ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 1 19 August 2010 JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST
Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in.
Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
You are expected 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 the 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 examination, fasten all your work securely together.
This document consists of 6 printed pages.
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A-PDF Merger DEMO : Purchase from www.A-PDF.com to remove the watermarkANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010 MATHEMATICS 9740 Higher 2 Paper 1 Calculator model: _____________________
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Anglo-Chinese Junior College
H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 3 of 6
1 The nth term of a sequence is given by un 1n 1 2n , for n 1. The sum of the first n terms is denoted by Sn. Use the method of mathematical induction to show that
1 1 1 2 n n n n
S for all positive integers n. [4]
2 The 3 flavours of puddings produced by a dessert shop are mango, durian and strawberry. A mango pudding requires 5g of sugar and 36ml of water. A durian pudding requires 6g of sugar and 38ml of water. A strawberry pudding requires 4g of sugar and 40ml of water. The puddings are sold in pairs of the same type at $1.60, $2.20 and $1.80 for mango, durian and strawberry respectively.
On a particular day, 754g of sugar and 5972ml of water were used to make the puddings and all the puddings made were sold except for a pair of strawberry puddings. The collection from the sale of puddings was $142.40. Formulate the equations required to determine the
number of each type of pudding made on that day. [4]
3 The diagram below shows the graph of y f( )x . The curve passes through the origin and
has a maximum point atA 4 , 4 and asymptotes x 2 and y 2.
Sketch on separate diagrams, the graphs of
(i) 1
f( )
y
x [3]
(ii) y f '( )x [3] showing clearly asymptotes, intercepts and coordinates of turning points where possible.
4 The complex number w has modulus 3 and argument2
3 . Find the modulus and argument of *
i
w , where w* is the complex conjugate of w. Hence express *
i
w in the form a ib , where
a and b are real, giving the exact values of a and b in non-trigonometrical form. [4]
Find the possible values of n such that * n i w is purely imaginary. [2] [Turn Over y x f( ) y x 2 2 4 , 4 A
5 Solve the equation z4 i 0, giving the roots in the form re , where i r 0 and
. [3]
The roots represented by z and 1 z are such that 2 arg z1 arg z2 0. Show z , 1 z and 2
1 2
z z on an Argand diagram. Deduce the exact value of arg z1 z . 2 [3]
6 An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables.
(i) Country A is a developing country and the economist found that G and t are can be modeled by the differential equation 1
2
dG G
dt . Given that, when t 0, G 0, find G in terms of t. [4]
(ii) Comment on the suitability of the above differential equation model to forecast the future
economic growth of Country A. [1]
(iii) Country B is a developed country and the economist found that G and t can be modeled by
the differential equation 1 2
dG G
dt .
Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2]
7 (i) Given that f x ecos1x, where 1
1
x , find f 0 , f 0 and f 0 . Hence write down the first three non-zero terms in the Maclaurin series for f x . Give the coefficients in terms of k
e , where k . [4]
(ii) Given that g x tanx secx , where x is sufficiently small for x and higher powers of x 3
to be neglected. Deduce the first three non-zero terms in the series expansion of g x .
Hence, show that f x e2g x 2e2 x e . 2 2 [3]
(iii) Explain clearly why it is inappropriate to state that af 2g a 2 2 2 2
a x e x dx a e x e dx ,
where a . [1]
8 (i) Show that 1! 2 1 2
1 ! 2 ! Ar Br C2 !
r r r r , where A, B and C are constants to be
found. [2]
(ii) Hence find 2
1 3 3 3 2 ! n r r r r . [3]
(iii) Give a reason why the series 2
0
3 3 3
2 !
r
r r
r converges, and write down its value. [2] y
Anglo-Chinese Junior College
H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1 Page 5 of 6
9 Relative to the origin O, two points A and B have position vectors given by a 3i j 3k
and b 5i 4j 3k respectively.
(i) Find the length of the projection of OA on OB . [2]
(ii) Hence, or otherwise, find the position vector of the point C on OB such that AC is
perpendicular to OB. [2]
(iii) Find a vector equation of the reflection of the line AB in the line AC. [3]
10(a) The first 2 terms of a geometric progression are a and b ( b < a ). If the sum of the first n
terms is equal to twice the sum to infinity of the remaining terms, prove that an 3b . n [3]
(b) The terms u u u1, ,2 3,... form an arithmetic sequence with first term a and having non-zero
common difference d.
(i) Given that the sum of the first 10 terms of the sequence is 105 more than 10u , find 5
the common difference. [3]
(ii) If u is the first term in the sequence which is greater than 542, find the range of 26
values of a. [3]
11 The region R in the first quadrant is bounded by the curve 2y2 a a x , where 2 a 0, and the line joining 2 ,0a and 0, a . The region S, lying in the first quadrant, is bounded by the curve 2y2 a a x and the lines 2 x 2a and y a .
