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 the proof properly. A number of students did not prove that P1 is true but proved that P2 is true instead. 5. a Express the binomial coefficient ቀ𝑛൅1 𝑛െ2 ቁ as a polynomial in n. ቀ𝑛൅1 𝑛െ2 ቁ ൌ ሺ௡ାଵሻ! ሺ௡ିଶሻ!ଷ! Note: Given in the formula booklet, ቀ𝑛 𝑛ቁ ൌ ௡! ሺ௡ି௥ሻ!௥! ൌ ሺ௡ାଵሻሺ௡ሻሺ௡ିଵሻ ଺ ൌ ൫௡మିଵ൯௡ ଺ ൌ ଵ ଺ ሺ𝑛ଷ െ𝑛 ሻ b Hence find the coefficient of 𝑥ହି௡ in the expansion of ቀ𝑥൅ ଵ ௫ቁ ௡ାଵ in terms of 𝑛.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 4 Qn Solution 𝑇௥ାଵ ൌ ቀ𝑛൅1 𝑟 ቁ 𝑥௡ାଵି௥ 𝑥ି௥ ൌ ቀ𝑛൅1 𝑟 ቁ 𝑥௡ାଵିଶ௥ General term formula (not in formula booklet. Memorize this.) 𝑥௡ାଵିଶ௥ ൌ𝑥 ହି௡ 𝑛൅1െ2 𝑟ൌ5െ𝑛 Equating the index of x 2𝑛 െ 4 ൌ 2𝑟 𝑟ൌ𝑛െ2 Sub. r into ቀ𝑛൅1 𝑟 ቁ Coefficient = ቀ𝑛൅1 𝑛െ2 ቁ ൌ ଵ ଺ ሺ𝑛ଷ െ𝑛 ሻ Use 5a. to solve this part. Remarks: Well attempted. 6. a Show that 𝑙𝑜𝑔௥మ 𝑥ൌ ଵ ଶ log௥ 𝑥 . 𝑙𝑜𝑔௥మ 𝑥ൌ ௟௢௚ೝ௫ ௟௢௚ೝ௥మ Change of base ൌ ଵ ଶ 𝑙𝑜𝑔௥𝑥 b Hence express 𝑦ൌ𝑙 𝑜 𝑔ଷ 𝑥൅𝑙 𝑜 𝑔ଽ 𝑥 in terms of 𝑙𝑛ሺ𝑥ሻ and describe the transformation that will map 𝑦ൌ𝑙 𝑛ሺ𝑥ሻ into 𝑦ൌ𝑙 𝑜 𝑔ଷ 𝑥൅𝑙 𝑜 𝑔ଽ 𝑥. 𝑦ൌ𝑙 𝑜 𝑔ଷ 𝑥൅𝑙 𝑜 𝑔ଽ 𝑥ൌ𝑙 𝑜 𝑔ଷ 𝑥൅ ଵ ଶ 𝑙𝑜𝑔ଷ𝑥 Using result in (a) ൌ ଷ ଶ 𝑙𝑜𝑔ଷ𝑥 ൌ ଷ ଶ௟௡ଷ 𝑙𝑛𝑥 Stretch parallel to y-axis, by factor ଷ ଶ௟௡ଷ. Remarks: well attempted. 7. a Consider the equations of the lines: 1 :l 11 3 21 , 01           r  and 2 :l 2 113 x y z   . Determine whether or not, 1l and 2l intersect.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 5 Qn Solution Check if lines are parallel: ൭ 3 െ1 1 ൱ ്𝑘 ൭ 3 1 െ1 ൱ The direction vectors are not parallel, 1l and 2l are not parallel. Check if lines intersects by equating the two lines: 11 3 2 3 9 3 3 21 1 11                                           Equating the first two equations and solving: 93 3   and 1    66 1     or 2  Check for consistency of the values: Sub. into third equation, LHS = 2  and RHS = 12   Since values of λ and μ are consistent, sub either values into the line equation to find the point of intersection. Point of intersection (-5, 0, 2) b Find a vector that is perpendicular to the two lines. Since the two lines intersects, they lie on the same plane. A vector perpendicular to the two lines is the normal vector of that plane. Normal Vector = 33 11 11        Cross product of the dir. vectors ൌ ൭ 1െ1 െሺ3 ൅ 3ሻ െ3 െ 3 ൱ ൌ ൭ 0 െ6 െ6 ൱ Also accept any other any vector parallel to ൭ 0 1 1 ൱. Remarks: well attempted.