Dhs 2011 Prelim

Name: Index Number: Class:

DUNMAN HIGH SCHOOL Preliminary Examination Year 6

M A T H E M A T I C S (H igher 2) Paper 1

Additional Materials: Answer Paper List of Formulae (MF15)

9740/01 13 September 2011

3 hours

READ THESE INSTRUCTIONS FIRST

Write your Name, Index Number and Class on all the work 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-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 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 [ J at the end of each question or part question.

At the end of the examination, attach the question paper to the front of your answer script.

The total number of marks for this paper is 100.

For teachers' use: Qn Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Total

Score Max

score 5 5 5 7 7 13 10 12 12 10 14 100

This document consists of 6 printed pages (including this cover page), [Turn over

Given that Y r 2 =^(n+l)C2fi + l), showthat Y ( 2 r + l)(2r + 3) = -Un2 + 18n + 23).

Hence find the exact sum of the 25* to the 75lh term of the series. [5]

The position vectors of the points P, Q and R relative to the origin O, are p, q and 2 1

- p - - q respectively. The point S with position vector s relative to O is such that O

is the midpoint of the line segment RS. Prove that 5s + 2p -q =0. [2] The point T lies on QS extended such that QS.QT = 4:5. Show that the points P, O and T are collinear and find PO:OT. [3]

The diagram shows the graph of y 2 = f (x) where f (x) = Ax4 + Bx3 + Cx2 + Dx + E. The curve y = f (x) has stationary points at x = 0 and x — 3.

y

(3,4)

/ = f

0 /4

(3, -4 )

Find the exact equation of the curve y = f (x). Hence find the area enclosed by the curve >• • f (x) in the given diagram, giving your answer to 1 decimal place. [5]

4(a) Given that f is a one-one function, determine if ff exists. Justify your answer. [ I ]

(b) The functions g and h are defined as follows:

g: x H-> In x. x £ 1,

h : x i - > x 2 - 2 j : + 2, x>0.

(i) Given that gh exists, define gh in a similar form and find the range of gh. [2]

(ii) Sketch, on the same diagram, the graphs of g, g _ 1 and g~'g. [3]

(Hi) State the range of values of x satisfying the equation g~'g(x) = gg~'(*). [1]

fflDHS 2011 2011 DHS Year 6 H2 Math Prellmlnaiy Examination Paper 1 [Turn over

5 (i) Sketch on the same diagram, the loci given by

(a) |z-5i | = 4, (b) | z - 5 - 5 i | = 3. [2]

Hence, or otherwise, find the complex numbers z that satisfy both (a) and (b). [2]

(ii) The complex numbers, z, and z2 satisfy the equations given in (i)(a) and (l)(b) respectively.

Given that 0 < B < it and a e R, a > 8, find the smallest value of 6 for which

0 = arg(z,-a-5i) = arg(z 2 -a-5i). [3]

6(a) Given that In >• » tan - 1 x, show that (1 + x2) — = y. x ' dx

By further differentiation of the above result, find the Maclaurin's series for e , a n '*, up to and including the term in x . [6]

- 2

2^

term in x". • [2]

Using the above expansion, or otherwise, (i) show that sinj: + 2sec2j: = 2 + .t + 2j: 2 if x is sufficiently small for x* and

higher powers of x to be neglected, [3]

(ii) find Y - ^ r r by substituting a suitable value of x. [2] r=\2

(b) Obtain the expansion of I 1 - — in ascending powers of x, up to and including the

© D H S 20t l 2011 DHS Year 6 H2 Malh Preliminary Examination Papar 1 [Turn over

4

7(a) The diagram shows the graph of y = f (x), which has turning points atA(-2,4)and fl(2,3).The horizontal and vertical asymptotes are y = 2 and x =-1 respectively.

(b)

fl(2,3) y = f W

/ y = 2

Oj

x=- 1 J Sketch, on separate diagrams, the graphs of

(i) y=-f(|jc|),

(ii) y 2 = f ( * ) ,

[2]

[3]

showing clearly all relevant asymptotes, intercepts and turning point(s), where possible.

y = g(x)

The graph of y = gU) above intersects the A:-axis at (a,0) and (/J,0), where a > -1 and fi>l. It has a turning point (0,-1) and a vertical asymptote x = -\.

y = g(*) undergoes two transformations in sequence: a translation of 1 unit in the positive y-direction, followed by a scaling of factor 2 parallel to the jr-axis. The resulting graph is y = h(jc).

Sketch, on separate diagrams, the graphs of y = h(.t) andv = ——, showing clearly g(x)

all relevant asymptotes, intercepts and turning point(s), where possible. [5]

© D H S 2011 2011 DHS Year 6 H2 Math Preliminary Examination Paper 1 [Turn over

5

8(a) Express —=• in partial fractions, Hence, find | — T dx. l-2x + x2 J 1-2X + X1

(b) Find

(>)

(ii) J

(Hi) J —

J e2 j rsin(e2 j :)ck,

sin - 1 * dr,

2

-dx.

[3]

[2]

[3]

[4]

The plane IT, has cartesian equation 5x + y = 2. The plane n 2 contains the line I

and is perpendicular to the plane with equation f3> rn with equation r = 2 0

,2; , 1 ,

f 5 ] f2̂ r = a -2 +0 -1 . The planes f l , and-flj meet in the line L .

