New test - February 23, 2016 [276 marks]

Transcription

1 New test - February 3, 016 [76 marks] The first three terms of a infinite geometric sequence are m 1, 6, m +, where m Z. 1a. Write down an expression for the common ratio, r. correct expression for r N1 6 r =, m 1 [N/A] m+ 6 1b. Hence, show that m satisfies the equation m + 3m 0 = 0. correct equation correct working correct working m () AG N0 [N/A] 6 m 1 m+ 6 6 m+ =, = (m + )(m 1) = 36 m 1 6 m m + m = 36, m + 3m = m 0 = 0 1c. Find the two possible values of m. valid attempt to solve N3 [N/A] (m + 8)(m 5) = 0, m = m = 8, m = 5 3± d. Find the possible values of r.

2 attempt to substitute any value of m to find r r = N3 [N/A] ,, r = e. The sequence has a finite sum. State which value of r leads to this sum and justify your answer. r = (may be seen in justification) 3 valid reason R1 N0 r < 1, 1 < < 1 3 Notes: Award R1 for r < 1 only if awarded. [N/A] 1f. The sequence has a finite sum. Calculate the sum of the sequence. finding the first term of the sequence which has r < 1 () u 1 = 9 (may be seen in formula) () u correct substitution of u1 and their r into 1, as long as r < 1 [ marks] 8 1, 6 3 S 9 =, 1 ( ) 3 7 S = (= 5.) N3 1 r [N/A] 7 a. Expand r as the sum of four terms. r= [1 mark]

3 7 r = (accept ) N1 r= [1 mark] This question proved difficult for many candidates. A number of students seemed unfamiliar with sigma notation. Many were successful with part (a), although some listed terms or found an overall sum with no working. 30 b. (i) Find the value of r. r= [6 marks] r (ii) Explain why cannot be evaluated. r= (i) METHOD 1 recognizing a GP u 1 =, r =, n = 7 () correct substitution into formula for sum () ( 1) e.g. S 7 = S 7 7 METHOD 1 = N recognizing = 3 r= r=1 r=1 recognizing GP with u 1 =, r =, n = 30 () correct substitution into formula for sum 30 ( 1) S 30 = 1 = () 30 r r= N (ii) valid reason (e.g. infinite GP, diverging series), and r 1 (accept r > 1 ) R1R1 N [6 marks] = ( + + 8) = The results for part (b) were much more varied. Many candidates did not realize that n was 7 and used 30 instead. Very few candidates gave a complete explanation why the infinite series could not be evaluated; candidates often claimed that the value could not be found because there were an infinite number of terms. th The n term of an arithmetic sequence is given by = 5 + n. u n Write down the common difference. 3a. [1 mark]

4 d = N1 [1 mark] The majority of candidates could either recognize the common difference in the formula for the n term or could find it by writing out the first few terms of the sequence. th th 3b. (i) Given that the n term of this sequence is 115, find the value of n. (ii) For this value of n, find the sum of the sequence. [5 marks] (i) 5 + n = 115 () n = 55 N (ii) = 7 (may be seen in above) () u 1 correct substitution into formula for sum of arithmetic series () e.g. = ( ), = ((7) + 5()), (5 + k) 55 S S 55 S 55 = 3355 (accept 3360) N3 [5 marks] 55 k=1 Part (b) demonstrated that candidates were not familiar with expression, "n term". Many stated that the first term was 5 and then decided to use their own version of the nth term formula leading to a great many errors in (b) (ii). th The diagram shows a circle of radius 8 metres. The points ABCD lie on the circumference of the circle. BC = 1 m, CD = 11.5 m, AD = 8 m, ADC = 10, and BCD = 73. Find AC. a.

