Edexcel Gce Core Mathematics C4 6666 Advanced Subsidiary Jun 2007 Mark Scheme

Mark Scheme (Results) Summer 2007 GCE

GCE Mathematics Core Mathematics C4 (6666)

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

June 2007 6666 Core Mathematics C4 Mark Scheme Question Number

Scheme

Marks

** represents a constant

f( x ) = (3 + 2 x )

1. (a)

=

1 27

−3

= (3)

−3

2x ⎞ ⎛ ⎜1 + 3 ⎟ ⎝ ⎠

−3

1 ⎛ 2x ⎞ = 1+ ⎜ 27 ⎝ 3 ⎟⎠

Takes 3 outside the bracket 1 to give any of (3)−3 or 27 .

−3

See note below.

⎧ ⎫ ( −3)( −4) ( −3)( −4)( −5) (* * x)2 + (* * x)3 + ...⎬ ⎨1 + ( −3)(* * x); + 2! 3! ⎭ ⎩

with * * ≠ 1

B1

Expands (1 + * * x )−3 to give a simplified or an unsimplified 1 + ( −3)(* * x) ;

M1;

A correct simplified or an un-simplified

{..........} expansion with

A1

candidate’s followed thro’

(* * x )

=

1 27

⎧ ⎫ ( −3)( −4) 2x 2 ( −3)( −4)( −5) 2x 3 2x (3) + ( 3 ) + ...⎬ ⎨1 + ( −3)( 3 ) + 2! 3! ⎭ ⎩

=

1 27

⎧ ⎫ 8x 2 80 3 − x + ...⎬ ⎨1 − 2x + 3 27 ⎩ ⎭

1 2x 8x 2 80x 3 = − ;+ − + ... 27 27 81 729

Anything that cancels to 1 − 2x ; 27

Simplified

8x 2 81

−

A1;

27

80x3 729

A1 [5] 5 marks

Note: You would award: B1M1A0 for =

1 27

⎫ ( −3)( −4) ( −3)( −4)( −5) ⎪⎧ 2x (2x)2 + (2x)3 + ...⎬ ⎨1 + ( −3)( 3 ) + 2! 3! ⎭ ⎩⎪

Special Case: If you see the constant 271 in a candidate’s final binomial expression, then you can award B1

because * * is not consistent.

6666/01 Core Maths C4 2 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number Aliter 1. Way 2

Scheme

Marks

f(x) = (3 + 2x)−3 1 27

or (3)−3 (See note ↓ )

B1

−3

⎧ −3 ⎫ ( −3)( −4) −5 −4 (3) (* * x)2 ⎪ ⎪(3) + ( −3)(3) (* * x); + ⎪ ⎪ 2! =⎨ ⎬ ( −3)( −4)( −5) −6 3 ⎪ ⎪ (3) (* * x) + ... + ⎪⎩ ⎪⎭ 3! with * * ≠ 1

Expands (3 + 2x) to give an un-simplified or simplified (3)−3 + ( −3)(3)−4 (* * x) ;

M1

A correct un-simplified or simplified

{..........} expansion with

A1

candidate’s followed thro’ ( * * x )

⎧ −3 ⎫ ( −3)( −4) −5 −4 (3) (2x)2 ⎪ ⎪(3) + ( −3)(3) (2x); + ⎪ ⎪ 2! =⎨ ⎬ ( 3)( 4)( 5) − − − ⎪ ⎪ (3)−6 (2x)3 + ... + ⎪⎩ ⎪⎭ 3! 1 1 ⎧⎪ 271 + ( −3)( 81 )(2x); + (6)( 243 )(4x 2 )⎫⎪ =⎨ ⎬ 1 + ( −10)( 729 )(8x 3 ) + ... ⎪⎩ ⎪⎭

1 2x 8x 2 80x 3 = − ;+ − + ... 27 27 81 729

Anything that cancels to 1 − 2x ; 27

Simplified

8x 2 81

−

A1;

27

80x3 729

A1 [5] 5 marks

Attempts using Maclaurin expansions need to be escalated up to your team leader. If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader.

Special Case: If you see the constant 271 in a candidate’s final binomial expression, then you can award B1

6666/01 Core Maths C4 3 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme 1

2.

