954/2 Oct 2009

954/2 Oct 2009

1

MATHEMATICS S/T PAPER 2 ANSWER SCHEME

Confidential 3. If x – y = z, express ddxy in terms of ddxz . Using the substitution

1. Solve the equation sin 3θ + sin 5θ = cos θ for values of θ between 0 and π.

x – y = z, solve the differential equation

[5].

dy dx

y = 0 when x = 0. x – y = z …(I) Differentiate (I) with respect to x

sin 3θ + sin 5θ = cos θ sin 5θ + sin 3θ = cos θ

 5θ + 3θ   5θ − 3θ  2sin   cos   = cos θ  2   2 

M1

2sin4θ cos θ = cos θ 2sin4θ cos θ – cos θ = 0 cos θ (2sin4θ – 1) = 0

M1 (2sin4θ – 1) = 0

1 cos θ = 0 , sin4θ = (both) 2 π π 5π 13π 17π θ= 4θ = , , , 2 6 6 6 6 π 5π π 13π 17π θ = , , , , (CAO) 24 24 2 24 24 0.1309, 0.6545, 1.571, 1.702, 2.225 0.131, 0.654, 1.57, 1.70, 2.23

dx dx

–

1–

dy dx

dy dx

dy dx

dx

= dz

A1

(CAO) (CAO)

dy dx

⇒

dx

B1

= 1– dz …(II) dx

=x–y

Substitute (I) and (II) ⇒ dz = 1– z

dx

B1

[6]

= dz

1– dz = z A1

= x – y, given that

M1

dx

1

∫ 1 − z dz = ∫ dx

M1

– ln(1– z) = x + C Given x = 0 , y = 0 then z = x – y = 0 – ln(1– 0) = 0 + C ⇒ C = 0 ∴– ln(1– z) = x – ln(1– (x – y)) = x ln(1– x + y) = – x or y = e– x+ x –1

A1 M1

A1

2. Express 6 sin x cos x – 8 cos2 x + 22 in the form A sin 2x + B cos 2x + C, with A, B and C are constants to be determined. Hence, find the largest and smallest value for

1 6 sin x cos x − 8 cos2 x + 22

chord BC is parallel to HK. Chords AD and BC produced meet at X. Prove that (i)∠ ∠ABC = ∠ADB,[3] (ii) ∆ABD and ∆CXD are similar. [4].

[6].

6 sin x cos x – 8 cos2 x + 22 = 3(2 sin x cos x) – 4(2 cos2 x) + 22 = 3(2 sin x cos x) – 4(cos 2x + 1) + 22 = 3sin2x – 4cos 2x + 18 Let g(x) = 3sin2x – 4cos 2x = rsin(2x –α α) , r>0, α= acute = r sin2x cosα α – r cos2x sinα α ∴ r cosα α = 3 … (I) and r sinα α= 4 …(II) (I)2+(II)2: r2 (cosα α)2 +r2 (sinα α )2 = 3 2 + 4 2 2 2 2 r [(cosα α) + (sinα α) ] = 25 r2 = 25 ⇒ r = 5 From (I) : 5 cosα α = 3 ⇒ cosα α=

4. In the diagram, HAK is a tangent to the circle ABCD. The

1 Let f(x) = 6sin x cos x − 8cos2 x + 22 1 1 = = 3sin 2 x − 4 cos 2 x + 18 5sin(2 x − 53.13° ) + 18 1 1 f(x)min= = 5 + 18 23 1 1 f(x)max= = − 5 + 18 13

α2

D

α3 H (i)

B1

K

A

Let ∠ABC = α1, ∠ADB =α2 , ∠HAB =α3

α1 = α3 (given BC//HK, a pair of alternate interior angle) α2 = α3 (angles in alternate segments)

B1 B1

∴ α1 = α2 ⇒ ∠ABC = ∠ADB

B1

B

C

α1 θ2

A1 A1

X

α1

B1

3 ⇒ α =53.13o 5

g(x) = 3sin2x – 4cos 2x = 5 sin (2x –53.13o) g(x)min= –5 g(x)max= 5

C

B

M1 A1

α3

α2 β2

θ3

β1 θ1 α4

X

D

H K Let A ∠XDC=α4, ∠CXD =θ1,∠ABD=θ2 , ∠XCD=β1, ∠BAD=β2, ∠KAD=θ3

954/2 Oct 2009 From part (i)

