TEMASEK JUNIOR COLLEGE, SINGAPORE JC 2 Mid-Year Examination 2017 Higher 2 MATHEMATICS 9758

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TEMASEK JUNIOR COLLEGE, SINGAPORE JC Mid-Year Eamination 07 Higher MATHEMATICS 9758 Paper 0 June 07 Additional Materials: Answer paper hours List of Formulae (MF6) READ THESE INSTRUCTIONS FIRST Write

1 TEMASEK JUNIOR COLLEGE, SINGAPORE JC Mid-Year Eamination 07 Higher MATHEMATICS 9758 Paper 0 June 07 Additional Materials: Answer paper hours List of Formulae (MF6) READ THESE INSTRUCTIONS FIRST Write your Civics group and name on all the work that you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-eact numerical answers correct to significant figures, or decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are epected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the eamination, fasten all your work securely together. This document consists of 7 printed pages. [Turn over

2 At the beginning of the year, Mr Lui invested a total of $ in three banks. Bank A pays a fied amount of interest of $ at the end of each year. Bank B pays a compound interest of.5% on the amount invested at the end of each year and Bank C pays an interest of 5% on the amount invested at the end of 5 years. The amount of money Mr Lui invested in Bank B is half of the amount he invested in Bank C. At the end of the first year, Mr Lui has a total of $40800 in the three banks. His investment grew to $4 609 at the end of the second year. Find the total amount of money Mr Lui has in the three banks at the end of 5 years, assuming that he did not make any withdrawal or further investment during the period. [5] [Solution] Let $a, $b, $c be the amount of money Mr Lui invested in Bank A, B and C respectively. a bc b 0.5c b0.5c 0 a.05bc a.05 bc 4609 Using GC, a = 0 000, b = , c = and = 00 Define the unknowns clearly. Avoid using A, B and C which are the names of the banks. It is more efficient to use GC to solve the simultaneous equations than algebraically. Total amount of money Mr Lui has at the end of 5 years 5 = $0000 5($00) (.05) $40000 (.05)($80000) $ In this contet (money), the non-eact amount should be rounded off to dec place unless otherwise stated. Marker s Comments This question was generally well done. However, some students did not correctly understand the conditions given in the contet and thus scored very few marks. Students who solved the simultaneous equation algebraically instead of using GC wasted precious time (at the epense of other questions) and often make mistakes. There is also a large number of students who have gotten the simultaneous equations correct but generated wrong answers from GC. Students need to be more careful when entering numbers in GC.

3 It is given that f( ), 0. Solve the inequality f( ) using an algebraic method. [4] Hence find the range of values of for which f. [], Multiply both sides by, 0, 0 Some students sketched this curve wrongly as 0 Thus, solution is or 0. For f, replace by. or 0 (No real solution 0 or 0 0 ) Many students forgot the condition 0. contributes no solution, and not NA or reject. Solution to 0 can be easily seen from Marker s Comments The first part was generally well done. and not from splitting into 0 and Many students gave a poor presentation of the solution to the Hence question. Student s solution: or 0 WHY? Is this replacement Replace by, part of the method for solving or 0 the first question? There should be a clear indication that you are solving the nd question by first writing the question f.

4 4 The vectors a and b are unit vectors such that = + (i) Find a b b a b. ab b b. [] (ii) Show that the angle between the vectors a and b is 0. [] (iii) Find the eact value of [Solution] ab b (i) Since is perpendicular to b, a b b b 0 a b. [] Note that cross product is not associative: ab b a bb (ii) abb ab ab b b ab b 0ab b ab abcos, where is the angle between a and b cos 0 (Shown) (iii) ab abab = a ab b Alternative solution a+b b Using cosine rule, a ab a b a b cos Take note that if use cosine rule, must draw diagram to see that the angle (opposite a b ) is 60 o and not 0 o.

5 5 Marker s Comments (i) Many students leave the answer as ab which is not accepted since question did not say leave answer in terms of a and b. Some students went on to wrongly assume that vector a and vector b are perpendicular. There are also many students who are not aware that cross product is not associative and they did by the WRONG method: X a0b ab b b a bb b 0b 0 (ii) Many did this part (ii) by working backward then they realise that part (i) answer should be zero. However, they are unable to eplain the reason why a b b b. Many 0 are also stuck at not knowing how to arrive at a b. (iii) Attempt to use cosine rule is common. However, many did not draw diagram and they use 0 o instead of 60 o. Mistakes such as b and ab a b are still common. In general this is a badly done question. Many students skip part (i) and (iii) with little working for part (ii).

