IB Questionbank Test · Web viewevidence of choosing cosine rule (M1) eg correct substitution into...

IB Questionbank Test

SL Week 2 Revision – Trigonometry Mark Scheme

1a. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

1b. [3 marks]

Markscheme

recognizing that sinθ is bounded (M1)

eg 0 ≤ sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

1c. [4 marks]

Markscheme

correct substitution into formula for infinite sum A1

eg

evidence of choosing an appropriate rule for cos 2θ (seen anywhere) (M1)

eg cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere) (A1)

eg

correct working that clearly leads to the given answer A1

eg

AG N0

[4 marks]

1d. [6 marks]

Markscheme

METHOD 1 (using differentiation)

recognizing (seen anywhere) (M1)

finding any correct expression for (A1)

eg

correct working (A1)

eg sin 2θ = 0

any correct value for sin−1(0) (seen anywhere) (A1)

eg 0, , … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values) (A1)

2θ = , 3 (accept values in degrees)

both correct answers A1 N4

Note: Award A0 if either or both correct answers are given in degrees.Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S∞ is greatest (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallestcorrect working (A1)

eg minimum value of 2 + cos 2θ is 1, minimum r =


eg

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere) (A1)

eg , 3, … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ (ignore additional values) (A1)

2θ = , 3 (accept values in degrees)

OR (using sinθ)

sinθ = ±1 (A1)

sin−1(1) = (accept values in degrees) (seen anywhere) A1

THEN

both correct answers A1 N4

Note: Award A0 if either or both correct answers are given in degrees.Award A0 if additional values are given.

[6 marks]

2. [6 marks]

Markscheme

METHOD 1 (using |p| |2q| cosθ)

finding p + q + r (A1)

eg 2q,

| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1

correct angle between p and q (seen anywhere) (A1)

(accept 60°)

substitution of their values (M1)

eg 3 × 6 × cos

correct value for cos (seen anywhere) (A1)

eg

p•(p + q + r) = 9 A1 N3

METHOD 2 (scalar product using distributive law)

correct expression for scalar distribution (A1)

eg p• p + p•q + p•r

three correct angles between the vector pairs (seen anywhere) (A2)

eg 0° between p and p, between p and q, between p and r

Note: Award A1 for only two correct angles.

substitution of their values (M1)

eg 3.3.cos0 +3.3.cos + 3.3.cos120

one correct value for cos0, cos or cos (seen anywhere) A1

eg

p•(p + q + r) = 9 A1 N3

METHOD 3 (scalar product using relative position vectors)

valid attempt to find one component of p or r (M1)

eg sin 60 = , cos 60 = , one correct value

one correct vector (two or three dimensions) (seen anywhere) A1

eg

three correct vectors p + q + r = 2q (A1)

p + q + r = or (seen anywhere, including scalar product) (A1)

correct working (A1)eg

p•(p + q + r) = 9 A1 N3

[6 marks]

3. [7 marks]

Markscheme

evidence of correctly substituting into circle formula (may be seen later) A1A1eg

attempt to eliminate one variable (M1)eg

correct elimination (A1)eg

correct equation (A1)eg

correct working (A1)eg

r = 4 (cm) A1 N2

[7 marks]

4a. [2 marks]

Markscheme

−0.394791,13

A(−0.395, 13) A1A1 N2

[2 marks]

4b. [1 mark]

Markscheme

13 A1 N1

[1 mark]

4c. [1 mark]

Markscheme

, 6.28 A1 N1

[1 mark]

4d. [3 marks]

Markscheme

valid approach (M1)

eg recognizing that amplitude is p or shift is r

(accept p = 13, r = 0.395) A1A1 N3

Note: Accept any value of r of the form

[3 marks]

4e. [3 marks]

Markscheme

recognizing need for d ′(t) (M1)

eg −12 sin(t) − 5 cos(t)

correct approach (accept any variable for t) (A1)

eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32

maximum speed = 13 (cms−1) A1 N2

[3 marks]

4f. [5 marks]

Markscheme

recognizing that acceleration is needed (M1)

eg a(t), d "(t)

correct equation (accept any variable for t) (A1)

eg

valid attempt to solve their equation (M1)

eg sketch, 1.33

1.02154

1.02 A2 N3

[5 marks]

5a. [3 marks]

Markscheme

valid approach to find area of segment (M1)

eg area of sector – area of triangle,

correct substitution (A1)

eg

area = 80 – 8 sinθ, 8(θ – sinθ) A1 N2

[3 marks]

