C4 June 2006 Mark Scheme

8/6/2019 C4 June 2006 Mark Scheme

1/26

June 2006

6666 Core Mathematics C4

Mark Scheme

QuestionNumber Scheme Marks

Differentiates implicitly to include either

dyky

dx or

dy3

dx . (Ignore

=

dy

dx.) M1

1. 6x 4ydx

dy+ 2 3

dx

dy= 0

Correct equation. A1

dy 6x 2

dx 4y 3

+=

+ not necessarily required.

Substituting x= 0 & y= 1 into an

equationinvolving dydx

;

to give 27or 2

7

dM1;A1 cso

At (0, 1),+

= =+

dy 0 2 2

dx 4 3 7

Uses m(T) to correctly find m(N). Canbe ft from their tangent gradient.

A1 oe.

Hence m(N) = 7

2or

27

1

y 1 = m(x 0) withtheir tangent or normal gradient;

or uses y= mx+ 1 with their tangent ornormal gradient ;

M1;

Either N: = 72

y 1 (x 0)

or N: 72

y x 1= +

Correct equation in the form+ + ='ax by c 0 ' ,

where a, b and c are integers.

A1 oecso

N: 7x + 2y 2 = 0 [7]

7 marks

Beware:dy 2

dx 7= does not necessarily imply the award of all the first four marks in this

question.So please ensure that you check candidates initial differentiation before awarding the first A1mark.

Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes thefinal line then they must be awarded A0.


2/26

Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) = , but obtains M0 if

they write y 1 (x 0) = . If they write, however, N: x = 0, then can score M1.

Beware: A candidate finding an m(T) = can obtain A1ft for m(N) = 0, and also obtains M1

if they write y 1 0(x 0) = or y = 1.

Beware: The final cso refers to the whole question.


3/26

QuestionNumber

Scheme Ma

Aliter

1. d x

d y

dx dx6x 4y 2 3 0

dy dy

= + =

Differentiates implicitly to include eitherdx

kxdy

ordx

2dy

. (Ignoredx

dy

=

.)

Correct equation.

M1

Way 2

dx 4y 3

dy 6x 2

+=

+ not necessarily required.

At (0, 1),dx 4 3 7

dy 0 2 2

+= =

+

Substituting x = 0 & y = 1 into anequationinvolving dx

dy;

to give 72

dM1

A1

Hence m(N) = 7

2or

27

1

Uses m(T) or dxdy

to correctly find m(N).

Can be ft using dxdy

1. .A1

Either N: = 72

y 1 (x 0)

or N: 72

y x 1= +

y 1 m(x 0) = with

their tangent, dxdy

or normal gradient;

or uses y mx 1= + with their tangent,dxdy

or normal gradient ;

M1;

N: 7x + 2y 2 = 0

Correct equation in the form+ + ='ax by c 0 ' ,


A1 ocso

7 m


4/26

tionNumber

Scheme arks

liter

1. =2

3y 3x 2x 5 0ay 3

) = + +22 9 3x 5

16 2 2x

( )+ + 23x 49 3

2 16 4x

( ) ( )

+ + +

12 23x 49

2 16

1x 3x 1

2

Differentiates using the chain rule;

Correct expression fordy

dx.

1),

= = =

12dy 1 49 1 4 2

dx 2 16 2 7 7

stituting x = 0 into an equationinvolving dydx

;

to give 27

or 27

o

m(N) = 7

2

Uses m(T) to correctly find m(N).Can be ft from their tangent gradient.

N: = 72

y 1 (x 0)

:27y x 1= +

y 1 m(x 0) = with

their tangent or normal gradient;or uses y mx 1= + with their tangent or normal

gradient

+ 2y 2 = 0orrect equation in the form + + ='ax by c 0 ' ,


ark


5/26

tionNumber

Scheme arks

. (a) +1 A(1 2x) B

nsiders this identityand eithersubstitutes 1

2x = ,

equatescoefficients or

solvessimultaneous

equations

lete

12;= = =3 1

2 21 B B

e x terms; 32

3 2A A= = 3 12 2

A ; B= =

orking seen, but A and B correctly stated award all threemarks. If one of A or B correctly stated give two out of thethree marks available for this part.)