6 6 4 (i) Show that ( r )! r! r! r. [] (ii) Find!! n! n n! n n (iii) Hence, find r! r r in terms of n. [] in terms of n. [] (iv) Determine with a reason if r! r eists. [] r [Solution] (i) ( r )! r! ( r ) r! r! r r! r! r (Shown) Alternatively by epansion: ( r)! r! ( r) r! r! r r! r! r! r! r (ii) n n n n!!!! n r n r r! r ( r)! r!!!!! 4!! n n n n n! n!!!!! Take note that the last term is given by r = n and not r = n. (n )!

7 7 n (iii) r! r r rn r n r n r r! r r! r r! r!! (n )! 5 (n )! 6 We are looking for n r! r which is same as r n n! 4! 4...! Thus any method to change the general term must keep the series the same. n (iv) r r n r!! 6 Since as n, n!, does not eist. the sum r! r r Marker s Comments (i) (ii) Very well done but there are still some students who do not know the meaning of factorial. Very well done ecept for few cases of ending with r = n. Students must be reminded to show cancellation in their working. There are some students who present their working!!!!... n! n! n! n!. Even though in the row form they showed cancellation in the working, this way of arranging the terms should be discouraged as not all students are able to see which terms are being cancelled out.

8 8 (iii) Very badly done. Mistake I Many students replace r by r in n r r! r. However this leads to mistakes such as: Did not change upper limit. Lower limit becomes r and students do not know how to handle it. There are some who successful find r! r method: n r n r n n r r n by the following correct alternative r r! r r! r n!!! r! r r r n!! 6 n r r! r n! 6 (Replace n by n) n! 6 Mistake II Another very common problem is that many students do not know how to do Hence after obtaining n r! r. The answer n!! is written down without any working. A r misconception is due to some students thinking that since lower limit is, so can just replace the! in ( )!! n as!. Other possibility is that the student obtain this answer without using Hence, but using MOD separately instead. Another common wrong answer is (n )!!

9 9 (iv) Very badly done. Many students do not understand the meaning of r! r eists. They write down n r conclude that r! r r eists.! and they can wrongly Poor presentation such as n! 66 is common. There are also a few students who state that there is no common ratio and so r! r does not eists, indicating poor understanding of sum to infinity of the r different types of series.

10 0 5 The parametric equations of a curve C are where 4 acos t, 0 t and a is a positive constant. 4 4 y asin t (i) Find the equation of the tangent to C at the point P with parameter p. [] The tangent at P meets the -ais and y-ais at point A and B respectively. (ii) Show that OA + OB depends only on a, where O is the origin. [] (iii) Find in terms of a, the area of the region bounded by C, the aes and the tangent of the curve at point P when OA = OB. [4] [Solution] d (i) 4acos t dt sint dy 4asin t dt cost Note: There is no need to 4 convert cos t to dy 8asin tcost = d 8acos tsint sin t (or cos t tan t ) cos 4t using double angle formula when differentiating. Quite a number of students perform this unnecessary step. Equation of the tangent at point P (i.e. when t = p) : 4 sin p 4 yasin p acos p cos p 4 4 ycos pasin pcos p sin p asin pcos p 4 4 ycos p sin pasin pcos p asin pcos p y cos p sin pasin pcos p cos p sin p ycos p sin p asin pcos p A number of students did not sub t = p when finding the equation of the tangent at point P.

(ii) When y = 0, sin p asin pcos p acos p Point A = acos p,0 When = 0, y asin p Point B = 0, asin p (iii) ycos p asin pcos p OA OB acos p asin p a cos p sin p a, depends only on a (Shown) a aa Area =

11 (ii) When y = 0, sin p asin pcos p acos p Point A = acos p,0 When = 0, y asin p Point B = 0, asin p (iii) ycos p asin pcos p OA OB acos p asin p a cos p sin p a, depends only on a (Shown) a aa Area = y d a asin t 8acos tsin tdt 4 8 a a (By GC) 6 8 a units 4 Marker s Comments Overall, this question was very badly done. Although most students knew how to find the gradient, they made mistakes with the differentiation of the trigonometry function. Many also struggled when they tried to simplify the trigonometry terms, especially when trying to find points A and B. Some failed to read the question carefully. They mistook points A and B to be the and y intercepts of the curve rather than the and y intercepts of the tangent. Part (iii) was mostly left unattempt. Those who attempt gave the wrong limits (t values) for the integral.

