Upload
vuanh
View
227
Download
1
Embed Size (px)
Citation preview
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 1
Requirement : Answer all questions
Total marks : 80
Duration : 2 hours
1. The equation of a curve is 22 4y x kx , where k is a constant, and the equation of a line
is 2 12y x .
(i) In the case where k = 6, find the coordinates of the points of intersection of the line
with the curve. [3]
(ii) Show that, for all values of k, the line intersects the curve at two distinct points. [2]
Solution:
(i) Solving 22 4y x kx and 2 12y x ,
22 2 16 0x k x ----- (1)
For k = 6,
2 2 8 0x x
2 4 0x x
2x or 4x
When 2x , 12 2 2 16y
When 4x , 12 2 4 4y
the coordinates of the points of intersection (‒2, 16) and (4, 4)
(ii) From equation (1),
The discriminant
2
2 4 2 16k
2
2 128 128 0k
Since the discriminant > 0, the equation 22 2 16 0x k x will always have two distinct
roots, for all values of k.
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 2
2. A rectangular block has a square base of side 6 2 cm and a height of h cm.
The volume of the rectangular block is 16 4 3 cm3. Without using a calculator, show
that h can be expressed as 3a b , where a and b are integers. [5]
Solution:
2
16 4 3
6 2
h
2
2 8 2 3
2 3 1
h
2
2 4 3
3 1
h
2
2 4 3
3 1
h
2 4 3
2 2 3h
4 3 2 3
2 3 2 3h
2
2
8 2 3 4 3 3
2 3
h
5 2 3h
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 3
3. (a) State the values between which each of the following must lie:
(i) the principal value of 1sin x ,
(ii) the principal value of 1cos x .
(b)
The figure shows part of the graph of cosx
y a cb
.
Find the value of each of the constants a, b and c. [3]
Solution:
(a) (i) 190 sin 90x
(ii) 10 cos 180x
(b)
3 1
22
c
1 3
12
a
Period = 12π
12
62
b
y
x O ‒3π
3
2
1
‒6π ‒9π ‒12π 3π 6π 9π 12π
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 4
4. (i) On the same axes sketch the curves 2 256y x and 22y x . [3]
(ii) Find the equation of the line passing through the points of intersection of the two
curves. [2]
Solution:
(i)
(ii) Solve 2 256y x and 22y x
2
22 256x x
44 256x x
4 64 0x x
3 64 0x x
0x or 3 64x
4x
Coordinates of the points of intersection are (0, 0) and (4, 32).
Gradient of the line = 8
Equation of the line, 8y x
y
x O
22y x
2 256y x
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 5
5. Express 2
2
2 4 31
6
x x
x x
in partial fractions. [5]
Solution:
2
2 2
2 4 31 2 192
6 6
x x x
x x x x
.
2
2 19 2 19
6 3 2
x x
x x x x
Let
2 19
3 2 3 2
x A B
x x x x
2 19 2 3x A x B x
When 3x , 25 5A
5A
When 2x , 15 5B
3B
2
2
2 4 31 5 32
6 3 2
x x
x x x x
.
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 6
6.
The diagram shows part of the curve 2y ax bx c , where 0a .
The curve touches the x-axis at A(p, 0) and at B(6,0) and has a maximum point at M(4, 6).
(i) Explain why p = 2. [1]
(ii) Determine the value of each of a, b and c. [4]
(iii) State the set of values of q for which the line y = q intersects the curve at four distinct
points. [2]
Solution:
(i) Since 0a , consider 2y ax bx c
Line of symmetry, 4x
4 6 4 2p
(ii)
Let 2 6y a x x
For 4, 6 ,
6 4 2 4 6a
6
4a
3
2a
(iii) 0 6q
y
x O A (p, 0)
2y ax bx c
B (6, 0)
M (4, 6)
y
x O A (p, 0)
2y ax bx c
B (6, 0)
M (4, ‒6)
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 7
7. At a given instant, a cyclist is at a cross roads O, travelling due west at a constant speed
of 5 m/s.
At the same instant a second cyclist is 100 m from O, travelling due north towards O at a
constant speed of 10 m/s. This situation is shown in Fig. 1.
The position, t seconds later, when the cyclists have reached points P and Q, is shown in
Fig. 2.
(i) Express OP and OQ in terms of t and hence show that the distance, s m, between the
two cyclist at time t is given by 2125 16 80s t t . [3]
(ii) Obtain an expression for d
d
s
t. [2]
(iii) Find the least distance between the two cyclists. [2]
Solution:
(i) 5OP t
100 10OQ t
2 2
5 100 10s t t
N • O 5 m/s
Q • 10 m/s
P •
Fig. 2
N • O 5 m/s
100 m
• 10 m/s
Fig. 1
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 8
2 225 10000 100 2000s t t t
2125 2000 10000s t t
2125 16 80s t t
(ii) 2 2125 16 80s t t
d
2 125 2 16d
ss t
t
d 125
8d
st
t s
2
125 8d
d 125 16 80
ts
t t t
(iii)
Let d
0d
s
t ,
2
125 80
125 16 80
t
t t
0 125 8t
8t
When 7t , d
0d
s
t , s is downward sloping.
When 8t , d
0d
s
t , s is a stationary point.
When 9t , d
0d
s
t , s is upward sloping.
When 8t , s is a locally minimum point.
2125 8 16 8 80 20000 44.7s (3 s.f.)
the least distance between the two cyclists is 44.7 m when 8t s.
