Post on 13-May-2018
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 1
1. Baby food is heated in a microwave to a temperature of 80°C. It subsequently cools in such
a way that its temperature, T °C, t minutes after removal from the microwave, is given by
ktAT e20 , where A and k are constants.
(i) Explain why A = 60. [1]
When t = 1 the temperature of the food is 65°C.
(ii) Find the value of k correct to 3 significant figures. [2]
A baby should only be given this food when the temperature of the food is less than 40°C.
(iii) Determine, with working, whether it is safe to give the food 4 minutes after removal
from the microwave. [2]
Solution :
(i) When 0t , 80T ,
0e2080 A
60A
(ii) When 1t , 65T ,
k e602065
k e60
45
k4
3ln
3
4lnk
288.0k
(iii) For 40T ,
40e6020 kt
3
1e kt
[Analysis]
Understand the modelling of cooling with exponential function.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 2
3
1ln kt
k
t3ln
3ln4ln
3ln
t
82.3t (3 s.f)
Therefore, when 4t , 40T . It is safe to feed the baby.
In Summary:
Modelling question similar to this will continue to be in the future paper.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 3
2. Given that 61132f 23 xxxx ,
(i) find the remainder when xf is divided by 2x , [2]
(ii) show that 2x is a factor of xf and hence solve the equation 0f x . [4]
Solution :
(i) Given that 61132f 23 xxxx ,
1262212162f
(ii) 062212162f , therefore 2x is a factor of xf .
Let 32261132 223 kxxxxxx
When 1x ,
32361132 k
536 k
7k
Therefore, 372261132 223 xxxxxx
061132 23 xxx
03722 2 xxx
03122 xxx
2x or 2
1x or 3x
In Summary:
A very typical question on R/F theorems.
[Analysis]
Applying Remainder and Factor Theorems.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 4
3. The area of a quadrilateral is 4813 cm2.
(i) In the case where the quadrilateral is a rectangle with width 33 cm, find, without
using a calculator, the length of the rectangle in the form 3ba cm. [4]
(ii) In the case where the quadrilateral is a square with side c32 cm, find, without
using a calculator, the value of the constant c. [3]
Solution :
(i) 33
4813
3333
3331613
22 33
333413
6
1231231339
6
327
36
1
2
9
(ii) 4813322
c
34133432 22
cc
34133412 2 cc
1c
[Analysis]
Simplifying surd.
In Summary:
Surd will still be in the future paper.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 5
4. The quadratic equation 0452 2 xx has roots α and β.
(i) Find the value of α2 + β2. [3]
(ii) Find the quadratic equation whose roots are α3 and β3. [5]
Solution :
(i) Given that 0452 2 xx , 2
5 2
2222 .
2222
222
52
44
25
4
12
(ii) SOR: 2233
33
2
523
2
53
8
515
8
125
POR: 8333 .
The quadratic equation is 088
52 xx or 06458 2 xx
[Analysis]
SOR: 2
5 ; POR: 2 .
In Summary:
As expected for the new cubic identity.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 6
5. (a) Express 34loglog2 22 xx as a quadratic equation in x and explain why there are
no real solutions. [5]
(b) Given that
08log
log2
x
y
y
x , express y as a power of x. [4]
Solution:
(a) Given that 34loglog2 22 xx , 4x
4log3log 2
2
2 xx
4log2log3log 22
2
2 xx
4log2loglog 2
3
2
2
2 xx
42loglog 3
2
2
2 xx
3282 xx
03282 xx
Taking discriminant, 06412864321482
Hence, 03282 xx has no real root.
(b) Given that
08log
log2
x
y
y
x , 0log xy , 1, yx , 0, yx
0log8log2
xy yx
0log8log
log2
x
x
yy
y
y
0log8log
12
xx
y
y
0log813 xy
8
1log
3xy
[Analysis]
Log rules, discriminant. base changing.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 7
2
1log xy
2
1
yx
2
1
xy
In Summary:
The last part in (ii) is a little non-routine.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 8
6.
The diagram shows a triangle ABC whose vertices lie on the circumference of a circle. The
triangle DEF is formed by tangents drawn to the circle at the points A, B and C.
