Additional Mat Hemati Cs

Additional Mat hemati cs

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… 1. The Quadratic equation 3x2 - 4x + 5 = 0 dy dx

• y = 3x2 + 4x y = 6x + 4

3. ∫ 6 x + 4 dx = ∫ 3 x + 4 x + c 2

4. sin x = 300 , 1500 →

5.

AB

1 = → 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 us e 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 0 0 is an an sw er , then 36 0 0 is al so an an swer !

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.

Rem emb er : you need to find g firs t ! 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 :

Sk etch y = |3-2x| first !!! y 6 5 4 3

Range :

2

1≤ y ≤6

1

0

3 2

4

x

F4

2. Quadratic equations: SP M 20 04, K 1, Q 4 Form t he q ua dr ati c equa ti on wh ich has th e roots – 3 a nd 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

x 2 – ( S.O .R ) x + ( P.O.R. ) = 0 =

S.O.R

b − 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

has

>0

b2 - 4ac = 0 b2 - 4ac

<0

**The straight line y = mx -1 is a tangent to the curve y = x2 + 2 …….

???

F4

3 Quadratic Functions : Quadratic Inequalities SPM 20 04, K1 , S5

Fin d the ra nge of v al ues of x f or whi ch x(x – 4) ≤ 1 2 x (x – 4) ≤ 12 x 2 – 4x – 12 ≤ 0 (x + 2)( x – 6) ≤ 0 -2

6

– 2 ≤ x ≤ 6

x

Back to BASIC

F4

Sol ve

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 Fa ctorisat ion

•

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. IND ICES

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. I NDIC ES

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. IN DIC ES

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. IND ICES

F4

Sol ve

2x

+

3 = 2 x+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

x2

5. IND ICES

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

Sol ve

2

2x

.5

4 .5 x

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. INDICE S & LOG ARITHM S

(Mi d-Yr 0 7) Solv e th e equa ti on

lo g 2 ( x  2 )  2  2 lo g 4 ( 4  x )

[ 4 mark s ]

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

5. INDICE S & LOG ARITHM S

Back to basic… … 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 4 log 3 3 1 = 2( )+ log 3 p log 4 p log a a = 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. mPQ= 3 4 4 mAB= −

y

K1

3

Midpoint of PQ = (4, -3) 4 ( y + 3) = − ( x − 4 ) 3 The equation :

or

Q K1

4x + 3y -7 = 0 4 7 y =− x + 3 3

P

O

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

Fr om a gi ve n s et of dat a, (e.g . The frequ ency distri buti on of marks o f a grou p of stu dents ) Stud ents shou ld be ab le to f in d …. • the m ean , mo de & med ian • Q1, Q3 an d IQR • the v ari ance & S.D eviati ons • Con stru ct a CFT and draw an og ive • Use the ogi ve to sol ve relate d pr oblems

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 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

A B

= ½ x 42 x 0.8 = 6.4 cm 2

Area of triangle OAC

= ½ x 42 x sin 0. 8 = 5.7388 cm

Area of shaded region

2

= 6.4 – 5.7388 = 0.661 2 cm2

0.8c

O

C K1

K1 In radians !!!!

K1 N1

F4

DIFFERENTIATION : Given that

3x + 1 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

Giv en that f(x) = x 3 + x 2 – 4x + 5 , fi nd th e va lue of f ” ( 1)

f’ (x ) = 4 2

3 x2 + 2 x –

f” (x ) = 6x + f” ( 1 ) =

8

F4

9 Differentiation : The second derivative Given that



g (x)  x  1 2



5

, find the value of g ” (1)

g’ (x) = 10x (x 2 + 1) 4 g’ ’ (x) = 40x (x 2 + 1)

3

. 2x

Ya ke ??

.

F4- 9

Given that .



g (x)  x  1

g’ (x) = 1) 4 g’’ (x) 10

2

10x



5

, find the value of g ” (-1)

d  uv  dx

(x + 2

= 10 x . 4(x 2 + 1)

3

.2x +( x 2 +1) 4 .

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 value of dx at the point (2, 16). Hence, find the small incr emen t in x which causes y to increase from 16 to 16.05.

dy dx

= 6x 2 – 2x = 20 ,

x =

K1

δy 2 d y ≈ δx dx 0 .0 5 ≈ 20 δx

δx = 0 .0 0 2 5

K1 N1

F5

Progr essi on s : A.P

A. P.

:

a, a+ d, a+2d,

& G. P a +3d , …

…..

Mos t im por tant is “d”

G.P.

:

a, ar, ar 2 ,

ar 3 , ……..

Mo st im por tant ” !!

is

“

r

F5

Progressions : G.P - Recurring Decimals SP M 20 04, P 1, Q 12 Ex press t he recurr ing d eci ma l 0.9 6969 6 … as a f ra ction i n the si mpl est for m.

(1)

x = 10 0x =

(2) – (1)

0. 96 96 96 … 96. 9 6 96 …. .

99x = 96 96 x99 =

32 33 =

(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

1-2

2. Correct axes and scale used

1

3. Plot all points correctly

1

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

2-4

F5

Linear Law Bear in mi nd t ha t … ...... 1. Sc ale mu st be un iform Y

2. Scale of both a xes may d efer : FO LLOW giv en in st ructi ons ! 3. Ho rizo ntal ax is sho uld s tar t f ro m 0 ! X

4. P lot

Vertical Axis

……… a gai nst

……….

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)

F5

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

F5

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

F5

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

F5

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

F5

5

TRIG ONOMET RIC FU NCT ION S

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

F5

5 TRI GONOMET RIC FUN CT IO NS

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

F5

5 TRI GONOMET RIC FUN CT IO NS

!

(Graph s)

(Us uall y Pap er 2, Que st ion 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

F5

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

F5

Find the number of ways the word BES TARI 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

F5

Two unb iase d dice ar e toss ed . Fi nd the pro ba bi li ty that the su m of the tw o numb ers obt ai ned is mo re 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

F5

PROB AB ILITY D IST RIB UTIONS 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

F5

PROB AB ILITY DIST RIB UTIONS 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

T5

f(z)

f(z)

=

-1.5 0

1

z

1 –

f(z)

– 0 1

z

0

1.5

z

F4

•

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 !