ACSI 2017 Promo Paper 2 ans

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 1 FINAL EXAMINATION 2017 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL Paper 2 Solutions with Marker’s Comments Qn Solution Marker’s Comments Section A 1) [6 marks] Solving with GDC (Solving with Linear Equations Function and Log Function) or alternative Most students used graphical method to solve. Final answer

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 2 Qn Solution Marker’s Comments 2) [7 marks] At the end of year 1 The total in bank account At the end of second year At the end of seven years Many students were unable to find the general expression for S7 Using nsolve on GDC (or plotting graphs) 3) [6 marks] (a) (b) Observe that has argument (giving bearing of ) and a modulus of 4, giving a distance of 4 km from A Very few students were able to make this observation. Hence the complex number representing B Coordinates of B is A majority of those who found the coordinates of B correctly did not use complex numbers to obtain the answer. 4) [5 marks] Consider first number odd, no of ways 4 23 3 108P× ×= Consider first number even, no of ways 4 22 2 48P× ×= Total number of ways = 108 48 156+= Most students were able to do this question.

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 3 Qn Solution Marker’s Comments Alternative method: No. of 4-digit numbers ending with 0 = 5P3 x 1 = 60 No. of 4-digit numbers ending with 2 or 4 = 4 x 4P2 x 2 = 96 Total number of ways = 60 +96 = 156 5) [5 marks] For an even function Generally well done. But a minority of students do not know the definition of an even function. Equating , 6) [7 marks] (a) Use GDC to plot and Many attempt to solve the problem manually, without using the GDC which can solve this problem graphically which is more efficient. or

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 4 Qn Solution Marker’s Comments (b) Plot graph of The graph is concave downwards where Many students confuse point of inflexion with stationary points, wrongly finding the range by setting '( ) 0fx < 7) [7 marks] (a) A possible unit vector Students did not follow instruction and did not find the unit vector and the plane equation in the requested form. Equation of plane (b) Possible equations of plane or Students not heeding the hence requirement and taking a much longer method to solve the problem.

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 5 Qn Solution Marker’s Comments Alternatively, 12xyz−+= or 6xyz−+= − 8) [7 marks] (a) Perimeter of inner hexagon = Many fail to see that the polygon is a hexagon and each triangle is equilateral and that means they can infer that the perimeter is 6r, and there is no need for lengthy proof. Perimeter of outer hexagon Many gave very unnecessary long proof for this. (b) Perimeter of enclosed circle = 6 2 43rr r π≤≤ Note: The objective is to find an estimate for π and the above represents the bound for the lower and upper bound for the perimeter of the circle Hence 3 23π≤≤ Note that the above simply mean that an estimate of π is between the given range. Section B 9) [18 marks] (a) Using ti Nspire, Many gave very unnecessary long working for this, and failed to use cpolyroots function on their Ti Nspire. The zeros are zi=− or 1zi= + or 2z= (b)(i) ( )sin cos cos sin *i i i izθθ θθ+= −= (ii) ( ) ( ) ( ) ( ) 3 2332sin cos sin 3sin cos 3sin cos cosi i iiθ θ θ θθ θθ θ+ = +++ Students use De Moivre’s theorem on ( ) 3 sin cosiθθ+ Failure to make use of b(i) to find sin 3θ ( ) ( ) ( )( ) ( ) 33 * cos sin cos3 sin 3iz i i i iθ θ θθ= − −+ − =− + sin 3 cos3iθθ= −− This step was often left out Comparing coefficients, 32sin 3 sin 3sin cosθ θ θθ−=−

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 6 Qn Solution Marker’s Comments ( ) 32sin 3sin 1 sinθθ θ= −− 3sin 3 3sin 4sinθθ θ= − (c)(i) () 0fz = ( )28 1 ik ze ππ+ = −= Students choose π− instead of π which is a principal angle 2 88 ki ze ππ += Considering conjugate roots, or otherwise 8i ze π± = 3 8i ze π± = 5 8i ze π± = 5 8i ze π± = (ii) Drawing an Argand Diagram Many students drew the Argand Diagram wrongly as a series of points or simply a closed polygon. (iii) Noting that , , and are conjugate pairs and the sum of all distinct roots equal to zero (see Argand diagram) Many variety of proofs including trigonometric identity types, which are less preferred given the “hence” nature of the question. Students should have observed from the Argand diagram that all 8 roots sum up to zero and they form four conjugate pairs. Hence using sum of conjugates,

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 7 Qn Solution Marker’s Comments 10) [ 17 marks] (a) (i) Sketch the possible triangles Students failed to conceive that the facts given about triangle ABC describe TWO possible triangles ABC, one larger than the other in terms of its area. This idea of two possible triangles is not at odds with the question, which many students thought to be in error because they can only find one. The problem here is a case of the ambiguous triangle that one may get employing sine rule. Using Sine Rule, Using Ti Nspire, or The two possible lengths of AB given by m or m Area of smaller triangle ABC = m2 (ii) Ratio of areas of smaller to larger triangle = Ratio (3sf) The ratio can be easily found by considering the ratio of lengths found in part (i). There is no need to evaluate area of larger triangle. (b) (i)

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 8 Qn Solution Marker’s Comments Many students wrongly assume the following a) CQ=QB=3.5 b) Triangle ABC is isosceles with AB=AC=10 c) Points A, O and Q are collinear Substituting Using nsolve, m Quicker Alternative Solution ( )11 10.4322 10 7 10.4322 10 sin 4022 2.44 r r × ++=× ×× ≈ (ii) Length of AR is ( ) 2.44442923014 6.716014tan 20 =° m Area of bounded region 21 1 1402 6.716014 2.44445 2.444452 2 180 9.116711 9.12 π×=×× × −× ×  = ≈ Area = 9.12 m2 Students who work on this manually tend to round off the values of r and AP early leading to inaccurate value of area. 11) [ 15 marks] (a) (i) and

ACS (Independent) / Mathematics Department / Mathematics HL / 2017 / Final Exam / Paper 2 / Solutions 9 Qn Solution Marker’s Comments Hence OB and AC are perpendicular (ii) is isosceles and OB is therefore the perpendicular bisector of AC, thus by similar triangles, OB bisects (b) (i) Consider any two position vectors on the line, say and and Line is in and Line is in Hence the line is the line of intersection of both planes. (ii) Vector normal to planes and is given by Equation of required plane perpendicular to planes and and passing through A(2,4,0) (iii) From