SUGGESTED SOLUTIONS FOR ADDITIONAL MATHEMATICS PAPER 2 4030/2 JULY/ AUGUST (G.C.E) EXAMINATIONS – 2021 PREPARED & COMPILED BY KACHAMA DICKSON. C 0970 295655/0966 295655/0955 295655 © 2021 – Copperbelt Mufulira Why should you fail mathematics when success is guaranteed? Page 2 of 12 QUESTION 1 2π₯π₯ + 3π¦π¦ + π§π§ = 17 ……………. (i) π₯π₯ − 3π¦π¦ + 2π§π§ = −8 …….………(ii) 5π₯π₯ − 2π¦π¦ + 3π§π§ = 5……….…… (iii) Step 1: Eliminate π¦π¦ from equations (i) and (ii) by adding the two equations. 3π₯π₯ + 3π§π§ = 9 π₯π₯ + π§π§ = 3…………………..(iv) Step 2: Eliminate π¦π¦ from equations (i) and (iii) by multiplying (i) by 2 and (iii) by 3 4π₯π₯ + 6π¦π¦ + 2π§π§ = 34 +15π₯π₯ − 6π¦π¦ + 9π§π§ = 15 19π₯π₯ + 11π§π§ = 49 ……………(v) Step 3: Solve equations (iv) and (v) simultaneously by substitution method. π₯π₯ = 3 − π§π§ 19π₯π₯ + 11π§π§ = 49 19(3 − π§π§) + 11π§π§ = 49 π₯π₯ = 3 − 1 π₯π₯ = 2 57 − 19π§π§ + 11π§π§ = 49 −8π§π§ = 49 − 57 −8π§π§ = −8 π§π§ = 1 Step 4: Substitute the values of π₯π₯ and π§π§ in equation (i) to find π¦π¦. 2(2) + 3π¦π¦ + 1 = 17 4 + 3π¦π¦ + 1 = 12 3π¦π¦ = 12 − 5 3π¦π¦ = 12 βΉ π¦π¦ = 4 Therefore ππ = ππ, ππ = ππ and ππ = ππ Ans Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 3 of 12 QUESTION 2 (a) ππππππ − ππππ < 3 3π’π’2 − 8π’π’ − 3 < 0 3π’π’2 + π’π’ − 9π’π’ − 3 < 0 π’π’(3π’π’ + 1) − 3(3π’π’ + 1) < 0 (3π’π’ + 1)(π’π’ − 3) < 0 The critical points are (3π’π’ + 1) = 0 or π’π’ − 3 = 0 1 π’π’ = − 3 or π’π’ = 3 1 −3 Factors (3π’π’ + 1) π’π’ − 1 (3π’π’ + 1)(π’π’ − 3) 3 1 1 π₯π₯ < − 3 − 3 < π’π’ < 3 + − − − π₯π₯ > 3 + + − + + 1 Therefore the range of ππ values lies between − 3 < π’π’ < 3 (b) −3π₯π₯ 2 + 24π₯π₯ − 38 = −3 οΏ½π₯π₯ 2 − 8π₯π₯ − 38 3 οΏ½ = −3 οΏ½π₯π₯ 2 − 8π₯π₯ + (−4)2 − = −3 οΏ½(π₯π₯ − 4)2 − 38 3 86 38 3 − 16οΏ½ − (−4)2 οΏ½ = −3 οΏ½(π₯π₯ − 4)2 — 3 οΏ½ = −ππ(ππ − ππ)ππ + ππππ as required ∴ the coordinates of the turning point are (ππ, ππππ) Ans Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 4 of 12 QUESTION 3 1 −3 (a) 133π₯π₯ −4 = οΏ½4οΏ½ (b) 133π₯π₯−4 = (4−1 )−3 log 3 (π₯π₯ + 25) = 2 + log 3 (2π₯π₯ − 1) log 3 (π₯π₯ + 25) − log 3 (2π₯π₯ − 1) = 2 π₯π₯+25 133π₯π₯−4 = 43 log 3 οΏ½2π₯π₯−1οΏ½ = 2 (3π₯π₯ − 4) lg 13 = 3 lg 4 π₯π₯ + 25 = 9(2π₯π₯ − 1) π₯π₯+25 log 133π₯π₯ −4 = lg 43 2π₯π₯−1 3π₯π₯ lg 13 − 4 lg 13 = 3lg 4 π₯π₯ + 25 = 18π₯π₯ − 9 3π₯π₯ lg 13 =3lg 4 + 4 lg 13 π₯π₯ = = 32 25 + 9 = 18π₯π₯ − π₯π₯ 3lg 4+4 lg 13 34 = 17π₯π₯ 3 lg 13 6.261953383 π₯π₯ = 1.873809642 ππ = ππ Ans ππ = ππ. ππππ Ans QUESTION 4 (a) Let ππ(π₯π₯) = π₯π₯ 3 + 3π₯π₯ 2 − ππππ − 24 If f(x) is divisible by π₯π₯ − 2, then (π₯π₯ − 3) is the factor of ππ(π₯π₯). π₯π₯ − 3 = 0 implies that π₯π₯ = 3 and ππ(3) = 0. f(3) = 33 + 3(3)2 − ππ(3) − 24 = 0 27 + 27 + 3ππ − 24 = 0 54 − 24 = −3ππ 30 = −3ππ ππ = −ππππAns (b) ππ(π₯π₯) = π₯π₯ 3 − 7π₯π₯ + 6 = 0 By trial and error, we have ππ(1) = 0, therefore (π₯π₯ − 1) is the factor of ππ(π₯π₯). By synthetic division, we have 1 1 1 0 −7 6 1 1 −6 1 −6 0 Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 5 of 12 π₯π₯ 2 + π₯π₯ − 6 π₯π₯ 2 − 2π₯π₯ + 3π₯π₯ − 6 (π₯π₯ − 2)(π₯π₯ + 3) ∴ (π₯π₯ − 1)(π₯π₯ − 2)(π₯π₯ + 3) = 0 π₯π₯ − 1 = 0 ππππ π₯π₯ − 2 = 0 ππππ π₯π₯ + 3 = 0 ππ = ππ, ππ = ππ ππππ ππ = −ππ Ans QUESTION 5 (a) (i) Joseph C4 × 4! × 1C1 × 1! 4 = 4P4 × 1 = 4×3×2×1 = 24 Ways (ii) MJ, S, Jos, Jam 4! × 2! 24 × 2 ππππ Ways (b) Possibilities Women(5) Men(8) 0 6 1 5 2 4 3 3 4 2 5 1 Number of ways = 5C0 × 6C5 + 5C1× 8C5 Number of ways = 6 + 280 Number of ways = 286 ways Ans Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 6 of 12 QUESTION 6 (a) 3 sin(60° − π₯π₯) = sin π₯π₯ 3 sin 60° cos π₯π₯ − 3 cos 60° sin π₯π₯ = sin π₯π₯ 1 √3 3 οΏ½ 2 οΏ½ cos π₯π₯ − 3 οΏ½2οΏ½ sin π₯π₯ = sin π₯π₯ 3√3 2 3√3 2 3√3 2 3 cos π₯π₯ − 2 sin π₯π₯ = sin π₯π₯ (Divide throughout by sinx 3 cot π₯π₯ − 2 = 1 1 6√3 π¦π¦ = tan−1 οΏ½ 10 οΏ½ 3 × tan π₯π₯ = 1 + 2 3√3 2 tan π₯π₯ π¦π¦ = 46.1 5 =2 Now tan is positive in 1st and 3rd quadrants 10tan π₯π₯ = 6√3 tan π₯π₯ = tan π¦π¦ = In 1st quadrants: 6√3 π₯π₯ = π¦π¦ = 46.1 In the 3rd quadrant: π₯π₯ = 180° + π¦π¦ = 180° + 46.1 10 6√3 10 ππ = ππππ. ππ, ππππππ. ππ Ans 5 (b) (π’π’)R cos(π₯π₯+∝) = 4 cos π₯π₯ − 3 sin π₯π₯ Now R cos(π₯π₯+∝) = R cos π₯π₯ cos ∝ − Rsin π₯π₯ sin ∝ 5 R cos π₯π₯ cos ∝ − Rsin π₯π₯ sin ∝ = 4 cos π₯π₯ − 3 sin π₯π₯ R cos π₯π₯ cos ∝ = 4 cos π₯π₯ 5 R cos ∝ = 4 and Rsin ∝ = 3 5 R2 cos2 ∝ + R2 sin2 ∝ = 42 + οΏ½3 οΏ½ R2 (cos2 ∝ + sin2 ∝) = 16 + R2 (1) = 5 and − Rsin π₯π₯ sin ∝ = − 3 sin π₯π₯ 144 + 25 9 25 9 2 ∝= 22.6° 5 ∴ 4 cos π₯π₯ − 3 sin π₯π₯ = 13 3 cos(π₯π₯ + 22.6°) (ii) Therefore the maximum value is 13. 