Mathematics T P1P2 Answer Schema_Pahang

CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2012

Instructions to candidates:

Answer all questions.

Answers may be written in either English or Bahasa Malaysia.

All necessary working should be shown clearly.

Non-exact numerical answers may be given correct to three significant figures,

or one decimal place in the case of angles in degrees, unless a different level of accuracy is

specified in the question.

Mathematical tables, a list of mathematical formulae and graph paper are provided.

This question paper consists of 5 printed pages.

950/1, 954/1 STPM 2012

Three hours

MATHEMATICS S

PAPER 1 (SET 1)

MATHEMATICS T

PAPER 1 (SET 1)

Percubaan Pahang STPM 2012 http://edu.joshuatly.com/ http://fb.me/edu.joshuatly

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CONFIDENTIAL*

2

Mathematical Formulae for Paper 1 Mathematics T / Mathematics S :

Logarithms :

ax

xb

ba log

loglog

Series :

)1(21

1

nnrn

r

)12)(1(61

1

2

nnnrn

r

22

1

3 )1(41

nnrn

r

Integration :

dxdxduvuvdx

dxdvu

cxfdxxfxf

)(ln)()('

cax

adx

xa

1

22 tan11

caxdx

xa

1

22sin1

Series:

N n where

,

21)( 221 nrrnnnnn bba

rn

ban

ban

aba

1,!

)1()1(!2

)1(1)1( 2

xxr

rnnnxnnnxx rn where

Coordinate Geometry : The coordinates of the point which divides the line joining (x1 , y1) and (x2 , y2) in the ratio m : n is

nmmyny

nmmxnx 2121 ,

The distance from ),( 11 yx to 0 cbyax is

22

11

ba

cbyax

Numerical Methods :

Newton-Raphson iteration for 0)( xf :

)(')(

1n

nnn xf

xfxx

Trapezium rule :

b

a nn yyyyyhdxxf ])(2[21)( 1210

nabhrhafyr

and where )(

Trigonometry : BAAABA sincoscossin)sin( BABABA sinsincoscos)cos(

BABABA

tantan1tantan)tan(

AAAAA 2222 sin211cos2sincos2cos

AAA 3sin4sin33sin

AAA cos3cos43cos 3


CONFIDENTIAL*

3

1. If A , B and C are arbitrary sets, show that CCBACBCA )()]'()'[( .[4 marks]

2. Using trapezium rule, with five ordinates, evaluate

2

1

21ln dxx , giving your answer correct

to three decimal places. [4 marks] 3. Solve the following inequalities.

(a) 02

15

1

xx [3 marks]

(b) 523 xx [3 marks]

4. Given sinxe y x , show that 2

22 2 0.d y dy y

dx dx . [6 marks]

5. The function f is defined by

21,2310,

)(2

xxxx

xf .

(a) Find )(lim

1xf

x and )(lim

1xf

x .

Hence, show that f is continuous at x = 1. [4 marks] (b) Given the function g is periodic with period 2 and )()( xfxg for 20 x ,

sketch the graph of g for 42 x . [3 marks]

6. Given the complex number 1,,3 2 iRpipz such that i

z34

2

is a real number,

(a) simplify the complex number 2z in terms of p , [1 marks] (b) determine the possible values of p. [4 marks] If p < 0 , find

(c) the real number i

z34

2

, [1 mark]

(d) the argument of z. [2 marks]


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4

7. The circle C has the equation 06012633 22 yxyx . The equation of the straight line l is 52 xy . (a) Find

(i) the coordinates of the centre of C and the radius of C. [3 marks] (ii) the distance from the centre of C to the straight line l. [2 marks]

(b) Given that the straight line l intersects the circle C at ),( 11 yxA and ),( 22 yxB such that

21 xx , find (i) the coordinates of the points A and B, (ii) the value of k if the points A, B and (0 , k) are collinear. [4 marks]

8. It is given that xx

xx

3

24 123 can be express as xx

xxf

3

21)( .