∫ 0

2x dx , with substitution u = 2 x 2 x (2 + 1)

du = 2 x.ln 2 dx

∫

Marks

⇒

dx 1 = x du 2 .ln 2

2x ⎛ 1 ⎞ dx = ⎜ ⎟ x 2 (2 + 1) ⎝ ln 2 ⎠

du dx

= 2x.ln 2 or or

∫ (u + 1) 1

2

k

du

du dx

= u.ln 2

1 u

du dx

( )

= ln 2

∫ (u + 1) 1

2

du

B1

M1 ∗

where k is constant

(u + 1)−2 →

⎛ 1 ⎞ ⎛ −1 ⎞ =⎜ ⎟+c ⎟⎜ ⎝ ln 2 ⎠ ⎝ (u + 1) ⎠

a(u + 1)−1

M1

(u + 1)−2 → − 1.(u + 1)−1

A1

change limits: when x = 0 & x = 1 then u = 1 & u = 2 1

∫ 0

2

2x 1 ⎡ −1 ⎤ dx = x 2 ln 2 ⎢⎣ (u + 1) ⎥⎦ 1 (2 + 1)

⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎢⎜ − 3 ⎟ − ⎜ − 2 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎣⎝

=

1 ln 2

=

1 6 ln 2

1 6ln2

or

1 ln4

Correct use of limits u = 1 and u = 2

depM1 ∗

−

A1 aef

1 ln8

or

1 2ln2

−

1 3n2

Exact value only!

[6]

Alternatively candidate can revert back to x … 1

∫ 0

1

2x 1 ⎡ −1 ⎤ dx = x 2 ln 2 ⎢⎣ (2 x + 1) ⎥⎦ 0 (2 + 1)

⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎢⎜ − 3 ⎟ − ⎜ − 2 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎣⎝

=

1 ln 2

=

1 6 ln 2

1 6ln2

or

1 ln4

Correct use of limits x = 0 and x = 1

depM1 ∗

−

A1 aef

1 ln8

or

1 2ln2

−

1 3ln2

Exact value only! 6 marks

If you see this integration applied anywhere in a candidate’s working then you can award M1, A1

There are other acceptable 1 or answers for A1, eg: 2ln8

1 ln64

NB: Use your calculator to check eg. 0.240449…

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25th June 2007 Version 8: THE FINAL VERSION

Question Number

3. (a)

Scheme

Marks

⇒ ddux = 1 ⎧⎪u = x ⎫⎪ ⎨ dv ⎬ 1 ⎩⎪ dx = cos 2 x ⇒ v = 2 sin 2 x ⎭⎪ (see note below)

Int =

∫

x cos 2 x dx =

=

1 2

1 2

x sin 2 x −

x sin 2x −

1 2

(−

1 2

∫

1 2

sin 2 x.1 dx

cos 2x ) + c

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

M1 A1

sin 2 x → − 21 cos 2 x or sin kx → − k1 cos kx

dM1

with k ≠ 1 , k > 0

=

1 2

x sin 2 x +

1 4

cos 2 x + c

Correct expression with +c

A1 [4]

(b)

∫

∫ (

x cos2 x dx =

x

cos2 x + 1 2

∫

Substitutes correctly for cos2 x in the given integral

) dx

∫

=

1 1 x cos 2 x dx + x dx 2 2

=

1⎛ 1 1 ⎞ 1 x sin 2 x + cos 2 x ⎟ ; + x dx ⎜ 2⎝2 4 ⎠ 2

=

M1

∫

1 1 1 x sin 2 x + cos 2 x + x 2 ( + c ) 4 8 4

1 ( their answer to (a)) ; 2

A1;

or underlined expression Completely correct expression with/without +c

A1 [3] 7 marks

Notes: (b)

Int =

∫ x cos 2x dx

=

1 2

x sin 2 x ±

∫

1 2

sin 2 x.1 dx

This is acceptable for M1

M1

= λ x sin 2 x ± λ sin 2 x.1 dx

This is also acceptable for M1

M1

⇒ ddux = 1 ⎪⎧u = x ⎪⎫ ⎨ dv ⎬ ⎩⎪ dx = cos 2 x ⇒ v = λ sin 2 x ⎭⎪ Int =

∫ x cos 2x dx

∫

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25th June 2007 Version 8: THE FINAL VERSION

Question Number Aliter 3. (b) Way 2

Scheme

∫

∫ (

x cos2 x dx =

x

cos2 x + 1 2

Marks

Substitutes correctly for cos2 x in the given integral …

) dx

M1

⇒ ddux = 1 ⎧⎪u = x ⎨ dv 1 1 1 ⎩⎪ dx = 2 cos 2 x + 2 ⇒ v = 4 sin 2 x +

∫(

1 4

x sin 2 x +

1 2

x2 −

=

1 4

x sin 2 x +

1 2

x 2 + 81 cos 2 x − 41 x 2 + c

1 2

… or u = x and

dv dx

= 21 cos 2 x +

1 2

x ) dx

=

1 4

sin 2 x +

⎫⎪ 1 ⎬ x⎪ 2 ⎭

1 ( their answer to (a)) ; 2

A1

or underlined expression

1 1 1 x sin 2 x + cos 2 x + x 2 ( + c ) 4 8 4

=

Completely correct expression with/without +c

A1 [3]

