Additional Mathematic s
How to Solve a Problem
Understand
Plan your
Do - Carry out
Check your
the Problem
Strategy
Your Strategy
Answers
•Which Topic / •Subtopic ? •What info has been given? •What is to be found?
•Choose suitable strategy •Choose the correct formula
•Carry out the calculations
•Is the answer reasonable?
•Graph sketching
• Any other methods ??
•Creating tables ...
PAPER 1 FORMAT Objective Test : No. of Questions : Total Marks : Duration : L.O.D. :
Additional Materials :
Short Questions 25 questions 80 2 hours L (15) , M(7-8), H(2-3) Scientific Calculators, Mathematical Tables, Geometrical sets.
PAPER 2 FORMAT
Subjective Questions No. of Questions : A (6), B (4/5), C (2/4)
Total Marks : 100
Duration
L.O.D
: 2 hours 30 minutes
: L (6) , M(4-5), H(4-5) Additional Materials : Scientific Calculators, Mathematical Tables, Geometrical sets.
Key towards achieving 1A … Read question carefully Follow instructions Start with your favourite question Show your working clearly Choose the correct formula to be used +(Gunakannya dengan betul !!!) Final answer must be in the simplest form The end answer should be correct to 4 S.F. (or follow the instruction given in the question) π ≅ 3.142
Kunci Mencapai kecemerlangan
Proper / Correct ways of writing mathematical notations Check answers! Proper allocation of time (for each question)
Paper 1 : 3 - 7 minutes for each question Paper 2 : Sec. A : 8 - 10 minutes for each question Sec. B : 15 minutes for each question Sec. C : 15 minutes for each question
Common Mistakes…
dy dx
1. The Quadratic equation 3x2 - 4x = 0 +5 • y = 3x2 + 4x y = 6x + 4
3. ∫ 6 x + 4 dx = ∫ 3 x + 4 x + c 2
4. sin x = 300 , → 0 150 AB 1 = 5. → PQ
3
x = 300 , 1500
AB
PQ
1 3
Kesilapan Biasa Calon … • f ' (x) wrongly interpreted as f – 1(x) and / or conversely
•
x2 = 4
⇒
•
x >4
⇒ x > ±2 •
2
⇒
•
⇒
x = 2
Common errors…
PA : PB then
=
2 : 3
2PA = 3 PB
Actually, … PA : PB =
2 : 3
PA PB
2 3
3 PA
=
=
2 PB
More mistakes …… 3 ( ) + ( ) = 2 ( ) + ( ) 2
3
[(
2
2
) + ( ) ] = 2 [( ) + ( ) ] 2
2
32 PA2
2
=
9 PA2 =
2
22 PB2 4 PB2
2
Common mistakes …
loga x + loga y = 0, then
xy =
0
It should be… xy = a0 = 1
Common mistakes …
loga (x – 3) = loga x – loga 3
2 x 2 = 1 x + y = 1 x
y
2 x 2 = 2 2x + y = 2 0 x + y = 0 x
y
0
Common mistakes …
loga x + loga y = 0, then So,
loga xy =
0
xy = 0
It should be… xy = a0 = 1
Common mistakes …
sin (x + 300) = ½ , then sin x + sin 300 = ½ …………………gone !
Do NOT use Sin(A+B) = sin A cos B + cos A sin B !
Correct way… …
sin (x + 30 ) = ½ , 0
then x +300 = 300 , 1500 So, x = 0 , 120 0
0
If 00 is an answer, then 3600 is also an answer !
?
sin (x + 30 ) = ½ , 0
then x +30 = 30 , 150 , 390 0
0
0
So, x = 00 , 1200 , 3600
0
Relationship between Functions and Quadratic Functions y
Domain
Codomain
X
Y
f(x) = x2 4
1
O
1
2
1
1
2
4
x
Image
Objec t (1, 1) , (2, 4). …. form ordered pairs and can be plotted to obtain a curve.
