutions to the equation 243x x 32+=. (1)0580/43/M/J/17© UCLES 2017[Turn over(d) (i) The equation 241x x x 32+-= can be solved by drawing a straight line on the grid.Write down the equation of this straight line.y = ..................................................[1] (ii) Use your graph to solve the equation 241x x x 32+-=.x = ............................ or x = ............................ or x = ............................[3] (e) The tangent to the graph of 24y x x 32=+ has a negative gradient when x k =. Complete the inequality for k ....................... 1 k 1 . (2)0580/43/M/J/17© UCLES 20174 (a) The diagram shows a solid metal prism with cross section ABCDE .BGFKJ DCAEH2 cm7 cm4 cm8 cm 4 cmNOT TOSCALE(i) Calculate the area of the cross section ABCDE .............................................cm 2 [6](ii) The prism is of length 8 cm.Calculate the volume of the prism.............................................cm 3 [1](b) A cylinder of length 13 cm has volume 280 cm3.(i) Calculate the radius of the cylinder..............................................cm [3] (ii) The cylinder is placed in a box that is a cube of side 14 cm.Calculate the percentage of the volume of the box that is occupied by the cylinder................................................% [3]© UCLES 2017[Turn over0580/43/M/J/175 (a) Haroon has 200 letters to post.The histogram shows information about the masses, m grams, of the letters.Mass (grams)mFrequencydensity(i) Complete the frequency table for the 200 letters.Mass (m grams)0 1m G 1010 1m G 2020 1m G 2525 1m G 3030 1m G 50Frequency5017[3](ii) Calculate an estimate of the mean mass.................................................g [4]0580/43/M/J/17© UCLES 2017(b) Haroon has 15 parcels to post.The table shows information about the sizes of these parcels.Size Small LargeFrequency96Two parcels are selected at random.Find the probability that(i) both parcels are large, (2)(ii) one parcel is small and the other is large. (3)(c) The probability that a parcel arrives late is 803.4000 parcels are posted.Calculate an estimate of the number of parcels expected to arrive late. (1)6(a) Describe fully the single transformation that maps shape A onto(i) shape B,...................................................................................................................................................... (2)(ii) shape C....................................................................................................................................................... (3)(b) Draw the image of shape A after rotation through 90° anticlockwise about the point (3, -1). [2]y=. [2] (c) Draw the image of shape A after reflection in 1f p.(d) Describe fully the single transformation represented by the matrix 3003.............................................................................................................................................................. (3)7 (a) Solve the simultaneous equations. You must show all your working.x y 2311+=x y 3550-=- x = ..................................................y = (4)(b) 12x x a x b 22-+=+^h Find the value of a and the value of b .a = ..................................................b = (3)(c) Write as a single fraction in its simplest form.x x x x 25132-+-+ (4)8 (a)The table shows the marks gained by 10 students in their physics test and their mathematics test.The first six points have been plotted for you.MathematicsmarkPhysics mark[2](ii) What type of correlation is shown in the scatter diagram? (1)(b) The marks of 30 students in a spelling test are shown in the table below.Mark012345Frequency245568Find the mean, median, mode and range of these marks.Mean = ..................................................Median = ..................................................Mode = ..................................................Range = (7)(c) The table shows the marks gained by some students in their English test.Mark 527591Number of students x4511The mean mark for these students is 70.3 .Find the value of x.= (3)x9ACQ B525 m872 m104°NOT TO SCALEABC is a triangular field on horizontal ground. There is a vertical pole BQ at B .AB = 525 m, BC = 872 m and angle ABC = 104°.(a) Use the cosine rule to calculate the distance AC .AC = ..............................................m [4] (b) The angle of elevation of Q from C is 1.0°.Showing all your working, calculate the angle of elevation of Q from A . (4)(c) (i) Calculate the area of the field.............................................. m2 [2] (ii) The field is drawn on a map with the scale 1 : 20 000.Calculate the area of the field on the map in cm2.............................................cm2 [2]10 = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30} A = { x : x is a multiple of 3} B = { x : x is prime} C = { x : x G 25}(a) Complete the Venn diagram.ABCᏱ[4] (b) Use set notation to complete the statements. (i) 26 ..................... B [1](ii) A + B = .....................[1](c) List the elements of B , (C + A ). (2)(d) Find (i) n(C ),...................................................[1] (ii) B B C n ,+l ^^h h ....................................................[1] (e)A C +^h is a subset of A C ,^h . Complete this statement using set notation.A C +^h ..................... A C ,^h [1]11 The table shows the first four terms in sequences A, B, C and D.Complete the table.BLANK PAGEPermission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at after the live examination series.Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.。
