Extra Features of This Book
CHAPTER Functions
1
sMART SCOPE 1.1 Functions Important Learning Standards Page CONCEPT MAP
3
Contains Learning 1.2 Composite • Explain function using graphical representations and notations. 5 The entire content of the
Standards (LS) that need to Functions 6 chapter is summarised in
be achieved in each chapter. • Determine domain and range of a function. the form of a concept map.
• Determine the image of a function when the object is given and
vice versa.
• Describe the outcome of composition of two functions. 9
9
• Determine the composite functions. 10
11
• Determine the image of composite functions when the object is 11
given and vice versa.
• Determine a related function when the composite function and
another function are given.
• Solve problems involving composite functions.
• Describe inverse of a function. 13 Fo4rm Additional Mathematics Chapter 7 Coordinate Geometry
13
1.3 Inverse Functions • Make and verify conjectures related to properties of inverse
functions.
Fo5rm • Determine the inverse functions. 15
Additional Mathematics Chapter 3 Integration
3.1 Integration as the Inverse of Example 1
Differentiation (a) If f(x) = 9x2 + 5x and f(x) = 18x + 5, find Coordinate Geometry
∫ ∫(b••)
Explaining the relation between (18x + 5) dx. • Range / Julat
differentiation and integration FRGueilnvaecttinoioynn=//H4Fxuu(b1nug–nsxig)a3 nand dy • Domain / Domain
dx = g(x), find g(x) dx.
AAGbrirvsooewlnutdddeiyxavga3rlua5–mexfu/=nGchat(mixo)bn, af/irnFrduajnahghs(xia)nndialxak.i
3CHAP. 1. Given y = x3 + 7, thus dy = 3x 2. ( ) ∫(c••) panah • Codomain / Kodomain Divisor of a Area of Polygon The Straight Line Equation of Locus
dx mutlak • Object / Objek Line Segment
•SoFluutnicotnio: n notation / Tatatanda fungsi • Image / Imej
Conversely, if given dy = 3x 2, thus y = x 3 + 7 will (a•) AGbivsoelnutf(exv) a=lu9exf2u+nc5txion graph • Composite function / Fungsi gubahan
be obtained. dx • Inverse function / Fungsi songsangan
Graf fufn(gxs)i=ni1la8ixm+u5tlak • Horizontal line test / Ujian garis mengufuk
2. The reverse process of this differentiation is ∫• VHeernticcea,l li1n8exte+st5/dUxji=anf(xg)aris mencancang n Parallel line Perpendicular line
called integration. 1 m P(x, y) Gradient of two Product of gradient
lines are the same, for two lines is –1,
∫ 18x + 5 dx = 9x2 + 5x A(x1, y1) B(x2, y2)
m1 = m2 m1m2 = –1
∫3. If ddx[f(x) = f (x), thus the integral f (x) with ∫Thus, (18x + 5) dx is 9x2 + 5x. nx1 + mx2 ny1 + my2
respect to x is f (x) dx = f(x). m+n m+n
(b) Given y = 4x(1 – x)3 P(x, y) = ,
Differentiation ddx[f(x)] = f (x) dy = g(x)
dx
∫Hence, g(x) dx = y CHAP
∫ g(x) dx = 4x(1 – x)3 7 5Triangle
∫Thus, g(x) dx is 4x(1 – x)3. QuadrilaterFaolrm n-sides of polygon
Integration ∫ f (x) dx = f(x) Chapter 1 Circular Measure Additional Mathematics
( )(c) d 5 1
Given dx 3–x = h(x) 1.1 Radian y Example 1 y CHAP. Area = 2 x1 x2 … xn x1
y1 y2 … yn y1
∫Hence, h(x) dx = y 1Convert Bth(xe2, ayn2)gle in the unit of radiaAn(x1t,oy1)the B(x2, y2)
Comparison between 5 Relating angle measurement in radian and degree. [Use π =C(3x.31, 4y32) ]
differentiation and integration ∫ h(x) dx = 3 – x degree (a) 1.15 radian 5π
bit.ly/2K4y3b0 1. In circular measures, the angle can be measured 6
∫Thus, h(x) dx is 5 x. (b) radian
3.1 3– in 2 units, which are x
(a) degree (°) and minute (ʹ). Solution:
Try question 1 to 5 in Formative Zone 3.1 (b) unit of radian (in or not in the terms of π). (aA)(x1, y1)π rad = 180° D(x4, y4) C(x3, y3) x
Area1.=1521raxyd11 1xy1811π800°° 1 x1 x2 x3 x4 x1
2. Radian involves the angle that related with the =x12.15x3× 3.142 Area =2 y1 y2 y3 y4 y1
=y12.15y3×
∫∫ ( ) ∫( ) ( ) ∫ ∫ ∫D321e...r2iGIvfifi2n(vxxeg)nI–=na7xndd2d3–xyde3dfd=xix4en.a2tinextCdre1–mf(Ixixn)2n=3tine–1gag2nrixdna3,dlyfien=fdinxi–t1e+2ixn31xtdeC2x21,g.a..rflicCanETTc1uldoahxlleacafiutnsmolcdarrptetohelrene,453p...2vwra4eiGGGllsulC(siiib1evvvceeee)oSnnndHfaiaIysddnFdndpaTx=xdtlxia(lxcfonyx(42lxgao2d)–1rg3+xae5.8nC2cx15=63dho9x+.an2u.dd7p6ssc=0ayxnitt)ne0axg==+gr8n=n(54+tax231s9.1).xhs,In.c(f–xidi+en)5i,ncd,fcteiifnfsiinsdgc,udS(xcuhh)r((dd2xaxs)xs.da–xnnC5.d(P(2ba)r≠L)d)Co2oExv1gll.exooaaggtrhmiaatehxyxpmyfo=l=sellloo2lo3wg5g.AianaxdTatrgxhadnh−.de+idetilisooulaomdngsmgiaaaiaalnyegnMyOurdlaetaemnctshigr5etcab7mhun°e.amdl1oti7fwcre'sardesihnaocnewwosfhtahileceFior4atcrhnmlOeeg.lreadin1iuradsdehgarveee
= 65.89°
(b) π rad = 180°
5π 5π 180°
6 rad = 6 × π
A==c1o65(5nx0s×–t°a1xn81t)02f°r+om(y a fixed point: A constant ratio from two fixed points:
– y1)2 = r2
(x – xx21))22 + (y – yy12))22 = m2
(x – + (y – n2
SPOTLIGHT PORTAL for algebraic functions Alternative Method
Substitute π =180° into the expression,
5π 5(180°) Always equidistant from two fixed points:
Solution: 6 = 6 = 150° (x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2
Find the 2(vxa72lu+=2e.3xoxTfSSf)utthx=eeneppicn2fto21ieol::alnocAM,w((hdbadi))axondinfngllw1footoiaggsitlitol5xnthohx6tw1treh4h=eiensse=tgp−peve2p−oaecswl6qtutueteooraoftoxiifonr:cndion.tndhseetxainnottf,exag.rfLaierlstotll.fooaggaa x = In degree and minute In radian
(a) log9 x = m Try question 1 in Formative Zone 1.1
1. Formula of integration: y = am …❶ 57° 17ʹ = 1 rad
(c) log2 y =
∫constants. n 4. ai1…sbrt❷oaahumdetai×samtahn+amennisec==eaanxxmsyyttreheaeosulferanegcmitrhelncoetfosrufacadhniuaassnoCgtHhflAeecPisarurcbcletle.enndgethd 126 Example 2
(a) a dx = ax + c where a and c are an Convert
Scan the QR code to ∫ 4(b) (a) 30° into radian unit, in term of π.
xn + 1 Solution: Step 3: Divide the term with the new ind(eax). ❶ × ❷: (b) 200° into radian unit.
n+1 xStep 4: A(bd)d ltohge5 cxon=s−ta2nt c with the integrals. [Use π = 3.142]
xn dx = + c such as n ≠ 1 and c(aa)re log9 27 = 9Fx or example, xn (m + n) mlog+a a = lllooogggaaa xy A Solution:
27 = 32x dxx===51n−x2n++ 1 loga x + loga n = xy
33 = 3 y = xy
svbuirdobewtoospereicwlsaetleebdasrittneoeodthr. e ∫270constants. 2x = 1 + c rr
52 am x O 1 radian B (a) 180° = π rad
= 1 3.1.1 3.2.1 (b3).2.❶2 ÷ ❷: an = y r
x =3 25
2 −ceHliqonreugc)nalalceolaae,gtmm˙,oa−−aAatnnnhO===gelBllexyrAooaiggBssduaa1i=ubxyxyrstOaeodnAfidat=enhdeOifcaBtibhr=coeluerat.rcthleengctehntorfe 30° = 30° × π
180°
π
1 5. of a = 6 rad
64 (m AB is
(c) log2 (x2 + 3x) = 2 (d) logx = −6 Alternative Method
x2 + 3x = 22
Substitute 180° = π into the expression, ALTERNATIVE METHOD
x2 + 3x − 4 = 0 1 = x−6 180°
(x + 4)(x − 1) = 0 64 30° = 6 = π rad
64−1 = x−6 6
x = −4 or x = 1 (26)−1 = x−6 Taxnh−gelmleoregi−nlaanytriao==dnillasoohnggi2aapwπyyxxibtrhaedtdwe=ege3rne6e0t°ihse
6. measurement of (b) 180° = π rad
loga
2−6 = x−6 200° = 200° × π
180°
x=2 3.142
200° = 200° × 180°
Try question 5 in Formative Zone 4.3 π rad = 180° Provide alternative
Example 276. The conversion of angle measurement in the degree = 3.49 rad
to the radian and vice versa are as follow:
Proving laws of logarithms Try question 2 and 3 in Formative Zone 1.1
1. The following is the laws of logarithms. Prove that loga xn = n loga x. × 180° solutions to certainCalculator
Solution: π
Recheck the answer in Example 2(b) by using
• Product rule Let m = loga x Radian Degree calculator,
loga xy = loga x + loga y
CALCULATOR • Quotient rule am = x × π questions.1. Press 2 0 0 × SHIFT EXP ÷ 1 8 0 =
(am)n = xn 180°
loga x = loga x – loga y 4(loga(mx)(nnlno) lglloooagggaaaama1axn.)1====.1llllooooggggaaaaFxxxxonnnnrm 2. The screen will display 3.490658504
y
• Power rule Additional Mathematics Chapter 5 Progressions
loga xp = p loga x
Example 30 217
Explains how to use 2. From the laws of logarithms, the following can Example 27 Find the sum of the first 8 terms of the For Example 31, the sum from 5th term to 9th term
a scientific calculator be derived:
in mathematics tahf(naoed)l lfoolow1llg4oin3,w g52in,=gge27.1o,.m 4…6e5t,ric progression. is S9 – S4.