[3]

(b)

+ 1 23 1

2 2(1 2x) (1 2x)

ving powers to top onany one of the two

expressions

2 3( 1)( 2) ( 1)( 2)( 3)

1 ( 1)( 2x); ( 2x) ( 2x) ...2! 3!

+ + + +

1 2x or 1 4x fromeither first or second

expansionsrespectively

2 3( 2)( 3) ( 2)( 3)( 4)( 2)( 2x); ( 2x) ( 2x) ...2! 3!

+ + + +

Ignoring 32

and 12,

any one correct

}.......... expansion.

oth }{.......... correct.

} }{+ + + + + + + + +2 3 2 3121 2x 4x 8x ... 1 4x 12x 32x ...

2 3x ; 0x 4x+ + 2 31 x ; (0x ) 4x + 1

[6]

marks


6/26

tionNumber

Scheme arks

liter

. (b) 2(3x 1)(1 2x) Moving power to top

ay 2

)

2

3

( 2)( 3)1 ( 2)( 2x) ; ( 2x)

2!1

( 2)( 3)( 4)( 2x) ...

3!

+ + +

+

1 4x ;Ignoring (3x 1) , correct

( )........... expansion

+ + + +2 31)(1 4x 12x 32x ...)

2 3 2 312x 36x 1 4x 12x 32x ...+ + Correct expansion

2 3x ; 0x 4x+ + 2 31 x ; (0x ) 4x + 1

liter

. (b) urin expansionay 3

+

1 23 12 2(1 2x) (1 2x)

Bringing bothpowers to top

2 33(1 2x) 2(1 2x) +

Differentiates to give2 3

a(1 2x) b(1 2x)

;2 33(1 2x) 2(1 2x) +

3 412(1 2x) 12(1 2x) +

4 572(1 2x) 96(1 2x) + Correct f (x) and f (x)

1 , f (0) 1 , f (0) 0 and f (0) 24 = = = =

2 3

f(x) 1 x; 0x 4x ...= + + +

2 3

1 x ; (0x ) 4x +

1


7/26

tionNumber

Scheme arks

liter

. (b)1 21

23(2 4x) (1 2x) + g powers to top on anyone of the two

expressionsay 4

1 2 3 2

4 3

( 1)( 2)(2) ( 1)(2) ( 4x); (2) ( 4x)

2!

( 1)( 2)( 3)(2) ( 4x) ...

3!

+ +

+ +

er 12

x or 1 4x from

either first or secondexpansionsrespectively

2 3( 2)( 3) ( 2)( 3)( 4)( 2)( 2x); ( 2x) ( 2x) ...2! 3!

+ + + +

Ignoring 3 and 12,

any one correct}{.......... expansion.

Both }{.......... correct.

}{2 3 2 31 12 2x 2x 4x ... 1 4x 12x 32x ...+ + + + + + + + +

2 3x ; 0x 4x+ + 2 31 x ; (0x ) 4x + 1


8/26

tionNumber

Scheme ark

. (a) Shaded = ( )

2

x

2

03sin dx

( )

=

2x2

12 0

3cos

rating ( )x23sin to give ( )x2

kcos

with k 1.Ignore limits.

( )

= 2

x2 0

6cos ( ) x26cos or ( )12

3 x2

cos .

[ ] [ ]= = + =6( 1) 6(1) 6 6 12 12 o

er of 12 with no working scores M0A0A0.)

(b) e ( )( ) ( )2 2

2 2x x2 2

0 0

3sin dx 9 sin dx

= = Use of 2V y dx= .

Can be implied. Ignore limits.

