C4 MS Jun 2006

GCE

Edexcel GCE

Core Mathematics C4 (6666)

June 2006

Core Mathematics C4 (6666)

Edexcel GCE

Mark Scheme (Final)

June 2006 6666 Pure Mathematics C4 Mark Scheme Question Number

1.

Scheme

⎧ dy ⎫ =⎬ ⎨ ⎩ dx ⎭

6x − 4y

dy dy +2−3 =0 dx dx

Marks

Differentiates implicitly to include either M1 ⎛ dy ⎞ dy dy = ⎟ .) or ±3 . (Ignore ⎜ ±ky dx dx ⎝ dx ⎠ A1 Correct equation.

⎧ dy 6x + 2 ⎫ = ⎨ ⎬ 4y + 3 ⎭ ⎩ dx

not necessarily required.

dy 0 + 2 2 At (0, 1), = = dx 4 + 3 7

Hence m(N) = −

−1 7 or 2 2 7

Either N: y − 1 = − 72 (x − 0) or N: y = − x + 1 7 2

N: 7x + 2y – 2 = 0

Substituting x = 0 & y = 1 into an equation ; dM1; involving dy dx to give 72 or −−72 A1 cso Uses m(T) to ‘correctly’ find m(N). Can be ft from “their tangent gradient”.

A1

oe.

y − 1 = m(x − 0) with ‘their tangent or normal gradient’; M1; or uses y = mx + 1 with ‘their tangent or normal gradient’ ; Correct equation in the form ' ax + by + c = 0 ' , A1 oe cso where a, b and c are integers.

[7] 7 marks

Beware:

dy 2 = does not necessarily imply the award of all the first four marks in this question. dx 7

So please ensure that you check candidates’ initial differentiation before awarding the first A1 mark.

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

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.

2 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme

Marks

Aliter Differentiates implicitly to include either

1.

⎧⎪ d x ⎫⎪ =⎬ ⎨ ⎩⎪ d y ⎪⎭

dx dx 6x − 4y + 2 −3=0 dy dy

±kx

M1 ⎛ dx ⎞ dx dx or ±2 . (Ignore ⎜ = ⎟ .) dy dy ⎝ dy ⎠ A1 Correct equation.

Way 2 ⎧ dx 4y + 3 ⎫ = ⎨ ⎬ 6x + 2 ⎭ ⎩ dy

not necessarily required.

dx 4 + 3 7 At (0, 1), = = dy 0 + 2 2

Hence m(N) = −

−1 7 or 2 2 7

Either N: y − 1 = − 72 (x − 0) or N: y = − 72 x + 1

N: 7x + 2y – 2 = 0

Substituting x = 0 & y = 1 into an equation involving dx ; dM1; dy to give 72 A1 cso Uses m(T) or

dx dy

to ‘correctly’ find m(N).

Can be ft using “ −1 . dx ”. dy

A1

oe.

y − 1 = m(x − 0) with ‘their tangent,

dx dy

or normal gradient’;

or uses y = mx + 1 with ‘their tangent, dx dy

M1;

or normal gradient’ ;

Correct equation in the form ' ax + by + c = 0 ' , A1 oe cso where a, b and c are integers.

7 marks

3 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Aliter 1.

Way 3

Scheme

Marks

2y 2 + 3y − 3x 2 − 2x − 5 = 0

( y + 34 ) y=

(

2

− 169 =

3 x2 2

dy 1 = dx 2

(

+x+

3x 2 2

+x +

3x 2 2

49 16

+x+

)− 49 16

)

5 2

3 4

− 21

( 3x + 1)

At (0, 1), dy 1 ⎛ 49 ⎞ = dx 2 ⎜⎝ 16 ⎟⎠

− 21

=

Hence m(N) = −

1⎛ 4⎞ 2 = ⎜ ⎟ 2⎝7⎠ 7

7 2

Either N: y − 1 = − 72 (x − 0) or N: y = − 72 x + 1

N: 7x + 2y – 2 = 0

Differentiates using the chain rule; M1; Correct expression for

dy . dx A1 oe

Substituting x = 0 into an equation involving dy ; dM1 dx −2 2 A1 cso to give or 7

−7

Uses m(T) to ‘correctly’ find m(N).

Can be ft from “their tangent gradient”.

A1

y − 1 = m(x − 0) with ‘their tangent or normal gradient’; M1 or uses y = mx + 1 with ‘their tangent or normal gradient’ Correct equation in the form ' ax + by + c = 0 ' , A1 oe where a, b and c are integers.