Gfinivdetnhethvaat lluoegf3o2r = 0.631 (b) 1.6, 3.2, 6.4, …
• lllooogggaaa 1=0 (a) log3 10 each of S9
• loga a=1 (b) log3 2.5
• ar = r S4
•
1 = –loga b (c) log3 45 T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9
b
• log 1 b = –loga b 6 Solution: S9 – S4
a 5
(d) log3 a = 14, r = 2 1 |r| 1
14 7
(a) =
4.3.1 4.3.2 Sum of the first 689 terms, S Example 32
8
calculations. 141 – 1 8 The term of a geometric progression is given
7 by Tn = 21 + n. Find
EXAMPLE = (a) the first term and the common ratio,
1 (b) the sum of the first 8 terms,
Examples with Fo5rm CHAP 1– 7 of the geometric progression.
complete solutions
to enhance students' Additional Mathematics Chapter 6 Trigonometric Functions 5 = 16.33 Solution:
Example 19 Example 20 (b) a = 1.6, r = 3.2 = 2 |r| 1 (a) First term, T = 21 + 1
1.6 1
=4
T
Given f(x) = 4 cos 2x for 0 < x < 2π. Sketch the graph of the following tr igonSoumme otrfi cthe first 8 terms, S 2nd term, 2 = 21 + 2
(a) State the period of the graph function y = f(x). 8
functions in the given range. = 8
Hence, state the number of cycle of the graph (a) y = sin x + 1 for 0 < x < 2π 1.6(28 – 1)
in the given range. = 2–1 Common ratio, r = 8 =2
(b) State the amplitude of the graph. 4
(c) Write the coordinates of the maximum and the (b) y = –2 cos x for 0 < x < 2π = 408
minimum points. (c) y = | tan x | for 0 < x < 2π Try question 21 in Formative Zone 5.2 Thus, the first term and the common ratio
(d) Sketch the graph of y = f(x). (d) y = | cos 2x | + 1 for 0 < x < 2π
(e) Using the same axes, sketch the graph of are 4 and 2 respectively.
function y = –|4 cos 2x| for 0 < x < 2π.
Solution: Solution: (b) Sum of the first 8 terms, S BRILLIANT TIPS
8
(a) y = sin x + 1 for 0 < x < 2π = 4(28 – 1)
2–1
1 Sketch the basic graph, y = sin x. Example 31
= 1 020
The graph moves 1 unit upwardIt ,iss ugcihvetnh atthat –9, 27, –81, … is a geometric Try question 25 to 27 in Formative Zone 5.2
translation progression. Find the sum from 5th term to 9th
Compare f(x) = 4 cos 2x with the basic cosine ( )2 0 term of the geometric progression.
function, f(x) = a cos bx + c. 1
2π .
2
(a) Period = π or 180°. Number of cycle, b = 2. Useful tips to helpExample33
(b) Amplitude, a = 4 y Solution: The sum of a geometric progression is given by
(c) Maximum point: (0, 4), (π, 4) and (2π, 4). Sn = 3(2n) − 3. Find
6 ( ) ( )CHAP. 3π 2 ᕢ a = −9, r = − 27 = –3
Minimum point: π , –4 and 2 , –4 y = sin x + 1 9 (a) the sum of the first 5 terms,
2
(d) To sketch graph function y = 4 cos 2x: 1 students solve(b) the 7th term of the geometric progression.
Number of class = 2 × 2 × 2 = 8 S = −9(−3)4 − 1)
O 4 x−3 − 1 Solution:
–1 (a) S5 = 3(25) − 3
Size of class interval = 2π = π –2π π –32–π ᕡ 2=π 180
8 4 y = sin x
= 93
x0 π π 3π 2π –2 S = −9((−3)9 − 1) (b) S6 = 3(26) − 3 problems in the related
2 2 9 −3 − 1
= 189
y 4 –4 4 –4 4 (b) y = –2 cos x for 0 < x < 2π = − 44 289
S7 = 3(27) − 3
Thus, the graph function of y = 4 cos 2x: = 381
1 Sketch graph of y = cos x. The sum from 5th term to 9th term subtopics.
2 Reflect the graph at 1 on x-ax=isSt9 o−m Sa4 ke 7th term, T7 = S − S
y = − 44 289 − 180 7 6
the graph of y = – cos x. = 381 − 189
= − 44 469
4 y = 4 cos 2x = 192
understanding of the 2 y Try question 22 to 24 in Formative Zone 5.2 Try question 28 to 30 in Formative Zone 5.2
chapters learned.
O x 2 94 ᕡ
–2 –2π π –32–π 2π
1 y = – cos x 5.2.3
–4 O x
–1 2π
(e) Steps in sketching the graph of y = –|4 cos 2x| –2π π –32–π
y = cos x ᕢ
1 y = |4 cos 2x| is a reflection of graph on –2
negative side of x-axis.
2 y = –|4 cos 2x| is a reflection of graph at 1 3 The value of a is –2. The maximum value is
on x-axis. (π, 2) and the minimum value is (0, –2) and
(2π, –2).
y
4 y
2
2 ᕣ y = –2 cos x
O 2–π π –32–π x 1 y = – cos x
–2 2π
O 2–π π –32–π 2π x
–1 ᕢ
–4 y = –|4 cos 2x|
Try question 3 in Formative Zone 6.3 –2
354 6.3.1
iv
TAGGING 'Try question ... in
Formative Zone ...” Chapter 2 Quadratic Functions Additional Mathematics Fo4rm
Example 6 Solving quadratic inequalities
The tagging is located at the Given the roots of quadratic equation 3 1. A quadratic inequality is a function where
end of the example guides 2 its degree is 2 and uses the inequality symbols
the students to answer the 2x2 + (p − 2)x + (q + 1) = 0 are −1 and − , find which are less than, greater than, less than or
corresponding questions in the values of p and q. equals and greater than or equal.
Formative Zone. CHAP
Solution: Inequality Symbol
Less than 2
Greater than
Sum of roots = −1 + − 3 Less than or equal to <
2 Greater than or equal to
−p−2 = − 5 Fo4rm
2 2
p−2 = 5 Chapter 10 Index Numbers Additional Mathematics
2 2
2. Solving a quadratic inequality means to find the
p−2 =5 range of the values of x that satisfies the inequality. 10.1
p =7
3. There are three methods to determine the
Product of roots − 3 range of values of x that satisfy the quadratic
= −1 2 inequality which are: 1. Find the index number or price index of the 7. The table below shows the price (RM) and the
(a) Graph sketching following quantity or price as 2014 is the base price indices of four items, P, Q, R and S used
q+1 = 3 (b) Number line year. C1 in making a dress.
2 2 (c) Tabulation
(a) Year Price of a tile (RM)
q+1 =3 4. The range of values of x can be obtained by Price (RM) Price index for
q =2 considering two types of quadratic inequalities. the year 2019
(a) (x − a)(x − b) 0 2017 44 Item Year Year based on the
Thus, the values of p = 7 and q = 2. (b) (x − a)(x − b) 0 P 2016 2019
2014 55 year 2016
Try question 9 to 11 in Formative Zone 2.1 3.40 x
125
(b) Number of visitor
Example 7 A Graph sketching method Year (thousand) Q 2.50 3.20 y
Given that one of the roots of quadratic 1. In graph sketching method, we have to consider 2019 24.7 R z 2.30 115 FORMATIVE ZONE
equation 2x2 − kx + 54 = 0 is three times the two forms of quadratic equations which are 2014
other root, find the values of k. y = (x − a)(x − b) and y = −(x − a)(x − b). 19 S 2.85 2.80 98.25
Solution:
Let the roots be α and 3α. 2. For quadratic equation y = (x − a)(x − b), the 2. The price of a school bag in the years 2019 and Find the values of x, y and z. C3
graph is as follows: 2017 were RM63.90 and RM45 respectively.