Aliter (b) Way 3

∫

x cos 2 x dx =

∫

∫ (

⇒ 2 x cos2 x dx −

⇒

∫

Substitutes correctly

)

x 2 cos2 x − 1 dx

∫ x dx =

1 2

x sin 2 x +

for cos 2x in

1 4

∫

=

M1

cos 2 x + c

1⎛ 1 1 ⎞ 1 x cos x dx = ⎜ x sin 2 x + cos 2 x ⎟ ; + x dx 2⎝2 4 ⎠ 2 2

∫ x cos 2x dx

1 1 1 x sin 2 x + cos 2 x + x 2 ( + c ) 4 8 4

1 ( their answer to (a)) ; 2

A1;

or underlined expression Completely correct expression with/without +c

A1 [3] 7 marks

6666/01 Core Maths C4 6 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number 4. (a) Way 1

Scheme

Marks

A method of long division gives,

2(4 x 2 + 1) 4 ≡ 2+ (2x + 1)(2 x − 1) (2 x + 1)(2x − 1)

A=2

B1

Forming any one of these two identities. Can be implied.

M1

4 B C ≡ + (2 x + 1)(2 x − 1) (2 x + 1) (2 x − 1)

4 ≡ B(2 x − 1) + C (2 x + 1) or their remainder, Dx + E ≡ B(2 x − 1) + C (2 x + 1)

Let x = − 21 , Let x = 21 ,

4 = − 2B

4 = 2C

⇒ B = −2

⇒ C =2

See note below either one of B = − 2 or C = 2 both B and C correct

A1 A1 [4]

Aliter 4. (a)

2(4 x 2 + 1) B C ≡ A+ + (2 x + 1)(2 x − 1) (2 x + 1) (2 x − 1)

Way 2 See below for the award of B1

2(4 x 2 + 1) ≡ A(2 x + 1)(2 x − 1) + B(2 x − 1) + C (2 x + 1) Equate x2, Let x = − 21 , Let x = 21 ,

8 = 4A

B1

Forming this identity. Can be implied.

M1

⇒ A=2

4 = − 2B

4 = 2C

decide to award B1 here!! … … for A = 2

⇒ B = −2

⇒ C =2

See note below either one of B = − 2 or C = 2

A1

both B and C correct

A1 [4]

If a candidate states one of either B or C correctly then the method mark M1 can be implied.

6666/01 Core Maths C4 7 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number 4. (b)

Scheme

∫

2(4 x 2 + 1) dx = (2 x + 1)(2x − 1)

Marks

∫ 2 − (2x + 1) + (2x − 1) dx 2

2

= 2 x − 22 ln(2 x + 1) + 22 ln(2 x − 1) ( + c )

Either p ln(2 x + 1) or q ln(2 x − 1) or either p ln 2 x + 1 or q ln 2 x − 1

A → Ax − ln(2 x + 1) + ln(2 x − 1) or − ln(2 x + 1) + ln(2 x − 1) 2 2

2 2

M1 ∗ B1 A1 cso & aef

See note below. 2

∫ 1

2(4 x 2 + 1) 2 dx = [ 2 x − ln(2 x + 1) + ln(2 x − 1)] 1 (2 x + 1)(2 x − 1)

= ( 4 − ln 5 + ln 3 ) − ( 2 − ln 3 + ln1)

Substitutes limits of 2 and 1 and subtracts the correct way round. (Invisible brackets okay.)

depM1 ∗

Use of correct product (or power) and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical expression.

M1

⎛9⎞ 2 + ln ⎜ ⎟ ⎝5⎠

A1

= 2 + ln 3 + ln 3 − ln 5

⎛ 3(3) ⎞ = 2 + ln ⎜ ⎟ ⎝ 5 ⎠

⎛9⎞ = 2 + ln ⎜ ⎟ ⎝5⎠

Or 2 − ln ( 59 ) and k stated as

9 5

[6]

.

10 marks

Some candidates may find rational values for B and C. They may combine the denominator of their B or C with (2x +1) or (2x – 1). Hence: Either b(2ax −1) → k ln(b(2 x − 1)) or a b (2 x +1)

→ k ln(b(2 x + 1)) is okay for M1.

Candidates are not allowed to fluke − ln(2 x + 1) + ln(2 x − 1) for A1. Hence cso. If they do fluke this, however, they can gain the final A1 mark for this part of the question.

To award this M1 mark, the candidate must use the appropriate law(s) of logarithms for their ln terms to give a one single logarithmic term. Any error in applying the laws of logarithms would then earn M0.

Note: This is not a dependent method mark.