α1 = α4 ∴ α4 = α2 β1 = β2 θ1 =θ2

2

Confidential

α1 = α3 = α2

p

(Exterior angle of cyclic quadrilateral) …..(∗)

B1 B1 B1

(Exterior angle of cyclic quadrilateral) …(☆) (If two angles of one triangle equal two angles

of another triangle, the remaining angles are equal) . . . (ℂ) [Can also use: θ1 = θ3 (given BC//HK) θ2 = θ3 (angles in alternate segments). So θ1 = θ2 ] B1

From (∗),(☆),(ℂ) : ∆ABD and ∆CXD are similar (AAA)

5. The population of an island at time t is p. The number of births per unit time is b, and the number of deaths per unit time is proportional to the population at that time. By taking p as a continuous variable, show that

dp dt

= b – kp, where k is a

positive constant. [2] Solve the differential equation with the condition that p = b when t = 0. Express p as a function of 5k

t.[5]Find the exact time when the population of the island is 9b .[2]Sketch the graph of p versus t.[2] 10 k

dp =b dt dp dp Number death perunit time: ∝p ⇒ = –kp , k>0 dt dt

b k

∴

= b – kp

t

1 −k dp = dt − k b − kp 1 ln(b–kp) = t + C −k

θ 25

60

VP= (60) + (25) 2

2

β

β

12

9

VQ= (9) 2 + (12)2

 60   9  –   25   12 

=

A1

= 65 kmh−1

= 15 kmh−1

 51 =   13 

M1

)=67.38o θ=tan−1( 60 25

9 )=36.87o θ=tan−1( 12

=51 i +13 j kmh−1

Magnitude , VPQ = (51)2 + (13)2 = 2770 , 52.63 , 52.6 kmh−1 α

α

A1

13

51

Given p = b when t = 0

14.30o

90 –α = 90–75.70=14.30o Let shortest distance =QP1=d Sine Rule d 22 = 0 sin14.30 sin 900 d = 5.434 km , 5.43 km

d 22 km

P Ship P VP= 65 kmh−1

Q Ship Q 9 VQ =   kmh−1 ɶ  12 

Ship P relative to Q

P

β

β 9

M1A1

12

VQ= (9) 2 + (12)2 M1 A1

A1

A1

A1

In the direction N 75.70o E

VPQ ɶ A

M1

M1

51 ) =75.70o Direction , α= tan−1( 13

P1

5k

1 1 ln(b–k b ) = 0 + C ⇒ C = ln( 4 b ) 5k 5 −k −k 1 1 ln(b–kp) = t + ln( 4 b ) 5 −k −k 1 1 – ln(b–kp) + ln( 4 b ) = t (Multiply –k ) 5 k k ln(b–kp) – ln( 4 b ) = –k t 5 5 ln [(b–kp)× ] = –k t …………(☆) 4b 5 5k b 4b –k t [( – p)] = e–k t ⇒ p = – e 4 4b k 5k 5 Given: p= 9b From (☆):–k t = ln[(b–k 9b )× ]=ln 81 10 k 10 k 4b 1 3 t = ln 8 or ln 2 k k

D1

moves with a velocity (9i +12j) kmh−1 , where i and j are unit vectors in the directions to the east and north respectively. At a particular instant , the position vector of ship Q relative to ship P is 22i km. Calculate the magnitude and direction of the velocity of ship P relative to ship Q and the shortest distance between ship P and ship Q. [6] At 0900 hours when ship P is directly north of ship Q, ship P changes its course but maintains its speed to intercept ship Q. Find the course that must be steered by ship P and the time taken to intercept ship Q. [8] Ship P Ship Q Ship P relative to Q  60  9 VPQ = VP − VQ VP =   kmh−1 VQ =   kmh−1 ɶ ɶ ɶ ɶ  25  ɶ  12 

B1

∫

∫

shape of curve

6. Ship P moves with a velocity of (60i +25j) kmh−1 and ship Q

θ

1

∫ b − kpdp = ∫ dt

D1

b 5k

Number birth perunit time:

dp dt

asymptote and p value

= 15 kmh−1 −1

θ=tan (

9 12

Q o

)=36.87

M1 A1

954/2 Oct 2009

3 P2

P1 y

VPQ ɶ A

Let P2Q = y y = tan 14.30o 22km y = 5.608 km

14.30

x→1

M1 A1

lim [ 54 x − 12 x 2 ]= 1 − x→1

b 1

= P( 12 < X < M1

[ 54 − 12 ] = 1–b b= 1 –

22 km

P

(ii) P( 12 < X <

(i) lim F(x) = F(1)

d

o

Confidential

3 1 = 4 4

A1

Q Sine Rule sin λ sin 36.87o = 15 65 o λ=7.96 Ship P has to steered in the direction S 7. 96o E. µ= 180o –36.87o –7.96o= 135.17o

