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 =
correct working (A1)
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
correct working (A1)
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
correct working (A1)
eg
A2 N3
[7 marks]
12a. [3 marks]
Markscheme
evidence of choosing sine rule (M1)
eg
correct substitution (A1)
eg
4.13102
A1 N2
[3 marks]
12b. [3 marks]
Markscheme
correct working (A1)
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
correct working (A1)
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
correct working (A1)
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)
correct working (A1)
eg
A1 N2
[3 marks]
17c. [2 marks]
Markscheme
correct substitution (A1)
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 )
recognizing need to find BC (M1)
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)
recognizing need to find BC (M1)
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
correct working (A1)
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)
valid approach (seen anywhere) (M1)
eg
0.502654
A1 N2
METHOD 2
attempt to use a coordinate to make an equation (M1)
eg
correct substitution (A1)
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
evidence of choosing cosine rule (M1)
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,
correct working (A1)
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
valid approach (seen anywhere) (M1)
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
evidence of choosing cosine rule (M1)
eg
correct substitution of their and into RHS (A1)
eg
11.7557
A1 N2
METHOD 2
evidence of choosing sine rule (M1)
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
correct working (A1)
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
correct working (A1)
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
correct working (A1)
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
evidence of choosing sine rule (M1)
eg
correct substitution (A1)
eg
9.42069
A1 N2
[3 marks]
31b. [3 marks]
Markscheme
evidence of choosing cosine rule (M1)
eg
correct substitution (A1)
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
correct substitution (A1)
eg
= (cm) A1 N2
[2 marks]
34b. [4 marks]
Markscheme
METHOD 1
valid approach (M1)
egfinding reflex angle,
correct angle (A1)
eg
correct substitution (A1)
eg
A1 N3
METHOD 2
correct area of small sector (A1)
eg
valid approach (M1)
egcircle − small sector,
correct substitution (A1)
eg
A1 N3
[4 marks]
Total [6 marks]
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