perak 2010 ( x + 7) – 2 papers/2010/perak spm... · perak 2010 paper 1 1. in diagram 1, set b...

PERAK 2010PAPER 1

1. In Diagram 1, set B shows the images of certain elements of set A.

(a) State the type of relation between set A and set B.

(b) Using the function notation, write a relation between set A and set B.

2. Given the function f(x) = x – 2 and hf(x) = x2 + 3x + 1.Find(a) h(x),(b) the value of hf(−2).

[(a) x2 + 3x + 3 (b) 7]

3. Given the composite function fg(x) = 2x2 – 3 and function g(x) = 3 – x2, find f(−2).

[7]

4. If the straight line y = 1 – mx touches the curve y = x2 – 4x + m at a point, determine the values of m.

[2, 10]

5. The equation of a curve is given by f(x) = 5 – (x + 3)2, find(a) the coordinates of the maximum

point,(b) the equation of the axis of

symmetry of the curve.[(a) (−3, 5) (b) x = −3]

6. Given y = 2x2 – 15, find the range of x if y + x ≥ 0.

[x ≤ −3, x ≥ 5

2]

7. Solve the equation 4x – 1 = 8x + 3.

[−11

4]

8. Solve the equation log2( x + 7) – 2 = log2 5x.

[7

19]

9. The first three terms of a geometric

progression are 5, 3

x,

20

9. Find the

value of x, given x > 0.[10]

10. Given that 99

m = 2.020202…

= 2 + a + b + …Find the value of a, b and m.

[a = 0.02, b = 0.0002, m = 200]

11. The first 3 terms of an arithmetic progression are 1 – 2k, k + 4 and 7k – 2.Find(a) the value of k,(b) the sum of the first 8 terms of the

arithmetic progression.[(a) 3 (b) 296]

12. The variables x and y are related by the equation y = pxq, where p and q are constants. A straight line graph is obtained by plotting log10 y against log10 x, as shown in Diagram 12.

(a) Convert y = pxq to linear form.(b) Find the value of q and log10 p.

[(a) log10 y = q log10 x + log10 p(b) q = 4, log10 p = −12]

13. Diagram 13 shows a triangle PQR and sector PQT with centre P.

Retyped by Mr. Sim KY116, Jalan Arang. Tel: 450116

1http://edu.joshuatly.com/ http://www.joshuatly.com/

It is given that PQ = QR = 8 cm and

∠ PQR = 1

2π radian. Using π =

3.142, find the area, in cm2, of the shaded region.

[6.864 cm2]

14. Diagram 14 shows two vectors, OAuuur

and PQuuur

.

Express

(a) OPuuur

in the form x

y

,

(b) PQuuur

in the form xi yj+% %

.

[(a)2

5

(b) 6 5i j− −% %

]

15. The vectors PQuuur

and RSuuur

are parallel.

If PQuuur

= 2 3x y−% %

and RSuuur

= (k – 1) x%

− 3

2y%

, where k is a constant. Find

(a) the value of k,(b) the ratio of PQ : RS.

[(a) 5 (b) 2 : 1]

16. P is a point on the line segment joining the points S(−12, 1) and T(3, 6) such

that SP = 1

4PT. Find the coordinates of

the point P.

[(−9, 2)]

17. Solve the equation 2 sec2 x = tan x + 5 for 0o ≤ x ≤ 360o.

[56.31o, 135o, 236.31o, 315o]

18. Given that 6

1

( )f x dx∫ = 7 and

3 62

1 3

( ) [ ( ) ]f x dx f x hx dx+ +∫ ∫ = 42,

Find the value of h

[5

9].

19. The gradient function of a curve is

2

1

(3 )x−+ k, where k is a constant.

Given that the tangent to the curve at the point (2, −5) is parallel to the x-axis. Find(a) the value of k,(b) the equation of the curve.

[(a) −1 (b) y = 1

3 x− − x – 4 ]

20. Find the gradient of the normal to the curve y = (2x + 3)2 when x = 1.

[−1

6]

21. It is given that y = 3

5u6 + 4 and u = 5x

– 3. Find dy

dxin terms of x.

