Solutions to O Level Add Math paper 2 2012 - korlinang · PDF fileSolutions to O Level Add...
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Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 1
1. The equation of a curve is xxy cos3sin2 for x0 .
(i) Write down expression for x
y
d
d and
2
2
d
d
x
y. [3]
(ii) Find the value of x for which the curve has a stationary point. [2]
(iii) Determine the nature of this stationary point. [2]
Solution :
(i) Given that xxy cos3sin2 ,
xxx
ysin3cos2
d
d
xxx
ycos3sin2
d
d2
2
(ii) When 0d
d
x
y,
xx sin3cos20
3
2tan x
55.2x rad or 70.5x rad (rejected)
(iii) When 55.2x rad,
55.2cos355.2sin2d
d2
2
x
y < 0, hence a maximum point.
[Analysis]
This question is on differentiation and nature of turning points.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 2
Alternative solution:
(i) Given that xxy cos3sin2 ,
xxy cos
13
3sin
13
213
sincoscossin13 xxy where 2
3tan
xy sin13 983.0 rad
xx
ycos13
d
d
xx
ysin13
d
d2
2
(ii) When 0d
d
x
y,
xcos130
2
x or
2
3 x
55.2x rad 70.5x rad (rejected)
(iii) When 55.2x rad,
2sin13
d
d2
2
x
y < 0, hence a maximum point.
In Summary:
Be very careful in handling the radian. The actual value of 2nd
derivative is not
as important, as we just need to know the sign of it only.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 3
2. (i) Given that xBxAxx coscoscos3cos , state the value of A and of B. [1]
(ii) Hence, or otherwise, solve the equation 0cos23cos xx for 1800 x . [5]
Solution :
(i)
2
3cos
2
3cos2cos3cos
xxxxxx
xxxx cos2cos2cos3cos
Hence, 2A , 2B
(ii) Given that 0cos23cos xx for 1800 x , so 36020 x
0coscos2cos2 xxx
012cos2cos xx
12cos2 x or 0cos x
2
12cos x 90x
principle angle, 1202x
240,1202x
120,60x
Alternative solution:
(i) xxxxxxx cossin2sincos2coscos3cos
xxxxx coscossin2cos2cos 2
xxx 2sin212coscos
xxcos2cos2
[Analysis]
Part (i) apply the sum of angle identity. Part (ii) is about solving a trigo equation.
In Summary:
Be very careful in handling equation in part (ii). Take note of the double angle
solution, because the range of 2x is larger.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 4
3. (i) Express 383
202 xx
in partial fractions. [4]
(ii) Find x
xxd
383
202
and hence evaluate xxx
d383
207
22
. [4]
Solution :
(i) Since 313383 2 xxxx ,
let 313383
12
x
B
x
A
xx
when 3x , 10
1
10
1
B ,by cover-up method
when 3
1x ,
10
3
33
1
1
A
3
10
1
13
10
3
383
12
xxxx
3
2
13
6
383
202
xxxx
(ii) x
xxd
383
202
x
xxd
3
2
13
6
x
xx
xd
3
12d
13
32
cxx 3ln213ln2 where c is an integrating constant
[Analysis]
Part (i) is a simple 2 parts decomposition. Part (ii) will need integration into ln function.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 5
xxx
d383
207
22
273ln213ln2 cxx
cc 32ln216ln237ln2121ln2
cc 5ln25ln210ln220ln2
2ln2
39.1 (3 s.f.)
In Summary:
Nothing complicated in this question, except the integration into ln function.
Students are reminded that exponential and log functions are VERY important.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 6
4.
Two circles C1 and C2, intersect at P and Q as shown in the diagram. The tangent to C1 at P
meets C2 at R and the tangent to C2 at P meets C1 at S. Prove that
(i) triangle PQR and SQP are similar, [3]
(ii) Show that 2QPQRQS . [2]
Points A and B lie on the circumferences of C1 and C2 respectively,
(iii) Prove that angle SAP = angle PBR. [3]
Solution :
(i) Consider PQR and SQP
PSQRPQ (Alternate Segment Theorem)
QPSQRP (Alternate Segment Theorem)
Therefore, PQR SQP (AA)
(ii) Since PQR SQP ,
PS
RP
QP
QR
SQ
PQ .
QP
QR
SQ
PQ
QRSQQPPQ 2QPQRQS
[Analysis]
Part (i) needs to find 2 congruent angles in both triangles. Part (ii) can be arranged into length ratio.
Part (iii) may need properties of cyclic quadrilaterals.
S
A
B
C1
R
P
Q
C2
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 7
(iii)
SAPQ and PBRQ are cyclic quadrilaterals.
