add maths seminar 2014 handout

SPM ADDITIONAL MATHEMATICS SEMINAR 2014: 3472

3472/2 SULIT

CEMPAKA

CHERAS INTERNATIONAL

SCHOOL 12 TH , APRIL, 2014

GEARING TOWARDS

EXCELLENCE IN SPM

ADDITIONAL MATHEMATICS

2014

SPM ADDITIONAL MATHEMATICS SEMINAR 2014

SPM ADDITIONAL MATHEMATICS SEMINAR 2014 3472

3472/2

GEARING TOWARDS EXCELLENCE IN ADDITIONAL MATHEMATICS SPM 2014

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

ALGEBRA

1 x =a

acbb2

42 �r�

2 am u an = a m + n 3 am y an = a m - n

4 (am) n = a nm 5 loga mn = log am + loga n

6 loga nm

= log am - loga n

7 log a mn = n log a m

8 logab = ab

c

c

loglog

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan��

11 Tn = ar n-1

12 Sn = rra

rra nn

��

��

1)1(

1)1(

, (r z 1)

13 r

aS�

f 1 , r <1

CALCULUS

1 y = uv , dxduv

dxdvu

dxdy

� 4 Area under a curve = ³b

a

y dx or = ³b

a

x dy

2 vuy , 2

du dvv udy dx dxdx v

� , 5 Volume generated = ³

b

a

y 2S dx or = ³b

a

x 2S dy

3 dxdu

dudy

dxdy

u

5 A point dividing a segment of a line

( x,y) = ,21¨©§

��

nmmxnx

¸¹·

��

nmmyny 21

6. Area of triangle =

)()(21

312312133221 1yxyxyxyxyxyx ��

1 Distance = 221

221 )()( yyxx ��

2 Midpoint

(x , y) = ¨©§ �

221 xx

, ¸¹·�

221 yy

3 22 yxr �

4 2 2

xi yjrx y

� �

�

GEOM ETRY


STATISTICS

TRIGONOMETRY

1 Arc length, s = rT

2 Area of sector , A = 212

r T

3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A

6 sin2A = 2 sinAcosA 7 cos 2A = cos2A – sin2 A = 2 cos2A-1 = 1- 2 sin2A

8 tan2A = A

A2tan1

tan2�

9 sin (Ar B) = sinAcosB r cosAsinB

10 cos (Ar B) = cos AcosB B sinAsinB

11 tan (Ar B) = BABA

tantan1tantan

Br

12 Cc

Bb

Aa

sinsinsin

13 a2 = b2 +c2 - 2bc cosA

14 Area of triangle = Cabsin21

1 x = N

x¦

2 x = ¦¦

ffx

3 V = N

xx¦ � 2)( =

2_2

xN

x�¦

4 V = ¦

¦ �

fxxf 2)(

= 22

xf

fx�

¦¦

5 M = Cf

FNL

m»»»»

¼

º

««««

¬

ª �� 2

1

6 1000

1 u PPI

7 1

11

wIwI

¦¦

8 )!(

!rn

nPrn

�

9 !)!(

!rrn

nCrn

�

10 P(A�B)=P(A)+P(B)-P(A�B)

11 p (X=r) = rnrr

n qpC � , p + q = 1

12 Mean , P = np 13 npq V

14 z = VP�x


Techniques of answering challenging questions in SPM Additional Mathematics

1. [Topic : Solution of Triangles, Paper 2, Section C – 10 marks] Cloned Question Solution by scale drawing is not accepted. Diagram below shows two triangles PQR and PSR such that 78oQPR� , 108oPSR� and 12PQ cm . PS is parallel to QR. All angles corrected to the nearest degree.

Given that the area of triangle PQR is 81 2cm ,

(a) Calculate (i) the length of PR in cm, (ii) the length of QR in cm, (iii) SPR� , (iv) the length of RS in cm. [8 marks]

(b) The point 'P lies on PR such that 'P Q PQ . (i) Sketch the triangle 'P QR . (ii) Find 'QP R� and 'P QR� . (iii) Calculate the area of triangle 'P QR . [5 marks]

Answer:

