THE KING S SCHOOL 03 Higher School Certificate Trial Eamination Mathematics Etension General Instructions Reading time 5 minutes Working time hours Write using black or blue pen Board-approved calculators may be used A table of standard integrals is provided at the back of this paper All necessary working should be shown in every question Total marks 70 Section I 0 marks Attempt Questions -0 Answer on the Multiple Choice Answer Sheet provided. Allow about 5 minutes for this Section. Section II 60 marks Attempt Questions -4 Answer in the eamination booklets provided, unless otherwise instructed. Start a new booklet for each question. Allow about hour 45 minutes for this Section. Disclaimer This is a Trial HSC Eamination only. Whilst it reflects and mirrors both the format and topics of the HSC Eamination designed by the NSW Board of Studies for the respective sections, there is no guarantee that the content of this eam eactly replicates the actual HSC Eamination. Y THSC Mathematics Et 083.doc of 5
Y THSC Mathematics Et 083.doc of 5 Student Number
Student Number Section I Questions 0 (mark for each question) Read each question and choose an answer A, B, C or D. Record your answer on the Answer Sheet provided. Allow about 5 minutes for this section. sin α cos α A) sin4α B) 4sinα cosα C) sinα D) sinα The point R divides the interval joining P ( 3, 6) and Q (6, 6) eternally in the ratio :. Which of these are coordinates of R? A) (3, ) B) (5, 8) C) (0, ) D) (, 8) 3 The polynomial 3 + 4 0 has roots α, β, and γ. What is the value of + +? α β γ A) 4 B) C) D) 4 Y THSC Mathematics Et 083.doc 3 of 5
4 Which of the following are solutions of 0 +? A) 0 B) 0 C) < D) <, 0 5 Which of the following could be the equation of the polynomial P()? y P() A) P() ( ) ( + ) B) P() ( + ) ( ) 3 C) P() ( ) ( + ) 3 D) P() ( + ) ( ) 3 Y THSC Mathematics Et 083.doc 4 of 5
6 What is the equation of the function shown in the graph below? π A) y cos 3 4 B) y C) y 4 cos π π cos 4 4 3 3 D) y cos 3 π 3 3 7 A particle is moving in a simple harmonic motion with displacement. Its acceleration is given by 4 + 3. What are the centre and the period of the motion? A) 3, π T B) 3, T π C) 3, 4 T π D) 3, 4 π T 8 What is the derivative of y tan? A) 4 + B) 4 + C) 4 + D) 3 + Y THSC Mathematics Et 083.doc 5 of 5
9 The points A, B and C lie on the circle with centre O. OA is parallel to CB. AC intersects OB at D and ODC α. What is the size of OAD in terms of α? A) α B) α 3 C) α 3 D) 3α 0 The trigonometric epression 3 cos θ + sin θ may be written in the form R cos(θ α). The values of R and α could be? (A) R 4 and α π 3 (B) R 4 and α π 6 (C) R and α π 3 (D) R and α π 6 End of Section I Y THSC Mathematics Et 083.doc 6 of 5
Section II Question 4 (5 marks each) Allow about hour 45 minutes for this section Question MARKS a) Differentiate e cos with respect to. d b) Find () 3 c) Solve <. 3 d) Use the substitution u + ln to evaluate 3 e d ( + ln ) 3. e) i) Find the coefficient of 5 in the binomial epansion of 5 + 3 5 n ii) What is the greatest value of n for which + 3 has a non-zero term in 5. f) In how many ways can 0 people be divided into two groups of 5 each? Y THSC Mathematics Et 083.doc 7 of 5
Question MARKS a) Consider the function f() + 4 i) Show that this function is odd. ii) Find the equation of the horizontal asymptote. iii) Find the coordinates of its stationary points and determine their nature. iv) Sketch the graph of this function. b) Let g() ln ( ) i) Find the domain of g(). ii) Show that g() is monotonic increasing. iii) Find the point of intersection of y g() and y. iv) On the same set of aes, sketch the graphs of y g() and y g (). c) i) Given that f() ( a) g() where f() and g() are polynomials, show that ( a) is a factor of f (). ii) Hence, or otherwise, find the values of p and q if 3 p + q 0 has a double root at. Y THSC Mathematics Et 083.doc 8 of 5
