Trig Practice 09 & Nov The diagram below shows a curve with equation y = 1 + k sin x, defined for 0 x 3π.

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1 IB Math High Level Year : Trig: Practice 09 & 0N Trig Practice 09 & Nov 0. The diagram below shows a curve with equation y = + k sin x, defined for 0 x. The point A, lies on the curve and B(a, b) is the maximum point. 6 (a) Show that k = 6. () (b) Hence, find the values of a and b. (Total 5 marks). (a) Show that arctan arctan. () (b) Hence, or otherwise, find the value of arctan () + arctan. (Total 5 marks). The diagram below shows two straight lines intersecting at O and two circles, each with centre O. The outer circle has radius R and the inner circle has radius r. diagram not to scale Consider the shaded regions with areas A and B. Given that A : B = :, find the exact value of the ratio R : r. (Total 5 marks). Let f be a function defined by f(x) = x arctan x, x. (a) Find f() and f( ). () (b) Show that f( x) = f(x), for x. () (c) Show that x f ( x) x, for x. (d) (e) Find expressions for f (x) and f (x). Hence describe the behaviour of the graph of f at the origin and justify your answer. Sketch a graph of f, showing clearly the asymptotes. (f) Justify that the inverse of f is defined for all x and sketch its graph. () (8) E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of

2 IB Math High Level Year : Trig: Practice 09 & 0N (Total 0 marks) 5. A triangle has sides of length (n + n + ), (n + ) and (n ) where n >. (a) Explain why the side (n + n + ) must be the longest side of the triangle. (b) Show that the largest angle, θ, of the triangle is 0. (5) (Total 8 marks) 6. Two non-intersecting circles C, containing points M and S, and C, containing points N and R, have centres P and Q where PQ = 50. The line segments [MN] and [SR] are common tangents to the circles. The size of the reflex angle MPS is α, the size of the obtuse angle NQR is β, and the size of the angle MPQ is θ. The arc length MS is l and the arc length NR is l. This information is represented in the diagram below. The radius of C is x, where x 0 and the radius of C is 0. (a) Explain why x < 0. x 0 (b) Show that cos θ =. 50 (c) (i) Find an expression for MN in terms of x. (ii) Find the value of x that maximises MN. (d) (e) Find an expression in terms of x for (i) α; (ii) β. The length of the perimeter is given by l + l + MN + SR. (i) Find an expression, b(x), for the length of the perimeter in terms of x. (ii) Find the maximum value of the length of the perimeter. (iii) Find the value of x that gives a perimeter of length 00. diagram not to scale () () () () (9) (Total 8 marks) E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of

3 IB Math High Level Year : Trig: Practice 09 & 0N 7. Consider the graphs y = e x and y = e x 5 sin x, for 0 x. (a) 5 On the same set of axes draw, on graph paper, the graphs, for 0 x. Use a scale of cm to 8 on your x-axis and 5 cm to unit on your y-axis. (b) Show that the x-intercepts of the graph y = e x n sin x are, n = 0,,,,, 5. (c) Find the x-coordinates of the points at which the graph of y = e x sin x meets the graph of y = e x. Give your answers in terms of. (d) (i) Show that when the graph of y = e x sin x meets the graph of y = e x, their gradients are equal. (ii) Hence explain why these three meeting points are not local maxima of the graph y = e x sin x. (6) (e) (i) Determine the y-coordinates, y, y and y, where y > y > y, of the local maxima of y = e x 5 sin x for 0 x. You do not need to show that they are maximum values, but the values should be simplified. (ii) Show that y, y and y form a geometric sequence and determine the common ratio r. (7) (Total marks) 8. The diagram below shows two concentric circles with centre O and radii cm and cm. The points P and Q lie on the larger circle and PÔQ = x, where 0 < x <. (a) Show that the area of the shaded region is 8 sin x x. (b) Find the maximum area of the shaded region. diagram not to scale () (Total 7 marks) 9. Consider the function f : x arccos x. (a) Find the largest possible domain of f. (b) Determine an expression for the inverse function, f, and write down its domain. () () (Total 8 marks) E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of