[18(5x – 3)5]

22. Find the number of different arrangement that can be formed using all the letters in the word HARMONI, if(a) none of the letters can be

repeated,(b) vowels and consonants are

arranged alternatively.[(a) 5040 (b) 144]

23. Ali, Bala and Chong will compete against each other in a badminton match. The probability that Ali will



beat Bala and Chong are 1

3 and

2

5

respectively. Calculate the probability that Ali will win at least once from the match played between them.

[3

5]

24. The standard deviation of a set of numbers m, 2m, 3m, 4m and 5m is p. Find the mean in terms of m and p.

[ 2 211m p− ]

25. Diagram 25 shows a standard normal distribution graph which represents the masses of a group of students in a school.

The probability of a student chosen at random from this group with a mass of between 46 kg and 60 kg is represented by the area of the shaded region. (a) Find the value of k.(b) The mass of students in the school

is normally distributed with a mean of m and a standard deviation of 10. Find the value of m.

[(a) k = 0.7 (b) 53]



PERAK 2010 PAPER 2

1. Solve the following simultaneous equations:

2x + y = 52x2 + xy – 2y2 = 37x – 98

[x = −2, y = 9, x = 3, y = −1]

2. The quadratic function f(x) = x2 + 8ax + b2 + 2 has a minimum value of 2. (a) By using the method of

completing the square, show that b = 4a.

(b) If the graph is symmetrical at x = 2 – a, find the value of a and b.

[a = −2

3, b = −

8

3]

3. Diagram 3 shows a triangular pattern.

(a) The first triangle is ABC, followed by ADE, AFG and so on. The measurement of the base and height of the next triangle is twice the measurement of the base and height of the previous triangle. Show and state the type of progression of the series formed by the area of the triangles, and hence find the related common difference or common ratio.

(b) Given AB = 2 cm and BC = 1 cm, find(i) the length of the base of the

eighth triangle,(ii) the sum of the area of the

first four triangles.[(a) r = 4 (b) (i) 256 cm (ii) 85 cm2]

4. (a) Sketch the graph of y = 3 cos x

for 0 ≤ x ≤ 2π .

(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the

equation 5 cos x = x for 0 ≤ x ≤ 2π .State the equation of the straight line and the number of solutions.

5. Diagram 5 shows the equilateral OPQR.

3PQ x=uuur

%, 7OP y=uuur

%and 9OR x=

uuur

%.

(a) Express in terms of x% and y%

(i) PRuuur

(ii) OQuuur

(b) Using OS hOQ=uuur uuur

and

PS kPR=uuur uuur

, where h and k are constants, find the value of h and k.

[(a) (i) 9 7x y−% %

(ii) 3 7x y+% %

(b) h = 3

4, k

= 1

4]

6. Table 6 shows the frequency distribution of the scores obtained by 500 students in an English essay writing competition.

Score Number of students20 – 29 4030 – 39 15040 – 49 16050 – 59 6060 – 69 50

Retyped by Mr. Sim Kwang Yaw 1

y = 3

5x , 3 solutions

http://edu.joshuatly.com/ http://www.joshuatly.com/

70 – 79 40(a) Without drawing an ogive,

calculate the median score.(b) State the modal class of the

distribution.(c) Calculate the standard deviation

of the distribution. Give your answer correct to three significant figures.

[(a) 43.25 (b) 40 – 49 (c) 13.6]

SECTION B7. Diagram 7 shows the shaded region A

enclosed by the curve y2 = 4(x + 1), y-axis and the straight line 3y = 18 – 2x. The curve and the straight line meet at point P.

(a) Find the y-coordinate of points P, Q and R.

(b) Calculate the area of the shaded region A.

(c) Find the volume generated, in terms of π , when the shaded region B is revolved through 360o

about the x-axis.