P2 λ

VP=65

VP2Q

µ

β=36.87

VQ=15 Q

Sine Rule VP 2 Q

65 sin135.170 sin 36.870 VP2Q =76.38 kmh−1 5.608km Dist an ce y Time = = = 0.0734 hour = 4 minutes 76.38kmh −1 Speed VP 2Q =

A1

X ~ Po(1) and P(X = x) =

8

P(X = x)

1 8

1 4

9 16

1 16

E(|X–7|) =

∑ [ x − 7 P( X = x )]

=

9 16

M1

A1

or 0.5625

8. Continuous random variable X has the following cumulative

0, 5 1 2 distribution function. F(x) =  4 x − 2 x , 1 − b ,  x (i) Find the value of b.[2] (ii) Hence , find P(

=[ 45 ]–[ 12 ] =

A1

3 10

e− µ µ x x!

=e

−1

(1) x x!

,

x= 0,1,2, …..

e−1 (1)3 3!

= 0.06131 , 0.0613 Y = number of cars stopping over at a petrol station is 5 minutes

M1

9 +|8–7| 1 =|5–7| 81 +|6–7| 14 +|7–7| 16 16

M1

4

A1

4

7

= [1– 4(15 ) ]–[ 54 ( 12 ) − 12 ( 12 ) 2 ]

(i) P(X = 3) =

distribution. 5 6 7 8 x 2 2 P(X = x) c + 2c 2c c c2 where c is a constant to be determined.Find the exact value of E(|X–7|) [5] Sum of all probabilities = 1 2c2 + c + c2 + 2c + c2 = 1 M1 4c2 + 3c – 1 = 0 M1 A1 (4c –1)(c +1) = 0 ⇒ ∴ c = 1 , c>0 6

M1

M1

7. A discrete random variable X has the following probability

5

)

Only this method accepted =F( 54 ) –F( 12 )

station is one car per minute. Find, correct to three decimal places, the probability (i) that exactly three cars stop at the petrol station in a period of one minute, [2] (ii) that more than two cars stop over at the petrol station in a period of five minutes. [4]. X = number of cars stopping over at a petrol station is one minute

Y ~ Po(5) and P(Y = y) =

e− µ µ y y!

=

e −5 (5) y y!

e −5 (5)0 0!

+

e−5 (5)1 1!

+

e −5 (5)2 2!

] = 1–e–5[

B1 M1 (5)0 0!

+

(5)1 1!

+

(5)2 2!

]

= 1–[0.00674+0.03369+0.08422] = 0.8753 , 0.8754 , 0.875

1 5 < X < )[3]. 2 4

M1 A1

10. A survey shows that 60% of the housewives have seen the advertisement of a new product in television. The probability that the housewife who has seen the advertisement buys the new product is 0.9, while the probability that the housewife who has not seen the advertisement buys the new product is 0.3. (i) Find the probability that a housewife buys the new product. [2] (ii) Find the probability that a housewife who buys the new product has seen the advertisement. [2] (iii) Using an appropriate approximation, find the probability that less than 10 out of 50 housewives who buys the new product, have not seen the advertisement. [6] A = housewife has seen the advertisement B= housewife buys the new product P(B|A)=0.9 B P(A∩B)=0.54 A P(A)=0.6 P(B’|A)=0.1 B’ P(A∩B’)=0.06

x < 0, 0 ≤ x < 1, . x ≥ 1.

B1 B1

, y= 0,1,2, …..

(i) P(Y > 2) = 1–P(Y < 2) = 1 – [P(Y =0)+P(Y =1)+P(Y =2)] = 1–[

x

5 4

)

9. On average, the number of cars stopping over at a petrol M1

A1

The time is at 0904 hours

5 4

P(A’)=0.4

P(B|A’)=0.3

B P(A’∩B)=0.12

P(B’|A’)=0.7

B’ P(A’∩B’)=0.28

A’

954/2 Oct 2009

4

(i) P(B) = P(A ∩ B ) + P(A’ ∩ B ) = (0.6 × 0.9) + (0.4 × 0.3) 33 =0.66 , 50 (ii) P(A|B) =