Find the price index of the school bag for the 8. The price indices of a tube of facial cleanser
Sum of roots = α + 3α year 2019 based on the year 2017. C1 in the years 2017 and 2014 based on the year
y>0 Questions2012 are 135 and 120 respectively. Calculate to test
3. The table below shows the price of two type
−−k = 4α a b x of food, vegetables and chicken in the year the price index in the year 2017 based on the
2 2015 and 2018.
k y<0 year 2014. C3
2
= 4α (a) The values of a and b are the roots of the
quadratic equation.
α = −k …❶ Type of food Price (RM) per kg students' understanding9. In a boutique, the total number of customers
8 …❷ (b) Thus, Vegetable who came to the boutique are recorded as
Product of roots = α(3α) (i) x a or x b when (x − a)(x − b) 0 shown in the table below. C3
(ii) a x b when (x − a)(x − b) 0 2015 2018
54 = 3α2 3.20 3.52
2 3. For quadratic equation = −(Fxo5−rma)(x − b), the at the endYear of each
graph is as follows:
Number of
Additionayl >M0athematics customers
y 2012 2014 2016
3 000 3 930
3α2 = 27 Chicken 6.00 7.20
Substitute ❶ into ❷. Chapter 5 Probability Distribution Calculate the price index of vegetable and
chicken in the year 2018 based on the year
Simul3at8ki6ko422n==H927OTS Questions 2015. C1 (hundred)
k2 = 576
SPM y<0 a b x subtopic.(a) Find the index number of the number of
EXAMINER’S customers that came to the boutique in CHAP
Paper 1 COMMENT
1. A bank auditor claimks =th2a4t creodr it ckar=d’s−24 2. (a) The values of a and b are the roots of the 4. The number of accident in the year 2017 is 10the year 2014 based on the year 2012.
bThineomdiiaalgdraismtribbue(tilboo)nwoTq(isfuh)hXauo.dswa,rastitchxeequgabrtaiwponhhe. nof(xa− a)(x − b) 0 1 840 cases while 1 564 cases in the year
2018. Calculate the index number of accident (b) Given the index number of the number of
occur in the year 2018 based on the year 2017. customers in the year 2016 based on the
balances are normally distributed with a mean Describe the answer. C2
of RM2 870 aTnrydqueastiovna9ritaon11cein FoorfmRatMive8Z1o0ne02.010. year 2014 is the same as the index number
P (X = x()ii) x a or x b when (x − a)(x − b) 0 in the year 2014 based on the year 2012.
What is the probability a randomly selected Estimate how many customers came at the
credit card holder has a card balance less than
RM2 500? 2.1.2C4 2.1.3 0.432 25 5. The number of workers in the year 2018 is boutique on 2016?
2 106 compared to 1 950 in the year 2015. After
Examiner’s comment: 0.288 three years, the index number of workers in the 10. During a dry season, water level in a lake is
Let X is the credit card’s balance 0.216 year 2020 based on 2018 is 132.6. In what year, 25 m on February 2019 and 27.3 m on March
m = 2 870, s2 = 810 000 the increment of workers is higher? C2
2019. C3
s = 900
So, X ~ N(2 870, 900). CHAP. 6. The index number of annual income of James (a) Find the water level of the lake on March
decreases by 6.2% from the year 2017 to the as February is a base month.
0.064 5 year 2018. If his annual income in the year 2017 (b) If the water level of the lake on April (March
Given X is less than RM2 500. 0 x was RM48 750, what is James’s annual income is a base month) were twice than (a), find
Standardised variable X to Z, 01 2 3 in the year 2018? C2
the water level of the lake in April 2019.
( )P(X X– m
, 2 500) = P Z , s Find C4
(a) the value of probability of ‘success’,
( )= 2 870 (b) P(1 < X , 3).
P Z , 500 – 2 201
900
= P(Z , –0.411) Examiner’s comment:
= P(Z . 0.411) (a) Let p is probability of ‘success’
= 0.3405 q is probability of ‘failure’
From the graph, P(X = 3) = 0.216
3C3p3qp30 = 0.216
From standard normal distribution table, = 0.216
f (z) p = 0.6
Thus, the value of probability of ‘success’
is 0.6.
(b) P(1 < X , 3) = P(X = 1) + P(X = 2)
= 0.288 + 0.432
SPM SIMULATION HOTS QUESTIONS–0.4110 = 0.72 SPM MODEL PAPER
z
1 Paper 2 Paper 1
Substract 3. In normally distributed, the mean and standard
z 1 deviation of length of fish is 11 inches and Time: 2 hours
4 inches respectively. C5
Provide a complete solutio0.4 ns 0.3409 4 Section A
(a) What is the percentage of the length of fish (64 marks)
are longer than 14 inches? Instruction: Answer all questions
with the examiner's commentsCalculator (b) If 200 fish are randomly selected, how many 1. (a) Diagram 1 shows a part of graph for a (ii)
Check the answer by using scientific fishes has the length less than 9 inches? function of y = f(x).
calculator.
Examiner’s comment: y
1. Press MODE MODE and choose 1 Let X is the length of fish y = f(x)
for the SPM Simulation HOTSwhichisSD. Given m = 11, s = 4, so X ~ N(11, 4)
2. Press SHIFT 3 and choose 3 (a) Given the length of fish is longer than
represents P(z . a). 14 inches, X . 14.
02 x
–3
Change variable X to Z
( )P(X X–
. 14) = P Z . s m
questions. 3. Insert 0.411 and the screen will display ( )=P Z . 14 – 11
0.34054 . 4
= P(Z . 0.75) SPM MODEL PAPER Diagram 1
= 0.2266 State whether the function of f
(i) is a discrete or continuous,
(ii) has an inverse function or not. 2. (a) Diagram 2 shows a graph of function f for
[2 marks] domain 0 < x < 4 and its inverse function
335 a f –1.
(b) A function f is derived by f : x ˜ x , x ≠ 0 y
A (4, 12)
such as a is a constant. Given f–1(2) = 2 , find
(i) the value of a, f
(ii) f17(8). [3 marks]
Answer: f–1 B
(a) (i)
Fo5rm SPM MODEL PAPER0 x
Additional Mathematics Chapter 7 Linear Programming (ii) –4
Diagram 2
Based on the graph, determine
(i) the domain of f–1,
(b) (i) (ii) the coordinate of point B on the graph
of f–1 that corresponding with the point
SPM format questions accordingAonthegraphoff. [2 marks]
Paper 2 Answer:
(i)
1. On the given grid below, show the region that 4. By using the given grid below,
C2 satisfies all the following inequalities. C3 (a) show that the region is bounded by all the to the(ii) latest SPM 2021
x > 3, y > 1 and x + y < 5 following inequalities.
x > 2, y > x and x + y < 6
y
7 y assessment format cover all the
6
Summative Zone 5 7 418
4 6
5 chapters in Forms 4 and 5.
3 4
3
CHAP. 2 2
levels of7 1 1
Questions of various –1 0 123 4 5 6 7 x
–1
–1 0 1234567 x
–1
thinking skill are provided to2. A green grocer sells bananas and apples. In one (b) The point P with coordinates (x, y) lies inside
C3 day, he sells the region R. x and y are integers.Write down
I up to 80 bananas, the coordinates of all the points of R whose
II up to 90 apples, coordinates are both integers.
evaluate the understandingIII notmorethan110fruits.
If x be the number of bananas sold and y be 5. The cost of a book is 50 cents and a pen is
the number of apples sold, show the region that
satisfies these inequalities and label the region ANSWERSC4 RM1.30. A student wants to buy x books and Complete answers
of each chapter. as R. y pens based on the following conditions: http://bit.ly/2ORHlke
3. The diagram below shows the R region that
C2 satisfied all three linear inequalities. I At least three pens must be bought.
II The total number of books and pens bought
must not more than 12.
y III The amount of money spent is at mFoOsRt M 4 9. f(x) 5. –54
10 RM10. 13 6. 3
8 Write down the three inequalities other Cthhaanpter 1 Functions 9 10
6R x > 0 and y > 0 that satisfy all the above
4 conditions. 7. g(x) = 3x – 2
2 1.1
0 12345 x 3 8. f(x) = x – 2
6. By using the same graph, draw all the inequali1ti.es(:a) The relation is a function x 9. (a) 1153
5 (b) 5
C3 y > x – 1, x > 2 and 2x + y > 8 because each object has –6 – 3 0
only one image. 2
the(b) The relation is not a 10. h = –3k
Shade the region that bounded by function because there is Range of f is 0 f(x) 13. 5
11. p = 12, q = 2
inequalities. Hence, state the minimum value of one object does not have 10. (a) q = 1, p =1
y in the region. any image. (b) 3
(c) –10 12. (a) x (b) x
2. (a) h(x) = |10 – x| or 2x + 1 3x + 1
h(x) = |x – 10| 11. (a) 3
Define all the three linear inequalities. (b) 525 (c) x (d) x
(b) h(x) = x2 – 1 (c) 2 2x + 1 3x + 1
12 , 11
3. (a) A function. 13. k = h = 2
(b) A function.
(c) Not a function. 12. (a) 3 or 1.7321
391 2 14. (a) RM13 400
4. (a) {a, b, c, d} 2 (b) RM449 675
FORM 4 ANSWER (b) {–2, 0, 2, 4} (b) or 1.4142
(c) a, b, c, d
(d) –2, 0, 2 13. (a) 1 (b) 2 1.3 ANSWER
5. (a) 7 14. 3 1. (a) –4 (b) –2 Complete answers are
(b) {–3, –2, 2, 7} (c) 4 (d) 2 provided. Scan the
15. (a) 3 (ii) 3 QR Code provided to get
6. (a) 0, 2, 6, 8 (b) (i) –12 2. (a) –2 (b) 4 the steps to the solution.