Question 3 MARKS a) Two circles C and C centred at P and Q with equal radii r intersect at A and B respectively. AC is a diameter in circle C and AD is a diameter in C. i) Show that ABC is congruent to ABD. ii) Show that PB AD. iii) Show that PQDB is a parallelogram. b) Use mathematical induction to prove that 3 n r ln ln r r + (n + )(n + ) for n. c) The velocity v ms of a particle moving along the ais is given by.5 0.5 v e i) Find the fastest speed attained by the particle. ii) After the particle reaches its maimum speed, you may consider 7 as the first approimation of the position at which its speed drops to ms. By using one application of Newton s method find a better approimation of for which the speed drops to ms. Y THSC Mathematics Et 083.doc 9 of 5
Question 3 (continued) MARKS d) A( 0, pt ) and B ( pt + p, y ) with p > 0 are two points on the number plane. The point C is chosen such that BC is parallel to ais, ACB 35 and BC p units. i) Show that the coordinates of C are ( pt, pt + pt ). ii) Show that the locus of C is a parabola and find the equation of its directri. Y THSC Mathematics Et 083.doc 0 of 5
Student Number Question 4 MARKS a) Two particles are moving in a simple harmonic motion along the ais. The displacements of the two particles at a time t are given by a cost + 3a and a sint. Find the closest distance between these two particles and the time at 4 which it first occurs. b) A helicopter H takes off vertically from C. Paula and Bob are watching the helicopter. Bob is due east of the helicopter H. He is at B the top of a building AB of height a metres. Paula is on the ground, at a point P, a metres away from A. The bearing of the building AB from her position at point P is 330. At the instant that Bob sees the helicopter at an angle of elevation of θ, its distance from him is 3a metres. 3a H B θ A a s C N P i) Show that the distance from Paula to the helicopter at this time, can be 3 epressed as s a 4 + 6 π sin θ 4 ii) It is known that at the time when θ is increasing at a rate of 0. radians per minute the angle of elevation θ is π radians. 5 At what rate is s changing at this time, give your answer correct to two decimal places. Y THSC Mathematics Et 083.doc of 5
Question 4 (continued) MARKS c) Joe is playing indoor soccer. He is to take a free kick from the origin. The opposing team has positioned some of its players 3 m from the ball. These players are.8m tall, but can jump an etra. m to defend their goal. The ceiling of the stadium is 8m above the floor. Joe will kick the ball with a velocity of 3 ms - at an angle of θ to the horizontal. Use the aes as shown, and assume there is no air resistance and the position of Joe s ball t seconds after being kicked is given by the equations: 3t cosθ and y 3tsinθ 5t² (Do NOT prove this) 8 y 6 4 3ms - 3m.m.8m m.m i) Show that the maimum height reached by the ball can be 69 sin θ epressed as h. 0 ii) Show that the cartesian equation for the trajectory of the ball 5 is y tan θ (+ tan² θ). 69 iii) Show that for the ball to pass over the defenders and below the ceiling the angle θ must be between 50⁰4 and 76⁰39. iv) The goal is. m high and m from the origin. For what possible values of θ will the ball pass above the defensive players, below the ceiling, and enter the goal directly? Y THSC Mathematics Et 083.doc of 5
STANDARD INTEGRALS n+ n d, n ; 0, if n < 0 n + d ln, > 0 e a d e a, a 0 a sin a d cos a, a 0 a cos a d sin a, a 0 a sec a d tan a, a 0 a sec a tan a d sec a, a 0 a a + d tan a a, a 0 a d sin, a > 0, a < < a a a d ln + a, > a > 0 + a d ln + + a NOTE : ln loge, > 0 Y THSC Mathematics Et 083.doc 3 of 5