(a) Determine )(xf . [1 mark]

(b) Express xx

x

3

21 as sum of partial fractions. [4 marks]

(c) Show that

2

1 3

24

54ln

29123 dx

xxxx . [4 marks]

9. Expand 2

4 2

)1(1

xax

in ascending powers of x up to and including the term in 3x . [4 marks]

Given that the first four terms in the above expansion are 1 + 2x + 4x2 + 6x3, find (a) the value of a , [2 marks] (b) the set of values of x for which the expansion is valid. [2 marks]

(c) the value of 4 17 correct to three significant figures by taking 81

x . [2 marks]


CONFIDENTIAL*

5

10. The equation of a curve is given .

(a) Show that there exists only one turning point and determine the nature of this point. [4 marks] (b) Determine the respective intervals for which the curve concaves downward and concaves

upwards. [4 marks] (c) Sketch the curve showing clearly its asymptote. [3 marks]

11. The polynomial 1 2346)( xbxaxxxp , where a and b are real constants has a factor (2x – 1). The derivative of )(xp with respect to x , )(' xp , leaves a remainder

76 when divided by (x + 1). (a) Find the values of a and b. [6 marks] (b) Explain why 1 is a zero of )(xp .

Hence, solve the equation 0)( xp . [4 marks]

(c) Find the set of values of x such that 0123

)(2

xxxp . [3 marks]

12. Given that

211120102

M and .164441514115

N

(a) Show that M is a non singular matrix (i.e the inverse of M is defined). [2 marks] (b) Determine the matrices N – 6M and M(N – 6M).

Hence, find the inverse of M. [5marks] (c) From the inverse of M in (b), find the adjoin of M [2 marks] (d) Use all the information above, solve the simultaneous equations

20x – 10z = –100 20y + 10z = 300 –10x + 10y + 20z = 200 [4 marks]

===END OF QUESTION PAPER===


CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP

PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN

PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN

PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP









PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN

JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP

PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR 2012

Instructions to candidates:

Answer all questions. Answers may be written in

either English or Malay.

All necessary working should be shown clearly.

Non-exact numerical answers may be given correct to

three significant figures, or one decimal place in the

case of angles in degrees, unless a different level of

accuracy is specified in the question.

Mathematical tables, a list of mathematical formulae

and graph paper are provided.

This question paper consists of 6 printed pages.

Question no.

Full Marks

Marks

1 5

2 12

3 7

4 10

5 6

6 10

7 6

8 6

9 10

10 9

11 12

12 7

100

954/2 STPM 2012

Three hours

MATHEMATICS T

PAPER 2 SET 1


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CONFIDENTIAL*

2

Mathematical Formulae for Paper 2 Mathematics T : Logarithms :

a

xx

b

b

alog

loglog

Series :

)1(2

1

1

nnrn

r

)12)(1(6

1

1

2

nnnrn

r

22

1

3 )1(4

1

nnrn

r

Integration :

dxdx

duvuvdx

dx

dvu

cxfdxxf

xf )(ln

)(

)('

ca

x

adx

xa

1

22tan

11

ca

xdx

xa

1

22sin

1

Series:

N n where

,

21)( 221 nrrnnnnn bba

r

nba

nba

naba

Coordinate Geometry : The coordinates of the point which divides the line joining (x1 , y1) and (x2 , y2) in the ratio m : n is

nm

myny

nm

mxnx 2121 ,

The distance from ),( 11 yx to 0 cbyax is

22

11

ba

cbyax

Maclaurin expansions

1,!

)1()1(

!2

)1(1)1( 2

xx

r

rnnnx

nnnxx rn where

...!

