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 3θcot 6θ , for –π ≤ θ ≤ π. [3] ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 3 6/10 LHS=45140-145m20 Seco+tano =casino-145in201:-(¥0-1%-38) =(4sinO+45m20)÷(' ÷&) =45m04-15in01× ,-Y}㱺Ml =45m06s0 =2( 2sinocs.sc/--2sin2OM1--RHS - l≤o≤i - ☆9/10 - ◦≤so≤6 1-(45in30-145in230 4 sec30-1tango/=cstbo ¥(25in601=cstGO *sin60=6+60 {sinoo=#sin60 sin260=2cssGO 1-css260=200560 D=Css260+26560- I MI Css6@= -2£É2(1) =-218N2- =- IIF css60=-1+V2Or cos60=-1-52 lrejectas-i≤cssGO60=65'(rz- 1) ✗=1.14rad(3s.f.)MI 60belongsinthe1st&4thquadrant- 60=1.14,LTL- I.14, - I.14, -2kt1.14, =1.14,5.14,- I.14,- 5.14 0=0.191,0.857,-0.191,-0.857 (3s. f.)¢At

Show that — x ln x = ax ln x + bx , where x > 0. and a and b are constants to be found.63(i) ………………………………..…………………………………………………………………………………………….O Level Additional Mathematics 4047/1 [1]ddx 53 5 (ii)Only by expressing ∫ ax ln x + bx dx = x ln x + c , where c is an arbitrary constant,53 5 36 find ∫ x ln x dx.5 3 [3] 1 1 4 3110 d DIKGInK'= &,3×6Inx =3%,xGlnx =3[KG(¥)+(Lux)(bks)] =18×5Cnx+3×5 I6k5thK}-1325 a=At 2-⊥othermethods= of6×5Ln✗3+3×5dk=KGInK}+C, ☆9110 f)(5inx}+§✗5DX= f-✗°In✗3tCz f)(5in×}+-12×5DX= f-✗°In✗3tCz 5×5inx}+-12×5_ 1-2×5dx=§KGInk}_ 5£25dk-1Cz =&KGIn✗3_ MI %)+<3+<2 = 1-6×6(nks- FyKG+(4

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 (iii) [2]Integrate ––––– – ––––– + ––––– with respect to x.252 – 3xx – 1x + 1 2(x + 3) O x y y = x ln x 5 3 ––e2 Diagram 4 (iii)Look at Diagram 4 below. The shaded area can be expressed in two different ways.One way in expressing the shaded area is first making x the subject, then expressing it as a bound integral below: ∫m n x dy, where m and n are constants. Find the values of m and n, and the shaded area below, all in 3 sig fig .[3]☐ 2- Suby=0: KJLn✗3=o k=0or ×=I 5%5inidk •㱻.. ÷,,]: , AI E.innEi- EiEi)- ⾨a)°inis_ izcii) =(¥4(3)G-in2)- a%g) - SubK=§: (÷- E) y:({5-inlez]} =¥ga- in2)- ¥g- ¥ asconstantsg-me- maybe" :{•wet a, =e°C' ¥2_ ¥)- ¥unity m= }ze5(1-Ln2)=4.27(3s.- f.I n=0 133-2(1-in2))(ez)- [°C' 2_ 㱺)- ¥] ,"" ""£"^" 7/10 [¥312- ÷,, + die ✗+I - k+÷=2J,,÷spdx- 5xd"+ fl- ¥,die = 251kt3)"DX+ §fxdx+fIdie- 25¥,die = 2[{ ]+§In(2-3>c)+x- 2in(x-11)+C MI =- .EE?iii-iEE+c

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 It is given that f’’(x) = 10 sin 3x – 6 tan 6x – 11. Given that the normal of the line at x = 0 is perpendicular to the x-axis, find and simplify an expression for f’(x). 2 2 (i) [3] (ii) ddx 2 [1] 5 Find an expression for –– ln (2– 2 sin nx), where n is a constant. f''(x)=tosin23×-6tan2Gx- 11 8110 =-5C-25m23>c)- 6(sector- l)- 11 =-5(1-2514231)+5- 6sec26ktG-11 =-50s6K+5-6sec26×+6-11 =-56567C-6sec26KMIfor5hmpuffingf"(x) fix)=f-56567C-6sec26Kdx =-5Ctsin6k)- Gctgtanbk)-1C, =- %sinbk-tanbk-ci.mg At1=0,F'(x)=0: - %Sino- tan0+4=0 4=0 f/(x)=- %sinGK- tan6KAt 5110 Ln(2-25h2nk) OR = d d-x.lu/2css2nk)=2(wsnk(-nsinnxlg--adILn2tlncss2nKcos-noc = &,Zlncssme =2(- nsink)cosnk =_ 2 < /- 2htannk =- 2ntannX- a-

………………………………..……………………………………………………………………………………………. O Level Additional Mathematics 4047/1 (iii)Hence, given that y = f(x) intersects the point ( — , — ), find an expression for f(x).π312 [3] _☆9/'° f/(x)=- Egg;nGx- EanGx f-2htannxd)⾨In(2-25in2n>c) - 2n/tannkdk=In(2-Zsinlnk) f-(K)=f-{SinGa- tan6KDX Jeanniedk=- InLn(2-Zsinznsc) = _ §C-toss6K)-1Cz+¥f-12tan6Kdie = I3.GUSGX+Cz1-Faln(2-25in26×1 fck)= %cosGx-1,1-2In(2-25in26k)-1cgMI subk=¥,fcxs= 1-2i {= Css217+¥2bn(2-25in2212)→(3 {= +¥In(2)+<3 ⊥= %- ¥22mi FCK)= %cosGx-1,1-2In(2-25m26>c)+%- ¥22µAt