C4 MS Jan 2008

Mark Scheme (Results) January 2008

GCE

GCE Mathematics (6666/01)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH

1

January 2008 6666 Core Mathematics C4 Mark Scheme Question Number

Scheme 0 0

x y

1. (a)

Marks

π

π

4

2

3π 4

π

1.844321332…

4.810477381…

8.87207

0 awrt 1.84432 B1 awrt 4.81048 or 4.81047 B1

0 can be implied

[2] Outside brackets awrt 0.39 or 12 × awrt 0.79 B1 1 × π4 or π8 2

(b) Way 1

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

1 π Area ≈ × ;× 0 + 2 (1.84432 + 4.81048 + 8.87207 ) + 0} 2 4

{

Correct expression inside brackets which all must be multiplied by their “outside A1 constant”.

=

π 8

× 31.05374... = 12.19477518... = 12.1948 (4dp)

12.1948 A1 cao [4]

Area ≈ π4 ×

Aliter (b) Way 2

{

0 + 1.84432 2

+

1.84432 + 4.81048 2

+

4.81048 + 8.87207 2

+

8.87207 + 0 2

}

which is equivalent to: 1 π Area ≈ × ;× 0 + 2 (1.84432 + 4.81048 + 8.87207 ) + 0} 2 4

{

=

π 4

× 15.52687... = 12.19477518... = 12.1948 (4dp)

π

(or awrt 0.79 ) and a divisor of 2 on all terms inside B1 brackets.

One of first and last ordinates, two of the middle ordinates M1 inside brackets ignoring the 2.

Correct expression inside brackets if 12 was to be A1 factorised out.

4

12.1948 A1 cao [4] 6 marks

1 π Note an expression like Area ≈ × + 2 (1.84432 + 4.81048 + 8.87207 ) would score B1M1A0A0 2 4

2

Question Number

Scheme

Marks

** represents a constant (which must be consistent for first accuracy mark) 1

2. (a)

(8 − 3x )

1 3

Takes 8 outside the bracket to give any of B1 1 (8) 3 or 2 .

1

⎛ 3x ⎞ 3 ⎛ 3x ⎞ 3 = ( 8 ) ⎜1 − ⎟ = 2 ⎜1 − ⎟ 8 ⎠ 8 ⎠ ⎝ ⎝ 1 3

1

⎧ ( 1 )(− 23 ) ( 1 )( − 23 )( − 53 ) ⎫ (** x) 2 + 3 (** x)3 + ...⎬ = 2 ⎨1 + ( 13 )(** x); + 3 2!

3! ⎭ ⎩ with ** ≠ 1

⎧ ( 1 )(− 32 ) 3 x 2 ( 13 )(− 23 )(− 53 ) 3 x 3 ⎫ = 2 ⎨1 + ( 13 )(− 38x ) + 3 (− 8 ) + (− 8 ) + ...⎬ 2!

3! ⎭ ⎩ = 2{1 − 18 x ; −

1 64

x2 −

5 1536

x 3 − ...}

Expands (1 + ** x) 3 to give a simplified or an M1; un-simplified 1 + ( 13 )(** x) ;

A correct simplified or an un-simplified {..........} expansion with A1 candidate’s followed through (** x )

Award SC M1 if you see

( 13 )(− 32 ) ( 1 )(− 32 )(− 53 ) (** x) 2 + 3 (** x)3 2!

3!

Either 2{1 − 18 x ........} or anything that A1; cancels to 2 − 1 x ;

1 1 2 5 3 = 2 − x; − x − x − ...

4 32 768

4

Simplified −

1 32

x − 2

5 768

x3

A1 [5]

(b)

1

(7.7) 3 ≈ 2 −

Attempt to substitute x = 0.1 into a candidate’s M1 binomial expansion.

1 1 5 (0.1) − (0.1) 2 − (0.1)3 − ...

4 32 768

= 2 − 0.025 − 0.0003125 − 0.0000065104166... = 1.97468099...

awrt 1.9746810 A1 [2] 7 marks

You would award B1M1A0 for ⎧ ( 1 )(− 2 ) (1 )(− 2 )(− 5 ) ⎫ = 2⎨1 +(13)(− 38x ) + 3 3 (− 38x )2 + 3 3 3 (−3x)3 + ...⎬ 2!

3! ⎭ ⎩

If you see the constant term “2” in a candidate’s final binomial expansion, then you can award B1.

because ** is not consistent.

Be wary of calculator value of ( 7.7 ) 3 = 1.974680822...

