Solutions to O Level Add Math paper 1 2014 - korlinang · PDF fileSolutions to O Level Add...
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Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 1
1. Given that the line cxy 2 is a tangent to the curve kxy 2 , where c and k are positive
constants, prove that k is a multiple of c. [4]
Solution :
Given cxy 2 and kxy 2 ,
kxcx 2
2
kxccxx 22 44
044 22 cxkcx
The discriminant 22444 ckc
222 16816 ckckc
082 ckk
08 ckk
Since 0k , 0k , we have
08 ck ck 8
Alternative solutions:
Let mck , 0m
mcxcx 2
2
mcxccxx 22 44
044 22 ccxmx
The discriminant 222 444 cmc
16816 22 mmc
0822 mmc
Since 0c , 0c and 0m we have
08 mm 8m
ck 8
[Analysis]
To prove that mck . A curve is tangent to a line. Discriminant is zero.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 2
Alternative solution:
Given cxy 2 ,
2d
d
x
y
Given kxy 2 ,
y
x
k
y
d
d2
y
k
x
y
2d
d
y
k
22
4
ky
kxk
2
4
16
kx
ckk
162
4
ck
8
ck 8
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 3
2. Find the coordinates of the stationary point of the curve x
xy4
for 0x , and determine
the nature of this stationary point. [5]
Solution :
Given x
xy4
,
2
41
d
d
xx
y
When 0d
d
x
y,
2
410
x
14
2
x
42x
2x
2x , 0x
42
42 y
the stationary point is 4,2
32
2 8
d
d
xx
y
When 2x , 012
8
d
d32
2
x
y, the turning point is a minimum.
[Analysis]
Finding turning points of a curve by differentiation, 0d
d
x
y.
2
2
d
d
x
y is to determine min/max
points.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 4
Alternative solution:
Given x
xy4
, 0x , 04
xxy
42 xyx
042 yxx
Taking discriminant,
0442
y
162 y
4y or 4y (rejected as 0y )
y has a minimum value of 4.
When 4y ,
0442 xx
022x
2x
the stationary point 4,2 is a minimum point.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 5
3. It is given that 7d
3
1
2 xax , where a is a constant.
(i) Find the value of xax d
6
2
2
. [3]
(ii) Express xbax d
3
1
2
in terms of the constant b . [2]
Solution :
(i) Given 7d
3
1
2 xax ,
73
26
33
27
1
3
3d
33
1
2
aaaaxxax
56783
268
3
208
3
8
3
216
2
6
3d
36
2
2
aaaaaxxax
(ii) bxbxbxaxxbax 271
37ddd
3
1
3
1
2
3
1
2
Alternative solution:
(i) Given 7d
3
1
2 xax ,
73
26
33
27
1
3
3d
33
1
2
aaaaxxax
73
26
a
26
21a
[Analysis]
Some technical aspect of definite integration.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 6
56873
208
26
21
3
8
3
216
26
21
2
6
326
21d
36
2
2
xxax
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 7
4. It is given that xx 112 32 .
(i) Find the exact value of x12 . [3]
(ii) Hence find the value of x corrected to 2 decimal places. [2]
Solution:
(i) Given that xxxx 323212 22 ,
xx 112 32
xx
12
32
2
xx 12 322
xxx 33212 1
3212 x
612 x
(ii) From (i), we have 612 x .
6lg12lg x
6lg12lg x
72.012lg
6lgx (2 d.p.)
[Analysis]
Part (i) is to find the value of x12 , not x. Then, find the value of x in Part (ii).
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 8
Alternative solution:
Given that xx 112 32 ,
xx 112 3lg2lg ,
3lg12lg12 xx
3lg3lg2lg2lg2 xx
2lg3lg3lg2lg2 xx
6lg12lg x --------- (a)
6lg12lg x
612 x
(ii) From (a),
72.012lg
6lgx (2 d.p.)
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 9
5.
The diagram shows points A, B and C lying on a circle. The point T is such that the lines TA
and TC are tangents to the circle. Given that angle ABC = angle ATC, prove that triangle
ACT is equilateral. [4]
Solution:
TCTA (property of tangents from an external point)
ATC is an isosceles
TACTCA
TACCBATCA (tangent chord theorem)
Given that ATCABC ,
TACATCTCA
ATC is an equilateral.
[Analysis]
From the “tangents from an external point”, we know TC = TA, any one of the 3 angles is 60°
will have made an equilateral.
A
B
T C
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 10
Alternative solution:
Given that ATCABC ,
Let the centre of the circle be O.
2AOC (angle at the centre is twice the angle at the circumference)
90TAOTCO (radius tangent)
180AOCATC (sum of angles in quadrilateral OATC)
1802
60
TCTA (property of tangents from an external point)
ATC is an isosceles
602
60180TACTCA
TACCBATCA
ATC is an equilateral.
A
B
T C
2 O
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 11
6.
In the diagram, M is the midpoint of the line joining the points 4,1A and 6,7B .
