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 diagram above indicating the x-intercepts, turning point and asymptote(s) clearly. (b) If the domain of another function ( )gx is [ ]0,8 and its range is [ ]4,11 , write down the domain and range of ( ) 1 gx whenever possible. 2 vertical asymptotes x = -2, x = 0[1 mark] 1 horizontal asymptote y = 0.5 [1 mark] Local max point (-1, -0.5) and shape [2 mark] Shape of 2 branches [1 mark] Fairly well done except for some missing out on the horizontal asymptote and the positions of the branches were inaccurate Domain [0,8] Range 11,11 4   Many gave range as 11,4 11   6 The box-and-whisker diagram below shows the amount of salt used in different types of snacks in 2 confectionery shops. Shop 1 has 240 snacks and Shop 2 has 320 snacks. Maximum [5 marks]

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 5 (i) State what the number 10 represents in Shop 1. (ii) Write down the approximate number of snacks whose amount of salt ranges from 2 to 20 grams in Shop 1. (iii) A health inspector visited the two shops to check the snacks in the shop. Snacks sold in the shop are required to have less than 10 grams of salt. State, giving a reason, which shop has more snacks that do not meet the requirement. Median Mode & mean were wrong answers 75 240100× 180= Well done Shop 2 has more snacks that do not meet the requirement because Shop 1 has 50% of 240 = 120 snacks while Shop 2 has 75% of 320 = 240 snacks what do not meet the requirement. Fairly well done except some could not give concrete reasons 7 A sequence 𝑢𝑢1, 𝑢𝑢2, 𝑢𝑢3, … is such that 𝑢𝑢1 = 1 4 and 𝑢𝑢𝑛𝑛+1 = 𝑢𝑢𝑛𝑛 − 6 (3𝑛𝑛+5)(3𝑛𝑛+8), for 𝑛𝑛 ∈ 𝑍𝑍+. Use the principle of mathematical induction to prove that for all 𝑛𝑛 ∈ 𝑍𝑍+, 𝑢𝑢𝑛𝑛 = 2 3𝑛𝑛+5. Maximum [7 marks] Let Pn be the proposition: 𝑢𝑢𝑛𝑛 = 2 3𝑛𝑛 + 5 , 𝑛𝑛 ∈ 𝑍𝑍+. To prove P1 is true: LHS = 𝑢𝑢1 = 1 4 Well done although some started with the proposition Amount of salt per snack in grams Amount of salt per snack in grams SHOP 1 SHOP 2

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 6 RHS = 2 3𝑛𝑛+5 = 1 4 Hence P1 is true. Assume Pn is true for n = k, 𝑢𝑢𝑘𝑘 = 2 3𝑘𝑘 + 5 , 𝑘𝑘 ∈ 𝑍𝑍+ To prove Pk+1 is true: 𝑢𝑢𝑘𝑘+1 = 2 3(𝑘𝑘 + 1) + 5 LHS = 𝑢𝑢𝑘𝑘+1 = 𝑢𝑢𝑘𝑘 − 6 (3𝑘𝑘+5)(3𝑘𝑘+8) = 2 3𝑘𝑘 + 5 − 6 (3𝑘𝑘 + 5)(3𝑘𝑘 + 8) = 2 3𝑘𝑘 + 5 �1 − 3 3𝑘𝑘 + 8� = 2 3𝑘𝑘 + 5 �3𝑘𝑘 + 5 3𝑘𝑘 + 8� = 2 3(𝑘𝑘 + 1) + 5 Since P1 is true and Pk is true, implies Pk+1 is true, thus for all 𝑛𝑛 ∈ 𝑍𝑍+, 𝑢𝑢𝑛𝑛 = 2 3𝑛𝑛+5 𝑢𝑢𝑘𝑘+1 = 𝑢𝑢𝑘𝑘 − 6 (3𝑘𝑘 + 5)(3𝑘𝑘 + 8) 8 The diagram shows the probability density function of a continuous random variable, X whose median is m. (a) Find the value of k and of m. (b) Hence state the value of its mode. Maximum [5 marks] k m 120 0

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 7 Area of triangle = Total probability = 1 120 12 k× ×= 1 60k = Well done. Since m is the median, 1 11 2 60 2m×× = 60m= Mode = 60 9 Consider the following system of equations 2260xyz++= 4 5 14 4xy z++ = ( )2 2 2 12xy z αβ−+− = − (a) Find the values of α and β for which the system has an infinite number of solution. (b) In the case where the number of solutions is infinite, find the general solution of the system of equations in parametric form. Maximum [6 marks] 2 2 6 0 22 6 0 4 5 14 4 0 1 2 4 2 2 2 12 0 4 8 12αβ α β      →    −−− − −   → � 2 2 6 0 0 1 2 4 0 0 𝛼𝛼 𝛽𝛽 + 4 � For infinite solutions, 0α = and 𝛽𝛽 = −4 Some equated the rows of zeros to the matrix that was not in REF. This qn should be attempted using row operations on the matrix. From 2260xyz++= and 24yz+= Let 𝑧𝑧 = 𝑡𝑡, where 𝑡𝑡 ∈ ℝ Make y and x the subject respectively, 42yt= − and 4xt= −−

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 8 SECTION B 10. [Maximum mark: 21]* a) A function f is defined by 21() 3 xfx −= , 5 4ax≤≤ . (i) Given that the range of f is 11 () 2fx−≤ ≤ , find the value of a . (ii) Find 1()fx− and state its range. Another function g is defined by () 5 xgx x= − , 𝑥𝑥 ∈ ℝ , 5x≠ . (iii) Find ()fg x , given that it exists. (iv) Sketch the