2019 CCHY MYE AMath 4047 P1 sol

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 1 of 12 1 Mid-Year Examination (2019) Secondary 4 Express / 5 Normal (Academic) Candidate Answer Key Name Register No Class Additional Mathematics Paper 1 4047 / 1 Date : 13th May 2019 Duration : 2 hours Additional Materials : READ THESE INSTRUCTIONS FIRST Write your name, index number and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total marks for this paper is 80. Setter : Poh Eng Hua Terence This paper consists of 12 printed pages, INCLUDING the cover page. For examiner’s use / 80 [Turn over

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 2 of 12 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x = a acbb 2 42  Binomial Expansion nba = na +    1 n 1na b +    2 n 2na 2b +    +    r n rna  rb +    + nb , where n is a positive integer and    r n =  !! ! rnr n  =  ! 11 r rnnn  2. Trigonometry Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A sin ( A  B ) = sin A cos B  cos A sin B cos ( A  B ) = cosA cos B  sin A sin B tan ( A  B ) = BA BA tantan1 tantan   sin 2A = 2 sin A cos A cos 2A = cos2 A - sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A tan 2A = A A 2tan1 tan2  Formulae for ABC A a sin = B b sin = C c sin a2 = b2 + c2 – 2bc cos A area of ABC = 1 2 ab sin C

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 3 of 12 1(i) 𝑓(𝑥)=4|4𝑥−6|−2|9−6𝑥| = 8|2𝑥−3|−6|3−2𝑥| = 8|2𝑥− 3|− 6|2𝑥− 3| =2|2𝑥−3| 1(ii) 2|2𝑥−3| =5 2𝑥−3=ହ ଶ or 2𝑥−3=−ହ ଶ 𝑥=ଵଵ ସ 𝑥=ଵ ସ =2ଷ ସ 2) 2𝑥ଶ−4𝑥+𝑐 =2𝑥+1 2𝑥ଶ−6𝑥+𝑐−1=0 𝑏ଶ− 4𝑎𝑐> 0,intersect at two points 36 − 8𝑐+ 8 > 0 8𝑐 < 44 𝑐 <5ଵ ଶ 3) AB = CD = 5 cm 𝐴𝐵 𝐴𝐷=tan𝜋 3 𝐴𝐷= 5 √3 sin ∠𝐴𝐸𝐷 𝐴𝐷 =sin𝜋 3 3√2 sin ∠𝐴𝐸𝐷= √3 2 3√2× 5 √3 sin ∠𝐴𝐸𝐷= 5 6√2 sin ∠𝐴𝐸𝐷= 5√2 12 ∠𝐴𝐸𝐷= sinିଵହ√ଶ ଵଶ (proven) 4) 7 √7−√2− 1 √7=7൫√7+√2൯ 7 − 2 −√7 7 =7 5√7+7 5√2−1 7√7

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 4 of 12 =ସସ ଷହ√7+଻ ହ√2 𝑏=44 35 ,𝑎= 7 5 5) 2𝑥ଶ−𝑥− 1 = (2𝑥+ 1)(𝑥− 1) If 2𝑥ଶ−𝑥− 1is a factor for 2𝑥ସ− 7𝑥ଷ+𝑎𝑥ଶ+ 13𝑥+𝑏, (𝑥− 1) and (2𝑥+ 1) are factors too. f(1)= 2(1)ସ− 7(1)ଷ+𝑎(1)ଶ+ 13(1) +𝑏 0 = 8 +𝑎+𝑏 ------- (1) f൬−1 2൰= 2(−1 2)ସ− 7൬−1 2൰ ଷ +𝑎൬−1 2൰ ଶ + 13(−1 2) +𝑏 0 = −51 2+𝑎 4+𝑏 22 =𝑎+ 4𝑏 --------- (2) Eqn (2) – eqn (1), 3𝑏= 30 b= 10, sub into eqn (1), 𝑎= −8 −𝑏 𝑎=−18 6) (i) Shape x and y-intercepts Turning pt 𝑦=|4𝑥+25| 36 6 𝑦=|36−𝑥ଶ| −6.25 −6 y x

