Ceramah Add Mth

How to Solve a Problem Understand the Problem Plan your Strategy Check your Answers Do - Carry out Your Strategy Choose suitable strategy Choose the correct formula Which Topic / Subtopic ? What info has been given? What is to be found? Carry out the calculations Graph sketching Creating tables ... Is the answer reasonable? Any other methods ??

PAPER 1 FORMAT Short Questions 25 questions 80 2 hours L (15) , M(7-8), H(2-3)‏ Scientific Calculators, Mathematical Tables , Geometrical sets. Objective Test : No. of Questions : Total Marks : Duration : L.O.D. : Additional Materials :

PAPER 2 FORMAT Subjective Questions No. of Questions : A ( 6 ), B ( 4 /5), C ( 2 /4)‏ Total Marks : 100 Duration : 2 hours 30 minutes L.O.D : 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 +( G unakannya 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)‏   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… 4. sin x = 30 0 , 150 0 y = 3x 2 + 4x y = 6x + 4 1. The Quadratic equation 3x 2 - 4x + 5 = 0 x = 30 0 , 150 0 5.

Kesilapan Biasa Calon … f ' ( x ) wrongly interpreted as f – 1 ( x ) and / or conversely x 2 = 4 x = 2 x 2 > 4 x > ± 2 x 2 = 4 x = 2 x 2 > 4 x > ± 2

Common errors… PA : PB = 2 : 3 then 2 PA = 3 PB

Actually, … PA : PB = 2 : 3 3 PA = 2 PB

More mistakes …… 3 2 PA 2 = 2 2 PB 2 9 PA 2 = 4 PB 2

Common mistakes … log a x + log a y = 0, then xy = 0 It should be… xy = a 0 = 1

Common mistakes … log a (x – 3) = log a x – log a 3 2 x x 2 y = 1 x + y = 1 2 x x 2 y = 2 0 2 x + y = 2 0 x + y = 0

Common mistakes … log a x + log a y = 0, then log a xy = 0 So, xy = 0 It should be… xy = a 0 = 1

Common mistakes … sin (x + 30 0 ) = ½ , then sin x + sin 30 0 = ½ ………………… gone ! Do NOT use Sin(A+B) = sin A cos B + cos A sin B !

Correct way… … sin (x + 30 0 ) = ½ , then x +30 0 = 30 0 , 150 0 So, x = 0 0 , 120 0 If 0 0 is an answer, then 360 0 is also an answer ! ?

sin (x + 30 0 ) = ½ , then x +30 0 = 30 0 , 150 0 , 390 0 So, x = 0 0 , 120 0 , 360 0

Relationship between Functions and Quadratic Functions ( 1 , 1 ) , ( 2 , 4 ). …. form ordered pairs and can be plotted to obtain a curve. Image Object f(x) = x 2 x O y 1 4 2 1 X Y 1 1 Domain Codomain 2 4

SPM 2003 Paper 1, Question 1 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 ] P = { 1, 2, 3} Q = {2, 4, 6, 8, 10} Answer 2 , 4 1 1 1

SPM 2003 Paper 1, Question 2 Answer or 0.4 2 2 25x 2 + 2 B1 : (5x+1) 2 – 2(5x+1) + 3 B1 : or g(x) = 3

SPM 2003 Paper 1, Question 3 (SPM 2005,Q5)‏ Answer 2.591, - 0.2573 ( both + 4 s.f. )‏ 3 B1 : 3x 2 – 7x – 2 = 0 Solve the quadratic equation 2x(x – 4) = (1- x)(x+2). Write your answer correct to four significant figures. (3 marks) B2 :

SPM 2003 Paper 1, Question 4 Answer p < -3, p > 5 (kedua-duanya)‏ 3 B1 : (1 – p) 2 – 4(1)(4) > 0 The quadratic equation x (x+1) = px – 4 has two distinct roots . Find the range of values of p. (3 marks)‏ B2 : (p + 3) (p – 5) > 0

SPM 2003 Paper 1, Question 5 Answer T = 8 V ½ 4 Given that log 2 T - log 4 V = 3, express T in terms of V. (4 marks)‏ B1 B2 B3

SPM 2003 Paper 1, Question 6 Solve the equation 4 2x – 1 = 7 x (4 marks)‏ Answer x = 1.677 4 B1 B2 B3 (2x – 1) log 4 = x log 7 2 x log 4 – log 4 = x log 7 2 x log 4 – x log 7 = log 4 x (2 log 4 – log 7 ) = log 4