5b. [3 marks]

Markscheme

setting their area expression equal to 12 (M1)

eg 12 = 8(θ – sinθ)

2.26717

θ = 2.27 (do not accept an answer in degrees) A2 N3

[3 marks]

6. [7 marks]

Markscheme

correct substitution into the formula for area of a triangle (A1)

eg 15 = × 8.1 × 12.3 × sin C

correct working for angle C (A1)

eg sin C = 0.301114, 17.5245…, 0.305860

recognizing that obtuse angle needed (M1)

eg 162.475, 2.83573, cos C < 0

evidence of choosing the cosine rule (M1)

eg a2 = b2 + c2 − 2bc cos(A)

correct substitution into cosine rule to find c (A1)

eg c2 = (8.1)2 + (12.3)2 − 2(8.1)(12.3) cos C

c = 20.1720 (A1)

8.1 + 12.3 + 20.1720 = 40.5720

perimeter = 40.6 A1 N4

[7 marks]

7a. [3 marks]

Markscheme

evidence of choosing sine rule (M1)

eg

correct substitution (A1)eg

9.57429

DB = 9.57 (cm) A1 N2

[3 marks]

7b. [3 marks]

Markscheme

evidence of choosing cosine rule (M1)

eg

correct substitution into RHS (A1)eg

10.5677

DC = 10.6 (cm) A1 N2

[3 marks]

8a. [2 marks]

Markscheme

valid approach to find k (M1)

eg 8 minutes is half a turn, k + diameter, k + 111 = 117

k = 6 A1 N2

[2 marks]

8b. [3 marks]

Markscheme

METHOD 1

valid approach (M1)eg a = radius

(A1)

a = −55.5 A1 N2

METHOD 2

attempt to substitute valid point into equation for f (M1)eg h(0) = 6, h(8) = 117

correct equation (A1)eg

a = −55.5 A1 N2

[3 marks]

8c. [3 marks]

Markscheme

valid approach (M1)eg sketch of h and

18.4630

t = 18.5 (minutes) A1 N3

[3 marks]

9a. [2 marks]

Markscheme

valid approach (M1)

eg (7, 4, 9) − (3, 2, 5) A − B

4i + 2j + 4k A1 N2

[2 marks]

9b. [2 marks]

Markscheme

correct substitution into magnitude formula (A1)eg

A1 N2

[2 marks]

9c. [4 marks]

Markscheme

finding scalar product and magnitudes (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR =

correct substitution of their values to find cos M1

eg cos

0.581746

= 0.582 radians or = 33.3° A1 N3

[4 marks]

9d. [2 marks]

Markscheme

correct substitution (A1)eg

area is 11.2 (sq. units) A1 N2

[2 marks]

9e. [3 marks]

Markscheme

recognizing shortest distance is perpendicular distance from R to line through P and Q (M1)

eg sketch, height of triangle with base


eg

3.72677

distance = 3.73 (units) A1 N2

[3 marks]

10a. [4 marks]

Markscheme

evidence of choosing the cosine rule (M1)

eg

correct substitution into RHS of cosine rule (A1)

eg

evidence of correct value for (may be seen anywhere, including in cosine rule) A1

eg

correct working clearly leading to answer A1

eg

AG N0

Note: Award no marks if the only working seen is or (or similar).

[4 marks]

10b. [3 marks]

Markscheme

correct substitution for semicircle (A1)

eg

valid approach (seen anywhere) (M1)

eg

A1 N2

[3 marks]

11. [7 marks]

Markscheme

attempt to find the area of OABC (M1)

eg

correct expression for area in one variable (A1)

eg

valid approach to find maximum area (seen anywhere) (M1)

eg

correct derivative A1

eg


eg

A2 N3

[7 marks]

12a. [3 marks]

Markscheme


eg


eg

4.13102

A1 N2

[3 marks]

12b. [3 marks]

Markscheme


eg, 18°,

correct substitution into area formula (A1)

eg

3.19139

A1 N2

[3 marks]

13a. [3 marks]

Markscheme

substituting M1

eg

(A1)

A1

AG N0

[3 marks]

13b. [3 marks]

Markscheme

substituting the value of (M1)

A1A1 N3

[3 marks]

13c. [3 marks]

Markscheme

attempt to find the gradient (M1)

eg


eg

y = x A1 N3

[3 marks]

13d. [2 marks]