1 cos2x2 2

2cos2x 1 2sin x gives sin x = =

( ) ( ) 1 cosx2 2x x2 2 2cos x 1 2sin gives sin = =

onsideration of the Half Angle

Formula for ( )2 x2sin or the

Double Angle Formula

for 2sin x

me

2

0

1 cos x9( ) dx

2

=

orrect expression for Volume

Ignore limits and .

( )2

0

9(1 cos x) dx

2

=

( )[ ]

2

0

9x sin x

2

=

tegrating to give ax bsin x ;

Correct integrationk k cos x kx k sin x

1 ;

[ ]

= 9

(2 0) (0 0)2

= =

9(2 )

29 2 or 88.8264

Use of limits to giveeither 9 2 or awrt 88.8

o

Solution must be completely

correct. No flukes allowed.


9/26

rks

tionNumber

Scheme arks

. (a) sint , ( )6y sin t= +

cost , ( )6dy

cos t

dt

= +

pt to differentiate both x and y wrt tto give two terms in cos

Correct dxdt and dydt

n t ,6

=

( )

( )

16 6 2

36 2

cosdy 1awrt 0.58

dx cos 3

+= = = =

s in correct way and substitutesfor t to give any of the four

underlined oe:Ignore the double negative ifcandidate has differentiated

sin cos

n t ,6

= 31x , y

2 2= = he point ( )

312 2, or ( )

12 , awrt 0.87

( )= 3 1 12 23

x

ng an equation of a tangent withtheir point and their tangentgradient or finds c and uses

y (their gradient)x " c "= + .

rect EXACT equation of tangentoe.

( )3 3 31 1

2 2 6 33

c c= + = =

3 3

3 3y x = +

(b) ( )6 6 6t sin t cos cost sin + = + f compound angle formula for sine.

2 2 2 2in t cos t 1 cos t 1 sin t+

sint gives ( )2cos t 1 x= Use of trig identity to find cost in

terms of x or 2cos t in terms of


10/26

x.

3 12 2

y sin t cos t = +

( )3 212 2y x 1 x= + AGSubstitutes for

6 6sin t, cos , cos t and sin to

give y in terms of x.

o

rks


11/26

tionNumber

Scheme arks

liter

. (a) sint , ( )6 6 6y sin t sin t cos cos t sin = + = + (Do not give this for part (b))

ay 2

cost ,6 6

dycos t cos sin t sin

dt =

mpt to differentiate x and y wrt t to

give dxdt

in terms of cos anddy

dtin

the form a cos t b sin t

Correct dxdt

anddy

dt

n t ,6

=

( )6 6 6 6

6

cos cos sin sindy

dx cos

=

3 1 14 4 2

3 3

2 2

1awrt 0.58

3

= = = =

Divides in correct way andsubstitutes for t to give any of

the four underlined oe:

n t ,6

=

31x , y

2 2= = e point ( )312 2, or ( )12 , awrt 0.87

( )= 3 1 12 23

x

inding an equation of a tangentwith their point and their

tangent gradient or finds cand uses


ect EXACT equation of tangentoe.

( )3 3 31 1

2 2 6 33c c= + = =

3 3

3 3y x = +


12/26

tionNumber

Scheme arks

liter

. (a) ( )3 21

2 2x 1 x+

ay 3

( ) ( )1223 1 1 1 x 2x

2 2 2

+

mpt to differentiate two terms usingthe chain rule for the second

term.

Correctdy

dx

( ) ( )1223 1 1 11 (0.5) 2(0.5)

2 2 2 3

+ =

Correct substitution of 12

x =

into a correctdy

dx

n t ,6

=

31x , y

2 2= = he point ( )312 2, or ( )12 , awrt 0.87

( )= 3 1 12 23

x

ng an equation of a tangent withtheir point and their tangentgradient or finds c and uses


rect EXACT equation of tangentoe.

