1 2 k = 6 AG N0. 2 b = 1 6( 1) = 7 A1 N2. METHOD 2 y = 0 M1. b = 1 6( 1) = 7 A1 N2

Transcription

1 Trig Practice Answer Key. (a) = + ksin 6 = k A k = 6 AG N0 (b) METHOD maximum sin x = a = A b = 6( ) = 7 A N METHOD y = 0 k cos x = 0 x =,,... a = A b = 6( ) = 7 A N Note: Award AA for (, 7). (a) area of AOP = r sin θ A (b) TP = r tan θ () area of POT = r(r tan θ) = r tan θ A (c) area of sector OAP = r θ A area of triangle OAP < area of sector OAP < area of triangle POT R r sin r r tan sin θ < θ < tan θ AG

2 . sin x sin x cos cos x sin () sin x cos cos x sin sin x sin sin x cos x sin x dividing by cos x and rearranging tan x = rationalizing the denominator tan x = 6 + A. tan Using tan θ = tan () tan tan tan θ + 8 tan θ = 0 A Using factorisation or the quadratic formula () tan θ = or AA 5. (a) METHOD let x = arctan tan x and y arctan tan y tan x tan y tan (x + y) = tan x tan y = so, x + y = arctan = AAG A A METHOD x y for x, y > 0, arctan x + arctan y = arctan if xy < xy so, arctan arctan arctan AAG METHOD an appropriate sketch e.g. IB Questionbank Mathematics Higher Level rd edition

3 correct reasoning leading to (b) METHOD arctan() + arctan() = RAG () = arctan arctan (A) Note: Only one of the previous two marks may be implied. arctan arctan = A N METHOD let x = arctan tan x = and y = arctan tan y = tan x tan y tan (x + y) = tan x tan y as x accept 0 x and y accept 0 y x y (accept 0 < x + y < ) Note: Only one of the previous two marks may be implied. () (R) so, x + y = A N METHOD x y for x, y > 0, arctan x + arctan y = arctan + if xy > xy () so, arctan + arctan = arctan + (A) Note: Only one of the previous two marks may be implied. = A N IB Questionbank Mathematics Higher Level rd edition

4 METHOD an appropriate sketch e.g. correct reasoning leading to RA 6. tan θ 5 sec θ 0 = 0 Using + tan θ = sec θ, (sec θ ) 5 sec θ 0 = 0 () sec θ 5 sec θ = 0 A Solving the equation e.g. ( sec θ + )(sec θ ) = 0 () sec θ = or sec θ = A θ in second quadrant sec θ is negative (R) sec θ = A N 7. sin x cos x cos x = 0 () cos x ( sin x ) = 0 (A) cos x = 0 sin x = A x = x =, AAA IB Questionbank Mathematics Higher Level rd edition

5 8. (a) (b) Note: Award A for shape. A for scales given on each axis. A Asymptotes x = 0, x =, x = Max,, Min, Note: Award A for shape A for asymptotes, A for one error, A0 otherwise. A for max. A for min. A5 sin x cos x (c) tanx + cot x cos x sin x sin x cos x sin x cos x sin x csc x A A AG (d) tan x + cot x csc x () Max is at, AA Min is at, 8 AA IB Questionbank Mathematics Higher Level rd edition 5

6 (e) csc x =.5 tan x 0.5 tan x cot x tan x tan x + cot x = tan x tan x tan x = 0 tan x tan x = 0 A ( tan x + )(tan x ) = 0 tan x = or A x = A Note: Award A0 for answer in degrees or if more than one value given for x. 9. area of triangle POQ = 8 sin 59 = 7. (A) area of sector = =.95 (A) area between arc and chord = = 5.5 (cm ) A [] 0. let the length of one side of the triangle be x consider the triangle consisting of a side of the triangle and two radii EITHER x = r + r r cos 0 = r OR x = r cos 0 THEN x = r A so perimeter = r A now consider the area of the triangle area = r sin 0 = r A P r A r = A r Note: Accept alternative methods IB Questionbank Mathematics Higher Level rd edition 6

7 . (a) PQ = 50 and non-intersecting R (b) a construction QT (where T is on the radius MP), parallel to MN, so that QTˆ M = 90 (angle between tangent and radius = 90 ) lengths 50, x 0 and angle θ marked on a diagram, or equivalent R Note: Other construction lines are possible. (c) (i) MN = 50 ( x 0) A (ii) maximum for MN occurs when x = 0 A (d) (i) α = θ x 0 = arccos 50 A (ii) β = α (= θ) A x 0 = cos 50 A (e) (i) b(x) = xα + 0β + 50 ( x 0) AAA = 0 0 x x x cos 0 cos 50 ( 0) 50 x 50 A (ii) maximum value of perimeter = 76 A (iii) perimeter of 00 cm b (x) = 00 () when x =. A [8] IB Questionbank Mathematics Higher Level rd edition 7