Add Maths June 2005

1
The Further Mathematics Network
Worked Solutions to Additional Mathematics
June 2005
2
Stationary points for
1
y 9x2 2x3 (1)
Differentiate
18x 6x2 0
3x x2 0
Simplify
Factorise
x(3 x) 0
Solve
x 0 or x 3
Sub. in (1) to find y
y 932 233 27
or y 902 203 20
Stationary points are
(3, 27) and (0, 0)
3
1
Stationary points for
(3, 27) and (0, 0)
y 9x2 2x3
Examine gradients either side of (3, 27)
Maximum
4
2
Function f(x) x3 4x2 5x 2
(i) When f(x) is divided by (x 2) remainder is
f(2)
f(2) (2)3 4(2)2 5(2) 2
8 16 10 2 36
(ii) (x 1) is a factor of f(x) ? f(1) 0
f(1) 13 412 51 2
1 4 5 2 0
? (x 1) is a factor of f(x)
(iii) f(x) 0 ? x3 4x2 5x 2 0
? (x 1)(px2 qx
r) 0
? (x 1)(x2 qx 2)
0
? (x 1)(x2 3x 2)
0
? (x 1)(x 1)(x
2) 0
? x 1 (repeated
root) or x 2
5
Let triangle be ABC with a 8, b 7 and c 12
3
Largest angle is opposite longest side, i.e. ?C
Cosine rule
? cos C 0.277 (to 3 s.f.)
? ? C 106? (to nearest degree)
6
4
Solve the equation 4 sin ? 3 cos ? for 0?
? 360?
Divide by cos ?
? 4 tan ? 3 ? tan ? 0.75 ?
? 36.9? or ? 36.9 ?
180 ? 216.9?
Solution set ? 36.9? or 216.9?
7
n 8, p 0.14 ? q 1 p 0.86
Binomial Distribution
5
Let X represent number of defective glasses X
B(8, 0.14)
P(X r) 8Cr0.14r0.868r
(a)
P(no defective glasses)
P(X 0)
0.868
0.299 (to 3 s.f.)
(b)
P(at least two defective glasses)
P(X 2)
1 P(X 0) P(X 1)
1 P(X 1)
1 0.868 8C10.140.867)
1 0.29921793 0.389679161
0.311 (to 3 s.f.)
(b) is greater than (a)
8
(a b)4 a4 4C1a3b 4C2a2b2 4C3ab3
b3
6
(i)
(ii)
Substituting x 1
0
1 4 6 4 1 0
9
Gradient function of curve
7
a bx
Integrate
y ax ½bx2 c
Curve passes through (0, 2), (1, 8) and (1, 2)
? 2 a0 ½b02 c
? 2 c (1)
? 8 a1 ½b12 c
? 8 a ½b c (2)
? 2 a(1) ½b(1)2 c
? 2 a ½b c (3)
From (1) c 2
? 6 2a
(2) (3) 8 2 a (a )
? a 3
? b 6
Sub. in (2)
8 3 ½b 2
? 3 ½b
10
y ax ½bx2 c
7
Equation of curve
? y 3x 3x2 2
11
8
Variable acceleration
v 27 ? t3 (1)
t 6?
(i) When car is at rest
v 0
Substitute t 6 in (1)
v 27 ? 63
? v 27 27 0 ms-1
(ii) To find distance in AB seconds, integrate
(1) between 0 and 6
s
? s (162 40.5) (0 0) 120.5 m
12
Pythagoras Theorem cos2? sin2?
1 ? cos2? 1 sin2?
9
1
sin ?
?
cos ?
2cos2? sin ? 2
(i)
Starting with equation
Substitute for cos2 ?
2(1 sin2?) sin ? 2
Expand bracket
2 2sin2? sin ? 2
Rearrange
2sin2? sin ? 0
13
9
To solve equation
for 0? ? 180?
2cos2? sin ? 2
for 0? ? 180?
Solve
2sin2? sin ? 0
(ii)
Factorise
sin ? (2 sin ? 1) 0
Either
sin ? 0
?
? 0? or 180?
Or
2 sin ? 1 0
sin ? ½
?
?
? 30? or 150?
14
y ?x2 2x 10
10
Parabola has equation
(i)
Differentiate
Substitute x 3
A
(3, 10)
Equation of tangent
y 2x c
Sub. x 3 , y 10 to find c
10 23 c
? c 10 6 4
Equation of tangent
y 2x 4
15
y ?x2 2x 10
10
Parabola has equation
(i)
Differentiate
Substitute x 3
A
(3, 10)
Equation of tangent
y 2x c
Sub. x 3 , y 10 to find c
10 23 c
? c 10 6 4
Equation of tangent
y 2x 4
16
y ?x2 2x 10
10
Parabola has equation
(ii)
Substitute x 0
(0, 10)
B
17
y ?x2 2x 10
10
Parabola has equation
(ii)
Substitute x 0
(0, 10)
Gradient of normal
1 ? (2) ½
B
Equation of normal
y ½ x c
Sub. x 0 , y 10 to find c
10 ½0 c
? c 10 0 10
Equation of normal
y ½ x 10
2y x 20
Rearrange
18
y ?x2 2x 10
10
Parabola has equation
(ii)
Substitute x 0
(0, 10)
Gradient of normal
1 ? (2) ½
B
Equation of normal
y ½ x c
Sub. x 0 , y 10 to find c
10 ½0 c
? c 10 0 10
Equation of normal
y ½ x 10
2y x 20
Rearrange
19
To find coordinates of point of intersection of
tangent and normal
10
(iii)
Equation of tangent
y 2x 4
C
y ½ x 10
Equation of normal
(4, 12)
B
Equate RHS
2x 4 ½ x 10
4x 8 x 20
Multiply by 2
Rearrange
3x 12
x 4
Divide by 3
Sub. x 3 in y 2x 4
y 12
Point of intersection
(4, 12)
20
To calculate the length of BC
10
(iv)
By Pythagoras Theorem
BC2 BD2 DC2
C
(4, 12)
? BC2 42 22 20
? BC v20
(0, 10)
B
? BC 4.47 (to 3 s.f.)
21
11
(i) Let x represent the number of components of
type X
Let y represent the number of components
of type Y