5 evidence of choosing cosine rule c = a + b ab cosc, C D + AD CD ADcosD correct substitution cos10, cos10 AC = 15.5 (m) N There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect. Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. b. (i) Find ACD. (ii) Hence, find ACB. [5 marks] (i) METHOD 1 evidence of choosing sine rule sin A a sin B b =, correct substitution sin ACD AD = sin D AC METHOD N evidence of choosing cosine rule sin ACD 8 = ACD = 30.0 sin c = a + b ab cosc correct substitution e.g. 8 = (11.5)( ) cosc ACD = 30.0 N (ii) subtracting their ACD from ACD, ACB = 3.0 N [5 marks]

6 There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect. Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. Find the area of triangle ADC. c. correct substitution area () area =.6 (m ) N 1 ΔADC = (8)(11.5) sin 10 There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect. Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and most could find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrong mode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle. (c) d. (d) Find the area of triangle ADC. Hence or otherwise, find the total area of the shaded rions. [6 marks] (c) correct substitution () area area =.6 (m ) N 1 ΔADC = (8)(11.5) sin 10 (d) attempt to subtract circle ABCD, r ΔADC ΔACB area 1 ΔACB = ( )(1) sin.98 correct working () ( ( )(1) sin.98, ) 1 shaded area is 8. (m ) N3 [ marks] Total [6 marks]

7 There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect. Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and most could find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrong mode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle. In part (d), most candidates recognized what to do and often obtained follow through marks from errors made in previous parts. Hence or otherwise, find the total area of the shaded rions. e. [ marks] attempt to subtract circle ABCD, r ΔADC ΔACB area 1 ΔACB = ( )(1) sin.98 correct working () ( ( )(1) sin.98, ) 1 shaded area is 8. (m ) N3 [ marks] Total [6 marks] There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect. Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. In part (d), most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.

8 The diagram below shows part of the graph of a function f. The graph has a maximum at A( 1, 5) and a minimum at B( 3, 1). The function f can be written in the form f(x) = p sin(qx) + r. Find the value of (a) 5a. p (b) q (c) r. [6 marks] (a) valid approach to find p amplitude, p = 6 p = 3 N = max min (b) valid approach to find q period =, q = q = N period (c) valid approach to find r axis = r = max+min N, sketch of horizontal axis, f(0) Total [6 marks] Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters p, q and r. The successful candidates were able to find the answers using the given points and their understanding of the different transformations. Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between q and the period of the function. There were also some candidates who showed working such as without explaining what b represented. b

9 p 5b. valid approach to find p amplitude, p = 6 p = 3 N = max min Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters p, q and r. The successful candidates were able to find the answers using the given points and their understanding of the different transformations. Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between q and the period of the function. There were also some candidates who showed working such as without explaining what b represented. b q 5c. valid approach to find q period =, q = q = N period Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters p, q and r. The successful candidates were able to find the answers using the given points and their understanding of the different transformations. Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between q and the period of the function. There were also some candidates who showed working such as without explaining what b represented. b r. 5d. valid approach to find r axis = r = max+min N, sketch of horizontal axis, f(0) Total [6 marks] p q r b b q

10 The following diagram shows a right-angled triangle, ABC, where sin A = a. Show that cosa = METHOD 1 approach involving Pythagoras theorem 5 + x = 13, labelling correct sides on triangle finding third side is 1 (may be seen on diagram) cosa = 1 AG 13 METHOD N0 approach involving sin θ + cos θ = 1 5 ( ) + cos θ = 1, x 5 + = correct working cos θ = cosa = 1 AG 13 N0 [N/A] 6b. Find cosa. correct substitution into cosθ correct working cosa = () N () [N/A] ( ), ( 1 ) 1, ( 1 ) , 1, ( ) 13

11 The following diagram shows a circle with centre O and radius 5 cm. The points A, rmb and rmc lie on the circumference of the circle, and AOC = 0.7 radians. 7a. Find the length of the arc ABC. correct substitution into arc length formula arc length = 3.5 (cm) N () [N/A] 7b. Find the perimeter of the shaded sector. valid approach , arc + r perimeter = 13.5 (cm) N [N/A] 7c. Find the area of the shaded sector. correct substitution into area formula 1 (0.7)(5) area = 8.75 (c m ) N () [N/A]

12 Let f(x) = p cos(q(x + r)) + 10, for 0 x 0. The following diagram shows the graph of f. The graph has a maximum at (, 18) and a minimum at (16, ). 8a. Write down the value of r. r = A N Note: Award for r =. [N/A] 8b. Find p. evidence of valid approach p = 8 N, distance from y = 10 [N/A] max y value -- y value 8c. Find q. valid approach period is,, substitute a point into their f(x) q = (, exact), 0.6 (do not accept drees) N [N/A] 1 360

