Download June 2005 - 6676 Pure P6

June 2005
6676 Pure P6
Mark Scheme
Question
Number
1
Scheme
Marks
(a)
6 x  10
8
6
x3
x3
B1
(b)
u1  5.2  5
B1
If result true for n = k , i.e. u k  5,
8
uk  1  6 
uk  3
8
 1 so u k  1  5
If u k  5, then
uk  3
Hence result is true for n = k + 1
Conclusion and no wrong working seen
(1)
M1A1
A1
(4)
[5]
2
(a)
(i) b  a is perpendicular to a (and b)
B1
a. b  a = |a| | b  a| cos 90 = 0 or equivalent
B1
(ii) a  b = a  c  a  (b – c) = 0
(2)
M1
As a  0 and b  c,
a is parallel to (b – c), so b – c =  a
(b) (i) If A non-singular, then A–1 AB = A–1 AC  B = C
 3 6  1 5   3 21

  

(ii) 
 1 2  0 1   1 7 
A1
(*)AG
(2)
M1A1 (2)
B1
 3 6  a b   3 21

  
 and finding two equations
Set 
 1 2  c d   1 7 
M1
Any non-zero values of a, b, c and d such that
a + 2c = 1 and b + 2d = 7.
A1
(3)
[9]
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Question
Number
Scheme
i
3
j
Marks
k
(a) Normal to plane is 2 0 3 = 6i + j – 4k
1 2 1
(or any multiple)
M1
(b) Equation of plane is 6x + y –4z = d
Substituting appropriate point in equation to give 6x + y – 4z = 16
[e.g. (1, 6,–1), (3, –2, 0), (3, 6, 2) etc.]
(c) p = –2
(d)
M1A1 (2)
A1
(2)
B1
(1)
Direction of line is perpendicular to both normals
i
j
k
M1
1 = –9i + 10j –11k
6 1 4
1 2
[Planes are: 6x + y –4z = 16,
x + 2y + z = 2]
M1A1
Finding a point on line
M1
a and b identified

  3   9 

  

Any correct equation of correct form e.g. r   6  x   10  = 0.
  7   11 


  
Alternative: Using equations of planes to find general point on line
Using equations of planes to form any two of
10x + 9y = 24, 11x – 9z = 30, 11y + 10z = –4
Putting in parametric form
 24  10  30  11 
,
e.g.   ,

9
9


a and b identified
Writing in required form; a correct equation
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1
M1
A1
M1
A1
A1 (5) [10]
(a)
Circle
M1
3
Correct circle.
A1 (2)
0
(centre (0, 3), radius 3)
4
A
(b) Drawing correct half-line passing as shown
B1
Find either x or y coord of A.
M1A1
3 2
3 2
+ (3 +
)i
2
2
[Algebraic approach, i.e. using y  3  x and equation of circle
will only gain M1A1, unless the second solution is ruled out,
when B1 can be given by implication, and final A1, if correct]
z= –
(c) |z – 3i| = 3 
2i

A1 (4)
M1
 3i = 3
A1
 | 2i – 3i | = 3
||
 |  – 2 /3 | = |  |
Line with equation u = 1/3
M1A1
A1 (5)
(x = 1/3)
[11]
Some alternatives:
2i
2 i ( x  iy )
2y
2x
 2
u 2
, v 2
M1A1
2
2
x  iy
x  y
x  y
x  y2
1
u ,
As x 2  y 2  6 y  0 ,
M1,A1A1
3
(i)  
(ii)  
=
2i
2 i{cos   i(1  sin  )}

3 cos   3i(1  sin  ) 3{cos 2   (1  sin  ) 2 }
2 1  sin    icos 
1
cos 
,   i
,
3
2  2sin 
3
1  sin 
So locus is line u =
1
3
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1A1
M1A1
A1
5
(a)
z n  e i n  (cos n + i sin n ),
Completion (needs to be convincing) z n 
z  n  e -i n  ( cos n – i sin n)
M1
1
zn
A1 (2)
 2 i sin n (*)AG
5
10
5
1
1

5
3
 3  5
 z   = z  5 z  10 z 
z
z
z
z

(b)
M1A1
1 
1 
1



=  z 5  5   5 z 3  3   10 z  
z
z 
z 



( 2 i sin  )5 = 32i sin5  = 2i sin 5 – 10i sin 3 + 20i sin 
 sin5  =
(c)
1
16
A1 (5)
( sin 5 – 5sin 3 + 10 sin  ) (*) AG
M1
Finding sin5  = ¼ sin
  0,  (both)
(sin4  = ¼)
M1A1
B1
 sin  = ±
1
M1
2
 

4
,
3
;
4
5
4
,
7
A1;A1 (5)
4
[12]
6
(a)
 d2 y 
1
 2  
 dx  0 4
y  y 1
1 y1  y 1
 dy 
(b)    1


 y1  y 1  0.1
2h
2
0.2
 dx  0
 d2 y 
y  2 y0  y 1
y  2  y 1
1
 2   1
 1

2
4
h
0.01
 dx  0
 y1  y1  2.0025
Adding to give y1  1.05125
d3 y
d2 y
d2 y
dy dy

8
x

4
x
4

3
2
2
dx dx
dx
dx
dx
 d3 y 
 d3 y 
3
3
Substituting appropriate vales  4  3      3   
2
8
 dx  0
 dx  0
y 
y 
1
1
1
(d) y = y0  y0 x  0 x 2  0 x 3  ... = 1 + x + x2 –
x3 + …
2
8
16
2!
3!
(c) Diff: 4(1  x 2 )
(e) 1.05119
B1
(1)
M1A1
M1
A1
M1A1 (6)
M1A1
M1A1 (4)
M1A1√(2)
A1
(1)
[14]
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
7
(a) Det = –12 –2(2k – 8) + 16 = 20 – 4k
(b) Cofactors
A–1
(*)
8  2k
 4

 8  2k 3k  16
 4
2

=
AG
4 

2 
 4 
 4

1
 8  2k
20  4k 
 4
M1A1 (2)
[A1 each error]
8  2k
3k  16
2
4 

2 
 4 
 3 2 4  0   0 

   
(c) Setting  2 0 2  2     2 
 4 2 3   1  1 

   
M1A1√ (6)
M1
 = –1
(d) Forming equations in x, y and z:
M1A3
A1
 3 2 4  x 
 x

 
 
 2 0 2  y   8 y 
 4 2 3  z 
z

 
 
–5x + 2y + 4z = 0, 2x + 2z = 8y, 4x + 2y – 5z = 0
(2)
M1
A1
Establishing ratio x: y: z : [x = 2y, x = z]
 2
 
Eigenvector ( k ) 1 
 2
 
M1
A1
(4)
[14]
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
6676 Pure P6
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics