MINISTRY OF EDUCATION
FIJI YEAR 12 CERTIFICATE EXAMINATION 2016
MATHEMATICS
Detailed Solutions
COPYRIGHT: MINISTRY OF EDUCATION, FIJI, 2016.
2.
SECTION A
[40 marks]
There are two parts to this section. Answer both parts.
MULTIPLE – CHOICE QUESTIONS
PART I
(20 marks)
Circle the letter which represents the best answer.
1
A
B
C
D
11
A
B
C
D
2
A
B
C
D
12
A
B
C
D
3
A
B
C
D
13
A
B
C
D
4
A
B
C
D
14
A
B
C
D
5
A
B
C
D
15
A
B
C
D
6
A
B
C
D
16
A
B
C
D
7
A
B
C
D
17
A
B
C
D
8
A
B
C
D
18
A
B
C
D
9
A
B
C
D
19
A
B
C
D
10
A
B
C
D
20
A
B
C
D
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
3.
1. The table shows the results for S = {2, 4, 6} under an operation 
If
2.
2  y = 4 then y is
A.
2
B.
4
C.
6
D.
8

2
4
6
2
6
4
2
4
4
2
4
6
2
4
6
When simplified 8  7  6 7 is equal to
8 7 6 7
A.
87 7
B.
14  2 7
C.
57
D.
28
Combining
the like
surds
= 87 7
3.
Which of the following expressions gives the size of angle θ ?
A
5
3
C
A.
4
cos  1  
3
B.
4
cos  1  
5
C.
3
cos  1  
4
D.
 3
cos  1  
5
4
B
 Adjacent 

Cos  
Hypotenuse


4
Cos   
5
4
5
  Cos  1  
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
4
4.
5.
6.
If a polynomial f (x) is divided by x  2, the remainder is
A.
f ( x  2)
B.
f ( x  2)
C.
f (2)
D.
f (2)
The gradient of the line parallel to y  2 x  5 is
A.
5
B.
2
Two lines are parallel if they have the same gradient.
y  2x  5
or y  2 x  5
y  5  2x
C.

1
2
General Form: y  mx  c
D.
–2
……gradient(m) = 2
When simplified
4  15
is equivalent to
1
A.
15  4
4  15
B.
15  4
=
C.
D.
7.
1
4 2  15
4  15
2
 4  15
=  4  15 or 15  4
1
=
y
The equation of the graph shown is
A.
y  x  1
2
 x  2
B.
y  x  1
2
 x  2
C.
y  x  1 x  2
The x-intercepts are -1 & 2
This is a positive graph.
2
-1
D.
4  15
 4  15
4  15
 4  15

x
0
1
y  x 1 x  2
2
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
2
5.
8.
The value of log 2 8 is
A.
2
B.
3
C.
4
D.
8
log 2 8
= log 8  log 2
=3
9.
10.
11.
If f ( x)  x  3 , then f (0) is equal to
f ( x)  x  3
A.
–3
B.
0
f (0)  0  3
C.
2
f (0)   3
D.
3
f (0)  3
The graph of y  f ( x) is obtained by reflecting the graph of y  f ( x)
in the
A.
y – axis.
B.
x – axis.
C.
line y = x.
D.
line y =  x.
The graph of y   f (x) is obtained by
reflecting the graph of y  f (x) in the x-axis.
The inverse of a graph is its reflection across
the line y = x.
The lim ( x 2  6 x) is equal to
x3
lim ( x 2  6 x)
x3
A.
3
B.
9
C.
18
D.
27
= (32  6(3))
= 27
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
6.
12.
An arithmetic sequence is defined as Tn  5n
The common difference of this sequence is
A.
2
B.
5
C.
2
D.
5
Tn  5n
Tn   5, 10, 15, 20, 25... 
10-5
=5
20-15
=5
Difference = 5
1
13.
 16 
The exact value of  
 25 
A.
B.
14.
2
is
4
5
 16 
 
 25 
2
5
C.
4
25
D.
8
25
1
2

16
25

16
25
Freda spins the pointer. The probability that it stops on an odd number is
2
1
Odd Numbers = {1, 3}
3
4
P(Odd Numbers) = P(1) + P(3)
A.
0.25
B.
0.50
C.
0.75
D.
1.00
P(Odd Numbers) =
1
1
+
4
4
P(Odd Numbers) =
1
or 0.5
2
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
7.
15.
16.
17.
The determinant of 2  2 transformation matrices which represent enlargements
give the scale factor for the
 
