2020 MFSS AM Prelim P1 Solution with Explanation

MATHEMATICS EXAMINATION INSTRUCTIONS TO CANDIDATES π π Brand / Model of Calculator For Examiner’s Use

2 2 (ܽ+ܾ )௡=ܽ௡+ ൫୬ ଵ൯ܽ௡ିଵܾ+൫୬ ଶ൯ܽ௡ିଶܾଶ+ ⋯+ ൫୬ ௥൯ܽ௡ି௥ܾ௥+ ⋯+ܾ௡ , n ൫௡ ௥൯= ௡! (௡ି௥)!௥!= ௡(௡ିଵ)…(௡ି௥ାଵ) ௥! 2 2 2 2 2 2 sin (    ∓ tan(ܣ±ܤ)=tanܣ±tanܤ 1 ∓tanܣtanܤ 2 – 2 2 – – 2 tan2ܣ=2tanܣ 1 −tanଶܣ  a2 b2 2–2 

3 1ݔଶ −2ݔ+5=0 α β . ߙଶ+ߚଶ . ߙଷ+ߚଷ= −22 . 1 ఈ ఉమ ఉ ఈమ . ߙ+ߚ =− −2 1 =2 ߚߙ=5 1=5 ߙଶ+ߚଶ=(ߙ+ߚ )ଶ −2ߚߙ =2ଶ −2(5) = −6 ߙଷ+ߚଷ=(ߙ+ߚ )(ߙଶ −ߚߙ+ߚ ଶ) =2(−6 −5) = −22 ߙ ߚଶ+ߚ ߙଶ=ߙଷ+ߚଷ ߙଶߚଶ = −22 5ଶ = −22 25 ߙ ߚଶ ൬ߚ ߙଶ൰= 1 ߚߙ=1 5 ݔଶ −(− ଶଶ ଶହ)ݔ+ ଵ ହ=0 ݔଶ+ ଶଶ ଶହݔ+ ଵ ହ=0

4 2 2(√3 − 1) ܽ√ܾ+ ܿa, b . 12(√3 − 1) 3 ݌+ 2ඥݍp q Area of triangle = 1 2 × 2൫√3 − 1൯ × 2൫√3 − 1൯ sin 60 = 1 2 (4)൫3 − 2√3 + 1൯ ቆ√3 2 ቇ = √3൫4 − 2√3൯ = 4√3 − 6 3 − 2√3 + 1 ቀ√ଷ ଶ ቁ Vol of prism = Area of triangle × height Height = 12൫√3 − 1൯ 4√3 − 6 = 6√3 − 6 2√3 − 3 × 2√3 + 3 2√3 + 3 = ൫6√3 − 6൯൫2√3 + 3൯ ൫2√3൯ ଶ − 3ଶ = 36 + 18√3 − 12√3 − 18 12 − 9 = 18 + 6√3 3 = 6 + 2√3 × ଶ√ଷାଷ ଶ√ଷାଷ 36 + 18√3 − 12√3 − 18 18 + 6√3

5 3ݕଶ = 6ݔݕ =ݔ య మݔ≥ 0 . logଷ(2ݔ− 17) = logଽ 81 − logଷݔ. ݕଶ = 6ݔ ݕ= ݔ య మ b logଷ(2ݔ− 17) = logଽ 81 − logଷݔ logଷ(2ݔ− 17) + logଷݔ= 2 logଷݔ(2ݔ− 17) = 2 2ݔଶ − 17ݔ= 9 2ݔଶ − 17ݔ− 9 = 0 (2ݔ+ 1)(ݔ− 9) = 0 ݔ= − 1 2 (rej),ݔ= 9 logଽ 81 = 2 logଷݔ(2ݔ− 17)

