1)
The point A has coordinates (7, 4).
The straight lines with equations x + 3y + 1 = 0 and 2x + 5y = 0
intersect at B.
(a)
(b)
Find the gradient of AB.
Hence show that AB is perpendicular to only one of these two lines.
(a)
B is the point of intersection of
and
Soln
x + 3y + 1 = 0
2x + 5y
=0
2x + 6y + 2 = 0
2x + 5y
=0
2x + 6y + 2 = 0
2x  5y
=0
y+2=0
(  2 both sides )
y = 2
2x + 5y
=0
2x + (5)(2) = 0
2x 10 = 0
( + 10 both sides )
2x = 10
(  2 both sides )
x=5
B is the point (5, 2)
( 2)
( 1)
Add
A is the point (7, 4)
B is the point (5, 2)
(b)
mAB =
y2  y 1
y2  y 1
=
2  4
57
=
6
2
=
6
2
mAB =
3
A line perpendicular to AB has gradient 1
3
If perpendicular m1  m2 = 1
i.e.
3  1 = 1
3
x + 3y + 1 = 0
2x + 5y = 0
3y = x  1
5y = 2x
y = 1 x  1
3
3
y = 2 x
5
m = 1
3
m = 2
5
AB is only perpendicular to
x + 3y + 1 = 0
2)
f(x) = x3  x2  5x  3
(a)
(i)
(ii)
Show that (x + 1) is a factor of f(x).
Hence or otherwise factorise f(x) fully.
(b)
One of the turning points of the graph of y = f(x) lies on the x-axis.
Write down the coordinates of this turning point.
Soln
(a) (i)
factor (x + 1)
root x = 1
1
1
1
(a) (ii)
f(x) =
=
=
f(x)
(b)
=
1
1
2
5
2
3
3
3
0
x3  x2  5x  3
(x + 1)( x2  2x  3)
(x + 1)(x + 1)(x  3)
(x  3)(x + 1)2
Stationary points
x-coordinates
=
x3  x2  5x  3
f’(x) =
f’(x) =
3x2  2x  5
(3x  5)(x + 1)
f(x)
Set f’(x) = 0
differentiate y = f(x) and set equal to zero
(3x  5)(x + 1) = 0
(3x  5) = 0 or (x + 1) = 0
3x = 5
x=5
3
or
x = 1
y-coordinates
f(x) = x3  x2  5x  3
x = 1
x=5
3
f(5/3) = (5/3)3  (5/3)2  5(5/3)  3
f(1) = (1)3  (1)2  5(1)  3
f(5/3) = 125  25  25  3
27
9
3
f(1) = 1  1 + 5  3
f(5/3) = 125  75  225  81
27
27 27 27
f(1) = 0
f(5/3) =  175  81
27 27
f(5/3) = 256
27
f( /3) =  9 /27
5
13
27
 9
243
The point ( 5/3 ,  913/27 )
is a stationary point
The stationary/turning point ( 1 , 0 )
lies on the x-axis
i.e. y = 0
The point ( 1 , 0 )
is a stationary point
3)
Find all the values of x in the interval 0  x  2 for which tan2(x) = 3.
Soln
tan2(x) = 3
( root both sides )
tan(x) =  3 =  3 (o)
1 (a)
All four quadrants
0
60

3
2
0
30

6
(o)
1
(a)
3
tan  = 3 = 3
3
1
tan (   ) = tan 2 = 3
3
3
tan ( +  ) = tan 4 = 3
3
3
tan (2  ) = tan 5 = 3
3
3
4 solutions
x =
 , 2 , 4 , 5
3 3 3
3
1800/
S
A
T
C
0/3600/2
4)
The diagram shows the graph of y = g(x).
(a)
Sketch the graph of y = g(x).
(b)
On the same diagram, sketch
the graph of y = 3  g(x).
y
(b, 3)
y = g(x)
(0, 1)
x
O
(a, 2)
Soln
y = g(x)
(a, 2)
(0, 1)
(b, 3)
y = g(x)
( reflect the graph of y = g(x)
in the x-axis )
( multiply the y-coordinates of
y = g(x) by 1 )
(a, 2)
(0, 1)
(b, 3)
y
(a, 5)
y = 3  g(x)
(a, 2)
(0, 2)
O
x
y = 3  g(x) = g(x) + 3
( add 3 to the y-coordinates of
y = g(x) )
(a, 5)
(0, 2)
(b, 0)
y = g(x)
(b, 0)
(0, 1)
(b, 3)
5)
A, B and C have coordinates (3, 4, 7), (1, 8, 3) and (0, 10, 1) respectively.
(a)
Show that A, B, and C are collinear.
(b)
Find the coordinates of D such that AD = 4AB.
Soln
(a) AB
=
b
=
1
8