4 IB Math High Level Year : Trig: Practice 09 & 0N 0. Let α be the angle between the unit vectors a and b, where 0 α. (a) Express a b and a + b in terms of α. (b) Hence determine the value of cos α for which a + b = a b.. (a) A particle P moves in a straight line with displacement relative to origin given by s = sin (t) + sin(t), t 0, where t is the time in seconds and the displacement is measured in centimetres. (i) Write down the period of the function s. (ii) Find expressions for the velocity, v, and the acceleration, a, of P. (iii) Determine all the solutions of the equation v = 0 for 0 t. (b) Consider the function f(x) = A sin (ax) + B sin (bx), A, a, B, b, x. Use mathematical induction to prove that the (n) th derivative of f is given by f (n) (x) = ( ) n (Aa n sin (ax) + Bb n sin (bx)), for all n +. () (Total 5 marks) (0) (8) (Total 8 marks). Triangle ABC has AB = 5cm, BC = 6 cm and area 0 cm. (a) Find sin Bˆ. () (b) Hence, find the two possible values of AC, giving your answers correct to two decimal places. () (Total 6 marks) E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of

5 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme Trig Practice 09 & Nov 0 - MarkScheme. (a) = + ksin 6 M = k k = 6 AG N0 (b) METHOD maximum sin x = M a = b = 6( ) = 7 N METHOD y = 0 M k cos x = 0 x =,,... a = b = 6( ) = 7 N Note: Award for (, 7). (a) METHOD let x = arctan tan x and y arctan tan y tan x tan y tan (x + y) = tan x tan y = M so, x + y = arctan = AG METHOD x y for x, y > 0, arctan x + arctan y = arctan if xy < xy M so, arctan arctan arctan AG METHOD an appropriate sketch M e.g. [5] E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of 8

6 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme correct reasoning leading to (b) METHOD arctan() + arctan = arctan arctan RAG (M) = arctan arctan () Note: Only one of the previous two marks may be implied. = N METHOD let x = arctan tan x = and y = arctan tan y = tan x tan y tan (x + y) = (M) tan x tan y as x accept 0 x and y accept 0 y x y (accept 0 < x + y < ) (R) Note: Only one of the previous two marks may be implied. so, x + y = N METHOD x y for x, y > 0, arctan x + arctan y = arctan + if xy > (M) xy so, arctan + arctan = arctan + () Note: Only one of the previous two marks may be implied. = N E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of 8

7 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme METHOD an appropriate sketch e.g. M correct reasoning leading to R. A = (R r ) B = r from A: B = :, we have R r = r M R = r () hence exact value of the ratio R : r is : N0. (a) f() = arctan = f ( ) arctan( ) (b) f( x) = x arctan( x) M = x + arctan x = (x arctan x) = f(x) AG N0 (c) as arctan x, for any x arctan x, for any x then by adding x (or equivalent) R we have x x arctan x x AG N0 (d) x f (x) = or x x x( x ) x x f (x) = or ( x ) ( x ) M [5] [5] E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of 8

8 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme (e) f (0) = f (0) = 0 EITHER as f (x) 0 for all values of x ((0,0) is not an extreme of the graph of f (or equivalent)) R OR as f (x) > 0 for positive values of x and f (x) < 0 for negative values of x R THEN (0, 0) is a point of inflexion of the graph of f (with zero gradient) N Note: Award for both asymptotes. for correct shape (concavities) x < 0. for correct shape (concavities) x > 0. (f) (see sketch above) as f is increasing (and therefore one-to-one) and its range is, f is defined for all x use the result that the graph of y = f (x) is the reflection in the line y = x of the graph of y = f(x) to draw the graph of f 5. (a) a reasonable attempt to show either that n + n + > n + or n + n + > n complete solution to each inequality M R (M) [0] (n ) ( n ) ( n n ) (b) cos θ = M (n )( n ) n n n = M (n )( n ) ( n )( n )(n ) = (n )( n ) = θ = 0 AG 6. (a) PQ = 50 and non-intersecting R (b) a construction QT (where T is on the radius MP), parallel to MN, [8] E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM of 8