The perpendicular bisector of AB intersects the line l at the point P. Given that line l is
parallel to the line 32 xy , find the coordinates of P. [6]
Solution:
Given that 4,1A and 6,7B ,
5,42
64,
2
71MM
Gradient of AB, 3
1
6
2
71
64
ABm
Gradient of PM, 31
AB
PMm
m , AB PM
Equation of PM,
435 xy
173 xy ---------- (1)
Equation of AP,
124 xy , AP // 32 xy
22 xy ---------- (2)
[Analysis]
Create equation for PA and PM, then solve for the intersection of these two lines.
A 4,1
O
M
x
y l
P B 6,7
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 12
Therefore, solve AP and PM,
173 xy and 22 xy
22173 xx
155 x
3x
8232 y
8,3P
Alternative solution:
Gradient of AB, 3
1
6
2
71
64
ABm
Gradient of PM, v
v
mm
AB
PM
31
,
( AB PM )
Gradient of AP, u
umAP
22
From the diagram,
3 vu vu 3 vu 262
132 vu
1326 vv
55 v
1v
213 u
8,3224,21 PP
A 4,1
O
M
x
y l
P
B 6,7
1
3
u
u2 v
v3
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 13
7.
The diagram above shows part of the graph of xy 4 . In each of the following cases
determine the number of intersections of the line cmxy with xy 4 , justifying your
answer.
(i) 1m and 2c . [2]
(ii) 2
1m and 0c . [2]
(iii) 2
1m and 2c . [2]
Solution:
From xy 4 , we get
When 0x ,
44 y
When 0y ,
x 40
4x
[Analysis]
Testing on the graph of modulus function and a straight line. Good graph sketching is
necessary.
O x
y
O x
y
4
4
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 14
(i) 1m and 2c . cxy
cxy intersects xy 4 ,
// 4 xy
One point of intersection.
(ii) 2
1m and 0c . xy
2
1
From the sketch,
there are 2 points of intersections.
(iii) 2
1m and 2c . cxy
2
1
When 0y ,
cx 2
10
42 cx
From the sketch, there is
no intersection.
O x
y
4
4
2
2 xy
4 xy
xy 4
O x
y
4
4
xy2
1
4 xy
xy 4
O x
y
4
4
2
22
1 xy
4 xy
xy 4
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 15
Alternative solution:
(i) 1m and 2c . cxy
When 4x , 44 xxy
4 xcx
4c , no solution
When 4x , xxy 44
xcx 4
42
4
cx
c 4
Since 2c , there is a solution.
There is 1 intersection.
(ii) 2
1m and 0c . xy
2
1
When 4x ,
42
1 xx
8x
When 4x ,
xx 42
1
3
8x
There are 2 points of intersections.
(iii) 2
1m and 2c . cxy
2
1
When 4x ,
42
1 xcx
43
28
cx
2c , but 2c , no solution.
When 4x ,
xcx 42
1
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 16
428 cx
2c , but 2c , no solution.
There is no intersections.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 17
8. The roots of the quadratic equation 0532 xx are and .
(i) Express 22 in terms of and . [1]
(ii) Find a quadratic equation whose roots are 3 and 3 . [6]
Solution:
Given that 0532 xx , 3 , 5 ,
(i) 32222222
(ii) 322233 ,
185333 233
12553333
0125182 xx
Alternative solution:
(i) Let ba 222
When 0 , 22 a 1a
When 1 , 1 , b2
2111 , 3b
3222
(ii) Given that 0532 xx , 3 , 5 ,
0532 xx
532 xx
xxx 53 23
[Analysis]
A question on SOR and POR, where (i) is a preparation for (ii).
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 18
xxx 55333
1543 xx
1543
1543
30433
18303433
12553333
0125182 xx
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 19
9. (i) Express the equation 1cosec5cot2 2 as a quadratic equation in cosec . [2]
(ii) Hence solve the equation 1cosec5cot2 2 for 3600 . [4]
(iii) State the number of solutions of the equation 1cosec5cot2 2 in the range
720720 . [1]
Solution:
(i) Given that 1cosec5cot2 2 , knowing 22 coseccot1 , we get
1cosec51cosec2 2
03cosec5cosec2 2
(ii) 03cosec5cosec2 2 , 3600
03cosec12cosec
012cosec or 03cosec
2
1cosec 3cosec
(Rejected as 1cosec ) 3sin
1
3
1sin
Therefore
3
1sin 1
5.19 or 5.1605.19180
(iii) For 720720 , there are 4 cycles. 824 solutions.
[Analysis]
(i) Apply 22 coseccot1 .
(ii) Solve quadratic equation .
(iii) Extend the range of solution.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 20
10. A particle travels in a straight line, so that t seconds after passing through a fixed point O ,
its acceleration, a m/s2, is given by
22
8
ta . The particle comes to rest when 2t .
Find
(i) an expression for the velocity of the particle in terms of t , [3]
(ii) the distance from O at which the particle comes to rest. [4]
Solution:
(i) Given that 22
8
ta ,
C
tt
tt
tV
2
8d
2
18d
2
822
When 2t , 0V ,
C
22
80
2C
2
82
tV
(ii) From 2
82
tV ,
20
2
0
2
0
2ln82d2
182d
2
82
ttt
tt
tS
2ln84ln84 S
2ln162ln84 S
55.1S (3s.f.)
