Answer for the three musketeers test amath

(ANSWERS)THREE MUSKETEERS EXAM 😈 Time: 2 hours 15 minutes Name: ……………………………. Marks: 90 Topics: the triple threat(trigonometry, differentiation, integration) good luck cus the paper isnt very lucky 90 ☆ =40m difficultyrating/10 (tougherquestions]

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………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 1(a) [2] (b) Answer all questions :) [1] Do not use a calculator for the whole of this question. It is given that cot (–B) < 0 and sec (–B) < 0, where 0 ≤ B ≤ 2π. Explain why sin — > 0.B2 Find the value of .–145cot [ 2 cos ( – —) ] (ii)Given that cot B = –– , find the exact value of cos ( ––––– ). 125 B – π2 [2] Sincetheprincipalvaluesofus- 'C-¥)are ◦◦ ≤css-'C-¥I≤180°,andcosless_ 'C-¥))<◦s css-'C-¥)liesinthe2ndquadrant. In est[css-'l- ¥-1]= tan(Zoss-' C-¥1)is 5 70s-'C-¥) =3 , = '- th"""- "- ¥" §flippeds 4 2tan(css-'C-¥)) Mi = I- C-¥12 21- ¥) = '_%_ -3-2 =7-✗(-2-3)16 =- ±7/10 24 Al - pusing◦◦≤≤180°is✗ as0°&180°donotmakesin >0 since- LotBLO, Secc- B)20 SincetanB>0andCssB<O, Bliesin +"° thezrd-qnadrant.TLB<270°)/(Te<B< )istB>◦ 1- <0cssC-B) Hence,Bzliesinthe2ndquadrant, ¥B>0 usC-B)<0 andinthe2ndquadrant,sineis tanB>0 CssB<0 positive. (90°<Bz<13501/CÉ<Bz<¥) Hence, 5inBz>0 cssc}- ¥) '' '"C-0)=cos0CstB= ¥ 2.Css(Iz- O)=sin0tanB= 5 -12 =cssc- (Ea- Ez)) n nosB=- ¥3 =sinBz 12 , css@()]=I- 25in2B-2at - ¥3=I- 25in2B-MI5 2 FE=13 ¥3=25in2 - I 25in> BI=¥3 s;n2Bz=¥ OR(sin2->0) sinBI= inBzbelongsinthe2ndquadrant)26 =5-→a, OR5€2626

………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 (c)By finding the exact value of cosec ( ––– ), find sec (–––).7π122 27π12 [3] 6/10 Used¥2= gg we24¥+I= used¥2 tan¥=8-453- i =[sin¥]2 I ta÷2=7-4V3 = (sin as¥+us¥sin¥)≥ tanz7¥= ¥fs✗¥¥%3I= =7+4☐ KEKE"CE"EI Sec-(¥2)=I+tan2C =1- =1+(7+453) (r¥E12 =8-3At = I 2-1+Fs+6 16 =I ✗¥¥f, m' 8-14nF = 128--645364- 48 = 128¥45= 8-4At

(a)Show, using the aid of a diagram, that ∫ 3√k – x dx = —– .k –k –––––22 3πk2 2 [2] (b)Hence, find in terms of k and/or π, the value of: (i) (ii) ∫ √4k – 4x –2π cos (–––) dx.0 k –––––––22 πx2k [2] ∫ √k – x + π tan (–––) dx.–k 0 –––––22 πx3k2 [2] (NCHS 2023 P2) ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 2 9.5/10 ☆ y=FEZ ¥32dx=31%5%2dx y2=K2- K2 =3×(areaÉe withradiusK) K2-1yd=K2 =3✗ ¥K2At⊥equationofa circleabovex-axis = 3T¥≥ "areaof semicircle (shown) O 'kmy-k k 1stPart✗→Capat1for l:)&Ciitonlyiftrigopartf- = f-2TicsSEEdiecorrectly→ MI =2C-¥414ksin × ]° J!2É- 2kCss(Idk2k K- =2J,?FÉdx - sTe s s1T¥)dx ' - 2C- ¥}- ksino)- 14k8in¥)]K =_ TLKZ=21- T¥)- (2¥)sin¥1]° I- '◦-4k) K =- ¥2k' -14k ATAT (w/steps1 fttan'l)diecorrectly→mi Jj"Fatdie+ftetanyIdk =(_ TLKZ 4-I- go.k-kkeiCI.sc)- I]die =_ T¥- [,, TeSeix- Kde =- " ¥- [3ktan 3k- KK]° - k =- " ¥- ¢3ktanO- O)- (3ktant¥1+KTC)]=- TLK' I1-(3kC-Fs)+KTL) =- IK? 4_,→ (w/steps)

(a)Prove that [2] (b)Hence, solve ——————— = sin 3θ + sin 3θ2 sec 3θ + tan 3θco