6666/01 Core Maths C4 8 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number 5. (a)

Scheme

Marks

If l1 and l2 intersect then:

⎛1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + λ ⎜ 1⎟ = ⎜ 3 ⎟ + µ ⎜ 1 ⎟ ⎜ −1⎟ ⎜0⎟ ⎜6⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i : 1 + λ = 1 + 2µ λ=3+ µ Any two of j : k: −1 = 6 − µ

(1) (2) (3)

(1) & (2) yields λ = 6, µ = 3 (1) & (3) yields λ = 14, µ =7 (2) & (3) yields λ = 10, µ =7 checking eqn (3), Either checking eqn (2), checking eqn (1),

-1 ≠ 3 14 ≠ 10 11 ≠ 15

Writes down any two of these equations correctly.

M1

Solves two of the above equations to find … either one of λ or µ correct

A1

both λ and µ correct

A1

Complete method of putting their values of λ and µ into a third equation to show a contradiction.

B1

or for example: this type of explanation is also allowed

checking eqn (3), LHS = -1 , RHS = 3 ⇒ Lines l1 and l2 do not intersect

for B1

. [4]

Aliter 5. (a) Way 2

k : −1 = 6 − µ ⇒

i: j:

1 + λ = 1 + 2µ λ=3+ µ

i: j:

λ = 14 λ = 10

µ=7

⇒ 1 + λ = 1 + 2(7) ⇒ λ = 3 + (7)

Either: These equations are then inconsistent Or: 14 ≠ 10 Or: Lines l1 and l2 do not intersect

Uses the k component to find µ and substitutes their value of µ into either one of the i or j component.

M1

either one of the λ ’s correct

A1

both of the λ ’s correct

A1

Complete method giving rise to any one of these three explanations.

B1 [4]

6666/01 Core Maths C4 9 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number Aliter 5. (a) Way 3

Scheme

Marks

If l1 and l2 intersect then:

⎛1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + λ ⎜ 1⎟ = ⎜ 3 ⎟ + µ ⎜ 1 ⎟ ⎜ −1⎟ ⎜0⎟ ⎜6⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i : 1 + λ = 1 + 2µ λ=3+ µ Any two of j : −1 = 6 − µ k:

(1) (2) (3)

Writes down any two of these equations

M1

either one of the µ ’s correct

A1

both of the µ ’s correct

A1

Complete method giving rise to any one of these three explanations.

B1

(1) & (2) yields µ = 3 (3) yields µ = 7 Either: These equations are then inconsistent Or: 3≠ 7 Or: Lines l1 and l2 do not intersect

[4] Aliter 5. (a) Way 4

i : 1 + λ = 1 + 2µ λ=3+ µ Any two of j : −1 = 6 − µ k:

(1) (2) (3)

Writes down any two of these equations

M1

µ =3

A1

RHS of (3) = 3

A1

Complete method giving rise to this explanation.

B1

(1) & (2) yields µ = 3 (3) RHS = 6 − 3 = 3 (3) yields −1≠ 3

[4]

6666/01 Core Maths C4 10 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

Marks Only one of either

5. (b)

⎛2⎞ uuur ⎜ ⎟ λ = 1 ⇒ OA = ⎜ 1 ⎟ & ⎜ −1⎟ ⎝ ⎠

⎛2⎞ ⎛5⎞ uuur ⎜ ⎟ uuur ⎜ ⎟ OA = ⎜ 1 ⎟ or OB = ⎜ 5 ⎟ or B1 ⎜ ⎟ ⎜ 4⎟ ⎝ −1⎠ ⎝ ⎠ A(2,1, − 1) or B(5,5, 4) .

⎛5⎞ uuur ⎜ ⎟ µ = 2 ⇒ OB = ⎜ 5 ⎟ ⎜ 4⎟ ⎝ ⎠

(can be implied)

⎛ −3 ⎞ ⎛5⎞ ⎛ 2 ⎞ ⎛3⎞ uuur ⎜ ⎟ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = OB − OA = ⎜ 5 ⎟ − ⎜ 1 ⎟ = ⎜ 4 ⎟ or BA = ⎜ −4 ⎟ ⎜ ⎟ ⎜ 4 ⎟ ⎜ −1⎟ ⎜ 5 ⎟ ⎝ −5 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Finding the difference between their

uuur uuur OB and OA . M1

(can be implied)

Applying the dot product formula between “allowable” vectors. See notes below.

uuur AB = 3 i + 4 j + 5k , d 1 = i + j + 0k & θ is angle

uuur AB • d 1 ⎛3 + 4 + 0⎞ = ± ⎜⎜ uuur ⎟⎟ AB . d 1 ⎝ 50 . 2 ⎠

cos θ =

cos θ =

M1

Applies dot product formula between

uuur d 1 and their ± AB.