13 8d. Solve f(x) = 7. valid approach line on graph at x = x = 11.5 (accept (11.5,7)) N y = 7, 8 cos( (x )) + 10 = 7 Note: Do not award the final if additional values are given. If an incorrect value of q leads to multiple solutions, award the final only if all solutions within the domain are given. [N/A] Let f(x) = sin(x + ) + k. The graph of f passes through the point (, 6). 9a. Find the value of k. METHOD 1 attempt to substitute both coordinates (in any order) into f correct working () sin = 1, 1 + k = 6 k = 5 N METHOD recognizing shift of left means maximum at 6 R1) recognizing k is difference of maximum and amplitude () k = 5 N [N/A] f ( ) = 6, = sin(6 + ) + k 6 1 9b. Find the minimum value of f(x). evidence of appropriate approach minimum value of sin x is 1, 1 + k, f (x) = 0, (, ) minimum value is N [N/A] 5

14 9c. Let g(x) = sin x. The graph of g is translated to the graph of f by the vector ( p ). q Write down the value of p and of q. p =, q = 5 (accept ( )) 5 N [N/A] Consider a circle with centre O and radius 7 cm. Triangle ABC is drawn such that its vertices are on the circumference of the circle. AB = 1. cm, BC = 10. cm and ACB = radians. 10a. Find BAC. Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working. Candidates may have their GDCs in dree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate. Ignore missing or incorrect units. evidence of choosing sine rule correct substitution sin A a sin A 10. = = BAC = sin B b sin () N [N/A] 10b. AC

15 Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working. Candidates may have their GDCs in dree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate. Ignore missing or incorrect units. METHOD 1 evidence of subtracting angles from correct angle (seen anywhere) attempt to substitute into cosine or sine rule correct substitution METHOD () evidence of choosing cosine rule correct substitution [5 marks] ABC = A C ABC = , 1.6, A (A) N3 71. M cos71., AC = sin 1.6 N3 AC = 13.3 (cm) a = b + c bc cosa 1. = b 10.b cos1.058 AC = 13.3 (cm) 1. sin [N/A] 10c. Hence or otherwise, find the length of arc ABC. [6 marks]

16 Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in subsequent parts. Accept answers that are consistent with their working. Candidates may have their GDCs in dree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate. Ignore missing or incorrect units. METHOD 1 valid approach cosaoc =, AOC = ABC correct working EITHER () () correct substitution for arc length (seen anywhere) l =, l = 17., subtracting arc from circumference OR attempt to find AOC reflex correct substitution for arc length (seen anywhere) l = , THEN METHOD N valid approach to find AOB or BOC choosing cos rule, twice angle at circumference correct working for finding one value, AOB or BOC () cosaob =, two correct calculations for arc lengths adding their arc lengths (seen anywhere) O A +OC AC OA OC = cosaoc, O = 1.6 AOC =.9 ( 1.8 ) r l, 1 = 17..9, 3.79, arc ABC = N AOB =.116,BOC = AB = (= ), (= ) raob + rboc, , 7( ) arc ABC = 6.5 (cm) M1 ()() Note: Candidates may work with other interior triangles using a similar method. Check calculations carefully and award marks in line with markscheme. [6 marks] [N/A] sin = m 11a. Let 100. Find an expression for cos100 in terms of m.

17 Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award the final FT mark. METHOD 1 valid approach involving Pythagoras e.g. sin x + cos x = 1, labelled diagram correct working (may be on diagram) e.g. m + (cos100) = 1, 1 m cos100 = 1 m N () METHOD valid approach involving tan identity e.g. tan = sin cos correct working e.g. cos100 = cos100 = m tan 100 sin 100 tan 100 () N While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is native. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write 1 m = 1 m. In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because sin 100 = m, then sin 00 = m. In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant. sin 100 = m tan b. Let. Find an expression for in terms of m. [1 mark] METHOD 1 tan100 = [1 mark] m 1 m m 1 m (accept ) N1 METHOD tan100 = [1 mark] m cos 100 N1