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 6 SECTION B Qn Solution 8. (i) a Expand and simplify ሺ𝑥െ1 ሻሺ𝑥ହ ൅𝑥 ସ ൅𝑥 ଷ ൅𝑥 ଶ ൅𝑥൅1 ሻ. ሺ𝑥െ1 ሻሺ𝑥ହ ൅𝑥 ସ ൅𝑥 ଷ ൅𝑥 ଶ ൅𝑥൅1 ሻ Expand the terms. ൌ𝑥 ଺ ൅𝑥 ହ ൅𝑥 ସ ൅𝑥 ଷ ൅𝑥 ଶ ൅𝑥െ𝑥 ହ െ𝑥 ସ െ𝑥 ଷ െ𝑥 ଶ െ𝑥െ1 ൌ𝑥 ଺ െ1 b Given that w is a root of the equation 𝑧଺ െ1ൌ0 which does not lie on the real axis in the Argand diagram, show that 𝑤ହ ൅𝑤 ସ ൅𝑤 ଷ ൅𝑤 ଶ ൅𝑤൅1ൌ0 . 𝑧଺ െ1ൌ ሺ𝑧െ1 ሻሺ𝑧ହ ൅𝑧 ସ ൅𝑧 ଷ ൅𝑧 ଶ ൅𝑧൅1 ሻ ൌ0 From 8ai, let z = x. 𝑧െ1ൌ0 or 𝑧ହ ൅𝑧 ସ ൅𝑧 ଷ ൅𝑧 ଶ ൅𝑧൅1ൌ0 Since w is a root which does not lie on the real axis, 𝑤്1 . Hence, 𝑤ହ ൅𝑤 ସ ൅𝑤 ଷ ൅𝑤 ଶ ൅𝑤൅1ൌ0 . Remarks: Some students were not able to see the link from a) to b). c Find the roots of the equation 𝑧଺ ൌ1 , giving your answers exactly in the form 𝑟𝑒௜ఏ, where െ𝜋 ൏ 𝜃 ൑ 𝜋 . 𝑧଺ ൌ1ൌ𝑒 ௜ሺ଴ାଶ௞గሻ Note that 𝑟ൌ |1| ൌ1 and argሺ1ሻ ൌ0 . 𝑧ൌ𝑒 ௜మೖഏ ల ൌ𝑒 ௜ೖഏ య , 𝑘ൌ0 , േ 1 , േ 2 , 3 𝑧ൌ𝑒 ଴,𝑒 ௜ഏ య ,𝑒 ି௜ഏ య ,𝑒 ௜మഏ య ,𝑒 ି௜మഏ య ,𝑒 ௜గ Alternatively, 𝑧଺ ൌ1ൌc o sሺ0൅2 𝑘 𝜋ሻ ൅𝑖 𝑠 𝑖 𝑛 ሺ 0൅2 𝑘 𝜋 ሻ 𝑧ൌc o s ଶ௞గ ଺ ൅𝑖 𝑠 𝑖 𝑛 ଶ௞గ ଺ , 𝑘 ൌ 0, േ1, േ2, 3 zൌc o s 0൅𝑖 𝑠 𝑖 𝑛 0 , c o s గ ଷ ൅𝑖 𝑠 𝑖 𝑛 గ ଷ ,c o s గ ଷ െ𝑖 𝑠 𝑖 𝑛 గ ଷ ,cos ଶగ ଷ ൅𝑖 𝑠 𝑖 𝑛 ଶగ ଷ , cos ଶగ ଷ െ𝑖 𝑠 𝑖 𝑛 ଶగ ଷ , c o s𝜋൅ 𝑖𝑠𝑖𝑛𝜋 Convert from Trigo to exponential form using zൌc o s 𝜃൅𝑖 𝑠 𝑖 𝑛 𝜃ൌ𝑒௜ఏ 𝑧ൌ𝑒 ଴,𝑒 ௜ഏ య ,𝑒 ି௜ഏ య ,𝑒 ௜మഏ య ,𝑒 ି௜మഏ య ,𝑒 ௜గ Remarks: Some found the wrong arg(1) or r. Use arg(z) for principal argument (and not Arg(z)). Sketch an Argand diagram if necessary to find these values accurately. Students who solved using trigo form were les successful and showed lengthy workings. (ii) Consider the complex numbers 𝑧ଵ ൌ1െ √3𝑖 and 𝑧ଶ ൌ2െ2 𝑖 a Find the exact value of the argument of ௭భ ௭మ .

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2019 Final Exam / Paper 1 / Solutions 7 Qn Solution 𝑧ଵ ൌ1െ √3𝑖 is in the 4th quadrant, argሺ𝑧ଵሻ ൌെ 𝑡 𝑎 𝑛ିଵ ቤ√3 1 ቤ ൌെ 𝜋 3 𝑧ଶ ൌ2െ2 𝑖 is in the 4th quadrant, argሺ𝑧ଶሻ ൌെ 𝑡 𝑎 𝑛ିଵ ቚ ଶ ଶቚ ൌെ గ ସ arg ቀ ௭భ ௭మ ቁ ൌa r gሺ𝑧ଵሻ െa r gሺ𝑧ଶሻ ൌെ గ ଷ ൅ గ ସ ൌെ గ ଵଶ Remarks: Knowledge of the trigo ratios of the ba