ACSI 2023 Promo Paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 1 FINAL EXAMINATION 2023 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL PAPER 2 ANALYSIS AND APPROACHES SECTION A Qn Solution Marks 1*. [7 marks] (a) 4 (b) 𝑢𝑢𝑛𝑛 = 2(4)𝑛𝑛−1 For 𝑢𝑢𝑛𝑛 < 10000 2(4)𝑛𝑛−1 < 10000 4𝑛𝑛−1 < 5000 𝑛𝑛 − 1 < lg(5000) lg(4) 𝑛𝑛 < lg(5000) lg 4 + 1 = 7.1438 The largest term is 𝑢𝑢7 = 2(46) = 8192 Alternative: From GDC, we generate a sequence with general term 𝑢𝑢𝑛𝑛 = 2(4)𝑛𝑛−1 The largest term smaller than 10000 is 𝑢𝑢7 = 8192 (as 𝑢𝑢8 = 32768 > 10000)

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 2 Qn Solution Marks Generally well attempted. However many candidates did not find the term 7U but give the answer s n=7 . (c) 𝑆𝑆𝑛𝑛 = 2(4𝑛𝑛−1) 4−1 = 2 3 (4𝑛𝑛 − 1) 𝑆𝑆𝑛𝑛 < 100000 2 3 (4𝑛𝑛 − 1) < 100000 4𝑛𝑛 − 1 < 150000 4𝑛𝑛 < 150001 𝑛𝑛 < lg(150001) lg 4 = 8.59 The largest possible value of 𝑛𝑛 is 8. Alternative: 𝑆𝑆𝑛𝑛 = 2(4𝑛𝑛−1) 4−1 = 2 3 (4𝑛𝑛 − 1) From GDC, we generate a sequence with general term 𝑆𝑆𝑛𝑛 = 2 3 (4𝑛𝑛 − 1) The largest possible value of 𝑛𝑛 is 8 (as 𝑆𝑆9 = 174762 > 100000) Well attempted. Qn Solution Marks 2*. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 3 Qn Solution Marks (a) log√𝑎𝑎 𝑚𝑚 = log𝑎𝑎 𝑚𝑚 log𝑎𝑎 √𝑎𝑎 = log𝑎𝑎 𝑚𝑚 1 2 log𝑎𝑎 𝑎𝑎 = 2 log𝑎𝑎 𝑚𝑚 = log𝑎𝑎 𝑚𝑚2 Well attempted. However, a number of students just write 2 2 ()log loga amm= without showing the proof and no mark is awarded for this. (b) log√2 𝑥𝑥 + 3 log𝑥𝑥 2 = 5 log2 𝑥𝑥2 + 3 log2 𝑥𝑥 = 5 Let 𝑦𝑦 = log2 𝑥𝑥 2𝑦𝑦 + 3 𝑦𝑦 = 5 2𝑦𝑦2 − 5𝑦𝑦 + 3 = 0 (2𝑦𝑦 − 3)(𝑦𝑦 − 1) = 0 𝑦𝑦 = 3 2 or 𝑦𝑦 = 1 Therefore 𝑥𝑥 = 2 3 2 or 𝑥𝑥 = 2 Very well done Qn Solution Marks 3*. [8 marks] (a) 𝜆𝜆+2 2 = 4.5 𝜆𝜆 = 9 − 2 = 7 Generally well done (b) Let 𝑥𝑥 = 0, 𝑦𝑦 = −4 −4 = 𝑎𝑎(−7)(−2) 𝑎𝑎 = − 4 14 = − 2 7 Generally well done

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 4 Qn Solution Marks (c) 𝑘𝑘𝑥𝑥 + 9 = − 2 7 (𝑥𝑥 − 7)(𝑥𝑥 − 2) −7𝑘𝑘𝑥𝑥 − 63 = 2(𝑥𝑥2 − 9𝑥𝑥 + 14) 2𝑥𝑥2 − 18𝑥𝑥 + 7𝑘𝑘𝑥𝑥 + 28 + 63 = 0 2𝑥𝑥2 + (7𝑘𝑘 − 18)𝑥𝑥 + 91 = 0 Since the line is a tangent to the curve, the discriminant is 0 (7𝑘𝑘 − 18)2 − 4(2)(91) = 0 (7𝑘𝑘 − 18)2 = 728 7𝑘𝑘 − 18 = ±√728 𝑘𝑘 = 18±√728 7 = 6.43 or −1.28 Generally well done Qn Solution Marks 4. [4 marks] 3 , 32α β αβ+= = − ( ) ( ) ( ) 22 2 3 2 2 3 22 4 43 12 pp p p qq q q α β αβ α β αβ + = −⇒ + = −  =− =− =⇒= −= =− Alternative: 2 2 2 2 2 3 5122 3 12 0 3, 12 yx ySubst x yy yy pq = =  − −=   −−= = −= −