SPM 2007 (???)‏ Answer 2 2 ( 2x – 1) = 2 3 x Solve the equation 4 2x – 1 = 8 x (3 marks)‏ 2 (2x – 1) = 3 x 4x – 1 = 3x x = 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. (3 marks)‏ Answer (a) 7 2 B1 1 (b) 252 (k + 3) – (k – 3) = (2k + 2) – (k + 3)‏ 6 = k – 1

SPM 2003 Paper 2 , Question 1 Solve the simultaneous equation 4x + y = - 8 and x 2 + x – y = 2 ( 5 marks)‏ Answer Make x or y the subject P1 N1 x = -2, -3 or y = 0 , 4 Eliminating x or y Solving the quadratic equation : y = 0 , 4 or x = -2, -3 N1 K1 K1

SPM 2003 Paper 2 , Question 2 The function f(x) = x 2 - 4kx + 5k 2 + 1 has a minimum value of r 2 + 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 = r 2 -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 – 4k 2 + 5k 2 + 1 N1 k = 0 , 4 (b) Equating (his) - (x – p) = 0 N1 r = -1, 3 Equating q ( q* = r 2 + 2k)‏ (k – 1) 2 = r 2 r = k – 1 Eliminating r or k by any valid method N1 K1 K1 N1 K1 K1

1. Functions F4 f(x) = x – 3, g(x) = 3x gf (1) = g [ f(1) ] = g [-2] = -6 f : x x - 3 , g : x 3 x , find gf(1) . 2. Given f : x x 2 - 2 . 1. Given Find the values of x which map onto itself. x 2 - 2 = x x 2 – x – 2 = 0 (x+1)(x-2) = 0 x = -1 , x = 2 f ( x ) = x

Functions : Inverse Functions T4 BAB 1 4. Given f (x) = 3 – 2x, find f -1. F4 Let f -1 (x) = y Then x = f (y) x = 3 – 2y Method 1 Method 2 Let f (x) = y Then 3 – 2x = y 3 – y = 2x

Functions : Applying the Idea of Inverse functions T4 BAB 1 F4 Method 2 ( No need f -1 )‏ = 8 Let f -1 (a) = 11 Then a = f (11)‏ 5. Given , find the value of a if f -1 (a) = 11 Let f -1 (x) = y Then x = f(y)‏ Method 1 (Find f -1 )‏ x = y = f -1 (a) = = 11 a = 8

Functions : Given composite function and one function, find the other function. T4 BAB 1 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 F4 6. Given find fg.

**Functions : To skecth the graphs of y = |f(x)| T4 BAB 1 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 !!! Range : 1 ≤ y ≤ 6 F4 x y 0 4 6 5 3 4 2 1

2. Quadratic equations: SPM 2004, K1, Q4 Form the quadratic equation which has the roots – 3 and ½ . x = – 3 , x = ½ (x+3) (2x – 1) = 0 2x 2 + 5x – 3 = 0 F4

2. Quadratic Equations x 2 – ( S.O.R ) x + ( P.O.R. ) = 0 a x 2 + b x + c = 0 F4 P.O.R. = S.O.R =

2. The Quadratic Equation : Types of roots The quadratic equation a x 2 + b x + c = 0 has 1. Two distinct roots if 2. Two equal roots if 3. No real roots if b 2 - 4ac b 2 - 4ac b 2 - 4ac > 0 < 0 = 0 ** The straight line y = mx -1 is a tangent to the curve y = x 2 + 2 ……. ??? F4

3 Quadratic Functions : Q uadratic Inequalities SPM 2004, K1, S5 Find the range of values of x for which x(x – 4) ≤ 12 F4 x (x – 4) ≤ 12 x 2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0 – 2 ≤ x ≤ 6 6 x -2

Solve x 2 > 4 Back to BASIC x 2 – 4 > 0 (x + 2 )(x – 2 ) > 0 x < -2 or x > 2 F4 x> ±2 ??? R.H.S must be O ! – 2 2

4. Simultaneous Equations Solve the simultaneous equations x + y =1 x 2 + 3y 2 = 7 Solve the simultaneous equations, give your answer correct to three decimal places. x + y = 1 x 2 +3y 2 = 8 Factorisation *** P = Q = R F4

Back to basic… … 3 2(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1 – x = 3 x = – 3 Betul ke ??? 5. INDICES F4 Solve ..