...!2

12

r

xxxe

rx

11...,1

...32

1ln

132

xr

xxxxx

rr


CONFIDENTIAL*

3

Mathematical Formulae for Paper 2 Mathematics T :

Numerical Methods : Newton-Raphson iteration for 0)( xf :

)('

)(1

n

n

nnxf

xfxx

Trapezium rule :

b

ann yyyyyhdxxf ])(2[

2

1)( 1210

n

abhrhafyr

and where )(

Correlation and regression :

Pearson correlation coefficient:

22

yyxx

yyxxr

ii

ii

Regression line of y on x :

y = a + b x

where

2

i

ii

xx

yyxxb

xbya

Trigonometry

BAAABA sincoscossin)sin(

BABABA sinsincoscos)cos(

BA

BABA

tantan1

tantan)tan(

AAAAA 2222 sin211cos2sincos2cos

AAA 3sin4sin33sin

AAA cos3cos43cos 3

2

BAcos

2

BAsin2BsinAsin

2

BAsin

2

BAcos2BsinAsin

2

BAcos

2

BAcos2BcosAcos

2

BAsin

2

BAsin2BcosAcos


CONFIDENTIAL*

4

1. Prove that sin + sin 3 + sin 5 + sin 7 = 16 sin cos 2

cos 2 2 . [5] 2. (a) In the diagram 2A, QAPK and RBP are straight lines and PT is a tangent. If KPT =72 and TPR = 65, find PQR and PRQ . [4]

Diagram 2A Diagram 2B (b) In the diagram 2B, PQ is a diameter of the circle and S is a point on the circumference of the circle with centre O. Perpendiculars from P and Q meet the tangent through S at points R and T. (i) Prove that PSR = OSQ. [3]

(ii) Prove that PSR and SQT are similar. [3] (iii) Prove that PR QT = RS ST . [2]

3. (a) If i + 3 j and 3 i + ( 8 + ) j are two parallel vectors find the possible values of . [3] (b) Points P , Q and R have positions vectors 3 i j , 2 i + 2 j and i 2 j . Calculate the scalar product of QP RP , hence find angle QPR . [4] 4. By using the substitution y = v x , show that the differential equation

y

dx

dy = 2 y x can be changed into the differential equation x

dx

dv + v

v 2)1( = 0.

Hence, solve this differential equation , given that y = 2 when x = 1 , show that

ln ( y x ) = xy

yx

2 . [10]


CONFIDENTIAL*

5

5. The following set of data shows the age ( to the nearest year) of a random sample of 36 people who registered their names to book Perodua Viva in January 2012.

28 35 31 32 31 25 34 26 37 33 30 29 30 36 48 44 32 37 41 37 39 60 51 41 37 37 31 28 43 32 35 41 30 37 50 43 (a) Copy and complete the stem plot below. 25 3 30 1 2 35 0 [2] (b) Draw a boxplot for this data and use your box plot to identify the ‘outliers’ . [4] 6. The height ( up to the nearest cm ) of 80 Form 6 male students in a school is given in the table below.

Height 150 154

155 159

160 164

165 169

170 174

175 179

180 184

Number of students

3 6 11 17 25 12 6

(a) Display this distribution on a histogram. Hence, estimate the median age. [4] (b) Calculate an estimate for the mean and standard deviation for the data in the above distribution. [5] (c) Explain why the mean and standard deviation are not necessarily the best statistical representation for this distribution. [1] 7.The probability distribution of a uniform discrete random variable , X is given below .

X 1 2 3 … … … n

P ( X = x ) n

1 n

1 n

1 … … … n

1

Show that E ( X ) = 2

1 ( n + 1 ) and Var ( X ) = 12

1 ( n 2 1 ) . [6]

8. The lifespan , in hours, of an electrical component is a random variable X with probability density function as follows

f ( x ) =

otherwise

xe

x

,0

0, 100

1 100

(a) Calculate the mean lifespan of the electrical components. [5] (b) Five electrical components are chosen at random and Y is the total lifespan of all the electrical components. Find E(Y). [1]


CONFIDENTIAL*

6

9. (a) Express 4 cos 3 sin in the form R cos ( ) , where R is a positive constant and an acute angle. Hence, solve the equation 4 cos 3 sin = 1 in the interval 0 360 . [4] (b) State the maximum and minimum values for y = 4 cos 3 sin and their corresponding angles. Sketch of the graph y = 4 cos 3 sin in the interval 0 360 . [4] Hence, solve 4 cos 3 sin 1 . [2]