18 While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is native. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write 1 m = 1 m. In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because sin 100 = m, then sin 00 = m. In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant. Let sin 100 = m. Find an expression for in terms of m. 11c. sin 00 METHOD 1 valid approach involving double angle formula e.g. sin θ = sin θcosθ sin 00 = m 1 m (accept m ( 1 m )) N Note: If candidates find cos100 = 1 m, award full FT in parts (b) and (c), even though the values may not have appropriate signs for the angles. METHOD valid approach involving double angle formula e.g. sin θ = sin θ cosθ, m m sin 00 = m tan 100 (= m cos100) tan 100 N While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is native. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write 1 m = 1 m. In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because sin 100 = m, then sin 00 = m. In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

19 Let f(x) = acos(b(x c)). The diagram below shows part of the graph of f, for 0 x 10. The graph has a local maximum at P(3, 5), a local minimum at Q(7, 5), and crosses the x-axis at R. Write down the value of 1a. (i) a ; (ii) c. (i) a = 5 (accept 5 ) N1 (ii) c = 3 (accept c = 7, if a = 5 ) N1 Note: Accept other correct values of c, such as 11, 5, etc. Part (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c. Find the value of b. 1b. attempt to find period e.g. 8, b = b = 8 period (exact),, [ 0.785, 0.786] (do not accept 5) N Finding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8. Find the x-coordinate of R. 1c.

20 valid approach e.g. f(x) = 0, symmetry of curve x = 5 (accept (5,0)) N Most candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve although follow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down the equations they entered into their GDC. The following diagram shows a circular play area for children. The circle has centre O and a radius of 0 m, and the points A, B, C and D lie on the circle. Angle AOB is 1.5 radians. Find the length of the chord [AB]. 13a.

21 Note: In this question, do not penalise for missing or incorrect units. They are not included in the markscheme, to avoid complex answer lines. METHOD 1 choosing cosine rule (must have cos in it) e.g. c = a + b ab cosc correct substitution (into rhs) e.g (0)(0) cos1.5, AB = cos1.5 AB = AB = 7.3 [7., 7.3] METHOD choosing sine rule N e.g. =, correct substitution e.g. sin A a AB sin 1.5 = sin B b AB sin O = 0 sin(0.5( 1.5)) AB = AB = 7.3 [7., 7.3] AO sin B N Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a rightangled triangle. A surprising number of candidates changed all angles to drees and worked with those, often leading to errors in accuracy. Find the area of triangle AOB. 13b. correct substitution into area formula e.g. 1 (0)(0) sin 1.5, 1 (0)( ) sin(0.5( 1.5)) area = (accept = 00, from using 7.3) area = 199 [199, 00] N1 Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a rightangled triangle. A surprising number of candidates changed all angles to drees and worked with those, often leading to errors in accuracy. Angle BOC is. radians. 13c. Find the length of arc ADC.

22 appropriate method to find angle AOC e.g correct substitution into arc length formula () e.g. ( 3.9) 0, arc length = arc length = 7.7 (7.6, 7.7] (i.e. do not accept 7.6) N Notes: Candidates may misread the question and use AOC =.. If working shown, award M0 then A0MR for the answer 8. Do not then penalize AOC in part (d) which, if used, leads to the answer However, if they use the prematurely rounded value of. for AOC, penalise 1 mark for premature rounding for the answer 8 in (c). Do not then penalize for this in (d). In part (c), some candidates misread the question and used. as the size of angle AOC while others rounded prematurely leading to the inaccurate answer of 8. In either case, marks were lost. Angle BOC is. radians. 13d. Find the area of the shaded rion. calculating sector area using their angle AOC () e.g. 1 (.38 )( 0 ), 00(.38 ), shaded area = their area of triangle AOB + their area of sector e.g , shaded area = (accept = 676 from using 199) shaded area = 676 [676, 677] N Part (d) proved to be straightforward and candidates were able to obtain full FT marks from errors made in previous parts. Angle BOC is. radians. 13e. [ marks] The shaded rion is to be painted red. Red paint is sold in cans which cost $3 each. One can covers 10 m. How much does it cost to buy the paint?