(i) Draw a sketch showing the regions R and S. [1]
(ii) Find, in terms of a, the volume of the solid formed when S is rotated completely about the
x-axis. [4]
(iii) By using a suitable translation, find, in terms of a, the volume of the solid formed when R is
rotated completely about the line x 2a . [4]
12 The curve C has the equation y 3x2x aax 2 where a is a constant.
(i) Find dydx and the set of values of a if the curve has 2 stationary points. [4]
(ii) Sketch the curve C for a = 1, stating clearly the exact coordinates of any points of
intersection with the axes and the equations of any asymptotes. [3]
Hence, find the range of values of k such that the equation 3x2 x 2 k x( 1) has exactly 2 real roots. [2]
13 The curve has the parametric equations
2 5 1 x t , 1 tan y t
(i) Sketch the curve for 2 t 2. [1]
(ii) Find the cartesian equations of the tangent and the normal to the curve at the point
wheret 1. [5]
(iii) Find the area enclosed by the x-axis, the tangent and the normal at the point where t = 1. [3] [Turn Over
14 The functions f and g are defined as follows: f :x sinx , , 2 2 x , g : 1 3 8 x x x , x .
(i) Sketch the graph of the function g, labeling clearly the exact values of the coordinates of turning point(s) and intersections with the axes, if any. [1]
State the range of the function g in exact values. [1]
(ii) Given that gf exists as a function. By considering the graphs of f and g, explain why
gf ( ) gf ( ) if
2 2. [2]
Hence what can be said about the function gf ? [1]
Without sketching the graph of gf , find the range of gf in the form ,a b , giving the exact
values of a and b. [1]
(iii) (a) Give a reason why fg does not exist as a function. [1]
(b) Find the greatest exact value of k for which fg is a function if the domain of g is
restricted to the interval 1, k . [2]
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 2 23 August 2010 JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST
Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in.
Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
You are expected 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 the 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 examination, fasten all your work securely together.
This document consists of 5 printed pages.
[Turn Over
9740 / 02
ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010 MATHEMATICS 9740 Higher 2 Paper 2 Calculator model: _____________________
Arrange your answers in the same numerical order.
Place this cover sheet on top of them and tie them together with the string provided.
Question no. Marks
1 2 3 4 5 6 7 8 9 10 11 12 13 14
/ 100
Index No:
Form Class: ___________
Anglo-Chinese Junior College
H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 3 of 5
Section A: Pure Mathematics [40 marks] 1 Find the exact value of
1 2 2 1 1 1 . x x e dx e [4]
2 The variable complex numbers z and w are such that z 2 i 3 and arg w 5 3i .
(i) Illustrate both of these relations on a single Argand diagram. [2]
(ii) State the least value of z w . [1]
(iii) Find the greatest and least possible values of arg z 3 , giving your answers in radians
correct to 3 decimal places. [4]
3 Find in terms of a, the range of values of x that satisfy the inequality ln 2x a 0
x ,
where a 1. [4]
4 (a) State the derivative of cos x . Hence, find 3 x5sinx dx 3 . [4]
(b) Find the exact value of
2 5 3
2 2
10
5
x dx , in the forma 2 b 3, using the
substitutionx 5 sec . [6]
5 The plane p1 has equation x 2y z 3. The line l1 has equation 1 1
2 4
x z
y .
(i) Show that the line l is parallel to, but not contained in the plane p1 1. [2]
(ii) Find the cartesian equation of the plane p2 which contains the line l1 and is perpendicular to
the plane p1. [3]
(iii) Find, in scalar product form, the vector equation of the plane p3 which contains the point
4,1, 1 and is perpendicular to both p1 and p2. [2]
Another line l2 which is parallel to the vector
2 0 3
intersects the line l1 at the
point A 1,0,1 .
(iv) Given that the line l2 meets the plane p1 at the point B, find the coordinates of B. [4]
(v) Find the sine of the acute angle between the line l2 and the plane p1, and hence, find the
length of the projection of the line segment AB on the plane p1, giving your answer in surd
form. [4]
Section B: Statistics [60 marks]
6 Mr Raju, who owns a supermarket wishes to find out what customers think about the goods that he sells. He has been advised that he should take a random sample of his customers for this purpose. State, with reasons, which of the following sampling procedures is preferable.
A. Select every 10th customer on each day in a typical week.
B. Select the first 20 customers on each day in a typical week [2]
7 One day, Pinocchio went shopping and bought a pair of size 30 Wood Shoes. The right shoes have lengths which are normally distributed with mean 20 cm and standard deviation 0.14 cm. The left shoes have lengths which are normally distributed with mean 20.1 cm and standard deviation 0.11 cm. The length of the right shoe is independent of the length of the left shoe.
When wearing the pair of shoes, Pinocchio takes six steps, heel to toe as shown in the diagram. Calculate the probability that the distance AB, from the back of the first step to the front of the sixth step, exceeds 120 cm.
[3] 8 On Ulu Island the weights of adult men and women may both be taken to be independent normal random variables with means 75kg and 65 kg and standard deviations 4 kg and 3 kg respectively.
Find the probability that the weight of a randomly chosen man and the weight of a randomly
chosen woman differ by more than 1 kg. [3]
Explain if this is equal to the probability that the difference in weight between a randomly chosen married woman and her husband is more than 1kg. [1]
9 Research has shown that before using an Internet service, the mean monthly family telephone costs is $72. A random sample of families which had started to use an Internet service was taken and their monthly telephone costs were :
$70, $84, $89, $96, $74
Stating a necessary assumption about the population, carry out a test at the 5% significance level, whether there is an increase in the mean monthly telephone costs. [5]
If the assumption stated above still holds, and if the standard deviation of the monthly telephone costs is $9.89, find the range of values of the mean monthly family telephone costs 0 that would lead to a reverse in the decision to the above test. [3]
10 Six overweight men registered at a slimming centre for a slimming programme. The following table records x, the height (to the nearest cm) and y, the weight (to the nearest 0.1 kg) of these six men.
