VIEWS: 495 PAGES: 36 CATEGORY: Algebra POSTED ON: 6/26/2012
PROGRAM PERMATA HARAPAN MATEMATIK TAMBAHAN (TINGKATAN 4) SMK JALAN EMPAT BANDAR BARU BANGI 25 & 26 JUN 2012 oleh: nklpunya PAPER 1 ( Jawab Semua Soalan ) Question 1: (a) { (1,p), (2,r), (3,s), (4,p) } √ 2 Any one pair correct √ A1 Common mistakes: * No bracket and wrong type of bracket * No comma * Wrong pairing * Wrong position of object-image Question 1: (b) many – to – one √ 1 Common mistakes: * one – to – many * many – to – one, one – to – many * fancy ways of writing the type of relation * mistaken as notation of relation Question 2: (a) 7 √1 (b) f(x) = x + 2 or f:x x+2 √1 Common mistakes: • Mistaken as the type of relation. • Weakness in deducing the f(x) from the diagram of the relation. Question 3: (a) 2, 4 √ 1 (b) { 1, 2, 3, 4 } √ 1 Common mistakes: * No bracket and wrong type of bracket * No comma * Listed objects as the answer or mistaken objects as the range Question 4: 2 (a) 3 √1 from: 3m – 2 = 0, substitute x as m (b) 0 ≤ f(x) ≤ 13 √ 2 Common mistakes: 13 (seen) √ A1 * 2 ≤ f(x) ≤ 13 * x = 0, f(x) = 2 *0 ≤ x ≤5 * Do not understand the meaning of range Question 5: (a) p = 8, q = −3 √ 3 (both correct) 4x + 8x – 3 √ A2 2 gf(x) = 4x 2 + px + q (2x + 1)2 + 2(2x + 1) – 6 √ A1 gf(x) = g(2x + 1) Weaknesses: * Not knowing that the question is about finding the composite function or do not understand in finding the composite function. * Not realise to compare the composite functions to get the value of p and q. Question 6: • Majority candidates did not include this term or did not know how to find. (a) 3/2 √ A1 –1 5 3x 1 (b) f , x 2 √2 (x) = 2x 1 y5 x √ A1 • Show the process of 3 2y finding the inverse Weakness: function * Do not know how to find the inverse function. Question 7: 5x 2 – 3x – 4 = 0 √ A1 (convert into general form of the quadratic equation) ( 3) ( 3)2 4(5)( 4) √ A2 or x use completing 2(5) the square method x = − 0.6434, x = 1 .243 √ 3 (both correct) Common mistakes: 1. 2. Answer not in four 3 3 2 4 x 5 x 4 significant figures x or not round-off 2x 5 correctly Question 8: Common mistakes: k<2 √3 • 36 - 4k -4(3) > 0 • 36 – 12k + 12 > 0 6 – 4(k + 1)(3) > 0 √ A2 2 • −12k > − 48 k > - 48/-12 6 – 4(k + 1)(3) 2 therefore: k > 2 or b2 – 4ac > 0 or a = (k+1), b = 6, c = 3 √ A1 Question 9: 2 x – 5x + 6 = 0 √ 3 Common mistakes: •α+β =−5 •α+β =5 4αβ = 6 • α + β = − 5/2 • αβ = 3 2(α + β) = 5 √ A2 (both) • x 