【独家带详解答案】IGCSE2019年数学真题卷1(060612)_20200830123606
0606/12 May/June 2019
2 hours
READ THESE INSTRUCTIONS FIRST
Write your centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80.
*0773973091*
Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education
ADDITIONAL MATHEMATICS Paper 1
Igcse 数学 历年真题 2
4400/4HEdexcel IGCSEMathematicsPaper 4HHigher TierFriday 11 June 2010 – AfternoonTime: 2 hoursMaterials required for examination Items included with question papersRuler graduated in centimetres and Nilmillimetres, protractor, compasses,pen, HB pencil, eraser, calculator.Tracing paper may be used.Instructions to CandidatesIn the boxes above, write your centre number, candidate number, your surname, initials andsignature.Check that you have the correct question paper.Answer ALL the questions. Write your answers in the spaces provided in this question paper.You must NOT write on the formulae page.Anything you write on the formulae page will gain NO credit.If you need more space to complete your answer to any question, use additional answer sheets.Information for CandidatesThe marks for individual questions and the parts of questions are shown in round brackets: e.g.<2>.There are 22 questions in this question paper. The total mark for this paper is 100.You may use a calculator.Advice to CandidatesWrite your answers neatly and in good English.This publication may be reproduced only in accordance withEdexcel Limited copyright policy.©2010 Edexcel Limited.Printer’s Log. No. N36905AIGCSE MATHEMATICS 4400FORMULA SHEET – HIGHER TIERAnswer ALL TWENTY TWO questions.Write your answers in the spaces provided.You must write down all stages in your working.1. Solve6 y – 9 = 3 y + 7y = ................................<Total 3 marks>2. The diagram shows two towns, A and B, on a map.<a> By measurement, find the bearing of B from A.....................................<2>C is another town.The bearing of C from A is 050︒.<b> Find the bearing of A from C.....................................︒<2><Total 4 marks>3. A spinner can land on red or blue or yellow.The spinner is biased.The probability that it will land on red is 0.5The probability that it will land on blue is 0.2Imad spins the spinner once.<a> Work out the probability that it will land on yellow......................................<2>Janet spins the spinner 30 times.<b> Work out an estimate for the number of times the spinner will land on blue......................................<2><Total 4 marks>4.Rosetta drives 85 kilometres in 1 hour 15 minutes.<a> Work out her average speed in kilometres per hour...................................... km/h<2>Rosetta drives a total distance of 136 kilometres.<b> Work out 85 as a percentage of 136................................. %<2>Sometimes Rosetta travels by train to save money.The cost of her journey by car is £12The cost of her journey by train is 15% less than the cost of her journey by car.(c)Work out the cost of Rosetta’s journey by train.£ ...................................<3><Total 7 marks>5.Calculate the value of x.Give your answer correct to 3 significant figures.x = ................................<Total 3 marks>6. A = {2, 3, 4, 5}B = {4, 5, 6, 7}(a)<i> List the members of A ⋂B......................................<ii> How many members are in A ⋃B?.....................................<2>ℰ = {3, 4, 5, 6, 7}P = {3, 4, 5}Two other sets, Q and R, each contain exactly three members.P ⋂Q = {3, 4}P ⋂R = {3, 4}Set Q is not the same as set R.(b)<i> Write down the members of a possible set Q......................................<ii> Write down the members of a possible set R......................................<2><Total 4 marks>7. Rectangular tiles have width <x + 1> cm and height <5x – 2> cm.Some of these tiles are used to form a large rectangle.The large rectangle is 7 tiles wide and 3 tiles high.The perimeter of the large rectangle is 68 cm.<a> Write down an equation in x...............................................................................................................<3><b> Solve this equation to find the value of x.x = ................................<3><Total 6 marks>8. Show that121 141 = 151 9. The depth of water in a reservoir increases from 14 m to 15.75 m.Work out the percentage increase.................................. %<Total 3 marks>10. Quadrilaterals ABCD and PQRS are similar.AB corresponds to PQ . BC corresponds to QR . CD corresponds to RS . Find the value of <a> xx = ...............................<2><b> yy = ...............................<1><Total 3 marks>11. Simplify fully 6x + 43x.....................................<Total 3 marks>12.(a) Find the equation of the line L ......................................<3><b> Find the three inequalites that define the unshaded region shown in the diagrambelow...................................... ..................................... .....................................<3> <Total 6 marks>13. <a> Solve x 2– 8x + 12 = 0.....................................<3><b> Solve the simultaneous equations y = 2x 4x – 5y = 9x = ................................ y = ................................<3><Total 6 marks>14.The area of the triangle is 6.75 cm 2. The angle x° is acute.Find the value of x .Give your answer correct to 1 decimal place.x = ................................<Total 3 marks>15. The unfinished histogram shows information about the heights, h metres, of some trees.<a> Calculate an estimate for the number of trees with heights in the interval4.5 < h ≤ 10.....................................<3><b> There are 75 trees with heights in the interval 10 < h ≤ 13 Use this information to complete the histogram.<2><Total 5 marks>16. A bag contains 3 white discs and 1 black disc.John takes at random 2 discs from the bag without replacement. <a> Complete the probability tree diagram. First disc Second disc<3>(b) Find the probability that both discs are white......................................<2>All the discs are now replaced in the bag.Pradeep takes at random 3 discs from the bag without replacement. (c) Find the probability that the disc left in the bag is white......................................<3><Total 8 marks>17. The diagram shows a sector of a circle, radius 45 cm, with angle 84°.Calculate the area of the sector.Give your answer correct to 3 significant figures.............................. cm 2<Total 3 marks>18.Calculate the length of AC .Give your answer correct to 3 significant figures................................ cm<Total 3 marks>19. A cone has slant height 4 cm and base radius r cm.The total surface area of the cone is 433π cm 2.Calculate the value of r .r = ................................<Total 4 marks>20. f<x > = <x – 1>2<a> Find f<8>.....................................<1>The domain of f is all values of x where x ≥ 7(a)Find the range of f......................................<2>xg<x> =x1<c> Solve the equation g<x> = 1.2.....................................<2><d> <i> Express the inverse function g –1 in the form g –1<x> = .......g –1<x> = ...................................<ii> Hence write down gg<x> in terms of x.gg<x> = ....................................<6><Total 11 marks> 21.In the diagram OA= a and OC= c.<a> Find CA in terms of a and c......................................<1>The point B is such that AB=1c.2<b> Give the mathematical name for the quadrilateral OABC......................................<1>The point P is such that OP= a + k c, where k ≥ 0<c> State the two conditions relating to a + k c that must be true for OAPC to be arhombus.<2><Total 4 marks> 22. <a> Work out 5.2 × 102 + 2.3 × 104Give your answer in standard form......................................<2>a × 102 +b × 104 =c × 104<b> Express c in terms of a and b.c = ................................<2><Total 4 marks>TOTAL FOR PAPER = 100 MARKSEND。
IGCSE math 数学试卷0580_s17_qp_22
This document consists of 12 printed pages.DC (RW/JG) 130075/2© UCLES 2017[Turn overCambridge International ExaminationsCambridge International General Certificate of Secondary Education*2343051529*MATHEMATICS 0580/22Paper 2 (Extended) May/June 20171 hour 30 minutesCandidates answer on the Question Paper.Additional Materials: Electronic calculator Geometrical instrumentsTracing paper (optional)READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use an HB pencil for any diagrams or graphs.Do not use staples, paper clips, glue or correction fluid.DO NOT WRITE IN ANY BARCODES.Answer all questions.If working is needed for any question it must be shown below that question.Electronic calculators should be used.If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.For r , use either your calculator value or 3.142.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.The total of the marks for this paper is 70.0580/22/M/J/17© UCLES 20171Write 0.071 64 correct to 2 significant figures. (1)2 The probability that Stephanie wins her next tennis match is 0.85 .Find the probability that Stephanie does not win her next tennis match. (1)3Change 6200 cm 2 into m 2............................................ m 2 [1]4Calculate .12038252+-. (1)5Work out 85 cents as a percentage of $2.03 ............................................. % [1]6 Factorise.x y1421- (1)7 Find the value of a b53- when a7= and b2=-. (2)8q°p°NOT TOSCALE 70°The diagram shows a straight line intersecting two parallel lines.Find the value of p and the value of q.p=................................................q= (2)9 Without using a calculator, work out 6521- .Show all the steps of your working and give your answer as a fraction in its simplest form. (2)0580/22/M/J/17© UCLES 2017[Turn over10 Solve.x x251-=+x= (2)11 (a) Write 0.0605 in standard form. (1)(b) Calculate ..0151104##, giving your answer in standard form. (1)12x°NOT TO SCALE7cm4cmCalculate the value of x.x= (2)0580/22/M/J/17© UCLES 20170580/22/M/J/17© UCLES 2017[Turn over13 Solve the inequality.n n 3115182-- (2)14 Work out. (a) 12532 (1)(b) 312-J L K K N P O O (1)15 Make q the subject of the formulap q 22=.q = (2)0580/22/M/J/17© UCLES 201716BAC(a) Using a straight edge and compasses only , construct the bisector of angle BAC . [2](b) Shade the region inside the triangle that is nearer to AC than to AB . [1]17ABCNOT TOSCALE30°110°8.15 mCalculate AC .AC = ........................................... m [3]0580/22/M/J/17© UCLES 2017[Turn over18 A rectangle has length 62 mm and width 47 mm, both correct to the nearest millimetre. The area of this rectangle is A mm 2.Complete the statement about the value of A .......................................... A 1G ......................................... [3]19 In a triangle PQR , PQ = 8 cm and QR = 7 cm. The area of this triangle is 17 cm 2.Calculate the two possible values of angle PQR .Angle PQR = ........................ or ........................ [3]20 Write as a single fraction in its simplest form.x x 32112--+ (3)0580/22/M/J/17© UCLES 201721 y is inversely proportional to x 1+. When x 8=, y 2=.Find y when x 15=.y = (3)22 Factorise completely. (a) t u 922- (2)(b)c d pc pd 242--+ (2)0580/22/M/J/17© UCLES 2017[Turn over23 (a) = {students in a class}P = {students who study physics} C = {students who study chemistry}The Venn diagram shows numbers of students.ᏱC11578P(i) Find the number of students who study physics or chemistry. (1)(ii) Find n P C k l ^h ..................................................[1] (iii) A student who does not study chemistry is chosen at random. Find the probability that this student does not study physics..................................................[1] (b) On the Venn diagram below, shade the region D E j l .ᏱED[1]0580/22/M/J/17© UCLES 201724CDXBA2.8 cm2 cm8 cm4 cmNOT TO SCALEIn the diagram, AB and CD are parallel. AD and BC intersect at X . AB = 8 cm, CD = 4 cm, CX = 2 cm and DX = 2.8 cm. (a) Complete this mathematical statement.Triangle ABX is ............................................ to triangle DCX . [1](b) Calculate AX .AX = ......................................... cm [2](c) The area of triangle ABX is y cm 2. Find the area of triangle DCX in terms of y .......................................... cm 2 [1]0580/22/M/J/17© UCLES 2017[Turn over25 (a) Simplify. x 161643^h (2)(b) p 25423=Find the value of p .p = (2)26E NOT TOSCALEA ,B ,C ,D andE lie on the circle. AB is extended toF .Angle AED = 140° and angle CBF = 95°.Find the values of w , x and y .w = ................................................ x = ................................................y = (5)Question 27 is printed on the next page.0580/22/M/J/17© UCLES 2017Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at after the live examination series.Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.27xNOT TO SCALEA is the point (-2, 0) andB is the point (0, 4).(a) Find the equation of the straight line joining A and B . (3)(b) Find the equation of the perpendicular bisector of AB . (4)。
igcse考试题0580_w02_qp_1
Paris
Answer.……………….……… [3]
15 (a) Write down the values of 20 = ……. , 21 = ……. , 22 = ……. , 23 = ……. , 24 = ……. [2]
(b)
Change
5 49
to a decimal. Write down your full calculator display.
4
10 Complete this diagram accurately so that it has rotational symmetry of order 3 about the point O.
For Examiner's
Use
O
[2]
11 An athlete’s time for a race was 43 .78 seconds.
–3 £ n < 3.
List all the possible values of n.
Answer.……………….……… [2]
6 B
54° A
NOT TO SCALE
AB and AC are tangents to the circle,
centre O.
O
Angle BAC = 54°.
(a) Write down the size of angle ABO.
OCTOBER/NOVEMBER SESSION 2002
Candidates answer on the question paper. Additional materials:
Electronic calculator Geometrical instruments Mathematical tables (optional) Tracing paper (optional)
【A-level数学】S1知识点及题型大放送
【A-level数学】S1知识点及题型大放送【A-level数学】S1知识点及题型大放送数学，数学，对，没错，就是数学！数学虐你千百遍，你也要待数学如初恋。
为什么？谁让数学如此重要呢？忍了吧！哈哈，今天索引留学小编跟各位分享的是A-Level数学--CIE S1的知识点和题型，赶紧一起来看看吧~~数据的中心及位置测量知识点：After completing this chapter you should be able to1.Recognise different types of data数据的分类2.Find the mean, median, quartiles and mode for a data set平均值，中位数,四分位数和众数3.Find range, interquartile range, variance and standard deviation for a data set全距，四分位距，方差和标准差4.Find mean and variance for a combining set of data找出几组数据之和的平均值和方差5.Construct and interpret diagrams to represent data—stem and leaf diagram, box-and-whisker plots, histogram, and cumulative frequency graphs考试主要题型1.找出几组数据的之和的平均值和方差2.根据频数分布表画出直方图3.根据已知数据画出茎叶图，盒式图等4.数据的图解表示及分析，以及根据图解写出平均值，方差，中位数等；5.比较两组数据的分布比较A-Level数学S1知识点1：Recognise different types of data There are two types of data: qualitative（定性变量）and quantitative （定量变量）.There are two types of quantitative variables, discrete variable and continuous variable.离散型随机变量可以用频数表表示Frequency distribution for discrete data连续型随机变量用频数表表示：Frequency distribution for continuous dataTo form a frequency distribution of the continuous data, group the information into classes or intervals. When the data is presented as a grouped frequency table, the specific data values are lost.A-Level数学S1知识点2：Find the mean, median, quartiles and mode for a data setA-Level数学S1知识点3：Find range, interquartile range, variance and standard deviation for a data set1.Range全距：highest value-lowest value2.Interquartile range四分位距：Q3-Q13.Variance and standard deviation方差和标准差Standard deviation gives a measure of the spread of the data in relation to the mean, of the distribution; it is calculated using all the values in the distributions.For each value x, calculate how far it is from the mean byfinding(x-mean).A-Level数学S1知识点4：Find mean and variance for a combining set of data1.In general, for two sets of data, x and y,2.In general, if each data value is increased by a constant aThe mean is increased by aThe standard deviation