12 6 Mr Ng s New Year resolution is to learn a new piece of piano music. To achieve that, he comes up with a practice schedule to learn a new piece. He practises for 5 minutes on Day, and he increases his practice duration by two minutes each day, until the duration reaches a maimum of one hour. He will then subsequently practise only hour each day. (i) Show algebraically that Mr Ng first practised for hour on Day 9. [] (ii) Show that Mr Ng practised for a total of 60N 4 minutes from Day to Day N inclusive, where N 9. [] Mr Ng sets a metronome at 00 Beats Per Minute (BPM) throughout his practices. He uses it only during his practices. However, the metronome is faulty. In the first minute, the metronome beats at 00 BPM as set by Mr Ng. For each subsequent minute of use, the number of beats decreases by %. (iii) The number of beats played by the metronome after the last minute of practice on Day N is 00k 60N 4, where N 9. State the value of k. [] The metronome is due for calibration when the metronome plays fewer than 99% of Mr Ng s setting of 00 BPM. (iv) Determine the day that the metronome is due for calibration. [4] [Solution] (i) On Day 8, Mr Ng would play for u8 5 (8 )() 59 minutes. Hence, Mr Ng would play for hour on Day 9. (ii) Method [Direct] From Day to Day 8, Mr Ng would have played a total of 8 S minutes. From Day 9 to Day N, Mr Ng would have played a total of 60N 860N 080 minutes Hence, he would have played a total of 60N N 4 mins(shown) Method [Indirect] The number of minutes short of hour played each day follows an AP: 5 (Day ),,, 5,, (Day 8) If he played for 60 minutes each day from Day to Day N, he would have played 60N minutes. He would have played a total of 60N 4 minutes. (Shown) Students should include a written eplanation to answer the question. Finding S 9 is incorrect as Day 9 (60 mins) does not follow the AP.

13 (iii) Actual number of beats (accuracy) after Day N N 4 Hence, k (iv) Method [use result in (iii)] To find the day when the accuracy falls below % of 00 BPM, no. of beats in original setting actual no. of beats beat 60N N N ln N 4 ln N least integer 9 N Hence, the metronome is due for calibration on Day 9. As is an eact value, students should not round it off. Since ln , the inequality sign must be changed when division is performed. Method [use the fact that the number of beats played after each minute follows a GP] Let n be the total number of minutes played. n ar n n ln 0.99 n ln n 0.06 From (ii), we know that Mr Ng played 60N 4 min on Day N. 60N N N 9 Some students who used this method mistook n to be the number of days instead. We cannot divide n by 60 to find the number of days because Mr Ng did not play for 60 minutes everyday. Marker s Comments The question was generally well attempted. Students were able to make use of the results they were required to show to construct their solutions. However, in doing so, students should be very careful of the possible logical lapses in their presentation. In fact, basic algebraic errors caused students to lose their marks even though they were able to answer the question.

4 7 (a) The diagram below shows the sketch of the curve C with equation y 4, y y = R O = = 0. Find the eact area of the shaded region R bounded by the curve C, the lines, y and the y-ais.

14 4 7 (a) The diagram below shows the sketch of the curve C with equation y 4, y y = R O = = 0. Find the eact area of the shaded region R bounded by the curve C, the lines, y and the y-ais. [] (b) The region bounded by the curve y, the line and the - and y- aes is rotated through 60 o about the y-ais. Find the eact value of the volume of the solid formed. [4] (c) A wine cask has a radius of 0 cm each at the top and bottom, and a radius of 40 cm at the middle. The height of the cask is m. To find the volume of the cask b by using the formula y d, we lay the cask on its side as shown in the a diagram below. Find a possible equation in the form y = f() for the arc AB of the cask. []