Student Number Multiple Choice Answer Sheet Section I Total marks (0) Attempt Questions -0 Allow about 5 minutes for this section Select the alternative A, B, C or D that best answers the question. Fill in the response oval completely. A B C D A B C D 3 A B C D 4 A B C D 5 A B C D 6 A B C D 7 A B C D 8 A B C D 9 A B C D 0 A B C D Y THSC Mathematics Et 083.doc 4 of 5
Student Number THE KING S SCHOOL 03 Higher School Certificate Trial Eamination Mathematics Etension Question Algebra and Number Differential Calculus Functions Geometry Integral Calculus Trigonometry Total -0 4 /, 3, 5, 6, 8 / 5 9 / 7 /, 0 / /0 c, e, f / 6 a / b / 3 d /3 /5 a, b, c /5 /5 3 b / 3 c(i) / c(ii), d / 6 a /4 /5 4 a, b, c /5 /5 Total /0 /9 /9 /5 /4 / /70 Y THSC Mathematics Et 083.doc 5 of 5
03 Etension Mathematics Solutions and Marking Guidelines A 4 D sin4α sinα cosα sinα cosα sin4α 0 0 + + + + B P( 3, 6) + 0 + From the table we can see that 0 Q(6, 6) when <, 0 Alternative method: 0 y 0 Hence, R (5, 8) 3 D + + α β γ βγ + αγ + αβ αβγ From the graph, 0, but as Hence, <, 0 5 B This graph has a double root at, hence (+)² is a factor. This graph has a triple root at, hence ( )³ is a factor. The only equation which could be the correct equation is P() (+)² ( )³ S & G Publishing 03 Mathematics Et Trial Solutions
6 B 7 C 8 C In A and D the domain is 3 that is which does not match 3 3 the given graph. The range in B of the equation y cos - is 0 y 4 Hence, B is the solution. The acceleration 4 + 3 4 ( ) The acceleration at the centre is 0. that is 0 so the centre is at Also, n 4 that is n as n > 0. Therefore, the period π. Hence, C is the solution. y tan - where u 9 B 0 D Let OAD Hence ACB (alternate angles, OA is parallel to CB) AOB (angle at centre is twice the angle at the circumference subtended by the same arc AB) α + (eterior angle of triangle OAD equals the sum of the two opposite interior angles) So α 3 Hence, R cos α 3 and R sin α R 4 R So tan α 3 4 + α π 6 S & G Publishing 03 Mathematics Et Trial Solutions
Question Suggested Marking Criteria Marks Question a) y e cos Using product rule where u e and v cos u e and v sin e cos e sin Indicate use of Product Rule. State the correct solution e (cos sin) b) d () d 4 d 4 4 Indicate use of sine inverse. Choose a correct substitution or equivalent 4 d sin + C State the correct solution. sin + C S & G Publishing 03 Mathematics Et Trial Solutions 3
Question Suggested Marking Criteria Marks c) so that is + + From the table: < < Begin a correct method. Alternative Method: ± ² - 0 From the graph, < < Create a correct epression without denominators. y ² State the correct solution. < < S & G Publishing 03 Mathematics Et Trial Solutions 4
Question Suggested Marking Criteria Marks d) e d ( + ln ) du 3 u u 3 du 3 Let u + ln du When e, u When, u Show du Find the new limits of integration. or equivalent. Show the correct solution. 3 8 ( ) ( ) e) i) 5 + 3 has T k+ 5 C k ( -3 ) 5-k ( ) k 5 C k -45 +3k k k 5 C k k -45 +5k To find the coefficient of 5, we let 45 +5k 5 we get k 0 Show an epression for T k+ (or equivalent). So the term needed is 5 C 0 0 5 Hence, the coefficient of 5 is 5 C 0 0 3 075 07 Show correct solution. 5 C 0 0 3 075 07 ii) 5 n + 3 has T k+ 5 C k ( -3 ) 5-k ( n ) k 5 C k -45 +3k k nk 5 C k k -45 +(3 + n )k To find the value of n such that the term in 5 is not a zero, we let: 45 + (3+n) k 5, so (3+n) k 50 n + 3 50 50 that is n 3 k k Hence, the maimum value of n occurs when k and this maimum value is n 47 State n 47 f) The first group of 5 can be selected in 0 C 5 ways. The other group of 5 can be formed in 5 C 5 ways. So the number of ways to get the two groups of 5 is 0 C 5 5 C 5 but the two groups formed are equal in size so they cannot be distinguished so we must divide by! Hence, the number of ways to form the two State the correct answer, 6 ways. S & G Publishing 03 Mathematics Et Trial Solutions 5