[7] 7 marks

4 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

2. (a)

Scheme

Marks Considers this identity complete and either substitutes x = 21 , equates M1 coefficients or solves simultaneous equations

3x − 1 ≡ A(1 − 2x) + B

Let x = 21 ;

3 2

−1 = B

⇒ B=

1 2

Equate x terms; 3 = − 2A ⇒ A = − 32

A = − 32 ; B =

1 2

A1;A1

(No working seen, but A and B correctly stated ⇒ award all three marks. If one of A or B correctly stated give two out of the three marks available for this part.)

(b)

[3]

Moving powers to top on any one of the two M1 expressions

f(x) = − 32 (1 − 2x)−1 + 21 (1 − 2x)−2 ⎧ ⎫ ( −1)( −2) ( −1)( −2)( −3) = − 32 ⎨1 + ( −1)( −2x); + ( −2x)2 + ( −2x)3 + ...⎬ 2!

3! ⎭ ⎩

Either 1 ± 2x or 1 ± 4x from either first or dM1; second expansions respectively

⎧ ⎫ ( −2)( −3) ( −2)( −3)( −4) + 21 ⎨1 + ( −2)( −2x); + ( −2x)2 + ( −2x)3 + ...⎬ 2!

3! ⎭ ⎩

Ignoring − 32 and 21 , any one correct {..........} expansion. A1

Both {..........} correct. A1

{

}

= − 32 1 + 2x + 4x 2 + 8x 3 + ... +

1 2

{1 + 4x + 12x

2

}

+ 32x 3 + ...

= −1 − x ; + 0x 2 + 4x 3

−1 − x ; (0x 2 ) + 4x 3

A1; A1 [6] 9 marks

Beware: In part (a) take care to spot that A = − 32 and B =

1 2

are the right way around.

Beware: In ePEN, make sure you aware the marks correctly in part (a). The first A1 is for A = − 32 and the

second A1 is for B =

1 2

.

Beware: If a candidate uses a method of long division please escalate this to you team leader.

5 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter 2. (b) Way 2

Scheme f(x) = (3x − 1)(1 − 2x)−2

( −2)( −3) ⎛ ⎞ ( −2x)2 + ⎜ 1 + ( −2)( −2x) ; + ⎟ 2! ⎟ = ( 3x − 1) × ⎜ ( −2)( −3)( −4) ⎜ 3 ( −2x) + ... ⎟⎟ ⎜ 3! ⎝ ⎠

Marks Moving power to top M1

1 ± 4x ; dM1; Ignoring (3x − 1) , correct (...........) expansion A1

= (3x − 1)(1 + 4x + 12x 2 + 32x 3 + ...)

= 3x + 12x 2 + 36x3 − 1 − 4x − 12x 2 − 32x3 + ... = −1 − x ; + 0x 2 + 4x 3

Correct expansion −1 − x ; (0x 2 ) + 4x 3

A1 A1; A1 [6]

Aliter 2. (b) Way 3

Maclaurin expansion f(x) = − 32 (1 − 2x)−1 + 21 (1 − 2x)−2

f ′(x) = − 3(1 − 2x)−2 + 2(1 − 2x)−3

Bringing both M1 powers to top Differentiates to give a(1 − 2x) −2 ± b(1 − 2x)−3 ; M1; A1 oe −3(1 − 2x)−2 + 2(1 − 2x)−3

f ′′(x) = − 12(1 − 2x)−3 + 12(1 − 2x)−4 f ′′′(x) = − 72(1 − 2x)−4 + 96(1 − 2x)−5

Correct f ′′(x) and f ′′′(x)

A1

∴ f(0) = − 1 , f ′(0) = − 1 , f ′′(0) = 0 and f ′′′(0) = 24 gives f(x) = − 1 − x; + 0x 2 + 4x 3 + ...

−1 − x ; (0x 2 ) + 4x 3

A1; A1 [6]

6 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter 2. (b)

Scheme

Marks

f(x) = − 3(2 − 4x)−1 + 21 (1 − 2x)−2

Moving powers to top on any one of the two M1 expressions

⎧ −1 ⎫ ( −1)( −2) −3 −2 (2) ( −4x)2 ⎪ ⎪(2) + ( −1)(2) ( −4x); + ⎪ ⎪ 2! = −3 ⎨ ⎬ ( −1)( −2)( −3) −4 3 ⎪ ⎪ (2) ( −4x) + ... + ⎪⎩ ⎪⎭ 3!

Either 21 ± x or 1 ± 4x from either first or dM1; second expansions respectively

Way 4

⎧ ⎫ ( −2)( −3) ( −2)( −3)( −4) + 21 ⎨1 + ( −2)( −2x); + ( −2x)2 + ( −2x)3 + ...⎬ 2!

3! ⎭ ⎩

Ignoring −3 and 21 , any one correct A1 .......... } expansion. {

Both {..........} correct. A1 = −3

{

1 2

}

+ x + 2x 2 + 4x 3 + ... +

1 2

{1 + 4x + 12x

2

}

+ 32x 3 + ...