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 9
8.
The diagram shows a triangle ABC in which the point A is (‒2, 6), the point C lies on
the x-axis and angle ABC is 90°. The equation of BC is 2 3 45y x .
(i) Find the coordinates of B [5]
(ii) Given that M is the midpoint of AC and that ABCD is a rectangle,
find the coordinates of M and of D. [3]
Solution:
(i) 2 3 45y x , gradient of BC = 3
2
Gradient of AB =2
3
The equation of AB ,
26
2 3
xy
2 22
3 3y x
2 22
2 3 453 3
x x
13 91
3 3x
7x
2 3 7 45y
12y
the coordinates of B (7, 12)
y
x O
A
(‒2, 6)
2 3 45y x
B
C
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 10
(ii) A(‒2, 6). Let 0y , 15x . C(15, 0)
M(2 15
2
,
6 0
2
)
M(6.5, 3)
Let D(a, b), B (7, 12)
7
6.52
a
6a
12
32
b
6b
D(6, ‒6)
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 11
9. The equation of a curve is 2
2
162y x
x .
(i) Find the coordinates of the stationary points of the curve. [5]
(ii) Use the second derivative test to determine the nature of each of these points. [3]
Solution:
(i) 3
d 322
d
yx
x x
Let d
0d
y
x ,
3
320 2x
x
4 16x
2x
When 2x , 2
2
162 2 6
2y
When 2x ,
2
2
162 2 6
2y
the coordinates of the stationary points are (2, ‒6) and (‒2, ‒6)
(ii) 3
d 322
d
yx
x x
2
2 4
d 962
d
y
x x
When 2x , 2
2
d0
d
y
x , it is a locally maximum point.
When 2x , 2
2
d0
d
y
x , it is a locally maximum point.
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 12
10. (i) Show that 2 3 3
d ln 1 2ln
d
x x
x x x x
. [3]
(ii) Integrate 3
ln x
x with respect to x. [3]
(iii) Given that the curve fy x passes through the point 3
1,4
and is such that
3
lnf
xx
x , find f x . [2]
Solution:
(i) let 2
ln xy
x , we get
2
22
12 ln
d
d
x x xy x
x x
4
d 2 ln
d
y x x x
x x
4
1 2lnd
d
x xy
x x
3
d 1 2ln
d
y x
x x
Therefore,2 3 3
d ln 1 2ln
d
x x
x x x x
(ii) since 2 3 3
d ln 1 2ln
d
x x
x x x x
2 3 3
d ln 1 2lnd d
d
x xx x
x x x x
2 3 3
d ln 1 2lnd d d
d
x xx x x
x x x x
3 3 2
2ln 1 d lnd d d
d
x xx x x
x x x x
3 3 2
2ln 1 d ln2 d d d
2 d
x xx x x
x x x x
3 2 2
ln 1 lnd
4 2
x xx c
x x x
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 13
(iii) Given that 3
lnf
xx
x and the curve fy x passes through the point
31,
4
.
3
lnf d d
xx x x
x
2 2
1 lnf
4 2
xx c
x x
2 2
1 ln
4 2
xy c
x x
At 3
1,4
,
3 1 ln1
4 4 2c
1c
2 2
1 ln1
4 2
xy
x x
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 14
11. (a) Prove that 2 1 sin
sec tan1 sin
. [4]
(b)
A public building has a large clock set on an exterior wall. The distance from the
centre of the clock, O, to the tip, M, of the minute hand is 80 cm.
The distance, d cm, of M from the vertical line through O is given by 80sind kt ,
where t is the time in minutes past the hour.
(i) Find the value of k in radians per minute. [1]
(ii) For how long in each hour is d > 40? [3]
Solution:
(a) R.H.S
21 sin
1 sin 1 sin
2
2
1 sin
1 sin
2
2
1 sin
cos
21 sin
cos
21 sin
cos cos
2
sec tan
XII
80 cm
O III
VI
IX •
d cm M
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 15
(b) (i) When t = 15, kt = 2
k = 30
(ii) For 80sin 40kt ,
80sin 40kt or 80sin 40kt
sin 0.5kt
1sin 0.5kt
From the sketch of the graph of siny kt ,
Principal angle, 1sin 0.56
kt
6
kt
or 5
6 6kt
5 2 30
206 6 30 3
t
By symmetry, d > 40, 2 20 2 40t minutes
y
kt O
‒1
1
0.5
‒π π 2π
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 16
12. It is given that f sin cos 2x dx x k x c , where c is a constant of integration,
and that 6
0
3f
4x dx
.
(i) Show that 1
2k . [1]
(ii) Find f x . [3]
(iii) Find the equation of the normal to the curve fy x at the point where 6
x
. [5]
Solution:
(i) 6
0
3f
4x dx
3
sin cos 2 64
0
x k x c
3
sin cos sin 0 cos06 3 4
k c k c
1 3
2 2 4
kc k c
1 3
2 2 4
k
1
2k
(ii) 1
f sin cos 22
x dx x x c
d 1
f sin cos 2d 2
x x x cx
f cos sin 2x x x
(iii) d
sin 2cos 2d
yx x
x
When 6
x
,
d
sin 2cosd 6 3
y
x
Add Math (4047/01)
Prepared by Mr Ang, Nov 2016 17
d 1 1 1
2d 2 2 2
y
x
cos sin 36 3
y
Gradient of the normal at 6
x
is ‒2.
3
2 6
yx
2 33
y x