SPM 2003 Paper 1, Question 1 P = { 1, 2, 3} Q = {2, 4, 6, 8, 10} The relationship between P and Q is defined by the set of ordered pairs { (1, 2), (1, 4), (2, 6), (2, 8)}. State the image of 1, The object of 2. [2 marks]
Answer
(a) 2 , 4 (a) 1
1 1
SPM 2003 Paper 1, Question 2 g : x → 5x + 1
Answer
(a)
2 5 B1 :
25x2 + 2 B1 :
h : x → x2 − 2x + 3
or 0.4 ( x −1) 5
2
or g(x) = 3
2 (5x+1)2 – 2(5x+1) + 3
SPM 2003 Paper 1, Question 3 (SPM 2005,Q5) Solve the quadratic equation 2x(x – 4) = (1- x)(x+2). Write your answer correct to four significant figures. (3 marks) Answer
2.591, - 0.2573 (both + 4 s.f.) B2 :
B1 :
− (−7) ± (−7) 2 − 4(3)(−2) 2(3) 3x2 – 7x – 2 = 0
3
SPM 2003 Paper 1, Question 4 The quadratic equation x (x+1) = px – 4 has two distinct roots. Find the range of values of p. (3 marks) Answer
p < -3, p > 5
(kedua-duanya)
B2 :
(p + 3) (p – 5) > 0
B1 :
(1 – p)2 – 4(1)(4) > 0
3
SPM 2003 Paper 1, Question 5
Given that log 2 T - log4 V = 3, express T in terms of V. (4 marks) Answer
4
T= 8V½ log 2 V log 2 T − =3 log 2 4
B1 T
log 2 V log 2 T − =3 2 log 2 T − log 2 V log 2
T V
1 2
= 3
1 2
=3
B3
V
B2
1 2
= 23
T = 8V
1 2
SPM 2003 Paper 1, Question 6 Solve the equation 42x – 1 = 7x
(4 marks)
Answer
4
x = 1.677 (2x – 1) log 4
= x log 7
2x log 4 – log 4
= x log 7
2x log 4 – x log 7 = log 4
B1 B2
x (2 log 4 – log 7 ) = log 4 x
log 4 2 log 4 log 7
or
0.6021 0.3591
B3
SPM 2007 (???) Solve the equation 42x – 1 = 8x
(3 marks)
Answer
22(2x – 1) = 23x 2(2x – 1) = 3x
4x – 1 =
x =
3x
1
4x – 2 = 3x x =
2
No !!!
SPM 2003 Paper 1, Question 7 The first three terms of an A.P. are k-3, k+3, 2k+2. Find (a) the value of k, (b) the sum of the first 9 terms of the progression. Answer
(3 marks)
2
(a) 7 (k + 3) – (k – 3) = (2k + 2) – (k + 3)
B1
6 = k–1
(b) 252
1
SPM 2003 Paper 2, Question 1 Solve the simultaneous equation 4x + y = - 8 and x2 + x – y = 2 (5 marks) Answer Make x or y the subject 8 y x 4
or
P1
y 8 4x
Eliminating x or y
K1
2
8 y 8 y y 2 or 4 4
x 2 x (8 4 x) 2
Solving the quadratic equation :
K1 N1
x = -2, -3 or y = 0 , 4 y = 0 , 4 or x = -2, -3
N1
SPM 2003 Paper 2, Question 2 The function f(x) = x2 - 4kx + 5k2 + 1 has a minimum value of r2 + 2k , with r and k as constants. •
By the method of completing the square, show that r=k–1 (4 marks)
•
Hence, or otherwise, find the value of k and the value of r if the graph of the function is symmetrical about the line x = r2 -1. (4 marks)
SPM 2003 Paper 2, Question 2 *** Answer 2(a) Writing f(x) in the form (x – p)2 + q (x – 2k)2 – 4k2 + 5k2 + 1 Equating
q
K1
( q* = r2 + 2k) (k – 1)2 = r2 r= k–1
(b) Equating (his) - (x – p) = 0
N1
N1 K1
2
b − 8 − y dy = 0 atau x = − + 2a 4 dx
k=0,4 r = -1, 3
K1
K1 N1
N1
Eliminating r or k by any valid method
F4
1. Functions 1. Given
f:x
x2 - 2 .