M1

Correct expression.

A1

7 10

7 10

or 0.7 or

7 100

but not

7 50 2

A1 cao [6] 10 marks

Candidates can score this mark if there is a complete method for finding the dot product between their vectors in the following cases: uuur Case 1: their ft ± AB = ± ( 3 i + 4 j + 5k )

Case 2: d 1 = i + j + 0k

Case 3: d 1 = i + j + 0k

and d 1 = i + j + 0k

and d 2 = 2i + j − 1k

and d 2 = 2(2i + j − 1k )

⎛3 + 4 + 0⎞ ⇒ cos θ = ± ⎜⎜ ⎟⎟ ⎝ 50 . 2 ⎠

⇒ cos θ =

uuur Case 4: their ft ± AB = ± ( 3 i + 4 j + 5k ) and d 2 = 2i + j − k

uuur Case 5: their ft OA = 2 i + 1j − 1k uuur and their ft OB = 5 i + 5 j + 4k

⎛6 + 4 − 5⎞ ⇒ cos θ = ± ⎜⎜ ⎟⎟ ⎝ 50 . 6 ⎠

⎛ 10 + 5 − 4 ⎞ ⇒ cos θ = ± ⎜⎜ ⎟⎟ ⎝ 6 . 66 ⎠

2 + 1+ 0

⇒ cos θ =

2. 6

4+2+0 2 . 24

Note: If candidate use cases 2, 3, 4 and 5 they cannot gain the final three marks for this part. Note: Candidate can only gain some/all of the final three marks if they use case 1.

6666/01 Core Maths C4 11 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Examples of awarding of marks M1M1A1 in 5.(b) Example

50 . 2 cos θ = ± ( 3 + 4 + 0 )

Marks M1M1A1 (Case 1)

2 . 6 cos θ = 3

M1M0A0 (Case 2)

2 . 24 cos θ = 4 + 2

M1M0A0 (Case 3)

6666/01 Core Maths C4 12 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

y = sin t

x = tan2 t ,

6. (a)

Marks

dx dy = 2(tan t ) sec 2 t , = cos t dt dt

Correct

± cos t their ddxt + cos t their ddxt

⎛ cos t ⎞ ⎜= ⎟ ⎝ 2 sin t ⎠ 4

dy cos t ∴ = dx 2 tan t sec 2 t

dx dy and dt dt

B1

M1 A1 [3]

(b)

x =1, y =

When t = π4 ,

1 2

(

The point 1,

(need values)

1 2

) or (1, awrt 0.71)

B1, B1

These coordinates can be implied. ( y = sin ( π4 ) is not sufficient for B1)

π

When t =

4

1 2

=

2.(1)

( )

T: y =

1 4 2

=

2.(1)

1 2

1 2

1 2

2

=

1 4 2

cos π4 dy = dx 2 tan π4 sec 2

1 2

=

1

T: y −

or

, m(T) =

1 4 2

x+

1 2

1 2

2.(1)(2)

=

4

1

=

4 2

3 4 2

or

1 4 2

y=

⇒ c =

x+

3 4 2

2 8

any of the five underlined expressions or awrt 0.18

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

( x − 1)

(1) + c

Hence T: y =

() 1

=

π

1 2

or

2 8

x + 3 82

−

1 4 2

y=

Correct simplified EXACT equation of tangent

= 2 8

B1 aef

M1

aef

A1 aef cso

3 4 2

x+

3 2 8

[5]

Note: The x and y coordinates must be the right way round.

A candidate who incorrectly differentiates tan2 t to give dx = 2 sec 2 t or ddxt = sec 4 t is then able to fluke the dt correct answer in part (b). Such candidates can (b) B1B1B1M1A0 cso. potentially get: (a) B0M1A1 Note: cso means “correct solution only”. Note: part (a) not fully correct implies candidate can achieve a maximum of 4 out of 5 marks in part (b).

6666/01 Core Maths C4 13 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number 6. (c)

Scheme

x = tan2 t =

sin2 t cos2 t

Marks

y = sin t

Way 1

x=

sin2 t 1 − sin2 t

x=

y2 1− y 2

Uses cos2 t = 1 − sin2 t

Eliminates ‘t’ to write an equation involving x and y.