. 0 . . 1 ;

(i) Find an equation for I l j in the form r • n = d.

(ii) Find a vector equation for L .

[3]

[3]

(iii) Show that 0, the acute angle between / and L is given by cosf? = [2]

The plane n , has cartesian equation x+ py + 2z = q , where p and q are positive constants.

(iv) If the three planes £"1,. FI2 and 113 do not have any point in common, show that p = I . If the perpendicular distance from the origin to f l j is 2 units, find the exact value of q. [4]

10 The terms in the arithmetic sequence ( 3n + \,n = 1.2,3,...) are grouped into sets such that the r l h bracket contains 2r terms as shown below:

{ 4,7 ), {10,13,16,19 ), { 22,25,28,31,34,37,40,43 )

(i) Show that the total number of terms in the first N brackets is 2(2* -1). [2]

(ii) Find the sum of the terms in the first N brackets. [2]

(iii) Show that the first term in the A/* bracket is 3(2W) - 2 and find also the last term in the NA bracket. [4]

(iv) Find the least value of N such that the sum of terms in the /V l h bracket is more [2]

[Turn over

than 10'2.

2011 DHS Year 6 H2 Math Prolimlnary Examination Paper 1

11(a) The graph of y = x2 + 2 for 0 £ x £ l , is shown in the diagram below. Rectangles,

each of width —, where n is an integer, are drawn under the curve. n

y .3

(i)

(ii)

w 9 1 1

is given by + — T . 4 2n 4n2

Given that ] T r 3 = -^-(n +1) 2 , show that the total area of all n rectangles, A,

[4]

Deduce the exact area under the curve for 0 < x < 1. [1]

(iii) Find the least value of n such that the area A is no less than 99% of the exact area under the curve. [2]

(b) The diagram shows the curves C\d Ci given by the equations y = —^--1 and 4x

y = 2-cos(5x2) respectively. C\s the x-axis at * = ™ and * = ând Ci has a

y-intercept at y = 1. The shaded region R is bounded by the curves C\, C% and the .r-axis. y

C 2 :y = 2-cos(5;c2)

0

- I • y=-

(i)

(ii)

State the coordinates of the points of intersection between the curves C\d Cz, correct to 3 decimal places. [1]

Find the volume of the solid formed when R is rotated through (I) 4 right angles about the .r-axis, [3] (II) 2 right angles about the y-axis. [3]

© D H S 2011

END OF PAPER 2011 DHS Year 6 H2 Math Prolimlnary Examination Paper 1

Name: Index Number: Class:

DUNMAN HIGH SCHOOL Preliminary Examination Year 6

M A T H E M A T I C S (Higher 2) 9740/02 Paper 2 19 September 2011

3 hours Additional Materials: Answer Paper

List of Formulae (MF15)

READ THESE INSTRUCTIONS FIRST

Write your Name, Index Number and Class on all the work 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-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 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 examination, attach the question paper to the front of your answer script.

The total number of marks for this paper is 100.

For teachers' use: Qn Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 0 9 Q10 Q11 Total

Score

Max score 8 9 10 13 5 5 8 10 10 11 11 100

This document consists of 6 printed pages (including this cover page).

©DHS 20 i i [TurnOver

2

Section A: Pure Mathematics [40 marks]

1 (a) A family of curves is given by y = Ae* +2sin x + B, where A and B are arbitrary

d 2y dy constants. Show that —— = 2(sin x + cos x). [3]

Ac2 dx (b) Use the substitution y = ux to find the particular solution of die differential

equation x— = 4x2 +2y 2 + y, given that v = 0 when x = —. Express your dc 8

answer in the form y = f (x). [5]

(a) Given that the complex numbers w and z satisfy the equations

vv*+2z = i and w + ( l - 2 i ) z =3 + 3i,

find w and z in the form a + bi, where a and b are real. [4]

(b) Solve the equation

iz3 = 2 - 2 i . [3]

Hence deduce the roots of the equation - iw 3 = 2 + 2i . [2]

[Give all answers exactly in the form re , where r > 0 and -it < & <, x. ]

3 A curve C has parametric equations

x~at-~. y = bt + -, t>0. t l

For a = 1 and b = 2.

(i) without finding the cartesian equation of C, find the area enclosed by C, the line x = 5, the y-axis and the *-axis, giving your answer correct to 2 decimal places,

[4] (ii) find the value(s) of I for which both x and y have the same rate of change with

respect to I. [3]

For a = 1 and b • 1, the curve C has two oblique asymptotes y = x and y = -x. (iii) By considering the curve of C, sketch the graph of y = f '(x). [3]

© O H S 2011 2011 DHS Year 6 H2 Math Preliminary Examinallon Paper 2 [Turn over

3

3 (a) Express in partial fractions. [1]

r(r + 3)

Hence find the sum of the series —+—+-—+...+ . [3] 4 10 18 M(«+3)

Give a reason why the series —+ — ' s convergent and state its exact value.

