NCHS Amath P1 Answer Sheet

NCHS 2023 AM Prelim P1 solutions for students Q1 (a) Area of triangle = ଵ ଶ ቀ ଵ ଶା√଺ቁ ଶ sin 60° = ଵ ଶ ቀ ଵ ସା଺ାସ√଺ቁ ቀ√ଷ ଶ ቁ = ቀ ଵ ଵ଴ାସ√଺ቁ ቀ√ଷ ସ ቁ = ቀ ଵ ଶቁ ቀ ଵ ହାଶ√଺ቁ ቀ√ଷ ସ ቁ =ቀ ଵ ହାଶ√଺ቁ ቀ√ଷ ଼ ቁ = ቀ√ଷ ଼ ቁ ቀ ଵ ହାଶ√଺ቁ ቀ ൫ହିଶ√଺൯ ହିଶ√଺ ቁ =ቀ√ଷ ଼ ቁ ቀ ହିଶ√଺ ଶହିଶସቁ =ቀ√ଷ ଼ ቁ ቀ ହିଶ√଺ ଵ ቁ = ହ√ଷ ଼ − √ଵ଼ ସ = ହ√ଷ ଼ − ଷ√ଶ ସ Q1(b) Mtd 1 ݐ݈݁ ݕ= √ݔ+ 3 ݕ= 1 ݕ− 5 6 ݕଶ = 1 − 5 6ݕ ݕଶ − 1 + 5 6ݕ= 0 6ݕଶ + 5ݕ− 6 = 0 (3ݕ− 2)(2ݕ+ 3) = 0 ݕ= 2 3 ݎ݋ ݕ= − 3 2 √ݔ+ 3 = 2 3 ݎ݋ √ݔ+ 3 = − 3 2 (݆݁ݎ) ݔ+ 3 = 4 9 ݔ= − 23 9 Or 6(ݔ+ 3) = 6 − 5√ݔ+ 3 6ݔ+ 18 − 6 = −5√ݔ+ 3 (6ݔ+ 12)ଶ = ൫−5√ݔ+ 3൯ ଶ 36ݔଶ + 144ݔ+ 144 = 25(ݔ+ 3) 36ݔଶ + 144ݔ+ 144 = 25ݔ+ 75 36ݔଶ + 119ݔ+ 69 = 0 (4ݔ+ 3)(9ݔ+ 23) = 0 ݔ= − 3 4 (݆݁ݎ)ݎ݋− 23 9 Q2(a) ݀ ݔ݀( ݔହ݈݊ݔଶ) =ݔହ ൬ 1 ݔଶ൰ (2ݔ)+ 5ݔ^4݈݊ݔଶ = 2ݔସ + 5ݔସ݈݊ݔଶ = 2ݔସ + 10ݔସݔ݈݊ 2(b) Method 1 නݔସݔ݈݊ ݔ݀= 1 10 න 10ݔସݔ݈݊+ 2ݔସ − 2ݔସݔ݀ =1 10 ቈݔହ݈݊ݔଶ − 2ݔହ 5 ቉ +ܿ =1 5ݔହݔ݈݊− 1 25ݔହ +ܿ 2(b) Method 2 න 10ݔସݔ݈݊+ 2ݔସݔ݀= ݔହ݈݊ݔଶ +ܿ 10 නݔସݔ݀ݔ݈݊+ න 2ݔସ݀ݔ= ݔହ݈݊ݔଶ +ܿ 10 නݔସݔ݀ݔ݈݊= ݔହ݈݊ݔଶ +ܿ− 2ݔହ 5 +ܿ1 =ݔହ݈݊ݔଶ − 2ݔହ 5 +ܿ2 නݔସݔ݀ݔ݈݊= (ݔହ݈݊ݔଶ) 10 −ݔହ 25 +ܿ3 ݎ݋ (ݔହݔ݈݊) 5 −ݔହ 25 +ܿ3