M1

M1

x (1 − y 2 ) = y 2 ⇒ x − xy 2 = y 2 x = y 2 + xy 2

y2 =

⇒

x = y 2 (1 + x )

Rearranging and factorising with an attempt to make y 2 the subject.

x 1+ x

x 1+ x

ddM1

A1 [4]

Aliter 6. (c) Way 2

1 + cot 2 t = co sec 2 t

=

Hence,

1+

Uses 1 + cot 2 t = co sec 2 t

1 sin2 t

Uses cos ec 2 t =

1 1 = 2 x y

Hence, y 2 = 1 −

1 (1 + x )

1 sin2 t

Eliminates ‘t’ to write an equation involving x and y.

or

x 1+ x

1−

1 (1 + x )

or

x 1+ x

M1

M1 implied

ddM1

A1 [4]

1 is an acceptable response for the final accuracy A1 mark. 1+ x1

6666/01 Core Maths C4 14 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number Aliter 6. (c)

Scheme

x = tan2 t

Marks

y = sin t

Way 3

1 + tan2 t = sec 2 t

Hence,

=

1 cos2 t

=

1 1 − sin2 t

1+ x =

Hence, y 2 = 1 −

M1

1 cos2 t

M1

Uses sec 2 t =

1 1− y 2

1 (1 + x )

Uses 1 + tan2 t = sec 2 t

Eliminates ‘t’ to write an equation involving x and y.

or

x 1+ x

1−

1 (1 + x )

or

x 1+ x

ddM1

A1 [4]

Aliter 6. (c) Way 4

y 2 = sin2 t = 1 − cos2 t

= 1−

1 sec 2 t

= 1−

1 (1 + tan2 t )

Hence, y 2 = 1 −

1 (1 + x )

Uses sin2 t = 1 − cos2 t

M1

1 sec 2 t

M1

Uses cos2 t =

or

then uses sec 2 t = 1 + tan2 t

x 1+ x

1−

1 (1 + x )

or

x 1+ x

ddM1

A1 [4]

1 is an acceptable response for the final accuracy A1 mark. 1+ x1

6666/01 Core Maths C4 15 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number Aliter 6. (c)

Scheme

x = tan2 t

Marks

y = sin t

Way 5

x = tan2 t

⇒ tan t =

x

(1 + x )

Draws a right-angled triangle and places

x t

1

Hence, y = sin t =

Hence, y 2 =

x

1+ x

M1

both x and 1 on the triangle

Uses Pythagoras to deduce the hypotenuse

Eliminates ‘t’ to write an equation involving x and y.

x 1+ x

x 1+ x

M1

ddM1

A1 [4] 12 marks

1 is an acceptable response for the final accuracy A1 mark. 1+ x1 There are so many ways that a candidate can proceed with part (c). If a candidate produces a correct solution then please award all four marks. If they use a method commensurate with the five ways as detailed on the mark scheme then award the marks appropriately. If you are unsure of how to apply the scheme please escalate your response up to your team leader.

6666/01 Core Maths C4 16 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

7. (a)

Marks

x

0

π

π 8

3π 16

π

16

y

0

0.445995927…

0.643594252…

0.817421946…

1

4

Enter marks into ePEN in the correct order.

0.446 or awrt 0.44600 awrt 0.64359 awrt 0.81742

B1 B1 B1 [3]

0 can be implied Outside brackets π 1 × 16π or 32 2

1 π Area ≈ × ; × 0 + 2 ( 0.44600 + 0.64359 + 0.81742 ) + 1} 2 16

{

(b) Way 1

=

B1

For structure of trapezium

{

}

rule ............. ; Correct expression inside brackets which all must be multiplied by h2 .

π × 4.81402... = 0.472615308... = 0.4726 (4dp) 32

for seeing 0.4726

M1

A1

A1 cao [4]

Area ≈ Aliter (b) Way 2

π 16

×

{

0 + 0.44600 2

+

0.44600 + 0.64359 2

+

0.64359 + 0.81742 2

+

}

0.81742 + 1 2

which is equivalent to:

Area ≈

=

1 π × ; × 0 + 2 ( 0.44600 + 0.64359 + 0.81742 ) + 1} 2 16

{

π 16

and a divisor of 2 on all

terms inside brackets. One of first and last ordinates, two of the middle ordinates inside brackets ignoring the 2. Correct expression inside brackets if 21 was to be factorised out.

π × 2.40701... = 0.472615308... = 0.4726 16

0.4726

B1

M1

A1

A1 cao [4]

Area =

1 2

×

π 20

In (a) for x =

× {0 + 2(0.44600 + 0.64359 + 0.81742) + 1} = 0.3781, gains B0M1A1A0 π 16

writing 0.4459959… then 0.45600 gains B1 for awrt 0.44600 even though 0.45600 is incorrect.