A.
area.
B.
length.
C.
volume.
D.
perimeter.
If M  ca bd , det M  M   ad  bc
The scale factor for area is the determinant
Area of image = determinant × area of object
A 2  2 matrix M represents the transformation reflection in the line y = 2x. Which
of the following transformations does M 1 , the inverse of M, represent?
x
2
A.
Reflection in the line y 
B.
Reflection in the line y  
C.
Reflection in the line y =  2x
D.
Reflection in the line y = 2x
3
3
1
1
Inverse of reflection in the line y = 2x
is reflection in the line y = 2x
x
2
If  f ( x) dx  5 , then   f ( x)  1 dx is equal to
A.
5
3
3
3
B.
6
1
1
1
C.
7
  f ( x)  1 dx =  f(x) dx +  1 dx
3
D.
8
[]
=5+ x
1
= 5 + [3 - 1]
=7
18.
If y  x 2  x , the instantaneous rate of change of y with respect to x when x = 2
is
A.
2
yx2 x
B.
4
C.
5
dy
 2x  1
dx
D.
6
at x = 2,
dy
 2( 2 )  1  5
dx
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
8.
19.
The graph of a circle is given below.
y
1
-1
0
x
1
-1
The equation of this circle is
20.
A.
x 2  y 2  1
B.
x2  y2  1
C.
x2  y2  2
D.
x2  y2  1
The general form of the equation of
a circle with centre (0,0) is
x 2  y 2  r 2 , where r = radius.
The graph of y  2 x  1 is shown below.
y
x
0
The y-intercept of this function is
A.
(0, 1)
B.
(0, 2)
C.
(0, 3)
D.
(0, 4)
y–intercept (make x = 0)
y  2x 1
y  20  1
y2
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS
9.
SHORT – ANSWER QUESTIONS
PART II
21.
n3
 ( n  1) = (1 - 1) + ( 2 – 1) + (3 - 1)
n1
= 0
+ 1
+ 2
Substitute and
evaluate
n3
 ( n  1) = 3
n1
22.
 b  b 2  4ac
x
2 x 2  1  5x
Equate to zero first
before finding the
values of a, b & c
2a
2 x 2  5x  1  0
a  2, b  5, c  1
x 
x
 5 
(5) 2  4  2  1
a = 2 b = -5 c = 1
2 2
5  17
4
x {
5  17 5  17
,
} 0r {2.28, 0.22}
4
4
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS
10.
SECTION A
(continued)
23.
a 2  b 2  c 2  2bc cos A
82  62  72  2  6  7  cos A
82  62  72  2  6  7  cos A
 21  84  cos A
 21
 cos A
 84
A  cos 1(0.25)
C
b=7
A
ϴ
Use Cosine Rule
a=8
c=6
B
θ = 75.52
24.
n
S  [2a  (n  1)d]
n 2
n
9801  [2  1  (n  1)2]
2
n
9801  [2  2n  2]
2
n
9801  [2n]
2
9801  n 2
a = first term = 1
Sn = sum of terms = 9801
d = common difference = 2
n  9801
n = __99__
25.
log
a =
1
7 = log 7 2
1
log 7
2
1
= log (21  3)
2
1
=
(log 21 - log 3)
2
=
n
log (x ) = nlog x
x
y
log ( ) = log x − log y
log
7 =
yx
2
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
11.
26.
2
1
sin 2 
2
30
30 
θ1
1
2  sin 1 ( )
2
2  30
2θ = 30, 150
Add 360 to each angle: 390, 510
θ = 15, 75
θ = 195, 255
4 5m  1
27.
2 9m


2
Expressing 4 as a power
with base 2
2(5m  1)
2 9m
210m  2
2
2 9m
When dividing, subtract
the powers if the bases
are the same.
(10m  2) - (9m)
4 5m  1
2 9m
28.
(a)
P(E) =
number of favourable outcomes
total number of possible outcomes
(b)
= 2m 2 or 4  2m
P (5 or 6) =
2
1
or or 0.33
6
3
(1 mark)
If an event E is certain to occur
then P(E) = 1
P( less than 7 ) = 1
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
12.
SECTION A
(continued)
29.
yy
1
x2
x
Equating the 2 equations
x 2  2 x  1
Inverse operations
x 2  2x  1  0
Factorise and equate to
Zero to find the value of x.
(x  1)( x  1 )  0
x  1
Substitute the value of x
to find the value of y.
y =  1+2
Coordinates ( -1, 1)
30.
x 2  2 x x  2 x 2  2 x 16