[2]

(b) A sequence w,, w2, i i j , . . . is such that u, = - and

-. for aline Z\ (n + l)(« + 2)

(i) Find the values of ult uy and u4. [2] (ii) Form a conjecture for un in terms of n. [1] (iii) Prove your conjecture for un using mathematical induction. [4]

Section B: Statistics [60 marks]

5 In a binomial probability distribution X, there are n trials and the probability of success for each trial is p. If n = 20 and P(A" <, 1) = 0.8, determine the value of p. Hence find the least value of a such that P(X <a)> 0.999. [5]

6 Find the number of ways 4 boys and 3 girls can be seated (i) in a row such that no two girls are next to each other, [2] (ii) at a round table with 10 identical seats, such that the 4 boys must sit together. [3]

7 [In this question give all answers correct to 3 places of decimals.]

A computer game consists of at most 3 stages. The probability that a player clears the first stage (i.e. successful) is 0.7. For each subsequent stage,

• the conditional probability that a player clears that stage, given that he cleared the preceding stage, is half the probability of success at the preceding stage,

• the conditional probability that a player clears that stage, given that he failed at the preceding stage, is the same as the probability of success at the preceding stage.

The game ends prematurely if a player fails to clear 2 consecutive stages.

Draw a probability tree diagram to represent the above information. [2]

• Find the probability that (i) the game ends prematurely, [I] (ii) a player clears exactly 2 stages, [3] (iii) a player clears the third stage given that he cleared exactly 2 stages. [2]

© DHS 2011 2011 DHS Year 6 H2 Math Preliminary Examination Paper 2 [Turn over

4

8 (a) Give a reason why it is advisable to plot a scatter diagram before interpreting a correlation coefficient calculated for a sample drawn from a bivariate distribution.

[1]

(b) Due to urbanisation in a country, the average density of birds (per km ) found in the city, x, and those in the forest, y, have changed over the years from 2000 to 2007. The findings are given in the table.

Year 2000 200) 2002 2003 2004 2005 2006 2007 City bird density {x) 75 79 86 97 110 125 131 136

Forest bird density (y) 400 398 395 392 380 351 331 303

According to experts, counting birds in the city is easier than counting those in the forest. It is desired to determine only the density of birds in the city and use it to estimate the density of birds in the forest.

(i) Calculate the product moment correlation coefficient between x and y. Comment on whether a linear model is appropriate. [2]

(ii) Draw a scatter diagram for the data, labelling the axes clearly. [2]

(iii) Another suggested model takes the form y = a + bx1 where a and b are constants. Explain whether this model provides a better fit than a linear model between x and y. [2]

(iv) Given that the city bird density for a particular year is 150, use an appropriate model to find the forest bird density for that year. Comment on the reliability of your prediction. [2]

If birds migrate from the forest to the city due to urbanisation, state a reason why the total bird density in the city and forest for each year from 2000 to 2007, is not a constant number. [1]

© D H S 2011 2011 DHS Year 6 H2 Math Preliminary Examination Paper 2 [Turn over

5

9 In a certain junior college, the marks (out of 100) scored by a JC 1 student in a Class Test, Common Test and Promotional Examination are denoted by C, 7" and S respectively. C, T and S may be modelled by normal distributions with means and standard deviations as shown in the table below.

Type of assessment Mean Standard deviation Class Test, C 68 a

Common Test, T 65 8 Promotional

70 10 Examination, S

70 10

(i) Given that P(C > 85) = 0.05, determine the value of a. [2]

In a particular year, a student sits for five Class Tests, a Common Test and a Promotional Examination. The average mark of the five Class Tests constitutes 20% of the overall assessment mark for the year. The Common Test and Promotional Examination constitute 20% and 60% of the overall assessment mark for the year respectively.

For the following parts, assume a =10,

(ii) Find the probability that the average mark scored by a student in the five Class Tests is more than 75. [3]

(iii) Find the probability that a student scores an overall mark of more than 80 for the year. [4]

(iv) State an assumption used in your calculations for (iii). [1]

10 Based on past data, the mean number of dengue cases per day was reported as 30. The authorities suspect that the recent hot and wet weather has increased the mean number of dengue cases per day. A random sample of 60 days is taken and the number of dengue cases per day, X, is summarised by

£ ( ; c - 8 ) = 1400, £ ( * - 8 ) 2 = 34800.

(i) Calculate unbiased estimates of the mean and variance of X. [2]

(ii) Test, at the 5% significance level, whether the mean number of dengue cases per . day has increased. [4]

Explain, in the context of the question, the meaning of the p-value. [ 1]

© D H S 2011 2011 DHS Year 6 H2 Main Preliminary Examination Paper 2 (Turn over

6

(iii) After the authorities have increased surveillance to eliminate mosquito breeding and stepped up public education, they claimed that the mean number of dengue cases per day has dropped below 30. The number of dengue cases occurring over 7 randomly selected days are recorded and the unbiased estimate of the population variance can be assumed to be the same as that obtained in (i).

State suitable null and alternative hypotheses, and also the assumption required for an appropriate significance test. [2]

Find the range of values for the sample mean x such that ..there is evidence to support the authorities' claim at the 5% significance level. 12J

11 The number of traffic offences occurring at a particular stretch of road between 7:30am to 9:30am on any day follows a Poisson distribution with mean A. Given that the probability of at least one traffic offence occurring between 7;30am to 9:30am on any day is 0.95, show that 2 = 3, correct to the nearest integer. [2]

For the subsequent parts, use X = 3.

