C4 MS Jan 2009 by The de Ferrers Academy

Mark Scheme (Results) January 2009

GCE

GCE Mathematics (6666/01)

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

January 2009 6666 Core Mathematics C4 Mark Scheme Question Number 1

(a)

Scheme

Marks

C: y 2 − 3 y = x3 + 8

⎧ dy ⎫ =⎬ ⎨ ⎩ dx ⎭

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

dy dy − 3 = 3x 2 dx dx

Correct equation. A1 A correct (condoning sign error) attempt to dy dy combine or factorise their ‘ 2 y − 3 ’. M1 dx dx Can be implied.

dy ( 2 y − 3) = 3 x 2 dx

3x 2 2y −3

dy 3x 2 = dx 2 y − 3

A1 oe (4)

(b)

Substitutes y = 3 into C. M1

y = 3 ⇒ 9 − 3(3) = x3 + 8

Only x = − 2

x3 = − 8 ⇒ x = − 2

(−2,3) ⇒

dy 3(4) dy = ⇒ =4 dx 6 − 3 dx

dy = 4 from correct working. dx Also can be ft using their ‘x’ value and y = 3 in the A1 dy 3x 2 correct part (a) of = dx 2 y − 3

(3) 1(b) final A1

then an answer of

Note if the candidate inserts their x value and y = 3 into

dy 3x = , dx 2 y − 3

dy = their x 2 , may indicate a correct follow through. dx

[7]

6666/01 GCE Mathematics January 2009

Question Number

Scheme 2

(a)

Area(R) =

Marks

3 −1 dx = ∫ 3(1 + 4 x) 2 dx (1 + 4 x) 0

∫ 0

Integrating 3(1 + 4 x) ⎡ 3(1 + 4 x) =⎢ 1 ⎢⎣ 2 .4

1 2

⎤ ⎥ ⎥⎦ 0

− 12

to give 1

± k (1 + 4 x) 2 .

Correct integration. A1 Ignore limits.

= ⎡ 32 (1 + 4 x) 2 ⎤ ⎣ ⎦0 1

(

Substitutes limits of 2 and 0 into a changed function and subtracts the M1 correct way round.

)

9 − ( 32 (1) )

3 2

−

9 2

3 2

= 3 (units) 2

3 A1 (4)

(Answer of 3 with no working scores M0A0M0A0.) 2

Use of V = π ∫ y 2 dx .

⎛ ⎞ 3 (b) Volume = π ⎜⎜ ⎟⎟ dx (1 4 ) + x ⎝ ⎠ 0

∫

Can be implied. Ignore limits and dx .

9 dx 1 + 4x

= (π ) ∫ 0

= (π ) ⎡⎣ 94 ln 1 + 4x ⎤⎦ 0

9 4

± k ln 1 + 4 x

ln 1 + 4x

9 4

Substitutes limits of 2 and 0 dM1 and subtracts the correct way round.

= (π ) ⎡⎣( 94 ln 9 ) − ( 94 ln1) ⎤⎦ Note that ln1 can be implied as equal to 0.

So Volume =

π ln 9

9 4

Note the answer must be a one term exact value. Note, also you can ignore subsequent working here.

π ln 9 or 92 π ln 3 or

18 4

π ln 3

Note that = 94 π ln 9 + c (oe.) would be awarded the final A0.

A1 oe isw (5)

[9]

6666/01 GCE Mathematics January 2009

Question Number 3

(a)

Scheme

27 x 2 + 32 x + 16 ≡ A(3 x + 2)(1 − x) + B(1 − x) + C (3x + 2) 2

x = − 23 , 12 − 643 + 16 = ( 53 ) B ⇒

x = 1,

20 3

= ( 53 ) B ⇒ B = 4

27 + 32 + 16 = 25 C ⇒ 75 = 25 C ⇒ C = 3

Equate x2:

27 = − 3 A + 9C ⇒ 27 = − 3 A + 27 ⇒ 0 = − 3 A ⇒ A=0

x = 0, 16 = 2 A + B + 4C ⇒ 16 = 2 A + 4 + 12 ⇒ 0 = 2 A ⇒ A = 0

Marks

Forming this identity M1 Substitutes either x = − 23 or x = 1 into their identity or equates 3 terms or substitutes M1 in values to write down three simultaneous equations. Both B = 4 and C = 3 A1 (Note the A1 is dependent on both method marks in this part.) Compares coefficients or substitutes in a third x-value B1 or uses simultaneous equations to show A = 0. (4)

