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 Equa