Topic 3 Review [82 marks]

Transcription

1 Topic 3 Review [8 marks] 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. Angle BOC is.4 radians. 1a. Find the length of arc ADC. appropriate method to find angle AOC correct substitution into arc length formula () ( 3.9) 0, arc length = arc length = 47.7 (47.6, 47.7] (i.e. do not accept 47.6) N Notes: Candidates may misread the question and use AOC =.4. If working shown, award M0 then A0MR for the answer 48. Do not then penalize AOC in part (d) which, if used, leads to the answer However, if they use the prematurely rounded value of.4 for AOC, penalise 1 mark for premature rounding for the answer 48 in (c). Do not then penalize for this in (d). 1b. Find the length of the chord [AB].

2 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) c = a + b ab cosc correct substitution (into rhs) (0)(0) cos1.5, AB = cos1.5 AB = AB = 7.3 [7., 7.3] METHOD choosing sine rule N =, correct substitution 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 Find the area of triangle AOB. 1c. correct substitution into area formula 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 Angle BOC is.4 radians. 1d. Find the area of the shaded region. calculating sector area using their angle AOC () 1 (.38 )( 0 ), 00(.38 ), shaded area = their area of triangle AOB + their area of sector , shaded area = (accept = 676 from using 199) shaded area = 676 [676, 677] N

3 1e. Angle BOC is.4 radians. The shaded region is to be painted red. Red paint is sold in cans which cost $3 each. One can covers 140 m. How much does it cost to buy the paint? dividing to find number of cans , cans must be purchased () multiplying to find cost of cans (3), 3 cost is 160 (dollars) N3 Let f(x) = cosx and g(x) = x 1. Find f ( ). a. f ( ) = cos = 1 N () Let f(x) = cosx and g(x) = x 1. b. Find (g f)( ). (g f)( ) = g( 1) = 1 N (= ( 1) 1) () c. Given that (g f)(x) can be written as cos(kx), find the value of k, k Z. (g f)(x) = (cos(x) ) 1 (= cos (x) 1) evidence of cos θ 1 = cosθ (seen anywhere) (g f)(x) = cos4x k = 4 N

4 Let f(x) = (sin x + cosx). 3. Show that f(x) can be expressed as 1 + sin x. attempt to expand (sin x + cosx)(sin x + cosx) correct expansion sin x + sin xcosx + cos x f(x) = 1 + sin x AG N0 ; at least 3 terms The diagram below shows part of the graph of f(x) = acos(b(x c)) 1, where a > 0. 4 The point P (,) is a maximum point and the point Q(, 4) is a minimum point. 3 4 Find the value of a. 4a. evidence of valid approach a = 3 max y value min y value N, distance from y = 1

5 The diagram below shows part of the graph of f(x) = acos(b(x c)) 1, where a > 0. 4 The point P (,) is a maximum point and the point Q(, 4) is a minimum point. 3 4 (i) 4b. Show that the period of f is. (ii) Hence, find the value of b. (i) evidence of valid approach finding difference in x-coordinates, evidence of doubling ( ) period = AG N0 (ii) evidence of valid approach b = b = N Given that 0 < c <, write down the value of c. 4c. [1 mark] c = 4 N1 [1 mark] 3x Let f(x) = + 1, g(x) = 4 cos( ) 1. Let h(x) = (g f)(x). x 3 Find an expression for h(x). 5a.

6 attempt to form any composition (even if order is reversed) correct composition h(x) = 4 cos( ) 1 3x x h(x) = g ( + 1) () 1 1 (4 cos( x + ) 1,4 cos( ) 1) 3 3x+ 6 N3 3x Let f(x) = + 1, g(x) = 4 cos( ) 1. Let h(x) = (g f)(x). x 3 5b. Write down the period of h. [1 mark] period is 4(1.6) N1 [1 mark] Write down the range of h. 5c. range is 5 h(x) 3 ([ 5,3]) N Show that 4 cosθ + 5 sin θ = sin θ + 5 sin θ a. attempt to substitute 1 sin θ for cosθ correct substitution 4 (1 sin θ) + 5 sin θ 4 cosθ + 5 sin θ = sin θ + 5 sin θ + 3 AG N0 6b. Hence, solve the equation 4 cosθ + 5 sin θ = 0 for 0 θ. evidence of appropriate approach to solve factorizing, quadratic formula correct working ( 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 () 4 θ = 3 A N3