= −1 − x ; + 0x 2 + 4x 3

−1 − x ; (0x 2 ) + 4x 3

A1; A1 [6]

7 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme 2π

3. (a)

Area Shaded =

∫ 3 sin (

x 2

Marks

) dx

0

Integrating 3 sin ( 2x ) to give

⎡ −3 cos ( 2x ) ⎤ =⎢ ⎥ 1 2 ⎣ ⎦0

2π

= ⎡⎣ − 6 cos ( 2x ) ⎤⎦

k cos ( 2x ) with k ≠ 1 . M1 Ignore limits.

−6 cos ( 2x ) or

2π 0

−3 1 2

cos ( 2x )

= [ −6( −1)] − [ −6(1)] = 6 + 6 = 12

A1 oe.

12 A1 cao

[3] (Answer of 12 with no working scores M0A0A0.)

2π

(b)

Volume = π

∫ (3 sin ( )) x 2

∫ 0

⎡NB : cos 2x = ±1 ± 2 sin2 x gives sin2 x = 1 − cos2x ⎤ 2 ⎣ ⎦ 1 − cos x⎤ 2 2 ⎡NB : cos x = ±1 ± 2 sin ( x ) gives sin ( x ) = 2 2 2 ⎣ ⎦

∴ Volume = 9( π)

=

=

9 ( π) 2

9 ( π) 2

⎛ 1 − cos x ⎞ ⎟ dx 2 ⎠

∫ ⎜⎝ 0

Use of V = π y 2 dx .

dx = 9π sin2 ( 2x ) dx

0

2π

∫

2π

2

M1

Can be implied. Ignore limits.

Consideration of the Half Angle Formula for sin2 ( 2x ) or the Double Angle Formula for sin2 x

M1 ∗

Correct expression for Volume A1 Ignore limits and π .

2π

∫ (1 − cos x) dx 0

[ x − sin x ] 0

2π

=

9π [(2π − 0) − (0 − 0)] 2

=

9π (2π) = 9 π2 or 88.8264…

2

Integrating to give ±ax ± b sin x ; depM1 ∗ ; Correct integration k − k cos x → kx − k sin x A1

Use of limits to give A1 cso either 9 π2 or awrt 88.8 Solution must be completely [6] correct. No flukes allowed.

9 marks

8 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question 3 Note: π is not needed for the middle four marks of question 3(b).

Beware: Owing to the symmetry of the curve between x = 0 and x = 2π candidates can find: π

•

∫

Area = 2 3 sin ( 2x ) dx

in part (a).

0

π

•

Volume = 2 π

∫ (3 sin (

x 2

))

2

dx

0

Beware: If a candidate gives the correct answer to part (b) with no working please escalate this response up to your team leader.

9 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme

y = sin ( t +

x = sin t ,

4. (a)

dx = cos t , dt

T: y −

or

3 2

=

3 2

π , 6

=

1 3

1 3

3 3

1 2 3 2

x=

=

π 6

1 3

Attempt to differentiate both x and M1 y wrt t to give two terms in cos and dy Correct dx A1 dt dt

)

= awrt 0.58

x+

3 1 , y= 2 2

The point

(

1 2

,

3 2

) or (

1 2

, awrt 0.87 )

B1

Finding an equation of a tangent with their point and their tangent gradient or finds c and uses dM1 y = (their gradient)x + " c " .

Correct EXACT equation of A1 oe tangent oe.

⇒ c= 3 3

Divides in correct way and substitutes for t to give any of the four underlined oe: A1 Ignore the double negative if candidate has differentiated

sin → − cos

( x − 21 )

( 21 ) + c

or T: ⎡ y = ⎣

)

dy = cos ( t + dt

π When t = , 6 π cos ( 6 + 6π ) = dy = dx cos ( 6π )

When t =

π 6

Marks

3 2

−

3 6

=

3 3

⎤ ⎦

[6] (b)

y = sin ( t +

π 6

) = sin t cos 6π

+ cos t sin 6π

Use of compound angle formula M1 for sine.

Nb : sin2 t + cos2 t ≡ 1 ⇒ cos2 t ≡ 1 − sin2 t

(1 − x )

∴ x = sin t gives cos t =

gives

∴y =

3 2

sin t + 21 cos t

y=

3 2

x+

1 2

Use of trig identity to find cos t in M1 terms of x or cos2 t in terms of x.

2

(1 − x ) 2

AG

Substitutes for sin t , cos 6π , cos t and sin 6π to A1 cso give y in terms of x.