[(a) 4, 6, 2 (b) 52

3 unit2 (c) 30π

unit2]8. Table 8 shows the values of two

variables, x and y, obtained from an experiment. Variables x and y are related by the equation y + 10 = hkx, where h and k are constants.

x 0.4 2.5 4.0 6.0 7.5 8.0y −7.6 −3.5 2.9 22.

256.5

73.9

(a) Based on the table, construct a table for the values of log10 (y + 10).

(b) Plot log10(y + 10) against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.2 unit on the log10 (y + 10)-axis.Hence, draw the line of best fit.

(c) Use the graph in (b) to find the value of(i) h(ii) k.

[(c) k = 1.6 h = 2.0]

9. The diagram shows an isosceles triangle ABC where AB = BC. The line BD is perpendicular to the line AC.

Given the gradient of the line AC is −2

3. Find

(a) the y-intercept of line AC,(b) the equation of line BD,(c) the coordinates of point A,(d) the area of triangle ABC.

[(a)20

3(b) 2y = 3x – 4 (c) (−2, 8)

(d) 26 unit2]

10. Diagram 10 shows sector ORUS with centre O and sector TRVS with centre T. Given OR = 10 cm, TR = 6.5 cm and ∠ ROS = 1.2 radian.[Use π = 3.124]

(a) Find the length, in cm, of chord RS.

Retyped by Mr. Sim Kwang Yaw 2http://edu.joshuatly.com/ http://www.joshuatly.com/

(b) Find the value of θ , in radian.(c) Calculate the perimeter, in cm, of

the shaded region.(d) Calculate the area, in cm2, of the

shaded region.[(a) 11.29 cm (b) 2.105 rad (c) 20.61 cm(d) 12.89 cm2]

11. (a) In a farm, it is found that the male and female chicks hatched from eggs are in the ratio 3 : 5. From a sample of 9 eggs chosen at random from this farm, calculate the probability that(i) 3 male chicks are hatched,(ii) at least 3 male chicks are

hatched.(b) The weight of the chicks has a

normal distribution with a mean of 65.75 g and a standard deviation of 5 g. Given that there are 2500 chicks in the farm, find(i) the probability that a chick

chosen at random has a weight of more than 60.5 g,

(ii) the number of chicks that has a weight of less than 60.5 g.

[(a) (i) 0.2640 (ii) 0.7183 (b) (i) 0.8531(ii) 367]

SECTON C12. Diagram 12 shows triangles ABC and

ACD. Given that AD = 10 cm, AC = 8 cm, ∠ ABC = 30o and ∠ BAC = 40o. The area of ∆ ABC is four times the area of ∆ ACD.

Calculate(a) the length on cm, of AB,(b) the area in cm2, of ∆ ACD,(c) the length, in cm, of CD.

[(a) 15.04 cm (b) 9.668 cm2 (c) 2.957 cm]

13. Table 13 shows the price indices and weightages for five types of food in the year 2009 based on the year 2007.

Food

Price Price index 2009 based 2007

Weightage2007 2009

A 1.50 x 140 7B 3.00 3.78 126 kC 4.50 5.40 120 5D 3.48 4.35 125 6E y 9.10 130 4

(a) Find the value of x and y.(b) Given the composite index for the

year 2009 based on 2007 is 129, find k.

(c) Calculate the price index of food C for the year 2011 based on the year 2007, if its price increases at the same rate as it increases from 2007 to 2009.

(d) Given the price of food D is estimated to increase by 10% from the year 2009 to 2010, while other foods remain unchanged. Calculate the composite index for the year 2010 based on the year 2007.

[(a) x = 2.10 y = 7.00 (b) 4 (c) 144(d) 131.9]

14. A furniture company produces tables and chairs to supply to schools. The information of their production is shown in the table below.

Table (one unit)

Chair (one unit)

Profit RM7 RM5Carpentry 3 hr 4 hrsPainting 2 hrs 1 hr

The company received a tender for Sekolah Berjaya to make x tables and y chairs. Based on the tender, the company has to consider the following constraints:I: The working hours for carpentry

must not exceed 2400 hours.