(b)

f ( x) =

4 3 + 2 (3 x + 2) (1 − x)

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

= 4(3 x + 2) −2 + 3(1 − x) −1 −2 = 4 ⎡ 2 (1 + 32 x ) ⎤ + 3(1 − x) −1 ⎣ ⎦

= 1(1 + 32 x ) + 3(1 − x) −1 −2

⎧ (−2)(−3) ( = 1 ⎨1 + (− 2)( 32x ); + 2! ⎩

⎫ ) + ...⎬ ⎭

3x 2 2

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

{

Either 1 ± (− 2)( 32x ) or 1 ± (−1)(− x) from either first dM1; or second expansions respectively Ignoring 1 and 3, any one A1 correct {..........} expansion. Both {..........} correct. A1

{

= 1 − 3x + 274 x 2 + ...} + 3 1 + x + x 2 + ...} = 4 + 0 x ; + 394 x 2

6666/01 GCE Mathematics January 2009

4 + (0 x) ;

39 4

A1; A1 (6)

Question Number (c)

Scheme

1.08 + 6.4 + 16 (6.76)(0.8) 23.48 2935 = = 4.341715976... = 5.408 676

Actual = f (0.2) =

Or 4 3 + (3(0.2) + 2) 2 (1 − 0.2) 4 2935 = + 3.75 = 4.341715976... = 6.76 676

Actual = f (0.2) =

Estimate = f (0.2) = 4 +

39 4

(0.2) 2

4.39 − 4.341715976... 4.341715976...

Attempt to find the actual value of f(0.2) or seeing awrt 4.3 and believing it is candidate’s actual f(0.2). Candidates can also attempt to M1 find the actual value by using A B C + + 2 (3x + 2) (3x + 2) (1 − x) with their A, B and C.

Attempt to find an estimate for f(0.2) using their answer to (b) M1

= 4 + 0.39 = 4.39

%age error =

Marks

their estimate - actual × 100 actual

× 100

= 1.112095408... = 1.1 % (2sf )

1.1% A1 cao (4) [14]

6666/01 GCE Mathematics January 2009

Question Number 4

(a)

Scheme

Marks

d 1 = − 2 i + j − 4k , d 2 = q i + 2 j + 2 k ⎧ ⎛ − 2⎞ ⎛ q ⎞ ⎪ ⎜ ⎟ ⎜ ⎟ As ⎨d1 • d 2 = ⎜ 1 ⎟ • ⎜ 2 ⎟ ⎪ ⎜ − 4⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎩

⎫ ⎪ ⎬ = (−2 × q) + (1 × 2) + (−4 × 2) ⎪ ⎭

d1 • d 2 = 0 ⇒ − 2q + 2 − 8 = 0

Apply dot product calculation between two direction vectors, M1 ie. (−2 × q ) + (1 × 2) + (−4 × 2) Sets d1 • d 2 = 0 and solves to find q = − 3

− 2q = 6 ⇒ q = − 3 AG

A1 cso (2)

(b) Lines meet where:

⎛ 11 ⎞ ⎛ −2 ⎞ ⎛ −5 ⎞ ⎛q⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ + λ ⎜ 1 ⎟ = ⎜ 11 ⎟ + µ ⎜ 2 ⎟ ⎜ 17 ⎟ ⎜ −4 ⎟ ⎜ p⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ i : 11 − 2λ = − 5 + q µ First two of j : 2 + λ = 11 + 2 µ k : 17 − 4 λ = p + 2µ (1) + 2(2) gives: (2) gives:

(3) ⇒ 17 − 4(5)

15 = 17 + µ

(1) (2) (3)

Need to see equations (1) and (2). Condone one slip. M1 (Note that q = −3 .) Attempts to solve (1) and (2) to dM1 find one of either λ or µ Any one of λ = 5 or µ = − 2 A1

⇒ µ = −2

2 + λ = 11 − 4 ⇒ λ = 5

Both λ = 5 and µ = − 2

Attempt to substitute their λ and µ into their k component to ddM1 give an equation in p alone.

