ACSI 2023 Promo Paper 1 ans

FINAL EXAMINATION 2023 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL PAPER 1 ANALYSIS AND APPROACHES NOTE: 1. Answers should not be in pencil. Answers in pencil are not acceptable. 2. Answers should be in the given box in section A, if insufficient space students need to indicate where the continuation is. Section A *1. [Maximum mark: 4] Solve the following equation 23𝑥𝑥−2 = 5(3−𝑥𝑥). Give your answer in the form ln 𝑎𝑎 ln 𝑏𝑏, where 𝑎𝑎 and 𝑏𝑏 are integers. Solution: 23𝑥𝑥−2 = 5(3−𝑥𝑥) 8𝑥𝑥 4 = 5(3−𝑥𝑥) 8𝑥𝑥 × 3𝑥𝑥 = 20 24𝑥𝑥 = 20 𝑥𝑥 = 𝑙𝑙𝑙𝑙20 𝑙𝑙𝑙𝑙24 Many students are not able to get this correct as they are not able to manipulate the laws of Logarithm and Indices well. Different kinds of mistakes are made.

2. [Maximum mark: 4] It is given that 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 3 2, 𝜋𝜋 2 < 𝑐𝑐 < 3𝜋𝜋 2 . Find the exact value of 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. Solution: 𝑐𝑐𝑠𝑠𝑙𝑙𝑐𝑐 = 2 3 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = − √5 3 Note its 2nd Quadrant 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑠𝑠𝑠𝑠𝑐𝑐 = − √5 2 Many students are not able to get the correct sign, they reject the -ve and accept the positive answer Or just give the positive answer. Some note that the range is 2 nd and 3rd quadrant but they did not use the fact sine is positive. *3. [Maximum mark: 7] (a) Show that �(2𝑟𝑟 − 1) 𝑠𝑠 𝑟𝑟=1 = 𝑙𝑙2 [3] (b) Hence, find the value of 11 + 13 + 15 + ⋯ + 99. [4] Solution: a) �(2𝑟𝑟 − 1) 𝑠𝑠 𝑟𝑟=1 = 2 � 𝑟𝑟 𝑠𝑠 𝑟𝑟=1 − � 1 𝑠𝑠 𝑟𝑟=1 = 2 �𝑙𝑙 2 (1 + 𝑙𝑙)� − 𝑙𝑙 = 𝑙𝑙(1 + 𝑙𝑙) − 𝑙𝑙 = 𝑙𝑙2 b) 11 + 13 + 15 + ⋯ + 99 = �(2𝑟𝑟 − 1) 50 𝑟𝑟=1 − �(2𝑟𝑟 − 1) 5 𝑟𝑟=1 (a)Many students are able to use the AP formula to find the sum. Some students use the suggested method. Quite a few uses Induction but did not necessarily score full as their Induction is not fully correctly presented. Technically the question did not ask for Induction. They should not use. (b) “Hence” must use the result in (a) otherwise strictly no marks awarded if the answer is correct. Various mistakes like 45

= 502 − 52 = 2475 instead of 50. 6 instead of 5. 502 = 250. *4. [Maximum mark: 7] The expansion of (𝑥𝑥 + ℎ)7, where ℎ is a non-zero rational number, can be written as 𝑥𝑥7 + 𝑎𝑎𝑥𝑥6 + 𝑏𝑏𝑥𝑥5 + 𝑐𝑐𝑥𝑥4 + ⋯ + ℎ7 where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 ∈ 𝑅𝑅. Given that 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐 are three consecutive terms of an arithmetic sequence, find the possible values of ℎ. Solution: (𝑥𝑥 + ℎ)7 = 𝑥𝑥7 + 𝑎𝑎𝑥𝑥6 + 𝑏𝑏𝑥𝑥5 + 𝑐𝑐𝑥𝑥4 + ⋯ + ℎ7 Equating 𝑥𝑥6 term, 𝐶𝐶1 7 × ℎ = 𝑎𝑎 Equating 𝑥𝑥5 term, 𝐶𝐶2 7 × ℎ2 = 𝑏𝑏 Equating 𝑥𝑥4 term, 𝐶𝐶3 7 × ℎ3 = 𝑐𝑐 Since 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐 are three consecutive terms of an arithmetic sequence, 𝑐𝑐 − 𝑏𝑏 = 𝑏𝑏 − 𝑎𝑎 2𝑏𝑏 = 𝑎𝑎 + 𝑐𝑐 2(21ℎ2) = 7ℎ + 35ℎ3 7ℎ(6ℎ) − 7ℎ(1 + 5ℎ2) = 0 Most of the students can give the coefficient except some are not able to give the correct values due to no calculator. Quite a few did not reject h=0. Some have obvious mistakes like h is a surd which is impossible as the question stated non- zero rational number. Some has calcu