new test - february 23, 2016 · 2015-07-10 · new test - february 23, 2016 [276 marks] [2 marks]...

New test - February 23, 2016 [276 marks]

[2 marks]1a.

The first three terms of a infinite geometric sequence are , where .

Write down an expression for the common ratio, .

Markschemecorrect expression for A1 N1eg

[2 marks]

Examiners report[N/A]

m − 1, 6, m + 4 m ∈ Z

r

r

r = , 6m−1

m+46

[2 marks]1b. Hence, show that satisfies the equation .

Markschemecorrect equation A1eg

correct working (A1)eg

correct working A1eg

AG N0[2 marks]


m + 3m − 40 = 0m2

= , =6m−1

m+46

6m+4

m−16

(m + 4)(m − 1) = 36

− m + 4m − 4 = 36, + 3m − 4 = 36m2 m2

+ 3m − 40 = 0m2

[3 marks]1c. Find the two possible values of .

Markschemevalid attempt to solve (M1)eg

A1A1 N3[3 marks]


m

(m + 8)(m − 5) = 0, m = −3± 9+4×40√2

m = −8, m = 5

[3 marks]1d. Find the possible values of .r

Markschemeattempt to substitute any value of to find (M1)eg

A1A1 N3

[3 marks]


m r

, 6−8−1

5+46

r = , r = −32

23

1e. [3 marks]The sequence has a finite sum.

State which value of leads to this sum and justify your answer.

Markscheme (may be seen in justification) A1

valid reason R1 N0eg

Notes: Award R1 for only if A1 awarded.

[2 marks]


r

r = − 23

|r| < 1, − 1 < < 1−23

|r| < 1

1f. [3 marks]The sequence has a finite sum.

Calculate the sum of the sequence.

Markschemefinding the first term of the sequence which has (A1)

eg

(may be seen in formula) (A1)

correct substitution of and their into , as long as A1

eg

A1 N3

[4 marks]


|r| < 1

−8 − 1, 6 ÷ −23

= −9u1

u1 ru1

1−r|r| < 1

= , S∞−9

1−(− )23

−953

= − (= −5.4)S∞275

2a. [1 mark]Expand as the sum of four terms.∑

r=4

72r

Markscheme (accept ) A1 N1

[1 mark]

Examiners reportThis question proved difficult for many candidates. A number of students seemed unfamiliar with sigma notation. Many weresuccessful with part (a), although some listed terms or found an overall sum with no working.

= + + +∑r=4

72r 24 25 26 27 16 + 32 + 64 + 128

2b. [6 marks](i) Find the value of .

(ii) Explain why cannot be evaluated.

Markscheme(i) METHOD 1

recognizing a GP (M1)

, , (A1)

correct substitution into formula for sum (A1)

e.g.

A1 N4

METHOD 2

recognizing (M1)

recognizing GP with , , (A1)

correct substitution into formula for sum

(A1)

A1 N4

(ii) valid reason (e.g. infinite GP, diverging series), and (accept ) R1R1 N2

[6 marks]

Examiners reportThe results for part (b) were much more varied. Many candidates did not realize that was and used instead. Very fewcandidates gave a complete explanation why the infinite series could not be evaluated; candidates often claimed that the value couldnot be found because there were an infinite number of terms.

∑r=4

302r

∑r=4

∞2r

=u1 24 r = 2 n = 27

=S27( −1)24 227

2−1

= 2147483632S27

= −∑r=4

30∑r=1

30∑r=1

3

= 2u1 r = 2 n = 30

=S302( −1)230

2−1

= 214783646

= 2147483646 − (2 + 4 + 8)∑r=4

302r

= 2147483632

r ≥ 1 r > 1

n 27 30

[1 mark]3a.

The n term of an arithmetic sequence is given by .

Write down the common difference.

th = 5 + 2nun

Markscheme A1 N1

[1 mark]

Examiners reportThe majority of candidates could either recognize the common difference in the formula for the n term or could find it by writing outthe first few terms of the sequence.

d = 2

th

3b. [5 marks](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.

Markscheme(i) (A1)

A1 N2

(ii) (may be seen in above) (A1)

correct substitution into formula for sum of arithmetic series (A1)

e.g. , ,

(accept ) A1 N3

[5 marks]

Examiners reportPart (b) demonstrated that candidates were not familiar with expression, "n term". Many stated that the first term was 5 and thendecided to use their own version of the nth term formula leading to a great many errors in (b) (ii).

th

5 + 2n = 115

n = 55

= 7u1

= (7 + 115)S55552

= (2(7) + 54(2))S55552

(5 + 2k)∑k=1

55

= 3355S55 3360

th

[3 marks]4a.

The diagram shows a circle of radius metres. The points ABCD lie on the circumference of the circle.

BC = m, CD = m, AD = m, , and .

Find AC.

8

14 11.5 8 A C =D 104∘ B D =C 73∘

Markschemeevidence of choosing cosine rule (M1)

eg ,

correct substitution A1

eg ,

AC (m) A1 N2

[3 marks]

Examiners reportThere was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer thequestion, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possiblecombinations 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 tothe fact that cyclic quadrilaterals is not part of the syllabus.

= + − 2ab cosCc2 a2 b2 C + A − 2 × CD × ADcosDD2 D2

+ − 2 × 11.5 × 8 cos10411.52 82 196.25 − 184 cos104

= 15.5

4b. [5 marks](i) Find .

(ii) Hence, find .