In (b) you can follow though a candidate’s values from part (a) to award M1 ft, A1 ft

Question Number

Scheme

6666/01 Core Maths C4 17 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

Marks

25th June 2007 Version 8: THE FINAL VERSION

π

7. (c)

Volume = (π )

∫(

π

∫( 4

tan x

)

2

4

dx =

0

(π ) ∫ tan x dx π

= (π ) ⎡⎣( − ln cos 4 ) − ( ln cos 0 ) ⎤⎦

or

= π ⎡ − ln ⎣

1

1 1

1 2

∫ tan x dx

ln(sec 0) = 0 may be

π

( ) − ln ( )⎤⎥⎦ = π ⎡⎣ln

dx or

The correct use of limits on a function other than tan x; ie x = π4 ‘minus’ x = 0 .

= (π ) ⎡⎣( ln sec π4 ) − ( ln sec 0 ) ⎤⎦

= π ⎡ln ⎢⎣

2

tan x → ln sec x or tan x → − ln cos x

or = (π ) [ − ln cos x ] 04

π

)

Can be implied. Ignore limits and (π )

0

= (π ) [ln sec x ] 04

or

tan x

M1

A1

dM1

implied. Ignore (π )

2 − ln1⎤ ⎦

( ) − ln (1)⎤⎦ 1 2

π ln 2 or π ln = π ln 2 or π ln

2 2

or

1 2

π ln 2 or −π ln

( ) or 1 2

π 2

ln ( 21 )

or

1 2

π ln 2 or −π ln or

π 2

2 2

( ) 1 2

A1 aef

ln ( 21 )

must be exact.

[4] 11 marks

If a candidate gives the correct exact answer and then writes 1.088779…, then such a candidate can be awarded A1 (aef). The subsequent working would then be ignored. (isw) Beware: In part (c) the factor of π is not needed for the first three marks. Beware: In part (b) a candidate can also add up individual trapezia in this way:

(

)

Area ≈ 21 . 16π ( 0 + 0.44600 ) + 21 . 16π ( 0.44600 + 0.64359 ) + 21 . 16π 0.64359 + 0.81742 + 21 . 16π ( 0.81742 + 1)

6666/01 Core Maths C4 18 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

dP = kP dt

8. (a)

and

Marks

t = 0, P = P0 (1) Separates the variables with

∫

dP = P

∫

∫P

dP

k dt

and

∫ k dt on either

M1

side with integral signs not necessary.

ln P = kt ; ( + c )

Must see ln P and kt ; Correct equation with/without + c.

A1

When t = 0, P = P0 ⇒ ln P0 = c

Use of boundary condition (1) to attempt to find the constant of integration.

M1

( or

P = Ae kt ⇒ P0 = A

)

ln P = kt + ln P0 ⇒ eln P = e kt + ln P0 = e kt .eln P0 Hence, P = P0 e kt

P = P0 e kt

A1 [4]

(b)

P = 2P0 & k = 2.5 ⇒ 2P0 = P0 e 2.5 t e 2.5t = 2 ⇒ ln e 2.5t = ln 2 or 2.5t = ln 2 …or e

kt

⇒ t =

= 2 ⇒ ln e 1 2.5

kt

= ln 2 or kt = ln 2

Substitutes P = 2P0 into an expression involving P Eliminates P0 and takes ln of both sides

M1

M1

ln 2 = 0.277258872... days

t = 0.277258872... × 24 × 60 = 399.252776... minutes t = 399min or

t = 6 hr 39 mins (to nearest minute)

awrt t = 399 or

6 hr 39 mins

A1 [3]

P = P0 e kt written down without the first M1 mark given scores all four marks in part (a).

6666/01 Core Maths C4 19 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

dP = λP cos λ t dt

8. (c)

and

Marks

t = 0, P = P0 (1) Separates the variables with

∫

dP = P

∫

∫P

dP

λ cos λ t dt

∫ λ cos λt dt on

M1

either side with integral signs not necessary.

ln P = sin λt ; ( + c ) When t = 0, P = P0 ⇒ ln P0 = c

( or

and

P = Ae sin λt ⇒ P0 = A

ln P = sin λ t + ln P0

)

Must see ln P and sin λ t ; Correct equation with/without + c.

A1

Use of boundary condition (1) to attempt to find the constant of integration.

M1

⇒ eln P = e sin λt + ln P0 = e sin λt .eln P0

Hence, P = P0 e sin λt

P = P0 e sin λt

A1 [4]

(d)

P = 2P0 & λ = 2.5

⇒ 2P0 = P0 e sin2.5 t

e sin2.5t = 2 ⇒ sin 2.5t = ln 2 …or … e λt = 2 ⇒ sin λ t = ln 2

t =

1 2.5

sin−1 ( ln 2 )

t = 0.306338477...

Eliminates P0 and makes

sin λ t or sin 2.5t the subject M1 by taking ln’s Then rearranges to make t the subject. (must use sin-1)

dM1

t = 0.306338477... × 24 × 60 = 441.1274082... minutes t = 441min or

t = 7 hr 21 mins (to nearest minute)

awrt t = 441 or

7 hr 21 mins

A1 [3] 14 marks

P = P0 e sin λt written down without the first M1 mark given scores all four marks in part (c).