10. A gold prospector examines 800 gold bearing ingots each month. The contents of each ingot can be assumed to be independent and the probability of one ingot containing gold is 0. 005. (a) The prospector is considered having a lucky month if 4 or more ingots examined by him contain gold. By using a suitable approximation , show that the probability a month chosen randomly is a lucky month is 0. 567 ( correct to 3 significant figures ). [4] (b) By using a suitable approximation , determine the probability that in 24 months chosen at random, there are more than 12 lucky months. [5] 11. (a) Diameters of a type of steel pipes produced in a factory are normally distributed with mean 0. 95 cm and standard deviation cm. If at least 88 % of the steel pipes produced have diameters which are less than 0. 98 cm , find the range of values of .[4] (b) Four runners A , B , C and D ran 100 m each . The time taken, in seconds , by each runner can be considered as independent observations from a normal distribution with mean 14 and standard deviation 0. 2 . A runner, E , ran 400 m. The time taken, in seconds, by E can be considered as an observation from a normal distribution with mean 58 and standard deviation 1. 0 and independent of the times taken by the other runners.

(i) Determine the probability that the time taken by runner E is less than four times the time taken by runner A . [4] (ii) Determine the probability that the time taken by runner E is less than 3 seconds more than that of the total time by all four runners A , B , C and D. [4]

12. At noon, an observer on a ship A sees another ship B which is 20 km away to the north of the observer’s ship A. Ship A is sailing eastwards at a speed of 20 km/h . Ship B is sailing in the directions S 30 E at a speed of 8 km/h. (a) Find the velocity of B relative to A. [3] (b) Find the closest distance between A and B. [2] (c) Find the time when the distance between the two ships is the closest. [2]


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1

Marking Scheme

PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2012

Mathematics T Paper 1 (954/1)/ Mathematics S Paper 1 (950/1)

Set 1

NO. SCHEME MARKS

1.

[(A’ C)(B’ C)] (A B C) = [(A’B’) C) (A B C) = [(A’ B’) (A B) C = [(A B)’ (A B)] C = C = C

B1 B1 B1

B1

[4 marks]

2.

h = 4

12 = 0.25

2

1

2 )1ln( dxx = 2

25.0 [ 0.6931+1.6094 + 2(0.9410+1.1787+1.4018)]

= 1.168

x y

1 0.6931 1.25 0.9410 1.5 1.1787 1.75 1.4018

2 1.6094

B1

B1

M1 A1

[4 marks]


```

2

NO. SCHEME MARKS

3.

(a) 02

15

1

xx

0)2)(5(

7

xx

The solution set = {x : x < -2 or x > 5, x R } (b) 523)5( xxx 235 xx and 523 xx x43 and 72 x

x43 and

27

x

}27

43:{ xx

B1

M1

A1

M1

M1

A1

[4 marks]

4. sinxe y x

cosx xdye e y x

dx

( ) cosx dye y x

dx

2

2( ) ( ) sinx xd y dy dy

e y e xdx dx dx

2

2

d y dy dyy y

dx dx dx

2

22 2 0.d y dy

ydx dx

M1A1

M1A1

M1 A1

[6 marks]

- +

+ + -

-2 5 + +


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3

NO. SCHEME MARKS

5.

(a) 1)23(lim)(lim

11

xxf

xx

1lim)(lim 2

11

xxf

xx

1)1(23)1( f )1()(lim

1fxf

x

is continuous at . (b)

B1

B1

M1

A1

D1 (curve)

D1

(line segment)

D1 (All correct)

[7 marks]

6.

(a) z2 = p2 − 9 − 6pi

(b) )34)(34(

)34](6)9[( 2

ii

ipip

[multiplied by conjugate complex]

ipppp

2524)9(3

2518)9(4 22

025

24)9(3 2

pp

0982 pp 1or9 pp

(c) 234

2

i

z

(d) 13tan)31arg( 1

i

)3tan( 1 = –1.893 rad

B1

M1

A1

M1

A1

A1

M1

A1

[8 marks]


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4

NO. SCHEME MARKS 7. (a)(i) 06012633 22 yxyx

2042 22 yyxx 20)2()2()1()1( 2222 yx 222 5)2()1( yx centre = )2,1( and radius = 5 units (a)(ii) distance from centre of C to straight line l

22 )2(1

5)2(2)1(

52 units (b)(i) 06012633 22 yxyx , 52 xy 06012)52(63)52(3 22 yyyy 0342 yy 0)3)(1( yy 3or1 yy 1or3 xx )3,1(,)1,3( BA

(b)(ii) 25

k

M1

A1A1

M1

A1

M1

M1

A1

B1

[9 marks]