Markscheme(i) METHOD 1

evidence of choosing sine rule (M1)

eg ,


eg

A1 N2

METHOD 2

evidence of choosing cosine rule (M1)

eg


e.g.

A1 N2

(ii) subtracting their from (M1)

eg ,

A1 N2

[5 marks]

A DC

A BC

=sin A

a

sin B

b=sin A DC

ADsin D

AC

=sin A DC8

sin 10415.516…

A D =C 30.0∘

= + − 2ab cosCc2 a2 b2

= + 15.516 − 2(11.5)(15.516…)cosC82 11.52 …2

A D =C 30.0∘

A DC 73

73 − A DC 70 − 30.017…

A B =C 43.0∘



[2 marks]4c. Find the area of triangle ADC.

Markschemecorrect substitution (A1)

eg area

area (m ) A1 N2

[2 marks]


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 tothe fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and mostcould find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrongmode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle.

ΔADC = (8)(11.5)sin 10412

= 44.6 2

4d. [6 marks](c) Find the area of triangle ADC.

(d) Hence or otherwise, find the total area of the shaded regions.

Markscheme(c) correct substitution (A1)

eg area

area (m ) A1 N2

[2 marks]

(d) attempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3

[4 marks]

Total [6 marks]

ΔADC = (8)(11.5)sin 10412

= 44.6 2

circle − ABCD π − ΔADC − ΔACBr2

ΔACB = (15.516…)(14)sin 42.9812

π(8 − 44.6336… − (15.516…)(14)sin 42.98)2 12

64π − 44.6 − 74.1

82.4 2


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 tothe fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and mostcould find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrongmode 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.

[4 marks]4e. Hence or otherwise, find the total area of the shaded regions.

Markschemeattempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3

[4 marks]

Total [6 marks]



In part (d), most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.

circle − ABCD π − ΔADC − ΔACBr2

ΔACB = (15.516…)(14)sin 42.9812

π(8 − 44.6336… − (15.516…)(14)sin 42.98)2 12

64π − 44.6 − 74.1

82.4 2

5a. [6 marks]

The diagram below shows part of the graph of a function .

The graph has a maximum at A( , ) and a minimum at B( , ) .

The function can be written in the form . Find the value of

(a)

(b)

(c) .

Markscheme(a) valid approach to find (M1)

eg amplitude ,

A1 N2

[2 marks]

(b) valid approach to find (M1)

eg period = 4 ,

A1 N2

[2 marks]

(c) valid approach to find (M1)

eg axis = , sketch of horizontal axis,

A1 N2

[2 marks]

Total [6 marks]

Examiners reportMany candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when theyattempted to substitute into the equation of the function with the parameters , and . The successful candidates were able to find theanswers 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 and the period of thefunction. There were also some candidates who showed working such as without explaining what represented.

f

1 5 3 −1

f f(x) = p sin(qx) + r

p

q

r

p

= max−min2

p = 6

p = 3

q

q = 2π

period

q = π

2

r

max+min2

f(0)

r = 2

p q r

q2π

bb

[2 marks]5b.

Markschemevalid approach to find (M1)

eg amplitude ,

A1 N2

[2 marks]


p

p

= max−min2

p = 6

p = 3

p q r

q2π

bb

[2 marks]5c.


eg period = 4 ,

A1 N2

[2 marks]


q

q

q = 2π

period

q = π

2

p q r

q2π

bb

[2 marks]5d. .


eg axis = , sketch of horizontal axis,

A1 N2

[2 marks]

Total [6 marks]

r

r

max+min2

f(0)

r = 2

p q r

q2π

bb

[2 marks]6a.

The following diagram shows a right-angled triangle, , where .

Show that .

MarkschemeMETHOD 1approach involving Pythagoras’ theorem (M1)eg , labelling correct sides on triangle

finding third side is 12 (may be seen on diagram) A1 AG N0

METHOD 2approach involving (M1)eg

correct working A1eg

AG N0

[2 marks]


ABC sin A = 513

cosA = 1213

+ =52 x2 132

cosA = 1213

θ + θ = 1sin2 cos2

+ θ = 1, + = 1( )513

2cos2 x2 25

169

θ =cos2 144169

cosA = 1213

[3 marks]6b. Find .

Markschemecorrect substitution into (A1)eg


A1 N2

[3 marks]


cos2A

cos2θ

1 − 2 , 2 − 1, −( )513

2( )12

13

2( )12

13

2( )5

13

2

1 − , − 1, −50169

288169

144169

25169

cos2A = 119169

[2 marks]7a.

The following diagram shows a circle with centre and radius .

The points , and lie on the circumference of the circle, and radians.

Find the length of the arc .

Markschemecorrect substitution into arc length formula (A1)eg arc length (cm) A1 N2[2 marks]


O 5 cm

A rmB rmC A C = 0.7O

ABC

0.7 × 5= 3.5

[2 marks]7b. Find the perimeter of the shaded sector.

Markschemevalid approach (M1)

eg

perimeter (cm) A1 N2

[2 marks]


3.5 + 5 + 5, arc + 2r

= 13.5

[2 marks]7c. Find the area of the shaded sector.

Markschemecorrect substitution into area formula (A1)eg

A1 N2[2 marks]


(0.7)(512

)2

area = 8.75 (c )m2

[2 marks]8a.

Let , for . The following diagram shows the graph of .

The graph has a maximum at and a minimum at .

Write down the value of .

Markscheme A2 N2

Note: Award A1 for . [2 marks]


f(x) = p cos(q(x + r)) + 10 0 ⩽ x ⩽ 20 f

(4,18) (16,2)

r

r = −4

r = 4

[2 marks]8b. Find .