[3] 9 marks 10 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter 4. (a) Way 2

Scheme

y = sin ( t +

x = sin t ,

π 6

Marks

) = sin t cos 6π

+ cos t sin 6π

(Do not give this for part (b)) Attempt to differentiate x and y wrt t to give dx in terms of cos dt and dy in the dt form ±a cos t ± b sin t

dx = cos t , dt

When t =

dy = cos t cos 6π − sin t sin 6π dt

π dy cos 6π cos 6π − sin 6π sin 6π = , cos ( 6π ) 6 dx

=

When t =

T: y −

or

3 2

=

3 2

1 3

or T: ⎡ y = ⎣

π , 6

=

1 3

3 4

1 4

3 2

x=

=

1 2 3 2

1

=

3

= awrt 0.58

x+

dy dt

(

1 2

,

3 2

)

or ( 21 , awrt 0.87 )

A1

B1

Finding an equation of a tangent with their point and their tangent gradient or finds c and uses dM1 y = (their gradient)x + " c " .

Correct EXACT equation of A1 oe tangent oe.

1 2

3 3

and

Divides in correct way and substitutes for t to give any of the A1 four underlined oe:

3 1 , y= 2 2

⇒ c=

dx dt

The point

(x − )

( 21 ) + c 3 3

−

Correct

M1

3 2

−

3 6

=

3 3

⎤ ⎦

[6]

11 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter

Scheme

y=

4. (a)

x+

3 2

1 2

Marks

(1 − x ) 2

Way 3

dy 3 ⎛ 1⎞⎛ 1⎞ 1 − x2 = + dx 2 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

(

) ( −2x )

dy 3 ⎛ 1⎞⎛ 1⎞ = + ⎜ ⎟ ⎜ ⎟ 1 − (0.5)2 dx 2 ⎝ 2 ⎠⎝ 2 ⎠

(

When t =

T: y −

or

3 2

3 2

=

1 3

or T: ⎡ y = ⎣

π , 6

=

1 3

x=

) ( −2(0.5)) =

x+

Correct substitution of x =

1

− 21

3

3 1 , y= 2 2

into a correct

The point

(

1 2

,

3 2

) or (

1 2

1 2

dy dx

, awrt 0.87 )

A1

B1

Finding an equation of a tangent with their point and their tangent gradient or finds c and uses dM1 y = (their gradient)x + " c " .

Correct EXACT equation of A1 oe tangent oe.

( x − 21 )

( 21 ) + c 3 3

Attempt to differentiate two terms using the chain rule for the second M1 term.

A1 Correct dy dx

− 21

⇒ c=

3 2

−

3 6

=

3 3

⎤ ⎦

3 3

[6] Aliter 4. (b)

x = sin t gives y =

3 2

sin t +

1 2

Substitutes x = sin t into the M1 equation give in y.

(1 − sin t ) 2

Way 2 Nb : sin2 t + cos2 t ≡ 1 ⇒ cos2 t ≡ 1 − sin2 t

cos t =

(

gives y =

1 − sin2 t

3 2

Use of trig identity to deduce that

)

cos t =

(1 − sin t ) .

2

M1

sin t + 21 cos t

Hence y = sin t cos 6π + cos t sin 6π = sin ( t +

π 6

)

Using the compound angle formula to prove y = sin ( t + 6π ) A1 cso

[3] 9 marks

12 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number 5. (a)

Scheme Equating i ;

0=6+λ

Marks λ = −6

⇒ λ = −6

B1 ⇒ d

Can be implied

Using λ = − 6 and

equating j ;

a = 19 + 4( −6) = − 5

equating k ;

b = −1 − 2( −6) = 11

For inserting their stated λ into either a correct j or k component M1 ⇒ d Can be implied. a = −5 and b = 11 A1

[3] With no working… … only one of a or b stated correctly gains the first 2 marks. … both a and b stated correctly gains 3 marks. (b)

uuur OP = ( 6 + λ ) i + (19 + 4λ ) j + ( −1 − 2λ ) k direction vector or l1 = d = i + 4 j − 2k uuur uuur OP ⊥ l1 ⇒ OP • d = 0

⎛ 6+λ ⎞ ⎜ ⎟ ie. ⎜ 19 + 4λ ⎟ ⎜ −1 − 2λ ⎟ ⎝ ⎠

⎛ 1⎞ ⎜ ⎟ •⎜ 4 ⎟ = 0 ⎜ −2 ⎟ ⎝ ⎠

Allow this statement for M1 uuur if OP and d are defined as above.