(i) Prove that angle DEF = 2 angle ABC. [4]
(ii) Make a similar deduction about angle DFE. [1]
(iii) Prove that 2 angle BAC = 180° + angle EDF. [3]
Solution:
(i)
ACEABC (Alternate Segment Theorem)
EAEC (Tangents from an external point)
Hence, AEC is an isosceles triangle.
Therefore, ABCACECAE
Therefore, AECABC 2180
DEFABC 1802180
Therefore, DEFABC 2
(ii)
Similarly, ACBDFE 2
[Analysis]
Properties of tangents to a circle are test in this question.
A
B D
C
F
E
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 9
(iii)
ACBABCBAC 180 (Sum of angles in a triangle)
22
180DFEDEF
BAC
DFEDEFBAC 3602
EDFBAC 1803602 (Sum of angles in a triangle)
EDFBAC 1802
In Summary:
Geometrical Proof is a weak link common to many students. Kids, do more
practice, it is not at all that difficult.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 10
7. A curve has the equation 432 xy . The point ( p, q ) is the stationary point on the
curve.
(i) Determine the value of p and of q. [4]
(ii) Determine whether y is increasing or decreasing
(a) for values of x less than p, [1]
(b) for values of x greater than p. [1]
(iii) What do the results of part (ii) imply about the stationary point? [1]
(iv) What is the value of 2
2
d
d
x
y at the stationary point? [2]
Solution:
(i) Given that 432 xy ,
134d
d 3 x
x
y
334d
dx
x
y
Let 0d
d
x
y,
3340 x
x 30
3x
23324y
Therefore, 3p , 2q
(ii)
(a) When 3x ,
034d
d 3 x
x
y, y is increasing.
(b) When 3x ,
[Analysis]
Question on differentiation with graph.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 11
034d
d 3 x
x
y, y is decreasing.
(iii) The turning point is a maximum point.
(iv)
1312d
d 2
2
2
xx
y
When 3x ,
013312d
d 2
2
2
x
y
In Summary:
Part (ii) may have been challenging for some students.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 12
8. A particle travelling in a straight line passes through a fixed point O with a speed of 8 m/s.
The acceleration, a m/s2, of the particle, t s after passing through O, is given by ta 1.0e .
The particle comes to instantaneous rest at the point P.
(i) Show that the particle reaches P when 5ln10t . [6]
(ii) Calculate the distance OP. [4]
(iii) Show that the particle is again at O at some instant during the fiftieth second after first
passing through O. [2]
Solution:
(i) Given that ta 1.0e ,
The velocity, tv t de 1.0
Cvt
1.0
e 1.0
where C is an integration constant.
When 0t , 8v .
C1.0
e8
0
2C
21.0
e 1.0
t
v
When 0v ,
21.0
e0
1.0
t
t1.0e2.0
t1.05
1ln
t1.05ln
5ln10t
[Analysis]
Kinematics with integration.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 13
(ii) Since 21.0
e 1.0
t
v ,
Let the distance, tst
d21.0
e 1.0
Dtst
21.0
e2
1.0
where D is an integration constant.
When 0t , 0s
D2
0
1.0
e0
100D
10021.0
e2
1.0
tst
When 5ln10t , OPs
1005ln1021.0
e2
5ln101.0
OP
1005ln1021.0
5
1
2OP
8.471005ln2020 OP m (3 s.f.)
(iii) When 50t ,
1005021.0
e2
501.0
s
5e
100s
674.0s (3 s.f.)
Since, when 5ln10t , 8.47s > 0 and when 50t , 674.0s < 0, the particle must have
return to point O, at least once.
In Summary:
Part (iii) may have given problem to some students.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 14
9. (i) Solve the equation 02sin2cos3 AA for 3600 A . [4]
(ii) On the same axes sketch, for 1200 x , the graphs of
16cos2
3 xy and xy 3sin
2
12 . [6]
(iii) Explain how the solutions of the equation in part (i) could be used to find
the x-coordinates of the points of intersection of the graphs of part (ii). [2]
Solution:
(i) Given that 02sin2cos3 AA for 3600 A
02sinsincos3 22 AAA
02sinsinsin13 22 AAA
01sinsin6 2 AA
01sinsin6 2 AA
01sin21sin3 AA
3
1sin A or
2
1sin A
3
1sin 1A
2
1sin 1A
Principal angles, 5.19A 30A
5.199A or 5.340A 30A or 150A
[Analysis]
Trigo equation and its related graph.