3 2(x – 1) . 3 (– 3x) = 1 3 2x – 2 +(– 3x) = 3 0 – x – 2 = 0 x = – 2 5. INDICES F4 Solve

or… 9 x-1 = 27 x 3 2(x – 1) = 3 3x 3 2x – 2 = 3 3x 2x – 2 = 3x x = – 2 5. INDICES F4 Solve

Solve 2 x + 3 = 2 x+2 5. INDICES 2 x + 3 = 2 x . 2 2 x = 0 2 x + 3 = 4 (2 x )‏ 3 = 3(2 x )‏ 1 = (2 x )‏ F4 Can U take log on both sides ??? WHY? In the form u + 3 = 4u

Solve the equation , give your answer correct to 2 decimal places . [ 4 marks ] 5. INDICES F4 9 (3 x ) = 32 + (3 x )‏ 8 (3 x ) = 32 3 x = 4 x = 1.26 (Mid-Yr 07)‏

Solve 2 2x . 5 x = 0.05 5. INDICES x = – 1 a m b m = (ab) m F4 4 x . 5 x = 20 x = You can also take log on both sides.

(Mid-Yr 07)‏ Solve the equation [ 4 marks ] 5. INDICES & LOGARITHMS F4 x – 2 = 4 (4 – x) x = 3.6

Back to basic… … Solve the the equation log 3 (x – 4) + log 3 (x + 4) = 2 log 3 (x-4)(x+4) = 2 x 2 – 16 = 9 x = 5 5. INDICES & LOGARITHMS F4

Back to basic… … Solve the equation log 3 4x – log 3 (2x – 1) = 1 SPM 2005, P1, Q8 F4 4x = 3(2x – 1)‏ = 6x – 3 2x = 3 x =

5 Indices and Logaritms : Change of base Given that log 3 p = m and log 4 p = n. Find log p 36 in terms of m and n. = 2 log p 3 + log p 4 log p 36 = log p 9 + log p 4 log a a = 1 F4 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)‏

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 . Coordinate Geometry

6. Coordinate Geometry 6.2.2 Division of a Line Segment Q divides the line segment PR in the ratio PQ : QR = m : n n m P(x 1 , y 1 )‏ R(x 2 , y 2 )‏ Q(x, y)‏ ● n m R(x 2 , y 2 )‏ P(x 1 , y 1 )‏ Q(x, y)‏ Q(x, y) =

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. P(x, y) = ● 1 2 N(6, 2)‏ M(3, 7)‏ P(x, y)‏ = = P(x, y) =

6. Coordinate Geometry m 1 .m 2 = –1 Perpendicular lines : P Q R 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 ] m CB = 2 m AB = – ½ Equation of AB is y = – ½ x + 4 At B, 2x – 1 = – ½ x + 4 x = 2, y = 3 So, B is the point (2, 3). x y O A(0, 4)‏ C Diagram 5 B ● ● ● y = 2x – 1

6. Coordinate Geometry Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ. m PQ = m AB = Midpoint of PQ = (4, -3)‏ The equation : 4x + 3y -7 = 0 K1 K1 N1 or P Q x y O

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

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 m : n = 1 : 2 Let P = (x, y)‏ ● 2 1 B(4, 8)‏ A(-2, 3)‏ P(x, y)‏ 3 x 2 + 3y 2 + 24 x – 8 y – 28 = 0

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)‏ ● 5 A(-2, 3)‏ P(x, y)‏ ● A(-2,3)‏ Let P = (x, y)‏ 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 = PB PA 2 = PB 2 (x+2) 2 + (y – 3) 2 = (x – 4) 2 + (y – 9) 2 x + y – 7 = 0 is the equation of locus of P. Note : This locus is actually the perpendicular bisector of AB A(-2, 3)‏ ● B(4, 9)‏ ● Locus of P ● P(x, y)‏

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) Calculate the area, in units 2 , 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 ] x y O A(-3, 4)‏ Diagram 3 C ● ● ● B(6, -2)‏

(SPM 2006, P2, Q9) : ANSWERS 9(a)‏ = 9 x y O A(-3, 4)‏ Diagram 3 C ● ● ● B(6, -2)‏ 3 2 9(b) K1 N1 Use formula correctly N1 K1 Use formula To find area

(SPM 2006, P2, Q9) : ANSWERS √ AP = 2PB AP 2 = 4 PB 2 (x+3) 2 + (y – 4 ) 2 = 4 [(x – 6) 2 + (y + 2) 2 x 2 + y 2 – 18x + 8y + 45 = 0 N1 9(c) (i)‏ K1 Use distance formula K1 Use AP = 2PB x y O A(-3, 4)‏ C ● ● ● B(6, -2)‏ 2 1 P(x, y)‏ ● AP =

