2023 CCHY 4NA Prelim EM P2 Sol

1 @CCHY 2023 4045/2 [Turn Over Solution to 2023 CCHY Preliminary Examination 4NA Maths 4045/2 S/N Solution 1(a) 1.8056 8.1349 13.8321+ = 0.082196021 1b(i) 0.0822 1b(ii) 0.08 2(a) 2 459 457 – 62 671 = 2.396786 x 106 = 2.40 x 106 (3sf) # 2(b) 2459457 62671 = 39.24394058 = 40 (nearest 10) # 3(a) P(–4,3), Q(2,15) gradient PQ = ( ) 15 3 24 − −− = 2 equation of PQ : y = 2x + c At (–4,3), 3 = 2(–4) + c c = 3 + 8 = 11 ∴ equation of PQ is y = 2x + 11 # 3(b) gradient of CD = 2 equation of CD : y = 2x + c At R(7,11), 11 = 2(7) + c c = 11 – 14 = –3 ∴ equation of CD is y = 2x –3 # 4(a) 3x + y = 12 (1) x – 2y = 11 (2) from (1), y = 12 – 3x (3) subst (3) into (2), x – 2(12 – 3x) = 11 x – 24 + 6x = 11 7x = 35 ÷7, x = 5

2 @CCHY 2023 4045/2 [Turn Over subst x = 5 into (3), y = 12 – 3(5) = –3 Hence, x = 5, y = –3 # 4(b) interior angle = ( ) o8 2 180 8 −× = 135° x° + x° + 135° = 180° (∠ sum of ∆) x° = oo180 135 2 − = 22.5° ∴ x = 22.5 # 5(a) 12m – 3m3 = 3m(4 – m2) = 3m(2 + m)(2 – m) # 5(b) 63 27xx+ = = 33 x x + 6 = 3x 6 = 2x ÷2, x = 3 # 5(c) 23 4 90xx+ −= x = ( ) ( )( ) ( ) 2 4 4 43 9 23 −± − − − = 4 124 6 −± x = 4 124 6 −+ or x = 4 124 6 −− = 1.18925 = –2.52259 = 1.19 (2dp) = –2.52 (2dp) ∴ x = –2.52 or x = 1.19 # 6a(i) y = k x 54 = k 9 = 3k k = 54 3 = 18 ∴ y = 18 x # 6a(ii) 72 = 18 x

3 @CCHY 2023 4045/2 [Turn Over x = 72 18 = 4 x = 42 = 16 # 6(b) Pattern for 1st no. of 2nd column 2nd no. of 2nd column 1st no. of 3rd column 1 x 2 – 1 = 1 1 + 2 = 3 13 2 + = 2, 22 2 x 2 – 1 = 3 3 + 2 = 5 35 2 + = 4, 42 3 x 2 – 1 = 5 5 + 2 = 7 57 2 + = 6, 62 4 x 2 – 1 = 7 7 + 2 = 9 79 2 + = 8, 82 (2n – 1) x (2n – 1 + 2) = (2n – 1)(2n +1) 2 12 1 2 nn−+ + = 2n ∴ the nth pattern = (2n – 1)(2n +1) = 4n2 – 1 # 1st pattern = 1 x 3 = 22 – 1 2nd pattern = 3 x 5 = 42 – 1 3rd pattern = 5 x 7 = 62 – 1 4th pattern = 7 x 9 = 82 – 1 7(a) x°= 42°(alt ∠s, AB // CD) ∴ x = 42 # Method 1 65° + ∠AFE + 42° = 180° (int ∠s, AB // CD) ∠AFE = 180° – 65° – 42° = 73° y° = ∠AFE (vert opp ∠s) = 73° # ∴ y = 73

4 @CCHY 2023 4045/2 [Turn Over Method 2 to find y 65° + ∠AFE + 42° = 180° (int ∠s, AB // CD) ∠AFE = 180° – 65° – 42° = 73° 73° + ∠EFH = 180° (adj ∠s on a st line) ∠EFH = 180° – 73°