(b) 7, 1, 10
(c) {7, 1, 10} 16. (a) (i) 5 (ii) 11
(d) 7 5 5 3. (a) –7 (b) –2
(e) 8 (b) – 4 and 12
4. (a) Has an inverse function
7. (a) Domain = {–2, 0, 2, 4} 17. (a) (i) 15 because each element in set
Codomain = {4, 6} 4 P matched with only one
Range = {4, 6} 3 element in set Q.
(ii) 4
(b) Domain of f is –1 x 3. (b) Does not have an inverse
Codomain of f is 1 f(x) 3. (b) – 31 and 33 function because from
Range of f is 1 f(x) 3. 4 4 horizontal line test, the
line cuts the graph on two
8. (a) f(x) 18. –3 and 9 points.
4
10 19. (a) 4 5. (a) Has an inverse function g.
(b) 0 g(x) 10 (b) Does not has an inverse
5 function g.
4 6. y
1.2 11
g(x)
–2 0 4 x 1. (a) fg(x) = 3x – 1 y=x
(b) f(x) 3 3 3 6
5 (b) gf(x) = x – 5
2. (a) f 2(x) = 36x – 7 3
(b) g2(x) = 16 + 9x
3 (c) gf(x) = 18x + 1 2 23 6 11 x
0 15 (d) fg(x) = 18x + 23 –2 0 g–1(x)
2 3. (a) 111 (b) 731 –2
4. (a) –4 17
x (b) – 2 Domain of g–1(x) is 2 x 11.
5 Range of g–1(x) is –2 g–1(x) 3.
434
v
CONTENTS
FORM 4
ReDvaistieon ReDvaistieon
Chapter 1 Functions 1 Chapter 6 Linear Law 106
1.1 Functions 3 6.1 Linear and Non-Linear
1.2 Composite Functions 9 Relations 108
1.3 Inverse Functions 13
Summative Zone 18 6.2 Linear Law and 114
Non-Linear Relations
Chapter 2 Quadratic Functions 20
6.3 Application of Linear Law 116
Summative Zone 121
2.1 IQnueaqduraaltiticieEs quations and 22 Chapter 7 Coordinate Geometry 125
7.1 Divisor of a Line Segment 127
2.2 Types of Roots of Quadratic
Equations 29 7.2 Parallel Lines and
Perpendicular Lines 130
2.3 Quadratic Functions 31
7.3 Areas of Polygons 135
Summative Zone 42
7.4 Equations of Loci 140
Chapter 3 Systems of Equations 45 Summative Zone 146
3.1 Systems of Linear Equations Chapter 8 Vectors 151
in Three Variables 47
3.2 Simultaneous Equations 8.1 Vectors 153
involving One Linear
Equation and One 8.2 Addition and Subtraction
of Vectors 159
Non-Linear Equation 50
8.3 Vectors in a Cartesian Plane 164
Summative Zone 56
Summative Zone 172
Chapter 4 Indices, Surds and Chapter 9 Solution of Triangles 177
Logarithms 58
9.1 Sine Rule 179
4.1 Laws of Indices 60 9.2 Cosine Rule 183
62 9.3 Area of a Triangle 186
4.2 Laws of Surds 68 9.4 Application of Sine Rule,
189
4.3 Laws of Logarithms 75 Cosine Rule and Area of 193
77 a Triangle
4.4 Applications of Indices, Summative Zone
Surds and Logarithms
Summative Zone
Chapter 5 Progressions 79 Chapter 10 Index Numbers 197
5.1 Arithmetic Progressions 81 10.1 Index Numbers 198
5.2 Geometric Progressions 91 10.2 Composite Index 202
Summative Zone 103 Summative Zone 210
vi
FORM 5
ReDvaistieon ReDvaistieon
Chapter 1 Circular Measure 215 Chapter 6 Trigonometric Functions 341
1.1 Radian 217 6.1 Positive Angles and 343
1.2 Arc Length of a Circle 218 Negative Angles
1.3 Area of Sector of a Circle 223 6.2 Trigonometric Ratios of 345
any Angle
1.4 Application of Circular
Measures 228 6.3 Graphs of Sine, Cosine and
Summative Zone Tangent Functions 351
233
6.4 Basic Identities 357
Chapter 2 Differentiation 237 6.5 Addition Formulae and
Double Angle Formulae 358
2.1 Limit and its Relation to 6.6 Application of 364
Differentiation 239 Trigonometric Functions 371
2.2 The First Derivative 242 Summative Zone
2.3 The Second Derivative 248
2.4 DApifpfelriceanttiioantioonf 249 Chapter 7 Linear Programming 375
Summative Zone 266 7.1 Linear Programming
Model 377
Chapter 3 Integration 268 7.2 Application of 384
Linear Programming 391
3.1 Integration as the
Inverse of Differentiation 270 Summative Zone
3.2 Indefinite Integral 270 Chapter 8 Kinematics of Linear
Motion 394
3.3 Definite Integral 273
3.4 Application of Integration 284
Summative Zone 291 8.1 Displacement, Velocity and
Acceleration as a
Function of Time 396
Chapter 4 Permutation and 8.2 Differentiation in Kinematics
Combination 295 of Linear Motion 402
4.1 Permutation 297 8.3 Integration in Kinematics
302 of Linear Motion 407
4.2 Combination 309
Summative Zone 8.4 Applications of Kinematics 409
of Linear Motion
Summative Zone 417
Chapter 5 Probability Distribution 311
5.1 Random Variable 313 SPM Model Paper 418
317
5.2 Binomial Distribution 324 Answer 434
338
5.3 Normal Distribution
Summative Zone
vii
Formulae
FORM 4 Area of a quadrilateral: Chapter 3
1 Area under a curve:
• –2 [(x1y2 + x2y3 + x3y4 + x4y1)
(x2y1 + x3y2 + x4y3 + x1y4)] ∫ ∫• b y dx or b x dy
Chapter 2 aa
• x = –b ± ! b2 – 4ac Chapter 8 Volume of revolution:
2a
• |~r | = ! x2 + y2 ∫ ∫• b πy2 dx or b πx2 dy
x~i + y~j aa
Chapter 4 • ~^r = x2+ y2
• am × an = am + n
• am ÷ an = am – n Chapter 9 Chapter 4
• (am)n = amn
• a = b = c • n Pr = n!
sin A sin B sin C (n – r)!
• ! a × ! b = ! ab
• a2 = b2 + c2 – 2bc cos A • nCr = n!
– r)!r!
• ! a ÷ ! b = ! ab • b2 = a2 + c2 – 2ac cos B (n
• loga mmnn==lologga amm–+lolgoagann • c2 = a2 + b2 – 2ab cos C Identical object:
mn = n loga m
• loga Area of triangle: • P = n!
• loga 1 1 a!b!c!…
• 2 ab sin C = 2 ac sin B
• loga b = logc b = 1 bc sin A Chapter 5
logc a 2
• P(X = r) = nCr prqn – r, p + q = 1
Heron’s formula: • Mean, m = np
Chapter 5 • ! s(s – a)(s – b)(s – c)
a+b+c
Arithmetic progression: s= 2 • s = ! npq
• Tn = a + (n – 1)d 1)d] Chapter 10 • Z = X–m
• Sn = n [2a + (n – s
2 • I = Q1 × 100
n Q0
• Sn = 2 [a + l] ∑WiIi
∑Wi Chapter 6
Geometric progression: • I =
• sin2 A + cos2 A = 1
• Tn = arn – 1
• Sn = a(rn – 1) = a(1 – rn) , r ≠ 1 FORM 5 • sec2 A = 1 + tan2 A
1 – r
r–1 • cosec2 A = 1 + cot2 A
a
• S∞ = 1 – r , |r| 1 Chapter 1 • sin 2A = 2 sin A cos A
Chapter 7 • Arc length, s = rq 1 • cos 2A = cos2 A – sin2 A
A point dividing a segment of a 2 = 2 cos2 A – 1
line: • Area of sector, A = r2q = 1 – 2 sin2 A
( )• (x, y) = nxm1 ++mnx2 , nym1 ++mny2 Chapter 2 • tan 2A = 2 tan A
1 – tan2 A
Area of triangle: dy
• 12 [(x1y2 + x2y3 + x3y1) • y = uv, dx = u ddxv + v ddux • sin (A B) = sin A cos B
cos A sin B
– (x2y1 + x3y2 + x1y3)]
• y = u , dy = v ddux – u ddxv • cos (A B) = cos A cos B
v dx v 2 sin A sin B
• ddxy = dy × du • tan (A B) = tan A tan B
du dx 1 tan A tan B
viii
CHAPTER Functions
1
1.1 Functions Important Learning Standards Page
3
1.2 Composite • Explain function using graphical representations and notations. 5
Functions 6
• Determine domain and range of a function.
1.3 Inverse Functions • Determine the image of a function when the object is given and
vice versa.
• Describe the outcome of composition of two functions. 9
9
• Determine the composite functions. 10
11
• Determine the image of composite functions when the object is 11
given and vice versa.
• Determine a related function when the composite function and
another function are given.
• Solve problems involving composite functions.
