SOTA 2022 Year 6 MAA HL Prelim Paper 1 Solutions

2022 Prelim MAA HL Paper 1 Mark Scheme Section A No Solutions Mark Scheme 1 When , , 2(a) Let the probability of getting an odd number and even number be x and y respectively. Probability of getting a 1 is . 2(b) Required probability 3(a) 3(b) ( )sin 2 d cos 2 2 yx x x c = éù=- +êúëû ò 10, 2xy== 11 22 c=- + 1c=cos 2 12 xy=- + 2 3cos 2yx=- + 2 1sin 2yx=+ 331xy+= 2yx= 91x=1 9x= 192 12 3 99 æö=´ ´ç÷èø 2 243= 3() 32 axfx x -= + 2 3x=- 23 a= 6a=

3(c) 4(a) 4(b) 4(c) Equation of tangent: or 4() 2fx x x=- 2 4() 2fx x ¢ =+ (2) 2 1 3f¢ =+ = ( )() ()hx g f x= ( )(2) (2)hg f= ()(2) 2hg= (2) 1h = ( )() ()hx g f x= ( )() () ()hx g fx f x¢¢ ¢=´ ( )(2) (2) (2)hg f f¢¢ ¢=´ ()(2) 2 (2)hg f¢¢¢=´ (2) 4 3h¢ =´ (2) 12h¢ = 11 2 ( 2 )yx-= - 12 23yx=-

5 Method 1 Sub in Method 2 6 Solve for a and b \ 1 13 8 15 0 7 28 4 x pp p p =- -- + - = = = 15 15 4 60 ab pab abp =- =- ´ =- 2 322 2 3 2 (1 ) ( ) 13 8 0 15 3, 5, 4 x px apx bpx ab px apx bpx abx px apx bpx ab xp x ap bp p x ab ap bp p x ab ab p ++ + + +++ + + + + ® ®++ = - ®++= - ®= - == -= 35 4 6 0abp=´ - ´= - 22 22 22 25 27 25 2 va b ab ab =++ =++ =+ .1 7 52 0 7 10 uv a b ab =´ + ´ -+ ´ - =- - 22 71 0 2 ab ab =+ += 7 5 1 5 b a =- =

7. Finding Roots leads to no value for m 8.a) 8.b) Using l’hopital Sub in 0 and show indeterminate Differentiate numerator and denominator. Product rule Show indeterminate 2nd time Differentiate Simplify and eliminate 𝑒! 2 2 4 40 4 4 24 43 2 22 3 xxm m k k m k m ab ab ++ = = += - = =- =± =± 222 1 mk m -± -= 1 (1 ) x x ex xe -- - 0 10lim(1 ) 0 x xx ex xe® -- =- 0 10lim 10 x xxx e xe e® - =+- 0 11lim 22 x xxxx e xe e e x® ==++ +

9. Finding integrating factor Integrating the RHS using parts 2 ln5dy y x dx x x+= 5 5ln 5xxee xò== 53 lnyx x xdx=ò 43 3 44 lnln 44 ln ()41 6 xx xxx d x d x xx x c =- =- + òò 44 5 5 ln ()41 6 ln 1 41 6 xx xyx c xCy xx x =- + =-+

Section B No Solutions Mark Scheme 10(a) (i) Method 1: Method 2: 10(a) (ii) 10(b) (i) 10(b) (ii) () l n ( )fx a x= 1 e(1)2f- = e 12fæö=ç÷èø eln 1 2aæö´=ç÷èø 2a=1 1 e(1)2 ee 2 2 f a a - = = = () l n ( 2)fx x= ln(2 )yx= 1e2 yx= 1 1() e2 xfx- = ln(54 ), ln( ), ln( ), ln(2 ) xp x q x x 3l n ( 2 ) l n ( 5 4 )dx x=- 23l n 54 xd x æö=ç÷èø 13l n 27d= 33 l n 3d=-ln 3d=-1ln3d æö=ç÷èø 1ln( ) ln(54 ) ln 3 1ln ln54 3 543 18 px x px x p p -= æö =ç÷èø = =

10(b) (iii) Method 1: Method 2: 10(b) (iv) 1ln( ) ln(54 ) 2 ln 3 1ln ln54 9 1549 6 qx x qx x q q -= æö =ç÷èø =´ = 5 512l n ( 5 4 ) 4l n23Sx éù=+êúëû 5 15l n ( 5 4) 2 l n3Sx éù=+êúëû 5 2 545l n3 xS æö= ç÷èø ()5 5ln 6Sx= 5 51ln(54 ) (ln 2 ln )23Sx xéù=+ +êúëû 5 5 54 2ln23 xxS ´éù=êúëû 2 5 5ln(36 )2Sx éù=ëû [ ]5 52l n (6 )2Sx= ()5 5ln 6Sx= () () 5 5 113 53 113 5335 l n 6 2 1ln 6 2 S S x x ¥= - = - ´´= = 1 26ex= 1 2e 6x=

No Solutions Mark Sche