180SQPSAP 180PQRPBR
PQRPBRSQPSAP
Since PQRSQP , from part (i) SQP PQR , therefore
PBRSAP
In Summary:
Need a keen sense of observation of the given diagram. When part (i) is solved,
the rest of the question becomes straight forward.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 8
5. The equation of a curve is xx BAy ee2 , where A and B are constants. The point P(0, 4)
lies on the curve and the gradient of the tangent to the curve at P is 1 .
(i) Find the value of A and of B. [4]
(ii) Using your values of A and B from part (i), obtain an expression for xy d and hence
evaluate xy d1
0 corrected to one decimal place. [4]
Solution:
(i) Given xx BAy ee2 ,
At P(0, 4),
BA4 --------- (1)
xx BAx
y ee2d
d 2
At P(0, 4), 1d
d
x
y
BA 21 --------- (2)
(1) + (2),
1A
3B
(ii)
xxy e3e2
xy d
xxx de3e2
xx xx de3de2
[Analysis]
Part (i) requires differentiation. Part (ii) simply find the integration and the value of the definite
integration.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 9
xx xx de3d2e
2
1 2
Cxx e3e2
1 2 where C is an arbitrary integrating constant
therefore xy d1
0
CC 0012 e3e
2
1e3e
2
1
CC 0012 e3e
2
1e3e
2
1
32
1e3e
2
1 12
1.5 (1 d.p.)
In Summary:
Application of differentiation and integration of exponential functions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 10
6. (a) (i) Given that ux 4
3
8 log4log , express u in terms of x. [3]
(ii) Find the value of x for which 4log
1log5log
3
3
8
2
4 xxx . [3]
(b) Solve the equation 152ee yy . [3]
Solution:
(a)
(i) Given that ux 4
3
8 log4log ,
2
2
2
3
2
3
2
2log
log4
2log
log ux
2
log4log 2
2
ux
ux 222 log2log8log2
ux 2
8
2
2
2 log2loglog
ux
2
2
2 log256
log
256
2xu
(ii) Given that 4log
1log5log
3
3
8
2
4 xxx where 0x and ( 0x or 5x )
4log
3log
2log
log
2log
5log
2
2
3
2
3
2
2
2
2
2 xxx
2
3loglog
2
5log 22
2
2
xxx
3loglog5log 2
2
2
2
2 xxx
3log5
log 22
2
2
x
xx
35
2
2
x
xx
22 35 xxx
[Analysis]
(a) Part (i) requires base change formula. Part (ii) simply find the values of the log equation.
(b) to solve a routine exponential equation.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 11
052 2 xx
0x or 5.2x
(rejected)
(b) Given that 152ee yy ,
015e2e2 yy
03e5e yy
3e y or 5e y
(rejected) 61.15ln y (3 s.f.)
In Summary:
Part (a) requires application of log rules, part (b) is just another exercise.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 12
7. (i) Find the value of a and of b for which 232 2 xx is a factor of
bxxaxx 234 32 . [6]
(ii) Using the value of a and b found in part (i), solve the equation
032 234 bxxaxx . [3]
Solution:
(i) Given that 212232 2 xxxx , let bxxaxxx 234 32f .
When 2x , 02f .
ba 2223222f234
ba 280 --------- (1)
When 2
1x , 0
2
1f
.
ba
2
1
2
1
2
13
2
12
2
1f
234
ba 4
3
2
10 --------- (2)
(1) – (2),
a4
5
2
150
a 60
6a
b 6280
4b
[Analysis]
212232 2 xxxx , apply factor theorem to find a and b. Part (ii) factorizes completely to
find the roots.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 13
(ii) Given that 046632 234 xxxx
223246632 22234 kxxxxxxxx
When 1x ,
2123246632 k
133 k
0k
223246632 22234 xxxxxxx
02232 22 xxx
2x or 2
1x or 22 x
or 2x or 2x
In Summary:
Without fail, factor theorem and remainder theorem are regulars in this paper.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 14
8. (i) Find the set of values of m for which the curve 22 xxy and the line 1 mxy do
not intersect. [4]
(ii) Sketch the curve 22 xxy , giving the coordinates of the maximum point and of the
points where the curve meets the x-axis. [3]
(iii) Find the number of solutions of the equation 12 2 mxxx when
(a) 1m , (b) 2
1m [4]
Solution:
(i) Given 22 xxy and 1 mxy ,
221 xxmx
0122 xmx
Discriminant of the equation is to be negative,
0422
m
422m
222 m
40 m
The set of values 40: mm
(ii) Given 22 xxy , when 0y ,
220 xx
xx 20
0x or 2x
Line of symmetry, 12
02
x
The locally maximum value, 112 y
The coordinate of the maximum point is (1, 1)
[Analysis]
(i) Solving simultaneous equations with negative discriminant.