23 dividing to find number of cans e.g., cans must be purchased () multiplying to find cost of cans e.g. 5(3), 3 cost is 160 (dollars) N3 [ marks] Most candidates had a suitable straty for part (e) and knew to work with a whole number of cans of paint. The following diagram shows the graph of f(x) = acos(bx), for 0 x. There is a minimum point at P(, 3) and a maximum point at Q(, 3). (i) Write down the value of a. 1a. (ii) Find the value of b. (i) a = 3 N1 (ii) METHOD 1 attempt to find period e.g., b =, b = (= ) N METHOD attempt to substitute coordinates e.g. 3 cos(b) = 3, 3 cos(b) = 3 b = (= ) b N

24 In part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing the amplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, there were many candidates who thought b was the period of the function, rather than. period Write down the gradient of the curve at P. 1b. [1 mark] 0 N1 [1 mark] In part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, many candidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given value of x, rather than simply stating that the gradient of a curve at a minimum point is zero. 1c. Write down the equation of the normal to the curve at P. recognizing that normal is perpendicular to tangent e.g. m 1 m = 1, m = 1, sketch of vertical line on diagram 0 x = (do not accept or y = ) N For part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involving native reciprocal gradients, rather than recognizing that the equation of the vertical line was x =. There were also candidates who had trouble expressing the correct equation of a line parallel to the y-axis.

25 The diagram below shows part of the graph of f(x) = acos(b(x c)) 1, where a > 0. The point P (,) is a maximum point and the point Q(, ) is a minimum point. 3 15a. Find the value of a. evidence of valid approach e.g. a = 3 max y value min y value N, distance from y = 1 A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. (i) 15b. Show that the period of f is. (ii) Hence, find the value of b. [ marks] (i) evidence of valid approach e.g. finding difference in x-coordinates, evidence of doubling e.g. ( ) period = AG N0 (ii) evidence of valid approach e.g. b = b = [ marks] N

26 A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was, either incorrectly adding the given / and 3/ or using the value of b that they found first for part (b)(ii). Given that 0 < c <, write down the value of c. 15c. [1 mark] c = N1 [1 mark] A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was, either incorrectly adding the given / and /3 or using the value of b that they found first for part (b)(ii). Let f(x) = (sin x + cosx). 16a. Show that f(x) can be expressed as 1 + sin x. attempt to expand e.g. (sin x + cosx)(sin x + cosx) ; at least 3 terms correct expansion e.g. sin x + sin xcosx + cos x f(x) = 1 + sin x AG N0 Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates were certainly helped by the fact that it was a "show that" question.

27 16b. The graph of f is shown below for 0 x. Let g(x) = 1 + cosx. On the same set of axes, sketch the graph of g for 0 x. N Note: Award for correct sinusoidal shape with period and range [0, ], for minimum in circle. More candidates had difficulty with part (b) with many assuming the first graph was 1 + sin(x) and hence sketching a horizontal translation of / for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor p correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation. 16c. The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the vector ( k ). 0 Write down the value of p and a possible value of k. p =, k = N

28 Part (c) certainly served as a discriminator between the grade 6 and 7 candidates. A Ferris wheel with diameter 1 metres rotates clockwise at a constant speed. The wheel completes. rotations every hour. The bottom of the wheel is 13 metres above the ground. A seat starts at the bottom of the wheel. Find the maximum height above the ground of the seat. 17a. valid approach 13 + diameter, maximum height = 135 (m) N Most candidates were successful with part (a). After t minutes, the height h metres above the ground of the seat is given by h = 7 + acosbt. 17b. (i) Show that the period of h is 5 minutes. (ii) Write down the exact value of b. = 5 = 60. b = (= 0.08) 5

29 A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of b also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at t = 0 thereby making the value of a native. Incorrect values of 61 were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges. 5 (b) 17c. (i) Show that the period of h is 5 minutes. (ii) Write down the exact value of b. (c) Find the value of a. (d) Sketch the graph of h, for 0 t 50. [9 marks] = 5 = 60. b = (= 0.08) 5 max 7 a = a = 61 a = 61 a = = 7 + acos( ) = 7 + acos() 13 = 7 + a a = 61

30 A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of b also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at t = 0 thereby making the value of a native. Incorrect values of 61 were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges. 5 17d. Find the value of a. METHOD 1 valid approach max 7, a = , 7 13 a = 61 (accept a = 61 ) () a = 61 METHOD N attempt to substitute valid point into equation for h = 7 + acos( ) 5 correct equation () 135 = 7 + acos(), 13 = 7 + a a = 61 N A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of b also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at t = 0 thereby making the value of a native. Incorrect values of 61 were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges. 5 17e. Sketch the graph of h, for 0 t 50. [ marks]

