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‘Answering For Success’ Program By: Cg. Shaifur Azura Suhaimi Date: 18 March 2009, Friday Place: Ridzuan Resort, Melawi, Bachok, Kelantan Objectives: Solve the problems involving all Form 4 Chapters Reference: SPM Paper 1 questions, 2008 Question 1.Diagram 1 shows the graph of the function f ( x) 2 x 1 , for the domains 0 x 5 . State the: a) the value of t, b) the range of f(x) corresponding to the given domain [3marks] Strategies 1. What is the meaning of ‘modulus’ or . The negative value becomes positive value. 2. What is the x-intercept and yintercept 3. Given domain for x, means the image of x can be found. y 1 o t 5 x Diagram 1 2. Given the functions g:x→ 5x + 2 and h : x → x2 – 4x + 3, find a) g -1(6) b) hg(x) [4marks] AFS Program @ S2fo2r a) Find the inverse functions. Use easy math method or 90’s method. b) Find the composite functions Solution 1. Find the x-intercept points, value of t can be found. a) x-intercept happened when y = 0, 2x 1 0 2x – 1 = 0 2x =1 1 1 x = , therefore t = 2 2 b) the range of f(x) : when x = 0 , f(x) = 2(0) 1 1 x = 5 , f(x) = 2(5) 1 9 the range of f(x) : 0 f ( x) 9 5x 2 4 g-1(6) = a) g ( x) 5 1 x2 g-1(x) = or Method 2: 5 x2 x2 = g -1(x) = 5 5 6 2 6 2 4 g-1(6) = g -1(6) = = 5 5 5 Question Strategies f(x) = 5 x+ 2 x - 2 f -1(x) = 5 3. Given the functions a) f(5) means when x = 5, what f(x) = x -1 and is the value of f(x)? g(x) = kx + 2, find: b) find composite function of a) f (5) gf(5), when gf(5) = 14, k = ? b) the value of k such that gf (5) = 14 [3marks] 4. It is given that -1 is one Substitute x = -1 into quadratic of the roots of the equation, so the value of p can be 2 quadratic equation x – 4 found. x – p = 0. Find the value of p. Solution 2 (b) hg(x) = h [g(x) ] = h [5x + 2] = [5x+2]2 – 4[5x+2] +3 = 25x2 +20x + 4 – 20x -8 + 3 = 25x2 -1 = (5x+1)(5x-1) 3 a) b) f(5) = 5 – 1 =4 gf(5) = g [f(5)] = g [4] = k(4) + 2 14 = 4k + 2 14 – 2 = 4k 4 k = 12 k = 12/4 k= 3 Method 1: (-1)2 – 4 (-1) – p = 0 1+4–p=0 5–p=0 p=5 Method 2 : Roots: α = -1 , β = p SOR = -1 + p , POR = -p Equations : x2 – (SOR)x + POR = 0 x2 – (p-1) x –p = 0 By comparison: x2 – 4 x – p = 0 x2 – (p-1) x –p = 0 AFS Program @ S2fo2r Question Strategies Solution By comparison: x2 – 4 x – p = 0 x2 – (p-1) x –p = 0 p–1=4 p = 4+1 p = 5 5. f (x) = p (x + q )2 + r, has a minimum value of -4. The equation of the axis of symmetry is 3. State: a) the range of the values of p b) the value of q c) the value of r [3marks] 6. Find the range of the values of x for (x – 3 )2 < 5 - x p (x + q )2 + r i) from the word minimum , we know that p must be positive numbers. ii) p = coefficient iii) x + q = 0 , equation of axis of symmetry, given x = 3 iv) r = max or min point depends on p 2 (x – 3 ) < 5 – x, i) make sure inequalities in the form of ax2 + bx + c = 0 ii) the words range & < symbol means solution must sketch the graph. iii) use the concept of linear inequalities. iv) (x – 3 )2 = expand the bracket a) p 1 b) (x + q ) = 0 x=-q q=-x q = - (3) q=-3 c) r = -4 ( minimum value ) x2 – 6x + 9 < 5 – x x2 – 6x + 9 -5 + x < 0 x2 - 5x + 4 < 0 (x – 1) (x - 4) < 0 x – 1= 0 , x–4=0 x =1 or x = 4 x= 0 1 Range of x : AFS Program @ S2fo2r 4 1< x < 4 Question 7. Solve the equation: 162x – 3 = 34x Strategies Look at the base, not in the same base. Must use