Download Mark Scheme - June 2002

EDEXCEL ADVANCED EXTENSION AWARD (9801) – JUNE 2002 PROVISIONAL MARK SCHEME
Question
Number
1.
Scheme
Marks
sin 5x + sin x  2 cos 2x sin 3x
cos 5 x + cos x  2 cos 2x cos 3x
use of one
M1
both correct
A1
Equation becomes: 0 = 2 cos 2x (sin 3x – cos 3x)
i.e. cos 2 x = 0
M1
tan 3x = 1
M1
cos 2x = 0  2x =
 3
 3
,
, ...  x = ,
, ...
2 2
4 4
1st solution
M1
tan 3x = 1  3x =
 5 9
 5 9
,
,
...  x =
,
,
, ...
4
12 12 12
4 4
1st solution
M1
a correct 2nd solution
A1
 x=
  5 3
,
,
,
12 4 12
4
all 4
A1
(8 marks)
2.
(1 – 4x)p =
p ( p  1)
p( p  1)( p  2)
p( p  1)( p  2)( p  3)
(4x)2 +
(4x)3 +
(4x)4...
2!
3!
4!
Equation:
p ( p  1)
p( p  1)( p  2)( p  3)
 42 =
 44
2!
4!
1=
i.e.
0 = 4p2 – 20p + 21
i.e.
0 = (2p – 3)(2p – 7)
i.e.
p=
( p  2)( p  3)  16
12
3
7
or
2
2
p  0 and p  1
p
cancel or factor p(p – 1)
at least x2 term
M1 (ignore x’s)
M1
A1
coefficient of x3 > 0  p(p – 1)(p – 2) < 0
so
attempt equation
M1
solving
M1
both
A1
x3 coefficient examined
M1
A1
7
3
p=
2
2
A1
(9 marks)
1
Question
Number
3.
Scheme
Marks
At (14, 1), t = 1
B1
dy
dy
 4t
1
=
, at t = 1
=   gradient of normal, = 3
2
3
d x 15  3t
dx
M1, M1, A1
Equation of normal is: y – 1 = 3(x – 14) [or y = 3x – 41]
M1
3 – 2t2 – 1 = 3(15t – t3 – 14)
Cuts C when:
M1
3t3 – 2t2 – 45t + 44 = 0
i.e.
(t – 1) is a factor

i.e.
simplified cubic = 0
A1
(t – 1) is a factor
M1
[]
A1
(t – 1)[3t2 + t – 44] = 0
(t – 1) (3t – 11)(t + 4) = 0
t=
M1
11
, 4
3
(11 marks)
4.
dy
dy
d
: 3x2 + 3y2 , 3x
 3y = 0
dx
dx
dx
M1 A1, M1
dy
= 0,  y = x2
dx
M1, A1
substitute back:
x3 + x6 – 3x  x2 = 48
x6 – 2x3 – 48 = 0
A1
(x3 – 8)(x3 + 6) = 0
M1
i.e.
 x = 2 and y = 4
so x3 = 8
1
A1
2
or x3 = 6  x =  6 3 and y = 6 3
A1
(10)
2
dy
dy
d
d2 y
d2 y
 dy 
again: 6x + 6y   + 3y2 2  3
 3x 2  3
=0
dx
dx
dx
dx
dx
 dx 
(x = 2, y = 4) < 0
(2, 4) is maximum
A1
check signs of y
M1
dy
2x
d2 y
=0
=
2
dx
x  y2
dx
1
2
(x =  6 3 , y = 6 3 ) > 0
M1
1
2
(  6 3 , 6 3 ) is minimum
A1
(4)
(14 marks)
Alternative
for last 4
marks
y
x
Sketch from top left to bottom right
M1
Min in 2nd quad, max in first quad
M1
1
2
 minimum at (  6 3 , 6 3 )
maximum at (2, 4)
2
A1
A1
(4)
EDEXCEL ADVANCED EXTENSION AWARD (9801) – JUNE 2002 PROVISIONAL MARK SCHEME
Question
Number
5.
(a)
Scheme
sin (cos x) = 0  cos x = 0 (, ...)
 x=


or 
2
2
so A is (
x = 0  y = sin (1) ,
(b)
Marks
(ignore others)