1

3

Question Number Aliter 2. (a) Way 2

Scheme

Marks

1

(8 − 3 x) 3 1

2 or (8) 3 (See note ↓ ) B1 1

Expands (8 − 3 x) 3 to give an un-simplified or M1; simplified

⎧ 1 ⎫ ( 1 )(− 32 ) − 53 −2 (8) (** x) 2 ⎪(8) 3 + ( 13 )(8) 3 (** x); + 3 ⎪ ⎪ ⎪ 2! =⎨ ⎬ ( 1 )( − 23 )(− 53 ) − 83 ⎪ (8) (** x)3 + ...⎪ + 3 ⎪⎩ ⎪⎭ 3!

1

−2

(8) 3 + ( 13 )(8) 3 (** x);

A correct un-simplified or simplified {..........} expansion with A1 candidate’s followed through (** x )

with ** ≠ 1

⎧ 1 ⎫ ( 1 )(− 23 ) − 53 −2 (8) (−3x) 2 ⎪(8) 3 + ( 13 )(8) 3 (−3x); + 3 ⎪ ⎪ ⎪ 2! =⎨ ⎬ ( 13 )(− 32 )(− 53 ) − 83 3 ⎪ + (8) (−3x) + ...⎪ ⎪⎩ ⎪⎭ 3!

Award SC M1 if you see ( 13 )(− 23 ) − 53 (8) (** x) 2 2! ( 1 )(− 32 )(− 35 ) − 83 + 3 (8) (** x)3 3!

1 = {2 + ( 13 )( 14 )(−3x) + (− 91 )( 321 )(9 x 2 ) + ( 815 )( 256 )(−27 x 3 ) + ...}

=2−

Anything that cancels to 2 − 1 x ;

1 1 2 5 3 x; − x − x − ...

4 32 768

4

or 2{1 − x ........}

A1;

1 8

Simplified − 321 x 2 −

5 768

x3

A1 [5]

Attempts using Maclaurin expansion should be escalated up to your team leader.

Be wary of calculator value of ( 7.7 ) 3 = 1.974680822...

1

4

If you see the constant term “2” in a candidate’s final binomial expansion, then you can award B1.

Question Number 3.

Scheme 2

⎛ 1 ⎞ ⎜ ⎟ dx = π a ⎝ 2x + 1 ⎠

∫

Volume = π

=π

b

b

∫ ( 2 x + 1)

−2

Marks

∫

b

a

1

( 2 x + 1)

2

dx

Use of V = π ∫ y 2 dx .

Can be implied. Ignore limits.

B1

dx

a

b

⎡ (2 x + 1) −1 ⎤ = (π ) ⎢ ⎥ ⎣ (−1)(2) ⎦ a b

= (π ) ⎡⎣ − 12 (2 x + 1) −1 ⎤⎦ a ⎡⎛ −1 ⎞ ⎛ −1 ⎞ ⎤ = ( π ) ⎢⎜ ⎟−⎜ ⎟⎥ ⎣⎝ 2(2b + 1) ⎠ ⎝ 2(2a + 1) ⎠ ⎦ =

=

=

Integrating to give ± p (2 x + 1) −1

M1

− 12 (2 x + 1) −1

A1

Substitutes limits of b and a and dM1 subtracts the correct way round.

π ⎡ −2a − 1 + 2b + 1 ⎤ 2 ⎢⎣ (2a + 1)(2b + 1) ⎥⎦

π⎡

⎤ 2(b − a ) ⎢ ⎥ 2 ⎣ (2a + 1)(2b + 1) ⎦

π (b − a )

π (b − a )

(2a + 1)(2b + 1)

(2a + 1)(2b + 1)

A1 aef [5] 5 marks

Allow other equivalent forms such as

πb −πa (2a + 1)(2b + 1)

or

−π (a − b) π (b − a) πb −πa or . or 4ab + 2a + 2b + 1 4ab + 2a + 2b + 1 (2a + 1)(2b + 1)

Note that π is not required for the middle three marks of this question.

5

Question Number Aliter 3.

Way 2

Scheme 2

⎛ 1 ⎞ ⎜ ⎟ dx = π a ⎝ 2x + 1 ⎠

∫

Volume = π

= π

b

b

∫ ( 2 x + 1)

−2

Marks

∫

b

a

1

( 2 x + 1)

2

Use of V = π ∫ y 2 dx .

dx

Can be implied. Ignore limits.

B1

dx

a

Applying substitution u = 2 x + 1 ⇒ ddux = 2 and changing limits x → u so that a → 2a + 1 and b → 2b + 1 , gives = (π )

∫

u −2 du 2

2 b +1

2 a +1

2b + 1

⎡ u −1 ⎤ = (π ) ⎢ ⎥ ⎣ (−1)(2) ⎦ 2 a + 1 = (π ) ⎡⎣ − u 1 2

−1

Integrating to give ± p u −1

2b + 1

⎤ ⎦ 2a + 1

− u 1 2

=

=

M1 A1

Substitutes limits of 2b + 1 and 2a + 1 and subtracts the correct dM1 way round.

⎡⎛ −1 ⎞ ⎛ −1 ⎞ ⎤ = ( π ) ⎢⎜ ⎟−⎜ ⎟⎥ ⎣⎝ 2(2b + 1) ⎠ ⎝ 2(2a + 1) ⎠ ⎦

=

−1

π ⎡ −2a − 1 + 2b + 1 ⎤ 2 ⎢⎣ (2a + 1)(2b + 1) ⎥⎦

π⎡

⎤ 2(b − a ) ⎢ 2 ⎣ (2a + 1)(2b + 1) ⎥⎦

π (b − a )

π (b − a )

(2a + 1)(2b + 1)

(2a + 1)(2b + 1)

A1 aef [5] 5 marks

Note that π is not required for the middle three marks of this question.

Allow other equivalent forms such as

πb −πa (2a + 1)(2b + 1)

or

−π (a − b) π (b − a) πb −πa or or .

4ab + 2a + 2b + 1 4ab + 2a + 2b + 1 (2a + 1)(2b + 1)

6

Question Number

4. (i)

Scheme

∫ ln (

x 2

∫ ln (

x 2

) dx

) dx

=

∫

Marks

⎧ x ⎪⎪u = ln ( 2 ) 1.ln ( 2x ) dx ⇒ ⎨ ⎪ dv = 1 ⎪⎩ dx

= x ln (

x 2

) − ∫ x.

du = dx

⇒

1 2 x 2

⇒ v =x

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

Correct expression. A1

dx

1 x

⎫ = 1x ⎪ ⎪ ⎬ ⎪ ⎪⎭

∫

An attempt to multiply x by a candidate’s ax or bx1 or 1x . dM1

= x ln ( 2x ) − 1 dx = x ln ( 2x ) − x + c

Correct integration with + c A1 aef [4]

(ii)

∫

π

2

π

sin 2 x dx

4

⎡ NB : cos 2 x = ±1 ± 2sin 2 x or sin 2 x = ⎣ =

∫

π

2

π

4

1 − cos 2 x 1 dx = 2 2

=

∫

π

2

π

( ±1 ±

1 2

cos 2 x ) ⎤ ⎦

(1 − cos 2 x ) dx

4

Integrating to give ± ax ± b sin 2 x ; a , b ≠ 0 dM1 Correct result of anything equivalent to 12 x − 14 sin 2 x A1

1⎡ 2 x − 12 sin 2 x ⎤⎦ π ⎣ 2 4 π

= 12 ⎡ ⎢⎣

(

π 2

)

− sin(2π ) −

(

π 4

−

sin

)⎥⎦

( π2 ) ⎤

2

= 12 ⎡⎣ ( π2 − 0) − ( π4 − 12 ) ⎤⎦ =

Consideration of double angle M1 formula for cos 2x

1 2

( π4 + 12 )

= π8 + 14

Substitutes limits of π2 and π4 and subtracts the correct way ddM1 round.

1 2

( π4 + 12 )

or

π 8

+ 14 or

π 8

+ 82

A1 aef , cso

Candidate must collect their π term and constant term together for A1 No fluked answers, hence cso.

[5]

9 marks

Note:

∫

ln ( 2x ) dx = (their v)ln ( 2x ) −

∫ ( their v ).( their ddux ) dx 7

for M1 in part (i).

Note

π 8

+ 14 = 0.64269...

Question Number Aliter 4. (i) Way 2

Scheme

∫ ln (

x 2

) dx

Marks

∫ ( ln x − ln 2) dx = ∫ ln x dx − ∫ ln 2 dx

=

∫ ln x dx = ∫

⎧ ⎪⎪u = ln x 1.ln x dx ⇒ ⎨ ⎪ dv = 1 ⎩⎪ dx

∫ ln x dx = x ln x − ∫ x.

1 x

du 1 ⎫ = dx x ⎪⎪ ⎬ ⇒ v =x ⎪ ⎭⎪ ⇒

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

dx

Correct integration of ln x A1 with or without + c

= x ln x − x + c

∫ ln 2 dx = x ln 2 + c Hence,

∫ ln (

x 2

) dx

Correct integration of ln 2 M1 with or without + c

= x ln x − x − x ln 2 + c

Correct integration with + c A1 aef [4]

Note:

∫ ln x dx = ( their v ) ln x − ∫ ( their v ).( their

8

du dx

) dx

for M1 in part (i).

