VIEWS: 870 PAGES: 15 CATEGORY: Algebra POSTED ON: 11/19/2012 Public Domain
ADDITIONAL MATHEMATICS SPM 2012 PAPER 1 ( PART ONE ) LAST KOPEK REVISION TRY TO ANSWER THESE QUESTIONS AND CHECK YOUR ANSWERS FROM THE FOLLOWING SCHEME OF ANSWER AND MARKING PAPER 1 1. Diagram shows part of the relation f(x) State f(x) f(x) = (x – 2)2 − 3 (a) the type of the relation, x (b) the range. . (2, -3) Answer: (a) Many-to-one √1 (b) f(x) ≥ -3 √ 1 2. Given the function f(x) = 3 – x, g(x) = px2 – q and gf(x) = 3x2 – 18x + 5. Find (a) gf(- 2), (b) the value of p and q. Answer: (a) f (-2) = 3 – (−2) = 5 √ M1 gf(- 2) = g(5) = 3(5)2 – 18(5) + 5 = −10 √ 2 (b) gf(x) = p(3 – x)2 – q = px2 – 6px + 9p - q √ M1 Compare to 3x2 – 18x + 5 p = 3, 9p – q = 5 9(3) – q = 5 q = 22 √ 2 Both correct 3. A function g is defined as g(x) = 7x – 4. Find (a) g -1, (b) the value of h if g -1(h) = 2 Answer: Using the principle: x+4 √1 cx – b (a) g -1 = ax + b = 7 c a h+4 (b) g -1(h) = =2 √ M1 7 h = 10 √2 4. A quadratic equation with roots p and q is x2 + mx + m = 0 where p, q and m are constants. Express p in terms of q. Answer: (x – p) (x – q) = 0 OR: Use SOR/POR x2 – (p + q) x + pq = 0 √ M1 method: Compare to x2 + mx + m = 0 SOR = p + q − (p + q ) = m and pq = m POR = pq pq = − (p + q) √ M2 pq + p = −q −q p= √3 1+q 5. Find the range of values of x for 4x2 ≥ 3 – 4x Answer: a = 4 > 0 minimum graph 4x2 + 4x – 3 ≥ 0 √ M1 Let 4x2 + 4x – 3 = 0 x (2x – 1) (2x + 3) = 0 -³ 2 ½ √ M2 x=½; x=−³2 ? √ M2 x≤−³2; x≥½ √3 6. Diagram shows the graph of a quadratic function y = f(x). The straight line y = −15 is a tangent of the curve y = f(x). (a) State the equation of the axis of symmetry of the curve. (b) Express f(x) in the form of (x + p) 2 + q, where p and q are constants. Answer: y −2 + 6 √ M1 the mid-point of the two y = f(x) (a) x = points on the curve that 2 intersect at x-axis x −2 0 6 y = −15 x=2 √2 (x + p)² + q x + p = 0 x = −p (axis of symmetry) (b) (x – 2)² − 15 √ 2 q = optimum value (either minimum or q = −15 √ M1 maximum based on the graph) 2 7. Solve the equation: 16x = 8 2−x Answer: √ M1 2 24(x) = √ M1 23(2 − x) 24x = 2 [2 −3(2 − x)] 24x = 21 – 6 + 3x Use the concept of the rules of 4x = – 5 + 3x √ M2 equation of indices x = − 5 √3 8. Solve the equation: 2 logx4 + logx8 = −7 Answer: √ M1 logx 42 + logx 8 = −7 Converting log to indices logx 16(8) = −7 √ M1 OR logx 16(8) = logx x −7 Use the concept of the rules of equation of logarithms. 128 = x −7 √M2 27 = x −7 OR 2 7 = ( 1 /x ) 7 (½ )−7 = x −7 x=½ √3 9. The nth term of an arithmetic progression is given by Tn = 12 – 3n. Find the common difference of the progression. Answer: a = T1 = 12 – 3(1) = 9 √ M1 T2 = 12 – 3(2) = 6 d = T2 − a =6–9 √ M2 = −3 √3 10. Given that 12, 6, 3 … is a geometric progression, find the sum of the first 7 terms after the 3 th. term of the progression Answer: OR: T4 = 12 ( ½ )3 = 12/8 a = 12 S7* = (12/8) [1 – ( ½ )7] r = ½ √ M1 = 2.977 S7* = S10 – S3 12 [1 – ( ½ )10] – 12 [1 – ( ½ )3] = √ M2 1–½ 1–½ = 2.977 √3 S7* = Sum from T4 to T10 or first 7 terms after the 3rd. term k 11. Given 0.471 + 0.000471 + 0.000000472 + … = ------- Find the value of k. 333 Answer: a = 0.471 r = 0.001 √ M1 0.471 k Use the concept of S∞ = = √ M2 the sum of infinity. 1 – 0.001 333 = 157 333 k = 157 √3 12. Diagram shows a circle with center O. The length of minor arc AB is 4.8 cm and the angle of minor sector BOC is 1 rad. Using π = 3.142, find (a) the value of θ, (b) the length, in cm, of the major arc AC. Answer: 4.8 (a) θ = = 0.75 rad. √1 A 6.4 O θ B (b) θAC (major) = 2π – (1 + 0.75) √ M1 = 4.534 rad. C SAC (major) = 6.4 ( 4.534 ) = 29.02 cm. √ 2 dy 13. Given that y = x3 (3x + 1)4, find the value of dx when x = −1. Answer: dy = (3x + 1)4 3x2 + x3 (4)(3x + 1)3 (3) √ M1 dx dy x = −1; = [3(−1) + 1]4 3(−1)2 + (−1)3 (4)[3(−1) + 1]3(3) dx √3 √ M2 = 144