```

5

NO. SCHEME MARKS 8. (a) xxf 3)(

(b) )1(

112

2

3

2

xx

x

xx

x

1)1(

1Let 22

2

x

CBx

x

A

xx

x

))(()1(1then 22 xCBxxAx 1,0When Ax 0:oftscoefficienequatingBy Cx 21:oftscoefficienequatingBy 2 BBAx

1

21)1(

122

2

x

x

xxx

x

(c)

2

1

2

1 23

24

1213123

dxx

x

xxdx

xx

xx

1lnln

23 2

2 2

1

xxx

)2ln023()5ln2ln6(

54ln

29

B1

B1

M1A1

A1

B1

M1

M1

A1

[9 marks]


```

6

NO. SCHEME MARKS 9.

.

32

322

4321

)(!3

)4)(3)(2()(!2

)3)(2()(!121)1(

xxx

xxxx

2

22241

2

411

)(!2

)43)(

41(

)(!1

41

1)1(

ax

axaxax

32

322

241

22

4 2

)214()

413(21

)4321)(411(

)1()1()1(

1

xaxax

xxxax

xaxx

ax

(a) 6214or4

413 aa

a = 4 (b) 14and1 2 xx

}21

21:{ xx

(c)

.).3(03.217

02771.217

32422.1)(1617

)(6)(4)(21)1(

)(41

6421)1(

41

4

4

287

4

3812

81

81

281

4 281

322

4 2

fs

xxxx

x

M1

A1 (either one

series)

M1 (multiply both correct series)

A1

M1 A1

M1

A1

M1

A1

[10 marks]


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7

NO. SCHEME MARKS 10. (a)

= - 0.794

Since there is only one real value of x, there is only one turning point

This point is a maximum point. (b)

and Curve concaves downward for { Curve concaves downward for }10:{ xx (c)

M1

A1

M1 A1

M1

M1

A1 A1

D1

Curve in 3rd quadrant with

max. pt.

D1 Curve with

point of inflexion (1,0) and concavity

in correct direction

D1

Curve approaches y-

axis as asymptote

[11 marks]

y

x 1

x = 0

x 1 0

+

+

+

+


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8

NO. SCHEME MARKS 11.

(a) 01 212

213

214

216

21

bap

12 ba 1223324)(' bxaxxxp

761)1(22)1(33)1(24)1(' bap 5323 ba 13a , 7b

(b) 1 2731346)( xxxxxp

1 is zero of 1 2731346)( xxxxxp

because 01 1 2)1(73)1(134)1(6)1( p )()1)(12()( xQxxxp 0)13)(1)(1)(12( xxxx

31 ,1 ,

21

x

(c) 0)13)(1(

)13)(1)(1)(12(

xx

xxxx

0)1)(12( xx and 1,31

xx

The solution set = }121:{ xx

M1

A1

M1

A1

M1A1

B1

M1 A1

A1

M1

M1

A1

[13 marks]

21 1


```

9

NO. SCHEME MARKS

12.(a)

(a) 4

1120

)1(2110

02112

2

M

Since 0M , M is not singular

(b) N – 6M =

422231

213 M(N – 6M) =

400040004

M(N – 6M) = 4I N – 6M = 4M-1I M-1 = 1/4 (N – 6M)

=

422231

213

41 =

121

21

21

43

41

21

41

43

(c) 1 MMMadjoin = N – 6M

=

422231

213

(d)