, 0), C is (  , 0)
2
2
M1
A1 both
B1
i.e. B is (0, sin (1))
dy
= cos (cos x) sin x
dx
M1
 dy 
x = 0 at B and   = 0,  B is a stationary point
 dx  0
A1
(3)
(2)
(c) For   0, sin   
cos x  0 for x  [ 0, /2]  sin (cos x)  cos x; equality when x =
Equation of BC is y =
 sin (1)
x + sin (1)
 /2
2

 convex line is below curve,  sin (1) 1 
 

2
attempt equation BC

x   sin (cos x)

equality when x = 0,

(d)
A1 cso; B1
M1
A1 cso

2
B1 (both)
(6)


2
2
=
cos
x
d
x
sin
x

 , =1

0

0
I<1
M1, A1 cso

2
 2 
1 

 sin (1) 1  x dx OR area of triangle = 2  2 sin (1), I > 4 sin (1)
0


M1, A1 cso (4)
(15 marks)
cso = correct solution only
3
EDEXCEL ADVANCED EXTENSION AWARD (9801) – JUNE 2002 PROVISIONAL MARK SCHEME
Question
Number
6.
(a)
Scheme
(3, 0)  m1 = 3 n1 , m2 = 3 n2 and
Marks
 n2 > n1
m2 > m1
M1
curves are symmetric and n1, n2 are both even
M1
(b) so n1 + n2 = 12  n1 = 2, n2 = 10; or n1 = 4, n2 = 8
A1; A1
(4)
(1 each
extra solution)
3

Area = 2
0
 m
2

 m1   x n1  x n2
  dx
M1
3

x n1 1
x n2 1 
= 2(m2  m1 ) x 


n1  1 n2  1 0

M1 A1 ft
smallest area when m1 and m2 are closest M1
i.e. n1 = 4, n2 = 8, m1 = 34, m2 = 38 A1
 8

35 39
4
= 23(3  3 ) 

 0
5
9


= 2  39 – 2  35 +
= 2 [39 – 37 –
8
5
2
5
 35 – 2  37
 35] OR
Use of correct limits
M1
combine powers o3 that differ by 1 or less
M1
16  37 –
(c) Gradient same   n1 x n1 1   n2 x n2 1
i.e.
n1
= x n2  n1
n2
1
2
n1 = 2, n2 = 10  x8 =
( 12 )2 >
1
5
or
8 1
5
=
4
1
5
or x =
1
5
1
5
712
5
OR
 35
A1
Equation based on dy/dx
M1
single x
M1
(8)
4 1
2
or x =
and
 35
n1
n2
OR x = (n2  n1)
n1 = 4, n2 = 8  x4 =
8
5
8 1
5
<
both cases
1
2
,  greatest x =
 12 
1
4
A1 ft
M1, A1
(5)
(17 marks)
ft = follow-through mark
4
EDEXCEL ADVANCED EXTENSION AWARD (9801) – JUNE 2002 PROVISIONAL MARK SCHEME
Question
Number
7.
Scheme
(a)
pq =
(b)
x3 +
p=
1
2
3
4
x
i.e. x =
1
2
1
2
1
2
or q =
1
2
= 0  (x 
or x2 +
1
2
(line 3)
1
2
)(x2 +
x + 1 = 0,
1
2
x + 1) = 0
identify; explain
B1; B1 (2)
attempt to divide
M1
correct quadratic
A1
discriminant = ( 12 )2 – 4
< 0  no real roots (so only root is x =
(c)
Marks
x =  is a root   3 +    = 0,
1
2
M1
)
A1 cso
i.e.  = 1 –  2 (  0)
M1, A1
x 3 + x    (x – )[x2 +  x + 1]
M1 [A1]
Discriminant of x2 +  x + 1 is  2 – 4
M1
 x =  is the only real root if  2 – 4 < 0,
(4)
i.e.  < 2 ()
A1 cso
(6)
(d) Student’s method: x(x2 + ) = 
 x =  or x2 +  = 
require
M1
 >0
2+1>0
=
cvs

1 5
2
5 1
<<2
2
or
2 <  < 
attempt cvs
M1
2 correct cvs
A1
5 1
2
A1, A1
(7)
(19 marks)
STYLE INSIGHT & REASONING
(a) S marks
For a novel or neat solution to any of questions 3—7. Apply once per
question in up to 3 questions
S2 if solution is fully correct in principle and accuracy
S6 (S2  3)
S1 if principle is sound but includes a minor algebraic or numerical slip
T mark
For a good and largely accurate attempt at the whole paper
T1
(7 marks)
() indicates final line given in the question paper; ft = follow-through mark; cso = correct solution only
5