(i) Given that the least square regression line of x on y line is x 103.6 0.726y , show that
the value of k to the nearest 0.1 kg is 80.3. Hence or otherwise, find the least square regression line of y on x in the form y ax b , giving the values of a and b to the nearest
3 decimal places. [5]
(ii) Based on the data given, use an appropriate regression line to predict the weight of an
overweight man who is 165 cm tall. [2]
(iii) Find the value of the product moment correlation coefficient between x and y and sketch
the scatter diagram of y against x. A particular man among the 6 men who registered for the slimming programme is unusually overweight. Indicate who this man is. [3]
Man A B C D E F
x (height in cm) 150 157 160 162 167 170
y (weight in kg) 65.1 73.2 85 k 80.9 89.9
Anglo-Chinese Junior College
H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2 Page 5 of 5
11 In a hotel, large number of cups and saucers are washed each day. The number of cups that are broken each day while washing averages 2.1. State in context, a condition under which a Poisson distribution would be a suitable probability model. [1]
Assume that the number of broken cups and saucers follow a Poisson distribution.
(i) Show that on any randomly chosen day, the probability that at least 3 cups are broken is
0.350 correct to 3 significant figures. [1]
The probability that there will be at least two days in n days with at least 3 broken cups is
more than 0.999. Find the least value of n. [3]
(ii) The number of saucers broken each day averages 1.6, independently of the number of cups broken. The total number of cups broken and saucers broken during a week of 7 days is denoted by T. State a possible model for the distribution of T. [2]
A random sample of 100 weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed
26. [3]
12 Fish are bred in large batches and allowed to grow until they are caught at random for sale. When caught, only 20% of the fish measure less than 8 cm long.
(i) What is the probability that the 10th fish caught is the sixth fish that is less than 8 cm long?
[2]
(ii) A large number, n, of fish are caught and the probability of there being 10 or fewer fish in
the catch which measures less than 8 cm long is at most 0.0227 . Using a suitable approximation, derive the approximate inequality
10.5 0.2n 0.8 n . [4]
Hence find the least possible number of fish to be caught. [2]
13 An automated blood pressure machine is being tested. Members of the public, %p of whom
have high blood pressure (hypertension), try it out and are then seen by a doctor. She finds that 80% of those with hypertension and 10% of those with normal blood pressure have been diagnosed as hypertensive by the machine. The probability that a randomly chosen patient who was diagnosed as hypertensive by the machine actually has hypertension is
3 2.
(i) Find the value of p [3]
(ii) Hence, find the probability that a randomly chosen patient does not have hypertension, given that the machine diagnosed him as having normal blood pressure. [2]
Comment on the usefulness of the machine. [1]
14 Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labeled ‘1’ and ‘2’, the three blue balls are labeled ‘1’, ‘2’ and ‘3’, and the five green balls are labeled ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’.
(i) Find the number of ways of choosing 2 balls of identical colour. [2]
