Math 5SN. Answer Key. Part A. Part B. The ordered pairs are (1, 10), (5, 8) as well as the ordered pair whose coordinates are (3, 9).

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1 EXMINTION #1 Math 5SN nswer Key Questions 1 to 10 4 marks or 0 marks Part Questions 11 to 15 4 marks each Part 11 The ordered pairs are (1, 10), (5, 8) as well as the ordered pair whose coordinates are (, 9). marks marks 1 Rounded to the nearest hundredth, the value of x is 0.7. Note: o not penalize students who did not round their answer to the nearest hundredth. 1 To the nearest degree, the angle measure is 8. Note: o not penalize students who did not round their answer. 14 π 5π The exact values of x are and The number of hours elapsed is 5.

2 16 Part Questions 16 to 5 4 marks each No marks are to be given if work is not shown. Examples of correct solutions are given. However, other acceptable solutions are possible. Example of an appropriate method Let x: number of lobsters y: number of crabs onstraints: x 0 y 0 y 5 y 60 x y x + y 140 Objective Function: R = 8.70x y Graph: y 140 Vertex R = 8.70x y (80, 60) 17 max (9., ) 159 (70, 5) 945 (0, 5) 6 E(0, 60) E x nswer: The maximum revenue this fisherman can expect to make is $17.

3 17 Example of an appropriate method Find the rate y = a b x y = 110 b x 85 = 110 b x 7.59 = b = b Find time that elapsed when 000 victims have been infected t = t = t = log log Find the year the vaccine will be offered = nswer: The population will be offered the vaccine in the year 00. Note: Students who have determined the rate have shown a partial understanding of the problem. 18 Example of an appropriate method 1 sinx sinx 1 sin x sinx sinx 1 sin x sinx cos x sinx 1 mark 1 mark cosx cosx 1 mark sinx 1 cotx cosx 1 mark

4 19 Example of an appropriate method a = 10 b = 8 Find the equation of the semi-ellipse centre (0, 0) x + y = (Students may use other centres.) Find the value of y at (7, y) y + = 1 64 y 64 = y =.64 y (-5.71 is rejected) nswer: The cameras are 5.71 m from the ground. 0 Equation of the circle in standard form x + 6x + y y = 6 (x ) + (y 1) = (x ) + (y 1) = 6 Each radius measures 6 units Since the three circles are congruent, the border forms an equilateral triangle. ccording to the diagram on the right, Note: djustments will have to be made for different labelling. O is a right triangle, m O = 6 units and m O = 0 m O 6 tan0 = = m m m = 10.9 m = m + m + m = Perimeter: P =.78 = 98.4 nswer: To the nearest hundredth of a unit, the border measures 98.4.

5 1 It is a parabola in the form of x = 4p(y k) (distance to the line (directrix) = distance to a point (focus)) y (15, 10) 5 Focus: M(0, 0) Vertex (0, 5) p = 5 Equation : x = -4p (y 5) x = -0 (y 5) a and b are the zeros of the function: x = -0 (0 5) x = 100 x = ± 10 M x nswer: Points and are 0 m apart.

6 EXMINTION # Questions 1 to 1 4 marks or 0 marks Section Section Questions 1 to 5 4 marks each No marks are to be given if work is not shown. Examples of correct solutions are given. However, other acceptable solutions are possible. 1 Example of an appropriate method x = number of road bikes y = number of mountain bikes (10, 70) x 0 y 0 x 10 y 45 x + y 80 y x Mtn ikes 60 (0, 60) 50 (10, 45) (15, 45) 40 0 Objective Function Max. Profit = 50x + 175y Road ikes Points (x, y) alculation Profit 1. (10, 45) 50(10) + 175(45) $ (15, 45) 50(15) + 175(45) $ (0, 60) 50(0) + 175(60) $ (10, 70) 50(10) + 175(70) $ nswer The maximum weekly profit is $

7 14 Example of an appropriate method y = a x b + c b = ( + 8) = 5 y (- 4 0) a = = = x ( 0 ) (c (-4)) (5 0) = c = 6 - nswer The rule of correspondence is y = - x Example of an appropriate method sin cot + cos = 1 cos sin + cos = 1 sin cos + cos = 1 cos + cos 1 = 0 ( cos 1)(cos + 1) = 0 1 cos = or cos = -1 = π or 5π or = π nswer π 5π, π, 16 Example of an appropriate method Let x = width of fenced-in plot in metres 5 x = length of fenced-in plot in metres rea of plot = length width = x(5 x) x(5 x) 50 5x x 50 -x + 5x 50 0 x 5x (x 5)(x 10) 0 Zeros x 5 = 0 or x 10 = 0 x =.5 x = 10 Zeros are.5 and 10 Width of plot.5 m nswer The smallest value of dimension x is.5 m.