the logarithms on both sides of equations. Solution Method 1: 162x – 3 = 34x 42(2x – 3) = 34x log 44x – 6 = log 34x (4x – 6) log 4 = 4x log 3 4 x 6 log 3 4x log 4 4x – 6 = 4x (0.7925) 4x – 6 = 3.17x 4x – 3.17x = 6 0.83x = 6 6 x= 0.83 x = 7.2289 Method 2: 7. Solve the equation: 162x – 3 = 84x AFS Program @ S2fo2r Look at the base, not in the same base. Must use the logarithms on both sides of equations ( 2 4 ) 2 x 3 ( 2 3 ) 4 x 28x – 12 = 2 12x 8 x – 12 = 12 x 4x = -12 x = -3 Strategies Solution Question 8. Given that log 4 x = log 23, find the value of x. 1. Look at the base, not in the same base. 2. Changing the base is needed to solve the question. log 4 x = log 2 3 log 2 x log 2 3 log 2 4 log 2 x log 23 log 2 2 2 log 2x = 2 log 2 3 log 2x = log 2 32 = log 2 9 x = 9 AFS Program @ S2fo2r Question Strategies Solution a) y y a) y-intercep t T(0,4) x y 1 3 4 T(0,4) x-intercep t S(3,0) o S(3,0) x 13.(a) Write down the equation of the straight line ST in the form x y 1 a b o x x y 1 ------------(1) a b i) Identify a = x-intercept, b = y-intercept ii) substitute a & b into formula (1) (4 y ) 2 (0 x) 2 (0 y ) 2 (3 x) 2 b) b) A point P (x, y) moves such that PS = PT. Find the locus of P. y y2 - 8y +16 + x2 = y2 + x2 – 6x + 9 -8y +16 + 6x -9 = 0 -8y + 7 + 6x = 0 6x – 8y + 7 = 0 y-intercept T(0,4) x-intercept P S(3,0) o AFS Program @ S2fo2r b) PT = PS x equation of locus : 6x – 8y + 7 = 0 or 3 7 y= x 4 8 Question Strategies Solution 18. Radius = 10cm O OP = PQ and OPR 90 Find: a) QOR in radians b) the area, in cm2, of the coloured region O 5 P 5 10cm a) cos POR 5 10 P R R = 0.5 POR = cos-1(0.5) = 60 60x3.142 i) OPR 90 , OP = PQ = 5cm QOR = rad 180 ii) Use the right angle triangle OQR = 1.0472 rad Q iii) Use circular measures formulae O 10cm P 10cm 5 R Q [4marks] AFS Program @ S2fo2r b) Area of coloured region = Area of sector QOR – Area of triangle POR 1 1 = OQxORx OPxPR 2 2 1 1 = (10) 2 (1.0472) (5) 10 2 5 2 2 2 = 52.36 – 21.651 = 30.709cm2 Question 19. Two variables, x and y, Strategies i) approximate change in y = y = 16x -2 ii) given x = 4, x 4 h 4 =h iii) use are related by the equation 16 y 2. x Express, in terms of h, the approximate change in y, when x changes from 4 to 4 + h, where h is a small value. [3 marks ] AFS Program @ S2fo2r Solution y dy y dx x dy y xx dx dy 32 x 3 dx = 32 x3 y dy xx dx = 32 xh x3 32h y = x3 32h y ( 4) 3 1 y h 2 Question Strategies Solution 14. The points (0, 3), (2, t ) and (-2, -1) are the vertices of a i) Find the area of a triangle. Use the area formula, but remember to triangle. Given that the area of the triangle is 4 unit2, find eliminate the modulus. Eg : the values of t. (0, 3 ) (2, t ) 2x 2 5 2x - 2 = 5 (-2, -1) Anti clock-wise reading Area = 1 0 -2 2 0 2 3 -1 t =4 3 1 (2t 6 6 2) 4 2 14 -2t = 8 14 -2t = 8 2t=6 t=3 AFS Program @ S2fo2r or or 14 -2t = - 8 2 t = 22 t = 11 Question 20. The normal to the curve y = x – 5x at point P is parallel to the straight line y = - x + 12. Find the equation of normal to the curve at point P. Strategies Solution i) use the Easy math method P(3, -6 ) 2 dy y y= x2-5x dx dy = 2x - 5 dx m normal=m straight-line =-1 equation of normal: m tan = 1 y - (-6) = -1 ( x - 3 ) y + 6 = -x + 3 y = -x - 3 dy dx =1 2x - 5 = 1 2x = 6 x= 3 when x = 3, y = (3)2-5(3) = -6 AFS Program @ S2fo2r Question 22. A set of seven numbers