Find the values of x which map onto itself.
f (x) = x x2 - 2 = x x2 – x – 2 = 0 (x+1)(x-2) = 0 x = -1 , x = 2 2. Given
f:x
x - 3 ,
g:x
f(x) = x – 3, g(x) = 3x gf (1) = g [ f(1) ] = g [-2] = -6
3x
, find
gf(1).
T4 F4BAB 1
Functions :
Inverse Functions
4. Given f (x) = 3 – 2x, find f -1. Method 1
Let f (x) = y -1
Then x = f (y) x = 3 – 2y
3− x y= 2 3− x −1 f ( x) = 2
Method 2
Let Then
f (x) = y 3 – 2x = y 3 – y = 2x
3− y x= 2 3− y −1 f ( y) = 2 3− x f −1 ( x ) = 2
T4 F4BAB 1
Functions :
Applying the Idea of Inverse functions
f :x→
5. Given
3 x −1 4 ,
Method 1 (Find f-1 )
Let f -1(x) = y Then x = f(y) x= y=
3y − 1 4 4x + 1 3
f-1(a) =
find the value of a if f -1(a) = 11
Method 2 ( No need f-1 )
Let Then
f -1(a) = 11 a = f (11) = 8
4a + 1 3
a= 8
= 11
T4 F4BAB 1
Functions : Given composite function and one function, find the other function.
6. Given
f : x 2 x and gf : x 2 x 2,
find fg.
Remember : you need to find g first ! f(x) =2 - x , gf(x) = 2x-2 Let f(x) = u Then u = 2 – x
or
x=2-u
g(u) = 2(2-u) – 2 = 2-2u g(x) = 2-2x
fg(x) = f(2-2x) = 2 - (2-2x) =
2x
T4 F4BAB 1
**Functions : To skecth the graphs of y = |f(x)| 7. Skecth the graph of y = |3-2x|+1 for domain 0 ≤ x ≤ 4 and state the corresponding range. Tips :
Sketch y = |3-2x| first !!!
y 6 5 4 3
Range :
2
1≤ y ≤6
1
0
3 2
4
x
F4
2. Quadratic equations: SPM 2004, K1, Q4 Form the quadratic equation which has the roots – 3 and ½ . x =
– 3 ,
x =
(x+3) (2x – 1) = 0 2x2 + 5x – 3 = 0
½
F4
2. Quadratic Equations
ax2 + bx + c = 0 c b x − − x + = 0 a a 2
x2 – ( S.O.R) x + (P.O.R.) = 0 =
b S.O.R − a
P.O.R. =
c a
F4
2. The Quadratic Equation : Types of roots The quadratic equation
ax2 + bx + c = 0
1. Two distinct roots 2. Two equal roots 3. No real roots if
if if
b2 4ac b2 4ac b2 4ac
has
>0 =0 <0
**The straight line y = mx -1 is a tangent to the curve y = x2 + 2 …….
???
F4
3 Quadratic Functions : Quadratic Inequalities SPM 2004, K1, S5
Find the range of values of x for which x(x – 4) ≤ 12 x (x – 4) ≤ 12 x2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0 -2
6
–2≤ x ≤ 6
x
Back to BASIC
F4
Solve
x2 > 4 x2 – 4 > 0 (x + 2)(x – 2) > 0
x> ±2 ??? R.H.S must be O !
–2
2
x < -2 or x > 2
F4
4. Simultaneous Equations •
Solve the simultaneous equations x + y =1 x2 + 3y2 = 7 Factorisation
•
Solve the simultaneous equations, give your answer correct to three decimal places. x +y=1 x2 +3y2 = 8
*** P = Q = R
− b ± b − 4ac 2a 2
F4
5. INDICES
Back to basic… …
Solve ..
9
x −1
1 . 27
x
32(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1 – x = 3 x= –3
=
1
Betul ke ???