S

12 cm

R P

Q

078

0108

Fong Hiue Mun

Fong Hiue Mun

Fong Hiue Mun

Fong Hiue Mun

Fong Hiue Mun

Fong Hiue Mun

a

Fong Hiue Mun

b

Fong Hiue Mun

c

Fong Hiue Mun

1a.i) (1/2) x a x b x sin c = 81(1/2) x PR x 12 x sin 78 =81PR = [81/ (6 sin 78)PR = 13.80 cmii) QR^2 = 12^2 + 13.80^2 - 2(12)(13.80) cos 78QR^2 = 265.58QR = 16.30 cmiii) (12/sin PQR) = (16.30 / sin 78)sin PQR = 0.7201PQR = 46˚iv) (RS / sin 46) = (18.30 / sin108)RS = 10.44 cm

Fong Hiue Mun

b.ii) QP’R = 180˚- 78˚ = 102˚P’QR = 180˚ - 102˚ - 46˚= 32˚iii) (1/2)(12)(16.30)sin 32= 51.83 cm


2. [Topic : Index Numbers, Paper 2, Section C – 10 marks] Table 4 shows the prices of four ingredients in the making of a type of cake.

Ingredient

Price per kilogram (RM)

Year 2012

Year 2013

A 1.20 p B 10.00 16.00 C 8.00 12.00 D q r

Table 4 (a) The price index of ingredient A in the year 2013 based on the year 2012 is 120. Calculate

the value of p [2 marks]

(b) The price index of ingredient D in the year 2013 based on the year 2012 is 190. The price per kilogram of ingredient D in the year 2013 is RM1.35 more than its price in the year 2012. Calculate the values of q and r.

[3 marks]

(c) The composite price index for the cost of making the cake in the year 2013 based on the year 2012 is 150. Calculate

(i) the price of a cake in the year 2012 if its price in the year 2013 is RM22.50, (ii) the value of s if the quantities of ingredients A, B, C and D are used in the ratio of

5 : s : 7 : 3 as shown in the pie chart above. [5 marks] Answer:

A

B A

C

D 5

S 7

3

Fong Hiue Mun

a) I<a> = (p/1.20) x 100 = 120p = RM 1.44b) I = (r/q) x 100 = 190 r = 1.9 q ————1️⃣1.9 q = q + 1.3r = q + 1.35 —-—2️⃣q = 1.444r = 1.9 (1.44)q = 1.50r = 2.744r = 2.85c.i) (22.5/p<12>) x 100 = 150p<12> = RM 15.00ii) [5(20) + 160s + 7(150) + 3(190)] ÷ (5 + s + 7 + 3)2220 + 160 s = 150 (15 +s)s = 3


3. [Topic : Linear Law , Paper 2, Section B, No.7 : 10 marks] Use graph paper to answer this question. Table 1 shows the values of two variables, x and y, obtained from an experiment. The

variables x and y are related by the equation1xy pq � , where p and q are constants.

x 3 4 5 6 7 8 y 6.3 10.7 18.6 30.9 53.7 100

Table 1 (a) Plot log y against (x − 1), by using a scale of 2 cm to 1 units on the (x − 1)-axis and 2 cm to

0.2 units on the log y axis. Hence, draw the line of best fit. [5 marks]

(b) Use your graph in (a) to find the value of (i) p (ii) q [5 marks]

]

Answer: (a)

Fong Hiue Mun

x-1

Fong Hiue Mun

log y

Fong Hiue Mun

2

Fong Hiue Mun

3

Fong Hiue Mun

4

Fong Hiue Mun

5

Fong Hiue Mun

6

Fong Hiue Mun

7

Fong Hiue Mun

0.80

Fong Hiue Mun

1.03

Fong Hiue Mun

1.27

Fong Hiue Mun

1.49

Fong Hiue Mun

1.73

Fong Hiue Mun

2.00

Fong Hiue Mun

Fong Hiue Mun

Fong Hiue Mun


4. [Topic : Coordinate Geometry, Paper 2, Section B : 10 marks] Solution by scale drawing is not accepted. Diagram 4 shows a trapezium OJKM. The line OJ is perpendicular to the line JK which intersects with y-axis at the point N. It is given that the equation of OJ is y = 2x and the equation of JK is 2y = px − 20.

Diagram 4 (a) Find (i) the value of p, (ii) the coordinates of J. [4 marks]

(b) Given JN : NK = 2 : 3, find (i) the coordinates of K. (ii) the equation of the straight line KM. [4 marks]

(c) A point A moves such that AJ = 2AK. Find the equation of the locus of A. [2 marks]

Answer:


5. [Topic : Circular Measure , Paper 2, Section B : 10 marks] Diagram 5 shows two circles. The larger circle has centre S and radius 9 cm. The smaller circle has centre T and radius 6 cm. The circles touch at point R. The straight line PQ is a common tangent to the circles at point P and point Q.