has a mean of 9. a) find x b) when a number k is added to this set, the new mean is 8.5. Find the value of k. Strategies i) ii) Solution use formula mean of ungrouped data. x x N a) x x 9 x N 7 x 9x7 = 63 b) 8.5 = xk 7 1 8.5 (8) = 63 + k k = 68 – 63 k=5 AFS Program @ S2fo2r FORM 5 TOPICS: REFER TO SPM EXAM 2008 USING THE SCAN/STRATEGY/FORMULAE/OPERATION/OUTPUT/RE-CHECK (S2FO2R ) CONCEPT TO SOLVE PROBLEMS Q 9. QUESTION It is given that the first four terms of a G.P are 3, -6, 12 and x. Find the value of x. SCAN- STRATEGY / FORMULA 1. G. P – automatically have a and r 2. Find r use T1 , T2 , T3 r= 10. The first three terms of an A.P are 46, 43 and 40. The n th term of this progression is negative. Find the least value of n. T2 T3 T1 T2 1. A. P- automatically have a and d 2. Find d use T1 , T2 , T3 d = T3 – T2 = T2 – T1 3. Tn = a + (n-1) d 11. In a G.P, the first term is 4 and the 1. G. P – automatically have a common ratio is r. Given that the sum to infinity of this progression is 16, find the value of r. AFS Program @ S2fo2r and r 2. a = 4 , d = r , S = 16 OPERATION / OUTPUT / RE-CHECK 6 x 3 12 x 6(4) x 24 a = 46 d = T3 – T2 = T2 – T1 d = 43 – 46 = -3 Tn = a + (n-1) d a + (n-1) d < 0 46 + (n-1) -3 < 0 46 -3n + 3 < 0 -3n < -49 49 n> 3 1 n > 16 3 n = 17 a 1 S r = 11 r 4 4 3 r= 16 1 r 4 a 1 1- r = 1 r 4 SCAN- STRATEGY / FORMULA OPERATION / OUTPUT / RE-CHECK 1. Look at the x-axis and y- y k 5x axis k 2. y-axis = log10y, means log y log x 5 log must be inserted on log y log k log 5 x both side of the given 3. Use S Q QUESTION 12. The variables of x and y are k related by the equation y x , 5 where k is a constant. Diagram 12 shows the straight line graph obtained by plotting log10y against x. k a) Express the equation y x in 5 its linear form used to obtain the straight line graph shown in diagram 12. b) Find the value of k. log10 y o (0, -2) Diagram 12 AFS Program @ S2fo2r x equation. 3. Refer the new equation to find k. log y x log 5 log k log y log 5 x log k log k = -2 k = antilog 10 (-2) k = 10-2 1 k= = 0.01 100 Q QUESTION 15. The vectors a and b are non-zero and non-parallel. It is given that (h + 3 ) a = ( k – 5 ) b , where h and k are constants. Find the value of a) h b) k (2 marks ) 16. Diagram 16 shows a triangle PQR. Q 6b P SCAN- STRATEGY / FORMULA 1. The words non-zero and non-parallel means : (h + 3 ) = 0 and ( k – 5 ) = 0 (h + 3 ) = 0 and ( k – 5 ) = 0 a) h+3=0 h =-3 b) k -5 = 0 k=5 1. Change the ratio QT : TR = 3 : 1 to a fraction, to know the length of QT and TR. QT 3 TR 1 QT 3 TR 1 T 4a R Q Diagram 16 The point T lies on QR such that QT : TR = 3 : 1. Express in terms of a and b : a) QR b) PT AFS Program @ S2fo2r OPERATION / OUTPUT / RE-CHECK a) QT = 6b P QT 3,TR 1, QR 4 3 4 6b 4a QR QR T 4a Diagram 16 QP PR QR b) PQ QT PT 6b + 3 (6b 4a) = 4 PT Q QUESTION SCAN- STRATEGY / FORMULA OPERATION / OUTPUT / RE-CHECK 9 6b ( b) 3a PT 2 3 PT 3a b 2 17. Given that sin θ = p, where p 1. Second quartile for the trigo function is is a constant and 90 180 . referred. Find in terms of p: 2. Use the triangle to a) cosec θ solve problems. b) sin2θ 3. change the given (3 marks) questions using trigo- 90 p 1 180 1-p2 formulae 1 , sin sin2θ = 2sinθcosθ 4. cosec θ = 1 sin 1 = p a) cosec θ = b) sin2θ = 2sinθcosθ = 2 p ( 1 p2 ) = -2p 1 p 2 ) AFS Program @ S2fo2r Q 21. QUESTION SCAN- STRATEGY / FORMULA Given that (6 x 1)dx px x c , where p and c are constants, find a) the value of p b) the value of c if 2 (6 x 1)dx =13 when x = 1 23. 