203010

211120102

z

y

x

or

200300100

2010101020010020

z

y

x

203010

z

y

x

M

203010

422231

213

41

z

y

x

015

5

x = -5. y = 15, z = 0

M1

A1

B1 B1

M1

A1

M1

A1

B1

M1

A1

A1

[13 marks]


CONFIDENTIAL*

Marking Scheme PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2012

Mathematics T Paper 2 ( 954 / 2 ) Set 1 Q Mark Scheme Marks 1

sin + sin 3 + sin 5 + sin 7

= = 2 sin 3 cos 2 + 2 sin 5 cos 2 = 2 cos 2 = 2 cos 2 = 4 cos 2 cos sin 4 = 4 cos 2 cos = 8 cos sin 2 = 8 cos = 16 sin = 16 sin

B1 Factor formula B1 Factor formula B1 Double angle B1 Double angle B1

Q Mark Scheme Marks

2 (a)

PAB = TPR = 65 [ Angles in alternate segment are equal ] PRQ = PAB = 65 [ The exterior angles of a cyclic quadrilateral

is equal to the opposite interior angle ]

B1 B1

APB = 180 ( 65 + 72 ) [ The sum of angles on straight line = 43 APK is 180 ] PBA = 180 ( 65 + 43 ) [ The sum of angles in triangle APB is 180 ] = 72 PQR = PBA = 72 [ The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle. ]

B1 B1


CONFIDENTIAL*

2

Q Mark Scheme Marks 2 (b) (i)

PSR = PQS [ Angles in alternate segments are equal ] PQS = OSQ [ Base angles of an isosceles triangle are the same. or OQS OQ = OS , radius of a circle . ] PSR = OSQ

B1 B1 B1

2 (b) (ii)

PRS = QTS = 90 [ All right angles are congruent . ] Let PSR = x SPR = 90 x [ The sum of angles in a triangle is 180 . ] From (i) OSQ = PSR = x QST = 90 OSQ = 90 x [ Tangent is perpendicular to the radius ] SQT = x [ The sum of angles in a triangle is 180 . ] SPR = QST = 90 x PSR = SQT = x All the corresponding angles are the same, therefore PSR and SQT are similar.

B1 equate 1st pair of angle B1 equate 2 pairs of angles B1

2 (b) (iii)

PSR and SQT are similar.

PR QT = RS ST

M1 A1


CONFIDENTIAL*

3

Q Mark Scheme Marks 3 (a)

i + 3 j = k [ 3 i + ( 8 + ) j ] i + 3 j = 3 k i + k ( 8 + ) j 3 k = k ( 8 + ) = 3

k = = 9 or = 1

B1 with constant k M1 2 equations A1

3 (b)

= ( 3 i j ) ( 2 i + 2 j ) = ( 3 i j ) ( i 2 j ) = i 3 j = 4 i + j

B1 any one correct

= ( i 3 j ) ( 4 i + j ) = 1 ( 4 ) + ( 3 ) ( 1 ) = 1

cos =

cos =

cos = = 85 36’

B1 Scalar product M1 A1


CONFIDENTIAL*

4

Q Mark Scheme Marks 4 y = v x ,

B1

y

v x = 2 ( v x ) x

v + x 2 v x + x = 0 v + x = 0 v + x = 0

x + = 0

M1 substitute A1

x =

=

=

=

=

+ 1 = ln x

+ 1 = ln x

+ 1 = ln x

+ ln x = 1 +

=

=

M1 separate M1 A1 correct partial fractions M1 correct integration M1 substitute limits M1 Substitute v = A1


CONFIDENTIAL*

5

Q Mark Scheme Marks

5 (a)

D1 Stemplot D1 Key

5 (b) Median = = =

Median = 35.5 = 31 = 41

B1

Lower Boundary = 31 1.5 ( 41 31 ) = 16 Upper Boundary = 41 1.5 ( 41 31 ) = 56

B1

Outlier is 60

D1 Box & whiskers B1


CONFIDENTIAL*

6


From the histogram , median = 170.25 or 170.5

D1 correct scale D1 correct rectangles M1 draw line A1

(b) mean = mean = 169.1875

standard deviation = = 7.3673

M1 , A1 B1 M1 A1

(c) The mean and standard deviation are not the best statistical representation because this distribution is negatively skwed.

