C4 MS Jun 2008 by The de Ferrers Academy

Mark Scheme (Results) Summer 2008

GCE

GCE Mathematics (6666/01)

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

June 2008 6666 Core Mathematics C4 Mark Scheme Question 1. (a)

Scheme x y

or y

Marks

0.4

0.8

1.2

1.6

0.08

0.32

0.72

1.28

e 1.08329 …

1.37713…

2.05443…

3.59664…

7.38906…

Either e0.32 and e1.28 or awrt 1.38 and 3.60 B1 (or a mixture of e’s and decimals) [1]

(b) Way 1

Outside brackets B1; 1 × 0.4 or 0.2 2 For structure of trapezium M1 rule [ ............. ] ;

1 Area ≈ × 0.4 ; ×⎡⎣ e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e2 ⎤⎦ 2

= 0.2 × 24.61203164... = 4.922406... = 4.922 (4sf) Aliter (b) Way 2

Area ≈ 0.4 × ⎡ e ⎣

+ e0.08 2

e0.08 + e0.32 2

e0.32 + e0.72 2

e0.72 + e1.28 2

4.922

e1.28 + e2 2

A1 cao [3]

⎤ 0.4 and a divisor of 2 on B1 ⎦ all terms inside brackets. One of first and last ordinates, two of the middle ordinates inside M1 brackets ignoring the 2.

which is equivalent to: 1 Area ≈ × 0.4 ; ×⎡⎣ e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e 2 ⎤⎦ 2

= 0.2 × 24.61203164... = 4.922406... = 4.922 (4sf)

4.922

A1 cao [3] 4 marks

1 Note an expression like Area ≈ × 0.4 + e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e 2 would score B1M1A0 2

Allow one term missing (slip!) in the

(

)

brackets for

The M1 mark for structure is for the material found in the curly brackets ie ⎡⎣ first y ordinate + 2 ( intermediate 2 ft y ordinate ) + final y ordinate ⎤⎦

Question Number 2. (a)

Scheme

Marks

⎧⎪u = x ⇒ ddux = 1⎫⎪ ⎨ dv x x⎬ ⎪⎩ dx = e ⇒ v = e ⎪⎭

∫ xe

Use of ‘integration by parts’ formula in the correct direction. M1 (See note.) Correct expression. (Ignore dx) A1

dx = x e x − ∫ e x .1 dx

= x e x − ∫ e x dx = x ex − ex ( + c )

Correct integration with/without + c A1 [3]

(b)

⎧⎪u = x 2 ⇒ ⎨ dv x ⎩⎪ dx = e ⇒

= 2 x ⎫⎪ ⎬ v = e x ⎭⎪

du dx

Use of ‘integration by parts’ formula M1 in the correct direction. Correct expression. (Ignore dx) A1

2 x 2 x x ∫ x e dx = x e − ∫ e .2 x dx

= x 2 e x − 2 ∫ x e x dx Correct expression including + c. (seen at any stage! in part (b)) A1 ISW You can ignore subsequent working.

= x2e x − 2 ( x e x − e x ) + c

⎧⎪= x e − 2 x e + 2e + c ⎫⎪ ⎨ x 2 ⎬ ⎩⎪= e ( x − 2 x + 2 ) + c ⎭⎪ 2 x

[3]

Ignore subsequent working 6 marks

Note integration by parts in the correct direction means that u and ddvx must be assigned/used as u = x and ddvx = e x in part (a) for example + c is not required in part (a). i i di t (b)

Question Number 3. (a)

Scheme

From question,

Marks dA = 0.032 seen B1 dt or implied from working.

dA = 0.032 dt

2π x by itself seen B1 or implied from working

dA ⎫ ⎧ 2 = ⎬ 2π x ⎨A = π x ⇒ dx ⎭ ⎩

dx dA dA = ÷ = dt dt dx When x = 2cm ,

Hence,

(b)

( 0.032 )

⎧ 0.016 ⎫ ; ⎨= ⎬ 2π x ⎩ π x ⎭ 1

0.032 ÷ Candidate's

dx 0.016 = dt 2π

dx = 0.002546479... (cm s-1) dt

awrt 0.00255 A1 cso [4]