II: The working hours for painting must not exceed 1000 hours.

III: The number of tables is at more twice the number of chairs.

(a) Write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy all the above constraints.

(b) Using a scale of 2 cm to 100 units on both axes, construct and shade the region R which satisfies all the above constraints.

(c) Using the graph constructed in (b), find(i) the maximum number of

chairs produced when 200 tables are produced,

(ii) the optimum combination of tables and chairs and hence the maximum profit of this tender.

[(a) 3x + 4y ≤ 2400, 2x + y ≤ 1000, y ≥ 1

2x

(c) (i) 450 (ii) 320 tables 360 chairs RM2420]

15. A particle starts traveling along a straight line from a fixed point, P which lies 3 m to the right of O. The velocity of the particle, v m s-1 is given by v = 2t – 8, where t is the time, in seconds, after leaving P.[Assume motion to the right as positive]Find(a) the acceleration, in m s-2, of the

particle,(b) the value of t when the particle is

momentarily at rest,(c) the displacement of the particle, in

m, from O when its velocity is −2 m s-1,

(d) the total distance travelled, in m, by the particle in the first 7 seconds.

[(a) 2 m s-2 (b) 4 s (c) −12 m(d) 25 m]


Peperiksaan Percubaan SPM 3472/1 Negeri Perak 2010

1

x

MARK SCHEME FOR PAPER 1 ADDITIONAL MATHEMATICS

Question No.

Working scheme and Marks Mark

allocation 1. (a) Many-to-one relation

(b) Semua calon akan diberi 1 markah percuma disebabkan masalah teknikal soalan

1 1

2. (a) 2 3 3x x B1: 2( ) 2 3 1h x x x

(b) 7

B1: ( 4)h

2 2

3. 7

B2: ( ) 3 2f x x

B1: 2 2(3 ) 2 3f x x or 1( ) 3g x x

3

4. m = 2, 10 (Both correct) Substituting into 2b − 4ac = 0

B2: 2( 4) 4(1)( 1) 0m m B1: 2x − 4x + m = 1 – mx

3

5. (−3, 5) (Any method) x = − 3

1 1

6. 3x or

5

2x [Both inequalities correct]

B2: The correct section is shaded with the x values shown or indicated.

−3 5

2

B1: 2 2x + x − 15 0

3

7. 11 3 or 2 or 2.75

4 4

B2 : 2x - 2 = 6x + 9 B1 : ( 2 2)x-1 = (2 3)2x+3

3



2

Question No.


allocation 8.

19

7

B2 : 45

7

x

x or

75

4

xx

B1 : log 2 7

25

x

x

or log 2

7

4

x

3

9. x = 10

B1:

209 3

53

x

x

2

10. a = 0.02 b = 0.0002 m = 200

B1: 0.02

1 0.01S

1 1 2

11. (a) k = 3 (b) S8 = 296 B1: a = –5 and d = 12

1 2

12. (a) 10 10 10log log logy q x p

(b) 4q 10log 12p

B1: 10 108 4(5) log or 0 4(3) logp p

1 1 2



3

Question No.


allocation 13. 6.864

B3: 32 25.136

B2:

1Area of 8 8 sin 90 and

21

Area of sector 8 8 2 4

PQR

PQT

B1:

1Area of 8 8 sin 90 or

21

Area of sector 8 8 2 4

PQR

PQT

4

14. (a)

5

2

(b) 6 5 i j

1

1

15. (a) 5

B2 : )1( k = 2 or 2

3 = 3 or )1( k = 2 or

2

3 = 3

or = 2 or = 2

1

B1 : )1( k x 2

3y = ( x2 y3 ) or

x2 y3 = )1( k x 2

3 y

(b) 2 : 1

3 1

16. P ( –9, 2 )

B2: 48 3 4 6

and 5 5

B1: 4( 12) 1(3) 4(1) 1(6)

or 5 5

3

17. x = 56.31 ,135 , 236.31 ,315 or 56 19 ', 236 19 ', 135 , 315

B3 : 56.31 and 236.31x or 135 and 315x

B2 : (2 tan 3)(tan 1) 0x x

B1 : 22(1 tan ) tan 5x x

4



4

Question No.


