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## Chapter 1 IntegrationChapter 2 Differential EquationsChapter 3 Numerical MethodsChapter 4 Conics

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# Preparation for Mosys 2 SM025

### Keywords: Aim for "A"

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 12

   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 12u 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 2xdx ux dv  e2xdx
0 du  dx v  e2x
1
 xe2x  e2x
dx
1

  3
 xe2x  e2x 0

     3e1  e1   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
C2 49  A 49 
3 9

A3 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  y2xe2x. 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  y2xe2x
dx

1 dy  xe  2 x dx
y2

 1 dy   xe  2 x dx
y2

 1   xe2x  e2x  C
y 24

1  2xe2x  e2x  4C y  2 xe  2 x 4  11
y4  e2x
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 10  00C
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  ln5  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  ln5  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  ln5  x2 

 f 0.6  20.6  ln 5  0.62  0.335  0
 f 0.8  20.8  ln 5  0.82  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  ln5  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  20.7  ln 5  0.72   0.7460
20.7 
 2 
5  0.72

 
0.7460     0.7458


 x2
 20.7460  ln 5  0.74602
20.7460
2 
5  0.74602

 
0.7458     0.7458


 x3
 20.7458  ln 5  00.74582 K1
20.7458
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 through0,1 1 2 f  c  0        1
passes through0,11 121 22 f  c  0      2

2 1 120  24 f  0 K1 Solving
substitute f  5 in eq1 for f & c
f  120  5
24 Solving
for g
110  c  0
c  11 J1

Centre g,f  is on line 3y  4x 11  0

substitute f  5 in eq3  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

 12  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

2in 1 y 112  y2  2y 1110 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
y 11y  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 28 p
x
F p  6
F (4,2)
Equation of parabola;

x  h2  4 py  k
x  42  4(6)y 8
x  42  24y 8

   (b) i 4 x2  4x  9 y2  4y  16 (b) ii C (2,2)
a2  9 b2  4
   4 x  22   22  9 y  22  22  16 c2  9 4  5
a3 b2 c 5
4x  22 16  9y  22  36  16
Vmajor (5,2),(1,2)
4x  22  9y  22  36 Vmin or (2,0), (2,4)
x  22  y  22  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