C4 MS Jan 2006

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  12

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