C4 MS Jan 2006

Page 1

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]


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