8
16
8
x2

x( x  2) 16

8
x2
Multiplying by the reciprocal
Factorizing the numerator
of the first expression
2
x( x  2) 16


8
x2
Cancellations
x2  2x x  2

= 2x
8
16
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
13.
SECTION B
LONG ANSWER QUESTIONS
[60 marks]
The six questions in this section are all compulsory. Each question is worth 10 marks. Show all
your working clearly as partial marks will be awarded for appropriate methods and partially correct
working.
QUESTION 1
3
x  x
A.
x
2
2
-7x  x
12x  x
-7x + 12
x  2 x 3  5 x 2  2 x  24
2
x (x + 2)
3
-
x +2x
2
-7x  2 x  24
2
-7x - 14x
-7x (x + 2)
12x + 24
_ 12x + 24
12 (x + 2)
2
–
0
Factors = ( x  3)( x  4)
B.
4 x 2  3x  p  0
b 2  4ac  0
32  4  4  p  0
9  16 p  0
General Rule for 2
distinct roots
Substitutions into
the formula
a=4, b=3, c=p
and simplify
 16 p  9
p
9
 16
When dividing by
negative number,
change the sign
values of p: p 
9
16
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
14.
SECTION B
C.
tan 30 
(continued)

h
10
Identify the sides
height = opposite side
10 10 3
h
or
or 5.77
3
3
30°=ϴ
10m = adjacent side
10
tan θ 

3
Use Tan θ 
Opposite
Adjacent
6
 10 

3

1
θ  tan 

 6 


θ  43.90
D.
1. The system is closed
because there is no foreign element
2. There is an identity element which is c
3. Each element has an inverse
element
inverse
a
b
b
a
Reminder:
Four conditions for a group:
closure, there is an
identity, the operation is
associative and every
element has an inverse.
c
c
4. The operation is Associative
( a  b)  c = a  (b  c)
c  c = a b
c=c
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
15.
QUESTION 2
A.
(i)
Substitute the values into the formula:
1 2
r 
2
1
3   6 2  
2
3  2

36
A
Area of
Sector 
1 2
r 
2
Area = 3  , radius = 6cm
Solve to find the value of angle ϴ.


6
rad or 30
Use the formula
(ii)
Length of Arc  S  r
S  r
A  6

Radius = 6cm, ϴ =
6

6
rad
S=  or 3.14 cm
B.
(i)
x – intercept = (2, 0)
y – intercept = (0, -2)
(ii) Horizontal asymptote: y = 2
(iii) Equation
Vertical asymptote: x = -2
Vertical asymptote: x = -2 so denominator contains x + 2  y 
?
x2
Horizontal asymptote is at y = 2 so the numerator must contain 2 x as a
2x
2x  k
2  y
term since using cover up rule
x2
x
y – intercept = - 2 so
k
= -2  k = - 4
2
y=
2x  4
x2
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
16.
SECTION B
C.
(continued)
θ
y = -3 cos   .
2
(i)
Amplitude
Amplitude = Height of graph above the x-axis
Amplitude = 3
(ii)
Period
360
period 
1
2
Period 
360
B
Period = 720 or 4  rad
(iii)
Full Graph for
3
θ
y  - 3cos   .
2
0
360°
720°
-3
y
….but interval is for 0

   360 (i.e. only from 0° to 360°)
3
0
-3
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
x
17.
SECTION B
(continued)
QUESTION 3
Direction
Name of
. transformation
Factor
A.
(i) Shear parallel to x-axis with shear factor 2
Invariant features are features that does not change
after a transformation
(ii) Area
(iii)
 1 m  0  2

    
 n p  1  1



















Coordinate Coordinate

Transformation 
 



 of Im age

  of Object 
Matrix


 