(i) Find the probability that there are at least 17 traffic offences occurring between 7:30am to 9:30am during weekdays (i.e. Monday to Friday) in a given week. [2]

(ii) Find the probability that the mean number (taken over a year i.e. 52 weeks) of traffic offences occurring between 7:30am to 9:30am during weekdays in a week is at least 13 but not more than 16. (2)

(iii) Find, using a suitable approximation, the probability that there are between 30 and 42 weeks in a year where there are at most 16 traffic offences occurring between 7:30am to 9:30am during weekdays in a week. [4]

(iv) Explain why it is not appropriate to use a Poisson distribution with mean 36 to model the number of traffic offences occurring in any one full day. [1]

END OF PAPER

© 0 H S 2 0 1 1 2011 DHS Year 6 H2 Malh Prolimlnary Examination Paper 2

H 2 M A T H S Y R 6 2 0 1 1

PRELIMINARY EXAMINATION P A P E R 1

SUGGESTED SOLUTIONS

£ ( 2 r + l ) ( 2 r + 3 ) r-l = £ ( 4 r 2 + 8 r + 3)

r - l

= 4 £ r 2 + | > + 3)

= 4 ^ ( n + l)(2« + l)j + ^ ( l l + 8n + 3)

= y ( / » + l)(2w + l)+^(8n+14)

= ^2(2n 2 +3n + l) + |(8 /7 + 14)j

= y(4« 2 +18n + 23) (shown)

f)(2r + l)(2r + 3) r - 2 J

= £ ( 2 r + l)(2r + 3) - £ ( 2 r + l)(2r + 3)

= y (4(75)2 +18(75) + 23) + y (4(24)2 +18(24) + 23)

= 596825-22072 = 574753

24

ODHS 2011 9740/2011 Year 6 H2 Math Prelim Paper Solutlons/OHS Page 2 of 12

C O

Qni; ^Suggested Sol u t |onî=$f»- v. jifoW-."

OS = -OR

s 2 l 5 5

5s = q-2p 5s + 2p - q = 0 (shown) By ratio theorem,

40T=5OS-OQ

Since OT can be expressed as a scalar multiple of OP and 0 is a common point, the points P, O and Fare collinear.

PO:OT = \:- = 2:\ 2

f(x) = Ax* + Bx* + Cx1 + Dx+E. (0,0): £ = 0 (4,0):256/J + 64S + 16C + 4D = 0 -(1) (3,16): 8M + 27fl + 9C + 3D = 16 -(2)

f 'W = 4^ 3 +35x 2 +2Cx + Z) f'(x) = 0 atx = 0 and A: = 3 D = 0 108/1 +275+ 6C = 0 -(3)

^256 64 16 r i 0 0 27

RREF 81 27 9 16 = 0 1 0 64 27

^ 108 27 6 0 ; 0 1 0

f ( x ) —

Required area - 2J*yj-^x* + %x3dx = 19.3 units2

ODHS 2011 9740/2011 Year 6 H2 Math Prelim Paper Solutlons/OHS Page 3 of 12

4(a) f is one-one => fex i s t s

R r , = D r

.-. ff"' exists. (b) 0)

gh:xi-»ln(x 2 -2A: + 2), x > 0

R, h=[0,«>) (ii)

(0.1)

(!!!)

© DHS 2011 9740/2011 Year 6 H2 Math Prelim Paper Solutlons/OHS Page 4 of 12

50)

Method 1 / = 4 2 - x 2 =3 2 - ( 5 - x ) 2

16-JC 2 = 9 - 2 5 - x 2 + 10*

10 5

Hence, 16 37. , „, 16 13.

IV. s w - — + — i and jy, sw.= — + — i 1 1 5 5 2 2 5 5

OR, v^=3.2 + 7.4i and w 2=3.2 + 2.6i Method 2 Cartesian equation for locus (a): x2 +{y-5)2 = 16 Cartesian equation for locus (b): (x - 5)2 + (y - 5)2 = 9 Solving the two equations simultaneously:

\0x-2S = l=>x = — 5

L 6 J 1 6 Y m 12 " I 5 J = 5

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper Solutloru/DHS Page 5 of 12

Using similar triangles, 3 a-5 4 a 3a = 4a-20=>o = 20

4 sin(/r-#) = —

20 n-8 = sin"' -

5 e = 2.94 rad (3 s.f.)

' ; i r V ' ; L ; : ^ ^ : ; L J i B : T q i i ^ i ' - i ' i mm-. „ : - ? 6(a) In y = tan"1 x

1 dy _ 1 yd? 1+x 2

(1 + x 2 ) — = y (shown) dc

dx2 dx dx „ 2 ,d \ . d2y „ d 2y _ dy d2y (1 + x 2)—4 + 2x—4 + 2 x — + 2 — = —4

dx5 dx2 dx2 dx dx2

dy . d2y d3y Whenx = 0 , y = l , ^ - 1 , - T y - 1 , ^ T - - 1

Maclaurin's series for c'""'* = I + x+—x 2 — 2 6

x 3 + -

(b)

= 1 + ^ + 2 ^ + -4

2 + ...

