2022年金中劍橋高一期末數(shù)學(xué)滿分沖刺寶典【解析版】

金中劍橋高一期末數(shù)滿分沖刺寶典

考點1:函數(shù)

Question1

f-(Zr+3)2forx>0

(a)Findtherangeoff.[1]

(b)Explainwhyfhasaninverse.[1]

(c)Findf1.[3]

(d)Statethedomainoff-1.[1]

(e)Giventhatg:x?->ln(x+4)forx>0,findtheexactsolutionoffg(x)=49.[3]

(0606_s20_qp_12)

5(a)f>9BlAllowybutnotx

5(b)Itisaone-onefunctionbecauseoftheBl

restricteddomain

5(c)x=(2y+3)2orequivalentMlForacorrectattempttofindthe

inverse

Vx-3MlForcorrectrearrangement

y=--------

,2

「=6-3AlMusthavecorrectnotation

2

5(d)x>9BlFTontheir(a)

5(e)f(ln(x+4))=49MlForcorrectorder

(2In(x+4)+3)2=49MlForcorrectattempttosolve,depon

previousMmark,asfarasx=

ln(x+4)=2

x=e2-4Al

Question2

f(x)=3+eAfbrxeR

g(x)=9.x-5fbrxGR

(a)Findtherangeoffandofg.⑵

(b)Findtheexactsolutionoff_1(x)=g'(x).[3]

(c)Findthesolutionofg2(x)=112.[2]

(0606_s20_qp_13)

1(a)f>3BlAllowybutnotx

geRBlAllowybutnotx

1(b)ln(.r-3)Bl

ln(x-3)=9MlForattempttoequateto9andsolve,

mustgetridofIn

x-3=e9

x=e9+3Al

1(c)9(9x-5)-5=ll2MlForcorrectorderofoperation

x=2Al

Question3

Thefunctionfisdefinedbyf(x)=ln(2x4-1)forx>0.

(a)Sketchthegraphofy=f(x)andhencesketchthegraphofy=f_1(x)ontheaxesbelow.[3]

匕

ox

Thefunctiongisdefinedbyg(x)=(x-4)2+1forxW4.

(b)(i)Findanexpressionforg_1(x)andstateitsdomainandrange.[4]

(ii)Findandsimplifyanexpressionforfg(x).

(iii)Explainwhythefunctiongfdoesnotexist.[1J

(0606_s20_qp_21)

11(a)B3Blforcorrectshapeofforf1

Blfbrsymmetry

Blfordrawnovercorrectdomain

Maximumof2marksifnotfully

correct

H(b)(i)1[±]Vx-l=y-4soiMl

gT[x)=4-\lx-\Al

[Range]g_|<4Bl

[Domain]x>1Bl

1l(b)(ii)ln(2[(x-4)2+l]+l)Ml

ln(2x2-16x+35)Al

H(b)(iii)Validexplanation,e.g.someoftheBl

valuesintherangeoffareoutsidethe

domainofg

Question4

f(x)=3e〃+1forxR

g(x)=x+1fbr.rR

(i)Writedowntherangeoffandofg.R]

(ii)Evaluatefg2(0).

(iii)Ontheaxesbelow,sketchthegraphsofy=f(.r)andy=f-I(x),statingthecoordinatesofthepoints

wherethegraphsmeetthecoordinateaxes.[3]

y

Ox

(0606_wl9_qp_ll)

5。）f>1BlMustbeusingcorrectnotation

geRBlMustbeusingcorrectnotation

5(ii)g(O)=l,g(l)=2MlForattemptatg2andcorrectorder

andattemptatf(2)

f(2)=164.8awrt165Al

5(iii)B3Blfbrcorrectfand(0,4),mustbein

firstandsecondquadrant

Blfbrcorrectf1and(4,0),mustbein

B--ifirstandfourthquadrant

Blfbry=xand/orsymmetryimplied,

by'matchingintercepts\No

intersection.

