Edexcel Gce Core 4 Mathematics C4 6666 Advanced Subsidiary Jun 2005 Mark Scheme

GCSE Edexcel GCE Core Mathematics C4 (6666)

Summer 2005

Core Mathematics C4 (6666)

Edexcel GCE

Mark Scheme (Results)

Final Version June 2005 6666 Core C4 Mark Scheme Question Number

Scheme

Marks

1

1.

( 4 − 9x)

1 2

⎛ 9x ⎞2 = 2 ⎜1 − ⎟ 4 ⎠ ⎝ ⎛ 12 ⎛ 9 x ⎞ 12 ( − 12 ) ⎛ 9 x ⎞ 2 = 2 ⎜1 + ⎜ − ⎟ + ⎜− ⎟ + ⎜ 1⎝ 4 ⎠ 1.2 ⎝ 4 ⎠ ⎝ 81 2 729 3 ⎛ 9 ⎞ = 2 ⎜1 − x − x − x + ... ⎟ 128 1024 ⎝ 8 ⎠ 9 81 729 3 = 2 − x, − x 2 , − x + ... 4 64 512

B1 1 2

( − 12 ) ( − 23 ) ⎛

3 ⎞ 9x ⎞ ⎜ − ⎟ + ... ⎟⎟ ⎝ 4 ⎠ ⎠

1.2.3

M1

A1, A1, A1 [5]

Note The M1 is gained for

1 2

( − 12 ) 1.2

( ... )

2

or

1 2

( − 12 ) ( − 32 ) 1.2.3

( ... )

3

Special Case

81 2 729 3 ⎛ 9 ⎞ x − x + ... ⎟ and goes no further If the candidate reaches = 2 ⎜1 − x − 128 1024 ⎝ 8 ⎠ allow A1 A0 A0

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme

⎛ dy ⎞ dy =0 2x + ⎜ 2x + 2 y ⎟ − 6 y dx ⎝ dx ⎠ dy =0 ⇒ x+ y =0 dx

2.

Marks

M1 (A1) A1 or equivalent

Eliminating either variable and solving for at least one value of x or y. y 2 − 2 y 2 − 3 y 2 + 16 = 0 or the same equation in x y = ±2 or x = ± 2 ( 2, − 2 ) , ( −2, 2 )

M1 M1 A1 A1 [7]

dy x + y = Note: dx 3 y − x Alternative

3 y 2 − 2 xy − ( x 2 + 16 ) = 0 y=

2 x ± √ (16 x 2 + 192 )

6 dy 1 1 8x = ± . dx 3 3 √ (16 x 2 + 192 )

dy =0 ⇒ dx

8x =±1 2 √ (16 x + 192 )

64 x 2 = 16 x 2 + 192 x=± 2 ( 2, − 2 ) , ( −2, 2 )

M1 A1± A1

M1

M1 A1 A1 [7]

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

3.

Scheme

(a)

Marks

5x + 3 A B = + ( 2 x − 3)( x + 2 ) 2 x − 3 x + 2 5 x + 3 = A ( x + 2 ) + B ( 2 x − 3)

Substituting x = −2 or x = 32 and obtaining A or B; or equating coefficients and solving a pair of simultaneous equations to obtain A or B. A = 3, B = 1

M1 A1, A1 (3)

If the cover-up rule is used, give M1 A1 for the first of A or B found, A1 for the second. (b)

∫

5x + 3 3 dx = ln ( 2 x − 3) + ln ( x + 2 ) 2 ( 2 x − 3)( x + 2 ) 6 ⎡ ... ⎤ = 3 ln 9 + ln 2 ⎣ ⎦2 2 = ln 54

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

M1 A1ft

M1 A1 cao

A1

(5) [8]

Question Number

4.

Scheme

∫(

1

1 − x2 )

1 2

dx =

∫(

1

1 − sin 2 θ )

3 2

cos θ dθ

Marks

Use of x = sin θ and

dx = cos θ dθ

∫

1 dθ cos 2 θ = ∫ sec2 θ dθ = tan θ

=

Using the limits 0 and

π 6

M1 A1 M1 A1

to evaluate integral

[ tan θ ] 0 π

6

=

M1

M1

1 ⎛ √3⎞ ⎜= ⎟ √3 ⎝ 3 ⎠

cao

A1 [7]

Alternative for final M1 A1

Returning to the variable x and using the limits 0 and

1 2

to evaluate integral

M1

1

⎡ ⎤2 1 ⎛ √3⎞ x ⎢ ⎥ = ⎜= ⎟ 2 ⎢⎣ √ (1 − x ) ⎥⎦ 0 √ 3 ⎝ 3 ⎠

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

cao

A1

Question Number

5.