• Describe inverse of a function. 13
13
• Make and verify conjectures related to properties of inverse 15
functions.
• Determine the inverse functions.
• Function / Fungsi • Range / Julat
• Relation / Hubungan • Domain / Domain
• Arrow diagram / Gambar rajah anak panah • Codomain / Kodomain
• Absolute value function / Fungsi nilai mutlak • Object / Objek
• Function notation / Tatatanda fungsi • Image / Imej
• Absolute value function graph • Composite function / Fungsi gubahan
• Inverse function / Fungsi songsangan
Graf fungsi nilai mutlak • Horizontal line test / Ujian garis mengufuk
• Vertical line test / Ujian garis mencancang
1
Fo4rm Additional Mathematics Chapter 1 Functions
CHAP Functions
1
Functions Composite Functions Inverse Functions
A relation where every fg(x) = f [g(x)] Properties of inverse function
object in the domain has (a) A function f that maps set
one and only image in fg
gf A to set B has an inverse
the codomain. x g(x) f[g(x)] function f –1 if f is a one-to-
one function.
Function notation gf(x) = g[f(x)] (b) fg(x) = x where x in the
f:x→y domain of g and gf(x) = x
f(x) = y gf where x in the domain of f.
fg (c) If two functions, f and g are
Vertical line test x f(x) g[f(x)] inverse to each other, then
(i) domain of f = range of
y
Vertical line test g and domain of g =
range of f,
f (ii) graph g is a reflection
0x of graph f.
(d) For any real numbers, a and
Graph is a function b, if the point (a, b) is on
the graph f, then the point
Absolute value function (b, a) is on the graph g.
y Horizontal line test
f(x) = |x| y
f
0x Horizontal line test
|x| = x–xififxx00 0x
Discrete function Continuous function Graph is a function
If f : x → y, then f –1 : y → x
f(x) f(x)
0x0 x
2
Chapter 1 Functions Additional Mathematics Fo4rm
1.1 Functions 7. If the line intersects the graph at only one CHAP
point, then the graph is a function.
Explaining function by using graphical 1
representation and notation y
1. Function is a special relation where for every Vertical line test
f
object in domain, there is one and only image
in codomain. 0x
2. For example, every element in set X has only
one image in set Y. 8. If the line intersects the graph at more than
one points, then the graph is not a function.
a • • e
b • • f Vertical line test y
c • • g f
d • • h
0x
Set X Set Y
3. A function f, which maps x to (mx + n) can be
written as
f : x → mx + n or f(x) = mx + n 9. An absolute value function is a function
that contains an algebraic expression within
where x is the object and f(x) or (mx + n) is the absolute value symbols.
image if x is under the function f.
10. The absolute value function of x is written as
4. Function is a relation of one-to-one or a relation f(x) = |x| and is defined by
of many-to-one.
(a) One-to-one relation x if x 0
f(x) f(x) = 0 if x = 0
–x if x 0
f x 1 • f • y 1
y 2 x 2 • • y 2 11. The graph of a linear absolute value function
y 1 has a V-shaped.
x
0 x 1 x 2 y
(b) Many-to-one relation f(x) = |x|
f(x) x
y = f(x) 0
x 1 • f 12. The discrete function is a function where the
x 2 • points on the graph is real, separated and not
y • y connected by a straight line or curve.
x1 0 x2 x f(x)
5. If given a graph of a relation, we can determine 0x
whether the graph is a function or not by using
a vertical line test.
6. The vertical line test can be tested by drawing a
line vertically on the graph.
1.1.1 3
Fo4rm Additional Mathematics Chapter 1 Functions
1CHAP 13. The continuous function is a function where Example 2
the points on the graph are connected by a
straight line of curve. By using function notation, express f in terms of
x for each of the following relations.
f(x) (a) f
–4 1
27
5 10
7 12
0x (b) x fy
Example 1 –2 • •1
–1 • •4
Determine whether the following relation is a 1•
function or not. Justify your answer. 2•
(a)
Solution:
3 • • 10 (a) f : x → x + 5 or f(x) = x + 5
• 16 (b) f : x → x2 or f(x) = x2
5 • • 60
• 62 Try question 2 in Formative Zone 1.1
7 •
Example 3
(b)
By using vertical line test, determine whether
–1 • •1 each of the following graphs is a function or not.
2• •4 (a) (b)
4• • 36
yy
6•
(c)
p • •3 2 x0 x
•6
q • •9 02
• 12 Solution:
r •
(a)
(d) y
10 • • 10 Vertical line test
•x
20 • •y 2
•z 02 x
30 •
The graph is a function.
Solution: (b)
y
(a) The relation is a function because each Vertical line test
object has only one image.
0x
(b) The relation is not a function because there
is one object does not have any image.
The graph is not a function.
(c) The relation is not a function because there
is one object that has two images. Try question 3 in Formative Zone 1.1
(d) The relation is a function because each
object has only one image.
Try question 1 in Formative Zone 1.1
4 1.1.1
Fo4rm Additional Mathematics Chapter 2 Quadratic Functions
Example 26 (iii) When a 0,
Express the quadratic function f(x) = 2(x − 4)2 − 8 f(x) x
in the form of f(x) = a(x − p)(x − q) where a, p a = –0.5
and q are constants, and q < p. Hence, state the 0
values of a, p and q.
2CHAP
Solution:
f(x) = 2(x − 4)2 − 8 Convert to a = –2 a = –1
= 2(x2 − 8x + 16) − 8 general form first
= 2x2 − 16x + 32 − 8 Then, convert to • the shape of the graph is ∩,
= 2x2 − 16x + 24 intercept form • the width of the graph is increasing
= 2(x2 − 8x + 12)
= 2(x − 2)(x − 6) when the value of a is increasing and
Thus, a = 2, p = 6 and q = 2. vice versa.
(b) Changing the values of h
Alternative Method (i) The horizontal movement of the
f(x) = 2(x − 4)2 − 8 graph and the position of axis of
= 2[(x − 4)2 – 22] symmetry will be affected.
Use a2 − b2 = (a + b)(a − b) (ii) When the value of h increases, the
a = (x − 4) and b = −2 graph will move to the right.
f(x) = 2(x − 4 − 2)(x − 4 + 2) (iii) When the value of h decreases, the
= 2(x − 6)(x − 2) graph will move to the left.
Try question 11 in Formative Zone 2.3 f(x)
h = –2 h = –1 h=1 h=2 h=3
Analysing and making generalisation about 0x
the changes of a, h and k in quadratic
functions f (x) = a(x – h)2 + k towards the (c) Changing the values of k
shape and position of the graphs (i) The vertical movement of the graph
and the maximum or minimum
1. The change of values of a, h and k in quadratic values of the graph will be affected.
functions f(x) = a(x – h)2 + k will affect the (ii) When the value of k increases, the
shape and position of the graph. graph will move upwards.
(a) Changing the value of a (iii) When the value of k decreases, the
(i) The shape and the width of the graph graph will move downwards.
will be affected.
(ii) When a 0, f(x)
k=2
f(x) a=1
a=2 a = 0.5 k=1
k = –1
0 x k = –2
k = –3
• the shape of the graph is ∪, 0x
• the width of the graph is decreasing
when the value of a is increasing and 2.3.3 2.3.4
vice versa.
36
Chapter 2 Quadratic Functions Additional Mathematics Fo4rm
Example 27 (c) The graph with the same shape moves
vertically 3 units upwards and the maximum
The diagram below shows the quadratic function value becomes 6. However, the axis of
f(x) = a(x − h)2 + k. symmetry remains unchanged.
f(x) f(x) CHAP
(2, 6)
(2, 3) 2
0 –1 (2, 3)
x
0 –1 x
f(x) = –(x – 2)2 + 6
Find the values of h, k and a. Hence, make f(x) = –(x – 2)2 + 3
generalisation on the effect of change in each
of the following values towards the shape and Try question 3 in Formative Zone 2.3
position of the graph.
(a) The value of a changes to −5. Sketching graphs of quadratic functions
(b) The value of h changes to −1. 1. The graph of a quadratic function
(c) The value of k changes to 6.
f(x) = ax2 + bx + c can be sketched based on the
Solution: following steps:
From the graph, we know that h = 2 and Determine the shape of the graph is
k = 3. By substituting the coordinates (0, −1), ∪ or ∩ by determining the value of a.
h = 2 and k = 3 into the quadratic equation
f(x) = a(x − h)2 + k, we get Determine the position of the graph
−1 = a(0 − 2)2 + 3 relative to the x-axis by determining the
−1 = 4a + 3
4a = −4 value of discriminant, b2 – 4ac.
a = −1
Thus, a = −1, h = 2 and k = 3. Determine the vertex point
(maximum or minimum point).