Markschemeevidence of valid approach (M1)eg , distance from

A1 N2[2 marks]


p

max y value -- y value2

y = 10

p = 8

[2 marks]8c. Find .

Markschemevalid approach (M1)eg period is , , substitute a point into their

, (do not accept degrees) A1 N2

[2 marks]


q

24 36024

f(x)

q = ( , exact)2π

24π

120.262

[2 marks]8d. Solve .

Markschemevalid approach (M1)eg line on graph at

(accept ) A1 N2[2 marks] Note: Do not award the final A1 if additional values are given. If an incorrect value of leads to multiple solutions, award the final

A1 only if all solutions within the domain are given.


f(x) = 7

y = 7, 8 cos( (x − 4)) + 10 = 72π

24

x = 11.46828

x = 11.5 (11.5,7)

q

[3 marks]9a.

Let . The graph of f passes through the point .

Find the value of .

MarkschemeMETHOD 1attempt to substitute both coordinates (in any order) into (M1)

eg

correct working (A1)

eg

A1 N2

[3 marks]

METHOD 2recognizing shift of left means maximum at R1)

recognizing is difference of maximum and amplitude (A1)

eg

A1 N2

[3 marks]


f(x) = sin(x + ) + kπ

4( , 6)π

4

k

f

f ( ) = 6, = sin(6 + ) + kπ

4π

4π

4

sin = 1, 1 + k = 6π

2

k = 5

π

46

k

6 − 1

k = 5

[2 marks]9b. Find the minimum value of .

Markschemeevidence of appropriate approach (M1)eg minimum value of is

minimum value is A1 N2[2 marks]


f(x)

sin x −1, − 1 + k, (x) = 0, ( , 4)f ′ 5π

4

4

9c. [2 marks]Let . The graph of g is translated to the graph of by the vector .

Write down the value of and of .

Markscheme A1A1 N2

[2 marks]


g(x) = sin x f ( )p

q

p q

p = − , q = 5 (accept ( ))π

4

−π

4

5

[3 marks]10a.

Consider a circle with centre and radius cm. Triangle is drawn such that its vertices are on the circumference of the circle.

cm, cm and radians.

Find .

MarkschemeNotes: 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 degree 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 (M1)

eg

correct substitution (A1)

eg

A1 N2[3 marks]


O 7 ABC

AB = 12.2 BC = 10.4 A B = 1.058C

B CA

=sin A

a

sin B

b

=sin A

10.4sin 1.058

12.2

B C = 0.837A

10b. AC






METHOD 1evidence of subtracting angles from (M1)eg

correct angle (seen anywhere) A1

attempt to substitute into cosine or sine rule (M1)correct substitution (A1)eg

A1 N3METHOD 2evidence of choosing cosine rule M1eg

correct substitution (A2)eg

A2 N3[5 marks]


π

A C = π − A − CB

A C = π − 1.058 − 0.837, 1.246, B 71.4∘

+ − 2 × 12.2 × 10.4cos71.4, =12.22 10.42 ACsin 1.246

12.2sin 1.058

AC = 13.3 (cm)

= + − 2bc cosAa2 b2 c2

= + − 2 × 10.4b cos1.05812.22 10.42 b2

AC = 13.3 (cm)

[6 marks]10c. Hence or otherwise, find the length of arc .ABC






METHOD 1valid approach (M1)eg ,


(A1)EITHERcorrect substitution for arc length (seen anywhere) A1eg

subtracting arc from circumference (M1)eg

ORattempt to find reflex (M1)eg

correct substitution for arc length (seen anywhere) A1eg

THEN A1 N4

METHOD 2valid approach to find or (M1)eg choosing cos rule, twice angle at circumference

correct working for finding one value, or (A1)eg ,

two correct calculations for arc lengths

eg (A1)(A1)adding their arc lengths (seen anywhere)

eg M1 A1 N4

Note: Candidates may work with other interior triangles using a similar method. Check calculations carefully and award marks in line

with markscheme.

[6 marks]


cosA C =O O +O −AA2 C2 C2

2×OA×OCA C = 2 × A CO B

13. = + − 2 × 7 × 7 cosA C, O = 2 × 1.24632 72 72 O

A C = 2.492 ( )O 142.8∘

2.492 = , l = 17.4, 14π ×l

7142.8360

2πr − l, 14π = 17.4

A CO

2π − 2.492, 3.79, 360 − 142.8

l = 7 × 3.79, 14π × 217.2360

arc ABC = 26.5

A BO B CO

A BO B CO

cosA B =O + −72 72 12.22

2×7×7A B = 2.116,B C = 1.6745O O

AB = 7 × 2 × 1.058 (= 14.8135), 7 × 1.6745 (= 11.7216)

rA B + rB C, 14.8135 + 11.7216, 7(2.116 + 1.6745)O O

arc ABC = 26.5 (cm)

[3 marks]11a. Let . Find an expression for in terms of m.sin = m100∘ cos100∘

MarkschemeNote: 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 awardthe final A1FT mark.

METHOD 1

valid approach involving Pythagoras (M1)

e.g. , labelled diagram

correct working (may be on diagram) (A1)

e.g. ,

A1 N2

[3 marks]

METHOD 2

valid approach involving tan identity (M1)

e.g.


e.g.