equal groups is 0 C 5 5 C 5! 6 S & G Publishing 03 Mathematics Et Trial Solutions 6
Question Suggested Marking Criteria Marks Question a) i) f () + 4 ( ) f (-) ( ) + 4 f () + 4 Hence f ( ) f (), so f () is odd. Show that f() is odd. ii) To find the horizontal asymptote take the limit as approaches infinity. Show the correct solution. Horizontal asymptote is y 0 0 So the horizontal asymptote is y 0 iii) f() let u and v ² + 4 + 4 u and v f () let f () 0 to find the possible stationary turning points,we get that is ±. when, y and, y 3 0 3 f () 0 0 f () Ma T.P. at (, ) Min T.P. at (, ) Find stationary point and determine its nature or find the value for both turning points. Find the second stationary point and determine its nature or find the y value for both turning points and determine their nature. Ma T.P. at (, ) Min T.P. at (, ) S & G Publishing 03 Mathematics Et Trial Solutions 7
Question Suggested Marking Criteria Marks iv) min (, ) Sketch a curve, correctly showing their stationary points. min (, ) Sketch a curve, correctly showing their curve approaching their asymptotes. b) i) g() ln( ) g() eists if ln ( ) eists that is if > 0 so > that is <. Hence, the domain of g() is < State the correct domain, g() is <. ii) g () + as <, > 0 that is g () > 0 as it is the sum of two positive terms. Therefore, g() is monotonic increasing throughout the domain of the function. Calculate the correct derivative. g () + Justify that the curve is monotonic increasing. g () > 0 for all as > 0 iii) To find the point of intersection of y and y g() solve g() solve ln ( ) ln ( ) 0 so and so y Hence, the point of intersection is (, ). Find the correct point of intersection. (, ) S & G Publishing 03 Mathematics Et Trial Solutions 8
Question Suggested Marking Criteria Marks iv) Since g() is a monotonic increasing function then to sketch the graph of g() we must first find g"() to determine the concavity of this curve. g"() for all <. Hence, the curve is strictly increasing and is concave up for all <. y g() Sketch y g() 3 0 3 4 5 y g - () Sketch y g-() (c) (i) f() ( a) g() f () ( )g() + g ()( a) Correct answer f () ( )[g() + g ()( )] ( a) is a factor of f () (ii) 8 4p + q 0 Finding equations to solve simultaneously 4p 0 q 4 Correct answer p 3 S & G Publishing 03 Mathematics Et Trial Solutions 9
Question Suggested Marking Criteria Marks Question 3 a) Show that ABC is congruent to ABD. i) In ΔABC and ΔABD, AC AD (diameters of equal circles) ABC ABD 90º (angles in semicircles with diameters AC and AD ) AB is common ΔABC ΔABD (RHS) ii) Let ADB α ACB α (corresponding angles of congruent triangles ABC and ABD are equal) PB PC ( equal radii of circle C ) PBC is isosceles ( equal sides) PBC α (base angles of isosceles ΔPBC are equal) ADB PBC α As these two angles are corresponding and equal hence PB (converse theorem of corresponding angles) Show PBC PCB. Show that PB AD. iii) PB (proved above ) PB QD (equal radii of equal circles) Hence PQDB is a parallelogram as it has one pair of opposite sides that are parallel and equal. Show that PQDB is a parallelogram. S & G Publishing 03 Mathematics Et Trial Solutions 0
Question Suggested Marking Criteria Marks n b) Prove that r ln ln for n r r + (n + )(n + ) For n LHS ln and RHS ln ln 3 3 3 Since LHS RHS, the statement is true for n Assume the statement is true for n k that is k ln + ln +... + ln 3 4 k + ln (k + )(k + ) Our aim is to prove the statement is true for n k+ that is k k + ln + ln +... + ln + ln 3 4 k + k + 3 ln ( k + )( k + 3) LHS k k + ln +... + ln + ln 3 k + k + 3 ln + k + ln (from assumption) ( k + )( k + ) k + 3 k + ln ( k + )( k + ) k + 3 ln RHS ( k + )( k + 3) Hence, if the statement is true for n k then it is also true for n k +. Now, the statement is true for n and hence by mathematical induction it is true for n, 3 and so on. Hence the statement is true for all positive integers n. Prove statement true for n. Assume true for nk, attempt to prove true for n k+ Satisfactory completion of proof. S & G Publishing 03 Mathematics Et Trial Solutions
