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 calculation error due to poor factorization.

7ℎ(6ℎ − 1 − 5ℎ2) = 0 Since ℎ ≠ 0 (Or reject ℎ = 0) 6ℎ − 1 − 5ℎ2 = 0 (5ℎ − 1)(ℎ − 1) = 0 ℎ = 1 5 or ℎ = 1 5. [Maximum mark: 9] (a) Use mathema�cal induc�on to prove that for any real number 𝑥𝑥, |x|<1, and any posi�ve integer 𝑙𝑙, (1 + 𝑥𝑥) 𝑠𝑠 ≥ 1 + 𝑙𝑙𝑥𝑥. [6] (b) Hence show that 0.9994 > 0.996. [3] Solution: Let Pn be the proposition such that (1 + 𝑥𝑥)𝑠𝑠 ≥ 1 + 𝑙𝑙𝑥𝑥, any positive integer 𝑙𝑙 and any real number 𝑥𝑥, |x|<1. LHS = 1 + 𝑥𝑥 =RHS Hence P1 is true. Assume Pk is true for some 𝑘𝑘 ∈ ℤ+ (1 + 𝑥𝑥)𝑘𝑘 ≥ 1 + 𝑘𝑘𝑥𝑥 To prove Pk+1 is true. LHS = (1 + 𝑥𝑥)𝑘𝑘+1 = (1 + 𝑥𝑥)𝑘𝑘(1 + 𝑥𝑥) ≥ (1 + 𝑘𝑘𝑥𝑥)(1 + 𝑥𝑥) = 1 + (𝑘𝑘 + 1)𝑥𝑥 + 𝑘𝑘𝑥𝑥2 ≥ 1 + (𝑘𝑘 + 1)𝑥𝑥 = RHS Since 𝑘𝑘𝑥𝑥2 ≥ 0 for real number |x|<1 and any positive integer 𝑘𝑘. Many students do not write their statements properly and hence loose marks. Various mistakes, eg instead of “Some”, students wrote “all”. The closing statements written by them shows that the students do not know the true meaning of the MI process. → means approaches so it is not acceptable. ⇒ means imply. If not sure should just write in words. Poor presentation of the working for the Pk+1 case. Students cannot write from L to R properly. Some students try to “force out” the final step of the proof but the link in between does not make sense.

Since P1 is true and Pk is true implies Pk+1 is true. By MI, (1 + 𝑥𝑥)𝑠𝑠 ≥ 1 + 𝑙𝑙𝑥𝑥, any positive integer 𝑙𝑙 and any real number 𝑥𝑥, |x|<1. Since (1 + 𝑥𝑥)𝑠𝑠 ≥ 1 + 𝑙𝑙𝑥𝑥, Let 𝑥𝑥 = −0.001 and 𝑙𝑙 = 4, Then (1 − 0.001)4 ≥ 1 + 4(−0.001) 0.9994 > 1 − 0.004 (remove “=” since 𝑙𝑙 ≠ 0 𝑐𝑐𝑟𝑟 1 𝑎𝑎𝑙𝑙𝑎𝑎 𝑥𝑥 ≠ 0 ) 0.9994 > 0.996 (shown) Marker is lenient here as in if the students did not explain or give reasons why = is removed they still can get full marks as the previous part is marked stricter. However if the students did not remove the = sign or some how their writing does not show the results required, they will be penalised. 6. [Maximum mark: 6] Consider the following function: The graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) is shown in the following diagram. The curve intersects the 𝑥𝑥- axis at (−2,0) and it has a stationary point at (4,0). The equations of the asymptotes are 𝑥𝑥 = 0 and 𝑦𝑦 = 2(𝑥𝑥 − 6). Sketch the curve 𝑦𝑦 = 1 𝑓𝑓(𝑥𝑥) , indicating clearly the equations of any asymptotes, coordinates of any intercepts with the axes and the coordinates of any stationary point(s).