V = π x 2 (5 x) = 5 π x3

V = π x 2 (5 x) or 5π x3

dV = 15 π x 2 dx B1 or ft from candidate’s V in one variable

dV = 15π x 2 dx ⎛ 0.016 ⎞ dV dV d x = × = 15π x 2 . ⎜ ⎟ ; {= 0.24 x} dt dx dt ⎝ πx ⎠ When x = 2cm ,

dA ; M1; dx

Candidate’s

dV = 0.24(2) = 0.48 (cm3 s −1 ) dt

dV dx × ; M1 dx dt

0.48 or awrt 0.48 A1 cso [4] 8 marks

Question Number

Marks

Scheme 3 x 2 − y 2 + xy = 4

4. (a)

( eqn ∗ )

Differentiates implicitly to include either

± ky ddyx or x ddxy . (Ignore

⎧ dy ⎫ dy ⎛ dy ⎞ = ⎬ 6x − 2 y + ⎜y + x ⎟= 0 ⎨ dx ⎜⎝ dx ⎟⎠ ⎩ dx ⎭

⎧ dy −6 x − y ⎫ = ⎨ ⎬ or x − 2y ⎭ ⎩ dx

Correct application

( 3x

⎧ dy 6x + y ⎫ = ⎨ ⎬ 2y − x ⎭ ⎩ dx

( )

(

dy dx

)

= )

of product rule B1

⎛ dy ⎞ − y2 ) → ⎜ 6x − 2 y ⎟ and dx ⎠ ⎝

( 4 → 0)

not necessarily required.

dy 8 −6 x − y 8 = ⇒ = dx 3 x − 2y 3

Substituting

dy 8 = into their M1 ∗ dx 3 equation.

giving −18 x − 3 y = 8 x − 16 y

giving

Attempt to combine either terms in x or terms in y together to give either dM1 ∗ ax or by.

13 y = 26 x

simplifying to give y − 2 x = 0 AG A1 cso

Hence, y = 2 x ⇒ y − 2 x = 0

[6] (b)

At P & Q, y = 2 x . Substituting into eqn ∗ gives 3 x 2 − (2 x) 2 + x(2 x) = 4 Simplifying gives, x 2 = 4 ⇒ x = ± 2

Attempt replacing y by 2x in at least one of the y terms in eqn ∗

Either x = 2 or x = −2

Both (2, 4) and (−2, − 4)

y = 2x ⇒ y = ± 4 Hence coordinates are (2, 4) and (−2, − 4)

[3] 9 marks

Question Number

Scheme

Marks

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

1 3x ⎞ −1 ⎛ −1 = (4 − 3 x) 2 = ( 4 ) 2 ⎜1 − ⎟ 4 ⎠ (4 − 3 x) ⎝

5. (a)

− 12

1 ⎛ 3x ⎞ = ⎜1 − ⎟ 2⎝ 4 ⎠

− 12

(4)

− 12

1 2

outside brackets B1

−1

Expands (1 + ** x) 2 to give a simplified or an un-simplified M1; 1 + (− 12 )(** x) ; (− 1 )( − 23 ) ⎡ ⎤ = 12 ⎢ 1 + (− 12 )(** x); + 2 (** x) 2 + ... ⎥ 2! ⎣ ⎦ with ** ≠ 1

(− 1 )(− 23 ) 3 x 2 ⎤ 1⎡ 1 + (− 12 )(− 34x ) + 2 (− 4 ) + ... ⎥ ⎢ 2⎣ 2! ⎦

A correct simplified or an unsimplified [ .......... ] expansion with candidate’s followed A1 through (** x )

Award SC M1 if you see (− 12 )(** x) +

1 2

= 12 ⎡⎣ 1 + 83 x ; +

27 128

x 2 + ... ⎤⎦

3 27 2 ⎧ 1 ⎫ x; + x + ...⎬ ⎨= + 256 ⎩ 2 16 ⎭

(b)

1 2

+ 4 + 32 x + = 4 + 2 x; +

3 16 27 32

27 128

x 2 + ... ⎤⎦

27 2 ⎡⎣ ........; 128 x ⎤⎦

A1 isw A1 isw

Ignore subsequent working [5]

Writing ( x + 8) multiplied by candidate’s part (a) M1 expansion.