2 3 Diagram 23 shows numbered cards. six 1. Integral f(x) with respect to x. 2. Compare the answer with given value. 5 6 7 8 9 Diagram 23 A four-digit number is to be formed by using four of these cards. AFS Program @ S2fo2r = 2x3 + x + c 2x3 + x + c = px3 x c p = 2 b) 2x3 + x + c = 13 when x = 1, substitute x = 1 into the equation 2(1)3 + (1) + c = 13 3 + c = 13 c = 13 – 3 c = 10 The words ‘is to be formed’ means use n 3 OPERATION / OUTPUT / RE-CHECK 6 x3 a) (6 x 2 1)dx = xc 3 pr . 1. use the permutation rule a) 6 p4 360 Or using ‘fill in the box’ method How many a) different numbers can be formed? b) different odd numbers can be formed? * refer to the notes 3,5,6,7,8,9 6,7,8,9 6 7 7,8,9 3 5 3 picked number 6 number to be picked 5,6,7,8,9 4 5 3 x 4 x 5 x 6 = 360 a) Different ways of numbers can be formed are 360. b) 3, 5, 7, 9 5p3=60 4p14 5p3x 4p1=60x4 = 240 Different odd numbers can be formed = 240 AFS Program @ S2fo2r Q QUESTION SCAN- STRATEGY / FORMULA 24. The probability of Sarah being chosen as a prefect is 3 5 while the probability of Aini being chosen is 7 . Find the 12 probability that a) neither of them is chosen as a school prefect b) only one of them is chosen as a school prefect (4 marks ) a) Use the Probability X= (Sarah or Aini is chosen as a prefect) 3 2 of Independent Sarah: p = , q = 5 5 events. Event A 7 5 Aini: p = , q = occurs doesn’t 12 12 depend on a) P(Both of them are not chosen) outcomes of event 2 5 = X A. 5 12 i) Refer to AND: 1 alphabet N = = 6 ii) D = Darab 2 events iii) P(AB) = P(A) x P(B) b) Use the Probability of Mutually exclusive events. Both events cannot occur at the same time. AFS Program @ S2fo2r OPERATION / OUTPUT / RE-CHECK i) Refer to OR: Alphabet OR = b) P ( One of them is chosen) ATAU, U 3 5 7 2 )( X ) =( X ii) T = Tambah 2 5 12 12 5 events 29 iii) P(AUB) = = P(A or B) = 60 P(A) + P(B) Q QUESTION SCAN- STRATEGY / FORMULA 25. The mass of group of students in a school have a normal distribution with a mean of 40kg and a standard deviation of 5kg. Calculate the probability that a student chosen at random from this group has a mass of 1. Found the words, ‘mean’ & ‘standard deviation’. We must refer to Probability Distributions. 2. What is a discuss subject? X = mass of group of students and 3. Change X to Z. a) more than 45kg b) between 35kg 47.8kg [4marks] AFS Program @ S2fo2r X Z= OPERATION / OUTPUT / RE-CHECK a) μ = 40, σ = 5 45 40 ) 5 = P ( Z > 1) = 0.1587* * use the Z-Value table or Calculator P(X > 45) = P ( Z b) P(35 < X < 47.8) = 35 40 47.8 40 P( Z ) 5 5 = P( -1< Z < 1.56 ) Q QUESTION SCAN- STRATEGY / FORMULA OPERATION / OUTPUT / RE-CHECK f(z) z -1 1.56 = 1 – P(Z< -1) – P ( Z > 1.56 )* = 0.7819 * Use calculator : follow all the symbol & operations from left to right. AFS Program @ S2fo2r NOTES: PERMUTATIONS & COMBINATIONS PERMUTATIONS & COMBINATIONS PERMUTATIONS. COMBINATIONS The order of the objects in the chosen set is taken into consideration. The order of the objects in the chosen set is NOT taken into consideration. AFS Program @ S2fo2r Q1: CHEMISTRY a) no digits is repeated 9 x 8 x 7 6 x 5 x 4 x 3 x 2 x 1 x number of ways = 9x8x7x6x5x4x3x2x1 = 362,880 b) all the letters if the