( ) 3 3 31 12 2 6 33 c c= + = =

3 3

3 3y x = +

liter

. (b) t gives ( )3 212 2y sin t 1 sin t= + ubstitutes x sint= into the equation

give in y.

ay 22 2 2 2in t cos t 1 cos t 1 sin t+

( )21 sin t= se of trig identity to deduce that

( )2cos t 1 sin t= .

3 12 2

s y sin t cos t= +

6 6y sin t cos cos t sin = + = ( )6sin t

+ the compound angle formula to

prove y = ( )6sin t+ o

rks


13/26

tionNumber

Scheme arks

. (a) ing i ; 0 6= + 6 = 6 = d

Can be implied6 = and

ing j ; a = 19 4( 6) 5+ =

For inserting their stated intoeither a correct j or k

componentCan be implied.

d

ng k ; b = 1 2( 6) 11 = a 5 and b 11= =

o workingly one of a or b stated correctly gains the first 2marks.

th a and b stated correctly gains 3 marks.

(b) ( ) ( ) ( )6 19 4 1 2+ + + + i j k

ion vector or l1 4 2= = + d i j k

1l OP 0 =

d

uuur

Allow this statement for M1

if OP and d

uuur

are defined asabove.

6 1

19 4 4 0

1 2 2

+

+ =

( )or x 4y 2z 0+ = w either of these two underlinedstatements

4(19 4 ) 2( 1 2 ) 0 + + = Correct equation76 16 2 4 0 + + + + = tempt to solve the equation in

84 0 4 + = = 4 =

( ) ( ) ( )6 4 19 4( 4) 1 2( 4) + + + i j k titutes their into an expression

for OPuuur

2 3 7+ +i j k 2 3 7+ +i j k or P(2, 3, 7)


14/26

tionNumber

Scheme arks

liter

(b) ( ) ( ) ( )6 19 4 1 2+ + + + i j k

ay 2( ) ( ) ( )6 0 19 4 5 1 2 11+ + + + + i j k

ion vector or l1 4 2= = + d i j k

OP AP OP 0 =uuur uuur uuur

this statement for

M1 if AP and OPuuur uuur

are defined asabove.

6 6

24 4 19 4 0

12 2 1 2

+ +

+ + =

derlined statement

)(6 ) (24 4 )(19 4 ) ( 12 2 )( 1 2 ) 0 + + + + + = Correct equation2 2 212 456 96 76 16 12 24 2 4 0+ + + + + + + + + + = ttempt to solve the

equation in

210 504 0 + =

( )0 24 0 6 4 + = = = 4 =

( ) ( ) ( )6 4 19 4( 4) 1 2( 4) + + + i j k titutes their into

an expression

for OPuuur

2 3 7+ +i j k 2 3 7+ +i j k or

P(2, 3, 7)


15/26

tionNumber

Scheme arks

. (c) 2 3 7+ +i j k

0 5 11 +i j k and OB 5 15= + +i j kuuur

( )2 8 4+ i j k , ( )PB 3 12 6= + i j kuuur

( )5 20 10 + i j k

racting vectors to find any two of

APuuur

, PBuuur

or ABuuur

; and both arecorrectly ft using candidates

OAuuur

and OPuuur

found in parts (a)and (b) respectively.

1

( )2 23 33 12 6 PB= + =i j kr uuur

( )5 52 22 8 4 AP= + =i j kr uuur

( )5 53 33 12 6 PB= + =i j kr uuur

( )3 32 22 8 4 AP= + =i j kuuur

( )2 25 55 20 10 AB= + =i j kr uuur

( )3 35 55 20 10 AB= + =i j kuuur

etc

atively candidates could say for example that

( )2 4 2+ i j k ( )PB 3 4 2= + i j kuuur

he points A, P and B are collinear.

23

AP PB=uuur uuur

or 52

AB AP=uuur uuur

or 53

AB PB=uuur uuur

or 32

PB AP=uuur uuur

or 25

AP AB=uuur uuur

or 35

PB AB=uuur uuur

A, P and B are collinearCompletely correct proof.