3 27 2 ⎛1 ⎞ ( x + 8) ⎜ + x + x + ... ⎟ 2 16 256 ⎝ ⎠ =

3 ⎣⎡ 1 + 8 x ; ... ⎦⎤

SC: K ⎡⎣ 1 + 83 x + 1 2

(− 12 )(− 23 ) (** x) 2 2!

Multiply out brackets to find a constant term, two x terms and two x 2 terms.

x 2 + ..... x 2 + .....

Anything that cancels to 33 2 A1; A1 4 + 2 x; x 32

33 2 x + ... 32

[4] 9 marks

Question Number 6. (a)

Scheme

Marks

Lines meet where:

⎛ −9 ⎞ ⎛2⎞ ⎛3⎞ ⎛3⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ + λ ⎜ 1 ⎟ = ⎜ 1 ⎟ + µ ⎜ −1⎟ ⎜ 10 ⎟ ⎜ −1⎟ ⎜ 17 ⎟ ⎜5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Any two of

i : − 9 + 2λ = 3 + 3µ j: λ =1 − µ k : 10 − λ = 17 + 5µ

(1) – 2(2) gives:

(2) gives:

⎛ −9 ⎞ ⎛2⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 0 ⎟ + 3⎜ 1 ⎟ ⎜ 10 ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠

−9 = 1 + 5µ

(1) (2) (3)

Need any two of these correct equations seen anywhere in part M1 (a). Attempts to solve simultaneous equations to find one of dM1 either λ or µ

⇒ µ = −2

Both λ = 3 & µ = − 2

λ =1−− 2 = 3

Substitutes their value of either λ or µ into the line l1 or l2 respectively. This mark can be ddM1 implied by any two correct components of ( −3, 3, 7 ) .

⎛3⎞ ⎛3⎞ ⎜ ⎟ ⎜ ⎟ or r = ⎜ 1 ⎟ − 2 ⎜ −1⎟ ⎜ 17 ⎟ ⎜5⎟ ⎝ ⎠ ⎝ ⎠

⎛ −3 ⎞ ⎜ ⎟ ⎜ 3 ⎟ or −3i + 3 j + 7k ⎜7⎟ ⎝ ⎠

⎛ −3 ⎞ ⎜ ⎟ Intersect at r = ⎜ 3 ⎟ or r = −3i + 3 j + 7k ⎜7⎟ ⎝ ⎠

Either check k: λ = 3 : LHS = 10 − λ

Either check that λ = 3 , µ = − 2 in a third equation or check that λ = 3 , B1 µ = − 2 give the same coordinates on the other line. Conclusion not needed.

= 10 − 3 = 7

(As LHS = RHS then the lines intersect.)

d1 = 2i + j − k ,

or ( −3, 3, 7 )

µ = − 2 : RHS = 17 + 5µ = 17 − 10 = 7

(b)

[6]

d 2 = 3i − j + 5k

⎛2⎞ ⎛3⎞ ⎜ ⎟ ⎜ ⎟ As d1 • d 2 = ⎜ 1 ⎟ • ⎜ −1⎟ = (2 × 3) + (1 ×− 1) + (−1 × 5) = 0 ⎜ −1⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ Then l1 is perpendicular to l2.

Dot product calculation between the two direction vectors: (2 × 3) + (1 ×− 1) + (−1 × 5) M1 or 6 − 1 − 5 Result ‘=0’ and A1 appropriate conclusion

[2]

Question Number 6. (c)

Scheme Equating i ;

− 9 + 2λ = 5

Marks

⇒ λ=7

⎛ −9 ⎞ ⎛ 2 ⎞ ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 0 ⎟ + 7⎜ 1 ⎟ = ⎜ 7⎟ ⎜ 10 ⎟ ⎜ −1⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ uuur ( = OA. Hence the point A lies on l1.)