Markscheme

subtracting -coordinates of and (in any order) (M1)

eg

correct working (must be in correct order) A1

eg

distance is AG N0

[2 marks]

13e. [6 marks]

Markscheme

METHOD 1

recognizing the toothed-edge as the hypotenuse (M1)

eg, sketch

correct working (using their equation of (A1)

eg

(exact), 212.132 (A1)

dividing their value of by (M1)

eg

33.7618 (A1)

33 (teeth) A1 N2

METHOD 2

vertical distance of a tooth is (may be seen anywhere) (A1)

attempt to find the hypotenuse for one tooth (M1)

eg

(exact), 8.88576 (A1)

dividing 300 by their value of (M1)

eg

33.7618 (A1)

33 (teeth) A1 N2

[6 marks]

14a. [6 marks]

Markscheme

evidence of summing to 1 (M1)

eg

correct equation A1

eg

correct equation in A1

eg

evidence of valid approach to solve quadratic (M1)

egfactorizing equation set equal to

correct working, clearly leading to required answer A1

eg

correct reason for rejecting R1

eg is a probability (value must lie between 0 and 1),

Note: Award R0 for without a reason.

AG N0

14b. [3 marks]

Markscheme

valid approach (M1)

egsketch of right triangle with sides 3 and 4,

correct working

(A1)

egmissing side

A1 N2

[3 marks]

14c. [6 marks]

Markscheme

attempt to substitute either limits or the function into formula involving (M1)

eg

correct substitution of both limits and function (A1)

eg

correct integration (A1)

eg

substituting their limits into their integrated function and subtracting (M1)

eg

Note: Award M0 if they substitute into original or differentiated function.

(A1)

eg

A1 N3

[6 marks]

15. [6 marks]

Markscheme

METHOD 1

evidence of choosing the sine rule (M1)

eg

correct substitution A1

eg

(A1)(A1)

correct working A1

eg

correct answer A1 N3

eg

METHOD 2 (using height of ΔPQR)

valid approach to find height of ΔPQR (M1)

eg

(A1)

A1

correct working A1

eg


eg

correct answer A1 N3

eg

[6 marks]

16. [7 marks]

Markscheme

correct application of (A1)

eg

correct equation without logs A1

eg

recognizing double-angle identity (seen anywhere) A1

eg

evaluating (A1)

correct working A1

egand , one correct final answer

(do not accept additional values) A2 N0

[7 marks]

17a. [2 marks]

Markscheme

recognizing at A (M1)

A is A1 N2

[2 marks]

17b. [3 marks]

Markscheme

METHOD 1

valid approach (M1)

eg

correct approach to find (A1)

eg

A1 N2

METHOD 2

recognizing is two times the direction vector (M1)


eg

A1 N2

[3 marks]

17c. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

17d. [5 marks]

Markscheme

METHOD 1 (vector approach)

valid approach involving and (M1)

eg

finding scalar product and (A1)(A1)

scalar product

substitution of their scalar product and magnitudes into cosine formula (M1)

eg

A1 N2

METHOD 2 (triangle approach)

valid approach involving cosine rule (M1)

eg

finding lengths AC and BC (A1)(A1)

substitution of their lengths into cosine formula (M1)

eg

A1 N2

[5 marks]

17e. [4 marks]

Markscheme

Note: Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

METHOD 1 (using cosine rule)

recognizing need to find BC (M1)

choosing cosine rule (M1)

eg

correct substitution into RHS A1

eg

distance is 9 A1 N2

METHOD 2 (finding magnitude of )


valid approach (M1)

egattempt to find or , or

correct working A1

eg

distance is 9 A1 N2

METHOD 3 (finding coordinates and using distance formula)


valid approach (M1)

egattempt to find coordinates of B or C, or

correct substitution into distance formula A1

eg

distance is 9 A1 N2

[4 marks]

18. [7 marks]

Markscheme

attempt to find the central angle or half central angle (M1)

eg, cosine rule, right triangle


eg

correct angle (seen anywhere)

eg (A1)

correct sector area

eg (A1)

area of triangle (seen anywhere) (A1)

eg

appropriate approach (seen anywhere) (M1)

eg, their sector-their triangle

22.5269

area of shaded region A1 N4

Note: Award M0A0A0A0A1 then M1A0 (if appropriate) for correct triangle area without any attempt to find an angle in triangle OAB.