A1 N2

[3 marks]

Examiners reportWhile many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of anangle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. Acommon algebraic error in part (a) was for candidates to write . In part (c), many candidates failed to use thedouble-angle identity. Many incorrectly assumed that because , then . In addition, some candidates didnot seem to understand what writing an expression "in terms of m" meant.

x + x = 1sin2 cos2

+ (cos100 = 1m2 )2 1 − m2− −−−−−√

cos100 = − 1 − m2− −−−−−√

tan = sincos

cos100 = sin 100tan 100

cos100 = m

tan 100

= 1 − m1 − m2− −−−−−√sin = m100∘ sin = 2m200∘

[1 mark]11b. Let . Find an expression for in terms of m.

MarkschemeMETHOD 1

(accept ) A1 N1

[1 mark]

METHOD 2

A1 N1

[1 mark]

sin = m100∘ tan 100∘

tan100 = − m

1−m2√m

− 1−m2√

tan100 = m

cos 100


= 1 − m1 − m2− −−−−−√sin = m100∘ sin = 2m200∘

[2 marks]11c. Let . Find an expression for in terms of m.

MarkschemeMETHOD 1

valid approach involving double angle formula (M1)

e.g.

(accept ) A1 N2

Note: If candidates find , award full FT in parts (b) and (c), even though the values may not have appropriatesigns for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula (M1)

e.g. ,

A1 N2

[2 marks]


sin = m100∘ sin 200∘

sin 2θ = 2 sin θcosθ

sin 200 = −2m 1 − m2− −−−−−√ 2m (− )1 − m2− −−−−−√

cos100 = 1 − m2− −−−−−√

sin 2θ = 2 sin θ cosθ 2m × m

tan 100

sin 200 = (= 2m cos100)2m2

tan 100

= 1 − m1 − m2− −−−−−√sin = m100∘ sin = 2m200∘

12a. [2 marks]

Let . The diagram below shows part of the graph of f , for .

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

(i) ;

(ii) .

Markscheme(i) (accept ) A1 N1

(ii) (accept , if ) A1 N1

Note: Accept other correct values of c, such as 11, , etc.

[2 marks]

Examiners reportPart (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c.

f(x) = acos(b(x − c)) 0 ≤ x ≤ 10

a

c

a = 5 −5

c = 3 c = 7 a = −5

−5

[2 marks]12b. Find the value of b .

Markschemeattempt to find period (M1)

e.g. 8 ,

(exact), , 0.785 [ ] (do not accept 45) A1 N2

[2 marks]

Examiners reportFinding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8.

b = 2π

period

0.785398…

b = 2π

8π

40.785, 0.786

[2 marks]12c. Find the x-coordinate of R.


e.g. , symmetry of curve

(accept A1 N2

[2 marks]

Examiners reportMost candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve althoughfollow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down theequations they entered into their GDC.

f(x) = 0

x = 5 (5 ,0))

[3 marks]13a.

The following diagram shows a circular play area for children.

The circle has centre O and a radius of 20 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].

MarkschemeNote: In this question, do not penalise for missing or incorrect units. They are not included in the markscheme, to avoid complexanswer lines.

METHOD 1

choosing cosine rule (must have cos in it) (M1)

e.g.

correct substitution (into rhs) A1

e.g. ,

A1 N2

[3 marks]

METHOD 2

choosing sine rule (M1)

e.g. ,


e.g.

A1 N2

[3 marks]

Examiners reportCandidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors inaccuracy.

= + − 2ab cosCc2 a2 b2

+ − 2(20)(20)cos1.5202 202 AB = 800 − 800 cos1.5− −−−−−−−−−−−−√

AB = 27.26555…

AB = 27.3 [27.2, 27.3]

=sin A

a

sin B

b=AB

sin O

AOsin B

=ABsin 1.5

20sin(0.5(π−1.5))

AB = 27.26555…

AB = 27.3 [27.2, 27.3]

[2 marks]13b. Find the area of triangle AOB.

Markschemecorrect substitution into area formula A1

e.g. ,

(accept , from using 27.3)

A1 N1

[2 marks]

Examiners reportCandidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors inaccuracy.

(20)(20)sin 1.512

(20)(27.2655504…)sin(0.5(π − 1.5))12

area = 199.498997… 199.75106 = 200

area = 199 [199, 200]

13c. [3 marks]Angle BOC is 2.4 radians.

Find the length of arc ADC.

Markschemeappropriate method to find angle AOC (M1)

e.g.

correct substitution into arc length formula (A1)

e.g. ,

(i.e. do not accept ) A1 N2

Notes: Candidates may misread the question and use . If working shown, award M0 then A0MRA1 for the answer 48.Do not then penalize in part (d) which, if used, leads to the answer

However, if they use the prematurely rounded value of 2.4 for , penalise 1 mark for premature rounding for the answer 48 in(c). Do not then penalize for this in (d).

[3 marks]

Examiners reportIn part (c), some candidates misread the question and used 2.4 as the size of angle AOC while others rounded prematurely leading tothe inaccurate answer of 48. In either case, marks were lost.

2π − 1.5 − 2.4

(2π − 3.9) × 20 2.3831853… × 20

arc length = 47.6637…

arc length = 47.7 (47.6, 47.7] 47.6

A C = 2.4OA CO 679.498…

A CO

13d. [3 marks]Angle BOC is 2.4 radians.

Find the area of the shaded region.

Markschemecalculating sector area using their angle AOC (A1)

e.g. , ,

shaded area = their area of triangle AOB + their area of sector (M1)

e.g. ,

(accept from using 199)

A1 N2

[3 marks]

Examiners reportPart (d) proved to be straightforward and candidates were able to obtain full FT marks from errors made in previous parts.