Question Number Aliter 4. (i) Way 3

Scheme

∫ ln ( u=

x 2

) dx ⇒

x 2

Marks

du 1 = dx 2

Applying substitution correctly to give

∫ ln (

x 2

) dx

∫

∫ ln (

= 2 ln u du

x 2

) dx

∫

= 2 ln u du

Decide to award 2nd M1 here!

∫ ln u dx = ∫1.ln u du ∫

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

∫

ln u dx = u ln u − u. u1 du

Correct integration of ln u A1 with or without + c

= u ln u − u + c

Decide to award M1 2nd M1 here!

∫ ln (

x 2

) dx

Hence,

= 2 ( u ln u − u ) + c

∫ ln (

x 2

) dx

= x ln ( 2x ) − x + c

Correct integration with + c A1 aef [4]

9

Question Number Aliter 4. (ii) Way 2

Scheme

∫

π

2

π

∫

sin x dx = 2

π

2

π

4

Marks

sin x.sin x dx

∫

I = sin 2 x dx

and

4

du ⎪⎧u = sin x ⇒ dx = cos x ⎪⎫ ⎨ dv ⎬ ⎪⎩ dx = sin x ⇒ v = − cos x ⎪⎭

{

∫

}

{

∫ (1 − sin

An attempt to use the correct M1 by parts formula.

∴ I = − sin x cos x + cos 2 x dx

∴ I = − sin x cos x +

∫ sin

2

}

x ) dx

2

{

∫

{

∫ }

}

∫

x dx = − sin x cos x + 1 dx − sin 2 x dx

∫

2 sin 2 x dx = − sin x cos x + 1 dx

For the LHS becoming 2I dM1

∫

2 sin 2 x dx = {− sin x cos x + x}

∫ sin ∴

∫

π

2

π

4

2

{

x dx = − 12 sin x cos x +

x 2

}

(

sin 2 x dx = ⎡ − 12 sin( π2 )cos( π2 ) + ⎢⎣

Correct integration A1 ( π2 ) 2

) − (−

1 2

sin( π4 ) cos( π4 ) +

( π4 ) 2

)⎤⎥⎦

= ⎡⎣(0 + π4 ) − (− 14 + π8 ) ⎤⎦ = π8 + 14

Substitutes limits of π2 and π4 and subtracts the correct way ddM1 round.

1 2

( π4 + 12 )

or

π 8

+ 14 or

π 8

+ 82

Candidate must collect their π term and constant term together for A1 No fluked answers, hence cso.

Note

10

π 8

+ 14 = 0.64269...

A1 aef cso

[5]

Question Number

Scheme x 3 − 4 y 2 = 12 xy

5. (a)

Marks ( eqn ∗ ) Substitutes x = −8 (at least once) into * to obtain a three term quadratic in y . M1 Condone the loss of = 0.

x = −8 ⇒ − 512 − 4 y 2 = 12(−8) y

− 512 − 4 y 2 = − 96 y 4 y 2 − 96 y + 512 = 0 y 2 − 24 y + 128 = 0 ( y − 16)( y − 8) = 0 y=

An attempt to solve the quadratic in y by either factorising or by the formula or by dM1 completing the square.

24 ± 576 − 4(128) 2

Both y = 16 and y = 8.

y = 16 or y = 8.

or ( −8, 8 ) and ( −8, 16 ).

A1 [3]

(b)

Differentiates implicitly to include either M1 ± ky ddyx or 12 x ddyx . Ignore ddyx = ...

⎧ dy ⎫ dy ⎛ dy ⎞ ; = ⎜ 12 y + 12 x = ⎬ 3x 2 − 8 y ⎨ ⎟ dx ⎝ dx ⎠ ⎩ dx ⎭

Correct LHS equation; A1; Correct application of product rule (B1)

⎧ dy 3x 2 − 12 y ⎫ = ⎨ ⎬ 12 x + 8 y ⎭ ⎩ dx

not necessarily required.

dy 3(64) − 12(8) 96 = = = − 3, dx 12(−8) + 8(8) −32 dy 3(64) − 12(16) 0 @ ( −8, 16 ) , = = = 0. dx 12(−8) + 8(16) 32

@ ( −8, 8 ) ,

Substitutes x = −8 and at least one of their dM1 y-values to attempt to find any one of ddyx .

