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)
M1, A1 M1
A1√
x
0
16
8
3 16
4
y
1
1.01959
1.08239
1.20269
1.41421
[7]
M1 A1 (2)
M1 for one correct, A1 for all correct
(b)
(c)
Integral =
1 1 1.4142 2(1.01959 ... 1.20269) 2 16 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 )
1
M1 A1
A1 cao (3)
M1 A1
(2) [7]
January 2006
6666 Core Mathematics C4 Mark Scheme
Question Number
3.
Scheme
Marks
u 2 1 , 2
M1
Uses substitution to obtain x = f(u) and to obtain u
Reaches
du const. or equiv. dx
M1
3(u 2 1) 2u udu or equivalent
Simplifies integrand to
Integrates to
1 2
A1
M1
2 3 3u 2 du or equiv.
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 } ] 1 2
[ 3x 2 x 1)
……………. – Uses limits 5 and 1 correctly; [42 – 26]
16
2
2 x 12
M1{M1A1}
3
M1A1√ M1A1
January 2006
4.
6666 Core Mathematics C4 Mark Scheme
Attempts V = x 2e2 x dx
M1
x 2e2 x xe2 x dx 2
=
=
[x
2
e 2 x xe 2 x e 2x dx 2 2 2
M1A1√ refers to candidates
= [
(M1 needs parts in the correct direction)
xe
2x
]
(M1 needs second application of parts)
M1 A1
M1 A1√
dx , but dependent on prev. M1
x 2 e 2 x xe 2 x e 2 x ] 2 2 4
Substitutes limits 3 and 1 and subtracts to give… [dep. on second and third Ms] = 134 e6 14 e2 or any correct exact equivalent. [Omission of loses first and last marks only]
3
A1 cao
dM1
A1 [8]
January 2006
6666 Core Mathematics C4 Mark Scheme
Question Number 5.
(a)
Scheme
Marks
Considers 3x2 16 A(2 x)2 B(1 3x)(2 x) C (1 3x) 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 3x)1 4(2 x)2
(M1, A1)
= 3(1 3x, 9 x 2 27 x3 ......) +
4 (2) x (2)(3) x (2)(3)(4) x (1 +….) 4 1 2 1.2 2 1.2.3 2 2
3
= 4 8x, 27 34 x 2 80 12 x3 ...
( M1 A1 )
A1, A1 (7)
Or uses (3x2 16)(1 3x)1 (2 x)2
M1
(3x 2 16) (1 3x, 9 x2 27 x3 )
(M1A1)
(2) x (2)(3) x (2)(3)(4) x ¼ (1 ) 1 2 1.2 2 1.2.3 2 2
= 4 8x, 27 34 x 2 80 12 x3 ...
3
(M1A1) A1, A1 (7) [11]
4
January 2006
6. (a)
(b)
6666 Core Mathematics C4 Mark Scheme
4 a 18,
M1 A1, A1 (3)
1 b 9
8 1 12 1 =0 14 1
M1
A1
8 12 14 0
dM1
( 2 )
Solves to obtain
M1, A1 Then substitutes value for to give P at the point (6, 10, 16)
(c)
OP =
(5)
(any form) M1
36 100 256
A1 cao (=
7. (a)
392 ) 14 2
(2) [10]
B1
dV 4 r 2 dr
(1)
M1,A1 (b)
(c)
dr dV dr Uses . dt dt dV
in any form,
2
V 1000(2t 1) dt and integrate to p (2 t 1) 1 , Using V=0 when t=0 to find c ,
V 500(1 (d)
1000 = 4 r 2 (2t 1) 2
(2)
= 500(2t 1)1 (c)
M1, A1 M1
(c = 500 , or equivalent)
1 ) 2t 1
A1 (4)
(any form) M1,
(i) Substitute t = 5 to give V,
3V 4
then use r 3
M1, A1
to give r , = 4.77
(3)
(ii) Substitutes t = 5 and r = „their value‟ into „their‟ part (b)
dr 0.0289 dt
( 2.90 x10 2 ) ( cm/s)
5
M1 A1
AG
(2) [12]
January 2006
8.
(a)
6666 Core Mathematics C4 Mark Scheme
Solves y = 0 cos t
1 2
to obtain t
3
5 3
or
(need both for A1)
Or substitutes both values of t and shows that y = 0
dx 1 2cos t dt
(b) 5
ydx (1 2cos t )(1 2cos t )dt
Area=
(c)
=
= 1 4cos t 2(cos 2t 1)dt =
3 4cos t 2cos 2tdt
=
3t 4sin t sin 2t
Substitutes the two correct limits t
(1 2 cos t ) dt 2
AG
3
= 1 4cos t 4cos 2 tdt
Area
5 3
3
3
3 terms (use of correct double angle formula)
5 and subtracts. and 3 3
= 4 3 3
6