allocation 18.

h = 9

5

B2: 353

6

3

3

hx

B1: 3

1

6

3

26

342)()( dxhxdxxfdxxf or

42)(6

3

26

1 dxhxdxxf or equivalent

3

19. (a) k = –1 Parallel to x-axis

B1: 0)3(

12

kx

or 0)23(

12

k

(b) 43

1

x

xy

B1: 1(3 )

( 1)( 1)

xy x c

or cx

xy

3

1

2 2

20. 1

6

B2: d

=6d

y

x

B1: 2d=3(2 3) 2

d

yx

x

3

21. 5d=18(5 3)

d

yx

x

B2: 5d 18= (5 3) 5

d 5

yx

x

B1: 5 5d 3 6 d d 3= or =5 or = 6(5 3)

d 5 d d 5

y y yu x

u u x

3



5

4 3 3 2 2 11

Question No.


allocation 22. (a) 5040

B1: 7! or 7 6 5 4 3 2 1 or 7P7

(b) 144

B1: or 4 3 3 2 2 1 1 or 4 3

4 3P P

2 2

23. 3

5

B2: 3 4 2

15 15 15 or

2 31

3 5

B1: 1 3 2 2 1 2

3 5 3 5 3 5

or 2 3

3 5

3

24. 2 211m p or

12 2 211m p

B2: 2 2 211x m p

B1: 2 2 2 2 2

2 2(2 ) (3 ) (4 ) (5 )

5

m m m m mx p

3

25. (a) k = 0.7 B1: (1 – 0.5160) 2 or 0.2420 (b) m = 53

B1: 7.010

46

m or 7.0

10

60

m

2 2


Peperiksaan Percubaan SPM Negeri Perak 2010 3472/2

1

MARK SCHEME FOR PAPER 2 ADDITIONAL MATHEMATICS SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2

1

xy 25 or 2

5 yx

[1m]

9837)25(2)25(2 22 xxxxx or

982

5372

2

5

2

52 2

2

y

yy

yy

[1m]

Solve quadratic equation : 0)2)(3( xx or 0)1)(9( yy

[1m]

x = 3, -2 or y = 9, -1 [1m]

y = -1, 9 or x = -2, 3 [1m]

2 (a) f(x) = x2 + 8ax + b2 + 2 Completing the square, x2 + 8ax + b2 +2

= x2 + 8ax + 22 )2

8()

2

8(

aa + b2 + 2

= x2 + 8ax + (4a)2 – (4a)2 + b2 + 2 = (x + 4a)2 – 16a2 + b2 + 2

[1m]

– 16a2 + b2 + 2 is the minimum value of f(x), – 16a2 + b2 + 2 hence = 2

– 16a2 + b2 + 2 = 2 [1m]

b2 = 16a2 – 2 + 2

b2 = 16a2

b2 = (4a)2

b = 4a (shown) [1m]

(b) f(x) has a minimum value when x = – 4a

– 4a = 2 – a [1m]

4a – a = –2

3a = –2

a =

2

3

[1m]

Substitute a =

2

3 ,

b = 4 (

2

3 )

b =

8

3

[1m]



2

3 (a) area of ABC =

1

2bh

area of ADE = 1

(2 )(2 )2

b h = 2bh

area of AGF = 1

(4 )(4 ) 82

b h bh

12

8 2

2

bh bh

bh bh

[1m]

Geometric Progression

[1m]

Common ratio, r = 4 [1m]

(b) T8 = 2 ( 28-1 )

[1m]

= 256 Note: 2 marks allocated for answer w/o working.