( or x + 4y − 2z = 0 )

Allow either of these two M1 underlined statements

∴ 6 + λ + 4(19 + 4λ ) − 2( −1 − 2λ ) = 0

6 + λ + 76 + 16λ + 2 + 4λ = 0 21λ + 84 = 0

Correct equation A1 oe Attempt to solve the equation in λ

⇒ λ = −4

uuur OP = ( 6 − 4 ) i + (19 + 4( −4)) j + ( −1 − 2( −4)) k uuur OP = 2 i + 3 j + 7 k

λ = −4

dM1 A1

Substitutes their λ into an uuur M1 expression for OP 2 i + 3 j + 7 k or P(2, 3, 7)

A1

[6] uuur Note: A similar method may be used by using OP = ( 0 + λ ) i + ( −5 + 4λ ) j + (11 − 2λ ) k and d = i + 4 j − 2k uuur OP • d = 0 yields 6 + λ + 4( −5 + 4λ ) − 2(11 − 2λ ) = 0 This simplifies to 21λ − 42 = 0 ⇒ λ = 2 . uuur OP = ( 0 + 2 ) i + ( −5 + 4(2)) j + (11 − 2(2) ) k uuur OP = 2 i + 3 j + 7 k 13 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter (b) Way 2

Scheme

Marks

uuur OP = ( 6 + λ ) i + (19 + 4λ ) j + ( −1 − 2λ ) k uuur AP = ( 6 + λ − 0 ) i + (19 + 4λ + 5 ) j + ( −1 − 2λ − 11) k direction vector or l1 = d = i + 4 j − 2k Allow this uuurstatement uuur for M1 if AP and OP are defined as above.

uuur uuur uuur uuur AP ⊥ OP ⇒ AP • OP = 0

⎛ 6+λ ⎞ ⎜ ⎟ ie. ⎜ 24 + 4λ ⎟ ⎜ −12 − 2λ ⎟ ⎝ ⎠

⎛ 6+λ ⎞ ⎜ ⎟ • ⎜ 19 + 4λ ⎟ = 0 ⎜ −1 − 2λ ⎟ ⎝ ⎠

underlined statement M1

∴ ( 6 + λ )(6 + λ ) + (24 + 4λ )(19 + 4λ ) + ( −12 − 2λ )( −1 − 2λ ) = 0

36 + 12λ + λ 2 + 456 + 96λ + 76λ + 16λ 2 + 12 + 24λ + 2λ + 4λ 2 = 0

Correct equation A1 oe Attempt to solve the dM1 equation in λ

21λ 2 + 210λ + 504 = 0 λ 2 + 10λ + 24 = 0

⇒ ( λ = −6 ) λ = − 4

λ = −4

A1

Substitutes their λ into an expression for M1 uuur OP

uuur OP = ( 6 − 4 ) i + (19 + 4( −4) ) j + ( −1 − 2( −4) ) k uuur OP = 2 i + 3 j + 7 k

2 i + 3 j + 7 k or A1 P(2, 3, 7)

[6] uuur Note: A similar method to way 2 may be used by using OP = ( 5 + λ ) i + (15 + 4λ ) j + (1 − 2λ ) k uuur and AP = ( 5 + λ − 0 ) i + (15 + 4λ + 5 ) j + (1 − 2λ − 11) k uuur uuur AP • OP = 0 yields ( 5 + λ )(5 + λ ) + (20 + 4λ )(15 + 4λ ) + ( −10 − 2λ )(1 − 2λ ) = 0 This simplifies to 21λ 2 + 168λ + 315 = 0 . λ 2 + 8λ + 15 = 0 uuur OP = ( 5 − 3 ) i + (15 + 4( −3)) j + (1 − 2( −3)) k uuur OP = 2 i + 3 j + 7 k

⇒ ( λ = −5 ) λ = − 3

14 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number 5. (c)

Scheme

Marks

uuur OP = 2 i + 3 j + 7 k uuur uuur OA = 0 i − 5 j + 11k and OB = 5 i + 15 j + k

uuur uuur AP = ± ( 2 i + 8 j − 4 k ) , PB = ± ( 3 i + 12 j − 6k ) uuur AB = ± ( 5 i + 20 j − 10 k ) uuur As AP = uuur or AB = uuur or AB = uuur or PB = uuur or AP = uuur or PB =

(3 i 5 2i 2( 5 3i 3( 3 2i 2( 2 5i 5( 3 5i 5( 2 3

Subtracting vectors uuurto find any two uuur uuu r of AP , PB or AB ; and both are M1; uuur correctly uuur ft using candidate’s A1 OA and OP found in parts (a) and (b) respectively.

uuur PB uuur + 8 j − 4k ) = 52 AP uuur + 12 j − 6k ) = 53 PB uuur + 8 j − 4k ) = 32 AP uuur + 20 j − 10k ) = 52 AB uuur + 20 j − 10 k ) = 35 AB etc… + 12 j − 6k ) =

uuur AP = uuur or AB = uuur or AB = uuur or PB = uuur or AP = uuur or PB =

2 3

2 3 5 2 5 3 3 2 2 5 3 5

±

uuur PB uuur AP uuur PB uuur AP uuur AB uuur AB

alternatively candidates could say for example that uuur uuur AP = 2 ( i + 4 j − 2k ) PB = 3 ( i + 4 j − 2k ) A, P and B are collinear A1 Completely correct proof.