Question5

(a)Theftmctionfissuchthatf(x)=ln(5x+2).forx>a.whereaisassmallaspossible.

(i)Writedownthevalueofa[1]

(ii)Hencefindtlierangeoff.[1]

(iii)Findf_1(x).statingitsdomain.[3]

(iv)Ontheaxes,sketchthegraplisofy=f(x)andy=f-I(x),statingtheexactvaluesofthe

interceptsofthecurveswithtliecoordinateaxes.[4]

*j

oX

7??—2

(b)Thefimctiongissuchthatg：x?->—4,forx>0.Solvetheequationg(x)=-2.[3]

0606_m22_qp_12

9(a)①-0.4Bl

9(a)(ii)f(x)GRoeBl

9(a)(iii)x=lii(5y+2)oeMlForacoiTectattempttofindthe

inverse

=5y+2oe

「（工）=子AlMustbeinthecollectfbnn

xeRBl

9(a)(iv)4Blfortwocorrectlysliaped

giaphsilltheconectqiiadiaiits

Blforacorrectgraphfbr

y=f(x)withcoiTectintercepts

Blfbraconectgiaphfor

-1

I7S-f*Cr)y=f(x)withconect

intercepts

_______________羋

1unpliea.exactinterceptsand

twopointsofintersection

9(b)MlForacoixectorderofoperations

2

(Y3_4j_4

/

￡\MlDeponpreviousMmarkfbra

conectattemptatasolution.

卜)4「4-2￡

Mustdealwithx2conectlyto

￡obtainthefinalsolution

1jadingtox2=8.

A=64Al

考點2：方程組、二次函數(shù)、二次方程、二次不等式

Question1

Findthevaluesofkforwhichtheliney—kx+3isatangenttothecurvey=2x2+4x+k—1.[5]

(0606_m20_qp_12)

2x2+4x+Zr-l=Ax+32Mlfbrattempttoequatethelineand

curveandsimplifytoa3term

2x2+(4-Jt)x+(^-4)=0quadraticequation=0

Alfbracorrectequation,allow

equivalentform

(4-kf=4x2x(左-4)MlUseofdiscriminantinanyform

r-164+48=02DcpMlonpreviousMmark,fbr

4=12,k=4attempttosolveaquadraticequation

Donotiswink

Alfbrboth

Alternative1

2-+4x+4-1=br+3(2Mlfbrattempttoequatethelineand

curveandsimplify

2x2+(4-^)x+(Jt-4)=0Alfbracorrectequation,allow

equivalentform

左=4x+4MlEquatinggradientsandsubstitutionto

2（沿2+（4/（平卜（*-4）=。obtainaquadraticequationinterms

ofk

A2-16A-+48=02)DepMlonpreviousMmark,fbr

k=\2andk=4attempttosolveaquadraticequation

Donotiswink

Alfbrboth

Alternative2

2x2+4x+k-\=kx+3(2Mlfbrattempttoequatethelineand

curveandsimplify

Alfbracorrectequation,allow

2x?+(4—女)x+(左一4)二0

equivalentform

k=4x+4MlEquatinggradientsandsubstitutionto

2X2-4X=0obtainaquadraticequationinterms

x=0,2ofxandsolutionofthisequationto

obtain2xvalues

k=4x+42)DepMlonpreviousMmark,for

k=\2andk=4substitutionoftheirxvaluestoobtain

Donotiswkvalues

Alfbrboth

Question2

(a)Write2r2+3x-4intheforma(x+6)?+c,wherea,bandcareconstants.[3]

(b)Hencewritedownthecoordinatesofthestationarypointonthecurvey=2X2+3X-4.[2]

(c)Ontheaxesbelow,sketchthegraphofy=|2x2+3x—4|,showingtheexactvaluesofthe

interceptsofthecurvewiththecoordinateaxes.[3]

y

(d)Findthevalueofkforwhich12x2+3x—41=khasexactly3valuesofx.[1]

(0606_s20_qp_13)