(a) The shape of the graph and the minimum
point remain unchanged. However, the Determine the intersection point between
width of the graph decreases. the graph and x-axis by solving the
equation f(x) = 0.
f(x)
Determine y-intercept by finding the
0x value f(0).
f(x) = –(x – 2)2 + 3
Plot all the points obtained on a Cartesion
f(x) = –5(x – 2)2 + 3 plane. Hence, sketch a smooth parabolic,
where symmetry with horizontal line that
. passess through the vertex of graph
(b) The graph with the same shape moves
horizontally 3 units to the left and the
equation of axis of symmetry becomes
x = −1. However, the maximum value
remains unchanged.
f(x)
(–1, 3) (2, 3)
0 –1 x
f(x) = –(x – 2)2 + 3
f(x) = –(x + 1)2 + 3
2.3.4 2.3.5 37
Fo4rm Additional Mathematics Chapter 4 Indices, Surds and Logarithms
4.1 Laws of Indices (c) 1 × 32 = 3−4 × 32
Simplifying algebraic expressions 81 = 3−4 + 2
involving indices using the laws of indices
= 3−2
=1
1. A number can be written in index notation or 32
index form as an where a is a base and n is an
index or power. =1
9
2. An algebraic expression can be simplified using
the laws of indices. 11
(d) 25 × 325 ÷ 16 = 25 × (25)5 ÷ 24
= 25 × 21 ÷ 24
3. The following indices should be known first = 25 + 1 − 4
before we simplified the algebraic expressions
CHAP using the laws of indices. = 22
=4
4 • a–1 = 1
(Negative index) Try question 1 in Formative Zone 4.1
a
• a0 = 1 (Zero index) Calculator
• 1 = n a (Fractional index) Based on Example 1(a), the algebraic expression
25 × 16−2 can be evaluated using calculator as follows:
an
1. Press 2 ^ 5 × 16 ^ (–) 2 =
• m = (n a )m (Fractional index) 2. The answer will be displayed as 0.125 .
3. Press SHIFT abc to convert the answer to
an
fraction 1 8 .
4. The following shows the laws of indices:
• an × am = an + m Example 2
• an ÷ am = an – m
• (an)m = anm Simplify the following algebraic expressions.
• (ab)n = anbn
3
• a n = an
b bn (b) 2n + 1 ÷ 42 × 164
(a) 32 × 273 − n
Example 1 (c) (2p)3 × p−2 (d) (31x23xy54y)2
pq3
By using the laws of indices, evaluate the Solution:
following algebraic expressions.
(a) 32 × 273 − n = 32 × (33)3 − n
(a) 25 × 16−2 (b) (5−3 ÷ 5−4)2 = 32 × 39 − 3n
(c) 1 × 32 1 = 32 + 9 − 3n
81
(d) 25 × 325 ÷ 16
= 311 − 3n
Solution: 33
(a) 25 × 16−2 = 25 × (24)−2 (b) 2n + 1 ÷ 42 × 164 = 2n + 1 ÷ (22)2 × (24)4
= 2n + 1 − 4 + 3
= 25 × 2−8 = 2n
= 25 + (−8) am × an = am+n (c) (2p)3 × p−2 = 23p3 × p−2
= 2−3 am ÷ an = am−n pq3 pq3
=1
= 8p3 × p−2
23 pq3
=1 = 8p3−2
8 pq3
(b) (5−3 ÷ 5−4)2 = (5−3)2 ÷ (5−4)2 = 8p
= 5−6 ÷ 5−8 pq3
= 5−6 − (−8) =8
= 52 q3
= 25
60 4.1.1
Chapter 4 Indices, Surds and Logarithms Additional Mathematics Fo4rm
(d) (3x3y4)2 = 32(x3)2(y4)2 (ambn)p = ampbnp Solution:
12x5y 12x5y
(a) 7x + 2 − 7x + 1 = 6
= 9x6y8 7x·72 − 7x·71 = 6
12x5y
7x(72 − 71) = 6
= 43x6 − 5y8 − 1 7x(49 − 7) = 6
7x(42) = 6
= 43xy7 7x = 6
42
Express the equation
7x = 1 of both sided in the
7 same base
Try question 2 in Formative Zone 4.1
7x = 7−1
x = −1 CHAP
4
Example 3 • If 7x = 7–1, then x = –1.
Simplify 2n + 2n + 2 − 2n + 1 in the form of k(2n) • If x7 = 47, then x = 4.
where k is a constant. Hence, determine the
value of k. (b) 81·(3x + 2) = 1 30 = 1
34·(3x + 2) = 1
Solution: 34 + x + 2 = 1
2n + 2n + 2 − 2n + 1 = 2n + 2n·(22 − 21) 36 + x = 3°
= 2n(1 + 22 − 21) 6 + x = 0
= 2n(1 + 4 − 2) x = −6
= 2n(3)
= 3(2n)
Thus, k = 3.
Alternative Method
Let u = 2n, then Change 1, 3 and 81 to a common base, that
2n + 2n + 2 − 2n + 1 = 2n + 2n·22 − 2n·21 is 3. Hence, equate the indices to solve for x.
= u + 4u − 2u
= 3u Try question 5 in Formative Zone 4.1
= 3(2n)
Example 5
Try question 3 and 4 in Formative Zone 4.1 Show that 5n + 3 − 5n − 5n + 2 is divisible by 11
for all positive integers of n.
Solving problems involving indices
1. When solving an equation involving indices, Solution:
5n + 3 − 5n − 5n + 2 = 5n·53 − 5n − 5n·52
we should consider the following statement: = 5n(53 − 1 − 52)
If am = an, then m = n or if am = bm, then = 5n(125 − 1 − 25)
= 5n(99)
a = b when a > 0 and a ≠ 1.
Since 5n(99) is divisible by 11. Thus,
Example 4 5n + 3 − 5n − 5n + 2 is divisible by 11.
Solve the following equations.
(a) 7x + 2 – 7x + 1 = 6 Try question 6 in Formative Zone 4.1
(b) 81·(3x + 2) = 1
4.1.1 4.1.2 61
Fo4rm Additional Mathematics Chapter 5 Progressions
5.2
1. Determine whether the following sequence is 9. The 5th and 8th term of a geometric
a geometric progression or not. C1
progression are 337 1 and 42 3 respectively.
(a) 18, 27, 40 1 , … 2 16
2 Find C2
(b) 125, 31 1 , 7 13 , … (a) the first term and the common ratio,
4 16
(b) the sum of the 7th and 8th term,
(c) 16, 2 3 , 9 , … of the geometric progression.
5 25
10. It is given that 3 and 243 are the 2nd term
(d) –36, 12 3 , –4.41, … and 6th term of a geometric progression
5 respectively. Find C2
(a) the first term and the common ratio,
2. Given that the sequence below is a geometric (b) the sum of the 2nd and 3rd term,
of the geometric progression.
progression. Determine the value of common
ratio. C1
5CHAP (a) –144, 72, –36, … 11. Find the number of terms of the following
geometric progression. C2
(b) –12, –30, –75, …
(a) 2, 3, 4 1 , …, 15 3
(c) 8, 1, 1 , … 2 16
8
15
(d) 15, 3, 0.6, … (b) 120, 60, 30, …, 16
3. Given that 3h, 10h – 6 and 25h + 12 are the (c) 12k, 36k, …, 8 748k
3 terms in a geometric progression, find the
possible values of h. C2 (d) 3 , 2 1 , …, 57 681
24 1 024
4. Find the possible values of k if 2k + 6, 8k – 8 12. It is given that 24, 18, 27 , … is a geometric
and 12k + 4 are the 3 terms of a geometric 2
progression. C2 progression. Determine which term in the
5. It is given that 4p – 2, 6p + 3 and 12p + 18 are geometric progression is 3 417 . C2
the 3 terms of a geometric progression. Find 2 048
the value of p. C2
13. Which term in the geometric progression 4,
12, 36, … is 26 244? C2
6. Given that 32, 8 and 2 are the first 3 terms of 14. Determine the term in the geometric
a geometric progression. Find C2
(a) the first term and the common ratio, progression 28, 7, 1 3 , … which is equal to
(b) the 8th term. 4
7 . C2
262 144
7. It is given that 2p, –21 3 and –38 22 are the 15. It is given that k, 18, 54 and h are the four
5 25 terms in a geometric progression. Find the
first 3 terms of a geometric progression. Find value of k and h. C2
C2
(a) the common ratio of the geometric 16. Given that 10, p, q, 80 are the first four terms
in a geometric progression, find the value of p
progression, and q. C2
(b) the value of p,
(c) the 15th term of the geometric
progression. 17. Given that u, 15, v, w are the first four terms in
a geometric progression, express v in terms of
8. It is given that 448, –896 and 1 792 are the 6th, u and w. C3
7th and 8th term of a geometric progression.
Find C2 18. Given that 4, 10, 25, … is a geometric
(a) the first term and the common ratio, progression, what is the first term which its
(b) the 11th term, value more than 1 000? C3
of the geometric progression.
98
Chapter 5 Progressions Additional Mathematics Fo4rm
19. It is given that 6, 7 1 , 9 3 , … is a geometric 30. It is given that the sum of the first n terms of a
28
geometric progression is Sn = 3 [32n − 1]. C2
progression. Find the minimum value of n if 8
the value Tn > 300. C3 (a) Find the sum of the first 5 terms.