• Each component of type X requires materials
  costing 18.
• Each component of type Y requires materials
  costing 11.
• Each week materials worth 200 are available.

18x 11y 200

• Each component of type X takes 7 man hours to
  finish.
• Each component of type Y takes 6 man hours to
  finish.
• Each week there are 84 man hours available.

7x 6y 84
x 2

• At least 2 type X components must be produced
  each week.

y 2

• At least 2 type Y components must be produced
  each week.

22
11
(ii) Region for 18x 11y 200
18x 11y 200
23
(ii) Region for 18x 11y 200 and 7x
6y 84
11
18x 11y 200
7x 6y 84
24
Region for 18x 11y 200, 7x 6y 84,
x 2, y 2
11
18x 11y 200
7x 6y 84
25
Region for 18x 11y 200, 7x 6y 84,
x 2, y 2
11
18x 11y 200
7x 6y 84
(9, 3)
Maximum value of 70x 50y is 780, when x 9
and y 3
26
12
(i) Circle equation
x2 y2 2x 4y 20 0
Recall A circle with centre (a, b) and radius
r has equation (x a)2 (y b)2 r2
x2 2x y2 4y 20
Rearrange
(x 1)2 1 (y 2)2 4 20
Complete square
(x 1)2 (y 2)2 25
Rearrange
? Circle has centre C (1, 2) and radius 5
27
12
(ii) Circle equation
x2 y2 2x 4y 20 0 (1)
Line equation
y x 2 (2)
From (2) sub. y x 2 in (1)
(4, 6)
B
x2 (x 2)2 2x 4(x 2) 20 0
x2 x2 4x 4 2x 4x 8 20 0
C
? 2x2 2x 24 0
? 2(x2 x 12) 0
A
(3, 1)
? 2(x 3)(x 4) 0
? x 3 or x 4
When x 4, y 6
When x 3, y 1
B (4, 6)
Line meets circle at
A (3, 1)
28
12
(iii) Circle Centre C (1, 2) and radius 5
AC BC 5 (radii of circle)
(4, 6)
AB2 72 72 98
B
Cosine rule
C
A
(3, 1)
? cos C 0.96 (to 3 s.f.)
? ? C 163.7? (to one d.p.)
29
13
(i) At A and B, y 0
B
A
? (x2 16) 0
? (x 4)(x 4) 0
? x 4 or x 4
? A is (4, 0) and B is (4, 0) and width of
river is 8 m
(ii) Depth of river at deepest point occurs
when x 0
? Depth of river is 3 m
30
13
(iii) To find shaded area
B
A
16
? Cross sectional area 16 m2
31
13
(iv) To find volume per minute
B
A
Multiply rate of flow by cross sectional area
Volume per minute 20 16 320 m3 per
minute
(v) Reasons why model might not be a good model

• Water does not usually flow at a constant rate
• The river bed will not be symmetric