Many students are not able to completely give the full graph and information. Some did not give the equations of the asymptotes. Many did not give y=0. The shapes are not correct and the min and intercepts is not clearly given. Did not label the x and y axes. 7. [Maximum mark: 10] In the following diagram, the points 𝐴𝐴, 𝐵𝐵, 𝐶𝐶 and 𝐷𝐷 are on the circumference of a circle with centre 𝑂𝑂 and radius 𝑟𝑟. 𝐴𝐴𝐶𝐶 is the diameter of the circle. 𝐵𝐵𝐶𝐶 = 𝑟𝑟, 𝐴𝐴𝐷𝐷 = 𝐶𝐶𝐷𝐷 and 𝐴𝐴𝐵𝐵� 𝐶𝐶 = 𝐴𝐴𝐷𝐷�𝐶𝐶 = 90°. (a) Show that cos�𝐵𝐵𝐴𝐴̂𝐷𝐷� = √3−1 2√2 . [5] (b) Hence or otherwise, show that the area of triangle 𝐴𝐴𝐵𝐵𝐷𝐷 = 3+√3 4 𝑟𝑟2. [5] 𝑦𝑦 = 0 𝑥𝑥 = −2 𝑥𝑥 = 4 𝑟𝑟 𝑟𝑟

Solution: (a) Since 𝐵𝐵𝐶𝐶 = 𝑟𝑟, triangle 𝐵𝐵𝑂𝑂𝐶𝐶 is an equilateral triangle. 𝐵𝐵𝐴𝐴̂𝑂𝑂 = 60° 2 = 30° 𝐵𝐵𝐴𝐴̂𝐷𝐷 = 30° + 45° cos�𝐵𝐵𝐴𝐴̂𝐷𝐷� = cos(30° + 45°) = 𝑐𝑐𝑐𝑐𝑐𝑐30°𝑐𝑐𝑐𝑐𝑐𝑐45° − 𝑐𝑐𝑠𝑠𝑙𝑙30°𝑐𝑐𝑠𝑠𝑙𝑙45° = 1 √2 × √3 2 − 1 √2 × 1 2 = √3 − 1 2√2 (b) Area of triangle 𝐴𝐴𝐵𝐵𝐷𝐷 = 1 2 × 𝐴𝐴𝐵𝐵 × 𝐴𝐴𝐷𝐷 × 𝑐𝑐𝑠𝑠𝑙𝑙𝐵𝐵𝐴𝐴̂𝐷𝐷 = 1 2 × √3𝑟𝑟 × √2𝑟𝑟 × 𝑐𝑐𝑠𝑠𝑙𝑙(30° + 45°) = 1 2 × √6𝑟𝑟2 × � 1 √2 × √3 2 + 1 √2 × 1 2 � = 3 + √3 4 𝑟𝑟2 Some students try to just “cheat” by stating the last expected answer but the steps before does not actually leads to the given answer. So the question is “do the students understand that their answer is wrong but are trying their luck, hoping the examiner does not see or the students truly are not able to manage their surds thinking that their answer is correct?”. (a) Not many students can do this question, some left it blank. Students does not realise that they can use the addition formulae and the special angles to do this. Some cosine rule twice and equate the length BD. Some still cannot give the values for special angles. (b) Similar method is expected for sin(75°) but many uses the Pythagoras method and could not prove the last step. Some just write down sin(75°)= 13 22 + without giving any explanation so penalised. Some find the area of 3 triangles to sum up. 8. [Maximum mark: 6] Consider the following functions: 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 𝑎𝑎𝑥𝑥 + 𝑏𝑏 𝑔𝑔(𝑥𝑥) = 𝑎𝑎𝑥𝑥2 + 2𝑏𝑏𝑥𝑥 − 2 , where 𝑎𝑎 and 𝑏𝑏 are positive integers.

It is given that the solution to 𝑓𝑓(𝑥𝑥) ≥ 𝑔𝑔(𝑥𝑥) is −5 ≤ 𝑥𝑥 ≤ 1, find the values of 𝑎𝑎 and 𝑏𝑏. Solution: 𝑓𝑓(𝑥𝑥) ≥ 𝑔𝑔(𝑥𝑥) 𝑥𝑥2 + 𝑎𝑎𝑥𝑥 + 𝑏𝑏 ≥ 𝑎𝑎𝑥𝑥2 + 2𝑏𝑏𝑥𝑥 − 2 (𝑎𝑎 − 1)𝑥𝑥2 + (2𝑏𝑏 − 𝑎𝑎)𝑥𝑥 − (2 + 𝑏𝑏) ≤ 0 Since 𝑥𝑥 = 1 and 𝑥𝑥 = −5 are roots, by Factor Theorem, (𝑎𝑎 − 1)(1)2 + (2𝑏𝑏 − 𝑎𝑎) − (2