GCEExaminations

AdvancedSubsidiary/AdvancedLevel

PureMathematicsModuleP5

PaperE

MARKINGGUIDE

Thisguideisintendedtobeashelpfulaspossibletoteachersbyprovidingconcisesolutionsandindicatinghowmarksshouldbeawarded.Thereareobviouslyalternativemethodsthatwouldalsogainfullmarks.

Methodmarks(M)areawardedforknowingandusingamethod.

Accuracymarks(A)canonlybeawardedwhenacorrectmethodhasbeenused.

(B)marksareindependentofmethodmarks.

WrittenbyRosemarySmith&ShaunArmstrong

SolomonPress

Thesesheetsmaybecopiedforusesolelybythepurchaser’sinstitute.

P5PaperE–MarkingGuide

artanhx=1ln1x=ln3ln1x

=2ln3=ln9 M1A1

2 1x 1x

1x1x

=91+x=99x M1

5

10x=8x=4

A1 (4)

(a) f(x)=2cos2xcosh2x2xcoshxsinhx M1A2or2cos2xcosh2xxsinh2x

(b) forS.P.,f(x)=0

f(0.3)=0.367;f(0.4)=0.131 M1A1

f(x)cont.overinterval,changeofsignrootoff(x)=0S.P. A1 (6)

t=tan(1x)dt=1sec2(1x)=1(1+t2) M1A1

2 dx 2 2 2

2π 3

3 1

0 54cosx

dx=0

1

5

4(1t2)

1t2

2

1t2

dt M1A1

= 3 2

0 5(1t2)4(1t2)

dt M1

= 32

0 9t2

dt A1

=[2arctan(t)]3 A1

3 3 0

1

3

=2[arctan( )arctan0]=2(π0)=πa=1

M1A1 (9)

3 36 9 9

y=acosh(x),dy=sinh(x) B1

1sinh()

2

xa

a dx a

a

A=a

2acosh(x)

dx M1A1

a

a

=a

=a

2acosh2(x)dx M1A1

a

a

a

a

1+cosh(2x)dx M1

=a[x+asinh(2x)]a A1

2 a a

=a[a+asinh2{a+asinh(2)}] M1

2 2

=a[2a+asinh2+asinh2]=a2(sinh2+2) A1 (9)

2 2

P5EMARKSpage2

SolomonPress

(a) s=2,dsd

=2 M1

dx=cosdxds d

x

=dxds

dsd

=2cos M1A1

0dx=0

2cosd M1

x=[2sin]x=2sin A1

0 0

dy=sindyds d

y

=dyds

dsd

=2sin

0dx=0

2sind M1

y=[2cos]y=2cos+2 A1

0 0

sin=x,cos=2y

2 2

4

(b)

sin2+cos2=1x2

(2y)2

+ 4

=1orx2+(y2)2=4 M1A1

y

2

O

x

circle,centre(0,2),radius2 B2 (11)

(a) coshxcoshy+sinhxsinhy

=1(ex+ex)(ey+ey)+1(exex)(eyey) M1

4 4

4

=1(ex+y+exy+eyx+exy+ex+yexyeyx+exy) A1

2

=1(ex+y+e(x+y)) M1

=cosh(x+y) A1

5coshx+4sinhxRcoshxcosh+Rsinhxsinh

5=Rcosh,4=Rsinh M1

R=(5242)=3 M1A1

5

tanh=4

4

M1

1 14 1

=artanh(

5)=

2ln

5=

14

5

2ln9=ln3 M1A1

5coshx+4sinhx=3cosh(x+)min.value=3 A1 (11)

SolomonPress

P5EMARKSpage3

2

(a) u=xn1,u=(n1)xn2;v=xex2,v=1ex2

M1A1

I=[1xn1ex2]1 11(n1)xn2ex2

dx A1

n 2 0 0 2

I=1e01(n1)1

xn2ex2

dx M1

n 2 2 0

In=1e1(n1)In2 A1

2

1

(b) I=

2

xex2

dx=[1ex2]1=1e1

M1A1

1 0

2 0 2 2

I3=1e12I1=1e(1e1)=1

M1A1

2 2 2 2 2 2

I5=1e14I3=1e2(1)=1e1 M1A1 (11)

2 2 2 2 2

(a) (mx+c)2=8x M1

m2x2+2mcx+c2=8x

m2x2+x(2mc8)+c2=0 A1

tangent“b24ac”=0so(2mc8)24m2c2=0 M1(mc4)2m2c2=0

m2c28mc+16m2c2=0 M1

8mc=16mc=2 A1

m

(b) mc=2y=mx+2istangenttox2+y2=2 M1

m

x2+(mx+2)2=2 M1

2

x2+m2x2+4x+4=2

m

m2

x2(1+m2)+4x+(42)=0 A1

m2

“b24ac”=0164(1+m2)(42)=0 M1

m2

2(1+m2)(21)=0

m2

2(21+2m2)=0

m2

m2+12=0 A1

m4+m22=0 M1

(m21)(m2+2)=0

m2=2(nosolns)orm=1 A1

m=1,c=2;m=1,c=2y=x+2andy=x2 M1A1 (14)

Total (75)

P5EMARKSpage4

SolomonPress

PerformanceRecord–P5PaperE

Questionno.

1

2

3

4

5

6

7

8

Total

Topic(s)

eqn.ininv.hyp.fns.

diff.hyp.fns.

integr.

t=tan1/2x

surfacearea

intrinsiccoords.

hyp.idents.

reductionformula

parabola,tangent

Marks

4

6

9

9

11

11

11

14

75

Student