169 R2 = οΏ½ R= 13 3 9 since R > 0 Rsin ∝ 5 5 tan ∝ = Rcos ∝ = 3 ÷ 4 = 12 5 ∝= tan−1 οΏ½12 οΏ½ Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 7 of 12 QUESTION 7 ππ (a) (i) Sn = 2 [2ππ + ππ − 1)ππ] 5 Comment: Multiply both sides by 2 and divide both sides by 5. S5 = 2 [2ππ + (5 − 1)ππ] 50 = 5 (2ππ + 4ππ) 2 10(2ππ + 4ππ) = 100 2ππ + 4ππ = 20 ππ + 2ππ = 10 ………………(i) Next 10 terms implies that 5 + 10 = 15 S15 = 50 + 325 = 375 S15 = 15 375 = 2 [2ππ + (15 − 1)ππ] 15 2 Comment: Multiply both sides by 2 and divide both sides by 15. (2ππ + 14ππ) 750 = 15(2ππ + 14ππ) 2ππ + 14ππ = 50 ππ + 7ππ = 25…………..(ii) Solving equations (i) and (ii) simultaneously gives ππ = 4 and ππ = 3 ∴ the first term is ππ and the common difference is 3. Ans (ii) S13 = 13 2 [(2(4) + (13 − 1)3] S13 = 6.5(8 + 12 × 3) S13 = 6.5(44) S13 =286 Ans (b) (i) Since the annual salary increases by a ratio of 10% = 0.1, this is a compound interest with p = 4000 and r = 0.1 ππ = ππ(ππ + ππ)ππ S = 4000(1 + 0.1)6 S = 4000(1.1)6 S = 7086.244 S = 708 Therefore the sum of the Doctor’s annual salary is K708.00 Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 8 of 12 (ii) ππ(1 + ππ)π‘π‘ > 1 000 000 4000(1.1)π‘π‘ > 1 000 000 (1.1)π‘π‘ > 250 Taking log both sides we have tlog1.1 > log250 π‘π‘ > log 250 = 57.93149199 log 1.1 ππ ≈ 58 years Ans QUESTION 8 ππ ππππ ππππ ππππππ cf 164 −168 166 27 556 ππ 8 1 328 2 20 448 8 169−173 171 29 241 18 3 078 5 26 338 26 174 −178 178 31 684 28 4 928 8 67 328 54 181 32 761 21 3 801 6 87 981 75 184 −188 186 34 596 5 930 1 72 980 80 Scores 179 −183 ∑ ππ = 80 Totals (a) Median = ππ = 2 80 ∑ ππππ = 14065 ∑ πππ₯π₯ 2 = 2 475 075 = 40. The first number to reach 40 in the cumulative frequency 2 column corresponds to the median class. i.e. 54. Therefore, the median class is 174 −178. (b) (i) (ii) Mean (π₯π₯Μ ) = = ∑ πππ₯π₯ 2 SD = οΏ½ ∑ ππππ ∑ ππ ∑ ππ 14065 80 − (π₯π₯Μ )2 2 475 075 SD = οΏ½ 80 =175.8125 Ans − (175.8125)2 SD = √30 938.4375 − 30 910.03516 SD = √28.40234375 SD = 5.329384932 SD = 5.33 Ans Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 9 of 12 QUESTION 9 (a) At the point of intersection A, the curve π¦π¦ = π₯π₯ 2 and π¦π¦ = 2π₯π₯ are equal. π¦π¦ = π¦π¦ When π₯π₯ = 0 π₯π₯ 2 = 2π₯π₯ π¦π¦ = 2(0) = 0 π₯π₯ 2 − 2π₯π₯ = 0 When π₯π₯ = 2 π₯π₯(π₯π₯ − 2) = 0 π¦π¦ = 2(2) π₯π₯ = 0 or π₯π₯ − 2 = 0 π¦π¦ = 4 π₯π₯ = 0 or π₯π₯ = 2 Therefore, the coordinates of the point A are (ππ, ππ) Gradient of each curve is found as follows: ππ (b) V = ππ ∫ππ (π¦π¦12 − π¦π¦22 ) ππππ 