(SPM 2006, P2, Q9) : ANSWERS 9(c) (ii) x = 0, y 2 + 8y + 45 = 0 b 2 – 4ac = 8 2 – 4(1)(45) < 0 So, the locus does not intercept the y-axis. Use b 2 – 4ac = 0 or AOM K1 K1 Subst. x = 0 into his locus N1 √ (his locus & b 2 – 4ac)

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 x 2 + y 2 + 4x + 6y + 5 = 0 4 [ (x+1) 2 + (y+2) 2 ] = (x -2 ) 2 + (y -1) 2 3x 2 + 3y 2 + 12x + 18y + 15 = 0 F4 K1 J1 N1

Statistics 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 F4 100 Total 2 36-40 10 31-35 13 26-30 16 21-25 27 16-20 20 11-15 12 6-10 f Marks

To estimate median from Histogram F5 10 20 30 40 50 60 70 80 0.5 20.5 40.5 60.5 80.5 100.5 Modal age = 33.5 Age Number of people 33.5 Graph For Question 6(b)‏

8. CIRCULAR MEASURE F4 CHAPTER 8 ‘ Radian’ ‘Degrees’ S = rθ (θ must be in RADIANS )‏ A = ½ r 2 θ Always refer to diagram when answering this question. θ

8. CIRCULAR MEASURE F4 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. C A B O 0.8 c Area of sector OABC = ½ x 4 2 x 0.8 = 6.4 cm 2 = ½ x 4 2 x sin 0.8 = 5.738 8 cm 2 Area of triangle OAC Area of shaded region = 6.4 – 5.738 8 = 0 . 6612 cm 2 K1 N1 K1 K1 In radians !!!!

DIFFERENTIATION : F4 Given that , find

9 Differentiation : The second derivative Given that f(x) = x 3 + x 2 – 4x + 5 , find the value of f ” (1)‏ f’ (x) = 3x 2 + 2x – 4 f” (x) = 6x + 2 f” ( 1 ) = 8 F4

9 Differentiation : The second derivative Given that , find the value of g ” (1) . g’ (x) = 10x (x 2 + 1) 4 F4 g’’ (x) = 40x (x 2 + 1) 3 . 2x Ya ke ??

g’ (x) = 10x (x 2 + 1) 4 F4- 9 g’’ (x) = 10x . 4(x 2 + 1) 3 .2x +(x 2 +1) 4 . 10 g’’ ( -1 ) = 10( -1 ) . 4[( -1 ) 2 + 1] 3 +[( -1 ) 2 +1) 4 . 10 = 800 Mid-year, Paper 2 Given that , find the value of g ” (-1) .

Differentiation : Small increments F4 Given that y = 2x 3 – x 2 + 4, find the value of at the point (2, 16). Hence, find the small increment in x which causes y to increase from 16 to 16.05. K1 K1 N1 = 6x 2 – 2x = 20 , x = 2

Progressions : A.P & G.P A.P. : a, a+ d , a+2 d , a+3 d , …….. Most important is “ d ” F5 G.P. : a, a r , a r 2 , a r 3 , …….. Most important is “ r ” !!

Progressions : G.P - Recurring Decimals SPM 2004, P1, Q12 Express the recurring decimal 0.969696 … as a fraction in the simplest form. F5 x = 0. 96 96 96 … (1)‏ 100x = 96. 96 96 ….. (2)‏ (2) – (1) 99x = 96 x = =

Back to basic… … Usual Answer : S 10 – S 5 = ……. ??? Correct Answer : S 10 – S 4 Progressions Given that S n = 5n – n 2 , find the sum from the 5 th to the 10 th terms of the progression. Ans :-54 F5

Linear Law 1. Table for data X and Y 2. Correct axes and scale used 3. Plot all points correctly 4. Line of best fit 5. Use of Y-intercept to determine value of constant 6. Use of gradient to determine another constant F5 Y X 1 1-2 1 1 2-4

Linear Law Bear in mind that …...... 1. Scale must be uniform 2. Scale of both axes may defer : FOLLOW given instructions ! 3. Horizontal axis should start from 0 ! 4. Plot ……… against ………. F5 Y X Vertical Axis Horizontal Axis

Linear law 0 2 4 6 8 10 12 x 0.5 1.0 1.5 2.5 2.5 3.0 3.5 4.5 Y x x x x x x F5 Read this value !!!!!