2 + 5x + 6 = 0 α + β = 5/2 √ A1 (both correct) αβ = 3/2 Question 10: 2 2 b – 4ac > 0 or 4 – 4(2)(m) √ A1 42 – 4(2)(m) > 0 √ A2 16 – 8m > 0 16 > 8m 8m < 16 Common mistakes: * − 8m > −16/−8 m<2 √3 m>2 • 4 – 4(2)(m) > 0 (careless) Question 11: 2 y = a(x + p) + q minimum or maximum value axis of symmetry q=−1√1 p = − 2 √ 1 from: x + p = 0 2 + p = 0 2 At (0,3): 3 = a(0 – 2) + (− 2) 5 = 4a a = 5/4 √ 1 Question 12: 2 5x – x – 4 ≤ 0 √ A1 x2 – 5x + 4 ≥ 0 2 Let x – 5x + 4 = 0 − + (x – 1)(x – 4) = 0 + − − + x x 1 4 + 1 1 − 4 4+ x = 1, x = 4 √ A2 √ A2 Then: x ≤ 1 or (x – 1)(x – 4) ≥ 0 x≥4√3 x ≥ 1, x ≥ 4 ? (based on the graph) Then: x ≤ 1, x ≥ 4 √ 3 Question 13: 2 f (x) = 5 + 4x – x 5 + 4x – x 2 ≤ 8 2 − x + 4x – 3 ≤ 0 √ A1 √ A2 x2 – 4x + 3 ≥ 0 1 3 x 2 Let x – 4x + 3 = 0 (x – 1) (x – 3) = 0 x = 1 , x = 3 ? √ A2 Then: x ≤ 1, x ≥ 3 √ 3 Question 14: Common mistakes: * = lg2 x lg2 x lg5 ÷ lg2 lg20 = 0.3 x 0.3 x 0.7 ÷ 0.3 log2 20 = lg2 √ A1 * = 0.3 + 0.3 + 0.7 – 0.3 lg (2 x 2 x 5) lg 2 2 lg 5 = or = lg 2 lg 2 lg 2 lg 2 lg 5 2 lg 2 lg 5 = = √ A2 lg 2 lg 2 √ A2 0.3 0.3 0.7 2(0.3) 0.7 = √ A3 = √ A3 0.3 0.3 13 = √4 3 Question 15: Wrong Concept: log5 125x 2 logx125x2 = √ A1 log5 x log5 5 3 x 2 log5 x log5 5 3 log5 x 2 = √ A2 log5 x 3log5 x 2 = 3log5 5 2log5 x log5 x = √ A3 log5 x 3 2k = √4 k Question 16: 2n 2 n .2 3 2 n 5 2 √ A1 Concept Mistakes: 2n + 3 − 2n + 2n + 5(2n – 1) n 5 2 8 - 1 √ A2 * 2n + 3 − 2n + 2n + 10n – 1 2 19 n * 2n + 23 − 2n + 10n + 5−1 2 2 n -1 2 (19) √ 3 Question 17: 34 or 35 √ A1 Wrong Concept: 34x – 4 = 35 + x (3 4 ) x - 1 or 35 3x 33x – 4 = 35 √ A2 3 4x - 1 4x – 4 = 5 + x 35 or 3x 3x – 4 = 5 √ A3 3x − 1 = 5 x=2 x=3 √4 Question 17 - Wrong Concept: (continue) 4x – x – 1 – 5 *=3 = 33x – 6 * log3 4x – 1 – x = log353 ( 34 ) x - 1 5 * x =3 3 4x – 4 x =5 Question 18: Line 1: 6x + 3y – 5 = 0 y = −2x + 5/3 m1 = −2 √ A1 Line 2: m2 = ½ x-intercept = a; y-intercept = −3 Then: −(−3) = ½ √ A1 a a=6 √3 Question 19: 3 1 5 3 √ A1 12 = ½ h 0 −h h 24 = (0 – h + 5h) – (−3h + 0 + h) √ A2 6h = 24 , 6h = − 24 h=4 h = − 4 √ 3 (Both) Question 20: N = 9, x = 12 Use the formula to find Σx mean of ungroup data: (a) 12 = Σx 9 x= N Σx = 108 √ 1 (b) N = 8, x = 11 Σx = 108 – p 108 − p 11 = √ B1 p = −30 √ 2 8 PAPER 2 – SECTION A & B Question 1: y = 2x – 1 √ K1 x 2 + 2(2x – 1) + x(2x – 1) = 5 √ M1 3x 2 + 3x – 7 = 0 3 3 2 4(3)( 7) x √ M1 2(3) Follow question’s directive – x = 1.107, − 2.107 √ A1 whether in 3 decimal places, y = 1.214, − 5.214 √ A1 or did not mention, then leave the answer in 4 significant figures Question 2 x + 3y + 10 = 0 ...