6666/01 Core Maths C4 20 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

dP = kP dt

and

Marks

t = 0, P = P0 (1) Separates the variables with

Aliter 8. (a) Way 2

∫ 1 k

dP = kP

∫

∫ kP

dP

1 dt

Must see

1 k

M1

ln P and t ;

Correct equation with/without + c.

When t = 0, P = P0 ⇒

1 k

∫ dt on either side

with integral signs not necessary.

ln P = t ; ( + c )

( or

and

1 k

P = Ae kt ⇒ P0 = A

ln P0 = c

)

Use of boundary condition (1) to attempt to find the constant of integration.

A1

M1

ln P = t + k1 ln P0 ⇒ ln P = kt + ln P0

⇒ eln P = e kt + ln P0 = e kt .eln P0 Hence, P = P0 e kt

P = P0 e kt

A1 [4]

Separates the variables with Aliter 8. (a) Way 3

∫ 1 k

dP = kP

( or

∫ dt on either side

M1

with integral signs not necessary. Must see

ln ( kP ) = t ; ( + c )

When

1 k

∫

∫ 1 dt

dP and kP

1 k

ln ( kP ) and t ;

Correct equation with/without + c.

t = 0, P = P0 ⇒ k1 ln ( kP0 ) = c

kP = Ae kt ⇒ kP0 = A

)

Use of boundary condition (1) to attempt to find the constant of integration.

A1

M1

ln ( kP ) = t + k1 ln ( kP0 ) ⇒ ln ( kP ) = kt + ln ( kP0 )

⇒e

ln( kP )

=e

kt + ln( kP0 )

= e kt .e

ln( kP0 )

⇒ kP = e kt . ( kP0 ) ⇒ kP = kP0 e kt

( or

kP = kP0 e kt

)

Hence, P = P0 e kt

P = P0 e kt

A1 [4]

Question Number

Scheme

6666/01 Core Maths C4 21 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

Marks

25th June 2007 Version 8: THE FINAL VERSION

dP = λ P cos λ t dt

t = 0, P = P0 (1)

and

Separates the variables with Aliter 8. (c) Way 2

∫

dP = λP

λ ln P = 1

∫

1

λ

Must see

λ sin λ t ; ( + c ) 1

P = Ae

ln P =

dP

1

λ

M1

either side with integral signs not necessary.

When t = 0, P = P0 ⇒

( or

∫ λP and ∫ cos λt dt on

cos λ t dt

sin λ t

1

λ

ln P and

1

λ

sin λ t ;

Correct equation with/without + c. 1

λ

ln P0 = c

⇒ P0 = A

)

Use of boundary condition (1) to attempt to find the constant of integration.

A1

M1

sin λ t + λ1 ln P0 ⇒ ln P = sin λ t + ln P0

⇒ eln P = e sin λt + ln P0 = e sin λt .eln P0 Hence, P = P0 e sin λt

P = P0 e sin λt

A1 [4]

P = P0 e kt written down without the first M1 mark given scores all four marks in part (a). P = P0 e sin λt written down without the first M1 mark given scores all four marks in part (c).

6666/01 Core Maths C4 22 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION

Question Number

Scheme

dP = λ P cos λ t dt

Marks

t = 0, P = P0 (1)

and

Separates the variables with Aliter 8. (c) Way 3

∫

dP = λP

∫

∫ λP and ∫ cos λt dt on dP

cos λ t dt

Must see 1

λ

ln ( λ P ) =

M1

either side with integral signs not necessary.

1

λ

sin λ t ; ( + c )

1

λ

ln ( λ P ) and 1

λ

sin λ t ; A1

Correct equation with/without + c. When t = 0, P = P0 ⇒

( or 1

λ

1

λ

ln ( λ P0 ) = c

λ P = Ae sin λt ⇒ λP0 = A )

ln ( λ P ) =

1

λ

Use of boundary condition (1) to attempt to find the constant of integration.

M1

sin λ t + λ1 ln ( λ P0 )

⇒ ln ( λ P ) = sin λ t + ln ( λ P0 )

⇒e

ln( λ P )

=e

sin λ t + ln( λ P0 )

= e sin λt .e

ln( λ P0 )

⇒ λ P = e sin λt . ( λ P0 )

( or

λ P = λ P0 e sin λt )

Hence, P = P0 e sin λt

P = P0 e sin λt

A1 [4]

• 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∗ . ft denotes “follow through” cao denotes “correct answer only” aef denotes “any equivalent form”

6666/01 Core Maths C4 23 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics

25th June 2007 Version 8: THE FINAL VERSION