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 5 of 12 (ii) Shape x and y-intercepts (iii) 4 solutions 7) (i) Midpoint of AB Gradient of AB = ହିଵ ଵିସ= 1ଵ ଷ Gradient of the perpendicular bisector = ିଵ ିర య = ଷ ସ Equation of perpendicular bisector, Sub in, (ii) Let y = 0, (1+4 2 ,5+1 2 ) (21 2,3) 𝑦=3 4𝑥+𝑐 (21 2,3) 3=3 4(21 2)+𝑐 𝑐 =3−15 8 =9 8 𝑦=3 4𝑥+9 8 0=3 4𝑥+9 8 𝑥=−3 2 𝐶(−3 2 ,0) Area of ∆𝐴𝐵𝐶,

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 6 of 12 8) (i) amplitude = 1, Period = 2𝜋 (ii) (iii) 2.5 = 3 − sin2𝑥 sin2𝑥= 0.5 sin𝛼= 0.5 𝛼=𝜋 6 0 ≤ 2𝑥≤ 4𝜋 2x is in first and second quadrant, 2𝑥=𝜋 6, 5 6𝜋, 13 6 𝜋, 17 6 𝜋 𝑥= 𝜋 12, 5 12𝜋, 13 12𝜋, 17 12𝜋 =ଵ ଶቚ−1.5 4 0 1 1 −1.5 5 0 ቚ unit2 =ଵ ଶቀ−1.5+20−1+ଵହ ଶቁ unit2 =13 unitଶ 1 4𝜋 1 2𝜋 3 4𝜋 𝜋 5 4𝜋 6 4𝜋 7 4𝜋 2𝜋 4 3 2

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 7 of 12 9) Volume = uniform surface area x length =ଵଶ× 𝑏 × ℎ × 8 ----------------- (1) The filled trough cross sectional surface is similar to the trough tan45° =ℎ𝑏2 b = 2ℎ sub into eqn (1) V =ଵଶ× 2ℎ × ℎ × 8 = 8ℎଶ 𝑚ଷ (i) Alternative, 𝐴𝑋 = 2 m 𝐴𝐵 = 4 m ௛ଶ=௕ସ 𝑏 = 2ℎ V =ଵଶ×2ℎ×ℎ×8 =8ℎଶ 𝑚ଷ 45o 2 m h m b m 45o 2 m h m b m A X B C tan45°=2𝐴𝑋

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 8 of 12 𝑑ℎ 𝑑𝑡=𝑑ℎ 𝑑𝑣×𝑑𝑣 𝑑𝑡 (ii) = 1 16ℎ×(−0.25) = −1 64ℎ− − − − − (1) 𝑣= 24 mଷ 24 = 8ℎଶ ℎ =√3 m , −√3 m (rej as h > 0) sub into eqn (1) ௗ௛ ௗ௧= −ଵ ଺ସ√ଷ = −√ଷ ଵଽଶ m/s or (−0.00902)m/s 10) (a) 2 + logଷ(1 −𝑥)ଶ =୪୭୥యଶହ଺ ୪୭୥యଽ logଷ9 + logଷ(1 −𝑥)ଶ =logଷ256 logଷ9 logଷ9(1 −𝑥)ଶ = logଷ√256 Compare value, 9(1 −𝑥)ଶ =√256 (1 −𝑥)ଶ =16 9 1 −𝑥= ±4 3 𝑥= −ଵ ଷ , 2ଵ ଷ (rej, log value > 0) b) 4௫ + 2௫ାଷ= 33 2ଶ௫+ 2௫ାଷ= 33 let 2௫ be 𝑢 𝑢ଶ+8𝑢−33=0 (𝑢+11)(𝑢−3) =0

CCHY Mid-Year Examination (2019) Additional Mathematics Sec 4E/5N Page 9 of 12 𝑢=−11 (𝑟𝑒𝑗), 𝑢=3 2௫ = 3 lg2௫ = lg3 𝑥= 1.58 11) Using guess and check, f(1) = 2(1)ଷ+ 9(1)ଶ+ 13(1)+ 6 = 30 f(−1)= 2(−1)ଷ+ 9(−1)ଶ+ 13(−1)+ 6 = 0 (𝑥+ 1) is a factor of f(x) 2𝑥ଶ+ 7𝑥+ 6 (𝑥+ 1)√2𝑥ଷ+ 9𝑥ଶ+ 13𝑥+ 6 −(2𝑥ଷ+ 2𝑥ଶ) 7𝑥ଶ+ 13𝑥 −൫7𝑥ଶ+ 7𝑥 ൯ 6𝑥 + 6 −(6𝑥+ 6)