= p + 2(−2)

⇒ p = 17 − 20 + 4 ⇒ p = 1

p =1

A1 cso (6)

(c)

⎛ 11 ⎞ ⎛ − 2⎞ ⎜ ⎟ ⎜ ⎟ r = ⎜ 2 ⎟ + 5⎜ 1 ⎟ ⎜ 17 ⎟ ⎜ − 4⎟ ⎝ ⎠ ⎝ ⎠

⎛ −5⎞ ⎛ − 3⎞ ⎜ ⎟ ⎜ ⎟ or r = ⎜ 11 ⎟ − 2 ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 2⎟ ⎝ ⎠ ⎝ ⎠

⎛ 1 ⎞ ⎜ ⎟ Intersect at r = ⎜ 7 ⎟ or ⎜ − 3⎟ ⎝ ⎠

Substitutes their value of λ or M1 µ into the correct line l1 or l2 . ⎛ 1 ⎞ ⎜ ⎟ ⎜ 7 ⎟ or (1, 7, − 3) ⎜ ⎟ ⎝ − 3⎠

(1, 7, − 3)

A1 (2)

6666/01 GCE Mathematics January 2009

Question Number

Scheme

uuur (d) Let OX = i + 7 j − 3k be point of intersection ⎛ 1 ⎞ ⎛ 9 ⎞ ⎛ −8 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AX = OX − OA = ⎜ 7 ⎟ − ⎜ 3 ⎟ = ⎜ 4 ⎟ ⎜ − 3 ⎟ ⎜ 13 ⎟ ⎜ −16 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Marks

uuur Finding vector AX by finding the uuur uuur difference between OX and OA . M1 uuur Can be ft using candidate’s OX .

uuur uuur uuur uuur uuur OB = OA + AB = OA + 2 AX ⎛9⎞ uuur ⎜ ⎟ OB = ⎜ 3 ⎟ + ⎜ 13 ⎟ ⎝ ⎠

⎛ −8 ⎞ ⎜ ⎟ 2⎜ 4 ⎟ ⎜ −16 ⎟ ⎝ ⎠

⎛9⎞ ⎜ ⎟ ⎜3⎟+ ⎜ 13 ⎟ ⎝ ⎠

⎛ −7 ⎞ uuur ⎜ uuur ⎟ Hence, OB = ⎜ 11 ⎟ or OB = −7 i + 11 j − 19 k ⎜ −19 ⎟ ⎝ ⎠

⎛ ⎞ uuur ⎟ ⎜ 2 ⎜ their AX ⎟ ⎜ ⎟ ⎝ ⎠

⎛ −7 ⎞ ⎜ ⎟ ⎜ 11 ⎟ or −7 i + 11 j − 19 k ⎜ −19 ⎟ ⎝ ⎠

dM1

or ( −7, 11, − 19 ) (3) [13]

6666/01 GCE Mathematics January 2009

Question Number

(a)

Scheme

Similar triangles ⇒

Uses similar triangles, ratios or trigonometry to find either one of these M1 two expressions oe.

r 16 2h = ⇒ r= h 24 3 2

1 1 ⎛ 2h ⎞ 4 π h3 V = π r 2h = π ⎜ ⎟ h = 3 3 ⎝ 3 ⎠ 27

Marks

Substitutes r =

2h 3

into the formula for the A1 volume of water V. (2)

(b) From the question,

dV =8 dt

dV 12 π h 2 4 π h 2 = = dh 27 9

dV 12 π h 2 4π h2 or = dh 27 9

Candidate’s

dh dV dV 9 18 = ÷ = 8× = 2 dt dt 4π h π h2 dh

When h = 12,

dV =8 dt

dV d V ÷ ; M1; dt dh

⎛ 12 π h 2 ⎞ 9 18 8÷⎜ oe A1 or ⎟ or 8 × 2 π h2 4π h ⎝ 27 ⎠ 1 18 or 8π 144 π

dh 18 1 = = dt 144 π 8π

Note the answer must be a one term exact value. Note, also you can ignore subsequent working 18 after . 144 π

A1 oe isw (5)

[7]

6666/01 GCE Mathematics January 2009

Question Number 6

(a)

Scheme

∫ tan

Marks

x dx

⎡ NB : sec 2 A = 1+ tan 2 A gives tan 2 A = sec2 A − 1 ⎤ ⎣ ⎦

The correct underlined identity. M1 oe

= ∫ sec 2 x − 1 dx Correct integration A1 with/without + c

= tan x − x ( + c )