F4
5. INDICES
Solve
9
x −1
1 . 27
x
32(x – 1) . 3 (– 3x) = 1 32x – 2 +(– 3x) = 30 –x–2 =0 x = –2
=
1
F4
5. INDICES
Solve
or…
9
9
x −1
1 . 27
=
x
= 27 32(x – 1) = 3 3x 32x – 2 = 33x 2x – 2 = 3x x = –2 x-1
x
1
5. INDICES
F4
Solve
2x
+
3 = 2x+2
2x + 3 = 2x . 22 2x + 3 = 4 (2x ) 3 = 3(2 ) x
1 = (2x ) x
=
0
Can U take log on both sides ??? WHY? In the form u + 3 = 4u
F4
x2
5. INDICES
Solve the equation 3 , 32 3 give your answer correct to 2 decimal places. [ 4 marks]
9 (3x)
x
= 32 + (3x)
8 (3x) = 32 3x = 4
x x
lg 4 lg 3
= 1.26
(Mid-Yr 07)
5. INDICES
F4
Solve
2
2x
.5
x
4 .5 x
x
20
= 0.05 =
x
=
x
=
1 2 0 1 2 0
–1
ambm = (ab)m You can also take log on both sides.
F4
5. INDICES & LOGARITHMS
(Mid-Yr 07) Solve the equation l o g 2 ( x 2 ) 2 2 l o g 4 ( 4 x ) [4 marks]
log 2 (4 x) log 2 ( x 2) 2 2 . log 2 4
log 2 ( x 2) 2 log 2 (4 x)
lo g 2 ( x 2 ) lo g 2 4 ( 4 x ) x–2 x
= =
4 (4 – x) 3.6
F4
Back to basic… …
5. INDICES & LOGARITHMS
Solve the the equation log3 (x – 4) + log3 (x + 4) = 2
log3(x-4)(x+4) = 2 x2 – 16 = 9 x = 5
F4
Back to basic… … Solve the equation log3 4x – log3 (2x – 1) = 1
4x log 3 1 2 x 1
4x 3 2x 1 4x = 3(2x – 1) = 6x – 3 2x = 3 3 x = 2
SPM 2005, P1, Q8
F4
5 Indices and Logaritms : Change of base Given that log3 p = m and log4 p = n. Find logp 36 in terms of m and n.
logp 36 = logp 9 + logp 4 = 2log p 3 + logp
1 4 log 3 3 = 2( )+ log 3 p log 4 p logaa =1
2 1 = + m n
K1 K1 K1
N1
Coordinate Geometry
Some extra vitamins 4u …
Coordinate Geometry Distance between two points Division of line segments : midpoints + the ratio theorem Areas of polygons Equation of straight lines Parallel and perpendicular lines Loci (involving distance between two points)
Coordinate Geometry
Note to candidates: Solutions
to this question by scale drawing will not be accepted.
Coordinate Geometry
Note to candidates: A
diagram is usually given (starting from SPM 2004). You SHOULD make full use of the given diagram while answering the question.
Coordinate Geometry
Note to candidates: Sketch
a simple diagram to help you using the required formula correctly.
6. Coordinate Geometry 6.2.2 Division of a Line Segment PQ : QR = m : n
Q divides the line segment PR in the ratio
n m P(x1, y1)
n Q(x, y)
● R(x2, y2)
m
Q(x, y)
P(x1, y1)
nx1 + m x2 ny1 + m y 2 , Q(x, y) = m + n m + n
R(x2, y2)
6. Coordinate Geometry (Ratio Theorem) The point P divides the line segment joining the point M(3,7) and N(6,2) in the ratio 2 : 1. Find the coordinates of point P.
1 2
N(6, 2)
1(3) + 2(6) 1(7) + 2(2) P(x, y) = , 2 +1 2 +1
● P(x, y)
M(3, 7)
= =
P(x, y) =
nx1 + m x m + n
2
,
ny
+ m y m + n 1
2
15 11 , 3 3
11 5, 3
6. Coordinate Geometry
Perpendicular lines : R
m1.m2 = –1
P
Q S
6. Coordinate Geometry (SPM 2006, P1, Q12) Diagram 5 shows the straight line AB which is perpendicular to the straight line CB at the point B. The equation of CB is y = 2x – 1 . Find the coordinates of B.