0+ m =2
m =2
0 + p =1
p =1
 1 m 1 3

    
 n p 1 1
n+p=1
n+1 = 1
n=0
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
18.
B.
(i) It has a common ratio of
(ii)
1
3
To find the nth term of a
Geometric Sequence,
use
8 1
1
T8  27   
 3
T  a  r n  1
n
8th term =
a
1 r
27
S 
1
1
3
S   40.5
(iii) S 
C.
Sum of all the terms of a
Geometric Sequence is
the Sum to infinity.
Consider the product of the gradients:
m1  m2  1
2 
1
2
 1
The product of the gradients
for perpendicular lines (lines
meeting at right angles) is
equal to -1.
 sides BC and AB are perpendicular
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
1
or 0.01
81
19.
SECTION B
(continued)
QUESTION 4
Mark
10
Frequency 1
A.
(i)
x
20
4
Mean
30
9
40
6
Mean =
Sum of all Scores
Number of Scores
10  1  20  4  30  9  40  6
1 4  9  6
Mean =  xf
f
Mean = 30
(ii)
Substitution into the given
formula (Standard Deviation)
and simplify.
f  x  x 2

s
f
s
s
1  10  30   4  20  30   9  30  30   6  40  30 
1 4  9  6
2
2
2
2
 202  4 102  902  6102
20
s
400  400  0  600
20
s
1400
20
Standard Deviation =
70 or 8.37
Turn Over
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
20.
B.
(i)
P ( x.  498.1)  P ( z 
xx

)
498.1  500
)
5
 P( z  0.38)
 0.5  0.148
 P( z 
-0.38
x
Probability = 0.648
Expected number = (Probability) ×(number of trials)
(ii)
0.648  2000
Expected Number = 1296
C.
(i)
Using Tree Diagram
Start
G
Y
(ii)
Y
S = { GY,
G
1
P(GY+YY) =  1
4
YG,
YY }
Y
+
3 2

4 3
Probability =
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
3
or 0.75
4
21.
SECTION B
(continued)
QUESTION 5
A.
y  2 x 3  9 x 2  12 x  5
(i)
d n
( x )  nx n  1
dx
dy
3

1
2
1
To
differentiate: multiply the coefficient by
 2  3x
 9  2x
 12  0
the power and then reduce the power by one
dx
dy
 6 x 2  18 x  12
dx
(ii)
6 x 2  18 x  12  0
6( x 2  3 x  2 )  0
At turning point, dy  0
dx
x 2  3x  2  0
x - 1x - 2  0
x - 1  0 x - 2  0
x=
1,
2
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© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
22.
(iii)
y  213  912  121+5
= 10
y  223  922  122+5
=9
Any horizontal line will
only cut this graph at
exactly one point when
the y-values are less
than 9 and greater
than 10.
10
9
1
Values of k : k > 10 or k < 9
2
B.
y '  ( x  2)( x  4)
Stationary Points at x = -2 and 4
y '  x2  2x - 8
 y'  0  ( x  2)( x  4)
y   ( x 2  2 x - 8)dx
x3
y
 x2  8x  C
3
x-intercept (from the graph) = (6, 0)
When x = 6 , y = 0
63
 62  8  6  C
3
C  12
0
x3
y
 x 2  8 x  12
3
© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
23.
SECTION B
(continued)
QUESTION 6
A.
Area.


4
Area   x3dx
k

x4 4
k

4
44
4

60  64


4  
Integrate
Substitute the two values
(4 and k in place of x) and
equate to 60 (Area)
Solve for k
k4
4
k4
4
k4
4
16  k 4
k=2

Volume is decreasing so make
dV
dV
  0.1t
negative:
dt
dt

At time = 0, Volume = 18m3
B.
dV
  0.1t
dt
V    0.1t dt
0.1t 2
V 
C
2
or V  0.05t 2  C
At t = 0, V = 18 so C = 18
V  0.05t 2  18
t2
Or V    18
20
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© MINISTRY OF EDUCATION, FY12CE 2016: MATHEMATICS.
24.
C. (i)
Total Surface Area = (Area of Base) + (Area of Sides)
12 = (x  x) + (4  (x  y))
x 2  4 xy  12
4 xy 12 - x 2
y  12 - x
4x
(ii)
2
Volume = l  w  h
Volume = x x y
V  x2 y
 12 - x 2 

 x2 
 4x 


3
12 x - x
x3
V 
or V  3 x 4
4
(iii)
V  3x -
x3
4
At maximum V'  0
dV
3
dx
3

3x 2
4
3x 2
0
4

3x 2
3
4
x2  4
x=2
THE END
____________________
COPYRIGHT: MINISTRY OF EDUCATION, FIJI, 2016.