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= −15 + 10 = −5 ∴ܤ(15, −5)(ݏℎ݊ݓ݋) Q8(b) ݐ݌݀݅݉ ܥܣ= ݐ݌݀݅݉ ܤܦ ൬2 + 1 2 , 9 − 18 2 ൰ = ൬ݔ+ 15 2 , −5 +ݕ 2 ൰ 3 =ݔ+ 15 ݔ= −12 9 − 18 = −5 +ݕ ݕ= −4 ܦ(−12, −4) ܤ ݏ݅ ݐℎ݁ ݐ݌݀݅݉ ݂݋ ܧܦ (15, −5) = ൬−12 +ݔ 2 , −4 +ݕ 2 ൰ 30 = −12 +ݔ ݔ= 42 −10 = −4 +ݕ ݕ= −6 ܧ(42, −6) ܽ݁ݎܣ= 1 2 ቚ 2 42 −18 −6 1 2 9 −18ቚ = 1 2 (−12 + 378 − 18 − 18 + 6 + 758) = 546 ݐ݅݊ݑݏଶ Q9 (a) (graph) ݒ= ݔ(݌ݔଶ +ݔݍ) ݒ= ݌ݔଷ +ݍݔଶ ݒ ݔଶ =ݔ݌+ ݍ ݀݁ݐݐ݋݈݌ ݏݐ݊݅݋݌ ݊݅݋݆ ݐ݅ݓℎ ܽ ݃݅ܽݎݐݏℎݐ ݈݁݊݅ ݐ݊݁݅݀ܽݎ݃= ݌ =10 − 1 3 − 0 = 3 ݍ= 1

Q9(b) ܸ ݔଶ = 3ݔ+ 1 ܸ= ݔ(3ݔଶ +ݔ) ݏݑ݅݀ܽݎ= 2ݔ ݁ݏܾܽ ܽ݁ݎܽ= ߨ(2ݔ)ଶ = 4ߨݔଶ 4ߨݔଶ = 3ݔଶ +ݔ (4ߨ− 3)ݔଶ −ݔ= 0 ݔ[(4ߨ− 3)ݔ− 1] = 0 ݔ= 1 4ߨ− 3 = 0.1045 = 0.105 Q9(c) ܸ= ݔ(4ߨݔଶ) = 4ߨݔଷ ݓܽݎ݀ ܸ ݔଶ = 4ݔߨ ܶℎ݁ ݁݉ݑ݈݋ݒ ݂݋ ݐℎ݁ ݎ݈݁݀݊݅ݕܿ ݏ݅ ݀݁ݏݏ݁ݎ݌ݔ݁ ݊݅ ݏ݉ݎ݁ݐ ݂݋ ݔ ݀݊ܽ ݐℎ݊݁ ݀݁݃݊ܽݎݎܽ݁ݎ ݋ݐ݊݅ ݐℎ݁ ݁݉ܽݏ ݎ݈ܽ݁݊݅ ݉ݎ݋݂ ݏܽ ݐℎ݁ ݊݋݅ݐܽݑݍ݁ ݂݋ ݐℎ݁ ݃݅ܽݎݐݏℎݐ ݈݁݊݅ ݊ݓܽݎ݀. ܶℎ݁ ݔ −݁ݐܽ݊݅݀ݎ݋݋ܿ ݂݋ ݐℎ݁ ݊݋݅ݐܿ݁ݏݎ݁ݐ݊݅ ݐℎ݁ݎ݋݂݁ݎ݁ ݏ݁ݒ݅݃ ݐℎ݁ ݁ݑ݈ܽݒ ݂݋ ݔ ݎ݋݂ ݓℎܿ݅ℎ ݐℎ݁ ݈݀݅݋ݏ ݏ݅ ܽ ݎ݈݁݀݊݅ݕܿ ݐ݅ݓℎ ℎ݃݅݁ℎݐ ℎ݂݈ܽ ݏݐ݅ ݁ݏܾܽ ݏݑ݅݀ܽݎ. ݉݋ݎܨ ݌ܽݎ݃ℎ,ݔ =0.1 ܶℎݏ݅ ݏ݅ ݁ݏ݋݈ܿ ݋ݐ ݐℎ݁ ݔ ݁ݑ݈ܽݒ ݀݁݊݅ܽݐܾ݋ ݊݅ ݐݎܽ݌ (ܾ). verify means to confirm if something is accurate ݔ 0 0.5 1 ܸ ݔଶ 0 6.3 12.6 Q10a ܽ= 2.5 cos ൬1 2ݐ൰ × 1 2 = 1.25 cos ൬1 2ݐ൰ ݎ݋5 4 cos ൬1 2ݐ൰ ݓℎ݊݁ ݐ= ߨ 2 , ܽ= 1.25 cos ൬ߨ 2 × 1 2൰ = 0.88388 = 0.884݉ ݏଶݎ݋ݎ݋ 5√2 8݉/ݏଶ Q10b ݓℎ݊݁ ݒ= ݋, 2.5 sin ൬1 2ݐ൰ = 0 sin ൬1 2ݐ൰ = 0 1 2ݐ= ߨ ݐ= 2ߨ ݏ= −2.5 cos ൬1 2ݐ൰ × 2 +ܿ =−5 cos ൬1 2ݐ൰ +ܿ ݓℎ݊݁ ݐ= 0,ݏ= 0, 0 = −5 cos(0) +ܿ ܿ= 5 ݏ= −5 cos ൬1 2ݐ൰ + 5 ݓℎ݊݁ ݐ= 2ߨ, ݏ= −5 cos(ߨ)+ 5 = −5(−1) + 5 = 10݉

