ADDITIONAL MATHEMATICS (UP1) FORM 5 MARKING SCHEME AND QUESTIONS
(2013)
There are 9 questions in this First Performance Test. Attempt all questions
1.
The first two terms of an arithmetic progression are 5 and 9. Calculate the least number of terms
that must be taken for their sum to
exceed 1000.
[6 marks]
Given that a  5 and T2  9
Therefore, d  9  5
d 4
1
n
Using Sn  (2a  (n  1)d )
2
n
(2(5)  (n  1)4)  1000
2
n(10  4n  4)  2000
1
4n 2  6n  2000  0
2n 2  3n  1000  0
1
3  9  4(2)(1000)
4
3  8009
n
4
n  21.623
n
n  21.623
n  22
2.
1
6
1
1
The gradient of the curve y 
a
 bx, x  0 at (– 2, 4) is – 8. Find the value of a and of b.
x
[6 marks]
a
y   bx, x  0
x
y  ax 1  bx
dy
 ax 2  b
dx
at x  2 and y = 4
a
 b  8
1
x2
a
 b  8
4
a  4b  32 ----- eq 1
a
4
 2b
2
a  4b  8 ----- eq 2
Eq1 + Eq 2
8b  40
b  5
a  8  4(5)
1
a  12
1
1
1
1
6
3.
Differentiate with respect to x the following:
2 8
1  4x 
 3x  4 
a)
2
b)
[10 marks]
a)
b)
2 8
1  4 x 
 3x
d
 8(1  4 x 2 )7 (8 x)
dx
d
 64 x (1  4 x 2 )7
dx
4.
a)
2
 4
1
d
  3x2  4  2
dx
1

d 1
  3x 2  4  2  6 x 
dx 2
d
3x
2

;x  
1
dx
3
 3x2  4  2
2
2
Find the approximate value of the following without using calculator.
3
a)
8.01
3
b)
1006
b)
1
1
let y  x 3
let y  x 3
dy 1
 x
dx 3
1
2

3
1
3x
2
3
 y dy

 x dx
1
dy
 x
dx
 0.01
When x  8,  y 
1
32
y
10
[10 marks]
To find the approximation of 1006  3
1
let x  8 and  x  0.01
y
2
1
1
To find the approximation of  8.01 3
Thus,  y 
Condition
correct
2
dy 1  23
 x
dx 3
1
For  x is small,
2
2
3 3
 0.01
1

1
 0.01  0.000833
12
1
1
let x  1000 and  x  6
1
 y dy
For  x is small,

 x dx
dy
Thus,  y    x
dx
1
 y  2 6
3x 3
1
When x  8,  y 
 0.01
2
3 3
3 10 
y
1
 6  0.02
300
1
When x  8, y   23  3  2
When x  8, y  103  3  10
Therefore the approximation is 2  0.000833
Therefore the approximation is 10  0.02
10.02
1
2.000833
1
1
10
The curve y  ax 2  bx  c has a maximum point at (2, 18) and passes through the point
(0, 10). Evaluate a, b and c.
2
y  ax  bx  c
dy
 2ax  b
1
dx
dy
For maximum point,
0
1
dx
2ax  b  0, at the point (2,18)
4a  b  0 ----- eq 1
1
The curve passes through (0,10)
10  c
1
18  4a  2b  10
1
2a  b  4 ----- eq 2
1
Eq 1  Eq 2 2a  0  4
1
a  2
1
5.
From eq 1, 4  2   b  0
b 8
6.
1
[10 marks]
10
1
Find the coordinates of the stationary points of the curve
 x  3
y
x
nature of each point.
2
x  3

y
,x  0
x
2
, x  0 and determine the
[10 marks]
2
dy x  2  x  31   x  3 (1)

1
dx
x2
dy  x  3  2 x  x  3

1
dx
x2
dy  x  3 x  3
x2  9

or
or1  9 x 2 ; x  0
2
2
dx
x
x
dy
For stationary points,
0
1
dx
 x  3 x  3  0
1
x  3 or x  3
For x  3
1
For x  3
2
d y 18

0
dx 2  33
1
minimum point
 3  3
y
1
d2y
18

3 0
2
dx
 3
1
maximum point
2
 3  3
y
0
3
Minimum point (3,0)
2
 12
3
maximum point (  3, 12)
1
1
10
7.
A rectangular sheep pen has one fixed side which is part of a long straight stone wall. The
remaining sides are to be made by using 80m of fencing. Find the dimensions of the rectangle
with the greatest possible area.
[10 marks]
2 x  y  80
y  80  2 x ----- eq 1
1
Area of Sheep pen, A  xy
1
A  x(80  2 x)
1
A  80 x  2 x 2
dA
 80  4 x
dx
1
For maximum area,
dA
0
dx
1
80  4 x  0
1
4 x  80
1
x  20m
y  80  2(20)
1
y  40m
1
Therefore, the dimension of the sheep pen is 20m by 40m
8.
3
Given that  f ( x) dx  7 and
0
a)
3
3
0
3
0

3
0
1
g ( x) dx  4 , find the value of the following:
 f ( x) dx  2 g ( x) dx
  g ( x)  2  dx
 f ( x) dx   g ( x) dx
b)
10
[2 marks]
[3 marks]
0
c)
0
3
3
0
[3 marks]
a)

3
0
3
f ( x) dx  2 g ( x ) dx
0
7  2(4)  15
b)
1 mark for working and 1 mark for answers
2
3
  g ( x)  2  dx
0

3
0
3
g ( x) dx   2 dx
3
4   2 x 0
1
4   6  0
10
c)

0
3
1
0
1
1
f ( x ) dx 
7  4  28

3
0
g ( x ) dx
2
1 mark for working and 1 mark for answers
8
9.
Diagram 1 shows a hemispherical container with a radius of
12 cm. It contains a liquid that evaporates easily. The liquid
level in the container, h cm, decreases at a rate of 0.001 cm s -1
.
a)
Show that the top surface area of the liquid, A cm2, is
given by A   (24h  h 2 )
b)
Calculate the rate of change of the top circular surface
area of the liquid at the instant when h = 6 cm
12 cm
Surface of liquid
h cm
Diagram 1
[10 marks]
1
dh
Given that
 0.001
dt
let r be the radius of the circular liquid surface
r 2  144  (144  24h  h 2 )
2
r  24h  h
A  r
12 cm
1
12 – h
2
2
A    24h  h
1
2
 shown
12
r
h cm
1
dA
 24  2 h
1
dh
dA dA dh


1
dt dh dt
dA
  24  2 h   (0.001) when h  6
dt
dA
  24  12    0.001
1
dt
dA
 0.012 or  0.0377 cm 2 s 1
dt
1
1
Some form of
sketch is shown
to obtain the
relationship
between h and r.
1
Dear teachers, the total marks for this paper is 80 marks. Please add the homework or
assignments marks (total 20) for the student to make it a grand total of 100 marks.
Thank you very much
David.
10