(ii) Find the number of ways of choosing 6 balls if it includes at least one ball of each colour. [4] (iii) A person arranged 3 balls in a row with the numbered sides facing him forming a 3-digit
number. Among these 3 balls, none of them are green. Find the number of possible 3-digit numbers facing that person.
[The number formed is independent of the colours of the balls used. i.e. the number 112 is counted as one number whether the colour of the ball labeled ‘2’ is red or blue.] [3]
1 An g lo -C h in es e Ju n io r C o ll eg e H 2 M at he m at ic s 97 40 2010 JC 2 P R E L IM M ar ki ng Sc he m e er 1 : 1 Le t P n de not e the s ta te m ent () ( ) 1 1 1 2 n n nn S + + =− For n = 1, L H S = () 2 2 11 11 1 Su == − = R H S = () ( ) 2 12 11 2 −= L H S = R H S ∴ P1 is tr ue . A ss um e Pk tr ue f or s om e k + ∈ ℤ , i.e . () ( ) 1 1 1 2 k k kk S + + =− Pr ov e tha t P k+1 is tr ue , i.e . () ( )( ) 2 1 12 1 2 k k kk S + + ++ =− () () () ( ) () () () () () [] () () ( ) 11 12 2 1 1 2 LH S 1 11 1 2 1 12 1 2 1 12 2 12 1R H S 2 kk k kk k k k SS u kk k k kk k k kk ++ ++ + + + == + + =− + − + + =− − + + =− − − ++ =− = Si nc e P1 is tr ue , a nd Pk is tr ue ⇒ Pk+1 is tr ue , by th e pr in ci pl e of m at he m at ic al in du ct io n, Pn is tr ue n + ∀∈ ℤ . 2 Le t x b e th e no o f m an go p ud di ng s p ro du ce . Le t y b e th e no o f du ri an p ud di ng s p ro du ce . Le t z b e th e no o f st ra w be rr y pu dd in gs pr od uc e. 56 4 75 4 xy z ++ = ……… ………( 1) 36 38 40 5972 xy z ++= …… … … … … …… … (2 ) 0. 8 1. 1 0. 9( 2) 142. 4 xy z ++ − = ………… ……. (3 ) So lv in g (1 ), (2 ) & (3 ) us in g G C , 46 , 42 an d 68. xy z == = 3 (i) 1 f() y x = 2− y x A’ (4 , 0 .2 5) 0. 5 y = 0 x = 2 3 (ii) f' ( ) yx = 4 1 *3 i w − = 2 ar g ar g( ) ar g( *) *2 3 6 i iw w ππ π − =− − = − − − = 11 3 1 3 1 co s si n *3 6 6 3 2 2 6 6 i ii i w ππ − =+ = + = + 1 co s si n *3 6 6 nn in n i w ππ − =+ * n i w − is p ur el y im ag in ar y, co s 6 n π =0 () 21 , 62 nk k ππ =+ ∈ ℤ , ( ) 32 1 , nk k ∴= + ∈ ℤ . 5 44 4 2 0 1 i zi z i z e π −= ⇒ = ⇒ = 2 4 2 , 2, 1, 0, 1 ik ze k π π + == − − 1 2 42 73 5 88 8 8 , 2, 1, 0, 1 ,, , ik ii i i ze k ze e e e π π ππ π π + −− == − − = z 1 z2 z2 + z2 x y 8 π 5 8 π O y x 2− A ’ (4 ,0 )
3 5 8 1 8 2 i i ze ze π π = = ()12 14 3 ar g 82 8 8 zz ππ π += + = 6 1 2 dG G dt + = 11 12 ln 1 0. 5 dG dt G Gt C = + += + ∫∫ 0. 5 0. 5 1 1 , w he re tC tC Ge GA e A e + += ± =− + =± Wh en 0 t = , 0 G = , 1 A = 0. 5 1 t Ge ∴= − + Ex am pl es o f po ss ib le c om m en ts : Th e m od el is n ot s ui ta bl e be ca us e … Th e ec on om is t is a ss um in g th at t ha t th er e ar e no f luc tu at io ns i n th e ec on om ic gr ow th in th e fut ur e. Th e ec on om is t is a ss um in g th at t he c ou nt ry w il l en joy p er pe tua l ec on om ic gr ow th in th e lo ng te rm . Th e ec on om is t is a ss um in g C ou nt ry A i s al w ay s e xp er ie nc in g po si ti ve a nd in cr ea si ng e co no m ic g ro w th in th e fu tu re . Fa ct or s af fe ct in g ec on om ic g ro w th r em ai ns u nc ha ng ed . In t he l on g te rm , C ou nt ry B i s ex pe ct ed t o be s ti ll in r ec es si on w it h an e co no mi c gr ow th d ec re as in g to w ar ds -1% . 7( i) () 1 co s f x xe − = () 2 f0 e π ⇒= () 1 co s 2 1 f 1 x xe x − − ′ = − () 2 f0 e π ′ ⇒= − t G 1− 0. 5 1, 1 t GB e B − =− + = 0 4 (i i) (i ii) ()
()
() ()()