31 N Note: Award for approximately correct domain, for approximately correct range, for approximately correct sinusoidal shape with cycles. Only if this last awarded, award for max/min in approximately correct positions. [ marks] A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of b also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at t = 0 thereby making the value of a native. Incorrect values of 61 were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges. 5 In one rotation of the wheel, find the probability that a randomly selected seat is at least 105 metres above the ground. 17f. [5 marks] setting up inequality (accept equation) h > 105, 105 = 7 + acosbt, sketch of graph with line y = 105 any two correct values for t (seen anywhere) t = 8.371, t = 16.68, t = , t = 1.68 valid approach M1,,, p = [5 marks] t 1 t 5 N ( ) 5 Part (e) was very poorly done for those who attempted the question and most did not make the connection between height, time and probability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novel approaches from the strongest of candidates using circles and angles to solve this part of question 10. Let sin θ =, where < θ < a. Find cosθ.

32 METHOD 1 evidence of choosing sin θ + cos θ = 1 correct working () e.g. cos 9 θ =, cosθ = ± 3, cosθ = cosθ = 3 13 N Note: If no working shown, award N1 for. METHOD approach involving Pythagoras theorem e.g. + x = 13, 3 13 finding third side equals 3 cosθ = 3 13 N () Note: If no working shown, award N1 for While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a native value. 18b. Find tanθ. [5 marks] correct substitution into sin θ (seen anywhere) () 13 e.g. ( ) ( ) correct substitution into cosθ (seen anywhere) () 3 13 e.g. ( ) ( ), ( ) 1, valid attempt to find tanθ e.g., correct working 13 ( 13) e.g.,, N Note: If students find answers for cosθ which are not in the range [ 1, 1], award full FT in (b) for correct FT working shown. [5 marks] ( )( ) ( 3 ) ( ) ()()( 3) tanθ = ( ) 3 1 ( ) ( ) 13

33 In part (b), many candidates incorrectly tried to calculate tanθ as tanθ, rather than using the double-angle identities. The following diagram shows the graph of f(x) = asin(b(x c)) + d, for x 10. There is a maximum point at P(, 1) and a minimum point at Q(8, ). Use the graph to write down the value of 19a. (i) a ; (ii) c ; (iii) d. (i) a = 8 N1 (ii) c = N1 (iii) d = N1 Part (a) of this question proved challenging for most candidates. 19b. Show that b =.

34 METHOD 1 recognizing that period = 8 correct working e.g. 8 =, b = b 8 () b = AG N0 METHOD attempt to substitute M1 e.g. 1 = 8 sin(b( )) + correct working e.g. sin b = 1 b = AG N0 Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that this period could be found using the given coordinates of the maximum and minimum points. 19c. Find f (x). evidence of attempt to differentiate or choosing chain rule e.g. cos (x ), f (x) = cos( (x )) (accept cos (x ) ) A N3 8 In part (c), not many candidates found the correct derivative using the chain rule. 19d. At a point R, the gradient is. Find the x-coordinate of R. [6 marks]

35 recognizing that gradient is e.g. correct equation e.g. = cos( (x )), correct working () e.g. cos 1 ( 1) = (x ) using cos 1 ( 1) = (seen anywhere) () e.g. f simplifying () e.g. = (x ) x = 6 [6 marks] (x) = m = (x ) N f (x) 1 = cos( (x )) For part (d), a good number of candidates correctly set their expression equal to, but errors in their previous values kept most from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here. The following diagram represents a large Ferris wheel, with a diameter of 100 metres. Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The wheel rotates at a constant rate, in an anticlockwise (counter-clockwise) direction. One revolution takes 0 minutes. 0a. Write down the height of P above ground level after (i) 10 minutes; (ii) 15 minutes. (i) 100 (metres) N1 (ii) 50 (metres) N1 1 Nearly all candidates answered part (a) correctly, finding the height of the wheel at and of a revolution. 3

36 Let h(t) metres be the height of P above ground level after t minutes. Some values of h(t) are given in the table below. 0b. (i) Show that h(8) = (ii) Find h(1). [ marks] (i) identifying symmetry with h() = 9.5 subtraction e.g. 100 h(), h(8) = 90.5 AG N0 (ii) recognizing period e.g. h(1) = h(1) h(1) =. [ marks] N While many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function for height, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table. 0c. Sketch the graph of h, for 0 t 0. 0 h 100