[7 marks]

19a. [2 marks]

Markscheme

attempt to find the difference of -values of A and B (M1)

eg

6.25 (hours), (6 hours 15 minutes) A1 N2

[2 marks]

19b. [2 marks]

Markscheme

attempt to find the difference of -values of A and B (M1)

eg

A1 N2

[2 marks]

19c. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

19d. [3 marks]

Markscheme

METHOD 1

period (seen anywhere) (A1)


eg

0.502654

A1 N2

METHOD 2

attempt to use a coordinate to make an equation (M1)

eg


eg

0.502654

A1 N2

[3 marks]

19e. [2 marks]

Markscheme

valid method to find (M1)

eg

A1 N2

[2 marks]

19f. [3 marks]

Markscheme

METHOD 1

attempt to find start or end -values for 12 December (M1)

eg

finds -value for second max (A1)

23:00 (or 11 pm) A1 N3

METHOD 2

valid approach to list either the times of high tides after 21:00 or the -values of high tides after 21:00, showing at least two times (M1)

eg

correct time of first high tide on 12 December (A1)

eg10:30 (or 10:30 am)

time of second high tide = 23:00 A1 N3

METHOD 3

attempt to set their equal to 1.5 (M1)

eg

correct working to find second max (A1)

eg

23:00 (or 11 pm) A1 N3

[3 marks]

20a. [2 marks]

Markscheme

correct substitution into arc length formula (A1)

eg

A1 N2

[2 marks]

20b. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

20c. [2 marks]

Markscheme

correct substitution into area formula (A1)

eg

A1 N2

[2 marks]

21a. [2 marks]

Markscheme

valid approach (M1)

eg, sketch of graph,

A1 N2

[2 marks]

21b. [2 marks]

Markscheme

valid approach (M1)

eg, period is

0.448798

, (do not accept degrees) A1 N2

[2 marks]

21c. [2 marks]

Markscheme

valid approach (M1)

eg

7.01045

7.01 (m) A1 N2

[2 marks]

22a. [2 marks]

Markscheme

valid method (M1)

eg

235° (accept S55W, W35S) A1 N2

[2 marks]

22b. [5 marks]

Markscheme

valid approach to find (may be seen in (a)) (M1)

eg

correct working to find (may be seen in (a)) (A1)

eg,

evidence of choosing sine rule (seen anywhere) (M1)

eg

correct substitution into sine rule (A1)

eg

146.034

A1 N2

[5 marks]

22c. [3 marks]

Markscheme


eg

correct substitution into right-hand side (A1)

eg

192.612

A1 N2

[3 marks]

22d. [5 marks]

Markscheme

valid approach for locating B (M1)

egBE is perpendicular to ship’s path, angle

correct working for BE (A1)

eg

valid approach for expressing time (M1)

eg

correct working equating time (A1)

eg

27.2694

27.3 (km per hour) A1 N3

[5 marks]

23a. [3 marks]

Markscheme

evidence of valid approach (M1)

egright triangle,


egmissing side is 2,

A1 N2

[3 marks]

23b. [2 marks]

Markscheme

correct substitution into formula for (A1)

eg

A1 N2

[2 marks]

24a. [4 marks]

Markscheme

(i) valid approach to find

eg

A1 N2

(ii) valid approach to find (M1)

eg

A1 N2

[4 marks]

24b. [1 mark]

Markscheme

correct approach A1

eg

has coordinates AG N0

[1 mark]

24c. [2 marks]

Markscheme

(i) A1 N1

(ii) any correct expression for the area involving A1 N1

eg

[2 marks]

24d. [5 marks]

Markscheme

METHOD 1 (using sine formula for area)

correct expression for the area of triangle ACD (seen anywhere) (A1)

eg

correct equation involving areas A1

eg

recognizing that (seen anywhere) (A1)

(seen anywhere) (A1)

correct approach using ratio A1

eg

correct ratio AG N0

METHOD 2 (Geometric approach)

recognising and have same height (A1)

eguse of for both triangles,

correct approach A2

eg and

correct working A2

eg

AG N0

[5 marks]

24e. [4 marks]

Markscheme

correct working (seen anywhere) (A1)

eg


eg

correct working to find -coordinate (A1)

eg

D is A1 N3

[4 marks]

25a. [6 marks]

Markscheme

(i) valid approach (M1)

eg

A1 N2

(ii) valid approach (M1)

egperiod is 12, per

A1

AG N0

(iii) METHOD 1

valid approach (M1)

eg, substitution of points

A1 N2

METHOD 2

valid approach (M1)

eg, amplitude is 6

A1 N2

[6 marks]