0) Since x is sufficiently small for x'and higher powers of x to be neglected. sin;r + 2sec2x sin x + 2 sec: x _ s i n x + 2 = sinx + 2(l + <an'x)

0 0 5 * = x + 2(Ux !) 2

+ - - 5 - = 2 + x + 2x2

* X + 2(l + x') = 2 + x + 2xJ

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper SoluUons/DHS Page 6 oM 2

(b) }

-2

x = - 2 !

0

J >

-1

x = a

a 0 • • X

x = p

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper Soluiions/DHS Page 8 of 12

8(a)

( b )

0)

l - 2 x + x 2 ( l - x ) 2

-1 1 •+-\-x ( l - x ) 2

f — i - T d . - f ^ . + . J J l - 2 x + r J 1-x ( 1 -

= In 11 —JC + -

(l-xf 1

'<!-*)

dx

• + C

j e2* sin(e2') dc = - j (2e2*) sin(e2*) dx

cos(e2*) •+C

(ii) dx f sin"1* dc = jrshr'x- f x—=====

= xsin''x + -J (-2x).( l-x 2 ) 2 dc

= xsin"'x + V l - x 2 +C (iii)

dx x -2x + 3

. 1 l + - 2 i - L - d x I = 1 +

x 2 - 2 x + 3 2x-2

dx x '-2x + 3 (x - l ) ' + 2 <•

= x + l n | x 2 - 2 x + 3 l — i t a n - ' ^ i + C 1 1 & 72

9(i) i m I f 2 l Normal of given plane is -2 X -1 = -5 l = - 5

, 0 ; J , .'J f 2 l f 5 1

n 2 = 5 X 0 = -1

J , . 1 ,

f 5 > f 5 ' n 2 : r - -1 2

.2,

-1

"5,

= 15-2-10 = 3

(ii) n , :5x + y = 2 n j : 5 x - y - 5 2 = 3 Let >• = r e K . Thus x = j ( 2 - r ) 5z = 5 [ ^ ( 2 - / ) ] - / - 3 = - 2 / - l

©DHS 2011 9740/2011 Year 6 H2 Malh Prelim Paper Solutlons/OHS Page 9 of 12

L:r = 0

<x + P

f-l\ 5 H

Oii)

cos 8 = { 06 -1 + 0-2 3 T i l 760 10

Oii)

cos 8 = Vl+O+lVl+25+4

3 T i l 760 10

(iv) For the the line

f_1l f 5 •

-2, Distanc

9 Vl + H

thr

/' 2, e fr

4

:e planes to have no point in common, nust be parallel to the n 3 •

= 0=>- l + 5/?-4 = 0=>/> = l

om r i j to the origin • 2 units

E S S 1 . 1 .1.1111. 11'1_ .. . - . J - ; f c 2 ' - ^ . ' f ! } t i i i i i L - . 1 . 1 . 1 ; . . 1 1 " 1 1 1 1 1 "1:1 io(0 Let A/be the total number of terms in the first N brackets.

M = 2 + 2 2+.... + 2A' = 2(2" -1)

( » ) S W = ^ [ 2 ( 4 ) + (M-1)(3)]

= ( 2 A ' - l ) ( 6 ( 2 w ) - l )

(iii) No. of terms till "first number in the bracket" = 2 " - l

The first number in the iV* bracket = 4 + (2* - l - I ) (3 ) = 3(2N)-2 The last number in the A1"1 bracket = 3A/ + 1

= 6.2"-5

Alternative First number in the (W+l) , h bracket • 3 ( 2 y + l ) - 2 Last number in the bracket = 3(2*+l) - 2 - 3

= 6.2"-5 (iv) (3(2")-2) + (6(2»)-5) v | 2

(9(2")-7)(2 A ')>10 1 2

Least N = 19

© D H S 2011 9740/2011 Year6H2 Math Prelim Paper SoJurjorvs/DHS Ptcjt 10 of 12

;•:' SuRgested Solutfohs 11(a)

(i) f r i V ,1 1 [ m 3 , l - +2 +— - +2

\n)

Total area of all n rectangles,

= i ( 0 + 2) + i n n

1 ( 3 ? , 1 + — - +2 + ... + — +2

n n I n J

2 + 2 + ... + 2 + ̂ - j ( l 3 + 23 +3 3 +... + ( r 7 - l ) 3 )

= 2 + -T («-ir

= 2 + - ! T ( n 2 - 2 n + l) Ay,2 y- I 4«

, 1 1 1 9 1 1 , . x

= 2 + - - — + — - = + — 7 (shown) 4 2/7 4« 2 4 2/7 4* 2 (ii)

Area under the curve = lim 9 _ J _ J _

4 2/7 + 4/72J = - unit

4 (iii)

For ^ - - L + _ L S 0 . 9 9 | - | 4 2» 4*2 U J

^- < 0.0225 2/7 4/72

Using GC, " f c w i • • t i . i m w n . - f c i m

»7^0.512 or H 3*21.7

Since » e Z + , least n = 22 (b)(1) By GC, (-0.340,1.162) and (0.340,1.162)