: PB 2 : 3=uuur

2:3 or 321: or 84 : 189 aef

allow SC 23

liter

. (c)5 6 , 15 19 4 or 1 1 2= + = + =

; 1 = g down any of the three underlined

equations.

ay 21 = for all three equations.

en 1 = , this gives =r 5 15+ +i j k

1 = for all three equations

or 1 = gives =r 5 15+ +i j k

B lies on l1. As stated in the question both Aand P lie on l1. A, P and B are collinear.

Must state B lies on l1

A, P and B are collinear

: PB 2 : 3=uuur

2:3 or aef


16/26

mar


17/26

tionNumber

Scheme arks

. (a)

x 1 1.5 2 2.5 3y 0 0.5 ln 1.5 ln 2 1.5 ln 2.5 2 ln 3

or y 00.2027325541

ln21.374436098

2 ln 3

Either 0.5 ln 1.5 and 1.5 ln 2.5or awrt 0.20 and 1.37

mixture of decimals and lns)

b)(i) ( ) }{1 0 2 ln2 2ln3 + + For structure of trapezium

rule{ }............. ;

3.583518938... 1.791759... = = 1.792 (4sf) 1.792 o

(ii)

( ) }{0.5 ; 0 2 0.5ln1.5 ln2 1.5 ln2.5 2ln3 + + + + Outside brackets

10.5

2

For structure of trapezium

rule{ }............. ;

6.737856242... 1.684464... = awrt 1.684

(c)ncreasing ordinates, the line segments at the topof the trapezia are closer to the curve.

ason or an appropriate diagramelaborating the correct reason.


18/26

tionNumber

Scheme arks

. (d) 2

du 1dx x

x2

n x

x 1 v x

=

=

se of integration by parts

formula in the correctdirection

2 21 xx lnx x dx

x 2

Correct expression

2 xx ln x 1 dx

2 2

ttempt to multiply at least

one term through by 1x

and an attempt to ...

2 2xx ln x x2 4

(+c) integrate;

correct integration

32 2

1

x xx ln x x

2 4

+

( ) ( )3 9 1 12 4 2 4ln3 3 ln1 1 + + stitutes limits of 3 and 1 and

subtracts.

3 3 32 4 4ln3 0+ +

32 ln3= AG 32 ln3 o

liter

. (d) 1)ln x dx x ln x dx ln x dx= ay 2

2 2x x 1dx ln x . dx

2 2 x

=

ct application of by parts

=

2 2x xln x

2 4 (+ c) Correct integration

1dx xln x x. dx

x

=

ct application of by parts

= x ln x x (+ c) Correct integration

) ( ) ( )

9

21 ln x dx ln3 2 3ln3 2 = 3

2 ln3=

AG

stitutes limits of 3 and 1 into

both integrands andsubtracts.


19/26

32ln3 o


20/26

tionNumber

Scheme arks

liter

. (d)) ( )

2

du 1dx x

x 1

2

nx

x 1 v

= =

se of integration by partsformula in the correctdirection

ay 3

) ( )2 2

1 x 1ln x dx

2 2x

Correct expression

)2 21 x 2x 1

lnx dx2 2x

+

Candidate multiplies outnumerator to obtain

three terms

)2

1 1 1lnx x 1 dx

2 2 2x

+

ultiplies at least one term

through by 1x

and then

attempts to ...

)2 21 x 1

lnx x ln x2 4 2

+

(+c)

integrate the result;

correct integration

( )32 2

1

x 1 x 1ln x x lnx

2 4 2

+

( ) ( )9 1 14 2 42ln3 3 ln3 0 1 0 + + stitutes limits of 3 and 1 and

subtracts.

31 12 4 4

2ln3 ln3 1 + + 32ln3= AG 3

2ln3 o

.