[3 marks]
y
mCB = 2
●
A(0, 4) ●B
mAB = – ½
y = 2x – 1 Diagram 5
Equation of AB is
At B, 2x – 1 = – ½ x + 4
x
O ●C
y= –½x+4
x = 2, y = 3 So,
B is the point (2, 3).
6. Coordinate Geometry Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ. y mPQ= 3 4 4 mAB= −
K1
O
3
Midpoint of PQ (4, -3) 4 = ( y + 3) = − ( x − 4 ) 3
The equation : or
y =−
Q K1
4x + 3y -7 = 0 4 7 3
x +
3
P
N1
x
6 Coordinate Geometry
TASK : To find the equation of the locus of the moving point P such that its distances from the points A and B are in the ratio m : n
(Note : Sketch a diagram to help you using the distance formula correctly)
6. Coordinate Geometry Find the equation of the locus of the moving point P such that its distances from the points A(-2,3) and B(4, 8) are in the ratio 1 : 2. (Note : Sketch a diagram to help you using the distance formula correctly)
A(-2,3), B(4,8) and Let P = (x, y)
B(4, 8)
PA 1 PB 2 2PA PB 4PA
2
PB
m:n=1:2
A(-2, 3)
2 1
2
4 ( x 2 ) 2 ( y 3 ) 2 ( x 4 ) 2 ( y 8 ) 2
3x2 + 3y2 + 24x – 8y – 28 = 0
●
P(x, y)
6. Coordinate Geometry Find the equation of the locus of the moving point P such that its distance from the point A(-2,3) is always 5 units. (≈ SPM 2005)
A(-2,3) Let P = (x, y)
A(-2, 3)
●
5
( x 2) 2 ( y 3) 2 52
●
P(x, y)
x 2 y 2 4 x 6 y 12 0
is the equation of locus of P.
6. Coordinate Geometry Find the equation of the locus of point P which moves such that it is always equidistant from points A(-2, 3) and B(4, 9).
Constraint / Condition : PA PA
B(4, 9)
●
= PB 2
= PB
2
(x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2
x + y – 7 = 0 is the equation of
A(-2, 3)
●
● P(x, y) Locus of P
locus of P. Note : This locus is actually the perpendicular bisector of AB
Solutions to this question by scale drawing will not be accepted. (SPM 2006, P2, Q9) Diagram 3 shows the triangle AOB where O is the origin. Point C lies on the straight line AB. A(-3, 4) ●
y
Diagram 3
C
●
O
x ●
B(6, -2)
(a) Calculate the area, in units2, of triangle AOB. [2 marks] (b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks] • A point P moves such that its distance from point A is always twice its distance from point B. (i) Find the equation of locus of P, (ii) Hence, determine whether or not this locus intercepts the y-axis. [6 marks]
(SPM 2006, P2, Q9) : ANSWERS
y
A(-3, 4) ●
3 C
●
9(a)
1 0 6 3 0 2 0 2 4 0
1 0 24 0 0 6 0 2
= 9
9(b)
N1
2(4) 3( 2) 2(3) 3(6) , 3 2 3 2 12 2 , 5 5
N1
Diagram 3
2 x
O
K1
●
Use formula To find area
K1
B(6, -2)
Use formula correctly nx1 + m x2 ny1 + m y 2 , m + n m + n
(SPM 2006, P2, Q9) : ANSWERS
A(-3, 4) ●
y
● P(x, y)
C
9(c) (i)
AP =
2 ●
K1 AP = 2PB
(x+3) + (y – 4 )
●
B(6, -2) Use distance formula
K1
AP2 = 4 PB2 2
x
O
[ x (3 )]2 ( y 4 )2
1
2
= 4 [(x – 6) + (y + 2) 2
x2 + y2 – 18x + 8y + 45 = 0
2
Use AP = 2PB
N1
√
(SPM 2006, P2, Q9) : ANSWERS
9(c) (ii) x = 0, y2 + 8y + 45 = 0
K1
Subst. x = 0 into his locus
b2 – 4ac = 82 – 4(1)(45) < 0
K1 Use b2 – 4ac = 0 or AOM
So, the locus does not intercept the y-axis.