(ii) (D

Since both curves are symmetrical about they-axis,

Volume required ro.34ooi r -i2 fo.:

-2nU [2 -cos(V)] dr + J (0,5 Jo.34001 C-V'l 8 * i

.4x2 . I

= 2.61 unit3 or 0.832n (3s.f.)byGC

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper Solutions/DHS Page 11 of 12

(II) y = 2-cos(5xJ) Sx2 =cos"'(2- y)

x 2=-icos-'(2-y) 4 ? - ' * '

x2 = 1

4(y + l)

• Volume required f f 1 -162 1 ,1.162 1 . = rr 1 dv- , —cos

[Jo 4CV+1) ' J l 5

= 0.567 unit3 or 0.180n (3 s.f.) by GC

© D H S 2011 9740/2011 Year 6 H2 Math Pre-Jm Paper SolutionsADHS Page 12 of 12

H 2 M A T H S Y R 6 2011

PRELIMINARY EXAMINATION P A P E R 2

S U G G E S T E D SOLUTIONS

1(a) y = y4ex + 2sinx+S dy r

=> — = Ae +2 cos* dx

d 2 y , x „ • =>—i- = At -2sinx dx2

From(l),

Aex = — - 2 cos x dx

Substitute (3) into (2),

^ = ( ^ - 2 c o s x | ~ : s n > dx2 \dx

- d )

-(2)

-(3)

d 2 y dy dx2 dx

Alternative:

2(sinx + cosx) (Shown)

y = Acx +2sinx + 5

=> — = Aex + 2cosx dx

d 2 y . x . • — j = /te -2sinx

dy RHS = - 2(sin x + cosx)

sfê* + 2cosx)-2(sinx + cosx)

= Aex -2sinx

d 2 y = LHS = — f (Shown)

dx2

© DHS 2011 9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS Page 1 of 14

1(b) dy du r n

y = ux =>-f- = « + x — - ( I ) at dc

Substitute (1) into dc., we have dw x I u + x^-1 = •'. •• - 2u2x2 + ux

2..2 dx

dx 1

4 + 2 i / 1 r 1

-dw

2 ; 2 H V 1 _ i u x = —7- tan —]~ + c

2V2 V2 1 - l y x = — 7 - tan —7-— + c

2V2 V2x

Substitute x = — and y = 0,

c- —

:.y = -Jlx tan 2>/2x- V2/r

Alternative

-Utan"'4- = 2x + C V2 v/2 1 -\ — = 2x + C

Substitute x = — and y = 0,

c—£

© D H S 2011 9740/2011 Year6 H2 Math PreJm Paper 2 Solutions/DHS Page 2 o( 14

- . . . ' •

2(a) w*+2z = i (1) w + (l-2i)z=3 + 3i ( 2 )

i - w * From (1), substitute z =—-— into (2)

w + ( l - 2 i ) ( ' " 2 V V * ) = 3 + 3i

2 w + w * ( 2 i - l ) M + 5i Let w = a + b\ 2w+w*(2i- l) = 4 + 5i 2(o + W) + (o-6i ) (2 i - l ) = 4 + 5i 2o + 2ii + 2ai-o + 2Z) + /3i = 4 + 5i (a + 2b) + {2a + 3b)\ A + 5\

2(a)

Alternative

From (2), substitute z = -^-J W into (1)

w*(l-2i) + 6 + 6i-2iv = 2 + i vv*(]-2i)-2»v = - 4 - 5 i Let w = a + bi w*(l -2 i ) -2w = - 4 - 5 i (a - bi)(\ 2i) - 2(a + M) = -4 - 5i (o + 26) + (2a + 3/;)i = 4 + 5i

2(a)

Compare real parts: a + 2b = 4 (3) Compare imaginary parts: 2a + 3b = 5 (4) Solve equations (3) and (4): a = -2 and 6 = 3

w = -2 + 3i z = (3 + 3 i M - 2 + 3i ) = 1 + 2 .

( l -2 i )

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper 2 Solullons/DHS Page 3 of 14

3/M . . ( In cos +1SU1

4 J I 4

2j2~ cos + isin 4 J I 4

= 8 se* 4- ) . k = -, k = - 1 , 0, 1

or = 8 V » X A = -1,0,1

orz, » 8 V " ^ , z 2 =8 i e I ("^,2 3 =

Conjugate on both sides of iz3 = 2 - 2 i ( i r 3 ) * = ( 2 - 2 i ) *

i*(z 3 )* = 2 + 2i

- i ( z * ) 3 = 2 + 2i

Alternative z 3 = - 2 - 2 i

tv3 =-2 + 2i = ( - 2 - 2 i ) ' v v 3 = ( 2 3 ) * = (z*)3

w = z*

Comparing - i ( z *f - 2 + 2i with - iw 5 = 2 + 2i => w = z1

H> = Z * =

or = z * =

, A = - l , 0,1

, A- = — 1, 0,1

orw, =8 'e v l 2 - \w 2 = 8 t e V 4 ; , n 3 = 8 V

© DHS 2011 9740/2011 Year 6 H2 Math Prelim Paper2 Sdutions/DHS Page 4 of 14

3(1) x = i—,y = 2t + -, t > 0. t t

Required area

= j > d x

-rHK1* = 31.39 unit2 (2d.p.)