(b) Express the n term of the geometric
20. The 2nd and 5th terms of a geometric progression, in term of n.
progression are 10 and 19 17 . Find the first 31. Given that the sum of the first n terms of the
32 geometric progression 2, −4, 8, … is 21 846,
term which is more than 100. C3 find the value of n. C2
21. Find the sum of first terms of the following 32. It is given the sequence 3, 1, 1 , … is a
geometric progression. C1 3
(a) 1, 2, 4, …, to n = 13 geometric progression. Find the value of
(b) 5, 15, …, to n = 14
(c) 12, –24, 48, …, to n = 12 n when the sum of the first n terms of the
(d) 3, 32, 33, …, to n = 8
geometric progression is 4 29 524 . C2
59 049
22. It is given that 7, –7, 14, … is a geometric
2 33. Determine the value of n when the sum of the CHAP
progression. Find the sum from 3rd term to 8th first n of the geometric progression 125, –25,
term of the geometric progression. C2 5, … is 104 104 . C2 5
625
23. Find the sum from 4th term to 9th term of the
geometric progression 4, –6, 9, … C2 34. It is given that the 2nd term and the 5th
term of a geometric progression are –27 and
24. It is given that 9, –18, 36, … is a geometric 1 respectively. Find C3
progression. Find the sum from 6th term to
18th term of the geometric progression. C2 (a) the first term and the common ratio,
(b) the sum of the first 13 terms,
25. It is given that the nth term of a geometric of the geometric progression.
progression is Tn = – 5 (–2)n. Find C2 35. Given that the 4th term and the 10th term
2
of a geometric progression are –16 and – 1
(a) the first term and the common ratio, 4
(b) the sum of the first 6 terms, respectively. Find C3
of the geometric progression. (a) the first term and the common ratio,
(b) the sum from the 5th term to the 12th
26. Given that the nth term of a geometric
progression is Tn = 33 – n, find the sum of the term,
first 10 terms. C2 of the geometric progression.
27. It is given that the nth term of a geometric 36. In a geometric progression, the 3rd term and
the sum of 4th and 5th term are 60 and 360
progression is Tn = 625 . Find the sum from respectively. Find C3
5n
the 5th term to the 12th term of the geometric (a) the first term and the common ratio,
progression. C2 (b) the sum of the first 11 terms,
of the geometric progression.
28. The sum of the first nth terms of a geometric
22n + 1 – 2 . Find
progression is given by Sn = 3 C2 37. Find the sum to infinity of the following
geometry progression. C2
(a) the sum of the first 7 terms,
(a) 120, 80, 53 1 , …
(b) the 8th term, 3
of the geometric progression. (b) 8, 2 2 , 1 7 , …
39
29. Given that the sum of the first nth terms of a
Sn 4(2n 1).
geometric progression is given by = – (c) 1 , 1 1 , 1 1 2, …
Find C2 5 52 52
(a) the sum from the 3rd to 7th term, (d) 3x, 1, 1 , …
(b) the nth term of the geometric progression 3x
in terms of n.
(in terms of x)
99
Fo4rm Additional Mathematics Chapter 6 Linear Law
SPM Simulation HOTS Questions
Paper 1
1. The variable x and y are related by equation y = mx2 − nx Divide both
y x side with x2.
x = mx2 – nx where m and n are constanx–yt2.
y y n
x3 = m − x
O x k+2 Compare to Y = mX + c.
y 1x ,
Hence Y = x3 , X = m = –n and c = m.
From the graph, c = k + 2
x m = k + 2
–3k O k = m − 2 …❷
To express m in terms of n, eliminate
Diagram (a) variable k.
y x–y2 n
As ❶ = ❷, m − 2 = 3
O x k+2
3m − 6 = n
m is the m = n + 6
subject 3
CHAP Ox 2. The variable x and y are related by an equation
y = 102x2 + 1. The diagram below shows a
6–3k straight line graph obtained by plotting
log10 y against x2.
Diagram (b)
Diagram (a) and diagram (b) show the straight log10 y
line graphs obtained by plotting the relations (p, 9)
from the equation. Express m in terms of n.
C4
Examiner's Comment: q
In diagram (a) O x2
x is the X-intercept. Find the values of p and q. C3
y
Then, reduce x = mx2 − nx to get x as the
variable of X. Examiner's Comment:
y Divide both Convert the equation y = 102x2 + 1 into the
x side with x. linear form, Y = mX + c.
= mx2 − nx y = 102x2 + 1 Taking logarithm
log10 y = log10 102x2 + 1 to the base of 10
y = mx − n to both side.
x2
log10 y = (2x2 + 1) log10 10
Compare to Y = mX + c. log10 y = 2x2 + 1
Hence Y = xy2, X = x, m = m and c = −n.
From the graph, c = −3k Compare to Y = mX + c.
−n = −3k Hence Y = log10 y, X = x2, m = 2 and c = 1.
k = n …❶ From the graph, q = y-intercept
3 = 1
In diagram (b) Gradient, m = 2
9 – 1 = 2
y is the y-intercept. p–0
x3
2p = 8
y y p = 4
Then, reduce x = mx2 − nx to get x3 as
variable of y. Thus, p = 4 and q = 1.
118
Chapter 6 Linear Law Additional Mathematics Fo4rm
3. The diagram below shows a part of graph y Paper 2
against x for equation y = pq–x, where p and q
are constant. C4 4. The table below shows the value of variables,
y x and y, obtain from an experiments. The
variables x and y are related by equation
y = pq –x hy – k
x= x where h and k are constants. C4
x 0.5 1.0 1.5 2.0 2.5
y 14 24 32 48 69
Ox (a) Plot a graph of y against x2 using a scale
of 2 cm to 10 units on y-axis and 2 cm to
(a) Find the possible equation of the line of 1 unit on x2-axis. Hence, draw the line of
best fit for the non-linear graph. best fit.
(b) Based on your answer in (a), find the value (b) From the graph from (b), find the values of
of p and q, if y-intercept is 2 and gradient (i) h and k,
is –3. (ii) x when y = 43.
Examiner's Comment:
(a) Build a table of x2.
Examiner's Comment: x 0.5 1.0 1.5 2.0 2.5 CHAP
(a) Convert the equation y = pq–x into the x2 0.25 1.0 2.25 4.0 6.25
6
linear form, Y = mX + c.
y 14 21 32 48 69
y = pq–x Taking logarithm to the Graph y against x 2
base of 10 to both side. y
log10 y = log10 p + log10 q–x 70
log10 y = log10 p – x log10 q 60
log10 y = –log10 q(x) + log10 p 50
Thus, the possible equation of line of 40
30
best fit is 20
log10 y = –log10 q(x) + log10 p. 10
0 1 2 3 4 5 6 x2
(b) Compare to Y = mX + c.
log10 y = –log10 q(x) + log10 p
aHnedncce=, Ylo=g1l0opg.10 y, X = x, m = – log10 q
When y-intercept = 2 (b) Convert the equation of x = hy – k into
linear form, Y = mX + c. x
log10 p = 2
p = 102 x = hy – k
x
= 100
When gradient, m = –3 x2 = hy – k
– log10 q = –3 hy = x2 + k
log10 q = 3
q = 103 y = 1 x2 + k
h h
= 1 000
Compare to Y = mX + c.
Thus, the values of p and q are 100 and
1 000 respectively. Hence, Y = y, X = x2, m = 1 and c = hk .
h
119
Fo4rm Additional Mathematics Chapter 8 Vectors
Paper 1
1. The diagram shows a quadrilateral ABCD. 5. ∼rqp∼∼===h23∼a∼∼aa +–+(∼b4h∼b– k)∼b where h and k are constants.
C2 → → → C2
It is given = h∼a, = k∼a, = h∼b and
AB CD CA
→
= 2∼a + (k + 8)∼b.
DB
A Use the information above to find the values of
h and k when∼r = 4p∼ – 2q∼.
C
6. Three points lie on a Cartesian plane which are
C2 the origin, O, K(–2, 6) and L(5, 7). Find
D (a) → in of x.
terms
B KL y
Find the values of h and k. (b) the unit vector in the direction of → .
KL
7. The points P, Q and R are collinear. It is given
C2 → →
It is given t13hp∼at–p∼2q∼=. 9∼i – 6∼j and q∼ = – 4∼j , v=a3lu∼i e–o2f∼jma.nd = (1 – m)∼i + 4∼j , determine
determine PQ QR
2. ∼i the
C2 → →
→ → 8. The diagram shows OP = ∼a and OQ = ∼b.
3. The diagram shows vector BA , vector B→C and C3
C2 BA = u∼
panodintB→DC lie on a square grid. It is given 4
8CHAP = ∼v. 3Q
2
P1
C –4 –3 –2 –1 0 1 2
A
On the same square grid, draw the line that
B →
represents the vector = 3∼a + ∼b.
D OR
9. The diagram shows two vectors → and →
C3 that are parallel to each other.
OB CD
Express in terms of u∼ and ∼v for each of the y
following vectors.
6
→
(a) 5D
BD
→
(b) 4
DC
3C
4. It is given that → = –3∼i + 5∼j and → = 2∼i – 7∼j, 2B
C2 find
OA OB 1
(a) → in terms of ∼i +∼j, –1 O x
(b) unit vector in the 12 34 56 78
AB
the direction of → . It is given that → = (k + 2)∼i + 3∼j and
(2k – 1)∼i + the value of k.
AB OB
→ = 4∼j. Find
CD
172
Chapter 8 Vectors Additional Mathematics Fo4rm
10. The diagram shows a triangle ABO. 13. The diagram shows a Cartesian plane where O
C3 C4 is Arif’s house, P is a shop and Q is a school.
B y
C Q
OD A
P
It is given that 2 → = 3C→O, D is the midpoint
→ → →
BC
AB AO DC
of OA, a=nd3∼u∼v. and = 4∼v. Express in
terms of u∼
x
–2 O 2 4 6 8 10
11. The diagram shows a parallelogram ABCD. The shortest distance between Arif’s house and
C3 Point Q is the intersection of the diagonals. the shop is 10 km while the shortest distance
between the shop and the school is 8 km.