9 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme 7. (a) so that QTˆ M = 90 (angle between tangent and radius = 90 ) M 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) (ii) maximum for MN occurs when x = 0 (d) (i) α = θ M x 0 = arccos 50 (ii) β = α (= θ) x 0 = cos 50 (e) (i) b(x) = xα + 0β + 50 ( x 0) 0 0 = x x x cos 0 cos 50 ( 0) 50 x 50 M (ii) maximum value of perimeter = 76 A (iii) perimeter of 00 cm b (x) = 00 (M) when x =. [8] A Note: Award for each correct shape, for correct relative position. (b) e x sin (x) = 0 (M) sin (x) = 0 x = 0,,,,, 5 5 x = 0,,,,, AG (c) e x = e x sin (x) or reference to graph sin x = M 5 9 x =,, 5 9 x =,, N (d) (i) y = e x sin x E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM 5 of 8

10 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme dy dx = e x sin x + e x cos x M y = e x dy dx = e x verifying equality of gradients at one point R verifying at the other two R (ii) dy since dx 0 at these points they cannot be local maxima R (e) (i) maximum when y = e x cos x e x sin x = 0 M arctan( ) arctan() arctan() x =,,,... maxima occur at arctan( ) arctan() arctan() x =,, (arctan()) e so y = sin(arctan ()) (= 0.696) (arctan() ) y = e sin(arctan () + ) e (arctan() ) (arctan() ) sin(arctan()) 0.5 y = e sin(arctan () + ) (arctan() ) e sin(arctan()) 0.00 N (ii) y for finding and comparing y and y y M e r = Note: Exact values must be used to gain the M and the. 8. (a) shaded area = area of triangle area of sector, i.e. (M) sin x x = 8 sin x x AG (b) EITHER any method from GDC gaining x. (M)() maximum value for given domain is 5. A OR da dx = 8cosx da set dx = 0, hence 8 cos x = 0 M cos x = x. hence A max = 5. [] [7] E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM 6 of 8

11 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme 9. (a) arccos x 0 arccos x (M) x accept x () since x (M) x accept x Note: Penalize the use of < instead of only once. (b) y = arccos x x cos y M f : x cos x 0. METHOD (a) 0 x a b a b a b a b cos = cos a b a b cos( ) = cos Note: Accept the use of a, b for a, b. (b) cos cos M cos α = 5 METHOD (a) a b = sin M a + b = sin cos Note: Accept the use of a, b for a, b. (b) cos 6 sin 9 tan cos 0 M cos α = cos 5. (a) (i) the period is (ii) ds v = dt = cos (t) + cos (t) (M) dv a = dt = sin (t) sin (t) (M) E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM 7 of 8 M [8] [5]

12 IB Math High Level Year : Trig: Practice 09 & 0N - MarkScheme (iii) v = 0 (cos (t) + cos (t)) = 0 EITHER cos (t) + cos (t) = 0 M ( cos (t) ) (cos (t) + ) = 0 () cos (t) = or cos (t) = t =, t = 5 7 t =, t, t t = OR t t cos cos = 0 M t t cos 0 or cos 0 t =, 5 7 t =,,, (b) P(n) : f (n) (x) = ( ) n (Aa n sin (ax) + Bb n sin (bx)) P(): f (x) = (Aa cos (ax) + Bb cos (bx)) M = Aa sin (ax) Bb sin (bx) = (Aa sin (ax) + Bb sin(bx)) P() true assume that P(k) : f (k) (x) = ( ) k [(Aa k sin (ax) + Bb k sin (bx)) is true M consider P(k + ) f (k+) (x) = ( ) k (Aa k+ cos(ax) + Bb k+ cos(bx)) M f (k+) (x) = ( ) k ( Aa k+ sin (ax) + Bb k+ sin (bx)) = ( ) k+ (Aa k+ sin (ax) + Bb k+ sin (bx)) P(k) true implies P(k + ) true, P() true so P(n) true n + R Note: Award the final R only if the previous three M marks have been awarded.. (a) area = BC AB sin B (M) 0 56sin B sin Bˆ 5 (b) cos B = ± (= ± ) or B =.8... and 8... () AC = BC + AB BC AB cos B (M) AC = or AC =.0 or [8] [6] E:\Desert\HL\ Trig\HL.TrigPractice09_0N.docx on 0/9/06 at :7 PM 8 of 8

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

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 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