Scheme

∫ xe

(a)

2x

1 2

dx = x e − 1 2

2x

Marks

∫

1 e 2 x dx Attempting parts in the right direction 2

1 4

= x e2 x − e2 x

M1 A1 A1

1

1 1 2 ⎡ 1 2x 1 2x ⎤ ⎢⎣ 2 x e − 4 e ⎥⎦ = 4 + 4 e 0

M1 A1 (5)

x = 0.4 ⇒ y ≈ 0.890 22 x = 0.8 ⇒ y ≈ 3.962 43

(b)

Both are required to 5 d.p

B1

.

(c)

I ≈ × 0.2 × [ ... 1 2

]

≈ ... × ⎡⎣ 0+7.389 06+2 ( 0.29836+.890 22+1.992 07+3.962 43) ⎤⎦ ft their answers to (b) ≈ 0.1× 21.675 22 ≈ 2.168 cao

Note

1 1 2 + e ≈ 2.097 … 4 4

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

(1)

B1 M1 A1ft

A1

(4) [10]

Question Number

6.

Scheme

(a)

Marks

dx dy = −2 cosec 2 t , = 4sin t cos t dt dt d y −2sin t cos t = ( = −2sin 3 t cos t ) dx cosec 2 t

both

M1 A1 (4)

At t = π4 , x = 2, y = 1

(b)

Substitutes t = π4 into an attempt at

both x and y

dy ⎛ 1⎞ to obtain gradient ⎜ − ⎟ dx ⎝ 2⎠

Equation of tangent is y − 1 = −

M1 A1

B1 M1

1 ( x − 2) 2

M1 A1

Accept x + 2 y = 4 or any correct equivalent (c)

Uses 1 + cot 2 t = cosec2 t , or equivalent, to eliminate t

(4)

M1

2

2 ⎛ x⎞ 1+ ⎜ ⎟ = y ⎝2⎠

y=

correctly eliminates t

A1

cao

A1

8 4 + x2

The domain is x …0

B1

An alternative in (c) 1

1

x x ⎛ y ⎞2 ⎛ y ⎞2 sin t = ⎜ ⎟ ; cos t = sin t = ⎜ ⎟ 2 2⎝ 2⎠ ⎝2⎠ y x2 y sin 2 t + cos 2 t = 1 ⇒ + × =1 2 4 2 8 Leading to y = 4 + x2

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

M1 A1 A1

(4) [12]

Question Number

7.

Scheme

(a) k component

Marks

2 + 4λ = −2 ⇒ λ = −1

Note µ = 2 Substituting their λ (or µ ) into equation of line and obtaining B B: ( 2, 2, − 2 )

Accept vector forms

M1 A1 M1 A1 (4)

⎛1⎞ ⎜ ⎟ ⎜ −1⎟ = √ 18; ⎜4⎟ ⎝ ⎠

(b)

⎛1⎞ ⎜ ⎟ ⎜ −1⎟ = √ 2 ⎜0⎟ ⎝ ⎠

⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −1⎟ ⋅ ⎜ −1⎟ = 1 + 1 + 0 ( = 2 ) ⎜4⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ 2 1 cos θ = = √ 18 √ 2 3

(c)

(d)

uuur uuur 2 uuur AB = −i + j − 4k ⇒ AB = 18 or AB = √ 18 uuur uuur 2 uuur BC = 3i − 3 j ⇒ BC = 18 or BC = √ 18 uuur uuur Hence AB = BC ¿

uuur OD = 6i − 2 j + 2k

both

B1

B1

cao

M1 A1 (4)

ignore direction of vector

M1

ignore direction of vector

M1

Allow first B1 for any two correct Accept column form or coordinates

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

A1

(3)

B1 B1 (2) [13]

Question Number

8.

Scheme

(a)

Marks

dV is the rate of increase of volume (with respect to time) dt − kV : k is constant of proportionality and the negative shows decrease (or loss) dV = 20 − kV ¿ giving These Bs are to be awarded independently dt

B1

B1 (2)

∫

(b)

1 dV = ∫ 1dt 20 − kV

separating variables

1 − ln ( 20 − kV ) = t ( +C ) k Using V = 0, t = 0 to evaluate the constant of integration 1 c = − ln 20 k 1 ⎛ 20 ⎞ t = ln ⎜ ⎟ k ⎝ 20 − kV ⎠ Obtaining answer in the form V = A + B e − kt 20 20 − kt 20 V= − e Accept (1 − e− kt ) k k k dV = 20 e − kt dt

(c)

Can be implied

dV = 10, t = 5 ⇒ 10 = 20 e − kt dt 75 At t = 10, V = ln 2

1 ⇒ k = ln 2 ≈ 0.139 5

M1 M1 A1 M1

M1 A1

(6)

M1 M1 A1

awrt 108

M1 A1 (5) [13]

Alternative to (b) Using printed answer and differentiating

dV = − kB e − kt dt

M1

Substituting into differential equation − kB e − kt = 20 − kA − kB e − kt 20 A= k Using V = 0, t = 0 in printed answer to obtain A + B = 0 20 B=− k

6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

M1 M1 A1 M1 A1

(6)