in unaltered.This is particular useful when finding the mean and standard deviation using and , where a is a constant.3.Finding the mean and standard deviation using anda.To find the mean,Find the mean of (x-a)Now add ab.To find the standard deviation,Find the standard deviation of (x-a)This is the same as the standard deviation of x.A-Level数学S1知识点5：Construct and interpret diagrams to represent data—stem and leaf diagram, box-and-whisker plots, histogram, and cumulative frequency graphs1.Stem and leaf diagram茎叶图A way of grouping data into intervals while still retaining the original data is to draw a stem-and-leaf diagram, also known as a stemplot.These are the marks of 20 students in an assignment:84 17 38 45 47 53 76 54 75 3266 65 55 54 51 44 39 19 54 72In stem-and-leaf diagram all the intervals must be of equal width, so it seems sensible to choose intervals 10-19, 20-29, 30-39,…,80-89 for this data, so tale the stem to represent the tens and the leaf to represent the units.You enter the numbers one by one. When all the numbershave been entered, you must arrange the entries in each row in numerical order with the smallest nest to the stem.Important: You must always give a Key to explain what the stem and leaf represent.Stem-and-leaf diagrams can be used to compare two sets of data by showing them together on a back-to-back stem-and-leaf diagram.2.HistogramGrouped data can be displayed in a histogram.Histograms resemble bar charts, but there are two important differences:There are no gaps between the bars.The area of the bar is proportional to the frequency that it represents.If the height of the bar is adjusted so that the area is equal to the frequency in that interval, thenSo , this is known as the frequency density.3.Box-and-whisker plots盒式图In a box-and-whisker plot the median and quartiles are shown, as well as the minimum and maximum values of the distribution. It gives a very good visual summary of a distribution and is particularly useful when comparing set of data.4.Cumulative frequency graphsCumulative frequency is the total frequency up to a particular item. Cumulative frequency is particularly useful when finding the median and quartiles.Cumulative frequency table can be illustrated on a cumulative frequency graph in which the cumulative frequencies are plotted against the upper class boundaries. The points are joined either with a curve or with straight lines.Values can be estimated from the graph, for example:(a)Estimate how many plants had a height less than 10.5 cm.From 10.5 on the x-axis draw a vertical line up to the curve.Now draw a horizontal line to the y-axis and read off the value(b)10% of plants had a height of at least x cm. estimate x.10% of 30=33plants had height of at least x cm, so 27 plants had a height less than x cm.From 27 on the y-axis draw a horizontal line to the curve.Now draw a vertical line down to the x-axis and read off the value.From the graph, 10% of the plants had a height of at least 16.5.更多A-level资讯，敬请关注哦~~。
0606_m19_qp_22IGCSE数学真题)2019年3月
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Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education
igcse考试
igcse考试在选课过程中，很多人会问，什么是IGCSE，什么是GCSE？你想在读A-level之前学习IGCSE吗？今天，我想详细谈谈。
什么是IGCSE？GCSE课程被公认为是连接高中课程（A-level课程）的最佳纽带，因此越来越多的国家开始采用这一课程体系。
英国现阶段有一套叫做GCSE的课程体系。
除英国外，其他国家的GCSE课程称为IGCSE（国际普通中等教育证书）。
完成GCE后，他们可以到世界上100多个国家学习。
如果你想申请剑桥大学、牛津大学和其他类似的英国大学，这个考试成绩是申请的重要参考之一。
IGCSE是一门两年制的课程。
当然，一些国际学校选择将两年制课程压缩为一年，自然教学时间安排会非常紧张。
如何选择IGCSE课程？数学和英语是必修课。
另外，您可以根据自己的兴趣和专业选择3-4门课程。
虽然IGCSE有60多个科目可供选择，但我们会发现频率较高的科目包括数学、英语、物理、化学、生物、经济、商业、历史、地理、会计学、ICT（信息和通信技术）、艺术（如音乐、绘画、戏剧）等。
IGCSE检查时间爱德思IGCSE每年夏季和冬季有两次考试，分别在1月、5月和6月进行。
剑桥剑桥考试局的IGCSE也是一年两次，分别在11月、12月和5月和6月。
每个IGCSE学生至少需要5张通行证才能毕业并升级到alevel。
考试分类IGCSE分为扩展型和核心型。
前者的最高分数是a*，考试大纲比后者更深入。
它更注重知识点的运用、逻辑思维的培养和举一反三的能力。
后者得分最高的是C，主要考察知识点的核心内容、认知、记忆和基本应用，以及对实际问题的解释。
【最新】igcse试卷-精选word文档 (16页)
本文部分内容来自网络整理，本司不为其真实性负责，如有异议或侵权请及时联系，本司将立即删除！== 本文为word格式，下载后可方便编辑和修改！ ==igcse试卷篇一：201X年IGCSE数学考试题1. Solve the equation(9)cos x + ?(1 ? sin 2x) = 0, in the interval 0? ? x < 360?.12(1?x)2. (a) For the binomial expansion of , ?x? < 1, in ascending powers of x,(i) find the first four terms,(ii) write down the coefficient of xn.(2)?xnnx?2(1?x)n?1(b) Hence, show that, for ?x? < 1, = .(2)(a?1)x?x2(an?1)x??2(1?x)n?1(c) Prove that, for ?x? < 1, , where a is a constant.(4)?5n?1?3nn?12(d) Hence evaluate .(2)3. f(x) = x3 ? (k + 4)x + 2k, where k is a constant.(a) Show that, for all values of k, the curve with equation y = f(x) passes through the point (2, 0).(1)(b) Find the values of k for which the equation f(x) = 0 has exactly two distinct roots.(5)Given that k > 0, that the x-axis is a tangent to the curve with equation y = f(x), and that the line y = p intersects the curve in three distinct points,(c) find the set of values that p can take. (5) ?n4.y(0, 4)The circle, with centre C and radius r, touches the y-axis at (0, 4) and also touches the line with equation 4y ? 3x = 0, as shown in Fig.1.(a) (i) Find the value of r.(8)(b)(4)31??(ii) Show that arctan + 2 acrtan = 1 Figure 1 ?. The line with equation 4x + 3y = q, q > 12, is a tangent to the circle. Find the value of q. 1dy2?(1?t). dt5. (a) Given that y = ln [t + ?(1 +t2)],show that =(3)The curve C has parametric equations12?(1?t), y = ln [t + ?