3
=
=
1 + 3
a
BC
=

c
3
0
4
10
7
1
=
10  8
37
13
=
4
2
4
2
2
2
2
AB = 2 BC

8
3
1
1
=
1
0+1
84
2
b
2
=
1
2
4
4
B
BC = 1 AB
2
Parallell / same direction / common point B  collinear
2
A
1
C
(b)
AD = 4 AB
4 AB
=
2
4
AD
4
=
d
=
x
4
y
z
4 AB
=
8
16
AD
=
16
x+3
y4
z7
x+3=8
y  4 = 16

z  7 = 16
x=5

y = 20
z = 9
D(5, 20, 9)

a
3

4
7
6)
Given that y = 3sin(x) + cos(2x), find dy .
dx
Soln
y = 3sin(x) + cos(2x)
dy = 3cos(x) + ( sin(2x) )  (2)
dx
dy = 3cos(x) 2sin(2x)
dx
7)
Find
Soln
=
=

2

2

2
4x + 1 dx
0
4x + 1 dx
0
(4x + 1)1/2
0
(4x + 1)
3
2
3/2
1
2
4
0
=
(4x + 1)
3/2
21
3
2
4
0
=
=
2
2 (4x + 1)3/2
12
(4x + 1)3/2
6
0
2
0
=
( (4(2) + 1) )3
6
=
(9)3  (1)3
6
6
=
33  13
6
6
=
26
6
=
41/3 units2
=
=
13
3
( (4x + 1) )3
6

( (4(0) + 1) )3
6
=
27  1
6
6
2
0
8)
(a)
Write x2  10x + 27 in the form (x + b)2 + c.
(b)
Hence show that the function g(x) = 1 x3  5x2 + 27x  2
3
is always increasing.
Soln
(a)
=
x2  10x + 27
2
(x  5)2
(x  5)2 + 2
(x  5)2 + 2 = x2  10x + 25 + 2
= x2  10x + 25
= x2  10x + 27
(b)
g(x) = 1 x3  5x2 + 27x  2
3
g’(x) = x2  10x + 27
g’(x) = (x  5)2 + 2
(x  5)2  25 for all x

(x  5)2 + 2  27 for all x

g’(x)  0 for all x

the function g(x) is always increasing
g’(x) = (x  5)2 + 2
g’(x) is a parabola with a minimum turning point at (5, 2).
There is no point of contact of g’(x) with the x-axis (i.e. no real roots).
Therefore g’(x)  0 for all x.
g’(x) is the gradient of g(x) and g’(x)  0 for all x.
Therefore the function g(x) always has positive gradient and is always increasing.
9)
Solve the equation log2(x + 1)  2log2(3) = 3
Soln
log2(x + 1)  2log2(3) = 3
log2(x + 1)  log2(3)2 = 3
log2(x + 1)  log2(9) = 3
log2 x + 1 = 3
 9 
23 = x + 1
9
8=x+1
9
(  9 both sides )
72 = x + 1
( 1 both sides )
71 = x
x = 71
10) In the diagram
angle DEC = angle CEB = xo and
angle CDE = angle BEA = 900.
CD = 1 unit ; DE = 3 units.
B
C
1
By writing angle DEA in terms
of xo, find the exact value of
cos (DEA).
x
D
o
A
xo
E
3
Soln
C
Pythagoras
10
1
xo
D
sin xo = 1
10
cos (DEA)
E
3
CE2 =
=
=
CE =
32 + 1 2
9+1
10
10
cosxo = 3
10
=
cos(2x + 90)0
=
cos(2x)0cos900  sin(2x)0sin900
cos900 = 0
=
sin900 = 1
0  sin(2x)0
sin(2x)0 = 2sinxocosxo
=
2sinxocosxo
=
2
=
3
5

1  3
10 10
=
6
10
10) The diagram shows a parabola passing
through the points (0, 0), (1, 6) and (2, 0).
(a)
The equation of the parabola is of the
form y = ax(x  b).
y
(2, 0)
O
Find the values of a and b.
(b)
This parabola is the graph of y = f’(x).
Given that f’(1) = 4, find the formula
for f(x).
(1, 6)
Soln
(a)
roots of the parabola are x = 0 and x = 2
factors are x and (x  2)
y = a(x)(x  2)
y = ax(x  2)
y = ax2  2ax
the point (1, 6) lies on y = ax2  2ax
( x = 1, y = 6 )
6 = a(12)  2a(1)
6 = a  2a
6 = a
a=6
y = ax(x  b)
y = 6x(x 2)
a = 6 and b = 2
x
(b)
y =
f’(x) = 6x(x  2)
f(x) =  6x(x  2) dx
=  6x2  12x dx
= 6x3  12x2 + c
3
2
f(x)
= 2x3  6x2 + c
f(1) = 4
4
= 2(13)  6(12) + c
4
= 26+c
4
= 4 + c
( add 4 both sides )
8 = c
c = 8
f(x)
= 2x3  6x2 + c
f(x)
= 2x3  6x2 + 8
f(x)
= 2(x3  3x2 + 4)