[Analysis]
Part (i) is when the time 2t , the particle has zero velocity. Part (ii) is to find the distance between
the particle when it is zero velocity from point O.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 21
11. Variables x and y are connected by the equation naxy , where a and n are constants.
Using experimental values of x and y, a graph was drawn in which yln was plotted on the
vertical axis against xln on the horizontal axis. The straight line which was obtained passed
through the points 5.2,6.1 and 1.1,3.2 .
Estimate
(i) the value of n and , to 2 significant figures, the value of a, [4]
(ii) the coordinates of the point on the line at which xy 2 . [4]
Solution:
(i) Given the straight line passed through the points 5.2,6.1 and 1.1,3.2 ,
At 5.2,6.1 , 6.1ln5.2 na ---------- (1)
At 1.1,3.2 , 3.2ln1.1 na ---------- (2)
(1) – (2) , n7.04.1
2n
6.12ln5.2 a
7.5ln a
7.5ea
300a (2 s.f.)
2300 xy
(ii) xy 2 xy ln2lnln .
Let yY ln and xX ln
XY 2ln
and XY 27.5
XX 27.52ln
0.52ln7.53 X
67.1X (3 s.f.) and 36.2669.127.5 Y (3 s.f.)
The coordinates of the point on the line at which xy 2 is 36.2,67.1 .
[Analysis]
(i) the equation of the straight line is xnay lnlnln .
(ii) xy 2 , xy ln2lnln intersects xnay lnlnln . To find yx ln,ln .
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 22
Alternative solution:
Given that 2300 xy and xy 2 ,
23002 xx
1503 x
313.5x and 626.10313.52 y
67.1ln x and 36.2ln y
The coordinates of the point on the line at which xy 2 is 36.2,67.1 .
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 23
12.
[The volume of a cone of height H and base radius R is given by HR2
3
1 ]
The diagram shows a hollow conical tank of height 30 cm and radius 15 cm. The tank is
held fixed with its circular rim horizontal. Water is then poured into the empty tank at a
constant rate of 20 cm3/s. After t seconds the depth of water is h cm.
(i) Show that the volume of water in the tank, V cm3, at time t is given by
12
3hV
. [2]
(ii) Find the rate of change of the depth when 5h . [4]
(iii) State, with a reason, whether this rate will increase or decrease as t increases. [2]
[Analysis]
(i) is about formulating an equation, the equation has no t and r.
(ii) applies chain rule to find t
h
d
d.
(iii) need to deduce the change in t
h
d
d, i.e.
2
2
d
d
t
h .
30 cm
h cm
15 cm
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 24
Solution:
(i) Given tV 20 ,
Let 30
15
h
r
2
hr
hrV 2
3
1
hh
V
2
23
1
12
3hV
(ii) Let 20d
d
t
V and
4d
d 2h
h
V ,
t
h
h
V
t
V
d
d
d
d
d
d
5h , t
h
d
d
4
520
2
5
16
d
d
t
h
02.1d
d
t
h (3 s.f.)
(iii) t
hh
d
d
420
2
2
80
d
d
ht
h
32
2 160
d
d
ht
h
When 5h ,
05
160
d
d32
2
t
h
t
h
d
d is at a decreasing rate.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 25
13.
The height above ground level, h m, of a capsule on the Singapore Flyer is modelled by the
equation, kth cos180 , where k is a constant and t is the time in minutes after starting
the ride at ground level. The total time to complete one revolution is 30 minutes.
(i) Explain why this model suggests that the height of the Singapore Flyer is 160 m. [1]
(ii) Show that the value of k is 15
radian per minute. [2]
It is possible for a person riding in a capsule to see a certain landmark, provided the capsule
is at least 100 m above ground level.
(iii) Find the length of time for which the landmark will be in view during one revolution.
[5]
[Analysis]
Need to relate the formula kth cos180 to the flyer. Consider cos180 h where
being the angle of rotation of the flyer. Then kt , k is the speed of rotation. The flyer
rotates 3602 in 30 minute.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Feb 2014, source of questions ©UCLES & MOE 2012 Page 26
Solution:
Make a sketch of kth cos180 .
(i) when 1cos kt , flyer moved by half
of the revolution,
1601180 h
(ii) The flyer takes 30 min to complete one
Revolution, 2 radians.
230 k
15
k radian/min
(iii) when 100h ,
100cos180 kt ktcos4
1
Consider 4
1cos kt
4
1cos 1kt
8235.1kt rad or 4597.4kt rad
70.88235.1 kt or 29.214597.4 kt
The duration of above 100 m is 6.1270.829.21 minutes
Alternative solution:
(iii) Consider 100h ,
100cos180 kt 4
1cos kt
4
1cos 1kt
8235.1kt rad
70.88235.1 kt , it takes 15 minutes to reach the top. So, 6.127.8152 minutes
30
h
t (min) 0
100
80
160
15