(ii) Sketch the graph by flipping up the negative part to the upper half.
(iii) Add lines to the sketch in part (ii) to determine the number of solutions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 15
(a) 1m , 1 xy (b) 2
1m , 1
2
1 xy
number of solutions = 2 number of solutions = 3
In Summary:
This is a typical curve sketching question. Remember to label your curve and
lines clearly.
x
(1, 1)
2 O
y 1x
22 xxy
–1
1 xy
12
1 xy
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 16
9.
In the diagram ABC is a structure consisting of a rod AB of length 10 cm attached at B to a
rod BC of length 24 cm so that angle ABC = 90°. Small rings at A and B enable A to move
along a vertical wire Oy and B to move along a horizontal wire Ox. Angle OAB = Ɵ and can
vary. The horizontal distance of C from the vertical wire Oy is L cm.
(i) Explain clearly why cos24sin10 L . [2]
(ii) Express L in the form cosR , where R > 0 and 900 . [4]
(iii) Find the greatest possible value of L and the value of at which this occurs. [3]
(iv) Find the value of for which L = 20. [2]
Solution:
(i) sin10OB cm CBx , cos24xBC
cos24sin10 xBCOBL
(ii) sinsincoscoscos RRR
sinsincoscoscos24sin10 RRL
24cos R and 10sin R
12
5
24
10tan
6.22
12
5tan 1
[Analysis]
Part (i) is to formulate the projected horizontal length. Part (ii) is to express it into cosR .
Part (iii) is to find the maximum value of L. Part (iv) is simply solving the trigo equation.
O x
y
B
A
C
10 cm
24 cm
O x
y
B
A
C
10 cm
24 cm
Cx
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 17
22222410sincos RR
26R
cos26L where
12
5tan 1
(iii) When 1cos , L is at its maximum.
The largest 26L when 1cos
0
6.22
(iv) When 20L ,
cos2620
13
10cos
7.39
13
10cos 1
3.626.227.39
In Summary:
Question on R formula is a regular in this paper.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 18
10.
The diagram shows a trapezium ABCD in which AD is parallel to BC and angle BCD = 90°.
The vertices of the trapezium are at the points A(8, k), B(3, 6), C(1, 2) and D. The line with
equation 253 xy passes through M, the mid-point of AB, and through N, which lies on
AD.
(i) Show that k = 11. [3]
(ii) Find the coordinates of N and of D. [6]
(iii) Find the area of the triangle AMN. [2]
Solution:
(i)
2
6,
2
11
2
6,
2
38 kkM
Since M lies on the line 253 xy ,
252
113
2
6
k
63350 k
11k
A(8, k)
O
253 xy
y
x
C
(1, 2)
(3, 6) B
M
N
D
[Analysis]
Part (i), mid-point M, is on the line. Part (ii) needs to find the equation of AD using gradient BC.
And also need to find the equation of CD. Part (iii) begins with finding the area of BNA.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 19
(ii)
11,8A AD // BC, gradient of BC = gradient of AD 213
26
Equation of AD ,
8211 xy
52 xy
At point N, 52 xy intersects 253 xy ,
25352 xx
305 x
6x
7512 y
7,6N
CD BC, gradient of CD 2
1
Equation of CD ,
12
12 xy
5.22
1 xy
At point D, 52 xy intersects 5.22
1 xy ,
5.22
152 xx
5.75.2 x
3x
156 y
1,3D
(iii)
length CD 5211322 units
length AN 201178622 units
Area of triangle BNA 52052
1 square units
Area of triangle AMN = 2
1 of Area of triangle BNA 5.25
2
1 square units
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 20
11.
The diagram shows part of the curve xy 45 , meeting the x-axis at the point A and the
line 1x at the point B. The normal to the curve at B meets the x-axis at the point C. Find
(i) the coordinates of C , [6]
(ii) the area of the shaded region. [6]
Solution:
(i) When 1x , 345 y , coordinates of 3,1B
xxx
y
45
2
45
4
2
1
d
d
When 1x ,
3
2
45
2
d
d
x
y
The gradient of the tangent at B is 3
2, therefore the normal at B is
2
3 .
Equation of BC ,
12
33 xy
5.42
3 xy
x
B
A C O
y 1x
xy 45
[Analysis]
Part (i) requires differentiation to find gradient then the normal equation. Part (ii) breaks the shaded
area into two regions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Nov 2011 Page 21
When 0y ,
5.42
30 x
3x
coordinates of 0,3C
(ii) When 0y ,
x450
4
5x
coordinates of
0,
4
5A
area of the shaded region
3132
145
1
4
5
dxx
34542
3
6
11
4
5
dxx
3456
11
4
52
3
x
30456
12
3
2
3
3276
1
35.4
5.7 square units