4 V = ππ ∫2 (2π₯π₯)2 − (π₯π₯ 2 )2 ππππ 4 V = ππ ∫2 (4π₯π₯ 2 − π₯π₯ 4 )ππππ V = ππ οΏ½ 4π₯π₯ 3 3 + 4(2)3 V = ππ οΏ½ 3 32 π₯π₯ 5 + V = ππ οΏ½οΏ½ 3 + 64 V = ππ οΏ½15 οΏ½ 5 15π₯π₯οΏ½ (2)5 32 5 5 οΏ½οΏ½ 4 2 οΏ½ − (0) ππ V = ππ ππππ π π cubic units Ans QUESTION 10 (a) V = 3π‘π‘ 2 − 2π‘π‘ + 2 ππ = ππππ ππππ , ππ ππ = ππππ (3π‘π‘ 2 − 2π‘π‘ + 2) ππ = 6π‘π‘ − 2 ππ = 6(1.5) − 2 ππ = ππ − ππ = ππππ/ππππ Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 10 of 12 (b) To find the distance, we integrate the velocity function. ππ = ∫ π£π£ ππππ S = ∫(3π‘π‘ 2 − 2π‘π‘ + 2) ππππ S = π‘π‘ 3 − π‘π‘ 2 + 2π‘π‘ + ππ Now S = 3 and t = 0 3=0+c ππ = 3 S = π‘π‘ 3 − π‘π‘ 2 + 2π‘π‘ + 3 m 2 (c) S = ∫1 (3π‘π‘ 2 − 2π‘π‘ + 2)ππππ S=οΏ½ 3π‘π‘ 3 − 3 2π‘π‘ 2 2 + 2π‘π‘οΏ½ 2 1 S = οΏ½23 − 22 + 2(2)οΏ½ − οΏ½13 − 12 + 2(1)οΏ½ S= 8–2 S = 6m Ans QUESTION 11 (a) V = ππππ 2 β 500 = πππ₯π₯ 2 β Divide both sides by ππππ 2 500 β = ππππ 2 (b) .A = 2ππππ 2 + 2ππππβ 500 A = 2πππ₯π₯ 2 + 2ππππ × ππππ 2 (c) A = ππππππππ + ππππ ππππ = ππππππ − 4ππππ − ππππππππ ππππ ππππππππ ππ ππππππππ ππππ =0 Hence shown Ans 4πππ₯π₯ 3 − 1000 = 0 4πππ₯π₯ 3 = 1000 π₯π₯ 3 = 1000 4ππ ππππ , at the stationary, ππππ = ππ π₯π₯ 3 = 79.57747155 A = 2ππ(4.30)2 + 1000 4.30 A = 36.98ππ + 232.5581395 A = 348.7 d2A dx 2 d2A dr 2 = 4ππ + 2000 π₯π₯ 3 at ππ = 4.30 2000 = 4ππ + (4.30)3 > 0 this is a minimum point Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 11 of 12 3 π₯π₯ = √79.57747155 π₯π₯ = 4.301270069 ππ ≈ 4.30cm Ans QUESTION 12 (a) π¦π¦ = ππ −2π₯π₯ − 3, at Q; π₯π₯ = 0 At P; π¦π¦ = 0 π¦π¦ = ππ −2(0) − 3 ππ −2π₯π₯ − 3 = 0 π¦π¦ = ππ 0 − 3 ππ −2π₯π₯ = 3 ln ππ −2π₯π₯ = ln 3 π¦π¦ = −2 ∴ Q (0, −2) Ans −2π₯π₯ ln ππ = ln 3 −2π₯π₯ = ln 3 π₯π₯ = − (b) ∴ the coordinates of P and Q are οΏ½− π¦π¦ = ππ 2π₯π₯ − 3 π₯π₯ π¦π¦ −1 4.4 −0.5 −0.3 0 0.5 ln 3 2 ln 3 2 and so P οΏ½− ln 3 2 , 0οΏ½ , 0οΏ½ and (0, −2) respectively. 1 −2 −2.6 −2.9 π¦π¦ = ππ −2π₯π₯ − 3 -2 -1 0 1 −2 (c) π¦π¦ = ππ 2π₯π₯ − 3 π¦π¦ + 3 = ππ 2π₯π₯ π₯π₯ = ln √5 + 2π₯π₯ Now ππ 2π₯π₯ = π¦π¦ + 3 5+2π₯π₯ = π¦π¦ + 3 Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021 Page 12 of 12 1 π₯π₯ = ln(5 + 2π₯π₯)2 1 π₯π₯ = 2 lnβ‘ (5 + 2π₯π₯) 2π₯π₯ = lnβ‘ (5 + 2π₯π₯) 2π₯π₯ ln (5+2π₯π₯) ππ = ππ 5 + 2π₯π₯ = ππ 2π₯π₯ 5 − 3 + 2π₯π₯ = π¦π¦ ππ = ππππ + ππ Ans Prepared by Kachama Dickson C/High Standards in Mathematics/ Mufulira/2021