(2.38…)( )12

202 200(2.38…) 476.6370614…

199.4989973… + 476.6370614… 199 + 476.637

shaded area = 676.136… 675.637… = 676

shaded area = 676 [676, 677]

13e. [4 marks]Angle BOC is 2.4 radians.

The shaded region is to be painted red. Red paint is sold in cans which cost each. One can covers . How much does it cost tobuy the paint?

$32 140 m2

Markschemedividing to find number of cans (M1)

e.g. ,

5 cans must be purchased (A1)

multiplying to find cost of cans (M1)

e.g. ,

cost is 160 (dollars) A1 N3

[4 marks]

Examiners reportMost candidates had a suitable strategy for part (e) and knew to work with a whole number of cans of paint.

676140

4.82857…

5(32) × 32676140

14a. [3 marks]

The following diagram shows the graph of , for .

There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .

(i) Write down the value of a .

(ii) Find the value of b .

Markscheme(i) A1 N1

(ii) METHOD 1

attempt to find period (M1)

e.g. 4 , ,

A1 N2

[3 marks]

METHOD 2

attempt to substitute coordinates (M1)

e.g. ,

A1 N2

[3 marks]

f(x) = acos(bx) 0 ≤ x ≤ 4

a = 3

b = 4 2π

b

b = (= )2π

4π

2

3 cos(2b) = −3 3 cos(4b) = 3

b = (= )2π

4π

2

Examiners reportIn part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing theamplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, therewere many candidates who thought b was the period of the function, rather than .2π

period

[1 mark]14b. Write down the gradient of the curve at P.

Markscheme0 A1 N1

[1 mark]

Examiners reportIn part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, manycandidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given valueof x, rather than simply stating that the gradient of a curve at a minimum point is zero.

[2 marks]14c. Write down the equation of the normal to the curve at P.

Markschemerecognizing that normal is perpendicular to tangent (M1)

e.g. , , sketch of vertical line on diagram

(do not accept 2 or ) A1 N2

[2 marks]

Examiners reportFor part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involvingnegative reciprocal gradients, rather than recognizing that the equation of the vertical line was . There were also candidateswho had trouble expressing the correct equation of a line parallel to the y-axis.

× = −1m1 m2 m = − 10

x = 2 y = 2

x = 2

[2 marks]15a.

The diagram below shows part of the graph of , where .

The point is a maximum point and the point is a minimum point.

Find the value of a .

Markschemeevidence of valid approach (M1)

e.g. , distance from

A1 N2

[2 marks]

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph.

f(x) = acos(b(x − c)) − 1 a > 0

P ( ,2)π

4Q( ,−4)3π

4

max y value−min y value2

y = −1

a = 3

15b. [4 marks](i) Show that the period of f is .

(ii) Hence, find the value of b .

Markscheme(i) evidence of valid approach (M1)

e.g. finding difference in x-coordinates,

evidence of doubling A1

e.g.

AG N0

(ii) evidence of valid approach (M1)

e.g.

A1 N2

[4 marks]

π

π

2

2 × ( )π

2

period = π

b = 2π

π

b = 2

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had troubleshowing that the period was , either incorrectly adding the given and or using the value of b that they found first for part(b)(ii).

π π/4 3π/4

[1 mark]15c. Given that , write down the value of c .

Markscheme A1 N1

[1 mark]

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had troubleshowing that the period was , either incorrectly adding the given and or using the value of b that they found first for part(b)(ii).

0 < c < π

c = π

4

π π/4 π/3

[2 marks]16a.

Let .

Show that can be expressed as .

Markschemeattempt to expand (M1)

e.g. ; at least 3 terms

correct expansion A1

e.g.

AG N0

[2 marks]

Examiners reportSimplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidateswere certainly helped by the fact that it was a "show that" question.

f(x) = (sin x + cosx)2

f(x) 1 + sin 2x

(sin x + cosx)(sin x + cosx)

x + 2 sin xcosx + xsin2 cos2

f(x) = 1 + sin 2x

16b. [2 marks]The graph of f is shown below for .

Let . On the same set of axes, sketch the graph of g for .

Markscheme

A1A1 N2

Note: Award A1 for correct sinusoidal shape with period and range , A1 for minimum in circle.

Examiners reportMore candidates had difficulty with part (b) with many assuming the first graph was and hence sketching a horizontaltranslation of for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor pcorrectly or from follow-through on their own graph, very few successfully found the value and direction for the translation.

0 ≤ x ≤ 2π

g(x) = 1 + cosx 0 ≤ x ≤ 2π

2π [0, 2]

1 + sin(x)π/2

16c. [2 marks]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 .

Write down the value of p and a possible value of k .

Markscheme , A1A1 N2

[2 marks]

( )k

0

p = 2 k = −π

2

Examiners reportPart (c) certainly served as a discriminator between the grade 6 and 7 candidates.

[2 marks]17a.

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. Thebottom of the wheel is metres above the ground.

A seat starts at the bottom of the wheel.

Find the maximum height above the ground of the seat.


eg ,

maximum height (m) A1 N2

[2 marks]

Examiners reportMost candidates were successful with part (a).

122 2.413

13 + diameter 13 + 122

= 135

17b. [2 marks]

After t minutes, the height metres above the ground of the seat is given by

(i) Show that the period of is minutes.

(ii) Write down the exact value of .

h

h = 74 + acosbt.

h 25

b

= 602.4

= 25

b = 2π

25(= 0.08π)

Examiners reportA surprising number had difficulty producing enough work to show that the period was ; writing down the exactvalue of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on theFerris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seenwith correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show keyfeatures in approximately correct locations and candidates lost marks for incorrect domains and ranges.

