ACSI 2020 Promo Paper 1 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 1 FINAL EXAMINATION 2020 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL Paper 1 SOLUTIONS SECTION A Solution 1 The lengths of two sides of a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The area of the triangle is 5 15 2 2cm . (a) Show that 15sin 4θ = (b) Find the two possible values of the length of the third side. Maximum [7 marks] 1 5 154 5 sin22 θ××× = 15sin 4θ = Well done Let x be the length of the third side. ( )( )2 22452 4 5 c o sx θ=+− Most did not realise 2 values for cosine

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 2 ( ) 224 15 1cos 44θ ±− = =± ( )( ) 12 22452 4 5 4x =+− ±  2 31 or 51x = 31 or 51 x= 2 9, m, n are consecutive terms of a geometric progression and 9, 2m and 3n are consecutive terms of an arithmetic progression. Given that mn≠ , find the value of m and of n. Maximum [6 marks] GP: 9, m, n 3mn⇒= ------------ (1) AP: 9, 2m, 3n 932 2 nm +⇒= --------(2) Subst (1) into (2) gives 12 9 3nn= + 2144 81 54 9n nn= ++ 2 10 9 0nn − += ( )( )9 10nn− −= 9, 1n= 3 9 9 (NA) , 3m= = Hence 3, 1mn= = Well done but some introduce ‘r’ and ‘d’, getting lost in the manipulation 3 Given that , find the value of x . Maximum [6 marks] ( ) ( ) ( ) ( )3 log 4 log 5 log ... 22 log 170222 2xxx x++ ++ ++ + += ( )3 4 5 ... 22 20log 170 2 x++++ + = Some got number of terms wrong �[𝑘𝑘 + 𝑙𝑙𝑙𝑙𝑙𝑙2(𝑥𝑥)] 22 𝑘𝑘=3 = 170, 𝑥𝑥

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 3 ( )20 3 22 20log 17022 x++ = 20log 802 x=− log 42 x=− 142 16x −= = while others did not realise 𝑙𝑙𝑙𝑙𝑙𝑙2𝑥𝑥 is repeated 20 times. 4 , and αβ γ are roots of the equation 32 7 20 50 0zz z+ + += . Without solving for , and αβ γ find the polynomial of degree 3 whose roots are 1, 1 and 1αβ γ−− − . Maximum [6marks] 32 7 20 50 0zz z+ + += ( )( )( )1 1 10yyyαβγ− +− +− + = ( )( )( )1 1 10yyy αβγ+− +− +− = Substitute 1zy= + into 32 7 20 50zz z+++ gives ( ) ( ) ( )321 7 1 20 1 50yy y+ + + + ++ 32 23 3 1 7 14 7 20 20 50yyy y y y= + + ++ + ++ + + 32 10 37 78yy y= + ++ Mostly well done except some used 1zy= − . Many calculated the coeff one by one. 5 The following diagram shows the graph of ()y fx= with x-intercepts at 2− and 0, minimum point (-1, -2) and asymptote 2y= . Maximum [7 marks]

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 4 (a) Sketch the graph of ( ) 1 fx in the same diagr