N1
√
(his locus
& b2 – 4ac)
F4
6. Coordinate Geometry : the equation of locus Given that A(-1,-2) and B(2,1) are fixed points . Point P moves such that the ratio of AP to PB is 1 : 2. Find the equation of locus for P.
2 AP = PB
2 ( x +1) 2 + ( y + 2) 2 = ( x − 2) 2 + ( y −1) 2 4[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2 3x2 + 3y2 + 12x + 18y + 15 = 0 x2 + y2 + 4x + 6y + 5 = 0
K1 J1
N1
F4
Statistics Marks
6-10 11-15 16-20 21-25 26-30 31-35 36-40 Total
f 12 20 27 16 13 10 2 100
From a given set of data, (e.g. The frequency distribution of marks of a group of students) Students should be able to find …. • the mean, mode & median • Q1, Q3 and IQR • the variance & S.Deviations • Construct a CFT and draw an ogive • Use the ogive to solve related problems
To estimate median from Histogram
F5
Graph For Question 6(b)
Number of people
80 70 60 50 40 30
20 10
33.5 0.5
40.5 20.5 Modal age = 33.5
60.5
80.5
100.5
Age
F4 CHAPTER 8
8. CIRCULAR MEASURE
‘Radian’ S =
‘Degrees’ rθ
θ
A =
(θ must be in RADIANS)
½ r2 θ
Always refer to diagram when answering this question.
F4
8. CIRCULAR MEASURE A
Diagram shows a sector of a circle OABC with centre O and radius 4 ∠ cm. Given that AOC = 0.8 radians, find the area of the shaded region. Area of sector OABC Area of triangle OAC Area of shaded region
B
0.8c
O
C
= ½ x 42 x 0.8 = 6.4 cm 2 = ½ x 42 x sin 0.8 = 5.7388 cm
2
= 6.4 – 5.7388 = 0.6612 cm2
K1
K1 In radians !!!!
K1 N1
F4
DIFFERENTIATION :
3 x + 1 Given that y = 4x + 5
, find
dy dx
dy (4 x 5)(3) (3 x 1)(4) 2 dx (4 x 5)
11 = 2 (4 x + 5)
d u dx v
F4
9 Differentiation : The second derivative
Given that
f(x) = x3 + x2 – 4x +
5 , find the value of f’ (x) = 4
3x2 + 2x –
f” (x) = 6x +
2 f” (1) =
f ” (1)
8
F4
9 Differentiation : The second derivative Given that ” (1) .
g (x) x 1 2
5
, find the value of g
g’ (x) = 10x (x2 + 1)4 g’’ (x) = 40x (x2 + 1)
3
. 2x
Ya ke ??
F4- 9
Given that ” (-1) .
g (x) x 1
g’ (x) = 1)4
2
5
, find the value of g
10x (x + 2
d uv dx
g’’ (x) = 10x . 4(x2 + 1) 3.2x +(x2+1)4. 10
g’’ (-1) = 10(-1) . 4[(-1)2 + 1] 3 +[(-1)2+1)4. 10 =
800
Mid-year, Paper 2
F4
Differentiation
:
Small increments dy
Given that y = 2x3 – x2 + 4, find the valuedxof at the point (2, 16). Hence, find the small increment in x which causes y to increase from 16 to 16.05.
dy dx
= 6x2 – 2x = 20 ,
x=
K1
δy 2 d y ≈ δx dx 0 .0 5 ≈ 20 δx
δx = 0 .0 0 2 5
K1 N1
Progressions : A.P & G.P F5
A.P.
:
a, a+d, a+2d,
a+3d ,
…….. Most important is “d”
G.P.
:
a, ar, ar2,
ar3, ……..
Most important ” !!
is
“
r
F5
Progressions : G.P - Recurring Decimals SPM 2004, P1, Q12 Express the recurring decimal 0.969696 … as a fraction in the simplest form. x = 100x = (2) – (1)
0. 96 96 96 … 96. 96 96 …..