25b

t = 0 61

17c. [9 marks](b) (i) Show that the period of is minutes.

(ii) Write down the exact value of .

(c) Find the value of .

(d) Sketch the graph of , for .

h 25

b

a

h 0 ≤ t ≤ 50

= 602.4

= 25

b = 2π

25(= 0.08π)

max − 74 |a| = 135−132

74 − 13

|a| = 61 a = 61

a = −61

135 = 74 + acos( )2π×12.525

135 = 74 + acos(π) 13 = 74 + a

a = −61

2


25b

t = 0 61

[3 marks]17d. Find the value of .

MarkschemeMETHOD 1

valid approach (M1)

eg , ,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1)

eg

correct equation (A1)

eg ,

A1 N2

[3 marks]


a

max − 74 |a| = 135−132

74 − 13

|a| = 61 a = 61

a = −61

135 = 74 + acos( )2π×12.525

135 = 74 + acos(π) 13 = 74 + a

a = −61

25b

t = 0 61

[4 marks]17e. Sketch the graph of , for .h 0 ≤ t ≤ 50

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]


2

25b

t = 0 61

[5 marks]17f. In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.

Markschemesetting up inequality (accept equation) (M1)

eg , , sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg , , ,

valid approach M1

eg , , ,

A1 N2

[5 marks]

Examiners reportPart (e) was very poorly done for those who attempted the question and most did not make the connection between height, time andprobability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novelapproaches from the strongest of candidates using circles and angles to solve this part of question 10.

105

h > 105 105 = 74 + acosbt y = 105

t = 8.371… t = 16.628… t = 33.371… t = 41.628…

16.628−8.37125

−t1 t2

252×8.257

502(12.5−8.371)

25

p = 0.330

[3 marks]18a.

Let , where .

Find .

sin θ = 213√

< θ < ππ

2

cosθ

MarkschemeMETHOD 1

evidence of choosing (M1)


e.g. , ,

A1 N2

Note: If no working shown, award N1 for .

METHOD 2

approach involving Pythagoras’ theorem (M1)

e.g. ,

finding third side equals 3 (A1)

A1 N2

Note: If no working shown, award N1 for .

[3 marks]

Examiners reportWhile the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle inthe second quadrant will have a negative value.

θ + θ = 1sin2 cos2

θ =cos2 913

cosθ = ± 313√

cosθ = 913

−−√cosθ = − 3

13√

313√

+ = 1322 x2

cosθ = − 313√

313√

[5 marks]18b. Find .

Markschemecorrect substitution into (seen anywhere) (A1)

e.g.

correct substitution into (seen anywhere) (A1)

e.g. , ,

valid attempt to find (M1)

e.g. ,

correct working A1

e.g. , ,

A1 N4

Note: If students find answers for which are not in the range , award full FT in (b) for correct FT working shown.

[5 marks]

tan2θ

sin 2θ

2 ( ) (− )213√

313√

cos2θ

−(− )313√

2 ( )213√

22 − 1(− )3

13√

21 − 2( )2

13√

2

tan2θ

2( )(− )2

13√

3

13√

−(− )3

13√

2 ( )2

13√

2

2(− )23

1−(− )23

2

(2)(2)(−3)

13

−913

413

− 12

( )13√ 2

−11813

− 1213

513

tan2θ = − 125

cosθ [−1, 1]

Examiners reportIn part (b), many candidates incorrectly tried to calculate as , rather than using the double-angle identities.tan2θ 2 × tanθ

19a. [3 marks]

The following diagram shows the graph of , for .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i) a ;

(ii) c ;

(iii) d .

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

Examiners reportPart (a) of this question proved challenging for most candidates.

f(x) = asin(b(x − c)) + d 2 ≤ x ≤ 10

a = 8

c = 2

d = 4

[2 marks]19b. Show that .b = π

4

MarkschemeMETHOD 1

recognizing that period (A1)

correct working A1

e.g. ,

AG N0

METHOD 2

attempt to substitute M1

e.g.

correct working A1

e.g.

AG N0

[2 marks]

Examiners reportAlthough a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize thatthis period could be found using the given coordinates of the maximum and minimum points.

= 8

8 = 2π

bb = 2π

8

b = π

4

12 = 8 sin(b(4 − 2)) + 4

sin 2b = 1

b = π

4

[3 marks]19c. Find .

Markschemeevidence of attempt to differentiate or choosing chain rule (M1)

e.g. ,

(accept ) A2 N3

[3 marks]

Examiners reportIn part (c), not many candidates found the correct derivative using the chain rule.

(x)f ′

cos (x − 2)π

4× 8π

4

(x) = 2πcos( (x − 2))f ′ π

42πcos (x − 2)π

4

[6 marks]19d. At a point R, the gradient is . Find the x-coordinate of R.−2π

Markschemerecognizing that gradient is (M1)

e.g.

correct equation A1

e.g. ,


e.g.

using (seen anywhere) (A1)

e.g.

simplifying (A1)

e.g.

A1 N4

[6 marks]

Examiners reportFor part (d), a good number of candidates correctly set their expression equal to , but errors in their previous values kept mostfrom correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

(x)f ′

(x) = mf ′

−2π = 2πcos( (x − 2))π

4−1 = cos( (x − 2))π

4

(−1) = (x − 2)cos−1 π

4

(−1) = πcos−1

π = (x − 2)π

4

4 = (x − 2)

x = 6

−2π

20a. [2 marks]

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 ananticlockwise (counter-clockwise) direction. One revolution takes 20 minutes.

