UNIT 2 MATHEMATICAL METHODS 2015 MASTER CLASS PROGRAM WEEK 11 EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

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1 UNIT MATHEMATICAL METHODS 015 MASTER CLASS PROGRAM WEEK 11 EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT SECTION 1 MULTIPLE CHOICE QUESTIONS Where possible, students should use the CAS to obtain solutions. QUESTION 1 QUESTION QUESTION Answer is B Answer is A Answer is C QUESTION 4 QUESTION 5 QUESTION Answer is E Answer is C (Remember to expand the brackets) Answer is C The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 1

2 QUESTION 7 Answer is E cos cos θ + sin θ = 1 θ = 1 sin θ = = 5 cos θ = ± 4 5 As the angle lies in the nd Quadrant, cosθ = 4 5 cos θ Is Negative. QUESTION 8 Answer is B x cos + = 0 a x cos = a Basic Angle: 1 Cos π = x π x π =π + nπ or =π+ + nπ a a x 5 π x 7 π = + nπ or = + n π a a 5aπ 7aπ x = + naπ or x = + naπ n = 1: 5 aπ 7a x = naπ = π or 7 aπ 5a x = naπ = π n = 1: 5 aπ 17a x = + naπ = π or 7 aπ 19a x = + naπ = π Alternatively : The School For Excellence 015 Unit Master Classes Maths Methods Exam Page

3 QUESTION 9 Answer is C Graphs must be 1:1 functions. QUESTION 10 Answer is B f will have an inverse function if it is a 1-to-1 function. Use technology to draw a graph of f. The smallest value of x at which f has a turning point is x Therefore, f is a 1-to-1 function over the domain x < QUESTION 11 Answer is D QUESTION 1 Answer is C Container is filled from the bottom up. Initial part of container height of water increases by the same amount per unit time. Therefore h vs t graph is an oblique line. Answer is B or C. Middle part of container is becoming wider. The height occupied by water will decrease per unit time. Gradient of curve is decreasing, therefore, shape of graph resembles: Top part of container is becoming narrower. The height occupied by water will increase per unit time. Gradient of curve is increasing, therefore, shape of graph resembles: QUESTION 1 Answer is B The School For Excellence 015 Unit Master Classes Maths Methods Exam Page

4 QUESTION 14 Answer is B dy dx = ax + b dy Substitute = 4 dx Substitute (1, ) into when = 1 x : 4 = a + b. (1) y = ax + bx : = a + b. () Solve equations (1) and () simultaneously: a = and b = 0. Alternatively: QUESTION 15 Answer is D QUESTION 1 Answer is D You cannot differentiate at the end points of a domain. QUESTION 17 Answer is A The function g (x) is obtained from f (x) through the following three transformations: A dilation from the x axis (or parallel to the y axis) by a factor of 5 which results in the minimum value being at (, 5). A reflection in the y axis which now means that the minimum is at (, 5). Finally there is a translation of 1 unit to the right which results in the minimum now being at ( + 1, 5) The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 4

5 QUESTION 18 Answer is E If y x x x = then dy dx =x 8x +7 At x =, dy = dx = Gradient of tangent is. Therefore, gradient of the normal is 1. As y = 1 at x = : Equation of the normal is: y 1= 1 (x ) y = x + y + x 5 = 0 QUESTION 19 Answer is C QUESTION 0 Answer is E The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 5

6 QUESTION 1 Answer is E QUESTION Answer is B The School For Excellence 015 Unit Master Classes Maths Methods Exam Page

7 SECTION EXTENDED ANSWER QUESTIONS QUESTION 1 a. (i) (ii) b. (i) 1 mark for letting the derivative equation = 0. No marks for derivative. 1 mark for the correct answer in exact form. The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 7

8 (ii) c. d. This question can only be solved using technology. (i) Maximum value can occur at the stationary point or end point. When x = 0, y =. 44 When x = 4, y =. 45 Stationary point: x = 1. 70, y = 9. 8 Maximum value occurs at the end point and is equal to y =. 45. (ii) Find f ( x) = 0 : x =. 94 (iii) f '() = The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 8

9 QUESTION a. (i) B represents the amplitude and reflection in the X axis (if it exists). Amplitude = half the distance between the maximum and minimum values = = 1.5 m No reflection exists, therefore, b = (ii) Substitute the point ( 0, ) into the equation: = a +1.5 cos a = 4.50 ( 0) (iii) Period π = = n π ( π ) 1.4 = 1.40 hours b. Convert minutes to hours. Find h when t = hrs: () π h () = cos 1.4 4π h () = cos = 5.9 m 1.4 πt c. (i) cos = πt 1.5cos = πt cos = Equating with Acos( Bt) = C results in π A= 1, B=, C = The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 9

10 (ii) 1st Quadrant Angle: Cos 1 π (0.5) = Solutions are to lie in the quadrants where cos is negative i.e. Quadrants and. πt π π = π, π πt π 4π =, 1.4 t = π 1.4, π 4π 1.4 π t = 4.1, 8. t = 4.1, 8.7 hours. (iii) The tide is.75 m high 4.1 hrs and 8.7 hrs after high tide. πt π 7 d. (i) h = cos = cos =. 1m (ii) Let ( x y ) (, 5.9) 1, 1 = Let ( x y ) (7,.1), = y y m = = = 0.4 m / hr x x 7 1 The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 10

11 QUESTION a. x b. f ( x) = 00e c. (i) y x = 00 e x 500 y = e 00 x 500 y log e = loge e 00 x 500 log e = y f 1 ( x) = log x e 00 Domain of f (x) ; x R Range of f (x) ; { y : y > 500} Domain of f 1 ( x ) ; { x : x > 500} Range of f 1 ( x 1 ) ; f ( x) R (ii) x' 0x 0 y' = y 1 (iii) x ' y ' = x ' y ' = + = The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 11

12 QUESTION 4 a. (i) (ii) b. Simplify the expression by substituting any known/given data: When x = 0, the derivative 1 f : = i.e. ( 0 ) = 1 ( x) = x x b ( 0) ( ) + b f + 1 = 0 b = 1 x x Find an antiderivative: f ( x) = + x + c = x x + x + c Solve for c by substituting the given values of x and y into the equation describing f ( x): ( 0 ) = f i.e. When x = 0, y = = c c = The equation describing ( x) x f is x + x +. The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 1

13 c. (i) (ii) The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 1

14 QUESTION 5 (a) (b) (c) (d) (e) (f) The School For Excellence 015 Unit Master Classes Maths Methods Exam Page 14

UNIT 2 MATHEMATICAL METHODS 2013 MASTER CLASS PROGRAM WEEK 11 EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

UNIT 2 MATHEMATICAL METHODS 2013 MASTER CLASS PROGRAM WEEK 11 EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS UNIT MATHEMATICAL METHODS 01 MASTER CLASS PROGRAM WEEK 11 EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES SECTION 1 MULTIPLE CHOICE QUESTIONS QUESTION 1 QUESTION QUESTION