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 15
(ii)
(iii)
Given that 16cos2
3 xy and xy 3sin
2
12 , the intersections
16cos2
33sin
2
12 xx
013sin2
16cos
2
3 xx
023sin6cos3 xx
Let Ax 3 , 36030 x , 3600 A
02sin2cos3 AA
From (i), the solutions of the equation are 5.199A , 5.340A , 30A , 150A
Therefore, 5.1993x , 5.3403x , 303x , 1503x
5.66x , 5.113x , 10x , 50x
1
O x
y
30
2
3
60 90 120
1
xy 3sin2
12
16cos2
3 xy
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 16
10. A circle, C1, has equation x2 + y2 + 4x – 6y = 36.
(i) Find the radius and the coordinates of the centre of C1. [3]
A second circle, C2, has a diameter AB. The point A has coordinates (5, 5) and the equation
of the tangent to C2 at B is 1543 xy .
(ii) Find the equation of the diameter AB and hence the coordinates of B. [4]
(iii) Find the radius and the coordinates of the centre of C2. [3]
(iv) Explain why the point (4, 6) lies within only one of the circles C1 and C2. [2]
Solution:
(i) Given that x2 + y2 + 4x – 6y = 36 ,
3664 22 yyxx
36996444 22 yyxx
94363222
yx
222732 yx
Centre 3,2 , radius 7 units
A second circle, C2, has a diameter AB. The point A has coordinates (5, 5) and the equation
of the tangent to C2 at B is 1543 xy .
(ii) Find the equation of the diameter AB and hence the coordinates of B. [4]
1543 xy
53
4 xy
Therefore, gradient of AB , 4
3
equation of the diameter AB, 54
35 xy
54
15
4
3 xy
4
5
4
3 xy
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 17
Let 4
5
4
35
3
4 xx
54
5
4
3
3
4 xx
4
25
12
25x
3x
1533
4y
1,3 B
(iii) radius of C2
52
515322
centre of C2
2
51,
2
53
2,12 C
(iv)
The distance between point (4, 6) and 3,21 C 4945362422
The point (4, 6) is inside circle 1C .
The distance between point (4, 6) and 2,12 C 2541261422
The point (4, 6) is outside circle 2C .
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 18
11. (a) Show that 312
1
12d
d
x
x
x
x
x. [3]
(b)
The diagram shows the line x = 5 and part of the curve
312
18
x
xy . The curve
intersects the x-axis at the point A. The line through A with gradient 1 intersects the
curve again at the point B.
(i) Verify that the y-coordinate of B is 2
3. [5]
(ii) Determine the area of the shaded region bounded by the curve, the line 5x , the
x-axis and the line AB. [4]
Solution:
(a) Let 12
x
xu .
2
2
1
2
1
2
1
12
2122
112
d
d
x
xxx
x
u
12
12
122
12
1
x
x
xx
12
12
12
2
1
x
x
xx
32
3
12
1
12
1
x
x
x
x
A
O
5x
x
y
B
312
18
x
xy
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 19
(b)
(i) When 0y ,
312
180
x
x where 2
1x
10 x
1x
0,1A
Equation of the line AB,
1 xy
Let
312
181
x
xx
0
12
181
3
x
xx
012
811
3
xx
01x or
012
81
3
x
112
83
x
8123x
64123x
412 x
2
5x
2
31
2
5y
2
3,
2
5B
(ii) area of the shaded region
x
x
xd
12
18
2
31
2
5
2
15
2
53
x
x
xd
12
18
8
95
2
53
Solutions to O Level Add Math paper 2 2014
By KL Ang, Jan 2015 Page 20
2
55
128
8
9
x
x
12
52
2
5
152
58
8
9
4
5
3
58
8
9
12
140
8
9
3
10
8
9
24
107
46.4 square units (3 s.f.)
In Summary:
A simple routine question on integration. Part (a) is to facilitate the integration.