— - 1 + -dr~ + / 2

r 2 -When:r = 0,0 = —

1 -=>/ = 1 (v />0)

Whenx = 5,5=-— t

1 -=>/•- 5,-1 = 0 Whenx = 5,5=-—

t =>r = 5.1926 (vr>0)

(ii) 1 dx , 1 x-t — — = 1+—

t dt t2

t dt l2

dy dx , 1 . 1 dt dt t2 t2

t = 42 since t > 0 (iii) 1 1

x = t—,y = t + -, t>0. t t

V y\

\ / y = f(*)

6


4(a) 1 1 r(r + 3) r r + 3

(by Cover-up rule)

4 10 18 "' «(n + 3) ~ £ f r ( r + 3)

3

3 f ^ r ( r + 3) 3 f ^ V r + 3

i l l _ / 3 l 1 A

1 + -

3

1

/ - 2 H + l

nAl n + 2

+y ~ ( / i + 3) ' + I + i 1 1 I _

2 3 «+l n+2 « + 3

11 I f 1 1 1 18 H » + I n + 2 n + 3

( 1 1 1 . As n ->co, + + \n + l n+2 n+3

1 1 1 J I 1 I . - + - + - + ... = 2 — + — + — K . . 2 5 9 U 10 18

= 2 S 1 11

r ( r + 3)~ 9 series is convergent since sum to infinity of series exists.

© DHS 2011 9740V2011 V « a i 6 H2 Math Prelim Psper 2 SoVJonj/DHS PageSol 14

4 ( b )

• lirâ:i^-n r?r.iii5;t, :iir:'

(i)

= u. + (n+ !)(/? +2) ' 2 n U„ «„+•

1 1 I 2 3

2 I 7 3 4

3 7 9 4 T

4 2 5

(ii) Conjecture: «. = * for « e Z + . « + l

(iii)

Let P„ be the proposition u„ = 2 " ~ for i ieZ*.

When« = I , LHS = w, = |

RHS = i t ! = 2= LHS l + l 2

.'. Pi is true.

Assume Pk is true for some k e Z + , i.e. wt =

We want to show that Pî is also true, i.e. _2{k + l) + \ + 3

" k + ] (/t + l) + l " k + 2

2k + 1 A + l

LHS = =wA +

2* + l (* + !)(* + 2)

1 k + l (k + \)(k + 2)

_ (2k + ])(k + 2) + \2+5k + 3 (k + \)(k + 2) " (A + 1XA + 2)

_ (Ar -+ t)(2A: + 3) (*+I)(*+2)

= 2A±1 = RHS k + 2

:. Pi is true => P*+i is true. Since Pi is true and P* is true => P*+i is true, by mathematical induction, P„ is true for for all n e Z + .

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper 2 Solutlons/OHS Page 7 of 14

X~B(20,p) P(A"^l) = 0.8 P(X = 0) + P(X = 1) = 0.8 ( l - / j ) 2 0 + 20p(l-/>)" = 0.8

V2-.8

isea X=Q.0«IVIIIa]| T=B.i

FromGC, p = 0.041412 = 0.0414

*~B(20,0.041412) P(X<a)>0.999=>P(XZa-From GC, P(X$ 4) = 0.99888 < 0.999 P(X< 5)= 0.99988 > 0.999 Thus the least value of a - 6.

1)> 0.999

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper2Sc!u1tons/t)HS Page 8 of 14

6(1) No. of ways to arrange the 4 boys =4! No. of ways to arrange the girls such that they are separated (after the boys are seated) = 5 ? 3

By (MP), total number of ways = 4! 5P 3 = 1440

Alternative '2 + 3 = 5'

3 4! = 3!xl0x4!= 1440

(ii) Consider the arrangement of <BBBB>, 3 G & 3 distinct "objects" A, B and C.

No. of ways to arrange these 7 entities at a round table = 6!

No. of ways to arrange the 4 boys in the group = 4!

Since A, B and C represent 3 identical empty seats, 6141

.-. No. of ways - - j p = 2880

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper 2 SoMlons/DHS Page 9 of 14

c lear

'I c l e a r

' I c l e a r

Stage 1 Stage 2 Stige 3

0) P(game ends prematurely) = (0.3)(0.3) = 0.09

(ii) P(clears exactly 2 stages) = (0.7)(0.35)(0.825) + (0.7)(0.65)(0.35) + (0.3)(0.7)(0.35) = 0.43488 = 0.435 (3 d.p.)

(iii) P(clears the third stage | cleared exactly 2 stages) (0.7)(0.65)(0.35) + (0.3)(0.7)(0.35)

0.43488 = 0.535 (3 d.p.)

3 D H S 2 0 1 1 9740/2011 Year B H2 Math Prelim Paper 2 Sokilions/CHS Page 10 ol 14

• 1*;'Sugficste'diSouitiojts^î.i-.:^^.-^kt.^î, 8(a) Anv one possible answer

1. Determine i f there's a relationship/trend/pattern between two variables. 2. Identify outliers or suspicious observations.

