Edexcel Gce Core 4 Mathematics C4 6666 Advanced Subsidiary Jan 2006 Mark Scheme

GCE Edexcel GCE

Core Mathematics C4 (6666)

January 2006

Core Mathematics C4 (6666)

Edexcel GCE

Mark Scheme (Results)

January 2006 6666 Core Mathematics C4 Mark Scheme Question Number 1.

Scheme

Marks M1

Differentiates to obtain :

6 x + 8 y dy dx − 2,

A1,

....................... + (6 x dy dx + 6 y ) = 0

+(B1)

⎡ dy 2 − 6 x − 6 y ⎤ ⎢ = ⎥ 6x + 8 y ⎦ ⎣ dx

Substitutes x = 1, y = – 2 into expression involving

dy dx

,

to give

dy dx

= − 108

Uses line equation with numerical ‘gradient’ y – (– 2) = (their gradient)(x – 1) or finds c and uses y = (their gradient ) x + " c" To give 5 y + 4 x + 6 = 0 (or equivalent = 0)

2. (a)

0

π

π

x

16

8

y

1

1.01959

1.08239

M1, A1 M1

A1√

3π 16

π

1.20269

1.41421

4 M1 A1 (2)

M1 for one correct, A1 for all correct

(b)

(c)

Integral =

1 π × × {1 + 1.4142 + 2(1.01959 + ... + 1.20269)} 2 16 ⎛ π ⎞ x 9.02355 ⎟ = 0.8859 ⎜= ⎝ 32 ⎠

Percentage error =

M1 gained for (± )

approx − 0.88137 ×100 = 0.51 % (allow 0.5% to 0.54% for A1) 0.88137 approx − ln (1 + 2 ) ln (1 + 2 )

[7]

M1 A1√ A1 cao (3)

M1 A1

(2) [7]

Question Number 3.

Scheme

Uses substitution to obtain x = f(u) and to obtain u

Reaches

Marks

M1

⎡ u 2 + 1⎤ ⎥ , ⎢ ⎣ 2 ⎦

du = const. or equiv. dx

M1

3(u 2 + 1) ∫ 2u udu or equivalent

Simplifies integrand to Integrates to

1 2

⎛

∫ ⎜⎝ 3u

2

+

A1 M1

3⎞ ⎟ du or equiv. 2⎠

M1 A1√

u 3 + 23 u

A1√ dependent on all previous Ms Uses new limits 3 and 1 substituting and subtracting (or returning to function of x with old limits)

M1

To give 16

A1

cso

[8]

“By Parts” Attempt at “ right direction” by parts

⎛ ⎜ ⎝

1 2

M1

⎞ ⎟ – { ∫ 3 (2 x − 1) dx } ] M1{M1A1} ⎠ 1 2

[ 3 x ⎜ 2 x − 1) ⎟

……………. – Uses limits 5 and 1 correctly; [42 – 26]

16

3

(2 x − 1)2

M1A1√ M1A1

4.

∫

M1

Attempts V = π x 2 e 2 x dx

⎡ x 2e2 x ⎤ − ∫ xe 2 x dx ⎥ ⎣ 2 ⎦

=π ⎢

(M1 needs parts in the correct direction)

x 2 e 2 x ⎛ xe 2 x e 2x ⎞ ⎜ = π[ dx ⎟⎟ −⎜ −∫ 2 2 2 ⎠ ⎝ M1A1√ refers to candidates

= π

[x

2

∫ xe

2x

]

(M1 needs second application of parts)

M1 A1

M1 A1√

dx , but dependent on prev. M1

e 2 x ⎛ xe 2 x e 2 x ⎞ ⎟] − ⎜⎜ − 2 4 ⎟⎠ ⎝ 2

Substitutes limits 3 and 1 and subtracts to give… [dep. on second and third Ms] = π ⎡⎣ 134 e6 − 14 e 2 ⎤⎦ or any correct exact equivalent. [Omission of π loses first and last marks only]

A1 cao

dM1

A1 [8]

Question Number 5.

(a)

Scheme

Marks

Considers 3x 2 + 16 = A(2 + x) 2 + B (1 − 3 x)(2 + x) + C (1 − 3 x) M1

and substitutes x = –2 , or x = 1/3 , or compares coefficients and solves simultaneous equations

A1, A1

To obtain A = 3, and C = 4 Compares coefficients or uses simultaneous equation to show B = 0.

B1 (4)

(b)

M1

Writes 3(1 − 3 x) −1 + 4(2 + x) −2

(M1, A1)

= 3(1 + 3 x, +9 x + 27 x + ......) + 2

3

2

3

4 (−2) ⎛ x ⎞ (−2)(−3) ⎛ x ⎞ (−2)(−3)(−4) ⎛ x ⎞ (1 + ⎜ ⎟+ ⎜ ⎟ + ⎜ ⎟ +….) 4 1 ⎝ 2⎠ 1.2 ⎝ 2 ⎠ 1.2.3 ⎝ 2⎠ = 4 + 8 x, + 27 34 x 2 + 80 12 x 3 + ...

( M1 A1 )

A1, A1 (7)

Or uses (3 x 2 + 16)(1 − 3 x) −1 (2 + x) −2

M1

(3x 2 + 16) (1 + 3x, +9 x 2 + 27 x3 +) ×

(M1A1)× 2

3

(−2) ⎛ x ⎞ (−2)(−3) ⎛ x ⎞ (−2)(−3)(−4) ⎛ x ⎞ ¼ (1 + ⎜ ⎟+ ⎜ ⎟ + ⎜ ⎟ ) 1 ⎝ 2⎠ 1.2 ⎝ 2 ⎠ 1.2.3 ⎝ 2⎠ = 4 + 8 x, + 27 34 x 2 + 80 12 x 3 + ...

(M1A1) A1, A1 (7) [11]

6. (a)

(b)

λ = −4 → a = 18,

µ =1→ b = 9

M1 A1, A1 (3)

⎛8 + λ ⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜12 + λ ⎟ • ⎜ 1 ⎟ =0 ⎜14 − λ ⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠

M1

∴ 8 + λ + 12 + λ − 14 + λ = 0

A1

Solves to obtain λ

dM1

( λ = −2 )

Then substitutes value for λ to give P at the point (6, 10, 16) (c)

OP = (=

7. (a)

36 + 100 + 256

M1, A1 (5) M1 A1 cao

392 ) = 14 2

(2) [10]

dV = 4π r 2 dr

B1 (1)

dr dV dr = . dt dt dV

1000 4π r 2 (2t + 1) 2

(b)

Uses

(c)

V = ∫ 1000(2t + 1) dt and integrate to p (2 t + 1) − 1 ,

in any form,

=

−2

Using V=0 when t=0 to find c ,

∴V = 500(1 − (d)

(any form)

M1,A1 (2)

= −500(2t + 1) −1 (+ c)

(c = 500 , or equivalent)

1 ) 2t + 1

(any form)

(i) Substitute t = 5 to give V,

⎛ 3V ⎝ 4π

then use r = 3 ⎜

⎞ ⎟ to give r , = 4.77 ⎠

(≈ 2.90 x10 − 2 ) ( cm/s) ∗

M1 A1 (4) M1, M1, A1 (3)

(ii) Substitutes t = 5 and r = ‘their value’ into ‘their’ part (b)

dr = 0.0289 dt

M1, A1

AG

M1 A1 (2) [12]

8.

(a)

Solves y = 0 ⇒ cos t =

1 2

to obtain t =

π 3

5π 3

or

(need both for A1)

M1 A1 (2)

Or substitutes both values of t and shows that y = 0

dx = 1 − 2 cos t dt

(b) 5π

π

=

3

∫

= 1 − 4 cos t + 4 cos 2 tdt

Area

5π 3

3

∫ ydx = ∫ (1 − 2 cos t )(1 − 2 cos t )dt

Area=

(c)

M1 A1

∫

= 1 − 4 cos t + 2(cos 2t + 1)dt =

∫ 3 − 4 cos t + 2 cos 2tdt

=

[3t − 4sin t + sin 2t ]

Substitutes the two correct limits t = = 4π + 3 3

∫ (1 − 2 cos t ) dt 2

∗

AG

B1 (3)

π

3

3 terms (use of correct double angle formula)

M1 M1

M1 A1

5π π and subtracts. and 3 3

M1 A1A1 (7) [12]