Question Suggested Marking Criteria Marks c) i) v.5 e 0.5 So 3 v v e using the product rule let u 3 and v 4 e u 6 and v e 3 e Calculate the correct function for the acceleration. d v e (3 ) d 6 e 3 e e ( 3 ) To find the fastest speed we let 0 So e ( 3 ) 0 that is when 0 or 3 0 3 4 0 + 0 Find the velocity when their derivative indicates a maimum stationary point. v v.388 ms From the table the maimum speed occurs when 3 Fastest speed v 3.5 e.5.388 ms ] S & G Publishing 03 Mathematics Et Trial Solutions
Question Suggested Marking Criteria Marks ii) We need to find v that is we need to solve.5 e 0.5.5 0.5 which is e 0.5 0.5 Let g() e as 7 is the first approimation of the root then 7 Correctly substitute into the equation for Newton s Method. Using the product rule.5 0.5 let u and v e 0.5 0.5 u 3 and v 0.5e 0.5 0.5 So g () 3 e.5 e 0.5 0.5 e ( 3 ) 0.5 Calculate the correct value d) i) 7 7 + 0.37088 So 7.37088 Note: when 7, v.85... and when 7.37088 v.0040... This shows that 7.37088 is a better approimation. for the second approimation. 7.37088 D Show the value at C is pt. The coordinate of C is of B less by p. of C pt + p p pt Now producing the horizontal line BC to meet the y ais at a right angle at D. We are given BCA 35º then ACD 45º (angle of a straight line BD) DAC 45 (angle sum of ADC) So ΔACD is right angled and isosceles. AD DC pt Hence, the y coordinate of C is the y coordinate of A plus pt. y of C pt² + pt Hence, the coordinate of are C ( pt, pt + pt). Show the y value at C is pt² + pt. S & G Publishing 03 Mathematics Et Trial Solutions 3
Question Suggested Marking Criteria Marks ii) To find the equation of the locus of C, we need to find a relationship between and y independent of t. Given that pt and y pt² + pt t so y p + p Correctly substitute the epression for t into the epression for y. y + y p + y + 4py ² + 4p ² + 4p + 4p² 4py + 4p² ( + p )² 4p (y + p) This is the equation of a parabola with verte at ( p, p) and focal length p. Hence, its directri has the equation y p. Find the equation of the directri. y p S & G Publishing 03 Mathematics Et Trial Solutions 4
Question Suggested Marking Criteria Marks Question 4 a) The distance between the particles can be epressed as D a cost + 3a a sint. a cost a sint can be epressed in the form r cos (t + α ) where r a + a a a and tan α that is α as α is an acute angle. a π Therefore D a cos t + + 3a 4 π As the minimum of a cos t + 4 is a then the minimum distance between the particles is D a + 3a a ( 3 ) and this occurs for first time when t Alternative method: D a cost + 3a a sint Find the epression for the distance between the particles. D a cost + 3a a sint. Differentiate the epression for the distance between the particles and find when it equals zero. To find the minimum distance between the particles we let 0 that is t t 0 Use nd derivative to determine ma or min. ) Determine the time when the distance between the particles is a minimum. t t positive Negative From the table the minimum distance occur for the first time when t D a +3a a a (3 ) Hence, the minimum distance between the points is (3 ) a. Correctly calculate the closest distance between the particles. d (3 ) a. S & G Publishing 03 Mathematics Et Trial Solutions 5