One gradient found. A1 Both gradients of -3 and 0 correctly found. A1 cso [6] 9 marks

11

Question Number Aliter 5. (b) Way 2

Scheme

⎛ ⎧⎪ dx ⎫⎪ dx dx = ⎬ 3x 2 − 8 y ; = ⎜12 y + 12 x ⎨ ⎜ dy dy ⎪⎩ dy ⎪⎭ ⎝

Marks Differentiates implicitly to include either ± kx 2 ddyx or 12 y ddyx . Ignore ddxy = ... M1

⎞ ⎟ ⎟ ⎠

Correct LHS equation A1; Correct application of product rule (B1)

⎧ dy 3x 2 − 12 y ⎫ = ⎨ ⎬ 12 x + 8 y ⎭ ⎩ dx

not necessarily required.

dy 3(64) − 12(8) 96 = = = − 3, dx 12(−8) + 8(8) −32 dy 3(64) − 12(16) 0 @ ( −8, 16 ) , = = = 0. dx 12(−8) + 8(16) 32

@ ( −8, 8 ) ,

Substitutes x = −8 and at least one of their y-values to attempt to find any dM1 one of ddyx or ddxy .

One gradient found. A1 Both gradients of -3 and 0 correctly found. A1 cso [6]

12

Question Number Aliter 5. (b) Way 3

Scheme

Marks

x 3 − 4 y 2 = 12 xy ( eqn ∗ )

4 y 2 +12 xy − x3 = 0 y=

y=

y=

−12 x ±

144 x 2 − 4(4)(− x 3 ) 8

−12 x ±

144 x 2 + 16 x 3 8

−12 x ± 4 9 x 2 + x 3 8

y = − 32 x ±

dy = − 32 ± dx

1 2

1 2

(9x

2

+ x3 )

1 2

A credible attempt to make y the subject and an attempt to differentiate either − 32 x

( 12 ) ( 9 x 2 + x3 ) ; (18 x + 3x 2 ) − 12

or

∴

dy = − 32 ± dx

dy 3 18(−8) + 3(64) =− ± 1 dx 2 4(9(64) + ( −512)) 2 =−

(9x

2

+x

3

)

1 2

1 2

( 12 ) ( 9 x 2 + x3 ) ; (18 x + 3x 2 ) − 12

Substitutes x = −8 find any one of

dy dx

M1

.

− 12 dy = − 32 ± k ( 9 x 2 + x3 ) ( g( x) ) dx

dy 3 18 x + 3x 2 =− ± 1 dx 2 4(9 x 2 + x3 ) 2

@ x = −8

1 2

.

A1

A1

dM1

3 48 3 48 ± =− ± 2 4 (64) 2 32 One gradient correctly found. A1 Both gradients of -3 and 0 correctly found. A1

dy 3 3 = − ± = −3, 0. dx 2 2

[6]

13

Question Number

6. (a)

Scheme

Marks

⎛2⎞ ⎛ 3⎞ uuur ⎜ ⎟ uuur ⎜ ⎟ OA = ⎜ 6 ⎟ & OB = ⎜ 4 ⎟ ⎜ −1⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ Finding the difference uuur uuur M1 ± between OB and OA .

⎛ 3⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = OB − OA = ⎜ 4 ⎟ − ⎜ 6 ⎟ = ⎜ −2 ⎟ ⎜ 1 ⎟ ⎜ −1⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Correct answer. A1 [2]

An expression of the form ⎛2⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 6 ⎟ + λ ⎜ −2 ⎟ or ⎜ −1⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠

⎛ 3⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 4 ⎟ + λ ⎜ −2 ⎟ ⎜1⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠

⎛2⎞ ⎛ −1 ⎞ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 6 ⎟ + λ ⎜ 2 ⎟ or ⎜ −1⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠

⎛ 3⎞ ⎛ −1 ⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 4⎟ +λ ⎜ 2 ⎟ ⎜1⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠

( vector ) ± λ ( vector )

M1 uuur uuur r = OA ± λ their AB or uuur uuur r = OB ± λ their AB or uuur uuur A1 r = OA ± λ their BA or aef uuur uuur r = OB ± λ their BA

( ( (

(b)

(

) ) )

)

( r is needed.)

[2] (c)

⎛ 0⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ l2 : r = ⎜ 0 ⎟ + µ ⎜ 0 ⎟ ⎜ 0⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠

⇒

⎛1⎞ ⎜ ⎟ r = µ ⎜0⎟ ⎜1⎟ ⎝ ⎠

uuur AB = d 1 = i − 2 j + 2k , d 2 = i + 0 j + k & θ is angle

cos θ =

cos θ =

cos θ =

(

uuur AB • d 2 uuur AB . d 2

=

) (

⎛ 1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −2 ⎟ • ⎜ 0 ⎟ ⎜ 2 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ (1) 2 + (−2) 2 + (2) 2 . (1) 2 + (0) 2 + (1) 2

)

1+ 0 + 2

Correct followed through expression or equation. A1

(1) + (−2) + (2) 2 . (1) 2 + (0) 2 + (1) 2 2

3 3. 2

2

⇒ θ = 45o or

π 4

Considers dot product uuur between d 2 and their AB. M1

θ = 45o or

or awrt 0.79.