() () 11 11 3 co s co s 2 2 22 3 co s co s 2 2 2 2 11 1 f 12 2 11 1 1 f 0 1 xx xx xe e x x xx ex e x e x π −− −− − − −− − ′′ =+ − − − −− ′′ =− − ⇒ = − () 2 2 22 f. .. 2 e xe e x x π ππ ∴= − + + ( ) () 1 22 2 gt an se c 1 ta n 1 1 1 ... co s 2 2 1 2 − =+ =+ ≈ + − = + + − − + ≈+ + xx x xx xx x x x x () () 2 2 2 22 2 2 2 22 fg 1 22 2 ( S ho w n) ex xe x e e x x e x ee x π ππ π π ππ +≈ − + + + + =+ Th e st at em en t () () 2 22 2 fg 2 aa aa xe x dx e x e dx ππ π −− +≈ + ∫∫ is in ap pr op ri at e as () () 2 22 2 fg 2 xe x e x e ππ π +≈ + o nl y w he n a i s s uf fi ci en tly s m al l. 8 (i) (ii) () () () ( ) ( ) () () 2 12 1 ! 1! 2 ! 12 2 2 1 2! 1 2! r rr rr r r rr r −+ ++ ++ − + + = + +− = + 1, 1, 1. AB C == = − () () () () 2 1 2 1 1 33 3 2! 1 3 2! 12 1 3 ! 1! 2 ! n r n r n r rr r rr r r rr = = = +− + +− = + =− + ++ ∑ ∑ ∑5 (i ii) () () () () () () 12 1 1! 2! 3! 12 1 2! 3! 4! 12 1 3! 4! 5! 12 1 4! 5! 6! 3 12 1 2! 1 ! ! 121 1! ! 1! 12 1 !1 ! 2 ! nn n nn n nn n −+ +− + +− + +− + = +− + −− +− + −+ +− + ++ ⋮ () () 11 1 3 2 1! 2 ! nn =− + ++ ()
(
)
() 2 0 2 1 33 3 2! 33 3 3 2! 2! n r n r rr r rr r = = +− + +− =− + + ∑ ∑(
)
() () 31 1 1 3 22 1! 2 ! nn =− + − + ++ As () () 11 , 0 an d 0. 1! n+ 2 ! n n →∞ → → + th e se ri es c on ve rg es to 0 . ∴ 9( i) (i i) L en gt h of th e pr oj ec ti on o f OA o n OB = 35 12 0 14 2 2 50 50 33 ⋅− = = o r 4 2 M et hod 1: Fro m (i ), 22 OC = 55 12 22 4 4 5 50 33 OC =− = − M et hod 2: Li ne O B : 05 04 03 λ =+ − r 53 5 3 41 4 1 33 3 3 AC O C O A λ λλ λ − =− = − −= − − − Si nc e AC O B ⊥ , A C B O 6 (i ii ) 53 5 41 . 4 0 33 3 2 5 λ λ λ λ − −− − = −− = 5 2 4 5 3 OC =− Si nc e 2 5 OC O B = , :2 :3 OC C B = 5 11 '4 55 3 OB OB =− =− − O r us e m idp oi nt th eo re m , () 1 ' 2 OC O B O B =+ 55 5 21 '2 2 4 4 4 55 33 3 OB O C OB =− = − − − = − − 53 20 11 '' 4 1 1 55 33 18 AB O B O A =− = − − − = − Ve ct or e qu at io n of li ne ' AB is 32 0 11 , 31 8 r αα =+ ∈ ℝ ɶ 10 (a) (b) (i ) GP : a = a , r = b a Th e su m to in fi ni ty o f th e re m ai ni ng te rm s, 1 n ar S r ∞ = − ()()
2 1 2 11 12 31 1 3 1 3 3 n n n nn n n n nn SS ar ar rr rr r r b a ba ∞ = − = −− −= = = = =7 ( ii ) [] [ ] 10 5 105 10 10 29 10 5 10 4 2 10 45 105 10 40 21 Su ad a d ad ad d =+ += + + += + + = 25 542 24( 21) 542 38 u a a ≤ +≤ ≤ a nd 26 542 25( 21) 542 17 u a a > +> > 17 38 a ∴< ≤ 11 (i) (ii) (iii) Wh en S is r ot at ed c om pl et el y a bo ut th e x-a xi s, () () () () 2 2 0 2 2 3 0 32 3 Re qu ir ed v ol um e 2 2 2 2 2 22 22 2 c u. u ni ts a a a aa a x dx ax a a a aa a ππ π π π π π =− − − =− − =− = ∫ Af te r a tr an sl at io n of 2 a u ni ts in th e n eg at iv e x-d ir ec tio n, Ne w e qu at io n is () 2 2 2 22 ( 2 ) y ya a x a x a =− + ⇒ = − Wh en R is r ot at ed c om pl et el y a bo ut th e lin e 2 xa = , () ( ) 2 2 2 0 5 3 2 0 33 3 12 R equi re d vol um e 2 3 44 3 5 44 8 c u. uni ts 3 5 15 a a y aa dy a y a a aa a ππ ππ ππ π =− − =− =− = ∫ 12 x y a 2a O R S ( ) 2 22 ya a x =− 8 (i ) (i i) () ( )() () () () 2 2 2 22 2 32 63 2 36 2 xa x y xa xa xa x ax dy dx xa xa x a xa ++ = + ++ − + + = + ++ − = + Fo r st at io na ry p oi nt s, () 22 0 36 2 0 dy dx xa x a = ++ − = Fo r 2 st at io na ry p oi nt s, () ( )( ) 2 2 2 2 64 3 2 0 24 24 1 aa a a −− > >− >− 2 is a lw ay s po si ti ve . a . a ∴∈ ℝ 2 32 4 32 11 xx yx xx ++ == − + ++ Po in ts o f in te rs ec ti on w it h th e ax es : (0 ,2 ) As ym pt ot es : 32 an d 1 yx x =− = − . Th e ra ng e of v al ue s of k s uc h t ha t t he e qu at io n 2 32 ( 1) xx k x ++ = + h as e xa ct ly 2 re al ro ot s : 1. 93 or 11. 9 kk >< − 13 (i)
