SJI 2024 Year 6 HL MAA Prelim Examination Paper 1 (MS)

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 1 of 11 Year 6 MAA HL Preliminary Examination 2024 Paper 1 (Mark Scheme) Qn Suggested Solutions Marks 1 SL 2.5 Composite Function SL 2.2 Informal Concept of Inverse Function [Maximum mark: 5] (a) (b) ( )( ) ( )( ) ( )( ) 2 2 118 5 18 5 3 xg f x g f x f x += = − = −  ( )( ) ( ) ( ) 2 22 1 5 2 2 1 5g f x x x x= + − = + + − ( )( ) 22 4 3g f x x x = + − Method 1 Using domain of f –1 equals range of f ( ) ( ) 1 2g f a − = is equivalent to ( )( )2a g f= ( ) ( ) 2 2 2 4 2 3 8 8 3 13a= + − = + − = Method 2 Finding the inverse of g ◦ f ( ) 222 4 3 2 1 5y x x x= + − = + − ( ) ( ) 1 5 1, 52 xg f x x − += − − ( ) ( ) 1 5 122 ag f a − += − = ( ) 2 2 2 1 5 18 5 13a= + − = − = Method 3 Finding f –1 and g –1 to find the inverse of g ◦ f ( ) ( ) 11 53 1, 0 and , 5 18 xf x x x g x x−− += −  = − ( ) ( ) ( )( ) 1 11 553 1 1, 518 2 xxg f x f g x x − −− ++= = − = − − ( ) ( ) 1 5 122 ag f a − += − = ( ) 2 2 2 1 5 18 5 13a= + − = − = M1 A1 AG M1 M1 A1 M1 M1 A1 M1 M1 A1

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 2 of 11 2 SL 1.9 The Binomial Theorem AHL 1.10 Extension of Binomial Theorem to Fractional and Negative Indices [Maximum mark: 5] (a) (b) ( ) 6 2 3 61 1 6 15 20 ...x x x x x− = − + − + + ( ) ( ) ( ) ( ) 1 2 23 23 1 14 14 1 3 1 3 5 1 2 2 2 2 21 4 4 4 ...2 2 3! 1 2 6 20 ... x x x x x x x x − =+ +       − − − − −             = + − + + + = − + − + ( ) ( )( ) 6 2 3 6 2 3 2 3 2 3 2 3 3 23 1 1 6 15 20 ... 1 2 6 20 ... 14 1 6 15 20 2 12 30 6 36 20 ... 1 8 33 106 ... x x x x x x x x x x x x x x x x x x x x x − = − + − + + − + − + + = − + − − + − + − − + = − + − + 1 1 14 1 or or 4 4 4x x x  −   A1 for the first 4 terms A1 for the first 4 terms M1 A1 A1 3 SL 4.5 Complementary Events SL 4.6 Use of Venn Diagram, Combined Events and Conditional Probability [Maximum mark: 6] Method 1 ( ) ( )P 1 P 1 0.4 0.6BB = − = − = ( ) ( )P 1 P 1 0.1 0.9A B A B  = −  = − = ( ) ( ) ( ) ( )P P P P 0.5 0.6 0.9 0.2A B A B A B = + −  = + − = Method 2 ( ) ( ) ( )P P P 0.4 0.1 0.3A B B A B    = −  = − = ( ) ( ) ( )P P P 0.5 0.1 0.4B A A A B    = −  = − = ( ) ( ) ( ) ( )P 1 P P PA B A B A B B A    = −  −  −  ( )P 1 0.1 0.3 0.4 0.2AB = − − − = or ( ) ( ) ( )P P P 0.5 0.3 0.2A B A A B  = −  = − = ( ) ( ) ( ) P 0.2 1P P 0.6 3 ABAB B = = = (M1) A1 for either ( )P B or ( )P AB (M1) A1 (M1) A1 for either ( )P AB  or ( )P BA  (M1) A1 M1 A1

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 3 of 11 4 SL 1.7 Laws of Logarithms SL 3.5 Exact Values of Trigonometric Ratios SL 3.6 Double Angle Identities for Sine and Cosine SL 3.8 Solving Trigonometric Equations in a Finite Interval [Maximum mark: 7] (a) (b) ( ) ( ) ( ) 2 42 2 log 1 sin 2 1log 1 sin 2 log 1 sin 2 log 4 2 xxx ++ = = + ( )42log 1 sin 2 log 1 sin 2xx+ = + ( ) ( )42log 1 sin 2 log 2 cosxx+= ( ) ( )22log 1 sin 2 log 2 cosxx+= 1 sin 2 2 cosxx+= 21 sin 2 2cosxx+= 2sin 2 2cos 1xx=− sin 2 cos 2xx= tan 2 1x= Basic Angle arctan1 4 == 2 4x = 8x = M1 A1 AG M1 for using part (a) M1 for double angle formula for cosine A1 for tan 2x = 1 A1 for basic angle A1