25b. [3 marks]

Markscheme

(i) A1 N1

(ii) A2 N2

[3 marks]

25c. [6 marks]

Markscheme

(i) METHOD 1 Using

recognizing that a point of inflexion is required M1

egsketch, recognizing change in concavity

evidence of valid approach (M1)

eg, sketch, coordinates of max/min on

(exact) A1 N2

METHOD 2 Using

recognizing that a point of inflexion is required M1

egsketch, recognizing change in concavity

evidence of valid approach involving translation (M1)

eg, sketch,

(exact) A1 N2

(ii) valid approach involving the derivative of or (seen anywhere) (M1)

eg, max on derivative, sketch of derivative

attempt to find max value on derivative M1

eg, dot on max of sketch

3.14159

max rate of change (exact), 3.14 A1 N2

[6 marks]

26a. [1 mark]

Markscheme

A1 N1

[1 mark]

26b. [3 marks]

Markscheme

correct expression for area (A1)

eg

evidence of equating their expression to (M1)

eg

A1 N2

[3 marks]

26c. [3 marks]

Markscheme

METHOD 1


eg

correct substitution of their and into RHS (A1)

eg

11.7557

A1 N2

METHOD 2


eg

correct substitution of their and (A1)

eg

11.7557

A1 N2

[3 marks]

27a. [3 marks]

Markscheme

(i) 3 A1 N1

(ii) valid attempt to find the period (M1)

eg

period A1 N2

[3 marks]

27b. [4 marks]

Markscheme

A1A1A1A1 N4

[4 marks]

28a. [5 marks]

Markscheme

METHOD 1

correct substitution into formula for area of triangle (A1)

eg


eg

(A1)

(A1)

A1 N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle (A1)

eg


eg

height of triangle is 3 A1

(A1)

A1 N3

[5 marks]

28b. [3 marks]

Markscheme

recognizing supplementary angle (M1)

eg

correct substitution into formula for area of sector (A1)

eg

A1 N2

[3 marks]

29a. [5 marks]

Markscheme

attempt to form composite in any order (M1)

eg


eg

correct application of Pythagorean identity (do not accept ) (A1)

eg

valid approach (do not accept ) (M1)

eg

A1 N3

[5 marks]

29b. [2 marks]

Markscheme

valid approach (M1)

egamplitude , sketch with max and min -values labelled,

correct range A1 N2

eg, from to 3

Note: Do not award A1 for or for “between and 3”.

[2 marks]

30a. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

30b. [3 marks]

Markscheme

choosing cosine rule (M1)

eg

correct substitution into RHS (A1)

eg

12.0651

12.1 (km) A1 N2

[3 marks]

30c. [2 marks]

Markscheme

correct substitution (must be into sine rule) A1

eg

17.0398

A1 N1

[2 marks]

31a. [3 marks]

Markscheme


eg


eg

9.42069

A1 N2

[3 marks]

31b. [3 marks]

Markscheme


eg


eg

1.51271

(radians) (accept 86.7°) A1 N2

[3 marks]

32a. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

32b. [3 marks]

Markscheme

correct substitution into equation (A1)

eg

valid attempt to solve for (M1)

eg,

1.47679

A1 N2

[3 marks]

32c. [3 marks]

Markscheme

recognize the need to find the period (seen anywhere) (M1)

egnext value when

correct value for period (A1)

eg

5.2 (min) (must be 1 dp) A1 N2

[3 marks]

33a. [1 mark]

Markscheme

amplitude is 3 A1 N1

33b. [2 marks]

Markscheme

valid approach (M1)

eg

period is 2 A1 N2

33c. [4 marks]

Markscheme

A1

A1A1A1 N4

Note: Award A1 for sine curve starting at (0, 0) and correct period.

Only if this A1 is awarded, award the following for points in circles:

A1 for correct x-intercepts;

A1 for correct max and min points;

A1 for correct domain.

34a. [2 marks]

Markscheme


eg

= (cm) A1 N2

[2 marks]

34b. [4 marks]

Markscheme

METHOD 1

valid approach (M1)

egfinding reflex angle,

correct angle (A1)

eg


eg

A1 N3

METHOD 2

correct area of small sector (A1)

eg

valid approach (M1)

egcircle − small sector,


eg

A1 N3

[4 marks]

Total [6 marks]

Printed for International School of Monza

© International Baccalaureate Organization 2019

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

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