Special Maths Exam Paper 2 November 2013 Solutions

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1 Special Maths Eam Paper 2 November 2013 Solutions Question One 1.1 sin θ = 4/5 > 0, 270 o < θ 360 o. If 4 and 5 are the lengths of sides of a right-angled triangle, with 5 the hpotenuse, then the third side is of length 3 b the theorem of Pthagoras. θ (3, 4) (i) sin 2θ = 2sin θ cos θ = 2( 4/5)(3/5) = (ii) cos 2θ = cos 2 θ sin 2 θ = (9 16)/25 = 7/25. (4) 1.2 (a) sin( 150 o ) = sin 150 o = sin(180 o 30 o ) = sin 30 o = 1 2 (b) cos ( 345 o ) = cos 345 o = cos (360 o 15 o ) = cos 15 o = cos (45 o 30 o ) = cos 45 o cos 30 o + sin 45 o sin 30 o = = OR: cos 2θ = cos 2 θ sin 2 θ = cos 2 θ (1 cos 2 θ) = 2cos 2 θ 1 implies cos 2 1+cos 2θ θ =. Hence cos 2 15 o 1+cos 30o = = 1+ 3/2 = 2+ 3, impling cos (15 o ) =

2 (c) sin( 160o ) = sin(160o ) = sin(180o 20 o ) cos 430 o cos(360 o +70 o ) cos 70 o sin 20o sin 20o = = = 1 cos 70 o sin 20 o since 70 o + 20 o = 90 o. (7) sin 3 3 sin 3cos cos 3 sin 2sin cos 1.3 (a) cos = = = sin cos sin cos sin cos sin cos 2. (3) sin +tan (b) LHS = sin (cos +1)/(cos ) (1+cos )/sin csc +cot = sin2 cos = sin(3 ) = sin(2) sin cos sin +(sin )/(cos ) = = 1/sin +(cos )/(sin ) (sin cos +sin )/(cos ) (1+cos )/ sin = = sin sin cos = sin tan = RHS. (3) 1.4 (a) 2cos = 3 implies cos = 3/2. Therefore = 30 o o n or = 330 o o n, n Z. (b) 3cos 2 + 5sin 1 = 0 implies 3(1 sin 2 ) + 5sin 1 = 0. (3) Therefore 3sin 2 + 5sin + 2 = 0 or 3sin 2 5sin 2 = 0 = (3sin +1)(sin 2). Thus sin = 1/3, whence = (19, 5 o +180 o )+ 360 o n = 199, 5 o o n or = 19, 5 o o n, n Z. (4) [24] Question Two 2.1 = cos( 3) = cos 3. Consider the table 120 o 90 o 60 o 30 o 0 o 30 o 60 o 90 o 120 o graph is as follows:. Thus the 2

3 Its period is 360o 3 = 120 o. (4) 2.2 Consider the diagram: A D O B C (a) In triangles AOC and BOD we have AOC = BOD (vert. opp. angles); ACO = BDO and CAO = DBO (alt. angles, AC DB). Thus AOC/// BOD. (b) B similarit in (a), AC = OC = 20 CO = 2/5, i.e. = 2/5. BD OD 50 DO AO (c) B similarit in (a), = CO AB BO = 2/5, impling = 2/5. BO DO BO Therefore 80 BO = 2/5, impling 1 = 2/5, giving = 7/5. Hence BO BO BO BO = , and AO = AB OB = /7 =. (6) Consider the diagram: A D C B So AC = b = 8cm, BC = a = 9cm, AB = c = 10cm. B cosine rule, a 2 = b 2 +c 2 2(bc)cos A, impling cos A = a2 b 2 c 2 = ( ) = 83. 2bc 2(8)(10) 160 So A = 58, 8 o. Therefore ADC = 180 o 58, 8 o 27 o = 94, 2 o. Now b the sine rule, 3