(1 + t2)], t ? ?. x = A student was asked to prove that, for t > 0, the gradient of the tangent to C is negative. The attempted proof was as follows:?1??t??y = ln ?x??tx?1???x??= ln= ln (tx + 1) ?ln xdy? dx tx?11xt?x = t1?= tx?1xt?(1?t2)(1?t2)t??(1?t2)= – = t??(1?t2) ? ? (1 + t2)dyAs (1 + t2) > 0, and t + ?(1 + t2) > 0 for t > 0, dx < 0 for t > 0.(b) (i) Identify the error in this attempt.(ii) Give a correct version of the proof.(6)(c) Prove that ln [?t + ?(1 + t2)] = ?ln [t + ?(1 + t2)].(3)(d) Deduce that C is symmetric about the x-axis and sketch the graph of C.(3)6. f(x) = x ? [x], x ? 0where [x] is the largest integer ? x.For example, f(3.7) = 3.7 ? 3 = 0.7; f(3) = 3 ? 3 = 0.(a) Sketch the graph of y = f(x) for 0 ? x < 4.(3)??f(x)dx(b) Find the value of p for which ?2= 0.18.(3)Given that 1g(x) = 1?kx,x ? 0,k > 0,and that x0 = is a root of the equation f(x) = g(x),(c) find the value of k. (2)p2(1)The root of f(x) = g(x) in the interval n < x < n + 1 is xn, where n is an integer.(e) Prove that2xn2 ? (2n ? 1)xn ? (n + 1) = 0. (d) Add a sketch of the graph of y = g(x) to your answer to part (a).(4)7. Triangle ABC, with BC = a, AC = b and AB = c is inscribed in a circle. Given that AB is a diameter of the circle and that a2, b2 and c2 are three consecutive terms of an arithmetic progression (arithmetic series),(a) express b and c in terms of a,(4)。
2014年Alevel 数学真题(精排版)
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9. In the first month after opening, a mobile phone shop sold 300 phones. A model for future sales assumes that the number of phones sold will increase by 5% per month, so that 300 × 1.05 will be sold in the second month, 300 × 1.052 in the third month, and so on. Using this model, calculate (a) the number of phones sold in the 24th month, (2) (b) the total number of phones sold over the whole 24 months. (2) This model predicts that, in the N th month, the number of phones sold in that month exceeds 3000 for the first time. (c) Find the value of N. (3)
北京爱迪(国际)学校必修一第三单元《指数函数和对数函数》检测题(包含答案解析)
一、选择题1．若()()22ln 1f x x x e =+≤≤(e 为自然对数的底数)，则函数()()22y f x f x =+⎡⎤⎣⎦的最大值为（ ） A ．6B ．13C ．22D ．332．集合{}1002,x x x x R =∈的真子集的个数为（ ）A ．2B ．4C ．6D ．73．已知定义在R 上的函数()f x 满足()()2f x f x +=，且当[)1,1x ∈-时，()2f x x =，若函数()log 1a g x x =+图象与()f x 的图象恰有10个不同的公共点，则实数a 的取值范围为（ ） A ．()4,+∞ B ．()6,+∞ C ．()1,4D ．()4,64．设log a m 和log b m 是方程2420x x -+=的两个根，则log a bm 的值为（ ）AB ．2C ．D ．±5．函数()212()log 4f x x =-的单调递增区间为（ ）.A ．(0,+∞)B ．(-,0)C ．(2,+∞)D ．(-,-2)6．已知正实数a ，b ，c 满足：21()log 2a a =，21()log 3b b =，2log c c 1=，则（ ） A ．a b c <<B ．c b a <<C ．b c a <<D ．c a b <<7．已知2log 0.8a =，0.7log 0.6b =，0.60.7c =，则a ，b ，c 的大小关系是（ ） A ．a b c <<B ．b a c <<C ．a c b <<D ．b c a <<8．已知函数 ()lg 2x xe ef x --=，则f (x )是（ ）A ．非奇非偶函数，且在(0，+∞)上单调递增B ．奇函数，且在R 上单调递增C ．非奇非偶函数，且在(0，+∞)上单调递减D ．偶函数，且在R 上单调递减 9．函数2()ln(43)f x x x =+-的单调递减区间是（ ）A ．32⎛⎤-∞ ⎥⎝⎦，B ．3,42⎡⎫⎪⎢⎣⎭C ．3,2⎡⎫+∞⎪⎢⎣⎭D ．31,2⎛⎤- ⎥⎝⎦10．函数()22x xxf x -=+的大致图象为（ ）A ．B ．C ．D ．11．已知0.22a =，0.20.4b =，0.60.4c =，则（ ）A ．a b c >>B ．a c b >>C ．c a b >>D ．b c a >>12．对数函数log (0a y x a =>且1)a ≠与二次函数()21y a x x =--在同一坐标系内的图象可能是（ ）A ．B ．C ．D ．二、填空题13．已知正实数a 满足8(9)a a a a =，则log 3a =____________. 14．()()2lg 45f x x x =--+的单调递增区间为______.15．设函数2()ln(1)f x x x =+，若()23(21)0f a f a +-<，则实数a 的取值范围为_____.16．已知43==m n k ，且20+=≠m n mn ，则k =______.17．已知12512.51000x y ==，则11x y＝_____.18．已知2312a b ==，则21a b+=_______． 19．如果()231log 2log 9log 64x x x f x =-+-，则使()0f x <的x 的取值范围是______. 20．若函数1log 12a y x ⎛⎫=+⎪⎝⎭在区间3,62⎡⎤-⎢⎥⎣⎦有最小值-2，则实数a =_______.三、解答题21．设函数()log (1)log (3)(0,1)a a f x x x a a =++->≠． （1）求函数()f x 的定义域（2）若(1)2f =，求函数()f x 在区间3[0,]2上的最大值． （3）解不等式：log (1)log (3)a a x x +>-． 22．已知函数()3lg3xf x x+=-． （1）求函数()f x 的定义域；（2）判断函数()f x 的奇偶性，并说明理由．23．已知函数f (x )＝log a (x ＋1)－log a (1－x )，a ＞0，且a ≠1. （1）求f (x )的定义域；（2）判断f (x )的奇偶性，并予以证明； （3）当a ＞1时，求使f (x )＞0的x 的取值范围. 24．若函数()()()331xf x k a b a =++->是指数函数（1）求k ，b 的值；（2）求解不等式()()2743f x f x ->-25．已知函数()log (1)a f x x =+，()log (1)(0a g x x a =->，且1)a ≠. （1）求函数()()f x g x -的定义域；（2）判断函数()()f x g x -的奇偶性，并说明理由；（3）当2a =时，判断函数()()f x g x -的单调性，并给出证明.26．（1）0160.25371.586-⨯-+-⎫⎛ ⎪⎝⎭（2）1324lg lg82493-+【参考答案】***试卷处理标记，请不要删除一、选择题 1．B 解析：B 【分析】先依题意求函数定义域，再化简函数，进行换元后求二次函数在区间上的最大值即可. 【详解】由21x e ≤≤及()2f x知221x e ≤≤，故定义域为[]1,e ，又()()()()()222222ln 2ln ln 6ln 61y f x f x x x x x x e =+=+++=++≤≤⎡⎤⎣⎦令[]ln 0,1t x =∈，则266y t t =++，易见y 在[]0,1t ∈上单调递增， 故当1t =时，即x e =时，max 16613y =++=. 故选：B. 【点睛】易错点睛：利用换元法求函数最值时，要注意函数的定义域，否则求得的易出错.2．D解析：D 【分析】分析指数函数2xy =与幂函数100y x=的图像增长趋势，当0x <时，有1个交点；当0x >时，有2个交点；即集合{}1002,x x x x R =∈有3个元素，所以真子集个数为3217-=【详解】分析指数函数2xy =与幂函数100y x =的图像增长趋势，当0x <时，显然有一个交点；当0x >时，当1x =时，110021>；当2x =时，210022<；故()1,2x ∈时，有一个交点；分析数据发现，当x 较小时，100y x=比2xy =增长的快；当x 较大时，2xy =比100y x =增长的快，即2x y =是爆炸式增长，所以还有一个交点.即2xy =与100y x=的图像有三个交点，即集合{}1002,x x xx R =∈有3个元素，所以真子集个数为3217-= 故选：D. 【点睛】结论点睛：本题考查集合的子集个数，集合A 中含有n 个元素，则集合A 的子集有2n 个，真子集有()21n-个，非空真子集有()22n-个.3．D解析：D 【分析】转化条件为函数()f x 是周期为2的周期函数，且函数()g x 、()f x 的图象均关于1x =-对称，由函数的对称性可得两图象在1x =-右侧有5个交点，画出图象后，数形结合即可得解. 【详解】因为函数()f x 满足()()2f x f x +=，所以函数()f x 是周期为2的周期函数， 又函数()log 1a g xx =+的图象可由函数log a y x =的图象向左平移一个单位可得， 所以函数()log 1a g x x =+的图象的对称轴为1x =-，当[)1,1x ∈-时，()2f x x =，所以函数()f x 的图象也关于1x =-对称，在平面直角坐标系中作出函数()y f x =与()y g x =在1x =-右侧的图象，数形结合可得，若函数()log 1a g x x =+图象与()f x 的图象恰有10个不同的公共点， 则由函数图象的对称性可得两图象在1x =-右侧有5个交点，则()()13log 415log 61a a a g g ⎧>⎪=<⎨⎪=>⎩，解得()4,6a ∈. 