[1m]

(c)

4

4

1 (4 1)

4 1S

[1m]

= 85

[1m]

4 (a)

range of y correct, 0 3y [1m]

graph drawn correctly

[1m]

all correct markings on x-axis [1m]

(b)

3

5y x [1m]

Straight line drawn correctly [1m]

Number of solutions = 3 [1m]

3 cosy x

3

5y x

3

0

2

3

2

2



3

5 (a) i) PR = 9x – 7y [1m]

ii) OQ = 3x + 7y [1m]

(b) OS = h OQ

= h (3x + 7y )

= 3hx + 7hy [1m]

PS = k PR

= k (9x – 7y )

= 9kx – 7ky [1m]

OS = OP + PS

= 7y + 9kx – 7ky [1m]

= 7y – 7ky + 9kx

= (7 – 7k ) y + 9kx

3h = 9k [1m]

h = 3k

7(3k) = 7 – 7k

k =

7 1

28 4 [1m]

h = 3 (

1

4)

h =

3

4 [1m]

6 (a) 39.5 or

250 190

160

(seen)

[1m]

250 19039.5 10

160

[1m]

25.43 [1m]

(b) modal class = 40 - 49 [1m]

(c) 22750 fx or 11270252 fx (seen) [1m]

22 1127625 22750

500 500

[1m]

185 (seen) [1m]

13.60 [1m]



4

7 (a) the y-coordinate of the point P = 4 [1m]

the y-coordinate of the point Q = 2 [1m]

the y-coordinate of the point R = 6 [1m]

(b) Area of the shaded region A =

6 42

4 2

18 3 11

2 4

ydy y dy

[1m]

OR

1

232 +

42

2

11

4y dy

=

6 42 3

4 2

3 19

4 12y y OR y y

[1m]

= (27 – 24) + (1

1

3 + 1

1

3)

[1m]

= 3 + 2

2

3

= 5

2

3 [1m]

(c) Volume of revolution of region B =

3

04 1x dx

[1m]

=

32

0

14

2x x

[1m]

= 30 [1m]

8 (a) x 0.4 2.5 4.0 6.0 7.5 8.0 log 10 (y+10) 0.380 0.813 1.111 1.508 1.823 1.924

[1m]

(b) Plot )10(log10 y against x (Correct axes and uniform scales)

[1m]

6 points are correctly plotted. Note: 5 or 4 points correctly plotted Award 1 mark

[2m]

Line of best fit

[1m]

(c) (i) kxhy 101010 loglog)10(log

[1m]

Use hc 10log

10log 0.300h

[1m]

h = 1.995 0.07 [1m]

(ii) Use km 10log

0.198 0.005m [1m]

k = 1.578 [1m]



5

0 1 2 3 4 5 6 7 8 x

log10 (y+10)

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0 Plot lg (y+10) against x (Correct axes and uniform scales) P1 6 points are correctly plotted N2 5 or 4 points correctly plotted N1 Line of best fit N1



6

9

(a) y-intercept = 20

3

[1m]

(b)

2. 1 or . 1

3AC BD BDm m m

[1m]

37 (6)

2c

[1m]

32

2y x

[1m]

(c) Equation of AC,

2 20

3 3y x -----( 1 )

Equation of BD, 3

22

y x -----( 2 )

Solve simultaneous equations (1) & (2):

2 20 32

3 3 2x x

[1m]

D ( 4, 4 )

[1m]

10 04 or 4

2 2

x y

[1m]

A ( –2, 8 )

[1m]

(d) 1

10(7) 6(8) ( 2)(0) (10(8) ( 2)(7) 6(0))2

[1m]

= 26 [1m]



7

10 (a) 2 10sin 0.6RS or 2 10sin 34.37RS [1m]

11.29RS cm [1m]

(b) 1 11.29

2 sin2 6.5

or 120.56 or 120 34' [1m]

2.104 rad [1m]

(c) 10 1.2 or

120.566.5 3.142

180

or 6.5 2.104 (seen) [1m]

120.5610 1.2 6.5 3.142

180

[1m]

25.68 cm [1m]

(d) 21

10 1.2 sin 68.752 or 21

6.5 2.104 sin120.562 (seen) [1m]