15 5 [Solution] (a) Area of the shaded region R = 0 d 4 = 4 d 0 (b) y y When, y 7 When 0, y. 4 4 Volume of the solid = dy = dy 4 7 y 7 9 = y dy 4 7 y 7 9 = y ln y y = 7 ln = ln or ln7 7 Use f'( ) f d n f( ) = n n C Common mistakes: Vol = 0 Vol = 0 dy d

16 6 (c) Since points (0, 40) and ( 50, 0) and (50, 0) lie on the arc AB, and the arc AB is symmetrical about the y-ais with maimum point at (0, 40), Method : let equation of the arc AB be: y a b When = 0, y = 40 b = 40 When = 50, y = 0 0 = 50 a + 40 Therefore, the equation of the arc AB is y 40 for Method : let equation of the arc AB be: y a b a. 50 When = 0, y = 40 a = When = 50, y = 0 b. 40 b 7 Therefore, the equation of the arc AB is y 7, for and 0 y Marker s Comments Part (i) was generally well done with only a minority in each class who failed to recognise n that the integral is of the form f'( ) f d and either went to use integration by parts or wrote their answer as ln 4. For Part (ii), a few students from each class wrote d 0, not knowing that the question asks for volume of rotation about y ais. Some made mistakes with the limits. For Part (iii), some students did not attempt this part. Some were able to make the correct guess of the type of equation (either parabola or ellipse) but were unable to find the coefficients for the equations.

17 7 8 (a) Showing your workings clearly, find the comple numbers w and z which satisfy the simultaneous equations wz i and w iz. [5] (b) The comple number v satisfies the equations vv and v arg. v (i) Find the modulus and argument of v. [] (ii) n Find the smallest positive integer n such that v is a real number. [] [Solution] (a) wz i --- () w iz w iz --- () Substitute () in (): z iz i iz zi 0 44(i)( i) 4 i i z 4i 4i 4i i i i i i z i z i i i i i i i i i w i w i +i i +i i Thus, or Alternative solution wz i --- () w iz w z --- () i Substitute () in (): w w i i w w 0 4 4() 4 i w i i.e., w i or w i i i z z i i i i i i i i i i i i

18 8 (b)(i) vv v v since v 0 v arg v arg v arg v argv argv (b)(ii) arg v 4 n v is a real number argv n nargv k k, where k n k 4 4k n, k 4() Thus smallest positive integer n 4. Marker s Comments For (a), some students attempt to solve this problem by first epressing w = a + bi and z = c + di, which results in a system of equations involving 4 unknowns. Many students who use this method did not manage to get the correct answers. This method is not recommended. For (b), students must know what an integer means. Some students leave their answer as 4 n which is clearly not an integer.

19 9 9 (a) The gradient of a curve C changes with respect to at a rate of where 0. Given that the line y is a tangent to C at the point e, find the equation of C where 0. [5] (b) A Ricatti equation is a first order differential equation that is quadratic in y. It is of the form dy f ( ) f ( y ) +f ( y ) f 4( ) d where f ( ), f ( ), f ( ), f 4 ( ) are functions of. dy 5 It is given that sin sin y + sin cos y d 4 By using the substitution u sin y, show that the given Ricatti equation can du 5 be reduced to u u+. Hence, find the particular solution of the given d 4 Ricatti equation given that y when, leaving your answer in terms of. 6 [7] [Solution] Common Mistakes dy (a) d y dy Rate of change of gradient =, d y, d d d d d dy ln k d Did not realise > 0 and left answer as d y ln k d When e, y dy ln e d ln k ( Note that d k equation of tangent line d y d y ln d Wrong formula when doing integration by part: ln dc ln dc ln c ln c Note: d ece d, c When = e, y e, e eln e c c e Equation of curve C is y ln e where > 0. Many students did not find the value of c.

20 0 (b) Given: u sin Differentiate wrt, du dy sin d d y cos y Given DE: dy 5 sin sin y + sin cos y d 4 dy 5 sin cos y sin y + sin y d 4 du 5 Therefore, u u () (Shown) d 4 du d 5 u u+ 4 du d u When tan u C, y and thus u sin 6 6 Hence, tan C C 6 u tan sin cosec y tan y tan Common Mistakes Did not put brackets at the right dy place. E.g. cos y, sin d Cannot differentiate u sin y correctly: E.g. d u ycos d, d u ycos dy Prove it in long method way. Wrong separable variable 5 method: d u d u u 4 Many students cannot do completing square correctly. u tan C 4 Did not make y as a subject. Many students solve the DE () wrongly as follow: When, y u 6 du dy 4 d d y 4 d = Marker s Comments (a) Badly done for this part. Many student do not know that rate of change of gradient d y is d. (b) Most students can prove the first part result but many of them proved it in a long way. Many students failed to solve the DE.