9 (i i) (i ii) ( ) () () () 22 2 2 2 2 2 -5 2t 1 a nd = 1 1 . 1+t 1 . 10 1 1 = 10 dy dx dt dt t t dy dy dt dx dt dx t t t t = + + = = − + + − Wh en 1, 1 5 t dy dx = =− Eq ua ti on o f ta ng en t :
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1 4 55 2 11 52 4 π y x π yx − =− − =− + + Gr ad ie nt o f no rm al is 5 Eq ua ti on o f n or m al :(
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4 5 5 2 25 5 24 π y x π yx − = − =− −(
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2 11 1 25 Ar ea = 5 22 4 5 2 4 4 13 1. 60 (3s .f ) 80 ππ π π ×+ − − × == 14 F un ct ion s Gi ve n :s in , fx x ֏ , 22 x ππ ∈− 5 10 (i ) (i i) an d :( 1) (3 ), 8 gx x x π +− ֏ x ∈ ℝ . (, ] 2 g R π =− ∞ From th e gr ap h of () yf x = , () ( ) ff αβ ≠ if 22 ππ αβ −≤ < ≤ . In f ac t, 1( ) ( ) 1 ff αβ −≤ < ≤ . Fr om th e gr ap h of () yg x = , () ( ) gf gf αβ < or () ( ) gf gf αβ ≠ 0 x y -1 1 3 (1 , ) 2 π 2 π 3π 8 () yg x = 0 x y 2 π 2 π − -1 1 α β ()f α ()f β () yf x = 0 x y 1 -1 2 π 38π () yg x =11 (i ii) (a) (iii) (b) if 1( ) ( ) 1 ff αβ −≤ < ≤ . gf is o ne -o ne ( or in cr ea si ng ). [0 , ] 2 gf R π = . (, ] [ . ] 22 2 g f RD ππ π =− ∞ ⊄ − = . Al te rn at iv e: 5 (4 ) [ (4 )] [ ] 8 fg f g f π == − is u nd ef in ed b ec au se 5 [, ] 82 2 f D ππ π −∉ − = . He nc e, fg d oe s n ot e xi st a s a f un ct io n. () 2 gx π = () ( ) 13 82 xx ππ +− = − ( )( ) 13 4 xx +− = − 2 23 4 xx−+ + = − 2 27 0 xx −− = 23 2 2 x ± = 18x=± Si nc e 1, x ≥ 18 x =+ . 11 8 k ≤≤ + Gr ea te st v al ue o f k is 18 + . 0 x y -1 1 3 (1 , ) 2 π () yg x = 2 y π =− 12
1 An g lo -C h in es e Ju n io r C o ll eg e H 2 M at he m at ic s 97 40 2010 JC 2 P R E L IM M ar ki ng Sc he m e er 2: 1 () () () () () () 1 2 21 1 1 1 2 22 21 2 1 1 1 2 1 1 2 21 2 1 22 1 1 2 42 2 1 11 11 11 22 22 1 41 2 x x xx xx xx xx ed x e ed x ed x ee ee e e ee e e − − −− − −− −− − − − =− − + − =− + + + =+ − + + ∫ ∫∫ 2 (i ) 2 (i i) 2 (i ii ) Le as t v al ue o f zw − = 1 Gr ea te st ( ) ar g 3 z + = 11 13 ta n si n 0. 82 6 (3 d p) 5 26 −− += Le as t ( ) ar g 3 z + = 11 13 ta n sin 0. 43 2 (3 dp ) 5 26 −− −= − 3 ln 2 0 a x x −≥ w he re 1 a > . 2 21 2 0 11 8 ) 1 1 8 ) 2 44 0 11 8 1 1 8 0 44 a x x xx a x aa xx x aa xo r x −≥ −− ≥ ++ − + −− ≥ −+ + + ≤< ≥ 4 (a) () ( ) 32 3 co s 3 si n d xx x dx =− x y i(2 ,1 ) O −3 (5 ,− 3) o 2 3 26 3 3 z w 2 53 32 3 33 3 2 3 32 3 3 33 si n (s in ) 11 co s co s 3 33 cos co s 3 1 co s si n 33 xx dx xx x dx xx x x dx x xx x dx x xx c = =− − − =− + =− + + ∫ ∫ ∫ ∫ 4 (b ) 5s ec 10 , 4 5s ec ta n 2 5, 3 xw he n x dx wh en x d π θθ π θθ θ θ == = == = () () [] 25 3 2 2 10 3 3 2 2 4 3 2 4 3 3 2 4 4 3 3 4 4 5 5s ec 5 5 se c ta n 1s ec 5t an 1c os 1 co t co s 5s in 5 11 1 co s 5s in 5 12 1 2 23 . . 51 5 5 15 xd x d d do r ec d ec ie a b π π π π π π π π π π π π θθ θ θ θ θ θ θ θθ θ θ θ θ θ − − − =− = = =− = − =− = = − ∫ ∫ ∫ ∫∫ 5 (i) (ii) 1 1 :2 3 1 pr ⋅= ɶ − 1 12 :0 1 , 14 lr λλ − =+ ∈ ℝ ɶ 21 1. 2 2 2 4 0 41 =+ −= − 11 0. 2 1 0 1 3 11 − =− + − ≠ − ⇒ l1 is p ar al le l t o p1 . ⇒ l1 is n ot c on ta in ed in p1 . A lt er nat iv e m et hod: 12 1 22 3 14 1 λ λ λ −+ =− ≠ +− i Si nc e no s ol ut io n fo r λ , ∴ l1 is pa ra lle l a nd n ot c on ta in ed to p1 12 3 21 3 2 14 1 − ×= − −
