ACSI 2019 Promo Paper 1 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 1 FINAL EXAMINATION 2019 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL P a p e r 1 S O L U T I O N S SECTION A Qn Solution 1. Find the value of k if ෍ 𝑘 2௥ ൌ5 ஶ ௥ୀ଴ . ∑ ௞ ଶೝ ൌ𝑘 ቀ1൅ ଵ ଶ ൅ ଵ ସ ൅⋯ ቁஶ ௥ୀ଴ Expand to find the terms ൌ𝑘 ቆ ଵ ଵିభ మ ቇ ൌ2 𝑘 GP, Sum to infinity: a = k , r = ½ 2𝑘 ൌ 5 => 𝑘ൌ2 . 5 Remarks: Well attempted 2. Consider the equation ሺℎെ1 ሻ𝑥ଶ െ𝑥 ℎ൅ሺℎ൅1 ሻ ൌ0 , where ℎ is a real constant. Find the set of values of h for which this equation has real roots.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 2 Qn Solution Since the equation has real roots, 𝑏ଶ െ4 𝑎 𝑐൒0 ℎଶ െ4 ሺ ℎെ1 ሻ ሺ ℎ൅1 ሻ൒0 ℎଶ െ4 ℎଶ ൅4൒0 ℎ൑ 4 3 െ 2 √3 ൑ℎ൑ 2 √3 Remarks: Well attempted. Common mistake: b2-4ac > 0 applies only for real and equal (repeated) roots. 3. a The sum of the first n terms of a sequence {un} is given by 𝑆௡ ൌ𝑛 ଶ ൅𝑟 ଶ𝑛െ ଷ ସ 𝑛, where 𝑛∈𝑍 ା and 𝑟∈𝑅 . Write down the value of 𝑢ଵin terms of r. 𝑢ଵ ൌ𝑆 ଵ ൌ1 ଶ ൅1 ଶ𝑛െ ଷ ସ ൌ ଵ ସ ൅𝑟 ଶ sum of one term = first term. b Prove that {un} is an arithmetic sequence, stating clearly its common difference. 𝑢௡ ൌ𝑆 ௡ െ𝑆 ௡ିଵ ൌ𝑛 ଶ ൅𝑟 ଶ𝑛െ ଷ ସ 𝑛െ ሺ𝑛െ1 ሻଶ െ𝑟 ଶሺ𝑛െ1 ሻ ൅ ଷ ସ ሺ𝑛 െ 1ሻ ൌ ቀ𝑛ଶ ൅𝑟 ଶ𝑛െ ଷ ସ 𝑛ቁ െ ሺ𝑛ଶ െ2 𝑛൅1ሻ െ𝑟 ଶ𝑛൅𝑟 ଶ ൅ ଷ ସ 𝑛െ ଷ ସ ൌെ ଻ ସ ൅2 𝑛൅𝑟ଶ 𝑢௡ െ𝑢 ௡ିଵ ൌ ቀെ ଻ ସ ൅2 𝑛൅𝑟ଶቁ െ ቀെ ଻ ସ ൅2 ሺ𝑛െ1 ሻ ൅𝑟 ଶቁ ൌ2ൌ common difference The common difference is a constant, hence {un} is an arithmetic sequence Alternative solution is to show that Sn satisfy the equation for sum of an AP where 1[]2 nn nSu u Ie. 𝑆௡ ൌ𝑛 ଶ ൅𝑟 ଶ𝑛െ ଷ ସ 𝑛ൌ ௡ ଶ ቀ2𝑛 ൅ 2𝑟ଶ െ ଷ ଶቁ ൌ ௡ ଶ ൬ቀ ଵ ସ ൅𝑟 ଶቁ ൅ ቀെ ଻ ସ ൅2 𝑛൅𝑟ଶቁ൰ Remarks: Common mistake, students only show U3 − U2 = U2 – U1 = 2 which shows that only U1, U2 and U3 are in AP.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 3 Qn Solution 4. Use Mathematical Induction to prove that ሺ1൅𝑥 ሻ௡ ൒1൅𝑛 𝑥 for ሼ𝑛: 𝑛 ∈ 𝑍ାሽ where 𝑥൐1 . Let Pn be the proposition: ሺ1൅𝑥 ሻ௡ ൒1൅𝑛 𝑥 for ሼ𝑛: 𝑛 ∈ 𝑍ାሽ where 𝑥൐1 . To prove P1 is true: LHS = 1൅𝑥 RHS = 1൅𝑥 Hence P1 is true. Assume Pn is true for some n = k, ሺ1൅𝑥 ሻ௞ ൒1൅𝑘 𝑥, 𝑘∈𝑍 ା To prove Pk+1 is true: ሺ1൅𝑥 ሻ௞ାଵ ൒1൅ ሺ𝑘൅1 ሻ𝑥 LHS ൌ ሺ1൅𝑥 ሻ௞ାଵ ൌ ሺ1൅𝑥 ሻ௞ሺ1൅𝑥 ሻ ൒ ሺ1൅𝑘 𝑥ሻሺ1൅𝑥 ሻ Sub. Pk+1 ൌ1൅𝑥൅𝑘 𝑥൅𝑘 𝑥ଶ Expand RHS terms ൒1൅𝑥൅𝑘 𝑥 (Since 𝑥൐1 , 𝑘𝑥ଶ ൒0 ) Justification for the inequality ൌ 1൅ ሺ𝑘൅1 ሻ𝑥 = RHS Since P1 is true and Pk is true, implies Pk+1 is true, by Mathematical Induction ሺ1൅𝑥 ሻ௡ ൒1൅𝑛 𝑥 for ሼ𝑛: 𝑛 ∈ 𝑍ାሽ where 𝑥൐1 is true. Remarks: Most students understood the method of proof using MI but a number of students do not write the statements or t