(b)

∫

(2)

1 ln x dx x3

⎧⎪u = ln x ⇒ ddux = 1x ⎨ dv −2 −3 ⇒ v = x−2 = ⎩⎪ dx = x

−1 2 x2

⎫⎪ ⎬ ⎭⎪

=−

1 1 1 ln x − − 2 . dx 2 2x 2x x

=−

1 1 ln x + 2 2x 2

=−

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

Use of ‘integration by parts’ formula in the correct direction. M1 Correct direction means that u = ln x . Correct expression. A1

∫

An attempt to multiply through k , n ∈ , n …2 by 1x and an xn attempt to ...

1 dx x3

… “integrate”(process the result); M1 correct solution with/without + c A1 oe

6666/01 GCE Mathematics January 2009

(4)

Question Number (c)

Scheme

∫

e3 x dx 1 + ex

⎧ du dx 1 dx 1 ⎫ x = ex , = x , = ⎨u = 1 + e ⇒ ⎬ dx du e du u − 1⎭ ⎩

∫

e 2 x .e x dx = 1 + ex

or =

∫

(u − 1) 2 du u

∫

u 2 − 2u + 1 du u

∫

Differentiating to find any one of the B1 three underlined Attempt to substitute for e 2 x = f (u ) , dx 1 = and u = 1 + e x their du e x M1* dx 1 3x or e = f (u ) , their = and du u − 1 u = 1 + ex .

(u − 1) 2 .e x 1 . x du u e

(u − 1)3 1 . du u (u − 1)

= u−2+

Marks

∫

(u − 1) 2 du u

An attempt to multiply out their numerator to give at least three terms and divide through each term by u dM1*

1 du u

Correct integration with/without +c A1

u2 − 2u + ln u ( + c ) 2

Substitutes u = 1 + e x back into their integrated expression with at least two dM1* terms.

(1 + e x ) 2 = − 2(1 + e x ) + ln(1 + e x ) + c 2 = 12 + e x + 12 e 2 x − 2 − 2e x + ln(1 + e x ) + c = 12 + e x + 12 e 2 x − 2 − 2e x + ln(1 + e x ) + c = 12 e 2 x − e x + ln(1 + e x ) − 23 + c = e − e + ln(1 + e ) + k 1 2

1 2

e 2 x − e x + ln(1 + e x ) + k

must use a + c and " − 32 " combined. A1 cso (7) [13]

6666/01 GCE Mathematics January 2009

Question Number 7

Scheme

Marks

(a) At A, x = −1 + 8 = 7 & y = (−1) 2 = 1 ⇒ A(7,1)

A(7,1)

B1 (1)

(b)

x = t − 8t ,

y=t ,

dx dy = 3t 2 − 8 , = 2t dt dt Their

dy 2t ∴ = 2 dx 3t − 8 At A, m(T) =

2( −1) 2 −2 −2 = = = 2 3(−1) − 8 3−8 5 −5

T : y − ( their 1) = mT ( x − ( their 7 ) ) or 1 =

2 5

(7) + c

⇒ c =1−

14 5

dy dt

divided by their

dx dt

Correct

dy dx

Substitutes for t to give any of the four underlined oe:

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

= − 95

Hence T : y = 52 x − 95 gives T : 2 x − 5 y − 9 = 0

(c)

2x − 5 y − 9 = 0

cso (5)

Substitution of both x = t 3 − 8t and M1 y = t 2 into T

2(t 3 − 8t ) − 5t 2 − 9 = 0

2t 3 − 5t 2 − 16t − 9 = 0 (t + 1) {(2t 2 − 7t − 9) = 0}

A realisation that ( t + 1) is a factor. dM1

(t + 1) {(t + 1)(2t − 9) = 0}

{t = −1 (at A)}

x = ( 92 ) − 8 ( 92 ) = 2

9 2

729 8

at B

− 36 =

441 8

= 55.125 or awrt 55.1

y = ( 92 ) = 814 = 20.25 or awrt 20.3 2

9 2

Candidate uses their value of t to ddM1 find either the x or y coordinate One of either x or y correct. A1 Both x and y correct. A1 awrt

Hence B ( 441 , 814 ) 8

(6) [12]

6666/01 GCE Mathematics January 2009