Q10c Q10d ݈ܽݐ݋ݐ ݁ܿ݊ܽݐݏ݅݀ =20 × 4 + න 0.25ݐ− 4ߨ ଺଴ ଵ଺గ ݐ݀ =80 + ቈ0.25ݐଶ 2 − 4ݐߨ቉ ଵ଺గ ଺଴ = 80 + 1 8 (60)ଶ − 4ߨ(60) − 1 8 (16ߨ)ଶ + 4ߨ(16ߨ) =91.845݉ =91.8݉ Q11a ݎ݋݂ ݕ ݃݊݅ݏܽ݁ݎܿ݊݅,ݕ݀ ݔ݀> 0 ݕ݀ ݔ݀= 3ݔଶ݁ିଵ ଶ௫ ൬− 1 2൰ + 6ݔ݁ିଵ ଶ௫ =݁ିଵ ଶ௫ ቆ− 3ݔଶ 2 + 6ݔቇ ݁ܿ݊݅ܵ ݁ିଵ ଶ௫ > 0 ݎ݋݂ ݔ∈ ℝ, − 3ݔଶ 2 + 6ݔ> 0 ݔ൬− 3 2ݔ+ 6൰ > 0 0 <ݔ< 4 Q11b 1st derivative test ݓℎ݊݁ ݔ= 0,ݕ= 3(0)ଶ݁ି ଵ ଶ(଴) = 0 ݔ 0ିଵ 0 0ା ݕ݀ ݔ݀ 0 0 >0 Sketch Second derivative test ܹℎ݊݁ ݔ= 0,ݕ݀ ݔ݀= 0. ∴ݔ= 0 ݏ݅ ܽ ݕݎܽ݊݋݅ݐܽݐݏ ݐ݌. ݀ଶݕ ݀ݔଶ =݁ିଵ ଶ௫(3ݔ+ 6) − 1 2݁ିଵ ଶ௫ ൬− 3 2ݔଶ + 6ݔ൰ ݓℎ݊݁ ݔ= 0, ݀ଶݕ ݀ݔଶ = 6 > 0 ∴ minݐ݌ ݓℎ݊݁ ݔ= 0,ݕ= 0 ∴ (0,0) is minݐ݌ Q11c 0 <ݔ< 4 Q11d ݓℎ݊݁ ݔ= 1,ݕ݀ ݔ݀= ݁ିଵ ଶ ൬− 3 2 + 6൰ = 9 2݁ିଵ ଶ ݐ݊݁݅݀ܽݎ݃= −2݁ ଵ ଶ 9 ݓℎ݊݁ ݔ= 1,ݕ= 3݁ିଵ ଶ, ݕ− 3݁ିଵ ଶ = −2݁ ଵ ଶ 9 (ݔ− 1) ݕ= − 2݁ ଵ ଶ 9ݔ+ 2݁ ଵ ଶ 9 + 3݁ିଵ ଶ ݎ݋ ݕ= −0.366ݔ+ 2.19

Q12a Q12b Q12c ஽஺ ா஼ = ஽ி ாி (corresponding sides of similar triangles are proportional) ܨܦ×ܥܧ= ܣܦ×ܨܧ ݕܤ ݐ݌݀݅݉ ݐℎ݉݁ݎ݋݁,ܨܦ =1 2ܥܣ 1 2ܥܣ×ܥܧ= ܣܦ×ܨܧ