= 0.8182, 0.818

9 11

,

D1 ( whiskers seen on graph paper) D1 ( all correct with outlier)

M1 A1

P ( A ∩ B ) 0.6 × 0.9 = P( B ) 0.66

M1

(iv) the distribution is skewed to the right / positively skewed

B1

A1

Q2 – Q1 < Q3 – Q2 or right box longer than left box or left box shorter than right box or median is closer to Q1

B1

(iii) X = number of housewives who buys the new product, have not seen the advertisement 2 ) X~B(50, 11 Mean = np = 50×

2 11

= 100 (>5) 11

Variance = npq = 50×

2 11

B1(both mean & variance)

9 = 900 × 11 121

B1

Suitable approximation : X~N ( 100 , 900 ) 11 121 Z=

X − 100 11 900 121

=

X − 100 11 30 11

M1 M1

P(X<10) = P(X<9.5) =P(

X − 100 11

<

30 11

9.5− 100 11 30 11

)

A1 (for0.15) A1

=P (Z<0.15) =0.5596 , 0.560

Confidential

12. A company sells two types of flour, brand A and brand B, in a 10 kg bag. The mass of brand A flour in each bag is normally distributed with mean of 10.05 kg and a standard deviation of 0.2 kg, whereas the mass of brand B flour in each bag is normally distributed with mean of 10.05 kg and a standard deviation of 0.05 kg. (i) Find the probability that a bag brand A flour and a bag brand B flour selected at random both are more than 10 kg. [6] (ii) If J represents the total mass of two bags of brand A flour and six bags of brand B flour, find the mean and variance of J. Hence, find the probability that the total mass of the eight bags of flour is less than 80 kg. [8] (i) A~N(10.05, 0.22) B~N(10.05, 0.052) A−10.05 0.2

10.05 ) > 10−0.2

11. The number of ships which anchor at a port every week for 26

P(A>10) == P (

paricular weeks are as follows. 32 28 43 21 35 19 25 45 35 32 18 26 26 26 38 42 18 37 49 63 23 40 20 29 (i) Display the data in a stemplot. (ii) Find the median and interquartile range. (iii) Draw a boxplot to represent the data. (iv) State the shape of the frequency distribution. Give a reason for your answer. [2] (i) The number of ships that anchor at a port 1 8 8 9 2 0 1 3 5 6 6 6 8 9 3 0 2 2 5 5 7 8 4 0 2 3 5 6 9 5 6 3 Key : 1|8 means 18 ships (ii) Median : r= 12 (26)=13

=P (Z> –0.25) = 0.5987 10.05 > 10−10.05 ) P(B>10) == P ( B−0.05 0.05

Median = r= r=

1 4 3 4

y13 + y14 2

(26)=6.5 (26)=19.5

=

30 + 32 2

= 31

Q1 =y7= 25 Q3 =y20= 40

o 25

31

D1 (box and his median)

40

B1

B1

49

63

=P (Z> –1) = 0.8413 P(both more than 10kg)= P(A>10) ×P(B>10) = 0.5987 ×0.8413 = 0.5037 (ii) Let J = A1 + A2 + B1 + B2 + B3 + B4 + B5 + B6 .

D1 D1 D1

M1 A1 M1 A1

E(J) = E(A1 + A2 + B1 + B2 + B3 + B4 + B5 + B6) Mean of J = 2 (10.05) + 6 ( 10.05) = 80.4

M1 A1

Var(J) = Var (A1 + A2 + B1 + B2 + B3 + B4 + B5 + B6) Variance of J = 2(0.2)2 + 6 (0.05)2 = 0.095

M1 A1 B1

J ~N(80.4, 0.095) and Z= J −80.4 0.095

P(J < 80 ) = P ( B1 B1 (both Q) M1 A1

Interquartile Range = Q3 – Q1 = 40 – 25 = 15 (iii) 1.5(Q3 – Q1)= 22.5 Q1–22.5 = 25–22.5 = 2.5 Q3+22.5 = 40+22.5 = 62.5 Any value ∉(2.5,62.5) is an outlier 63 is an outlier On graph paper,

18

30 46 [2] [4] [3]

M1 A1

J −80.4 0.095

< 80−80.4 ) 0.095

= P ( Z < –1.2978) or P ( Z < –1.298) = P ( Z > 1.2978) or P ( Z > 1.298) = 0.09718 , 0.09714 , 0.0972 , 0.0971

13. 14.

M1 A1 A1