4 sin A sin 94,2o =, impling DC = DC Consider the diagram: C A B 8sin 58,8o sin 94,2 o = 6, 9cm. (5) So A = 110 o, BC = a = 9m, AC = b = 6m. Now Area = 1 ab sin C. 2 Also b the sine rule, sin A = sin B, impling sin B = b sin A 6 sin 110o =. a b a 9 So B = 38, 8 o. Thus C = 180 o 110 o 38, 8 o = 31, 2 o. Hence Area = 1(9)(6)sin 31, 2 14, 2 0m2. (5) 2.5 (a) (-1,4) P( 1, 4) N(3, 10) M(, 5) Now PN = PM implies (3+1) 2 +(10 4) 2 = (+1) 2 +(5 4) 2. Therefore = ( + 1) 2 + 1, impling ( + 1) 2 = 51. So = 0, impling = 2± 4+4(50) 2 = 2± = 1 ± 51 6, 14 or (3) [23] 4

5 Question Three 3.1 Consider the diagram: A( 3, 4) O T( 4, 3) M(1/2, 1/2) 5 B(4, 3) (a) To find points of intersection of the line L : + = 1 and the circle, solve their eqns simultaneousl: So we have 2 +(1 ) 2 = 25, impling = 0 = 2( 2 12) = 2( 4)( + 3), thus = 3 or = 4, and = 4 or = 3. Therefore A = ( 3, 4) and B = (4, 3). (b) AB = = 7 2; (c) M = ( 4 3 ) = (1/2, 1/2);, (d) Slope of OM is M OM = 1/2 1/2 = 1 and slope of AB is M AB = This implies M OM M AB = 1, thus OM AB. (e) ( 4) 2 + ( 3) 2 = = 25, thus T lies on the circle. 3 4 = (f) Slope of OT is 3 0 = 3/ (g) tangent at T is perp. to OT, impling that the slope of the tangent is 1 3/4 = 4/3. Therefore the eqn of the tangent to the circle at T is 5

6 + 3 = (4/3)( + 4), giving = (10) 3.2 (a) lim ( 1)(+2) 1 = lim 1 1 = = 3. 1 (b) lim = lim 2 2 ( 2) = 1. (4) f () = lim h 0 f(+h) f() h = lim h 0 1 lim h 0 h h(1 h)(1 ) = lim h 0 1 (1 h) = lim h 0 = h(1 h)(1 ) 1 = 1. Thus f (3) = 1 = (1 h)(1 ) (1 ) 2 (1 3) 2 1 h 1 1 h 1. (4) = /2 +5 implies the slope of the tangent to the curve is m = f () = (6/2) 1/2 = 6(16)+8/64+3/2 = 94, 375 when = 4. Now when = 4, = 2/64 4/16+6(2)+ 5 = 16, Therefore the eqn of the tangent is 16, = 94, 375( 4), giving = 94, The normal is perp. to the tangent, impling the slope of the normal 1 94,375 is. Thus the eqn of the normal at = 4 is 16, = (1/94, 375)( 4), giving = (1/94, 375) + 16, 8 = (0, 0106) + 16, 8. (5) Question Four 4.1 Discuss and draw the graph of = f() = Soln. -intercept: = 9. Critical points: solve f () = 0. Thus = 0 = 6(+2)( 1), impling = 2 or = 1. f( 2) = 11 and f(1) = 16. Thus the turning points are ( 2, 11) and (1, 16). Local ma/min: We draw up a table that shows how f () changes signs around the turning points: ( 1) ( + 2) f = 6( 1)( + 2) [23]

7 Thus region of increase is given b f > 0, impling < 2 or > 1 Region of decrease is given b f < 0, impling 2 < < 1 Local ma is at ( 2, 11) and the graph of f is concave down around this point; Local min is at (1, 16) and the graph of f is concave up around this point. Point of inflection: Solve f () = 0, impling = 0, giving = 1/2 and = 5/2. So the point of inflection is ( 1/2, 5/2) Sketch: ( 2, 11) (.5, 2.5) (1, 16) (10) 4.2 Consider the diagram with r = and h =. = h 7