99x = 96
96 x= 99
32 = 33
(1) (2)
Back to basic… …
F5
Progressions Given that Sn = 5n – n2 , find the sum from the 5th to the 10th terms of the progression.
Usual Answer :
S10 – S5 = ……. ??? Correct Answer :
S10 – S4 Ans :-54
F5
Linear Law 1. Table for data X and Y Y
2. Correct axes and scale used 3. Plot all points correctly 4. Line of best fit X
5. Use of Y-intercept to determine value of constant 6. Use of gradient to determine another constant
1-2 1 1 1
2-4
F5
Linear Law Bear in mind that …...... 1. Scale must be uniform Y
2. Scale of both axes may defer : FOLLOW given instructions ! 3. Horizontal axis should start from 0 ! X
4. Plot ……… against
Vertical Axis
……….
Horizontal Axis
Linear law
F5 Y 4.5
x
3.5
x 3.0
x 2.5
x
2.5
1.5
x
1.0
x 0.5
Read this value !!!!! 0
2
4
6
8
10
12
x
F5
INTEGRATION 1. 2.
∫ ( 3x +1) ∫ ( 2 − 3x )
4
4
dx
1 3. ∫ dx 4 ( 3x + 1)
−2 4. ∫ dx 4 ( 3x + 1)
dx
=
=
=
=
(3x + 1) 5 +c 15 (2 − 3x)5 +c −15 1 +c 3 − 9(3 x +1)
2 +c 3 9 ( 3x + 1)
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INTEGRATION SPM 2003, P2, Q3(a) 3 marks
dy Given that dx
= 2x + 2 and y = 6 when x = – 1, find y in terms of x. Answer:
dy = 2x + 2 dx
y
= =
x = -1, y = 6: Hence
6 c
(2 x 2)dx x2 + 2x + c = 1 +2 + c = 3
y = x2 + 2x + 3
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INTEGRATION SPM 2004, K2, S3(a) 3 marks The gradient function of a curve which passes through A(1, -12) is 3x2 – 6 . Answer:
dy dx
Find the equation of the curve. =
3x2 – 6
2 (3 x 6)dx y =
x = 1, y = – 12 :
= x3 – 6x + c – 12 = 1 – 6 + c c = –7
Hence
y = x3 – 6 x – 7
Gradient Function
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Vectors : B
Unit Vectors Given that OA = 2i + j and OB = 6i + 4j, find the unit vector in the direction of AB
AB = OB - OA = ( 6i + 4j ) – ( 2i + j ) = 4i + 3j
A
4 2 + 32
l AB l = = 5 Unit vector in the direction of AB =
1 (4i + 3 j ) 5
K1
N1
K1
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Parallel vectors Given that a and b are parallel vectors, with a = (m-4)i +2 j and b= -2i + mj. Find the the value of m. a=kb a= b (m-4) i + 2 j = k (-2i + mj) m- 4 = -2k
1
mk = 2
2
m=2
K1
K1 N1
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5
TRIGONOMETRIC FUNCTIONS
Prove that tan2 x – sin2 x = tan2 x sin2 x tan2 x – sin2 x =
sin 2 x − 2 kos x
sin 2x
K1
sin 2 x − kos 2 x sin 2 x = kos 2 x sin 2 x( 1 − kos 2 x ) = kos 2 x
K1
= tan 2 x sin 2 x
N1
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5 TRIGONOMETRIC FUNCTIONS
Solve the equation
2 cos 2x + 3 sin x - 2 = 0
2( 1 - 2sin2 x) + 3 sin x - 2 = 0
K1
-4 sin2 x + 3 sin x = 0 sin x ( -4 sin x + 3 ) = 0 sin x = 0 x = 00, 1800, 3600
,
K1
sin x = N1
3 4
x = 48.590, 131.410
N1
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5 TRIGONOMETRIC FUNCTIONS
(Graphs)
(Usually Paper 2, Question 4 or 5) -
WAJIB ! 1. Sketch given graph : (2003) y = 2 cos
3 x 2
(4 marks) , 0 x 2
(2004) y = cos 2x for 00 x 1800 (2005) y = cos 2x , 0 x 2 (2006) y = – 2 cos x , 0 x 2
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PERMUTATIONS AND COMBINATIONS
Find the number of four digit numbers exceeding 3000 which can be formed from the numbers 2, 3, 6, 8, 9 if each number is allowed to be used once only.