故选：D. 【点睛】关键点点睛：解决本题的关键是函数的周期性、对称性及数形结合思想的应用.4．D解析：D 【分析】利用换底公式先求解出+log log m m a b 、log log m m a b ⋅的结果，然后利用换底公式将log a bm 变形为1log log m m a b-，根据+log log m m a b 、log log m m a b ⋅的结果求解出log log m m a b -的结果，则log a bm 的值可求.【详解】因为log log 4log log 2a b a b m m m m +=⎧⎨⋅=⎩，所以114log log 112log log m mm m a b a b⎧+=⎪⎪⎨⎪⋅=⎪⎩ ，所以log +log 4log log 1log log 2m m m m m m a b a b a b ⎧=⎪⋅⎪⎨⎪⋅=⎪⎩，所以log +log 21log log 2m m m m a b a b =⎧⎪⎨⋅=⎪⎩， 又因为11log log log log a m m bmm aa b b==-，且()()22log log =log log lo +42g log m m m m m m a b a b b a -⋅=-，所以log log m m a b -=所以log 2a bm ==±，故选：D. 【点睛】关键点点睛：解答本题的关键是在于换底公式的运用，将log a bm 变形为1log log m m a b-，再根据方程根之间的关系求解出结果.5．D解析：D 【分析】求出函数的定义域，根据对数型复合函数的单调性可得结果. 【详解】函数()212()log 4f x x =-的定义域为()(),22,-∞-+∞，因为函数()f x 是由12log y u =和24u x =-复合而成，而12log y u =在定义域内单调递减，24u x =-在(),2-∞-内单调递减，所以函数()212()log 4f x x =-的单调递增区间为(),2-∞-， 故选：D. 【点睛】易错点点睛：对于对数型复合函数务必注意函数的定义域.6．B解析：B 【分析】a ､b ､c 的值可以理解为图象交点的横坐标，则根据图象可判断a ，b ，c 大小关系.【详解】因为21()log 2a a =，21()log 3b b =，2log c c 1=， 所以a ､b ､c 为2log y x =与1()2x y =，1()3xy =，y x =-的交点的横坐标，如图所示：由图象知： c b a <<. 故选：B 【点睛】本题主要考查对数函数，指数函数的图象性质以及函数零点问题，还考查了数形结合的思想方法，属中挡题.7．C解析：C 【解析】因为22log 0.8log 10a =<=，0.70.7log 0.6log 0.71b =>=，0.6000.70.71c <=<=，所以a c b <<，故选C.8．A解析：A 【分析】本题考查函数的奇偶性和和单调性的概念及简单复合函数单调性的判定. 【详解】要使函数有意义，需使0,2x x e e -->即21,1,x xx e e e >∴>解得0;x >所以函数()f x 的为(0,);+∞定义域不关于原点对称，所以函数()f x 是非奇非偶函数；因为1,xxx y e y ee-==-=-是增函数，所以2x xe e y --=是增函数，又lg y x =是增函数，所以函数()lg 2x xe ef x --=在定义域(0,)+∞上单调递增.故选：A 【点睛】本题考查对数型复合函数的奇偶性和单调性，属于中档题.9．B解析：B 【分析】先求函数的定义域，再利用复合函数的单调性同增异减，即可求解. 【详解】由2430x x +->得2340x x --<，解得：14x -<<，2()ln(43)f x x x =+-由ln y t =和234t x x =-++复合而成，ln y t =在定义域内单调递增，234t x x =-++对称轴为32x =，开口向下， 所以 234t x x =-++在31,2⎛⎫- ⎪⎝⎭ 单调递增，在3,42⎡⎫⎪⎢⎣⎭单调递减， 所以2()ln(43)f x x x =+-的单调减区间为3,42⎡⎫⎪⎢⎣⎭，故选：B 【点睛】本题主要考查了利用同增异减求复合函数的单调区间，注意先求定义域，属于中档题10．B解析：B 【分析】根据函数为奇函数排除C ，取特殊值排除AD 得到答案. 【详解】 当()22x xx f x -=+，()()22x x xf x f x ---==-+，函数为奇函数，排除C ； 2221(2)22242f -=<=+，排除A ； 3324(3)22536f -==+，4464(4)224257f -==+，故()()34f f >，排除D.故选：B. 【点睛】本题考查了函数图象的识别，意在考查学生的计算能力和识图能力，取特殊值排除是解题的关键.11．A解析：A 【解析】分析：0.20.4b =， 0.60.4c =的底数相同，故可用函数()0.4x f x =在R 上为减函数，可得0.60.200.40.40.41<<=．用指数函数的性质可得0.20221a =>=，进而可得0.20.20.620.40.4>>．详解：因为函数()0.4xf x =在R 上为减函数，且0.2<0.4 所以0.60.200.40.40.41<<= 因为0.20221a =>=． 所以0.20.20.620.40.4>>． 故选A ．点睛：本题考查指数大小的比较，意在考查学生的转化能力．比较指数式的大小，同底数的可利用指数函数的单调性判断大小，底数不同的找中间量1，比较和1的大小．12．A解析：A 【分析】由对数函数，对a 分类，01a <<和1a >，在对数函数图象确定的情况下，研究二次函数的图象是否相符．方法是排除法． 【详解】由题意，若01a <<，则log a y x =在()0+∞，上单调递减， 又由函数()21y a x x =--开口向下，其图象的对称轴()121x a =-在y 轴左侧，排除C ，D.若1a >，则log a y x =在()0+∞，上是增函数， 函数()21y a x x =--图象开口向上，且对称轴()121x a =-在y 轴右侧，因此B 项不正确，只有选项A 满足. 故选：A ． 【点睛】本题考查由解析式先把函数图象，解题方法是排除法，可按照其中一个函数的图象分类确定另一个函数图象，排除错误选项即可得．二、填空题13．【分析】利用已知式两边同时取以e 为底的对数化简计算再利用换底公式代入计算即可【详解】正实数a 满足两边取对数得即故解得故故答案为：【点睛】本题解题关键是对已知指数式左右两边同时取以e 为底的对数化简计算 解析：716-【分析】利用已知式两边同时取以e 为底的对数，化简计算ln a ，再利用换底公式ln 3log 3ln a a=代入计算即可. 【详解】正实数a 满足8(9)aaa a =，两边取对数得8ln ln(9)aaa a =，即ln 8ln(9)a a a a =，故()ln 8ln9ln a a =+，解得16ln ln 37a =-，故ln 3ln 37log 316ln 16ln 37a a ===--.故答案为：716-. 【点睛】本题解题关键是对已知指数式左右两边同时取以e 为底的对数，化简计算得到ln a 的值，再结合换底公式即突破难点.14．【分析】由复合函数的单调性只需求出的增区间即可【详解】令则由与复合而成因为在上单调递增且在上单调递增所以由复合函数的单调性知在上单调递增故答案为：【点睛】本题主要考查了复合函数的单调性对数函数的单调 解析：(]5,2--【分析】由复合函数的单调性，只需求出245t x x =--+的增区间即可. 【详解】令245t x x =--+，则()()2lg 45f x x x =--+由lg y t =与245t x x =--+复合而成，因为lg y t =在(0,)t ∈+∞上单调递增，且245(0)t x x t =--+>在(5,2]x ∈--上单调递增，所以由复合函数的单调性知，()()2lg 45f x x x =--+在(5,2]x ∈--上单调递增.故答案为：(]5,2-- 【点睛】本题主要考查了复合函数的单调性，对数函数的单调性，二次函数的单调性，属于中档题.15．【分析】根据已知可得为奇函数且在上单调递增不等式化为转化为关于自变量的不等式即可求解【详解】的定义域为是奇函数设为增函数在为增函数在为增函数在处连续的所以在上单调递增化为等价于即所以实数的取值范围为解析：1(1,)3- 【分析】根据已知可得()f x 为奇函数且在R 上单调递增，不等式化为()23(12)f a f a <-，转化为关于自变量的不等式，即可求解. 【详解】()f x 的定义域为R ，()()))ln10f x f x x x +-=+==，()f x ∴是奇函数，设,[0,)()x u x x =∈+∞为增函数，()f x 在[0,)+∞为增函数，()f x 在(,0)-∞为增函数， ()f x 在0x =处连续的，所以()f x 在R 上单调递增，()23(21)0f a f a +-<，化为()23(12)f a f a <-，等价于2312a a <-，即213210,13a a a +-<-<<， 所以实数a 的取值范围为1(1,)3-. 故答案为: 1(1,)3- 【点睛】本题考查利用函数的单调性和奇偶性解不等式，熟练掌握函数的性质是解题的关键，属于中档题.16．【分析】根据对数和指数的关系将指数式化成对数式再根据对数的运算计算可得【详解】解：故答案为：【点睛】本题考查对数和指数的关系对数的运算属于基础题 解析：36【分析】根据对数和指数的关系，将指数式化成对数式，再根据对数的运算计算可得. 【详解】 解：43m n k ==4log m k ∴=，3log =n k20m n mn +=≠211n m ∴+=，1log 4k m =，1log 3k n = 2log 3log 41k k ∴+=2log 3log 41k k ∴+=()log 941k ∴⨯=36k ∴=故答案为：36 【点睛】本题考查对数和指数的关系，对数的运算，属于基础题.17．【分析】根据指数与对数之间的关系求出利用对数的换底公式即可求得答案【详解】∵∴∴∴故答案为：【点睛】本题考查了指数与对数之间的关系掌握对数换底公式:是解本题的关键属于基础题解析：13【分析】根据指数与对数之间的关系,求出,x y ,利用对数的换底公式,即可求得答案. 【详解】∵12512.51000x y ==， ∴12512.51000100011log 1000,log 1000log 125log 12.5x y ====，∴1000100011log 125,log 12.5x y ==， ∴1000111log 103x y -==. 故答案为：13. 【点睛】本题考查了指数与对数之间的关系.掌握对数换底公式:log log log c a c bb a=是解本题的关键.属于基础题.18．【分析】根据指对互化先计算出的结果然后计算的结果由此即可计算出的结果【详解】因为所以所以所以故答案为：【点睛】关键点点睛：解答本题的关键是利用指对互化将化为对数形式然后根据对数运算法则完成计算 解析：1【分析】根据指对互化先计算出,a b 的结果，然后计算11,a b 的结果，由此即可计算出21a b+的结果. 【详解】因为2312a b ==，所以23log 12,log 12a b ==，所以121211log 2,log 3a b==，所以1212121212212log 2log 3log 4log 3log 121a b +=+=+==， 故答案为：1. 【点睛】关键点点睛：解答本题的关键是利用指对互化将2312a b ==化为对数形式，然后根据对数运算法则完成计算.19．【分析】可结合对数化简式将化简为再解对数不等式即可【详解】由由得即当时故；当时无解综上所述故答案为：【点睛】本题考查对数化简公式的应用分类讨论求解对数型不等式属于中档题解析：81,3⎛⎫⎪⎝⎭【分析】可结合对数化简式将()f x 化简为()1log 2log 3log 4x x x f x =-+-，再解对数不等式即可 【详解】由()2323231log 2log 9log 641log 2log 3log 4x x x x x x f x =-+-=-+-31log 2log 3log 41log 8x x x x =-+-=+，由()0f x <得81log 03x -<，即8log log 3xx x >， 当1x >时，83x <，故81,3x ⎛⎫∈ ⎪⎝⎭；当()0,1x ∈时，83x >，无解 综上所述，81,3x ⎛⎫∈ ⎪⎝⎭故答案为：81,3⎛⎫⎪⎝⎭【点睛】本题考查对数化简公式的应用，分类讨论求解对数型不等式，属于中档题20．