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 4 of 11 5 AHL 1.15 Proof by Mathematical Induction involving Inequalities and Factorial [Maximum mark: 7] Let P(n) be the statement that ( ) ( ) 2 2 ! 2 ! nnn  for n . When n = 0 , LHS = 0! = 1 , RHS = 20 (0!)2 = 1(1) = 1 Thus, LHS ≥ RHS. Therefore, P(0) is true. Assume that P(k) is true for some value of k for k , i.e. ( ) ( ) 2 2 ! 2 ! kkk  for k . When n = k +1 , ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 2 2 2 21 2 1 ! 2 2 ! 2 2 2 1 2 ! 2 2 2 1 2 ! 2 2 1 2 ! 2 1 1 2 ! 2 2 1 ! 1 ! 2 1 ! k k k k k k k k k k k k k k k k k k k k k k k k+ + = + = + +   + +  + + = + + = + + =+ Therefore, P(k +1) is true. Since P(0) is true and P(k) is true implies that P(k+1) is true for some k , by mathematical induction, ( ) ( ) 2 2 ! 2 ! nnn  for n . A1 M1 A1 M1 for inductive step A1 for 2k+1 ≥ k+1 or 2k2 + 3k + 1 ≥ 2k2 + 2k + 1 or 4k2 + 6k + 2 ≥ 2k2 + 4k + 2 A1 A1 awarded only if at least 5 of preceding marks are obtained 6 SL 5.5 Integration with Boundary Condition SL 5.10 Integration by Inspection and Integral of 1/x [Maximum mark: 8] ( ) ( ) 22 21 42dd2 2 1 2 2 2 1 ax axy f x x x x x x x − −= = = − + − + ( ) 2ln 2 2 12 ay f x x x C= = − + + When x = 0 or x = 1 , y = ln 2 gives C = ln 2 When x = 0.5 , y = 0 gives a = 2 ( ) ( ) 22ln 2 2 1 ln 2 or ln 2 2 1 ln 2f x x x x x= − + + − + + or ( ) ( ) 22ln 4 4 2 or ln 4 4 2 or equivalentf x x x x x= − + − + Note: Award maximum of M1 A1 A1 M1 A1 M0 A0 A0 to candidates who assume that a = 2 M1 A1 for ln with modulus A1 for + C M1 A1 M1 A1 A1

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 5 of 11 7 AHL 4.14 Mean and Median of Continuous Random Variable AHL 5.16 Integration by Parts [Maximum mark: 8] Let m be the median of X. ( ) 1P0 2Xm  = gives 0 1sin d , 0 36 6 2 m x x m  =   0 1cos 62 m x −=   11 cos 62 m−=  1cos , 0 362 m m =   1arccos6 2 3 m ==  m = 2 Mean of X = ( ) ( ) 33 00 E d sin d 66X x f x x x x x  ==  ( ) 3 3 00 E cos cos d 66X x x x x     = − +          ( ) 3 0 66E sin 6Xx    ==   Since 3  , therefore 66 3  , i.e. 6 2  Hence the mean of X is less than the median of X. M1 with correct limits A1 A1 M1 for E(X) with correct limits M1 A1 for integrating by parts A1 R1 for stating π > 3 AG 8 SL 2.2 Inverse Function SL 5.11 Area Between Curves [Maximum mark: 9] (a) Method 1 Using f (x) = f –1 (x) = x Solving ( ) ( ) 1f x f x −= is equivalent to solving ( )f x x = 21 3 , 24 x x x x− + =  2 4 12 4x x x− + = 2 8 12 0xx−+= ( )( )2 6 0xx− − = x = 2 or x = 6 M1 M1 for solving quadratic equation A1 A1

Year 6 MAA HL Prelim Exam 2024/P1 Mark Scheme Page 6 of 11 (b) Method 2 Finding f –1 ( ) ( ) 2211 3 2 244y f x x x x= = − + = − + ( ) 1 2 2 2 , 2f x x x− = + −  Solving ( ) ( ) 1f x f x −= ( ) 21 2 2 2 2 2 , 24 x x x− + = + −  ( ) ( ) 12 2 1 2 2 24 xx− = − ( ) ( ) 12 22 8 2xx− = − ( ) ( ) 13 222 2 8 0xx − − − = Since x ≥ 2 , x = 2 or x = 6 Method 1 ( )( ) ( ) 6 2 6 2 2 6 2 2 6 23 2 2 Area 2 d or equivalent 12 3 d 4 12 3 2 d 4 123 12 22 18 36 18 6 4 3 1 165 or units33 x f x x x x x x x x x x x x =−  = − − +   = − + − = − + −  = − + − − − + −   =    Method 2 ( ) ( )( ) ( ) ( ) 6 1 2 6 2 2 6 1 22 2 63 232 2 Area d or equivalent 12 2 2 3 d 4 12 2 1 d 4 4 1 1 23 2 12 f x f x x x x x x x x x x x x x x −=−  = + − − − +   = − − + − = − − + −    2 32 2Area 6 18 18 2 233 1 165 or units33    = − + − − − + −       = M1 for finding f –1 M1 for solving equation A1 A1 M1 for twice area or integral A1 FT for integral with correct limits A1 for correct integration M1 for substitution of limits A1 FT from their limits M1 for difference of functions A1 FT for integral with correct limits A1 for correct integration M1 for substitution of limits A1 FT from th