(b)(i) From GC, r = -0.930 (3 s.f.)

r = - 0.930 is close to - l => a strong negative linear correlation between x and y-=> linear model is appropriate

(U) y i ton —

i«n —

.Mf> -

un -

320 -

300 -

a

D a

L a 1 1 1 | 1 1 1 ,1 v

(U)

1 1 1 1 1 1 1 1 ^ x 7H 80 90 I f" 110 " 0 I V I 140

(iU) Anv one possible answer 1. The scatter diagram shows that as x increases, y decreases (but at an

increasing rateVgraph is concave downwards. For a linear model, the rate of decrease is constant.

2. Between x andy, r ~- 0.930 while between x1 andy, r -- 0.952 is closer to - 1 .

the model y-a + bx2 is better.

(iv) From GC, when x = 150, y = 446.87 - 0.006788(150)2

= 294 (3 s.f.)

Since city bird density x - 150 lies outside the given data range, the model may not be valid, hence the prediction for forest bird density may not be reliable.

Any one possible answer (or anv logical answer)

(iv)

1. Immigration / emigration of birds from/to other countries. 2. Births/deaths/diseases 3. Not constant due to sampling variation.

© D H S 2011 9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS Page 11 of 14

9(i) C~N(68 ,« 2 )

r~N(65t8a) S~N(70,102)

P(C>85) = 0.05

8 5 - 6 8 = ,.6449 a

or = 10.335 = 10.3 (3 s.f.)

(») C~N(68,102)

P(C > 75) = 0.058762 = 0.0588 (3 s.f.) (iii) Let X = 0.2C + 0.27/ + 0.65

E(*) = 0.2(68) + 0.2(65) + 0.6(70) = 68.6

Var(^) = 0.2 2j^y j + 0.22(82) + 0.62(102) = 39.36

*«-N(68.6,39.36) ?{X > 80) = 0.034601 = 0.0346 (3 s.f.)

(iv) Cor C, T and 5 are independent.

© D H S 2011 9740/2011 Years H2 Mah Prelim Paper 2 Soluliors/DHS Page 12 of 14

10(1) An unbiased estimate of population mean n ,

x = £ ^ ~ 8 ) + 8 = 31.333 = 31.3 (3 s.f.) 60

An unbiased estimate of population variance <r2:

= - 8 ) 2 - ( I ( 6 Q " 8 ) ) ] = 3 6 ' 1 5 8 • 3 6 - 2 ( 3 S - F - ) •

(ii) Let A* represent the number of dengue cases per day with population mean /J.

To test H 0 : n = 30 against Hi'.p > 30

One-tail test at 5% significance level.

Since sample size n = 60 is large, by Central Limit Theorem, , „ - 36.158^ under Ho, X ~ N 30, approx.

Using Z-test, p-value = 0.042978 - 0.0430 (3 s.f.) (from GC).

Since p-value = 0.0430 < 0.05, we reject Ho and conclude that there is sufficient evidence at 5% level of significance that the mean number of dengue cases has increased.

The p-value is the lowest level of significance for which the null hypothesis of mean number of dengue cases being 30, will be rejected. Alternative: The p-value is the probability of obtaining a test statistic at least as extreme as 31.3, assuming that Ho is true.

(iii) To test H 0 : /i = 30 against H | i / i < 30

One-tail test at 5% significance level. Since sample size n = 7 is small, assuming X is normally distributed with o 2

unknown, y—in

under Ho, T- . . . . r(6). V36.158/7 ;

Using P(T < O • 0.95, tt = l .943 (MFl 5) To reject Ho at 5% level of significance,

/ - 3 ° <-l.943 V36.158/7 x< 25.584 x< 25.6 (3 s.f.)


Let X be the number of traffic offences occurring at a particular stretch of road between 7:30am to 9:30am on any day. i.e. X-Po(X)

P(*>1) = 0.95 1-P(* = 0) = 0.95 P(* = 0) = 0.05

e _ / l = 0.05 A = 3 (nearest integer)

(0 Let Y be the number of traffic offences occurring between 7:30am to 9:30am during weekdays in a given week, i.e. r-Po(15)

P(y^l7) = 1-P(v<;l6) = 0.33588 = 0.336 (3 s.f.)

(ii) Since n = 52 is large, by Central Limit Theorem,

Y - ^ f ' ^ ' ^ l approximately

P(I3<F< 16) = 0.969 (3 s.f.) (iii) Let W be the no. of weeks (out of 52 weeks) where there are at most 16 traffic

offences occurring between 7:30am to 9:30am during weekdays, i.e. W-B(52,1-0.33588)

W-B(52,0.66412)

Since n = 52 is large, np = 34.534 > 5, «(1 - p ) = 17.466 > 5, :.W ~ N (34.534, It.599) approximately

P(30 < W < 42) = P(30.5<(T<41.5) (c.c. applied) = 0.861 (3 s.f.)

(iv) A Poisson distribution with mean 36 is not appropriate as the mean number of offences occurring at a different 2-hour period of a day is unlikely to be the same as A = 3, and hence the mean number of offences occurring in a full day is unlikely to be 12/1 = 36 as well.

© D H S 2011 9740/2D11 Year 6 H2 Math Prelim Papor 2 Solutlons/DHS Page 14 o1U

Documents

Dhs 2011 Prelim