then the points A, P and B are collinear. uuur uuur ∴ AP : PB = 2 : 3

2:3 or 1: 32 or 84 : 189 aef B1 oe allow SC 32 [4]

Aliter

5. (c)

Way 2

At B; 5 = 6 + λ , 15 = 19 + 4λ or 1 = −1 − 2λ or at B; λ = − 1 gives λ = − 1 for all three equations. or when λ = − 1 , this gives r = 5 i + 15 j + k

Writing down any of the three M1 underlined equations. λ = − 1 for all three equations A1 or λ = − 1 gives r = 5 i + 15 j + k

Hence B lies on l1. As stated in the question both A and P lie on l1. ∴ A, P and B are collinear. uuur uuur ∴ AP : PB = 2 : 3

Must state B lies on l1 ⇒ A1 A, P and B are collinear 2:3 or aef B1 oe

[4] 13 marks Beware of candidates who will try to fudge that one vector is multiple of another for the final A mark in part (c).

15 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme

Marks

6. (a) x y or y

1 0 0

1.5 0.5 ln 1.5 0.2027325541…

2 ln 2 ln2

2.5 1.5 ln 2.5 1.374436098…

3 2 ln 3 2 ln 3

Either 0.5 ln 1.5 and 1.5 ln 2.5 or awrt 0.20 and 1.37 B1 (or mixture of decimals and ln’s) [1]

{

For structure of trapezium rule {.............} ; M1;

1 I1 ≈ × 1× 0 + 2 ( ln 2 ) + 2ln 3} 2

(b)(i)

=

1 × 3.583518938... = 1.791759... = 1.792 (4sf) 2

(ii) I2 ≈

=

Outside brackets

{

1 × 0.5 ; × 0 + 2 ( 0.5 ln1.5 + ln 2 + 1.5 ln 2.5 ) + 2ln 3} 2

1.792

A1 cao

1 × 0.5 2

B1;

For structure of trapezium M1 rule {.............} ;

1 × 6.737856242... = 1.684464...

4

awrt 1.684 A1

[5] With increasing ordinates, the line segments at the top of the trapezia are closer to the curve.

(c)

Reason or an appropriate diagram B1 elaborating the correct reason.

[1] Beware: In part (b) candidate can add up the individual trapezia:

(b)(i) I1 ≈ (ii)

1 2

( 0 + ln 2 ) + 21 (ln 2 + ln 3 )

I2 ≈ 21 . 21 ( 0 + 0.5ln1.5 ) + 21 . 21 ( 0.5ln1.5 + ln 2 ) + 21 . 21 ( ln 2 + 1.5 ln 2.5 ) + 21 . 21 (1.5ln 2.5 + 2ln 3 )

16 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

6. (d)

Scheme = ⎧⎪u = ln x ⇒ du dx ⎨ dv ⎪⎩ dx = x − 1 ⇒ v =

x 2

⎛ x2 ⎞ I=⎜ − x ⎟ ln x − ⎝ 2 ⎠

⎞ 1 ⎛ x2 − x ⎟ dx ⎜ x⎝ 2 ⎠

⎛ x2 ⎞ =⎜ − x ⎟ ln x − ⎝ 2 ⎠

∫

⎫⎪ ⎬ − x ⎪⎭

1 x 2

∫

Marks Use of ‘integration by parts’ formula in the M1 correct direction Correct expression A1 An attempt to multiply at least one term through by 1 and an attempt to ... x

⎛x ⎞ ⎜ 2 − 1⎟ dx ⎝ ⎠

⎛ x2 ⎞ ⎛ x2 ⎞ =⎜ − x ⎟ ln x − ⎜ − x ⎟ (+c) ⎝ 2 ⎠ ⎝ 4 ⎠

… integrate; M1; correct integration A1

3

⎡ ⎛ x2 ⎤ ⎞ x2 − x ⎟ ln x − + x⎥ ∴I = ⎢ ⎜ 4 ⎠ ⎣⎝ 2 ⎦1

= ( 32 ln3 − = 32 ln 3 +

9 4

3 4

+ 3 ) − ( − 21 ln1 − +0−

1 4

+ 1)

= 32 ln3 AG

3 4

Substitutes limits of 3 and ddM1 1 and subtracts.

3 2

ln 3

A1 cso

[6] Aliter

6. (d)

∫ (x − 1)ln x dx = ∫ x ln x dx − ∫ ln x dx

Way 2

∫

x ln x dx =

=

∫

ln x dx

x2 ln x − 2

∫

x2 ⎛ 1 ⎞ . dx 2 ⎜⎝ x ⎟⎠

x2 x2 ln x − (+ c) 2 4

Correct integration A1

⎛ 1⎞ x. ⎜ ⎟ dx ⎝x⎠

Correct application of ‘by M1 parts’

= x ln x −

∫

= x ln x − x (+ c) 3

∴

∫

Correct application of ‘by M1 parts’

( x − 1) ln x dx =

(

9 2

ln3 − 2 ) − ( 3ln3 − 2 ) = 32 ln 3 AG

1

Correct integration A1 Substitutes limits of 3 and ddM1 1 into both integrands and subtracts.