4(a)2"1史B3Blfor2

3

l4j8Blfor-

4

Blfor--

8

4(b)B23

Blfbr——orFTontheir-b

㈠制4

41

Blfbr-----orFTontheirc

8

4(c)BlForshapewithmaxin2ndquadrant

4

*Bl-3±741

Forx-intercepts-----------

BlForj^-interceptof4andcusps

4(d)41BlFTontheirc

T

Question3

(a)Write9x2-12r+5intheformp(x—q)2+r,wherep,qandrareconstants.[3]

(b)Hencewritedownthecoordinatesoftheminimumpointofthecurvey=9x2—12x+5.[1]

(0606_s20_qp_21)

2(a)B3Blforeachofp,q,rcorrectin

9口一g1+1occorrectlyformattedexpression;allow

correctequivalentvalues

IfBOthenSC2for+lor

SCIfbrcorrectvaluesbutother

incorrectformat

2(b)BlFTtheir(a)

Question4

Findthevaluesofkforwhichtheliney-kx—1andthecurvey=3.r2+8x+5donotintersect.

[6]

(0606_s20_qp_21)

3f+8x+5=fcr-7Ml

3X2+(8-^)X+12[=0]soiAl

(8-%)2-4(3)(12)Ml

A2-166-80*0Ml

Criticalvalues:Al

-4and20soi

-4<Z20AlAlternativemethod:

Mlfbrk=6x+8oe

Mlfbry=(6x+8)x—7

Mlfbr3X2+8X+5=(6X+8)X-7

Alfbrx=±2

Alfbrk-4,A:=20

Alfor-4<A:<20

Question5

Findthevaluesofkfbrwhichtheliney=x-3intersectsthecurvey=k2x2+5kx+1attwo

distinctpoints.[6]

(0606_s20_qp_22)

3x-3=k2x2+56+1Ml

k2x2+(5Zr-l)x+4=0soiAl

(5*-1)2-4(/)(4)Ml

9公一10左+1*0Ml

Criticalvalues:Al

14

—and1soi

9

Al

k<-Qxk>1

9

Question6

Fiudthevaluesofksuchthattheline/=9kx+1doesnotmeetthecurvey=Zx2+3x(2A+1)+4.

[5]

0606_m22_qp_12

9kx+1=Ax2+3(2^+l)x+4.leadingtoMlForequatingthetwoequations

andattempttoobtaina3tenn

2

kx+x(3-3k)+3[=0]qiiadiaticequationequatedto

zero.

(3-3A-)2-(4X3/T)oeMlDeponpreviousMmarkfor

attempttousethediscriininant

illanyfbnn

3左2-io上+3oeMlDeponpreviousMmarkfbr

simplificationtoa3tenn

qiiadiaticexpressionintennsof

k

AlForboth

Criticalvalues3aud-

3

AlMaikthefilialanswer

-4<^<3

考點3：器和根式

Question1

DONOTUSEACALCULATORINTHISQUESTION.

Findthepositivesolutionoftheequation（5+4/7*+（4—2/7）x—1=0,givingyouranswerin

theforma+b/7,whereaandbarefractionsintheirsimplestform.[5]

（0606_s20_qp_ll）

-（4-2>/7）+^（4-2X/7）2-4（5+4>/7）（-1）MlForcorrectuseofquadraticfbnnula,

allowinclusionof±untilfinalanswer

2（5+4⑺

-（4-2V7）+^16+28-16A/7+20+16V7MlForattempttosimplifydiscriminant,

mustseeattemptatexpansionand

2（5+4⑺

subsequentsimplification

-（4-2>/7）+8

2（5+4"）

4+2近2+77AlForeither

、-2（5+4⑺、-（5+4"）

2+>/75-46MlForattempttorationalise,mustsee

attemptatexpansionandsubsequent

5+45-4V7

、一（何simplification

10+5V7-8V7-28

x=--------------------------

25-112

6不Al

x=—+—

2929

Question2

DONOTUSEACALCULATORINTHISQUESTION.