6 4 ୡ୭ୱ ఏ ଵାୱ୧୬ ఏ+ ଵାୱ୧୬ ఏ ୡ୭ୱ ఏ =2secߠ. ୡ୭ୱଶ ఏ ଵାୱ୧୬ଶ ఏ+ ଵାୱ୧୬ଶ ఏ ୡ୭ୱଶ ఏ =tanଶ2ߠ−2 0 ≤ߠ≤6 . ܵܪܮ=cosߠ 1+sinߠ+1+sinߠ cosߠ =cosଶߠ+(1+sinߠ)ଶ cosߠ(1+sinߠ) =cosଶߠ+1+2sinߠ+sinଶߠ cosߠ(1+sinߠ) =2+2sinߠ cosߠ(1+sinߠ) =2(1+sinߠ) cosߠ(1+sinߠ) =2 cosߠ =2secߠ=ܵܪܴ 2+2sinߠ cos2ߠ 1+sin2ߠ+1+sin2ߠ cos2ߠ= tanଶ2ߠ−2 2sec2ߠ=tanଶ2ߠ−2 2sec2ߠ=secଶ2ߠ−1 −2 secଶ2ߠ−2sec2ߠ−3=0 (sec2ߠ−3)(sec2ߠ+1)=0 sec2ߠ=3 sec2ߠ=−1 cos2ߠ=1 3 cos2ߠ=−1 Basic ∠=1.2309 2ߠ=1.2309,2ߨ−1.2309, 1.2309+2ߨ,2ߨ−1.2309+2ߨ2ߠ=ߨ ,ߨ +2ߨ ߠ=0.61545,2.5261,3.7570,5.6677ߠ=1.5707,4.7123 ߠ=0.615,2.53,3.76,5.67ߠ=1.57,4.71(3sf) secଶ2ߠ−1 1 if not shown) Basic ∠=1.2309 0.615,2.53,3.76,5.67 1.57,4.71

7 5ݕ= |ܽ(ݔ− ℎ)ଶ +݇|ܽ> 0 . (1, 0) (3, 0) (ℎ, −݇) . ℎ= 2 . a . ݕ= ݉ ݕ= |ܽ(ݔ− ℎ)ଶ +݇| h ℎ = ଵାଷ ଶ = 2 . h oݕ= ܽ(ݔ− ℎ)ଶ +݇, Sub (0, 9), 9 =ܽ(0 − 2)ଶ +݇ 9 = 4ܽ+ ݇− − − (1) Sub (1, 0), 0 =ܽ(1 − 2)ଶ +݇ 0 =ܽ+ ݇− − − (2) (1) − (2): 3ܽ= 9 ܽ= 3 ݇= −3 ݕ= ܽ(ݔ− ℎ)ଶ +݇ 0 <݉< 3

8 6 4ݔ− 3ݕ− 1 = 0 (4, 5) . ݔ= 13 (4, 5) 4ݔ− 3ݕ− 1 = 0 . aܽ> 0 aܽଶ + 24ܽ− 256 = 0 . . (10, 8) . pݔ= ݌. 4ݔ− 3ݕ− 1 = 0 ݉௣௘௥௣௘௡ௗ௜௖௨௟௔௥ = − 1 4 3 = − 3 4 Sub (4,5), 5 = − 3 4 (4) +ܿ ܿ= 8 Perpendicular line: ݕ= − ଷ ସݔ+ 8 ݉௣௘௥௣௘௡ௗ௜௖௨௟௔௥ Centre:ݕ= − 3 4ܽ+ 8 Distance bet centre & (4,5) = 13 −ܽ ඨ(ܽ− 4)ଶ + ൬− 3 4ܽ+ 8 − 5൰ ଶ = 13 −ܽ ܽଶ − 8ܽ+ 16 + ൬3 − 3 4ܽ൰ ଶ = (13 −ܽ)ଶ ܽଶ − 8ܽ+ 16 + 9 − 9 2ܽ+ 9 16ܽଶ = 169 − 26ܽ+ ܽଶ 9 16ܽଶ + 27 2ܽ− 144 = 0 ܽଶ + 24ܽ− 256 = 0 ට(ܽ− 4)ଶ + ቀ− ଷ ସܽ+ 8 − 5ቁ ଶ ට(ܽ− 4)ଶ + ቀ− ଷ ସܽ+ 8 − 5ቁ ଶ = 13 −ܽ

9 ܽଶ + 24ܽ− 256 = 0 (ܽ− 8)(ܽ+ 32) = 0 ܽ= 8,ܽ= −32 (rej) Radius of ܥ= 13 − 8 = 5 units Centre of ܥ= ൬8, − 3 4 (8) + 8൰ = (8, 2) Equation of ܥ:( ݔ− 8)ଶ + (ݕ− 2)ଶ = 25 ܽ= 8 = ඥ(10 − 8)ଶ + (8 − 2)ଶ = 6.3245 , . v݌< 3,݌> 13

10 7 ܯ= ܯ଴݁ି௞௧ Sub ݐ= 0,ܯ= 150, 150 =ܯ଴ Sub ݐ= 30,ܯ= 90, 90 = 150݁ି௞(ଷ଴) ݁ି௞(ଷ଴) = 0.6 ݇= − ୪୬ ଴.଺ ଷ଴ = 0.017027 Sub ݐ= 10