Program Aim For “A”
Mathematics SM025
Prepared by : Lim Hwee Cheng
Kolej Matrikulasi Johor
Topic 1 : Integrations
2
3
(a) Substitution Method
u m2 1 u m2 1 When u m2 1
du 2m m2 u 1 m 3,u 4
dm m 0, u 1
m4 u 12
3 1
m5
u du 1 m2 dx 1 4 u2 2u 1 u2 du
2m
m5 0 21
1 m4 1 4 u 5 3 1 du
2 2 1 2
2u 2 u2
u du
1 1 7 5 3 4
u 2
2 u 12u 2 du 1 2u 2 u2
2 5 3
7
2 2 2 1
(exact value)
1 7 2 5 1 (4) 3 1 2 1 8448
7 5 3 2 7 5 3 105
(4) 2 (4) 2
(b) Integration By Part
3 xe 2xdx ux dv e2xdx
0 du dx v e2x
1
xe2x e2x
dx
1
3
xe2x e2x 0
3e1 e1 e2
4 e2
e
e3 4 (Shown)
e
5
6
6x2 x 7 A B C
(4 3x)(1 x)2 4 3x 1 x (1 x)2
6x2 x 7 A(1 x)2 B(4 3x)(1 x) C(4 3x)
when x 1 when x 4 when x 0
14 7C 3 7 A 4B 4C
7 3 4B 8
C2 49 A 49
3 9
A3 4B 4
B 1
6x2 x 7 3 1 2
(4 3x)(1 x)2 4 3x 1 x (1 x)2
7
6x2 x 7 3 1 2
(4 3x)(1 x)2 4 3x 1 x (1 x)2
1 1
0
0
6x2 x 7 dx 3 1 2 dx
(4 3x)(1 x)2 4 3x 1 x (1 x)2
3ln 4 3x ln1 x 2 1
3 1 x 0
ln1 ln 2 1 ln 4 ln1 2
ln 2 1 ln 4 2 8
1 ln 4
2
1 ln 2
Topic 2 : Differential Equations
9
(a) Find the general solution of the differential equation
dy y2xe2x. Give your answer in the form y f (x).
dx
(b) Find the particular solution of the differential equation
dy xy 1 x2 , given that y 1 when x 0.
dx x2
1
10
(a) dy y2xe2x
dx
1 dy xe 2 x dx
y2
1 dy xe 2 x dx
y2
1 xe2x e2x C
y 24
1 2xe2x e2x 4C y 2 xe 2 x 4 11
y4 e2x
4C
(b)
dy xy 1 x2 y 1 x2 x x3 C
dx 1 x2 3
x dx 1 ln 1 x2 when y 1; x 0
I.F e 1 x2 e2 1 x2
1 x 2 dy xy 1 x2 1 x2 1 10 00C
dx x C 1
1 2
d y 1 x2 1 x2 y 1 x2 x x3 1
dx 3
y 1 x2 1 x2 dx y 1 x x3 1
1 x2 3
12
Solve the differential equation x dy 3y cos2 x
dx x2
6 Marks
13
x dy 3y cos2 x dy 3y cos2 x
dx x2 dx x x3
IF e3 1dx e3ln x eln x3 x3 K1 J1 Finding IF
x
x3dy x33y x3 cos2 x
dx x x3
Multiply IF on both sides of the Eq
d x3 y cos2 x K1 and simplify
dx
d x3 y dx cos2 xdx
dx
x3 y 1 cos 2x dx K1 Double angle formula
2 K1 Integrate RHS
J1
x3 y 1 x 1 sin 2x c 14
24
y 1 sin 2x c
2x2 4x3 x3
Topic 3 : Numerical Methods
15
Show that the equation 2x ln5 x2 has a root that lies
between 0.6 and 0.8.
By using the Newton-Raphson method, determine the root of
the equation 2x ln5 x2 correct to three decimal places.
.