π 4

or awrt 0.79

A1 cao [3]

This means that cosθ does not necessarily have to be the subject of the equation. It could be of the form 3 2 cos θ = 3.

14

Question Number

6. (d)

Scheme

Marks

⎛2⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ If l1 and l2 intersect then: ⎜ 6 ⎟ + λ ⎜ −2 ⎟ = µ ⎜ 0 ⎟ ⎜ −1⎟ ⎜ 2⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ i: 2 + λ =µ j : 6 − 2λ = 0 k : − 1 + 2λ = µ

Either seeing equation (2) written down correctly with or without any other equation or seeing equations M1 (1) and (3) written down correctly.

(1) (2) (3)

Attempt to solve either equation (2) or simultaneously solve any two of dM1 the three equations to find …

(2) yields λ = 3 Any two yields λ = 3, µ = 5

either one of λ or µ correct. A1

⎛2⎞ ⎛ 1 ⎞ ⎛ 5⎞ ⎛1⎞ ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 6 ⎟ + 3 ⎜ −2 ⎟ = ⎜ 0 ⎟ or r = 5 ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜ −1⎟ ⎜ 2 ⎟ ⎜ 5⎟ ⎜1⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 5⎞ ⎜ ⎟ ⎜ 0 ⎟ or 5i + 5k ⎜ 5⎟ A1 cso ⎝ ⎠ Fully correct solution & no incorrect values of λ or µ seen earlier.

[4] Aliter 6. (d) Way 2

⎛ 3⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ If l1 and l2 intersect then: ⎜ 4 ⎟ + λ ⎜ −2 ⎟ = µ ⎜ 0 ⎟ ⎜1⎟ ⎜ 2⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i: 3 + λ =µ j : 4 − 2λ = 0 k : 1 + 2λ = µ

Either seeing equation (2) written down correctly with or without any other equation or seeing equations M1 (1) and (3) written down correctly.

(1) (2) (3)

Attempt to solve either equation (2) or simultaneously solve any two of dM1 the three equations to find … either one of λ or µ correct. A1

(2) yields λ = 2 Any two yields λ = 2, µ = 5

⎛ 3⎞ ⎛ 1 ⎞ ⎛5⎞ ⎛1⎞ ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 4 ⎟ + 2 ⎜ −2 ⎟ = ⎜ 0 ⎟ or r = 5 ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜1⎟ ⎜ 2 ⎟ ⎜5⎟ ⎜1⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 5⎞ ⎜ ⎟ ⎜ 0 ⎟ or 5i + 5k ⎜ 5⎟ A1 cso ⎝ ⎠ Fully correct solution & no incorrect values of λ or µ seen earlier.

[4]

11 marks Note: Be careful! λ and µ are not defined in the question, so a candidate could interchange these or use different scalar parameters.

15

Question Number Aliter 6. (d) Way 3

Scheme

Marks

⎛2⎞ ⎛ −1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ If l1 and l2 intersect then: ⎜ 6 ⎟ + λ ⎜ 2 ⎟ = µ ⎜ 0 ⎟ ⎜ −1⎟ ⎜ −2 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 − λ =µ

(1)

j : 6 + 2λ = 0 k : − 1 − 2λ = µ

(2) (3)

i:

Either seeing equation (2) written down correctly with or without any other equation or seeing equations M1 (1) and (3) written down correctly.

Attempt to solve either equation (2) or simultaneously solve any two of dM1 the three equations to find …

(2) yields λ = −3 Any two yields λ = −3, µ = 5

either one of λ or µ correct. A1

⎛2⎞ ⎛ −1 ⎞ ⎛ 5 ⎞ ⎛1⎞ ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 6 ⎟ − 3 ⎜ 2 ⎟ = ⎜ 0 ⎟ or r = 5 ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜ −1⎟ ⎜ −2 ⎟ ⎜ 5 ⎟ ⎜1⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 5⎞ ⎜ ⎟ ⎜ 0 ⎟ or 5i + 5k ⎜ 5⎟ A1 cso ⎝ ⎠ Fully correct solution & no incorrect values of λ or µ seen earlier.

[4]

Aliter 6. (d) Way 4

⎛ 3⎞ ⎛ −1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ If l1 and l2 intersect then: ⎜ 4 ⎟ + λ ⎜ 2 ⎟ = µ ⎜ 0 ⎟ ⎜1⎟ ⎜ −2 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3 − λ =µ

(1)

j : 4 + 2λ = 0 k : 1 − 2λ = µ

(2) (3)

i:

Either seeing equation (2) written down correctly with or without any other equation or seeing equations M1 (1) and (3) written down correctly.