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 5 Qn Solution Marks Generally well done but a number of students made the careless mistake of giving the answer of p as 3 instead of -3. A number of students did not use the sum and product of roots to solve for p and q . They solve the equation to find the roots and use it to find the new equation to obtain the values of p and q . Qn Solution Marks 5. [4 marks] (a) ( ) ( ) 1arcsin arccosec 1let arcsin 1sin 1 sin cosec arccosec . xx x x x x x α α α α α  ≡  = = = = = Alternative: Accept constructed triangle as working Very poorly done. Most students cannot provide the proof (b) ( )( ) 1 1 sin arccosec 2 . 1sin arcsin 2 11 2 .2 ye ey ye ey − − =   =  = = ( ) 1arccosec 2 arcsiny e =  Generally well done, but many students who could not do part (a) did not bother to do part (b).

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 6 Qn Solution Marks 6. [9 marks] (a) t values as shown in graph. Alternative: 2sin(4 ) 2 2sin(4 ) 2 sin(4 ) 1 3 7 11 154 ,, , 22 2 2 3 7 11 15,, ,88 8 8 t tt t t t t ππ π π ππ π π +=− =− =− = = Generally well done, but a number of students did not sketch the graph as part of the working.

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 7 Qn Solution Marks (b) Using GDC, () 2ht t = + is the upper limit Hence 22 k−≤ ≤ Alternative: 2sin(4 ) sin(4 ) 2 11 2 2 2. t ttk kt k k +=+ = −≤ ≤ −≤ ≤ Well done but a number of students give the upper limit as 1.98… instead of 2 (c) Most intersections is when line is center of the range of k. k = 0, most intersections = 9. Alternative: sin(4 ) 0 02 0 t k k = = = most intersections = 9. Well done

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 8 Qn Solution Marks 7. [7 marks] (a) ( ) ( ) ( ) 11 11 1tan 51 1 1 tan 5 tan tan 155 tan 1 tan55 0.727 c= 1.73 z i x iy i x iy y x yx yx yx m π π ππ ππ −+= + −+ = −+ + += − += − = −−   = +−−     = − Very poorly done. Many students do not understand the question at all. (b) Gradient is tan of the required angle: ( ) 8arg 1 .5 3 15zi ππ π−+ = + = Poorly done. Qn Solution Marks 8. [9 marks] (a) ( ) 21 13 21 21 13 21 16 22 61 5 4 7 25 1 4 , . 27 r µµ −−     ×−  −      = −×  −  − =−−− + − = −−   = +∈ −  

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 9 Qn Solution Marks Many careless mistakes with the cross product. Some students used a longer method involving dot product and system of equations to work out the cross vector and wasted time. (b) 2 1 414 min 10 3 7 m a x 1 031 1 9 7 3 3 3 2 ab ab ab += −  = ++ = =− −= += += ≤+≤ Badly done. Did not realize the maximum length is created by a vector in the same direction as the original and the minimum length is created by a vector in the Opposite direction. (c) ( ) min when 10 3 10 10 3 2210 101 133 22 b ka b ka k k b or =− = = = = −   =−−  −  Badly done. Similar comment to above. Note this is an IB exam style question. Section B Qn Solution Marks 9 [25 marks] 9(a) () 12 2 12 () 2 x x x x fx f x even − − − = + =+= ∴ A significant number of students thought that even function meant a function divisible by 2. Even function does not imply even ‘number’. 9(b) 11(1) 2 2 2.5 ( 1) 2.5 f f −= += −=

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 5 / 2023 Final Exam / Paper 2 / Solutions 10 1, 1 for 2.5. not one-one and hence no inverse. xy= −= ∴ Alternative: Using horizontal line test, a horizontal y = 3 intersects the graph twice, hence the function is not one-one and inverse does not exist. Or using symmetry of the even function to deduce x has multiple values for some values of y. Poorly done. It is not enough to state no inverse. Must indicate the reason for no inverse by either providing a counter example (one value) or using horizontal line test and show graphically that there are 2 values of x that correspond to one value of y. note also the function just need to be one to many at one point. The entire graph need not be always many to one. 9(c) 1.17 or -1 1 or 1.17x xx<− < < >