8 17 Example of an appropriate method V(t) = -5 t V(t) 5-5 t t t 8-5 t 8 1. If t 8 0, then t 8 = t 8; if t 8 < 0, then t 8 = -(t 8) t (t 8) 1. t 9. -t t -6.8 t =.4 nswer The police officer should be on duty for.4 hours. (ccept hours and 4 minutes) 18 Example of an appropriate method log 5 (x 1) + log 5 (x + ) 1 = 0 log 5 (x 1)(x + ) = 1 (x 1)(x + ) = 5 1 x + x = 5 x + x 8 = 0 (x )(x + 4) = 0 x = 0 or x + 4 = 0 x = or -4 Sum of logs = log of the product -4 is an extraneous root nswer The solution of the equation is.

9 19 Example of an appropriate method Step 1: etermine centre and radius for each circle. entre Radius LG (-, ) 4 units SM (5, -) units Step : raw radii and ; draw ; raw E, perpendicular to and parallel to, forming rectangle E and right E. ( -, ) E (5, -) Step : In right E, m = ( 5 (- ) ) + (- ) = 8 + (- 4) = = 80 m E = m m E = 4 = units ( m E = m = units) m E = 80 4 = 76 Step 4: m = m E ( rect. E) = uits nswer The distance from point to point is approximately 8.7 units. 0 Example of an appropriate method Since is equilateral, m = 1 60 = 0 ( is an angle bisector) m OY = 5 units (Given) m O = 10 units (Hypotenuse = twice length of short side of triangle) m Q = 4 ( segment passing through the centre of a circle drawn perpendicular to a chord bisects the chord) m OQ = m OY = 5 (Radii of circle are congruent) x P O y Q In right triangle OQ, m O ( m OQ) ( m Q) = 5 4 = = (Pythagorean Theorem) m = m O + m O = 10 + = 1 units nswer The measure of is 1 units.

10 1 Example of an appropriate method Use exponential model: $ where y = (1 + i) 9 4 i = annual interest rate x = time, in years, since 1 st birthday y = accumulated investment ($) Investment Income ($) $ Substituting (9, ), we get = (1 + i) Time since 1 st birthday (yrs.) = (1 + i) = 1 + i i i 0.08 or 8% = i lternate solution let y = (1 + i) x where x = number of years since 5 th birthday Substituting (5, ), we get = (1 + i) x i = or 8% nswer The annual fixed interest rate is approximately 8%. (ccept 8.0% as well as 8%)

11 Example of an appropriate method etermine location of foci of the ellipse: c = a b c = c = 64 c = ±8 oordinates of foci at (±8, 0) Since circles have a radius of units, coordinates of centres are (-8, ) and (8, ) oordinates of points and are (-8 +, ) and (8, ) or (-6, ) and (6, ) oordinates of point are (0, -6) Equation of parabola in canonic form (x h) = 4c(y k) Substituting coordinates of points and into the equation to obtain value of c (x, y) = (6, ) (h, k) = (0, -6) 6 = 4c( -6) 6 = 4c(8) 6 = c 6 9 c = = or Therefore coordinates of centre of dark circle = 0, = 0, - 4 or (0, ) nswer The coordinates of the centre of the dark circle are 0, - 4 or (0, ). 8 Step 1: Find the height of the rocket at 5 seconds (50, 500) 500 H(5) = 00 5 = 00(5) = 1000 m (6, y) Height Step : Find the equation of the nd stage (m) H (t) = a x h + k Using (h, k) = (5, 1000), we get H (t) = a x Substituting (50, 500), we get 500 = a = a = 5a a = 1500 a = 00 Therefore, H (t) = 00 x (5, 1000) Time since launch (sec.)