7 The following diagram represents a large Ferris wheel at an amusement park. The points P, Q and R represent different positions of a seat on the wheel. The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes. A seat starts at the lowest point P, when its height is one metre above the ground. Find the height of a seat above the ground after 15 minutes. 7a. valid approach 15 mins is half way, top of the wheel, d + 1 height = 101 (metres) N After six minutes, the seat is at point Q. Find its height above the ground at Q. 7b. evidence of identifying rotation angle after 6 minutes 5 1, of a rotation, 7 5 evidence of appropriate approach drawing a right triangle and using cosine ratio correct working (seen anywhere) 5 x 50 cos =, 15.4(508 ) evidence of appropriate method M1 height = radius height = 35.5 (metres) (accept 35.6) N The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t c)) c. Find the value of b and of c. [6 marks]

8 METHOD 1 evidence of substituting into b = correct substitution period = 30 minutes, b = 0.09 ( ) 15 substituting into h(t) N h(0) = 1, h(15) = 101 correct substitution METHOD N 1 = 50 sin ( c) c = 7.5 evidence of setting up a system of equations two correct equations 1 = 50 sin b(0 c) + 51, 101 = 50 sin b(15 c) + 51 attempt to solve simultaneously evidence of combining two equations b = 0.09 ( ) 15 [6 marks] b = 30 period, c = 7.5 NN The height of the seat above ground after t minutes can be modelled by the function h(t) = 50 sin(b(t c)) d. Hence find the value of t the first time the seat is 96 m above the ground. evidence of solving h(t) = 96 equation, graph t = 1.8 (minutes) A N3 Consider the following circle with centre O and radius r. The points P, R and Q are on the circumference, POQ = θ, for 0 < θ <. Use the cosine rule to show that PQ = rsin θ. 8a.

9 correct substitution into cosine rule P Q = r + r (r)(r) cos(θ), P Q = r r (cos(θ)) substituting 1 sin θ for cosθ (seen anywhere) working towards answer P Q = r r (1 sin θ) () P Q = r r + 4r sin θ recognizing r r = 0 (including crossing out) (seen anywhere) P Q = 4r sin θ, PQ = 4r sin θ PQ = rsinθ AG N0 Let l be the length of the arc PRQ. 8b. Given that 1.3PQ l = 0, find the value of θ. PRQ = r θ correct set up (seen anywhere) () 1.3 rsin θ r (θ) = 0 attempt to eliminate r correct equation in terms of the one variable θ 1.3 sin θ θ = θ = 1. (accept 70.0 (69.9)) N3 () Consider the function f(θ) =.6sin θ θ 8c., for 0 < θ <. (i) Sketch the graph of f. (ii) Write down the root of f(θ) = 0.

10 (i) N3 Note: Award for approximately correct shape, for x-intercept in approximately correct position, for domain. Do not penalise if sketch starts at origin. (ii) θ = 1. N1 Use the graph of f to find the values of θ for which l < 1.3PQ. 8d. evidence of appropriate approach (may be seen earlier) M θ <.6sin θ, 0 < f(θ), showing positive part of sketch 0 < θ < < θ = 1. (accept θ < 1. ) N1 Consider the triangle ABC, where AB =10, BC = 7 and CAB = 30. 9a. Find the two possible values of ACB. Note: accept answers given in degrees, and minutes. evidence of choosing sine rule sin A a = sin B b correct substitution sin θ sin 30 =, sin θ = ACB = 45.6, ACB = 134 N1N1 Note: If candidates only find the acute angle in part (a), award no marks for (b).

11 9b. Hence, find ABC, given that it is acute. attempt to substitute their larger value into angle sum of triangle 180 ( ) ABC = 15.6 N International Baccalaureate Organization 015 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for East Mecklenburg High School