Substitutes candidate’s λ = 7 into the line l1 and finds 5 i + 7 j + 3k . B1 The conclusion on this occasion is not needed. [1]

(d)

uuur Let OX = − 3i + 3 j + 7k be point of intersection Finding the difference between uuur their OX (can be implied) and uuur OA . ⎛ ⎛ −3 ⎞ ⎛ 5 ⎞ ⎞ M1 uuur ⎜⎜ ⎟ ⎜ ⎟⎟ AX = ± ⎜ ⎜ 3 ⎟ − ⎜ 7 ⎟ ⎟ ⎜⎜ 7 ⎟ ⎜ 3⎟⎟ ⎝⎝ ⎠ ⎝ ⎠⎠

⎛ −3 ⎞ ⎛ 5 ⎞ ⎛ −8 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AX = OX − OA = ⎜ 3 ⎟ − ⎜ 7 ⎟ = ⎜ −4 ⎟ ⎜ 7 ⎟ ⎜ 3⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

uuur uuur uuur uuur uuur OB = OA + AB = OA + 2 AX ⎛ 5⎞ uuur ⎜ ⎟ OB = ⎜ 7 ⎟ + ⎜ 3⎟ ⎝ ⎠

⎛ −8 ⎞ ⎜ ⎟ 2 ⎜ −4 ⎟ ⎜ 4⎟ ⎝ ⎠

⎛5⎞ ⎜ ⎟ ⎜7⎟ + ⎜ 3⎟ ⎝ ⎠

⎛ ⎞ uuur ⎟ ⎜ 2 ⎜ their AX ⎟ ⎜ ⎟ ⎝ ⎠

⎛ −11⎞ ⎜ ⎟ ⎜ −1 ⎟ or −11i − j + 11k ⎜ 11 ⎟ ⎝ ⎠

⎛ −11⎞ uuur ⎜ uuur ⎟ Hence, OB = ⎜ −1 ⎟ or OB = −11i − j + 11k ⎜ 11 ⎟ ⎝ ⎠

dM1

or ( −11, − 1, 11) [3] 12 marks

Question Number 7. (a)

Scheme

Marks

2 2 A B ≡ ≡ + 2 4− y (2 − y )(2 + y ) (2 − y ) (2 + y )

Forming this identity. NB: A & B are not assigned in M1 this question

2 ≡ A(2 + y ) + B(2 − y ) Let y = −2,

2 = B ( 4) ⇒ B =

1 2

Let y = 2,

2 = A( 4) ⇒ A =

1 2

giving

1 2

(2 − y )

Either one of A =

1 2

(2 + y )

(2 − y )

(If no working seen, but candidate writes down correct partial fraction then award all three marks. If no working is seen but one of A or B is incorrect then M0A0A0.)

1 2

or B = 1 2

(2 + y )

1 2

, aef A1 cao [3]

Question Number 7. (b)

Marks

Scheme

∫

2 dy = 4 − y2

∫

1 2

(2 − y )

Separates variables as shown. Can be implied. Ignore the B1 integral signs, and the ‘2’.

1 dx cot x

1 2

(2 + y )

dy =

∫

tan x dx ln(sec x) or − ln(cos x) Either ± a ln(λ − y ) or ± b ln(λ + y )

∴ − 12 ln(2 − y ) + 12 ln(2 + y ) = ln(sec x) + ( c )

B1 M1;

their ∫ cot1 x dx = LHS correct with ft

for their A and B and no error A1 with the “2” with or without + c

y = 0, x =

π 3

{0 = ln 2 + c

⇒ − ln 2 + 1 2

1 2

ln 2 = ln

( ( )) + c 1 cos π3

Use of y = 0 and x = π3 in an integrated equation containing c M1* ;

}

⇒ c = − ln 2

− 12 ln(2 − y ) + 12 ln(2 + y ) = ln(sec x) − ln 2 Using either the quotient (or product) or power laws for M1 logarithms CORRECTLY.