3 ln 3 A1 cso 2

[6] 17 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter 6. (d)

Scheme

Marks

⇒ du = 1x ⎫⎪ ⎧⎪u = ln x dx ⎨ dv ( x − 1)2 ⎬ ⎪⎩ dx = ( x − 1) ⇒ v = 2 ⎪⎭

Way 3

I=

=

=

=

( x − 1)

2

2

( x − 1)

ln x − 2

ln x −

2

( x − 1)

2

ln x −

2

( x − 1) 2

2

∫

( x − 1)

Use of ‘integration by parts’ formula in the M1 correct direction

2

2x

Correct expression A1

dx

Candidate multiplies out numerator to obtain three terms…

∫

x 2 − 2x + 1 dx 2x

∫

1 ⎞ ⎛1 ⎜ 2 x − 1 + 2x ⎟ dx ⎝ ⎠

⎛ x2 ⎞ 1 ln x − ⎜ − x + ln x ⎟ (+c) 2 ⎝ 4 ⎠

… multiplies at least one term through by 1x and then attempts to ... … integrate the result; M1; correct integration A1

3

⎡ ( x − 1)2 ⎤ x2 1 ln x − + x − ln x ⎥ ∴I = ⎢ 2 4 2 ⎢⎣ ⎥⎦ 1

= ( 2ln3 −

9 4

+ 3 − 21 ln3 ) − ( 0 −

= 2ln 3 − 21 ln 3 +

3 4

+

1 4

−1

1 4

+ 1− 0)

= 32 ln3 AG

Substitutes limits of 3 and ddM1 1 and subtracts.

3 2

ln 3

A1 cso

[6]

Beware:

∫ 2x dx can also integrate to 2 ln 2x 1

1

Beware: If you are marking using WAY 2 please make sure that you allocate the marks in the order they appear on the mark scheme. For example if a candidate only integrated lnx correctly then they would be awarded M0A0M1A1M0A0 on ePEN.

18 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number By substitution Aliter u = ln x ⇒ 6. (d) Way 4 I=

=

∫ ( e − 1) .ue u

∫

(

Scheme du dx

u

=

1 x

du

Correct expression Use of ‘integration by parts’ formula in the M1 correct direction

)

u e2u − eu du

⎛1 ⎞ = u ⎜ e2u − eu ⎟ − ⎝2 ⎠

Marks

⎛1

∫ ⎜⎝ 2 e

2u

⎞ − eu ⎟ dx ⎠

⎛1 ⎞ ⎛1 ⎞ = u ⎜ e2u − eu ⎟ − ⎜ e2u − eu ⎟ (+c) ⎝2 ⎠ ⎝4 ⎠

Correct expression A1 Attempt to integrate; M1; correct integration A1

ln3

1 ⎡1 ⎤ ∴ I = ⎢ ue2u − ueu − e2u + eu ⎥ 4 ⎣2 ⎦ ln1

= ( 92 ln3 − 3ln3 − 94 + 3 ) − ( 0 − 0 − 41 + 1) = 32 ln 3 +

3 4

+

1 4

−1

= 32 ln3 AG

Substitutes limits of ln3 ddM1 and ln1 and subtracts.

3 2

ln 3

A1 cso

[6] 13 marks

19 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number 7. (a)

Scheme

From question,

Marks

dS =8 dt

dS = 12x dx

S = 6x 2 ⇒

dx dS dS = ÷ dt dt dx

=

2 8 ;= 3 12x x

⇒

(k = 32 )

Candidate’s

dS =8 dt

B1

dS = 12x dx

B1

8 dS dS ; ÷ dt dx 12x

M1; A1oe

[4] (b)

dV = 3x 2 dx

V = x3 ⇒

dV dV dx = × dt dx dt 1

As x = V 3 , then

dV = 3x 2 dx

⎛ 2 ⎞ = 3x 2 . ⎜ ⎟ ; = 2x ⎝ 3x ⎠ 1 dV = 2V 3 dt

AG

Candidate’s

1

dV dx ; λx × dx dt

Use of x = V 3 , to give

1 dV = 2V 3 dt

B1 M1; A1 A1

[4] Separates the variables with (c)

∫V

dV 1 3

∫V

dV

∫ 2 dt

=

1 3

or

∫ ∫ 2 dt on the other side. −1

V 3 dV on one side and

B1

integral signs not necessary.