Q3(a) −9 − 2ݔଶ − 4ݔ =−2ݔଶ − 4ݔ− 9 = −2(ݔଶ + 2ݔ)− 9 = −2[(ݔ+ 1)ଶ − 1] − 9 = −2(ݔ+ 1)ଶ − 7 ݔܽ݉ ݁ݑ݈ܽݒ= −7 Q3(b) Hence, −9 − 2ݔଶ − 4ݔ= −2 −݇ ݁ܿ݊݅ݏmax = −7 −2 −݇< −7 ݇> 5 Otherwise, −2ݔଶ − 4ݔ+ ݇− 7 = 0 ݐ݊ܽ݊݅݉݅ݎܿݏ݅݀> 0 (−4)ଶ − 4(−2)(݇− 7) > 0 16 + 8݇− 56 > 0 8݇− 40 > 0 ݇> 5 Q4 (ݔ+ 1)(ݔଶ − 1) =ݔଷ 3 ݔଷ +ݔଶ −ݔ− 1ඥ3ݔଷ + 0ݔଶ + 0ݔ+ 2 −(3ݔଷ + 3ݔଶ − 3ݔ− 3) −3ݔଶ + 3ݔ+ 5 3ݔଷ + 2 (ݔ+ 1)(ݔଶ − 1) = 3 + 3ݔ+ 5 − 3ݔଶ (ݔ+ 1)ଶ(ݔ− 1) 3ݔ+ 5 − 3ݔଶ (ݔ+ 1)ଶ(ݔ− 1) =ܣ ݔ− 1 +ܤ ݔ+ 1 +ܥ (ݔ+ 1)ଶ 3ݔ+ 5 − 3ݔଶ =ܣ(ݔ+ 1)ଶ +ܤ(ݔ− 1)(ݔ+ 1) +ܿ(ݔ− 1) ܾݑݏ ݔ= 1, 3 + 5 − 3 =ܣ(2)ଶ 4ܣ= 5 ܣ= 5 4 ܾݑݏ ݔ= −1, −3 + 5 − 3 =ܥ(−2) −1 = −2ܥ ܥ= 1 2 ݃݊݅ݎܽ݌݉݋ܿ ݐ݂݂݊݁݅ܿ݅݁݋ܿ ݂݋ ݔଶ : − 3 =ܣ+ ܤ −3 = 5 4 +ܤ ܤ= − 17 4 ∴ 3 + 5 4(ݔ− 1) − 17 4(ݔ+ 1) + 1 2(ݔ+ 1)ଶ

Q5(a) log଻ݔ+ log଻ݔଶ log଻ 7ଶ = 6 log଻ݔ+ 2 log଻ݔ 2 = 6 log଻ݔ+ log ଻ݔ= 6 2 log ଻ݔ= 6 log଻ݔ= 3 ݔ= 7ଷ = 343 Q5(b) ݔଶ + 8ݔ+ 15 > 0 ݀݊ܽ ݔ+ 4 > 0 (ݔ+ 5)(ݔ+ 3) > 0 ݀݊ܽ ݔ> −4 ݔ> −3,ݔ< −5 ݀݊ܽ ݔ> −4 ∴ݔ> −3 Q6(a) sin 2ߠ= 2 sinߠcosߠ =2ܿ(−√1 −ܿଶ) =−2ܿ√1 −ܿଶ Q6(b) cos(ߠ+ 30°) = cosߠcos 30° − sinߠsin 30° = ቀ−ඥ1 −ܿଶቁ ቆ√3 2 ቇ − (ܿ)൬1 2൰ = −1 2ܿ− √3 − 3ܿଶ 2 Q6(c) −(180° −ߠ) or ߠ− 180° Q7(a) 4 sinߠ+ 4 sinଶߠ secߠ+ tanߠ =4 sinߠ(1 + sinߠ)÷ ൬ 1 cosߠ+ sinߠ cosߠ൰ = 4 sinߠ(1 + sinߠ)÷ ൬1 + sinߠ cosߠ൰ = 4 sinߠ(1 + sinߠ)× cosߠ 1 + sinߠ =4 sinߠcosߠ =2(2 sinߠcosߠ) =2 sin 2ߠ Q7(b) 2 sin 2ߠ= 0.7 sin 2ߠ= 7 20 −ߨ≤ ߠ≤ߨ −2ߨ≤ 2 ߠ≤ 2ߨ ߙ= sinିଵ 7 20 = 0.35757 2ߠ ݏ݅ ݊݅ ݐℎ݁ 1ݐݏ ݀݊ܽ 2݀݊ ݐ݊ܽݎ݀ܽݑݍ. 2ߠ= 0.35757,ߨ − 0.35757, −(ߨ+ 0.35757), −(2ߨ − 0.35757) ߠ= 0.179, 1.39, −1.75, −2.96

Q8(a) ݐ݊݁݅݀ܽݎ݃ ݂݋ ܥܦ= tan 45° = 1 ݐ݊݁݅݀ܽݎ݃ ݂݋ ܥܤ= −1 ݕ− 9 ݔ− 1 = −1 ݕ− 9 = −ݔ+ 1 Equation of BC: ݕ= −ݔ+ 10 ݕ+ 18 ݔ− 2 = 1 ݕ+ 18 =ݔ− 2 Equation of AB: ݕ= ݔ− 20 −ݔ+ 10 =ݔ− 20 2ݔ= 30 ݔ= 15 ݕ=