2 21 1

6.5 2.104 sin120.56 10 1.2 sin 68.752 2 [1m]

212.86 cm [1m]

11 (a) (i) p =

8

3 , q =

8

5 [1m]

63

3

9

8

5

8

3

C

[1m]

0.2640

[1m]

(ii)

72

2

9

8

5

8

3

C or

81

1

9

8

5

8

3

C or

90

0

9

8

5

8

3

C

[1m]

1

72

2

9

8

5

8

3C

81

1

9

8

5

8

3C

90

0

9

8

5

8

3

C

or 01455.007858.01886.01 or 1 – P(x=2) – P(x=1) – P(x=0)

[1m]

0.7183 or 0.7182

[1m]

(b) (i)

60.5 65.75( )

5P z

or ( 1.05)P z or 1 – 0.1469 [1m]

0.8531 [1m]

(ii) 2500 x (1 - 0 .8531) or 2500 x 0.1469

[1m]

367 [1m]



8

12 (a) in ABC

180 30 40ACB

= 110 [1m]

in ABC , using the sine rule

AB AC

Sin ACB Sin ABC

8

110 30

AB

Sin Sin

[1m]

AB = 15.035 cm or 15.04 cm [1m]

(b) Area of ACD

1

4 Area of ABC

1 1

4 2AC AB Sin BAC

[1m]

1 18 15.035 40

4 2Sin

[1m]

9.664 cm2 or 9.668 cm2 [1m]

(c)

19.664

2AC AD Sin CAD

18 10 9.664 or 9.668

2Sin CAD

[1m]

0.2416 or 0.2417Sin CAD

13.98 13.99CAD [1m]

in ACD , using cosine rule

2 2 2 2CD AC AD AC AD Cos CAD

2 28 10 2 8 10 13.98Cos [1m]

= 8.7392

2.956 or 2.957CD cm [1m]



9

13 (a) x = 2.10 [1m]

y = 7.00 [1m]

(b) I =

w

IW

129 =

(140 7) 126 (120 5) (125 6) (130 4)

22

k

k

[1m] [1m]

2850 +126k = 2838 + 129k

k = 4 [1m]

(c) I D =

120 120 6.48100

100 4.50or

[1m]

= 144 [1m]

(d) I D =

125 110

100

= 137.5 [1m]

I =

(140 7) (126 4) (120 5) (137.5 6) (130 4)

26

[1m]

= 131.9 [1m]

14 x: number of tables, y: number of chairs (a) 3x + 4y ≤ 2400

[1m]

2x + y ≤ 1000 [1m]

y ≥ x + 20 [1m]

(b) Refer to graph paper. - correct scale for both axes

[1m]

- all 3 lines correctly drawn Award 1 mark for 2 lines correctly drawn

[2m]

- region marked correctly

[1m]

(c) (i) 450 [1m]

(ii) 320 + 5 tables and 360 + 5 chairs [1m]

320 ( RM 7) + 360 (RM 5)

RM 4040 [1m]



10

100 200 300 400 500 600 700 800

1000 900 800 700 600 500 400 300 200 100

R

x

y

3x + 4y = 2400

2x + y = 1000

y = x + 20

7x - 5y = k



11

15 (a) 2 [1m]

(b) When momentarily at rest, 2t −8 = 0 [1m]

t = 4 [1m]

(c) s = 2 8t dt

= 2t − 8t +c When t =0 s = 0 c = 0

[1m]

s = 2t − 8t. When v = − 2 ms−1, t = 3

When t =3, s = 23 −8(3) = − 15 m

[1m]

Distance of the particle from O = − 15 – (− 3) = − 12 m

[1m]

(d) When t =4, s = 24 −8(4) = − 16 m [1m]

When t =7, s = 27 − 8(7) = − 7 m [1m]

Total distance travelled = 16 + (16 – 7) [1m]

25 m [1m]


perak 2010 ( x + 7) – 2 papers/2010/perak spm... · perak 2010 paper 1 1. in diagram 1, set b...

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