Question Suggested Marking Criteria Marks b) i) B 3a θ H G A C Show that BG 3a cosθ cos θ sin θ AC BG 3a cosθ HG 3a sinθ HC a + 3a sinθ A 60º 3acosθ C a 30º Show that HC a + 3a sinθ P Using the cosine rule in triangle APC, we get: PC² 4a² + 9a² cos²θ a² cosθ cos60º 4a² + 9a²cos²θ 6a²cosθ Using Pythagoras theorem in triangle PCH, we get: PH² PC² + HC² s² 4a²+9a²cos²θ 6a²cosθ + a² + 6a²sinθ + 9a²sin²θ 4a² +9a²(cos²θ + sin²θ) 6a²cosθ + a² + 6a²sinθ 4a² + 6a²sinθ 6a²cosθ a² (4 + 6sinθ 6cosθ) 6sinθ 6cosθ can be epressed in the form r sin( θ - α) where r 6 + 6 6 6 and tan α that is α as α is 6 an acute angle. So s² a² (4 + 6 sin(θ )) Hence s a Show that s a S & G Publishing 03 Mathematics Et Trial Solutions 6
Question Suggested Marking Criteria Marks ii) s a Find. s Therefore but So ( rate for any θ ) As π θ then 5 Calculate the correct epression for. 0..7765 0.. a 0. 0.4a units per minute. c) i) y 3tsinθ 5t² So y 3sinθ 0t To find the time taken by the ball to reach the maimum height we let y 0, we get: 3sinθ 0t 0 3sinθ 0t t sinθ Now, the maimum height is y ma 3 sinθ sinθ 5 sin²θ sin²θ sin²θ sin²θ Show that the maimum height can be epressed as h 69sin θ. 0 S & G Publishing 03 Mathematics Et Trial Solutions 7
Question Suggested Marking Criteria Marks ii) 3t cosθ so t As y 3tsinθ 5t² then y 5 y tanθ Hence y tanθ Show that the cartesian equation is y tanθ (+tan²θ). S & G Publishing 03 Mathematics Et Trial Solutions 8
Question Suggested Marking Criteria Marks iii) For the ball to pass under the ceiling, if the maimum height y ma < 8 that is sin²θ < 8 sin²θ < sin θ < But θ is an acute angle so sin θ sin θ < Now as sin θ is an increasing function for θ between 0º and 90º then θ < 76º 39 A The ball must not be stopped by the defenders. This occur if when 3, y > 3 that is 3tanθ > 3 507 tanθ 45 45tan²θ 507 > 0 45 tan²θ 507tanθ + 55 < 0 First we solve 45tan²θ 507tanθ + 55 0, we get: Calculate the solution for the maimum height to be less than 8m. θ < 76º 39.0, 0.04556.. As y 45 tan²θ 507 tanθ + 55 is a parabola below the tan ais when is between.0 and 0.04556 So, the solution of 45 tan²θ 507 tanθ + 55 < 0 is.0 < tan θ < 0.04556 As tan θ is an increasing function for θ between 0º and 90º then 50º4 < θ < 84º9 B Now, for the ball to pass over the defenders, but under the ceiling, it must satisfy both conditions A and B Hence 50º4 < θ < 76º 39 Calculate the solution for the ball to stay above the defenders. 50º4 < θ < 76º 39 S & G Publishing 03 Mathematics Et Trial Solutions 9
Question Suggested Marking Criteria Marks iv) For the ball to enter the goal When, y <. that is tanθ <. tanθ <. 08 tanθ 70 70tan²θ 354.9 < 0 70 tan²θ 08 tan θ + 074.9 > 0 First,we solve: 70 tan²θ 08 tan θ + 074.9 0 we get: tan θ 0.707986 or.0868. As y 70 tan²θ 08 tan θ + 074.9 is a parabola above the tan ais when As tan θ is an increasing function for θ between 0º and 90º then θ 35º8, θ 64º38 Now, For the ball to enter the goal directly, it. must be above ground at that is When, y > 0 so tanθ > 0 08 tanθ 70 70 tan²θ > 0 70 tan²θ 08 tan θ + 70 < 0 First, we solve: 70 tan²θ 08 tan θ + 70 0 we get: tanθ or As y 70 tan²θ 08 tan θ + 70 is a parabola below the tan ais when < tanθ < As tan θ is an increasing function for θ between 0º and 90º then º37 < θ < 67º3 3 Calculate the solution for the ball to enter the goal. θ < 35º8 or θ > 64º38 and/or º37 < θ < 67º3 State the correct final range of angles for θ. 64º38 < θ < 67º3 (continues on net page) S & G Publishing 03 Mathematics Et Trial Solutions 0
Question Suggested Marking Criteria Marks (continued) Hence all of three conditions must hold simultaneously for a goal to be scored. above defenders, under ceiling under top edge of the goal under top edge of the goal enter goal above floor level 3 0 º37 35º8 50º4 64º38 67º3 76º 39 90 It can be seen on the above diagram that the largest range of values for which all of the conditions hold is 64º38 < θ < 67º3. S & G Publishing 03 Mathematics Et Trial Solutions