37 In part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles. However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs were constructed of line sments, rather than a curve. Given that h can be expressed in the form h(t) = acosbt + c, find a, b and c. 0d. [5 marks] evidence of a quotient involving 0, or 360 to find b b e.g. = 0, b = (= ) (accept b = 18 if working in drees) N 0 10 a = 50, c = 50 A N3 [5 marks] b = For part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accurate sketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that the position on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidates recognized that the value of a would be native. Candidates should be aware that while working in drees may be acceptable, the expectation is that radians will be used in these types of questions. 1a. Show that cosθ + 5 sin θ = sin θ + 5 sin θ + 3. attempt to substitute 1 sin θ for cosθ correct substitution e.g. (1 sin θ) + 5 sin θ cosθ + 5 sin θ = sin θ + 5 sin θ + 3 AG N0 In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that" question, working backward from the given answer is not a valid method. 1b. Hence, solve the equation cosθ + 5 sin θ = 0 for 0 θ. [5 marks]

38 evidence of appropriate approach to solve e.g. factorizing, quadratic formula correct working e.g. ( sin θ + 3)(sin θ + 1), (x + 3)(x + 1) = 0, sin x = 5± 1 correct solution sin θ = 1 (do not penalise for including sin θ = 3 () θ = 3 [5 marks] A N3 In part (b), many candidates seemed to realize what was required by the word hence, though some had trouble factoring the quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution to the equation sin θ = 1, and there were a few who did not realize that the equation sin θ = 3 has no solution. The diagram below shows a plan for a window in the shape of a trapezium. Three sides of the window are m long. The angle between the sloping sides of the window and the base is θ, where 0 < θ <. a. Show that the area of the window is given by y = sin θ + sin θ. [5 marks] evidence of finding height, h e.g. sin θ = h, sin θ () evidence of finding base of triangle, b e.g. cosθ = b, cosθ () attempt to substitute valid values into a formula for the area of the window e.g. two triangles plus rectangle, trapezium area formula correct expression (must be in terms of θ ) e.g. ( 1 cosθ sin θ) + sin θ, 1 ( sin θ)( + + cosθ) attempt to replace sin θ cosθ by sin θ M1 e.g. sin θ + ( sin θ cosθ) y = sin θ + sin θ [5 marks] AG N0 As the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) was generally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in a clear and organized manner. Some tried a "working backwards" approach, earning no marks.

39 b. Zoe wants a window to have an area of 5 m. Find the two possible values of θ. [ marks] correct equation e.g. y = 5, sin θ + sin θ = 5 evidence of attempt to solve e.g. a sketch, sin θ + sin θ 5 = 0 θ = ( 9.0 ), θ = 1.5 ( 71. ) N3 [ marks] In part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and instead attempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a second solution outside the domain. c. John wants two windows which have the same area A but different values of θ. Find all possible values for A. [7 marks] recognition that lower area value occurs at θ = finding value of area at θ = e.g. sin( ) + sin( ), draw square A = () recognition that maximum value of y is needed A = () < A < 5.0 (accept < A < 5.19 ) A N5 [7 marks] A pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or the maximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident from candidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken into consideration for future paper writing. 3x Let f(x) = + 1, g(x) = cos( ) 1. Let h(x) = (g f)(x). x 3 Find an expression for h(x). 3a. attempt to form any composition (even if order is reversed) correct composition 3x +1 h(x) = cos( ) 1 3 3x h(x) = g ( + 1) () 1 1 ( cos( x + ) 1, cos( ) 1) 3 3x+ 6 N3

40 The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient. Write down the period of h. 3b. [1 mark] period is (1.6) N1 [1 mark] The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient. Write down the range of h. 3c. range is 5 h(x) 3 ([ 5,3]) N The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient. Let f(x) = 3 sin x + cosx, for x. Sketch the graph of f. a.