Diagram 5 Given that ∠PSR = θ radians, calculate

(a) the value of θ, [2 marks]

(b) the length of the minor arc QR, [3 marks]

(c) the area of the shaded region. [5 marks]

[Use π = 3.142] Answer:


6. [Topic : Differentiation & Integration, Paper 2, Section A : 5 – 8 marks] Cloned Question A curve has a gradient function kx2 − 8x, where k is a constant. The tangent to the curve at the point (3, 26) is parallel to the straight line y − 30x − 3 = 0. The curve has two turning points when x = p. Find the (a) value of k, [2 marks] (b) values of p, [3 marks] (c) equation of the curve. [3 marks]

Answer:

7. [Topic : Logarithm, Paper 1 : 2 – 4 marks] Given that 4log 3 D and 4log 5 E , express 8log 180 in terms of D and E . Answer: [4 marks]


8. [Topic : Indices and Logarithm, Paper 1 : 2 – 4 marks] Find the values of x that satisfies the equation 2 15 5 6 0x x�� . [5 marks] Answer: 9. [Topic : Functions, Paper 1 : 2 – 4 marks or Paper 2 : 5 – 8 marks]

Given the functions 1 1( )3

xf x� � and 2: 9 6 7gf x x xo � � , find

(a) ( )f x , (b) ( )g x , (c) ( )fg x [8 marks] Answer:


10. [Topic : Quadratic Equation, Paper 1 : 2 – 4 marks] Given the roots of the quadratic equation 22 5 1x x � are D and E .

(a) Form the quadratic equation whose roots are 1DE

� and 1ED

� ,

(b) Show that 34 23 5D D � [6 marks] Answer: 11. [Topic: Quadratic Functions, Paper 1: 2–4 marks or Paper 2: 5–8 marks] Cloned Question Diagram 11 shows the graph of a quadratic function 2( ) 2 5 2f x x x � � that can be expressed in the form of 2( ) 2 2( )f x q x p � � , where c, p and q are constants. Given

41( , )8

p is the maximum point of the graph. Find the

(a) values of c, p and q , (b) maximum value of f(x) and the corresponding value of x , (c) equation of axis of symmetry. [8 marks] Answer:

( )f x 41( , )8

p

c 0 x

Diagram 11


12. [Topic: Statistics, Paper 1: 2-4 marks or Paper 2, Section A: 5-8 marks] Cloned Question The number of articles read by each pupil is given by

1 2 3 16, , ,...,x x x x and

2( ) 640x x� ¦ and the sum of the squares of the number of articles read by each pupil is 1040. Find (a) the standard deviation, [2 marks] (b) the mean number of articles read by each pupil, [2 marks] (c) the total number of articles read, [2 marks] If the number of articles read by each pupil is doubled, find (d) the new mean, [2 marks] (e) the new variance. [2 marks] Answer: 13. [Topic: Arithmetic Progression, Paper 1: 2 - 4 marks or Paper 2 Section A: 5 - 8 marks]

The sum of the first n terms of an arithmetic progression is given as [3 2]2nnS n � . Find

(a) the first term , [2 marks] (b) the common difference , [3 marks] (c) the thn term . [2 marks] Answer:


14. [Topic: Geometric Progression, Paper 1: 2 - 4 marks or Paper 2 Section A: 5 - 8 marks] A rope is cut into n parts. The length of each part of the rope increases and form a geometric progression. The length of the eighth part of the rope is eight times the length of the fifth part of the rope. (a) Calculate the common ratio. [2 marks] If the total length of the rope is 1275 cm and the first part of the rope is 5 cm, calculate (b) the value of n , [2 marks] (c) the length of the last part of the rope. [2 marks] 15. [Topic: Integration, Paper 1: 2 - 4 marks or Paper 2 Section A: 5 - 8 marks]

Given2

2 1xyx

�

, show that 2

2 ( 1)(2 1)

dy x xdx x

�

�.

Hence, find the value of 22

21

2 2 3[ ](2 1)x x dx

x� ��³

[6 marks]

END OF QUESTION PAPER

Prepared by: Mr. Chew Hock Hin SMK ST. MARY, K.L.

add maths seminar 2014 handout

Documents