1 ⎛ 2+ y⎞ ⎛ sec x ⎞ ln ⎜ ⎟ = ln ⎜ ⎟ 2 ⎝2− y⎠ ⎝ 2 ⎠ ⎛2+ y⎞ ⎛ sec x ⎞ ln ⎜ ⎟ = 2ln ⎜ ⎟ ⎝ 2 ⎠ ⎝2− y⎠ ⎛2+ y⎞ ⎛ sec x ⎞ ln ⎜ ⎟ = ln ⎜ ⎟ ⎝ 2 ⎠ ⎝2− y⎠

Using the log laws correctly to obtain a single log term on both dM1* sides of the equation.

2 + y sec 2 x = 2− y 4 Hence,

sec 2 x =

8 + 4y 2− y

sec 2 x =

8 + 4y 2− y

A1 aef [8] 11 marks

Question Number 8. (a)

Scheme At P (4, 2 3) either 4 = 8cos t

or 2 3 = 4sin 2t

⇒ only solution is t = π3 where 0 „ t „

Marks

4 = 8cos t or 2 3 = 4sin 2t

t = π3 or awrt 1.05 (radians) only

π 2

stated in the range 0 „ t „

[2]

x = 8cos t ,

(b)

dx = − 8sin t , dt

y = 4sin 2t

Attempt to differentiate both x and y wrt t to give ± p sin t and M1 ± q cos 2t respectively

dy = 8cos 2t dt

Correct

At P,

( )

−1

Uses m(N) = −

1 3

1 . dM1* their m(T)

Uses y − 2 3 = ( their mN )( x − 4 ) or finds c using x = 4 and dM1* y = 2 3 and uses y = (their m N ) x + " c " .

N: y − 2 3 = − 3 ( x − 4 )

dy dt

You may need to check candidate’s substitutions for M1* Note the next two method marks are dependent on M1*

⎧ ⎫ 8 ( − 12 ) 1 ⎪ ⎪ = = = awrt 0.58 ⎨ ⎬ 3 3 ⎪ ( −8 ) 2 ⎪ ⎩ ⎭

N: y = − 3 x + 6 3

and

Divides in correct way round and attempts to substitute their value of t (in degrees or M1* radians) into their ddyx expression.

dy 8cos ( 23π ) = dx −8sin ( π3 )

Hence m(N) = − 3 or

dx dt

y = − 3x + 6 3

A1 cso AG

2 3 = − 3 ( 4) + c ⇒ c = 2 3 + 4 3 = 6 3

so N: ⎡⎣ y = − 3x + 6 3 ⎤⎦ [6]

Scheme

Question

A = ∫ y dx = ∫ 4sin 2t. ( −8sin t ) dt π

A = ∫ −32sin 2t.sin t dt = π

∫

dx dt M1 dt correct expression A1 (ignore limits and dt )

attempt at A =

8. (c)

Marks

∫ −32 ( 2sin t cos t ) .sin t dt π

Seeing sin 2t = 2sin t cos t M1 anywhere in PART (c).

∫ −64.sin

t cos t dt

Correct proof. Appreciation of how the negative sign affects the limits. A1 AG Note that the answer is given in the question.

∫ 64.sin

t cos t dt

[4] (d)

{Using substitution u = sin t ⇒ ddut = cos t } {change limits: when t = π3 , u = 23 & when t = π2 , u = 1 } π

⎡ sin 3 t ⎤ 2 A = 64 ⎢ ⎥ ⎣ 3 ⎦ π3

⎡ u3 ⎤ A = 64 ⎢ ⎥ ⎣3⎦

k sin 3 t or ku 3 with u = sin t M1 Correct integration A1 ignoring limits.

3 2

Substitutes limits of either ( t = π2 and t = π3 ) or

( u = 1 and u = )

⎡ 1 ⎛ 1 3 3 3 ⎞⎤ A = 64 ⎢ − ⎜⎜ . . . ⎟⎟ ⎥ ⎣⎢ 3 ⎝ 3 2 2 2 ⎠ ⎥⎦

3 2

subtracts the correct way round.

64 −8 3 3

64 ⎛1 1 ⎞ A = 64 ⎜ − 3⎟ = −8 3 3 ⎝3 8 ⎠

(Note that a =

64 3

and dM1

A1 aef isw [4] Aef in the form a + b 3 , with awrt 21.3 and anything that cancels to a = 643 and b = − 8.

, b = − 8) 16 marks