Download Jan 2006 - 6673 Pure P3

January 2006
6673 Pure Mathematics P3Mark Scheme
Question
Number
Scheme
 1
(1      2 x  
 2
1 (a)
= 1  x, 
Alternative
Marks
1
3
1
3
5
( )( )
( )( )( )
2
2
2  2 x   2
2
2  2 x 3 +….)
1.2
1.2.3
3 2 5 3
x  x  ...
2
2
A1, A1
(4)
May use McLaurin f(0)=1 and f (0)  1 to obtain 1st two terms
1+x
Differentiates two further times and uses formula with correct factorials to give
3 2 5 3
x  x
2
2
(b)
2
(100  200 x)
 12
M1 (corr bin coeffs)
M1 (powers of –2x)
M1 A1
M1
A1
(4)
1
1
 100 2 (1  2 x) 2 . So series is
1
10
(previous series)
M1A1 ft
(2)
Uses f(2) = 0 to give
16 – 4 + 2a + b = 0
M1 A1
Uses f(-1) = 6 to give
-2 – 1 +-a + b = 6
M1 A1
Solves simultaneous equations to give
a = -7, and b = 2
M1 A1 A1
(7)
3 (a)
Uses circle equation
( x  4)  ( y  3)  ( 5)
Multiplies out to give x 2  8 x  16  y 2  6 y  9  5 and thus
x 2  y 2  8 x  6 y  20  0 (*)
2
Alternative
2
2
M1
A1
A1
(3)
Or states equation of circle is x 2  y 2  2 gx  2 fy  c  0 has centre (-g, -f) and so
g = - 4 and f = - 3
2
Uses g 2  f 2  c  r 2 to give c  32  42  5 , i.e. c = 20
M1 A1
x 2  y 2  8 x  6 y  20  0
A1
(3)
(b)
y = 2x meets the circle when
x  (2 x)  8 x  6(2 x)  20  0
M1
5 x  20 x  20  0
A1
M1 A1
2
2
2
Solves and substitutes to obtain x = 2 and y = 4. Coordinates are (2, 4)
Or Implicit differentiation attempt , 2 x  2 y
dy
Uses y = 2x and
=2 to give 10x – 20 = 0.
dx
Thus x =2 and y =4
dy
dy
86  0
dx
dx
(4)
M1 A1
M1 A1
(4)
January 2006
6673 Pure Mathematics P3Mark Scheme
Scheme
Question
Number
4.(a)
Marks
f  (x)=( x 2  1)  1x  ln x  2 x
M1 A1
f  (e) = (e 2  1)  1e  2e = 3e + 1e
M1 A1
(4)
(b)
x3
x3
1
 x) ln x   (  x) dx
3
3
x
3
2
x
x
= (  x) ln x   (  1)dx
3
3
(
M1 A1
e
 x3

x3
=  (  x) ln x  (  x) 
9
 3
1
10
2 3
= 9e  9
5. (a)
(b)
A1
M1 A1
(5)
9  4x2
18
 1 
, so A = - 1
2
9  4x
(3  2 x)(3  2 x)
B1
Uses 18 = B ( 3– 2x) + C ( 3+2x ) and attempts to find B and C
M1
B = 3 and C = 3
Or
Uses 9  4 x 2 = A( 9  4 x 2 )+ B ( 3– 2x) + C ( 3+2x ) and attempts to find A, B and
C
A1 A1
A = -1, B = 3 and C = 3
A1, A1, A1
Obtains Ax 
Substitutes limits and subtracts to give 2 A 
or
M1
(4)
B
C
ln(3  2 x)  ln(3  2 x)
2
2
= -2 +3ln5
(4)
–2 +ln125
M1 A1
B
C
ln(5)  ln( 15 )
2
2
M1 A1ft
A1
(5)
January 2006
6673 Pure Mathematics P3Mark Scheme
Question
Number
6 (a)
Scheme
dC
 kC ; rate of decrease/negative sign; k constant of proportionality/positive
dt
Marks
B1
(1)
constant
dC
 k  dt
C
 ln C   kt  ln A
 C  Ae kt

(b)
M1
M1
A1
(3)
(c)
At t = 0 C  C0 ,
 A  C0
and at t = 4 C  101 C0 ,  101 C0  C0 e

1
10
e
4 k
and 4k  ln
1
10
4 k
,
,  k  ln10
1
4
B1
M1
M1, A1
(4)
7
(a)
(b)
Solves 9  2  1 or 7  2  1 to give   4 so p = 3
Solves 9  2  7 or 7    6 to give   1 so q = 5
6i+ 6 j+ 3k  9 so unit vector is 19 ( 6i +6j +3k)
M1 A1
M1 A1
(4)
M1 A1
(2)
(c)
cos 
2  2  2  1  1 2
3 3
 cos  89
(d) Write down two of 9  2  3  2  , 7  2  2   or 7    3  2 
Solve to obtain   1 or   2
Obtain coordinates ( 5, 3, 5)
M1 A1
A1
(3)
B1 B1
M1 A1
A1
(5)
January 2006
6673 Pure Mathematics P3Mark Scheme
Question
Number
8(a)
Scheme
dx
 3a sin 3t ,
dt
When x = 0, t =
dy
 a cos t
dt

therefore
Marks
dy
cos t

dx 3sin 3t
B1
6
3
Gradient is 
6
M1
Line equation is ( y  12 a )  
(b)
Area beneath curve is
3
( x  0)
6
M1 A1
(6)
 a sin t (3a sin 3t )dt
M1
3a 2
(cos 2t  cos 4t )dt
2 
3a 2 1

 sin 2t  14 sin 4t 
2 2
M1
=

3 3a 2
6
16
1 a
Area of triangle beneath tangent is   3a =
2 2
2
3a 3 3a 2
3a 2
Thus required area is
=
4
16
16
Uses limits 0 and
N.B.
M1 A1
M1 A1
M1 A1
to give
3a 2
4
M1 A1
A1
(9)
The integration of the product of two sines is worth 3 marks (lines 2 and 3 of
scheme to part (b))
If they use parts
 sin t sin 3tdt   cos t sin 3t   3cos 3t cos tdt
  cos t sin 3t  3cos 3t sin t   9sin 3t sin tdt
8I = cost sin3t – 3 cos3t sint
M1
M1 A1