Thepoint(1—6p)liesonthecurvey=10+：".Findtheexactvalueofp,simplifyingyour

answer.x[5]

(0606_s20_qp_22)

Blorrationalises

Squares:(1-V5)=1-石-石+5

10+2石(1+V5)2

111'X1

(1一扃(1+疝

Rationalises,e.g.Blorsquares

10+2石6+2石(1+⑹2=1+V?+石+5

6-2756+20

Multipliesout,e.g.MlMultipliesout

60+206+12b+4(5)10+2石6+2^5_

36-20L-?"+而]

60+20后+12逐+4(5)

(l-5)2

5+2石A2Alfbrk+2Mor5+k也

Question3

DONOTUSEACALCULATORINTHISQUESTION.

/\C-ir，121

(a)Simplify-^=-.[2]

(b)Simplify1?一givingyouranswerasafractionwithanintegerdenominator.[4]

71+/33+2/5

(0606_s20_qp_23)

5(a)V128_V64x2Ml

V72-736x2

.,.質(zhì),[16

orsimplifriest0

,.4Al

correctcompletionto—

5(b)3+25/3-5/3(1+5/3)Ml

(1+73)(3+25/3)

百Ml

3+2百+3G+6

石9-5陋Ml

9+5為9-5百

973-15..Al

---------orequivalentt

Alternativemethod

Mlfbr

1-5/35/3(3-25/3)

(1+⑸(1-⑸(3+2⑸(3-2⑸

rI-A/J36-6

Mlfbr--------------------

1-39-12

Mlfbrwritingwithacommon

denominator

9>/3-15...

At1lfor---------orequivalent

Question4

.r\rSY—I

Giventhat7x49=1and5x1253=25Jcalculatethevalueofxandofj^.[5]

(0606_wl9_qp_ll)

XMlEorexpressingthetermsonthelefthand

Either7rx72ror49%x49vor55vx52vsideofeitheroneofthe2equationsin

5xtermsofpowersof7,49,5or25

or25yx25y

XAl

Tx72V=7°or492x49)'=49°

Al

5"X5"'=5-2or252x25,=25T

leadingtox+2v=0MlForattempttosolvetwolinearequations,

and5x+2y=-2withintegercoefficientsandconstants,in

tennsofxandy

11Al

x=—一,y=-

24

Question5

Solvetheequation(3—573)x2+(273+5)x-1=0,givingyoursolutionsiuthefbnna+b\/3,

whereaandbarerationalnumbers.[6]

0606_m22_qp_12

x=MlFortheuseofthequadratic

-(2^+5)±^(2V3+5)2-4(3-5>/3)(-l)fbnnula

2(3-5@

X=MlForexpansionofthesquare

-(2C+5)±J12+2(A/J+25+12-20Groot,mustseeatleast4tenns

2（3-5百）

-12-202-2百AlForboth

x=----------oe,x=-----------oe

2(3-5@2(3-5碼

-12-2百3+56MlForattempttorationaliseatleast

.v=.oeoneoftheirsolutions(mustbe

2(3-5^)3+5V3

similarstnictiue)

2-25/33+5>/3Sufficientdetailmustbeseen,at

orx=xoeleast3tennsinthenumerator

2(3-56)3+5。

withanattempttosimplify

1百AlMusthavesufficientdetail

---r----

22shown

AlMusthavesufficientdetail

IT-33shown

Question6

-2/—

(a)Giventhat-----------r=paqhrc,findthevalueofeachoftheconstantsa,bandc.[3]

￡2

(b)Solvetheequation3x^-8x7+5=0.[4]

0606_w21_qp_ll

4(a)7Bl

a=—

2

b=lBl

1Bl

c=—

6

4(b)2V2\22

3./-5苫-1=0Mlforrecognitionofaquadraticinx5

\?\?DepMlforsolutionandaconectattempt

togetatleastonesolutionfbrx

3.59Al

1