INTEGRATION SPM 2003, P2, Q3(a) 3 marks Given that = 2 x + 2 and y = 6 when x = – 1 , find y in terms of x. F5 Answer: = 2x + 2 y = = x 2 + 2x + c x = -1, y = 6: 6 = 1 + 2 + c c = 3 Hence y = x 2 + 2x + 3

INTEGRATION SPM 2004, K2, S3(a) 3 marks The gradient function of a curve which passes through A(1, -12) is 3x 2 – 6 . Find the equation of the curve. F5 Answer: = 3x 2 – 6 y = = x 3 – 6x + c x = 1, y = – 12 : – 12 = 1 – 6 + c c = – 7 Hence y = x 3 – 6 x – 7 Gradient Function

Vectors : Unit Vectors Given that OA = 2 i + j and OB = 6 i + 4 j , find the unit vector in the direction of AB AB = OB - OA = ( 6 i + 4 j ) – ( 2 i + j ) = 4 i + 3 j l AB l = = 5 Unit vector in the direction of AB = F5 A B K1 N1 K1

Parallel vectors Given that a and b are parallel vectors, with a = (m-4) i +2 j and b = -2 i + m j . Find the the value of m. a = k b (m-4) i + 2 j = k (-2i + mj) m- 4 = -2 k m k = 2 1 2 a = b m = 2 F5 K1 N1 K1

Prove that tan 2 x – sin 2 x = tan 2 x sin 2 x sin 2 x tan 2 x – sin 2 x = 5 TRIGONOMETRIC FUNCTIONS F5 K1 N1 K1

Solve the equation 2 cos 2x + 3 sin x - 2 = 0 5 TRIGONOMETRIC FUNCTIONS F5 sin x ( -4 sin x + 3 ) = 0 sin x = 0 , 2( 1 - 2sin 2 x ) + 3 sin x - 2 = 0 -4 sin 2 x + 3 sin x = 0 sin x = x = 0 0 , 180 0 , 360 0 x = 48.59 0 , 131.41 0 K1 N1 K1 N1

5 TRIGONOMETRIC FUNCTIONS (Graphs)‏ (Usually Paper 2, Question 4 or 5) - WAJIB ! F5 1. Sketch given graph : (4 marks)‏ (2003) y = 2 cos x , (2004) y = cos 2x for (2005) y = cos 2x , (2006) y = – 2 cos x ,

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 = 4 . 4. 3. 2 = 96 3, 6, 8, 9 F5 PERMUTATIONS AND COMBINATIONS

Vowels : E, A, I C onsonants : B, S, T, R Arrangements : C V C V C V C No. of ways = 4 ! 3 ! = 144 Find the number of ways the word BESTARI can be arranged so that the vowels and consonants alternate with each other [ 3 marks ] F5

Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4. n(S) = 6 x 6 = 36 Constraint : x + y > 4 Draw the line x + y = 4 We need : x + y > 4 F5 Dice B, y 4 1 5 6 2 3 Dice A, x 2 3 4 5 1 6 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X P( x + y > 4) = 1 – =

The Binomial Distribution r = 0, 1, 2, 3, …..n n = Total number of trials q = probability of ‘failure’ p = Probability of ‘success’ r = No. of ‘successes’ F5 PROBABILITY DISTRIBUTIONS Mean = np Variance = npq p + q = 1

The NORMAL Distribution F5 PROBABILITY DISTRIBUTIONS Candidates must be able to … determine the Z -score Z = use the SNDT to find the values (probabilities) z f ( z ) 0 0.5 0

T5 z f(z) 0 1.5 z z f(z) 0 -1.5 1 = – 1 f(z) 0 1 –

Index Numbers Index Number = Composite Index = Problems of index numbers involving two or more basic years. F4

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 y ≤ 2x 12. The ratio of the quantity of Q ( y ) to the quantity of P ( x ) should not exceed 2 : 1 x ≥ y + 10 11. x must exceed y by at least 10 y - 2x >10 13. The number of units of model B ( y ) exceeds twice the number of units of model A ( x ) by 10 or more. x + y > 40 10. The sum of x and y must exceed 40 x + y ≥ 50 9. The sum of x and y is not less than 50 3x - 2y ≥ 18 8. The minimum value of 3x – 2y is 18 x + 2y ≤ 60 7. The maximum value of x+ 2y is 60 y ≥ 35 6. The minimum value of y is 35 x ≤ 100 5. The maximum value of x is 100 y ≥ 2x 4. The value of y is at least twice the value of x x ≤ y 3. x is not more than y x ≤ 80 2. x is not more than 80 x ≥ 10 1. x is at least 10 Ketaksamaan Maklumat

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