Write down the height of P above ground level after

(i) 10 minutes;

(ii) 15 minutes.

Markscheme(i) 100 (metres) A1 N1

(ii) 50 (metres) A1 N1

[2 marks]

Examiners reportNearly all candidates answered part (a) correctly, finding the height of the wheel at and of a revolution.1

234

20b. [4 marks]

Let metres be the height of P above ground level after t minutes. Some values of are given in the table below.

(i) Show that .

(ii) Find .

Markscheme(i) identifying symmetry with (M1)

subtraction A1

e.g. ,

AG N0

(ii) recognizing period (M1)

e.g.

A1 N2

[4 marks]

Examiners reportWhile many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function forheight, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table.

h(t) h(t)

h(8) = 90.5

h(21)

h(2) = 9.5

100 − h(2) 100 − 9.5

h(8) = 90.5

h(21) = h(1)

h(21) = 2.4

[3 marks]20c. Sketch the graph of h , for .0 ≤ t ≤ 40

0 ≤ h ≤ 100

Examiners reportIn 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 wereconstructed of line segments, rather than a curve.

[5 marks]20d. Given that h can be expressed in the form , find a , b and c .

Markschemeevidence of a quotient involving 20, or to find b (M1)

e.g. ,

(accept if working in degrees) A1 N2

, A2A1 N3

[5 marks]

Examiners reportFor part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accuratesketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that theposition on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidatesrecognized that the value of a would be negative. Candidates should be aware that while working in degrees may be acceptable, theexpectation is that radians will be used in these types of questions.

h(t) = acosbt + c

2π 360∘

= 202π

bb = 360

20

b = 2π

20(= )π

10b = 18

a = −50 c = 50

[2 marks]21a. Show that .

Markschemeattempt to substitute for (M1)


e.g.

AG N0

[2 marks]

Examiners reportIn part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidateswho 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.

4 − cos2θ + 5 sin θ = 2 θ + 5 sin θ + 3sin2

1 − 2 θsin2 cos2θ

4 − (1 − 2 θ) + 5 sin θsin2

4 − cos2θ + 5 sin θ = 2 θ + 5 sin θ + 3sin2

[5 marks]21b. Hence, solve the equation for .4 − cos2θ + 5 sin θ = 0 0 ≤ θ ≤ 2π

Markschemeevidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

correct working A1

e.g. , ,

correct solution (do not penalise for including (A1)

A2 N3

[5 marks]

Examiners reportIn part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring thequadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solutionto the equation , and there were a few who did not realize that the equation has no solution.

(2 sin θ + 3)(sin θ + 1) (2x + 3)(x + 1) = 0 sin x = −5± 1√4

sin θ = −1 sin θ = − 32

θ = 3π

2

sin θ = −1 sin θ = − 32

[5 marks]22a.

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is , where .

Show that the area of the window is given by .

Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g. ,

attempt to replace by M1

e.g.

AG N0

[5 marks]

Examiners reportAs the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) wasgenerally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in aclear and organized manner. Some tried a "working backwards" approach, earning no marks.

2 m θ 0 < θ < π

2

y = 4 sin θ + 2 sin 2θ

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

[4 marks]22b. Zoe wants a window to have an area of . Find the two possible values of .

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

, A1A1 N3

[4 marks]

Examiners reportIn part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and insteadattempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a secondsolution outside the domain.

5 m2 θ

y = 5 4 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856 ( )49.0∘ θ = 1.25 ( )71.4∘

22c. [7 marks]John wants two windows which have the same area A but different values of .

Find all possible values for A .

Markschemerecognition that lower area value occurs at (M1)

finding value of area at (M1)

e.g. , draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

Examiners reportA pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or themaximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident fromcandidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken intoconsideration for future paper writing.

θ

θ = π

2

θ = π

2

4 sin( ) + 2 sin(2 × )π

2π

2

A = 4

A = 5.19615…

4 < A < 5.20 4 < A < 5.19

[3 marks]23a.

Let , . Let .

Find an expression for .

Markschemeattempt to form any composition (even if order is reversed) (M1)

correct composition (A1)

A1 N3

[3 marks]

f(x) = + 13x

2g(x) = 4 cos( ) − 1x

3h(x) = (g ∘ f)(x)

h(x)

h(x) = g ( + 1)3x

2

h(x) = 4 cos( ) − 1+13x

2

3(4 cos( x + ) − 1,4 cos( ) − 1)1

213

3x+26

Examiners reportThe majority of candidates handled the composition of the two given functions well. However, a large number of candidates haddifficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidatesknew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the periodand range, but this approach was not the most efficient.

[1 mark]23b. Write down the period of .

Markschemeperiod is A1 N1

[1 mark]


h

4π(12.6)

[2 marks]23c. Write down the range of .

Markschemerange is A1A1 N2

[2 marks]


h

−5 ≤ h(x) ≤ 3 ([−5,3])

[3 marks]24a.

Let , for .

Sketch the graph of f .

f(x) = 3 sin x + 4 cosx −2π ≤ x ≤ 2π

Markscheme

A1A1A1 N3

Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct , A1 forapproximately correct position of graph, (y-intercept , maximum to right of y-axis).

[3 marks]

Examiners reportSome graphs in part (a) were almost too detailed for just a sketch but more often, the important features were far from clear. Somegraphs lacked scales on the axes.