或2【分析】根据复合函数的单调性及对数的性质即可求出的值【详解】当时在为增函数求得即;当时在为减函数求得即故答案为:或【点睛】本题考查复合函数单调性对数方程的解法难度一般解析：12或2 【分析】根据复合函数的单调性及对数的性质即可求出a 的值. 【详解】当1a >时, 1log 12a y x ⎛⎫=+ ⎪⎝⎭在3,62⎡⎤-⎢⎥⎣⎦为增函数,min33log 1-224a y f ⎛⎫⎛⎫=-=-+= ⎪ ⎪⎝⎭⎝⎭,求得-214a =,即=2a ; 当01a <<时, 1log 12a y x ⎛⎫=+ ⎪⎝⎭在3,62⎡⎤-⎢⎥⎣⎦为减函数,()()min 6log 31-2a y f ==+=,求得-24a =,即1=2a . 故答案为:12或2. 【点睛】本题考查复合函数单调性,对数方程的解法,难度一般.三、解答题21．（1）(1,3)-；（2）2；（3）答案见解析. 【分析】 （1）由1030x x +>⎧⎨->⎩得解定义域（2）由(1)2f =求得2a =．化简 22()log (1)4f x x ⎡⎤=--+⎣⎦，求得函数单调性得解（3）分类1a >和01a <<讨论得解 【详解】 （1）由1030x x +>⎧⎨->⎩得13x ，所以函数()f x 的定义域为(1,3)-．（2）因为(1)2f =，所以log 42(0,1)a a a =>≠，所以2a =．22222()log (1)log (3)log [(1)(3)]log (1)4f x x x x x x ⎡⎤=++-=+-=--+⎣⎦，所以当(1,1]x ∈-时，()f x 是增函数；当(1,3)x ∈时，()f x 是减函数， 故函数()f x 在(1,3)-上的最大值是2(1)log 42f ==．（3）当1a >时1330x x x +>-⎧⎨->⎩解得13x x >⎧⎨<⎩不等式解集为：{|13}x x <<当01a <<时1310x xx +<-⎧⎨+>⎩解得11x x <⎧⎨>-⎩不等式解集为：{|11}x x -<<【点睛】简单对数不等式问题的求解策略(1)解决简单的对数不等式，应先利用对数的运算性质化为同底数的对数值，再利用对数函数的单调性转化为一般不等式求解．(2)对数函数的单调性和底数a 的值有关，在研究对数函数的单调性时，要按1a >和01a <<进行分类讨论．22．（1）()3,3-；（2）()f x 为奇函数，证明见解析. 【分析】（1）利用对数式的真数大于零求解出不等式的解集即为定义域；（2）先判断定义域是否关于原点对称，若定义域关于原点对称，分析()(),f x f x -之间的关系，由此判断出()f x 的奇偶性. 【详解】 （1）因为303xx+>-，所以()()330x x -+<， 所以{}33x x -<<，所以()f x 的定义域为()3,3-； （2）()f x 为奇函数，证明：因为()f x 的定义域为()3,3-关于原点对称，且()()1333lg lg lg 333x x x f x f x x x x --++⎛⎫-===-=- ⎪+--⎝⎭， 所以()()f x f x -=-，所以()f x 为奇函数. 【点睛】思路点睛：判断函数()f x 的奇偶性的步骤如下：（1）先分析()f x 的定义域，若()f x 定义域不关于原点对称，则()f x 为非奇非偶函数，若()f x 的定义域关于原点对称，则转至（2）；（2）若()()f x f x =-，则()f x 为偶函数；若()()f x f x -=-，则()f x 为奇函数. 23．（1）{x |－1＜x ＜1}；（2）f (x )为奇函数；证明见解析；（3）(0，1). 【分析】（1）根据真数大于零，列出不等式，即可求得函数定义域；（2）计算()f x -，根据其与()f x 关系，结合函数定义域，即可判断和证明； （3）利用对数函数的单调性，求解分式不等式，即可求得结果. 【详解】（1）因为f (x )＝log a (x ＋1)－log a (1－x )，所以1010x x +>⎧⎨->⎩解得－1＜x ＜1.故所求函数的定义域为{x |－1＜x ＜1}. （2）f (x )为奇函数.证明如下：由（1）知f (x )的定义域为{x |－1＜x ＜1}，且f (－x )＝log a (－x ＋1)－log a (1＋x )＝－[log a (x ＋1)－log a (1－x )]＝－f (x ). 故f (x )为奇函数.（3）因为当a ＞1时，f (x )在定义域{x |－1＜x ＜1}上是增函数，由f (x )＞0，得11x x+-＞1，解得0＜x ＜1. 所以x 的取值范围是(0，1). 【点睛】本题考查对数型复合函数单调性、奇偶性以及利用函数性质解不等式，属综合中档题. 24．（1）2,3k b =-=；（2）{}2x x <-. 【分析】（1）根据指数函数的定义列出方程，求解即可； （2）根据指数函数的单调性解不等式即可； 【详解】解：（1）∵函数()()()331xf x k a b a =++->是指数函数∴31,30k b +=-= ∴2,3k b =-= （2）由（1）得()()1xf x aa =>，则函数()f x 在R 上单调递增()()2743f x f x ->-2743x x ∴->-，解得2x <- 即不等式解集为{}2x x <-； 【点睛】本题主要考查了根据函数为指数函数求参数的值以及根据指数函数的单调性解不等式，属于中档题.25．（1）(1,1)-；（2）是奇函数，理由见解析；（3）单调递增，证明见解析. 【分析】（1）由对数有意义的条件列出不等式组1010x x +>⎧⎨->⎩，解之即可；（2）由（1）知，函数()()f x g x -的定义域关于原点对称，再根据函数奇偶性的概念进行判断即可；（3）当2a =时，函数()()f x g x -单调递增.根据用定义证明函数单调性的“五步法”：任取､作差､变形､定号､下结论，即可得证. 【详解】 （1）10x +>，10x ->，11x ∴-<<，∴函数()()f x g x -的定义域为(1,1)-.（2）由（1）知，函数()()f x g x -的定义域关于原点对称，()()log (1)log (1)log (1)log (1)[()()]a a a a f x g x x x x x f x g x ---=-+-+=--+=--，∴函数()()f x g x -是奇函数.（3）当2a =时，函数()()f x g x -单调递增.理由如下： 当(1,1)x ∈-时，1()()log 1a x f x g x x+-=-， 设1211x x -<<<， 则2121211222112121211211111[()()][()()]log log log (?)log 11111aa a ax x x x x x x x f x g x f x g x x x x x x x x x +++-+-----=-==---+-+-，1211x x -<<<，2121x x x x ∴->-+，21122112110x x x x x x x x ∴+-->-+->， ∴21122112111x x x x x x x x +-->-+-，即211221121log 01ax x x x x x x x +-->-+-， 2211()()()()f x g x f x g x ∴->-，故当2a =时，函数()()f x g x -单调递增. 【点睛】本题考查函数的单调性与奇偶性的判断､对数的运算法则，熟练掌握用定义证明函数单调性和奇偶性的方法是解题的关键，考查学生的逻辑推理能力和运算求解能力，属于中档题. 26．（1）110；（2）13lg5lg 222- 【分析】（1）利用指数幂的运算法则即得解； （2）利用对数的运算法则即得解. 【详解】（1）原式1111323334422()12223()33⨯=⨯+⨯+⨯-2108110=+=（2）原式153222124lg lg 2lg(57)273=-+⨯11(5lg 22lg 7)4lg 2(lg5+2lg7)22=--+ 11(5lg 22lg 7)4lg 2(lg5+2lg7)22=--+ 31lg 2lg522=-+【点睛】本题考查了指数与对数运算，考查了学生概念理解，数学运算能力，属于基础题.。
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Paper 11
Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). Mark for a correct result or statement independent of method marks.
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The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.
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ቤተ መጻሕፍቲ ባይዱ
© Cambridge International Examinations 2015
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Mark Scheme Cambridge International AS/A Level – May/June 2015
Syllabus 9709
Paper 11
The following abbreviations may be used in a mark scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no “follow through” from a previous error is allowed) Correct Working Only - often written by a “fortuitous” answer Ignore Subsequent Working Misread Premature Approximation (resulting in basically correct work that is insufficiently accurate) See Other Solution (the candidate makes a better attempt at the same question) Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance)
® IGCSE is the registered trademark of Cambridge International Examinations.
Page 2
Mark Scheme Cambridge International AS/A Level – May/June 2015
Syllabus 9709
Penalties MR–1 A penalty of MR–1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through “ marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B marks in the case of premature approximation. The PA–1 penalty is usually discussed at the meeting.
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When a part of a question has two or more “method” steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.