3 (i ii ) (i v) (v) 2 31 3 :2 0 2 3 0 1 4 11 1 pr −− − ⋅= = + + = i ɶ 2 :3 2 4 px y z −+ + = 3 24 2 :1 1 1 8 1 4 5 41 4 pr ⋅= ⋅= + − = − ɶ 3 2 :1 5 4 pr ⋅= ɶ 2 12 :0 0 , 13 lr µµ − =+ ∈ − ℝ ɶ 1 1 :2 3 1 pr ⋅= ɶ − () 12 1 0. 2 3 2 5 3 1 po in t is 1, 0, 2 13 1 µ µµ µ −+ =⇒ − + =⇒ = ∴ − −− B 21 11 5 si n 0 2 13 6 78 31 θ =⋅ = −− Le ng th o f th e pr oj ec ti on o f AB o n p1 = co s AB θ = 2 25 53 53 1 01 13 31 8 78 78 6 6 3 −= = = − 6 Pro ce du re A is p re fe ra bl e as it is u nb ia se d. Ea rl y cu st om er s m ay n ot b e ty pi ca l c us to m er s in g en er al . 7 Le t R b e th e r. v fo r th e le ng th o f a ri gh t s ho e an d L f or th e le ft s ho e R N (2 0, 0 .1 4 2) an d L N (2 0. 1 , 0 .1 1 2) M et hod 1 X = R + L N (4 0. 1, 0 .0 31 7) 3X N (4 0. 1 x 3, 0 .0 31 7 x 3 2) P(X > 1 20 ) = 0. 71 3 M et hod 2 R + L N (4 0. 1, 0. 03 17 ) P(R + L > 3 120 ) = 0. 713 8 Le t M a nd W b e th e rv f or th e w ei gh t o f an a du lt m an and w om an r es pe ct iv el y. M N (7 5, 4 2) a nd W N (6 5, 3 2) W M ~ N (-10 , 5 2) P( 1 > − M W ) = P (W – M > 1 ) + P( W – M < -1 ) = 0 .9 78 Or P ( 1 > − M W ) = 1 P ( 1 < − M W ) = 1 – P( -1 < W – M < 1) = 0 .9 78 4 No , (i ) T he w ei gh t o f a hu sb an d an d w if e m ay n ot b e in de pen de nt Or ( ii ) R an do m ne ss is n ot th er e ( a ra nd om ly c ho se n w om en b ut s po us e is n ot ra nd om ly c ho se n) Or ( ii i) Di st ri bu ti on o f w ei gh t o f m ar ri ed wo m an is d if fe re nt f ro m d is tr ib ut io n of ad ul t w om an . Et c 9 Te le ph on e co st s ar e as su m ed to b e no rm al ly d is tr ib ute d. To te st H0 : µ = 72 ag ai ns t H 1 : µ > 72 at 5% le ve l of s ig ni fi ca nc e Un de r H0 , T = n s x 0 _ µ − t( 5-1) Te st s ta ti st ic s : T= n s x 0 _ µ − = 82 .6 72 10 .66 77 / 5 − = 2. 2219 p v al ue = P (T > 2 .2 21 9) = 0 .0 45 2 < 0 .0 5 Re je ct H0 a t t he 5 % le ve l o f si gn if ic an ce . W e co nc lu de th at th er e is s uf fi ci en t evi de nc e a t t he 5% le ve l of s ign if ic an ce th at th er e is e vi de nc e o f a n in cr ea se in me an mo nt hly c os ts . To te st H0 : µ = 0 µ ag ai ns t H 1 : µ > 0 µ a t 5 % le ve l o f si gn if ic an ce Un de r H0 , Z = n x 89.9 0 _ µ − N (0 ,1 ) Te st s ta ti st ic s : Z = n x σ µ0 _ − = 0 82 .6 9. 89 / 5 µ − = (82. 6 - 0 µ ) 89.9 5 Do n ot r ej ec t H 0 if P (Z > ( 82 .6 -0 µ ) 89.9 5 ) > 0 .0 5 (8 2. 6 - 0 µ ) 89.9 5 < 1. 6448 5. ... ... ... .( 1) µ0 > 7 5. 3 10 (i) 16 1 x = ( fr om c al cu la to r o r co m pu ta ti on )
5 (i i) (i ii ) wh en 161 x = , 103. 6 0. 726 xy =+ (1 61 10 3. 6) /0 .7 26 y =− 79. 06336088 = us in g yy n =∑ 1 79. 06 336088 (6 5. 1 73. 2 85 80. 9 89. 9) 6 k =+ + + + + 80. 3 k = Us e G. C . t o fi nd r eg re ss io n li ne o f y o n x: 97 .593 1. 097 yx =− + Us e y o n x l in e t o pr ed ic t w ei gh t. Wh en 16 5 x = , 97. 593 1. 09 7( 165 ) y =− + 83. 4 y = ( 1 d .p .) – u si ng 3 d .p . o f a a nd b to c om pu te . or 83. 5 y = u si ng f ul l a cc ur ac y o f a a nd b to c om pu te . Us in g G. C ., 0. 893 r = C is u nu su al ly o ver w ei gh t. 11 (i) Br eak ag es o ccu r ra nd om ly o r Br eak ag es o ccu r in de pe nd en tl y or M ean n um be r of b re ak ag es i s a co ns ta nt Le t A b e th e r. v fo r th e nu m be r of b ro ke n cu ps p er da y A Po (2 .1 ) P( A≥ 3) = 1 – P (A ≤ 2) = 0. 350369 = 0. 350 (3 si g fi gs ) Le t X b e th e r. v fo r th e nu m be r of d ay s w it h a le as t 3 b ro ke n cu ps o ut o f n cu ps . X y 65 .1 73. 2 80 .3 80 .9 85. 0 89. 9 150 15 7 160 162 167 170 x 6 (i i) P( X ≥ 2) > 0. 999 1 – P( X ≤ 1) > 0. 999. ... ... ... ..( 1) P( X ≤ 1) < 0. 001 Us in g GC , n P( X ≤ 1) 21 0. 00 145 22 0. 00 09 84 le as t n is 2 2 T Po (2 .1 x7 + 1 .6 x 7 ) T Po (2 5. 