B1

Q Mark Scheme Marks

7

E(X) = = 1 + 2 + 3 + … + n

=

=

=

M1 A1

E( ) = = + + + … +

=

=

=

M1 A1

Var ( X ) =

=

=

=

=

M1 A1


CONFIDENTIAL*

7


E ( X ) =

= =

= +

=

= = 0 = 100

M1 M1 By Parts M1 Correct integration M1 correct Limits A1

8 (b)

X represent the lifespan of one electrical component Y = + + + + E ( Y ) = 5 E (X) = 5 ( 100 ) = 500

B1


4 cos 3 sin R cos ( + )

4 cos 3 sin R cos cos R sin sin

R cos = 4

R sin = 3

tan =

= 36 52’

R =

R = 5 4 cos 3 sin 5 cos ( + 36 52’ )

M1 correct R & A1

4 cos 3 sin = 1 5 cos ( + 36 52’ ) = 1

cos ( + 36 52’ ) = + 36 52’ = 78 28’ , 281 32’ = 41 36’ , 244 40’

M1 A1


Admin

eduQR

CONFIDENTIAL*

8

9 (b)

y = 4 cos 3 sin , y = 5 cos ( + 36 52’ )

is 5 when + 36 52’ = 360 = 323 8’

is 5 when + 36 52’ = 180 = 143 8’

B1 B1

D1 correct shape D1 max & min points

The solution set is

D1 draw line y = 1 B1


X represents the number of ingots containing gold. X B ( 800 , 0.005 ) x = 0 , 1 , 2 , 3 , …, 800 Poisson Approximation : = 4 P ( lucky month ) = P ( X 4 ) = 1

= 1

= 1 = 0. 567 ( correct to 3 significant figures ).

B1 B1 M1 A1

10 (b)

Y represents the number of lucky months. Y B ( 24 , 0.567 ) y = 0 , 1 , 2 , 3 , …, 24 Normal Approximation : mean = 13.608 , variance = 5.892264 P ( Y > 12 ) = = = 1 0.3242 = 0.6758

B1 B1 M1 continuity correction M1 standardize A1


CONFIDENTIAL*

9


X N ( 0.95 , ) P ( X < 0.98 ) 0.88

0.88

0.0255

B1 M1 standardize M1 A1

11 (b) (i)

Four runners A , B , C , D ran 100 m A , B , C , D N ( 19 , ) One runner , E N ( 58 , )

P ( E < 4 A ) = P ( E 4 A < 0 )

=

= = 0.0592

E ( E 4 A ) = 58 4 ( 14 ) = 2 Var ( E 4 A ) = + 16 ( ) = 1.64

B1 correct mean & variance B1 …P ( E < 4A) M1 standardize A1

11 (b) (ii)

E ( A + B + C + D ) = 4 ( 14 ) = 56 Var ( A + B + C + D ) = 4 ( ) = 0.16

=

= = 1 0.1768 = 0.8232

= 58 56 = 2 = =1.16

B1 correct mean & variance B1 … P ( ...< 3 ) M1 standardize A1


CONFIDENTIAL*

10

Q Mark Scheme Marks

12 (a)

=

, or

The velocity of B relative to A is 17.44 km/h .

D1 diagram B1

sin = Acute angle = 83 25’ = 180 83 25’ = 96 35’ The velocity of B relative to A is in the direction of S 66 35’ W or [ 246 35’ ]

B1

12 (b) 12 (c)

The closest distance = 20 sin 66 35’ between A and B = 18.35 km Time taken = = 0.4559 hour = 27.35 minutes Time when the distance between the two ships is the closest is 12.27 pm.

M1 A1 M1 A1


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Mathematics T P1P2 Answer Schema_Pahang