12 11 seconds after the firing of the second stage (at 5 sec.) is 6 sec. Step : Find the image of 6 in H H (6) = = m nswer Rounded to the nearest metre, the height of the rocket 11 seconds after firing of the nd stage was 1995 m. 5 The equation of the circle is (x + 5) + (y ) = 9, and the equation of the ellipse is ( ) x 5 y + = Solution Total length of pipes = Perimeter of P = m P + m + m P y Step 1: Ellipse: centre is (5,0). Since a = 5, the vertices are (0,0) and (10,0) m = 10 Step : ircle: centre is (-5, ) and radius is P x Step : P + P = P + 9 = (10-5) + (0 ) P + 9 = P = 15 Step 4: In P, tan P = 15 m Q = m + m Q = 15 In Q, tan Q = 15 m P = 11.1 m Q = 11.1 In P, m P =.6 Step 5: P = (10)(15) cos.6 m P = 6.9 Step 6: Perimeter = = 1.9 nswer Rounded to the nearest tenth, the total length of the irrigation pipes is 1.9 m.

13 EXMINTION # Section Questions 1 to 1 4 marks or 0 marks

14 Section 14 Rounded to the nearest tenth u + v is marks or 0 marks Section 16 cot P + tan P = cosec P sec P cos P sin P + sin P cos P cos P + sin P sin P cos P cos P + sin P sin P cos P cosec P sec P 17 cos x sin x = 0 (1 sin x) sin x = 0 sin x sin x = 0 -sin x sin x 1 = 0 sin x + sin x + 1 = 0 (sin x + 1)(sin x + 1) = 0 sin x + 1 = 0 sin x + 1 = 0 sin x = -1 sin x = -1-1 sin x = x = 7π 11π, 6 6 x = π nswer The values of x in the domain are 11π π and. 6

15 18 x: number of supervisors y y: number of staff workers (10, 0) (14, 16) onstraints : x 0, y 0 x + y 0 x + y 18 x 6 (6, 1) (10, 8) x 14 y 8 y x (14, 8) x Vertices of polygon of constraints: (10, 8) 10(500) + 8(1500) = $ (6, 1) 6(500) + 1(1500) = $9 000 (10, 0) 10(500) + 0(1500) = $ (14, 16) 14(500) + 16(1500) = $7 000 (14, 8) 14(500) + 8(1500) = $ The minimum cost is $ nswer The town should hire 6 supervisors and 1 staff workers in order to minimize its costs. 19 Let t: time after 1995 (years) V(t): value of the car ($) V(t) = (r) t where r is the rate at which the value declines and = (r) = (r) r r 0.8 When V(t) = (0.8) t 5000 (0.8) t ln t ln( 0.8) t 6.7 nswer The value of the car falls below $5000 when it is 6.7 years 6 years 9 months.

17 0 Example of an appropriate solution Since width of rectangle is 60 cm, minor axis measures 60 cm. So b = 0 cm. If c represents distance between centre and focal point, then c + 0 equals the length of the semimajor axis, a. 0 cm F. 1 F. 0 cm 60 cm Solving for c: a = b + c (c + 0) = 0 + c c + 40c = c 40c = 500 c = 1.5 Therefore length of plywood is (c + 0) = ( ) = 65 rea of plywood: = 900 nswer Rounded to the nearest cm, the area of the plywood is 900 cm. 1 Example of an appropriate solution x = 4py, sub in (4, -) p = m = - 4 able (0, 0) E (0.6, y) x = -16 y, let x = 0.6 y = (4, -) - 4 Therefore, m E = ( ) = Triangle E is right angled. Therefore, m = (1.658) = 1.4 m nswer Each cable is 1.4 m in length.

18 Example of an appropriate solution y-axis (4, 1) (x, 11) (0, 5) x-axis The square root function must be in the form: y = a x + k Substituting (0, 5) we get: 5 = a 0 + k So k = 5 Substituting (4, 1) we get: 1 = a So a = 4 So the function is: y = 4 x + 4 t the ring, y = = 4 x + 5 So x =.5 ut.5 represents the radius of the gold ring in centimetres. So the circumference is: = π (.5) 14.1 cm Therefore the gold ring will cost: 14.1 = 8.6 cents. nswer Rounded to the nearest cent, the cost of the gold ring is 8 cents.