(1) (y + 4) (y + 10) = 0 y = − 4, √ M1 4 12 + + 1 = 0 ...(2) y = −10 √ A1 x y Sub. y (4): 4y + 12x + xy = 0 ...(3) y=−4 x=2 x = − 3y – 10 ...(4) √ K1 y = −10 Sub. (4) into (3) x = 20 √ A1 4y + 12(− 3y – 10) + (−3y – 10)y = 0 √ M1 −3y 2 – 42y – 120 = 0 3y 2 + 42y + 120 = 0 Question 3(a): (a) x = y 2 - 2 √ M1 Let f -1(x) = y y2 = x + 2 f (y) = x f -1 (x) = √ x + 2 √ A1 Let x = an integer value, f -1 (x) = the value after substitution, √ M1 an object has two images or it’s a one-to-many relation or the function undefined. Therefore the invest function does not exist. √ A1 Question 3(b): Let the first function or the ‘inner’ function (b) Let h(x) = y g(y) = 6x + 7 From: g[h(x)] = 3x + 4 g(y) = 3y + 4 From: g(x) = 3x + 4 3y + 4 = 6x + 7 √ M1 y = 2x + 1 h(x) = 2x + 1 √ A1 Question 4: Axis of symmetry, from x value of the coordinate given, (2, k) (a) x = 2 √ A1 (b) f(x) = x2 + 2hx + h2 – h2 – 5 √ M1 = (x + h) – h – 5 2 2 √ K1 h = −2 √ A1 From: x + h = 0 x = 2; 2 + h = 0 k = −h – 5 2 = −4 – 5 minimum value = k From the y value of = −9 √ A1 coordinate given, (2, k) Question 4(c): From: f(x) = x2 + 2hx – 5 • a = 1 > 0; a minimum graph f(x) • y-intercept = 5 • axis of symmetry, x = 2 x=2 • minimum value = −9 √ N1 • minimum point = (2, −9) x 0 2 √ N1 −5 −9 (2, −9) √ N1 Question 5: f x fx fx2 ( midpoint ) 7 133 931 123823 8 138 1104 152352 13 143 1859 265837 10 148 1480 219040 8 153 1224 187272 4 158 632 99856 Σ f = 50 - Σfx = 5 890 Σfx2= 1 048 180 √ M1 √ M1 Based on the formula of mean and variance, prepare the above table for a much easier procedure of solving the problem Question 5: (continue) √ M1 5 890 Σfx (a) mean = = 117.8 √ A1 x= Σf 50 1 048 180 (b) variance = (117.8)2 √ M1 50 = 7 086.76 √ A1 2 Σfx2 σ = − x2 Σf (c) Std. dev. = 84.18 √ A1 σ = √ σ2 Question 6: (a) y–3 6–3 = √ K1 0 – 6 14 – 6 y=¾ √ A1 then; A = (0, ¾) 12 – ¾ 15 Gradient AE, mAE = = √ M1 3–0 4 Eqn. AE: y – 12 = 15 ( x – 3 ) 4 y= 15 x + ¾ √ A1 4 Question 6: (continue) 2(x) + 3(3) E (3, 12) (b) 6 = 2+3 2 B (6, 3) √ M1 (either one) 3 D (x, y) 2(y) + 3(12) 3= 2+3 x= 21 , y = − 21 √ A1 (either one) 2 2 D( 21 , − 21 ) √ A1 2 2 Question 6: (continue) 6 14 21/2 6 √ K1 (C) Area ∆BCD = ½ 3 6 −21/2 3 = ½ (36 + 147 + 63/2) – (−63 + 63 + 42) √ M1 = 86.25 √ A1 TAMAT SELAMAT BERJAYA TERIMA KASIH