first letter is a vowel 2 x 8 x 7 x 6 x 5 x 4 E,I 2P1=2 number of ways = 2x8x7x6x5x4x3x2x1 = 80640 OR number of ways = 2P1 X 8P8 = 2 X 40320 = 80640 AFS Program @ S2fo2r 8P8=40320 x 3 x 2 x 1 Q2: 5 Digit Numbers Are To Be Formed Using 0, 1, 2, 3, 4 a) no digits is repeated 4 x 4 x 3 2 x x 2,3,4 3,4 1,2,3,4 0,2,3,4 cannot choose'0' number of ways = 4x4x3x2x1 = 96 4 or 4p1=4 x 4 x 3 4p4= 24 number of ways = 4p1 x 4p4 = 96 AFS Program @ S2fo2r x 2 1 ways of arrangement 4 numbers to be chosen x 1 b) the digits are allowed to repeat 0,1,2,3,4 0,1,2,3,4 4 5 5 5 x 5 x x x 0,1,2,3,4 1,2,3,4 0,1,2,3,4 cannot choose'0' number of ways = 4x5x5x5x5 = 2500 or 4 4p1=4 x 5 x 5 x 5 (5p1)4= 625 number of ways = 4p1 x (5p1)4 = 2500 AFS Program @ S2fo2r x 5 numbers to be chosen ways of arrangement numbers to be chosen c) even numbers are obtained with no repeat of digits Case 1: 0 is chosen to fill the last number ways of arrangement 4 numbers to be chosen 1,2,3,4 x 3 x 2,3,4 2 x 1 x even number will end with 0,2,4: now "0" is selected 4 3,4 number of ways = 4x3x2x1x1 = 24 ways of arrangement 3 numbers to be chosen 1,3,4 x 3 0,1,3 x 2 x 1,3 1 3 number of ways = 3x3x2x1x2 = 36 Total number of ways = 24 + 36 = 60 AFS Program @ S2fo2r 1 x 2 even number will end with 2,4: now 2 or 4 is selected(2p1=2) n pr Q3: 1, 2, 3, 4, 5, 6 3 digit even numbers formed, exceed 200 i) 2 3 4 x 3, 5 x 5, 1, 4, 6 2, 4, 6 - even ways of arrangement number to be chosen ways of arrangement = 2 x 4 x 3 = 24 3 ii) x 2, 4, 6 2 4 x 5, 1, 3, 6 4, 6 - even ways of arrangement = 3 x 4 x 2 = 24 AFS Program @ S2fo2r ways of arrangement number to be chosen Q4: Words: FORMAT, n = 6 2 x i) 4 5 O, A x x F, R, M, T, A 3 R, M, T, A M, T, A ways of arrangement = 2 x 5 x 4x 3 = 120 or x x x O, A 2p1 5p3 ways of arrangement = 2p1 x 5p3 = 120 AFS Program @ S2fo2r ways of arrangement an alphabet to be chosen Q5: Word : VOWELS , n = 6 a) vowels are separated C C V C C V V C = CONSONANT V V = VOWEL V arranging vowels = 5p2 = 20 arranging consonant = 4! = 4p4=24 total number of ways = 20 x 24 = 480 b) the consonant are next to each other V W x L x S x x 4! 3! total number of ways = 4! x 3! = 144 AFS Program @ S2fo2r x Q5: c) each arrangement starts and ends with a consonant V x x x C x C L, S, O, E arranging of consonant = 4p2 = 12 arranging of others = 4p4 @ 4! = 24 total number of ways = 12 x 24 = 288 AFS Program @ S2fo2r W x Q6: How many different odd numbers can be formed using digits 3, 4, 5, 6, 7? How many of these are more than 50,000? a) the last digit ends with 3, 5, 7 x x x x 4, 5, 6, 7 3, 5, 7 4! or 4p4 = 24 selected number 3p1=3 no. of ways = 3 x 24 = 72 b) case 1: the first digit is '5' and '7' 2 2 x 5, 7 2p1 x x 4, 6, 7 3p3 start with '5' or '7' = 2 x 3p3 x 2 = 24 AFS Program @ S2fo2r > 50,000 x 3, 7 2p1 Q6: b) case 2: the first digit is '6' and end with odd number 3 1 x x x 6 3p3 = 6 start with '6' = 1 x 3p3 x 3 =1x6x3 = 18 total number of ways = 24 + 18 = 42 AFS Program @ S2fo2r x 3, 5, 7 3! Q7: 7 students are lining up for a photo session. In how many ways can they be arranged if: a) the tallest is in the middle 5 6 x x 1 4 3 x 2 x x 1 x T no. of ways = 1 x 6p6 = 720 b) the two shortest are at the end of the line and the tallest is in the middle 2 x x s, s 2p1 no of ways = 2 x 4p4 x 1 = 48 AFS Program @ S2fo2r x 1 T x x x 1 s