8 Area of the clinder is A = 2π 2 +2π and V = area of base height = π 2. Therefore A = 2π 2 + 2π V = 2π V. Min area occurs at π 2 critical points of A. Consider A () = 0, then 4π 2 V = 0, impling 2 4π 3 = 2V. Therefore = 3 V 2π h = V/π2 r = V π 3 = gives minimum surface area. Therefore V π(v/2π) = 2 1. (6) 4.3 (a) Initial velocit is v(0) = s (0) = 200 5(2)t = (0) = 200m/s; (b) At maimum height, the shell stops, impling v = 0 = t, giving t = 20 sec, which is the time taken to reach the ma height. Thus ma height = 200(20) 5(400) = 2000 m. (c) When the shell hits the ground, s = 0. Thus 200t 5t 2 = 0 = 5t(40 t), impling t = 40 sec. So the shell was in the air for 40 seconds. (d) Total distance is = 4000 meters. (e) Acceleration is a = dv = dt 10m/s2. On wa up: the speed decreases b 10 m/s per sec. On wa down: the speed increases b 10 m/s per sec. (7) 4.4 M = Median = (620+1) -th term = 310,5th term = a 310+a Q 1 = Lower quartile = (620+1) -th term = 155,25th term = a 155+a Q 3 = Upper quartile = 3(620+1) -th term = 3(155,25)th term = a 465+a 466 (3) 4 2 [26] 8

9 Question Five 5.1 Consider the diagram below: A( 3, 2) (0, 0) B(1, 4) (i) Reflection about the -ais is given b (, ) (, ). Hence A = ( 3, 2) and B = (1, 4). B (1, 4) (0, 0) A ( 3, 2) (ii) Reflection about the line = is given b (, ) (, ). Hence A = ( 2, 3) and B = (4, 1). 9

10 A ( 2, 3) (0, 0) B (4, 1) (iii) Rotation about the origin through 90 o is given b (, ) (, ). Hence A = (2, 3) and B = ( 4, 1). A (2, 3) (0, 0) B ( 4, 1) (iv) Rotation about the origin through 180 o followed b right horizontal shift b 3 units is given b (, ) ( 3, ). Hence A = (0, 2) and B = ( 4, 4). 10

11 B ( 4, 4) (0, 0) A (0, 2) (8) 5.2 The graph of = 1 + 3sin(2 90 o ) is obtained from the graph of = sin as follows: (i) = 1 + 3sin 2( 45 o ); (ii) first shrink the graph of = sin horizontall b a factor 1/2, i.e. halve the distances of points from the origin, (iii) then shift the resulting graph horizontall to the right b 45 o, (iv) thereafter magnif the resulting graph verticall b a factor 3, (v) finall shift the resulting graph verticall upward b one unit. Thus (0, 0) (0, 0) (45 o, 0) (45 o, 0) (45 o, 1) (90 o, 1) (45 o, 1) (90 o, 1) (90 o, 3) (90 o, 4) (180 o, 0) (90 o, 0) (135 o, 0) (135 o, 0) (135 o, 1) (270 o, 1) (135 o, 1) (180 o, 1) (180 o, 3) (180 o, 2) (360 o, 0) (180 o, 0) (225 o, 0) (225 o, 0) (225 o, 1) First consider the graph of = sin :

12 (5) 5.3 In ascending order, the numbers are : 19, 31, 32, 33, 33, 44, 56, 62, 74, 88, thus the median is M = = The lower quartile is Q 2 1 = 31.5 and the upper quartile is Q 3 = 68. Therefore the interquartile range is IQR = Q 3 Q 1 = = See the bo-and-whisker diagram below: (4) 5.4 From the table below, we compute the cumulative frequenc table: Time spent 0 t < t < t < t < t < 50 Frequenc

13 Cumulative frequenc table: Time spent t < 0 t < 10 t < 20 t < 30 t < 40 t < 50 Cum Frequenc n = 41. The -ais represents Cumulative Frequenc, while the -ais represents Time Time in minutes (Vertical ais: 1 unit = cum frequenc 4; Horizontal ais: 1 unit equals 5 minutes) (b) Median M = th item = 21st item = 31 (c) Q 1 = th item = 10.5 item = 22 4 Q 3 = 3( 41+1 )- th item = (10.5 3)th item = 31.5 st item = 37 4 Therefore IQR = Q 3 Q 1 = = 15. (8) 13 [25]