No. of ways
=
3. 2 3, 6, 8, 9
4 . 4. = 96
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Find the number of ways the word BESTARI can be arranged so that the vowels and consonants alternate with each other [ 3 marks ]
Vowels : E, A, I Consonants : B, S, T, R Arrangements : C V C V C V C No. of ways
= =
4! 3 ! 144
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Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4. Dice B, y
n(S) = 6 x 6 = 36
6
X
X
X
X
X
X
5
X
X
X
X
X
X
4
X
X
X
X
X
X
3
X
X
X
X
X
X
2
X
X
X
X
X
X
1
X
X
X
X
X
X
1
2
3
4
5
6
Constraint : x + y > 4 Draw the line x + y = 4 We need : x + y > 4 P( x + y > 4) = 1 – Dice A, x
5 = 6
6 36
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PROBABILITY DISTRIBUTIONS The Binomial Distribution
P ( X =r ) = nC r ( p ) r ( q ) n −r r = 0, 1, 2, 3, …..n
p+q=1
n = Total number of trials r = No. of ‘successes’ p = Probability of ‘success’ q = probability of ‘failure’
Mean Variance
=
np =
npq
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PROBABILITY DISTRIBUTIONS The NORMAL Distribution Candidates must be able to …
f(z)
determine the Z-score
Z =
x
use the SNDT to find the values (probabilities)
00
0.5
z
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f(z)
f(z)
=
-1.5 0
1
z
1 –
f(z)
– 0 1
z
0
1.5
z
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•
Index Numbers Index Number =
H1 I= × 100 H0 _
∑ wI ∑w
•
Composite Index =
•
Problems of index numbers involving two or more basic years.
I=
Solution of Triangles • • • • •
The Sine Rule The Cosine Rule Area of Triangles Problems in 3-Dimensions. Ambiguity cases (More than ONE answer)
Motion in a Straight Line Initial
displacement, velocity, acceleration... Particle returns to starting point O... Particle has maximum / minimum velocity.. Particle achieves maximum displacement... Particle returns to O / changes direction... Particle moves with constant velocity...
Motion in a Straight Line
Question involving motion of TWO particles. ... When both of them collide / meet ??? … how do we khow both particles are of the same direction at time t ??? The distance travelled in the nth second. The range of time at which the particle returns …. The range of time when the particle moves with negative displacement Speed which is increasing Negative velocity Deceleration / retardation
Linear Programming To answer this question, CANDIDATES must be able to ..... form inequalities from given mathematical information draw the related straight lines using suitable scales on both axes recognise and shade the region representing the inequalities solve maximising or minimising problems from the objective function (minimum cost, maximum profit ....)
Linear Programming Maklumat 1. x is at least 10 2. x is not more than 80 3. x is not more than y 4. The value of y is at least twice the value of x 5. The maximum value of x is 100 6. The minimum value of y is 35 7. The maximum value of x+ 2y is 60 8. The minimum value of 3x – 2y is 18 9. The sum of x and y is not less than 50 10. The sum of x and y must exceed 40 11. x must exceed y by at least 10
Ketaksamaan x ≥ 10 x ≤ 80 x ≤ y y ≥ 2x x ≤ 100 y ≥ 35 x + 2y ≤ 60 3x - 2y ≥ 18 x + y ≥ 50 x + y > 40 x ≥ y + 10
12. The ratio of the quantity of Q (y) to the quantity of P (x) should not exceed 2 : 1
y ≤ 2x
13. The number of units of model B (y) exceeds twice the number of units of model A (x) by 10 or more.
y - 2x >10
Selamat maju jaya !