(−2π, 4) (2π, 4)(0, 4)

24b. [3 marks]Write down

(i) the amplitude;

(ii) the period;

(iii) the x-intercept that lies between and 0.

Markscheme(i) 5 A1 N1

(ii) (6.28) A1 N1

(iii) A1 N1

[3 marks]

Examiners reportA number of candidates had difficulty finding the period in part (b)(ii).

−π

2

2π

−0.927

[3 marks]24c. Hence write in the form .

Markscheme (accept , , ) A1A1A1 N3

[3 marks]

Examiners reportA number of candidates had difficulty writing the correct value of q in part (c).

f(x) p sin(qx + r)

f(x) = 5 sin(x + 0.927) p = 5 q = 1 r = 0.927

[2 marks]24d. Write down one value of x such that .

Markschemeevidence of correct approach (M1)

e.g. max/min, sketch of indicating roots

one 3 s.f. value which rounds to one of , , , A1 N2

[2 marks]

Examiners reportThe most common approach in part (d) was to differentiate and set . Fewer students found the values of x given by themaximum or minimum values on their graphs.

(x) = 0f ′

(x)f ′

−5.6 −2.5 0.64 3.8

(x) = 0f ′

[2 marks]24e. Write down the two values of k for which the equation has exactly two solutions.

Markscheme , A1A1 N2

[2 marks]

Examiners reportPart (e) proved challenging for many candidates, although if candidates answered this part, they generally did so correctly.

f(x) = k

k = −5 k = 5

24f. [5 marks]Let , for . There is a value of x, between and , for which the gradient of f is equal to thegradient of g. Find this value of x.

g(x) = ln(x + 1) 0 ≤ x ≤ π 0 1

MarkschemeMETHOD 1

graphical approach (but must involve derivative functions) M1

e.g.

each curve A1A1

A2 N2

METHOD 2

A1

A1

evidence of attempt to solve M1

A2 N2

[5 marks]

Examiners reportIn part (f), many candidates were able to get as far as equating the two derivatives but fewer used their GDC to solve the resultingequation. Again, many had trouble demonstrating their method of solution.

x = 0.511

(x) =g′ 1x+1

(x) = 3 cosx − 4 sin xf ′ (5 cos(x + 0.927))

(x) = (x)g′ f ′

x = 0.511

25a. [6 marks]

The graph of , for , is shown below.

There is a minimum point at (0, −3) and a maximum point at (4, 7) .

Find the value of

(i) p ;

(ii) q ;

(iii) r.

y = p cosqx + r −5 ≤ x ≤ 14

Markscheme(i) evidence of finding the amplitude (M1)

e.g. , amplitude

A1 N2

(ii) period (A1)

A1 N2

(iii) (A1)

A1 N2

[6 marks]

Examiners reportMany candidates did not recognize that the value of p was negative. The value of q was often interpreted incorrectly as the period butmost candidates could find the value of r, the vertical translation.

7+32

= 5

p = −5

= 8

q = 0.785 (= = )2π

8π

4

r = 7−32

r = 2

[1 mark]25b. The equation has exactly two solutions. Write down the value of k.

Markscheme (accept ) A1 N1

[1 mark]

Examiners reportIn part (b), candidates either could not find a solution or found too many.

y = k

k = −3 y = −3

[1 mark]26a.

The diagram below shows a quadrilateral ABCD with obtuse angles and .

AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , , , .

Use the cosine rule to show that .

Markschemecorrect substitution A1

e.g. ,

AG

[1 mark]

A CB A CD

B C =A 30∘ A C =B x∘ A C =D y∘

AC = 41 − 40 cosx− −−−−−−−−−√

25 + 16 − 40 cosx + − 2 × 4 × 5 cosx52 42

AC = 41 − 40 cosx− −−−−−−−−−√

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b).

[2 marks]26b. Use the sine rule in triangle ABC to find another expression for AC.

Markschemecorrect substitution A1

e.g. ,

(accept ) A1 N1

[2 marks]

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b). Equally as many did not take the cue from theword "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set upan equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usuallyunsuccessful.

=ACsin x

4sin 30

AC = 4 sin x12

AC = 8 sin x 4 sin x

sin 30

26c. [6 marks](i) Hence, find x, giving your answer to two decimal places.

(ii) Find AC .

Markscheme(i) evidence of appropriate approach using AC M1

e.g. , sketch showing intersection

correct solution , (A1)

obtuse value (A1)

to 2 dp (do not accept the radian answer 1.94 ) A1 N2

(ii) substituting value of x into either expression for AC (M1)

e.g.

A1 N2

[6 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive 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 recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

8 sin x = 41 − 40 cosx− −−−−−−−−−√

8.682… 111.317…

111.317…

x = 111.32

AC = 8 sin 111.32

AC = 7.45

26d. [5 marks](i) Find y.

(ii) Hence, or otherwise, find the area of triangle ACD.

Printed for Colegio Aleman de Barranquilla

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme(i) evidence of choosing cosine rule (M1)

e.g.


e.g. , ,

A1 N2

(ii) correct substitution into area formula (A1)

e.g. ,

area A1 N2

[5 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive 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 recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

cosB = + −a2 c2 b2

2ac

+ −42 42 7.452

2×4×4= 32 − 32 cosy7.452 cosy = −0.734…

y = 137

× 4 × 4 × sin 13712

8 sin 137

= 5.42

new test - february 23, 2016 · 2015-07-10 · new test - february 23, 2016 [276 marks] [2 marks]...

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