21 0 (a) The graph of y f( ) is shown below. The curve has a maimum point at A(6,4) and aial intercepts at B (0,4) and C (,0). The lines and y are the vertical and horizontal asymptotes respectively. y B(0, 4) A(6, 4) 0 C(, 0) (i) Sketch the graph of y, giving the equations of any asymptotes, the f( ) coordinates of any turning points and the points of intersection with the aes. [] y f f 4. [] (ii) By sketching the graph of Solve the inequality (b) ( y b) The graph below shows the curve with equation, where a, b and a c c are positive constants. Find the values of a, b and c. [] y O Describe a sequence of transformation which would transform the graph of ( y b) y onto the graph of. [] a c

22 [Solution] (a) (i) y 0 (a) (ii) By sketching out the graph of y f( ), range of values of is or 0 or. (b) By observation, b. ( y ) Substitute (, ) into, a c we have a a since a is positive. Equations of oblique asymptotes: y c c a ( y ) curve has equation Step : Scaling of factor parallel to -ais Step : Scaling of factor parallel to y-ais Step : Translation of units in the positive y-direction

23 Marker s Comments (a)(i) The sketching of the y graph is generally well-attempted, with the main f( ) issue being that A and B must be level on the coordinate scale in order to achieve maimum marks. (a)(ii) Many students sketched the graph of y f( ) wrongly which resulted in them having confusion with the inequalities. Most students fail to notice that the inequality holds when 0. Students who used and instead of or for their solutions were penalised. (b) Descriptions of transformations were allowed if reasonable as there were many variations. For eample, the epected translation was presented as a move or shift, and a scaling parallel to the y-ais was presented as a scaling up the y-ais. Students lose marks by attaching units to scale factors, and by getting the sequence of transformations wrong.

24 4 The function f is given by (i) for 0, f for 0. Define f in a similar form. [4] y f and y f on the same diagram. Without the use of graphing calculator, find the eact solution(s) of the equation f f. [5] (ii) Sketch the graphs of, 5. Another function g is given by g( ) ln5 (iii) If the domain of f is now restricted to the interval [,0], show that the composite function gf eists. Find an epression for gf, stating its domain and range in eact form. [4] [Solution] for 0, (i) f for 0. Let y y y Since 0, y. Also, let y y 4 Hence, for, f 4 for.

25 5 (ii) At the intersection,. f( ) f ( ) 0 Since 0,. Also, 4 4( ) or Hence, the eact solutions of f f are, 4.

26 6 (iii) Domain of f restricted to [,0] R f [,] Since [, ] R f D g (,5), gf eists D D [,0] gf f gf ( ) ln 5, 0. From the graph of y gf ( ), R gf [ln,ln 4] Marker s Comments In general, students should improve on their solution (set notations, etc) for function questions. (i) Most students successfully considered the different domains and were able to epress in terms of y in order to find the inverse function. However, a number of students were unable to determine the domain of inverse function. (ii) The graphs were not drawn properly in general. Many students did not include details of the graph (part parabolic in shape, aial intercepts, symmetry between (iii) graphs). The part involved the finding the solution of f f was not well done. A number of students were not aware that they can solve the question by equating. Students who ended up with solutions of more, showed f complete ignorance of the graphs they had drawn in the earlier part ( intersections). A handful of students were penalised for writing out the solution in coordinates, indicating the fact that they did not understand what they were solving. Many problems arose when students were unable to write out the domain and range of the functions accordingly and show that the composite function eists. A number of students were not paying attention to the question, resulting in loss of marks, not stating the rule, domain and range of the composite function.

ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT. Paper 1 18 August 2015 JC 2 PRELIMINARY EXAMINATION Time allowed: 3 hours

ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 9740 / 0 Paper 8 August 05 JC PRELIMINARY EXAMINATION Time allowed: hours Additional Materials: List of Formulae (MF5) READ THESE