41 N3 Note: Award for approximately sinusoidal shape, for end points approximately correct (, ) (, ), for approximately correct position of graph, (y-intercept (0, ), maximum to right of y-axis). Some graphs in part (a) were almost too detailed for just a sketch but more often, the important features were far from clear. Some graphs lacked scales on the axes. b. Write down (i) the amplitude; (ii) the period; (iii) the x-intercept that lies between and 0. (i) 5 N1 (ii) (6.8) N1 (iii) 0.97 N1 A number of candidates had difficulty finding the period in part (b)(ii). Hence write f(x) in the form p sin(qx + r). c. f(x) = 5 sin(x ) (accept p = 5, q = 1, r = 0.97 ) N3 A number of candidates had difficulty writing the correct value of q in part (c).

42 Write down one value of x such that f (x) = 0. d. evidence of correct approach e.g. max/min, sketch of f (x) indicating roots one 3 s.f. value which rounds to one of 5.6,.5, 0.6, 3.8 N The most common approach in part (d) was to differentiate and set maximum or minimum values on their graphs. f (x) = 0. Fewer students found the values of x given by the Write down the two values of k for which the equation f(x) = k has exactly two solutions. e. k = 5, k = 5 N Part (e) proved challenging for many candidates, although if candidates answered this part, they generally did so correctly. f. Let g(x) = ln(x + 1), for 0 x. There is a value of x, between 0 and 1, for which the gradient of f is equal to the gradient of g. Find this value of x. [5 marks]

43 METHOD 1 graphical approach (but must involve derivative functions) M1 e.g. each curve x = METHOD A N g (x) = 1 x+1 f (x) = 3 cosx sin x (5 cos(x )) evidence of attempt to solve g (x) = f (x) M1 x = [5 marks] A N In part (f), many candidates were able to get as far as equating the two derivatives but fewer used their GDC to solve the resulting equation. Again, many had trouble demonstrating their method of solution. The graph of y = p cosqx + r, for 5 x 1, is shown below. There is a minimum point at (0, 3) and a maximum point at (, 7). Find the value of 5a. (i) p ; [6 marks] (ii) q ; (iii) r.

44 (i) evidence of finding the amplitude e.g., amplitude = 5 (ii) period = 8 (iii) 7+3 p = 5 [6 marks] 8 N () q = (= = ) r = r = 7 3 () N N Many candidates did not recognize that the value of p was native. The value of q was often interpreted incorrectly as the period but most candidates could find the value of r, the vertical translation. 5b. The equation y = k has exactly two solutions. Write down the value of k. [1 mark] k = 3 (accept y = 3 ) N1 [1 mark] In part (b), candidates either could not find a solution or found too many. The diagram below shows a quadrilateral ABCD with obtuse angles ABC and ADC. AB = 5 cm, BC = cm, CD = cm, AD = cm, BAC = 30, ABC = x, ADC = y. 6a. Use the cosine rule to show that AC = 1 0 cosx. [1 mark] correct substitution e.g cosx, cosx AC = 1 0 cosx AG [1 mark]

45 Many candidates worked comfortably with the sine and cosine rules in part (a) and (b). 6b. Use the sine rule in triangle ABC to find another expression for AC. correct substitution AC sin x e.g. =, AC = 8 sin x (accept ) N1 sin 30 1 AC = sin x sin x sin 30 Many candidates worked comfortably with the sine and cosine rules in part (a) and (b). Equally as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usually unsuccessful. 6c. (i) Hence, find x, giving your answer to two decimal places. (ii) Find AC. [6 marks] (i) evidence of appropriate approach using AC M1 e.g. 8 sin x = 1 0 cosx, sketch showing intersection correct solution 8.68, () obtuse value () x = to dp (do not accept the radian answer 1.9 ) N (ii) substituting value of x into either expression for AC e.g. AC = 8 sin AC = 7.5 N [6 marks] Equally as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recovered and earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode and rounded prematurely throughout this question often resulting in accuracy penalties. 6d. (i) Find y. (ii) Hence, or otherwise, find the area of triangle ACD. [5 marks]

46 (i) evidence of choosing cosine rule e.g. cosb = a+ c b ac correct substitution e.g., 7.5 = 3 3 cosy, cosy = 0.73 y = 137 N (ii) correct substitution into area formula 1 e.g. sin 137, 8 sin 137 area = 5. N [5 marks] () Equally as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recovered and earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode and rounded prematurely throughout this question often resulting in accuracy penalties. International Baccalaureate Organization 016 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Colio Aleman de Barranquilla