Attempt to solve either equation (2) or simultaneously solve any two of dM1 the three equations to find … either one of λ or µ correct. A1

(2) yields λ = −2 Any two yields λ = −2, µ = 5

⎛ 3⎞ ⎛ −1 ⎞ ⎛ 5 ⎞ ⎛1⎞ ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ l1 : r = ⎜ 4 ⎟ − 2 ⎜ 2 ⎟ = ⎜ 0 ⎟ or r = 5 ⎜ 0 ⎟ = ⎜ 0 ⎟ ⎜1⎟ ⎜ −2 ⎟ ⎜ 5 ⎟ ⎜1⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 5⎞ ⎜ ⎟ ⎜ 0 ⎟ or 5i + 5k ⎜ 5⎟ A1 cso ⎝ ⎠ Fully correct solution & no incorrect values of λ or µ seen earlier.

[4] 11 marks

16

Question Number

Scheme 1 ⎤ ⎡ ⎢ x = ln ( t + 2 ) , y = t + 1 ⎥ , ⎣ ⎦

7. (a)

Marks

⇒

dx 1 = dt t + 2

Must state

dx 1 = dt t + 2

Area = Area( R) =

∫

ln 4 ln 2

1 dx ; = t +1

∫

2 0

⎛ 1 ⎞⎛ 1 ⎞ ⎜ ⎟⎜ ⎟ dt ⎝ t + 1 ⎠⎝ t + 2 ⎠

∫

2 0

1

∫ t + 1 dx .

M1;

Ignore limits. ⎛ 1 ⎞ ⎛ 1 ⎞

∫ ⎜⎝ t + 1 ⎟⎠ × ⎜⎝ t + 2 ⎟⎠ dt . Ignore limits.

Changing limits, when: x = ln 2 ⇒ ln 2 = ln(t + 2) ⇒ 2 = t + 2 ⇒ t = 0 x = ln 4 ⇒ ln 4 = ln(t + 2) ⇒ 4 = t + 2 ⇒ t = 2 Hence, Area( R ) =

B1

changes limits x → t so that ln 2 → 0 and ln 4 → 2

A1 AG

B1

1 dt (t + 1)(t + 2) [4]

(b)

⎛ ⎞ 1 A B + ⎜ ⎟= ⎝ (t + 1)(t + 2) ⎠ (t + 1) (t + 2)

A B with A and B found M1 + (t + 1) (t + 2)

1 = A(t + 2) + B(t + 1) Let t = −1, 1 = A (1)

⇒ A =1

Finds both A and B correctly.

Can be implied. A1 (See note below)

Let t = −2, 1 = B ( −1) ⇒ B = − 1

∫

2 0

1 dt = (t + 1)(t + 2)

∫

2 0

1 1 dt − (t + 1) (t + 2) Either ± a ln(t + 1) or ± b ln(t + 2)

= [ ln(t + 1) − ln(t + 2) ] 0

2

Both ln terms correctly ft. A1

= ( ln 3 − ln 4 ) − ( ln1 − ln 2 )

Substitutes both limits of 2 and 0 ddM1 and subtracts the correct way round.

= ln 3 − ln 4 + ln 2 = ln 3 − ln 2 = ln (

Takes out brackets.

dM1

3 2

)

ln 3 − ln 4 + ln 2 or ln ( 34 ) − ln ( 12 ) or ln 3 − ln 2 or ln ( 32 )

A1 aef isw

(must deal with ln 1)

[6]

Writing down

1 1 1 = + means first M1A0 in (b). (t + 1)(t + 2) (t + 1) (t + 2)

Writing down

1 1 1 = − means first M1A1 in (b). (t + 1)(t + 2) (t + 1) (t + 2) 17

Question Number

Scheme x = ln ( t + 2 ) ,

7. (c)

y=

Marks

1 t +1

Attempt to make t =… the subject M1 giving t = e x − 2 A1

ex = t + 2 ⇒ t = ex − 2

y=

Eliminates t by substituting in y dM1 1 giving y = x A1 e −1

1 1 ⇒ y= x x e − 2 +1 e −1

[4] Aliter 7. (c) Way 2

t +1 =

1 1 1− y ⇒ t = −1 or t = y y y

Attempt to make t =… the subject M1

y (t + 1) = 1 ⇒ yt + y = 1 ⇒ yt = 1 − y ⇒ t =

⎛1 ⎞ x = ln ⎜ − 1 + 2 ⎟ ⎝y ⎠

or

1− y y

⎛1− y ⎞ + 2⎟ x = ln ⎜ ⎝ y ⎠

Giving either t =

1 1− y −1 or t = y y

A1

Eliminates t by substituting in x dM1

⎛1 ⎞ x = ln ⎜ + 1⎟ ⎝y ⎠ ex =

1 +1 y

ex − 1 =

y=

1 y

1 e −1

giving y =

x

1 e −1 x

A1 [4]