19 EXMINTION #4 Section Questions 1 to 1 4 marks or 0 marks

20 Questions 14 to 16 Section 14 π π 5π 7π The values of x belonging to 0, and that satisfy the equation are 0,,, π and Note Give marks if the student identifies 4 of the 5 values. Give marks if the student identifies of the 5 values. educt 1 mark if the student expresses the values in degrees. 15 The scalar product of vectors u and v is marks or 0 marks Questions 17 to 5 Section 17 Example of an appropriate method Height at which the rockets explode h 1 (6) = -1.5(6 4) + 00 = 150 metres Value of parameter d Since the two rockets explode at the same time and at the same height, h (6) = = 150 d 6 5 = 100 d 6 = 4 d 6 = 16 d d = 16 6 = 8 nswer The value of parameter d is 8.

21 18 Example of an appropriate method Start-up time - x = 1 - x 6 = -15 x 6 = 5 x 6 = 5 or x 6 = -5 x = 11 or x = 1 Since the start-up time is located to the left of the vertex on the graph, we must use the value of x that is less than 6. The air conditioning system starts up 1 hour after sunrise. Shutdown time - x = 0 - x 6 = x 6 = 16 x 6 = or x 6 = x = or x = Since the shutdown time is located to the right of the vertex on the graph, we must use the value of x that is greater than 6. The air conditioning system stops 4 hours after sunrise. Time during which air conditioning system is in operation 4 1 = 10. hours nswer On that day, the air conditioning system was in operation for 10 hours and 0 minutes or approximately 10. hours.

22 19 Example of an appropriate method Value of parameter i o = 000 (t) = o (1 + i) t 66 = 000(1 + i) 1.1 = (1 + i) 1.1 = 1 + i 0.1 = i (t) = 000(1.1) t Value after 10 years (10) = 000(1.1) nswer Emily's investment will be worth $ after 10 years. 0 Example of an appropriate method Form of the equation of the hyperbola ( x h) ( y k) a Parameters h and k + b = 1 The coordinates of are (0, 0). The coordinates of are (1, 0), because the circle is 1 units in diameter. The coordinates of point are (6, 0), because it is the midpoint of. Since point is the centre of the hyperbola, h = 6 and k = 0. Parameter a and b b = m E = 1 therefore, b = 6 The distance c between the foci is 0 units; therefore, c = 10. a + b = c a + 6 = 10 a = 8 nswer The equation of the hyperbola with vertices and E is ( x 6) 64 + y 6 = 1.

23 1 Example of an appropriate method iagram of the situation y b c 500 m x a Value of b y The distance c between each focus and the centre of the ellipse is 50 m. The triangle shown on the right is an isosceles triangle. Hence, b = 50 m. Value of a a = b + c a = a = m istance between Luke and the track a c = metres b 45 c = 50 m x nswer Rounded to the nearest tenth, the shortest distance between Luke and the track is 10.6 metres. Example of an appropriate method 1 ( + ) MN = M + N = + = = 1 ( + ) PO = P + O = + = = Vectors MN and PO are therefore parallel and of equal length. Quadrilateral MNOP is therefore a parallelogram.

24 Example of an appropriate method Polygon of constraints y (60, 90) x Function to be maximized Profit = cx + cy where c represents the cost of renting a campsite for a tent. Vertex Profit = cx + cy (0, 0) 0 (150, 0) 150c (60, 90) 40c maximum (0, 110) 0c Value of c Maximum profit 40c = $4800; therefore c = $0 nswer It cost $0 per day to rent a campsite for a tent. It cost $40 per day to rent a campsite for a trailer.

Mathematics 5 SN TRIGONOMETRY PROBLEMS 2., which one of the following statements is TRUE?, which one of the following statements is TRUE?

Mathematics 5 SN TRIGONOMETRY PROBLEMS 1 If x 4 which one of the following statements is TRUE? A) sin x > 0 and cos x > 0 C) sin x < 0 and cos x > 0 B) sin x > 0 and cos x < 0 D) sin x < 0 and cos x