6 Marks
16
Given 2x ln 5 x2
Algebraic Method Subtitute x = 0.6 & x = 0.8 K1
2x ln 5 x2 0
Let f x 2x ln5 x2
f 0.6 20.6 ln 5 0.62 0.335 0
f 0.8 20.8 ln 5 0.82 0.128 0
There is a change in signs. J1 conclusion
Therefore, the root lies between x 0.6 and x 0.8
17
f x 2x ln5 x2 f x 2 5 2x K1 Differentiate f
x2 K1
x0 0.7 Initial root ( any value 0.6 < x < 0.8)
x1
0.7 20.7 ln 5 0.72 0.7460
20.7
2
5 0.72
0.7460 0.7458
x2
20.7460 ln 5 0.74602
20.7460
2
5 0.74602
0.7458 0.7458
x3
20.7458 ln 5 00.74582 K1
20.7458
2 Stopping criteria
5 0.74582
since x3 x2 0.0000
18
The root x 0.746 (3dp) J1
19
h 1 0.2 y 1
5 x ln 2x
x y0 1.4427
x0 1 y1 0.9519
x1 1.2
x2 1.4 y2 0.6937
x3 1.6 y3 0.5373
x4 1.8
x5 2 y4 0.4337
Total
y5 0.3607
1.8034 2.6166
2 x 1 dx 0.2 1.8034 2(2.6166
ln 2x 2
1
20
0.704(3 d.p)
Topic 4 : Conics
21
The centre of a circle lies on the line 3y 4x 11 and the circle
intersects the y-axis at the point (0,1) and (0,11).
a) Find the equation of this circle.
b) Find the possible values of such that the circle passes
through the point (+1,6).
c) Find the coordinates of the points where the circle meets the
line y x 11 0
12 Marks22
(a) General Equation x2 y2 2gx 2 fy c 0 K1
passes through0,1 1 2 f c 0 1
passes through0,11 121 22 f c 0 2
2 1 120 24 f 0 K1 Solving
substitute f 5 in eq1 for f & c
f 120 5
24 Solving
for g
110 c 0
c 11 J1
Centre g,f is on line 3y 4x 11 0
substitute f 5 in eq3 3 f 4g 11 3 K1
15 4g 11
g 1
23
Equation of circle x2 y2 2x 10 y 11 0 J1
(b) x2 y2 2x 10 y 11 0 K1
K1
circle passes through 1,6
12 36 2 1 60 11 0
2 2 1 36 2 2 60 11 0
2 36 0
6 6 0
6 J1
24
(c) x2 y2 2x 10 y 11 0 1
Circle meets the line y x 11 0
x y 11 2
2in 1 y 112 y2 2y 1110 y 11 0 K1
y2 22 y 121 y2 2 y 22 10 y 11 0 Solving eq
2 y2 34 y 132 0 (1) & (2)
y2 17 y 66 0 Solving
quadratic
y 11y 6 0 K1
y 11or 6
x 0 or 5
25
points of intersection 0,11 or 5,6 J1 J1
26
(a) V (h, k) F (h, k p) y 27
V (4,8), F (4,2) V (4,8) x
V x 27
p 28 p
x
F p 6
F (4,2)
Equation of parabola;
x h2 4 py k
x 42 4(6)y 8
x 42 24y 8
(b) i 4 x2 4x 9 y2 4y 16 (b) ii C (2,2)
a2 9 b2 4
4 x 22 22 9 y 22 22 16 c2 9 4 5
a3 b2 c 5
4x 22 16 9y 22 36 16
Vmajor (5,2),(1,2)
4x 22 9y 22 36 Vmin or (2,0), (2,4)
x 22 y 22 1 Foci(2 5,2),(2 5,2)
94
28
(b) iii y
V3 (2,0) x
x
xxx xx
V2(1,2) F2 (2 5,2) C(2,2) F1(2 5,2) V1(5,2)
x
V4 (2,4)
29
30