∫V

− 31

∫ 2 dt

dV =

Attempts to integrate and …

2 … must see V 3 and 2t; M1; Correct equation with/without + c. A1

2 3

3 2

V = 2t (+c)

3 2

(8) 3 = 2(0) + c ⇒ c = 6

Hence: 3 2

Use of V = 8 and t = 0 in a changed equation containing c ; M1 ∗ ; A1 c=6

2

3 2

(16 2 )

2 3

2

V 3 = 2t + 6 = 2t + 6

⇒

12 = 2t + 6

Having found their “c” candidate … … substitutes V = 16 2 into an depM1 ∗ equation involving V, t and “c”.

giving t = 3.

t = 3 A1 cao [7] 15 marks

20 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number Aliter 7. (b) Way 2

Scheme 1

Marks 2

2

x = V 3 & S = 6x 2 ⇒ S = 6V 3

dS dV 1 31 −1 = 4V 3 or = V dV dS 4 dV dS dV = × dt dt dS

1 2 ⎛ 1 ⎞ ;= = 8. ⎜ = 2V 3 AG − 31 ⎟ − 31 V ⎝ 4V ⎠

S = 6V 3

B1

dS dV 1 31 −1 = 4V 3 or = V dV dS 4

B1

Candidate’s

1 dS dV ; 2V 3 × dt dS

M1; A1

In ePEN, award Marks for Way 2 in the order they appear on this mark scheme.

[4] Aliter

7. (c)

∫ 2V

dV 1 3

∫

∫ 1 dt

=

Separates the variables with dV 1 − 31 or V dV oe on one 1 3 B1 2 2V

∫

side and

∫ 1 dt on the other side.

integral signs not necessary.

Way 2

∫

1 V 2

− 31

dV =

( 21 ) ( 32 ) V 3 4

Attempts to integrate and …

2 … must see V 3 and t; M1; Correct equation with/without + c. A1

= t (+c)

Use of V = 8 and t = 0 in a changed equation containing c ; M1 ∗ ; A1 c=3

2 3

(8) = (0) + c ⇒ c = 3

Hence:

3 4

2 3

∫ 1 dt

3 4

(16 2 )

2 3

2

V3 = t + 3

=t+3

⇒

6=t+3

Having found their “c” candidate … … substitutes V = 16 2 into an depM1 ∗ equation involving V, t and “c”.

giving t = 3.

t = 3 A1 cao [7]

Beware: On ePEN award the marks in part (c) in the order they appear on the mark scheme.

21 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Scheme Number Aliter similar to way 1. dV V = x3 ⇒ = 3x 2 (b) dx Way 3 dV dV dS dx ⎛ 1 ⎞ = × × = 3x 2 .8. ⎜ ⎟ ; = 2x dt dx dt dS ⎝ 12x ⎠ 1

As x = V 3 , then

1 dV = 2V 3 dt

Marks dV = 3x 2 dx

Candidate’s

dV dS dx ; λx × × dx dt dS 1

Use of x = V 3 , to give

AG

1 dV = 2V 3 dt

B1 M1; A1 A1

[4] Aliter

Separates the variables with (c)

∫V

dV 1 3

=

∫V

dV

∫ 2 dt

1 3

or

∫ ∫ 2 dt on the other side. −1

V 3 dV on one side and

B1

integral signs not necessary.

Way 3

∫V

− 31

dV =

∫ 2 dt Attempts to integrate and …

2

2 3

V =

2 3

4 3

… must see V 3 and 34 t; M1; Correct equation with/without + c. A1

t (+c)

Use of V = 8 and t = 0 in a changed equation containing c ; M1 ∗ ; A1 c=4

(8) = (0) + c ⇒ c = 4 4 3

2

Hence: V 3 =

(16 2 )

2 3

=

4 3

4 3

t+4

t+6

⇒

8=

4 3

t+4

Having found their “c” candidate … … substitutes V = 16 2 into an depM1 ∗ equation involving V, t and “c”.

giving t = 3.

t = 3 A1 cao [7]

• • • • •

Beware when marking question 7(c). There are a variety of valid ways that a candidate can use to find the constant “c”.

In questions 7(b) and 7(c) there may be “Ways” that I have not listed. Please use the mark scheme as a guide of how the mark the students’ responses.

In 7(c), if a candidate instead tries to solve the differential equation in part (a) escalate the response to your team leader.

IF YOU ARE UNSURE ON HOW TO APPLY THE MARK SCHEME PLEASE ESCALATE THE RESPONSE UP TO YOUR TEAM LEADER VIA THE REVIEW SYSTEM. Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. depM1 ∗ denotes a method mark which is dependent upon the award of M1∗ .

22 6666/01 Core Maths C4 June 2006 Advanced Subsidiary/Advanced Level in GCE Mathematics