21/26

tionNumber

Scheme arks

liter bstitution

. (d) du 1dx xx =

ay 4

)u ue 1 .ue du Correct expression

( )2u uu e e du se of integration by parts

formula in the correctdirection

2u u 2u u1 1e e e e dx2 2

Correct expression

2u u 2u u1 1e e e e2 4

(+c)

Attempt to integrate;

correct integration

ln3

2u u 2u u

ln1

1 1ue ue e e

2 4

+

( ) ( )9 9 12 4 4ln3 3ln3 3 0 0 1 + + stitutes limits of ln3 and ln1

and subtracts.

3 3 12 4 4ln3 1+ + 3

2 ln3= AG32ln3 o

ar


22/26

tionNumber

Scheme ark

. (a) question,dS

8dt

= dS

8dt

=

2 dS 12xdx

= dS

12xdx

=

23dS dS 8 ;

dt dx 12x x = = ( )23k = Candidates

dS dS

dt dx ;

8

12x 1o

(b) 2dV 3xdx = 2dV 3x

dx=

2dV dx 23x . ; 2xdx dt 3x

= =

Candidates

dV dx

dx dt ; x 1

13V , then

13

dV2V

dt= AG Use of

13x V= , to give

13

dV2V

dt=

(c) = 2 dt arates the variables with 13

dV

Vor

13V dV

on one side and2 dt on the other side.

integral signs not necessary.

= dV 2 dt Attempts to integrate and

2t (+c) must see23V and 2t;

orrect equation with/without + c.

2(0) c c 6= + = e of V = 8 and t = 0 in a changed

equation containing c ; c 6= A1

:233

2V 2t 6= +

aving found their c candidate

)23

2 2t 6= + 12 2t 6= + substitutes V 16 2= into anequation involving V, t and c.

1

t = 3. t = 3 o


23/26

ar


24/26

tionNumber

Scheme ark

liter

. (b)1 2

32& S 6x S 6V= = 23S 6V=

ay 2134V

or

13

dV 1V

dS 4=

13

dS4V

dV

= or

13

dV 1V

dS 4=

13

1 13 3

dS dV 1 28. ; 2V

dt dS 4V V

= = =

AG Candidates

dS dV

dt dS ;

132V 1

In ePEN, award Marks for2 in the order they appear on

this mark scheme.

liter

. (c) 1 dt= Separates the variables with

13

dV

2V or13

1V dV

2

oe on one

side and 1 dt on the otherside.

ay 2 integral signs not necessary.1

3 dV 1 dt=

Attempts to integrate and 23V t= (+c) must see

23V and t;


(0) c c 3= + = of V = 8 and t = 0 in a changed

equation containing c ; c 3= A1

:233

4V t 3= +

Having found their ccandidate

)23

2 t 3= + 6 t 3= + substitutes V 16 2= into anequation involving V, t and c.

1

t = 3. t = 3 o

.


25/26

tionNumber

Scheme ark

liter r to way 1.

(b) 2dV 3xdx

= 2dV 3xdx

=

ay 3

2dV dS dx 13x .8. ; 2xdx dt dS 12x

= =

Candidates

dV dS dx

dx dt dS ; x 1

13V , then

13

dV2V

dt= AG Use of

13x V= , to give

13

dV2V

dt=

liter

(c) = 2 dt arates the variables with 13

dV

Vor

13V dV

on one side and2 dt on the other side.

ay 3 integral signs not necessary.

= dV 2 dt Attempts to integrate and

4

3

t (+c) must see

23V and 4

3t;


43(0) c c 4+ =

e of V = 8 and t = 0 in a changedequation containing c ; c 4=

A1

:23 4

3V t 4= +

aving found their c candidate

)23 4

3t 6= + 4

38 t 4= + substitutes V 16 2= into an

equation involving V, t and c.1

t = 3. t = 3 o


26/26

Documents

C4 June 2006 Mark Scheme