9) M et hod 1: Si nc e n is la rg e, _ T (2 5. 9, 100 9. 25 ) ap pro x by c en tra l l im it th eo re m P( _ T≤ 2 6) = 0 .5 78 ( to 3 s ig f ig ) M et hod 2: 100 w ee ks , Y Po (2 59 0) λ = 2 59 0 > 10 . N or m al a pp ro x to P oi ss on Y N (2 59 0, 25 90 ) ap pr ox P( Y ≤ 2600) = P ( Y < 2600 .5) ( W it h cc ) = 0. 578 (t o 3 si g fi g) 12 (i) (ii) Le t X b e th e r. v fo r th e nu m be r of f is h w hi ch m ea su re s le ss th an 8 c m lo ng . 95 4 5 141 () ( ) ( ) 55 5 C = 0 .0 03 30 ( to 3 s f) X B (n ,0 .2 ) Si nc e n la rg e an d p= 0 .2 , X N (0 .2 n, (0 .2 )( 0. 8) n) a pp ro x P( X ≤ 10) ≤ 0. 0227 P( X < 1 0. 5) ≤ 0. 0227 P( Z < n n 4. 0 2. 0 5. 10 − ) ≤ 0. 02 27 n n 4. 0 2. 0 5. 10 − ≤ -2 .0 00 92 9. ... ... ... (1 ) 10. 5 – 0. 2n ≤ -0 .8 00 37 2 n He nc e 10 .5 – 0 .2 n ≤ -0 .8 n a pp ro x Met hod 1 Us in g GC Y1 = 1 0.5 -0.2 x + 0 .8 x →Ta bl e A ns : 91 M et hod 2 : U se G C a nd g ra ph
7 M et hod 3 : He nc e (1 0. 5 – 0. 2n ) 2 ≥ (-0 .8 n ) 2 He nc e 4n 2 4 84 n + 1 10 25 ≥ 0 app rox……( 2) Fr om G C : n ≥90 .6 or n ≤ 3 0. 4 n ≥ 91 or n ≤ 30 ( N A b ec aus e doe s not s at is fy ( 1) ) Le as t n = 9 1 13 (i) (ii) Le t H b e th e ev en t t ha t t he m em be r of p ub lic h as h ype rt en si on Le t D b e th e ev en t t ha t t he m ac hi ne d ia gn os ed h yp er te ns io n (i ) P( H /D ) = 3 2 = ) ( ) ( D P D H P ∩ 3 2 = )1 .0 )( 1( 8. 0 8. 0 k k k − + ( 1) k = 0. 2 p % = 2 0 % p = 2 0 P( H ’/ D ’) = ) ( ) ( D P D H P ′ ′ ∩′ = )9 .0 )( 1( 2. 0 )9 .0 )( 1( k k k − + − = 0. 94 7 Ex am pl es o f po ss ib le c om m en ts : (1 ) If it d oe s no t fi nd y ou h yp ert en si ve th en y ou c an b e re as on ab ly c on fi de nt th at yo ur b lo od p re ss ur e is n or ma l. (2 ) I f it d ia gn os es h yp er te ns io n, th en y ou s ho ul d co ns ul t y our doc tor f or f ur th er te st s. (3 ) An y ot he r lo gi ca l c om m en ts wi th r ef er en ce to th e co nt ex t o f th e qu es tio n 14 (i) (ii) C as e 1: 2 re d ba lls -2 1 2 = C as e 2: 2 bl ue b al ls -3 3 2 = C as e 3: 2 gr ee n ba ll s - 5 10 2 = N o. o f w ay s = 13 10 14 ++ = C as e 1: N o re d bal l. C ho os e 6 ba ll s fr om a to ta l o f 8 ( bl ue a nd g re en ) b al ls : 8 6 . C as e 2: N o bl ue b al l. C ho os e 6 ba ll s ro m a to ta l o f 7 ( re d an d gr een ) H H’ D D D ’ D’ 0. 8 0.2 0. 1 0. 9 k 1-k 8 (i ii) b al ls : 7 6 [ N ot e: W e ca n’ t e xc lu de g re en b al ls b ec au se to ta l n um be r of re d an d bl ue b al ls is o nl y 5. ] N o. o f w ay s = 10 8 7 21 0 (28 7) 175 66 6 −+ = − + = A lt er nat ive M et hod : Cas e Gr ee n Bl ue Re d No . o f wa ys 1 4 1 1 532 30 41 1 ×× = 2 3 2 1 53 2 60 321 ×× = 3 3 1 2 53 2 30 31 2 ×× = 4 2 3 1 53 2 60 321 ×× = 5 2 2 2 532 20 222 ×× = 6 1 3 2 53 2 5 13 2 ×× = To ta l 17 5 E xc lu di ng th e gr ee n ba lls , w e o nl y ha ve 1 , 1 , 2 , 2 , 3 . S in ce w e a re ig no ri ng th e co lo ur s of t he b al ls , w e ar e fo rm in g 3 -d ig it n um be rs f ro m th e 5 d ig its 1 , 1 , 2 , 2 , 3 . C as e 1: A ll 3 di gi ts ar e di st in ct . T he 3 d ig its a re 1 , 2 , 3 a nd th e n um be r of w ay s of ar ra ng in g th em a re 3 ! C as e 2: 2 di gi ts ar e id en ti ca l. S te p 1 : C ho os e 2 d ig it s th at a re id en tic al ( 1, 1 o r 2, 2) : 2 S te p 2 : C ho os e a d ig it f ro m th e re m ai ni ng d ig its ( 1, 3 o r 2, 3) : 2 S te p 3 : A rr an ge th e 3 ch os en d ig it s in a r ow: 3! 3 2! = N o. of w ay s = () ( ) 3! 3! 2 2 6 12 18 2! += + =