(d)

Domain : x > 0

x > 0 or just > 0 B1

[1] 15 marks

18

Question Number Aliter 7. (c) Way 3

Scheme

Marks Attempt to make t + 1 = … the subject M1

e x = t + 2 ⇒ t + 1 = e x −1

giving t + 1 = e x −1

A1

Eliminates t by substituting in y dM1 1 giving y = x A1 e −1

1 1 y= ⇒ y= x e −1 t +1

[4] Aliter 7. (c) Way 4

Attempt to make t + 2 =… the subject M1 1 1+ y Either t + 2 = + 1 or t + 2 = A1 y y

1 1 1+ y ⇒ t + 2 = + 1 or t + 2 = t +1 = y y y ⎛1 ⎞ x = ln ⎜ + 1⎟ ⎝y ⎠

or

⎛1+ y ⎞ x = ln ⎜ ⎟ ⎝ y ⎠

Eliminates t by substituting in x dM1

⎛1 ⎞ x = ln ⎜ + 1⎟ ⎝y ⎠ ex =

y=

1 +1 y

⇒ ex − 1 =

1 y

1 e −1

giving y =

x

1 e −1 x

A1 [4]

19

Question Number

Scheme dV = 1600 − c h dt

8. (a)

(V = 4000h

dV = 1600 − k h , dt

Either of these statements M1

dV = 4000 dh

⇒)

dV dh 1 = 4000 or = dh dV 4000

dh 1600 − c h 1600 c h = = − = 0.4 − k h dt 4000 4000 4000 Convincing proof of

or

M1

dV dt dV dh

dh dh dV = × = dt dV dt Either,

or

Marks

dh 1600 − k h 1600 k h = = − = 0.4 − k h dt 4000 4000 4000

dh dt

A1 AG

[3]

(b)

When h = 25 water leaks out such that

dV = 400 dt

400 = c h ⇒ 400 = c 25 ⇒ 400 = c(5) ⇒ c = 80 From above; k =

c 80 = = 0.02 as required 4000 4000

Proof that k = 0.02

B1 AG [1]

Aliter (b) Way 2

400 = 4000k h ⇒ 400 = 4000k 25 Using 400, 4000 and h = 25 ⇒ 400 = k (20000) ⇒ k =

400 20000

= 0.02

or

h = 5 . Proof that k = 0.02

B1 AG [1]

(c)

dh = 0.4 − k h ⇒ dt

∴ time required =

dh

∫ 0.4 − k

∫

100

∫

100

0

time required =

0

h

∫

∫

= dt

Separates the variables with dh and dt on either side M1 oe 0.4 − k h with integral signs not necessary.

∫

÷ 0.02 1 dh ÷ 0.02 0.4 − 0.02 h 50 dh 20 − h

Correct proof A1 AG [2]

20

Question Number

Scheme

∫

8. (d)

50

100

20 − h

0

Marks

dh with substitution h = (20 − x) 2

dh dh = 2(20 − x)(−1) or = − 2(20 − x) dx dx h = (20 − x) 2 ⇒

∫

50 dh = 20 − h

Correct

dh dx

B1 aef

h = 20 − x ⇒ x = 20 − h

∫

20 − x dx or x M1 20 − x ±λ dx 20 − (20 − x) where λ is a constant ±λ

50 . − 2(20 − x) dx x

= 100

∫

∫

∫

x − 20 dx x

20 ⎞ ⎛ = 100 ⎜ 1 − ⎟ dx x ⎠ ⎝

∫

± α x ± β ln x ; α , β ≠ 0 100 x − 2000ln x

= 100 ( x − 20ln x ) ( + c )

M1 A1

change limits: when h = 0 then x = 20 and when h = 100 then x = 10

∫

100

0

or

∫

100

0

50 10 dh = [100 x − 2000ln x ] 20 20 − h 50

(

)

(

)

100

dh = ⎡100 20 − h − 2000ln 20 − h ⎤ ⎣ ⎦0 20 − h

= (1000 − 2000ln10 ) − ( 2000 − 2000ln 20 ) = 2000ln 20 − 2000ln10 − 1000 = 2000ln 2 − 1000

Correct use of limits, ie. putting them in the correct way round Either x = 10 and x = 20 ddM1 or h = 100 and h = 0 Combining logs to give...

2000ln 2 − 1000 A1 aef or −2000ln ( 12 ) − 1000 [6]

(e)

Time required = 2000ln 2 − 1000 = 386.2943611... sec = 386 seconds (nearest second) = 6 minutes and 26 seconds (nearest second)

6 minutes, 26 seconds B1 [1] 13 marks

21