BC Express the vector of Arif’s house to the school
in terms of ∼a + ∼b.
Q
14. The points A, B and C are collinear. It is given
AD → →
C3 that =m3i∼xs a+co2∼ynsatanndt, = (1 – m)∼x + 4∼y
such AB BC
that find
It is given that → = 2∼x + 3∼y and → = 6∼x – ∼y ,
find C→Q. (a) the value of m,
DA AB (b) the ratio of AB : BC.
CHAP
8
12. The diagram shows → = 8∼i – h∼j and 15. The diagram shows a regular octagon.
→ C3
OB
BC
C3 = 3∼i – 2∼j. F E
p∼
y GD
B q∼
HC
∼r B
A
C (a) State the vectors that are equal.
(b) Express → in terms of p∼, q∼ and∼r.
FA
O x
It is given | → | = 10 units, find 16. A toy truck was moved at a constant velocity
C3 from point O to point P. It is given that
OB
6O→sPeco=nd(3s6, ∼ifin–d 24∼j) m and the time taken is
(a) the value of h, (a) the velocity of the toy truck,
(b) coordinates of C. (b) the speed of the toy truck.
173
Fo4rm Additional Mathematics Chapter 8 Vectors
Clone The diagram shows a regular decagon with Clone The diagram shows vectors → , → and →
17. 20. AC AB AD
C3 center O.SPM SPM SPM C3 drawn on a square grid with sides of 1 unit.
SPM
GF A
D
HE
IO D
GC C
B
AB
(a) Express → + → + → + → as a single (a) It is given that → = p∼ and → = q∼,
vector. → of mp∼ m
BC CH HJ FE AB AC
AD
→ t=h∼ex,dO→ecDag=on∼y express in terms + nq∼, where
(b) It is given and the length and n are constants.
of OE is 2 units. Find (b) On the diagram above, mark and label
each side of
→ → → →
the unit vector in the direction of , in point E such that + = 3 .
terms of ∼x and ∼y. ED DE AC AB
Clone → → → 21. A school bus depart from Kampung Merbau
with aSaupjoasnitaioUn tavmecatowr it(h40∼ia p+os3it0i∼jo)n km to
18. The diagram shows vectors QP , QS and QR C4 SMK vector
C3 drawn on a square grid with sides of 1 unit. (c2o5n∼ista+nt16∼vje)lokcmit.y The school bus travels with
and take 1 hour back and
forth to take and drop off students. Find the
velocity of the vector.
QR
8CHAP S 22. 7 + –6 represents the
5 8 t is in seconds
C4
The equation u∼ = t
P path of a moving particle where
and the distance is in meter. Find
→
(a) Find |– |. → → (a) the initial position of the particle,
PQ (b) the velocity of the particle,
QP QS (c) the speed of the particle.
(b) It is given that of ∼x a(in=i)d ∼x∼yQ→,aRn.d = ∼y,
express in terms
→
(i) 23. In an archery competition, an arrow hits
PS
Clone aqI∼tr=eisc–o∼giniv–set8an∼jnttashn,adetxK→Kpr(L2es,=smnp∼),i–nLh(tneq∼r,mw7h)s,eorp∼fe = 7∼i + 4∼j, C4 3 points on archery’s board. The points have
m, n and h position vectors that marked E, F and G are
19. m. –a∼rje, c5∼oi ll–in3e∼jara.nd 11∼i – 6∼j. Show that E, and
C3 ∼i F
G
174
Chapter 9 Solution of Triangles Additional Mathematics Fo4rm
9.2
1. The diagram below shows a triangle ABC. 5. The diagram below shows a quadrilateral ABCD.
C 8.1 cm D
3.8 cm
x
B A 72° C
4.2 cm 5.7 cm 9.6 cm 6.3 cm
29° B
Given that AB = 9.6 cm, BC = 6.3 cm,
A
Given that AB = 4.2 cm, AC = 5.7 cm and CD = 3.8 cm, AD = 8.1 cm and ∠BCD = 72°.
Find C3
∠BAC = 29°, find the value of x. C2 (a) the length of BD,
(b) ∠ADB.
2. The diagram below shows a triangle PQR. 6. The diagram below shows a triangle KMN.
Given that KLM is a straight line.
11.2 cm R 2.7 cm
Q LM
124° 14.1 cm
y
8.1 cm K 9.8 cm 11.3 cm
x
P N
Given that PQ = 8.1 cm, QR = 11.2 cm and If KL = 14.1 cm, LM = 2.7 cm, MN = 11.3 cm
∠PQR = 124°, find the value of y. C2 and LN = 9.8 cm, find the value of x. C3
3. The diagram below shows a triangle ABC. 7. In a shooting practice, Syafiq was required to
shoot on two boards practice such as board A
B is 50 meters and board B is 75 meters from a
shooting point respectively. If the angle formed
5.7 cm between boards A and B at the shooting point CHAP
is 32°, find the distance between board A and
A 6.3 cm board B. C4 9
7.4 cm θ
C
Board
B
Given that AB = 5.7 cm, BC = 6.3 cm and Board 75 m
AC = 7.4 cm, find the value of θ. C2
A
50 m 32°
4. The diagram below shows a triangle KLM. Shooting
point
K
15.1 cm 8. Zaki drove a boat 25 km to the Nibong's
jetty at the bearing of 118° from Sekinchan's
11.2 cm θL lighthouse. Then, he drove the boat 38 km to
M 6.7 cm Redang Island at the bearing of 232° from the
Nibong's jetty. Calculate the distance between
Given that KL = 15.1 cm, LM = 6.7 cm and Sekinchan's lighthouse and Redang Island.
KM = 11.2 cm, find the value of θ. C2 C5
185
Chapter 9 Solution of Triangles Additional Mathematics Fo4rm
5. The diagram below shows the position of 6. The diagram below shows the position of two
rambutan tree (R), durian tree (D), mango tree ports, A and B and a ship sailing in an ocean.
(M) and a hut in an orchard.
Rambutan 17 m Mango
Durian
13 m 9m 67°
Port
32°
A
Hut 25 km Port
B
Given that the distance of durian tree and The distance from port A to port B is 25 km.
mango tree from the hut are the same. Find the Find the distance between the ship and
distance between durian tree and rambutan port B. C3
tree. C4
SPM Simulation HOTS Questions (b) Length of BD can be find by using
cosine rule.
Paper 2
1. The diagram below shows a quadrilateral BD2 = CD2 + BC2 − 2(BC)(CD) cos ∠BCD
= 62 + 72 − 2(6)(7) cos 72.25°
ABCD. = 59.39
6 cm B BD = 59.39
C 53° = 7.71 cm
7 cm CHAP
D (c) Using sine rule, 9
10 cm
A sin ∠BAD = sin 53°
7.71 10
The area of ΔBCD is 20 cm2 and ∠BCD is an
acute angled. Calculate C4 sin ∠BAD = sin 53° × 7.71
(a) ∠BCD, 10
(b) length BD,
(c) ∠ADB, and = 0.6157
(d) the area, in cm2, of quadrilateral ABCD. ∠BAD = 38°
Thus,
∠ADB = 180° − 53° − 38°
= 89°
Examiner's Comment: (d) Area ΔABD = 12(10)(7.71) sin 89°
(a) Given ΔBCD = 20 cm2
12(6)(7) sin ∠BCD = 20 2 = 38.54 cm2
sin ∠BCD = 20 42
× Area of equilateral ABCD
= Area ΔABD + Area ΔBCD
= 0.9524 = 38.54 + 20
= 58.54 cm2
∠BCD = 72.25°
191
Chapter 10 Index Numbers Additional Mathematics Fo4rm
Example 11 If the composite index of the four items in the
year 2019 based on the year 2015 was 136,
The table below shows the price indices for the calculate the value of p.
four ingredients, A, B, C and D used in making a
cake in 2017 based on 2014. Solution:
Ingredient ABCD ∑Iiwi = 115(2) + 150(3) + 135p + 140(1)
= 820 + 135p
Price index in 2017 80 140 150 100 ∑wi = 2 + 3 + p + 1
based on 2014 =6+p
∑∑Iwiwi i = I–
Bar chart below represents the relative amount 820 + 135p = 136
of the ingredient A, B, C and D, used in making
a cake. 6+p
820 + 135p = 136(6 + p)
Weightage 820 + 135p = 816 + 136p
p = 4
140
120 Try question 5 in Formative Zone 10.2
80 Ingredients
60
0A B CD
Calculate the composite index for the cost of Solving problems involving index numbers
making a cake in the year 2017 based on the and composite index
year 2014.
Solution: Example 13
∑Iiwi == 80(140) + 140(80) + 150(120) + 100(60) The table below shows the price indices of
46 400 four raw materials, P, Q, R and S which used to
produce a tin of drink A in a factory.
∑wi = 140 + 80 + 120 + 60
= 400 Raw Price index in 2019
material based on 2017
I– = ∑Iiwi 120
∑wi P 145
= 46 400 Q 130
400
R
= 116
S
Thus, the composite index for the cost of making
a cake in the year 2017 based on the year 2014 Pie chart represents the proportion of the raw CHAP
was 116. material used in the drink.
10
Try question 4 in Formative Zone 10.2
Example 12 PQ
140° 85°
The table below shows the price indices and 20°
their corresponding weightage of four items, E, 115° R
F, G and H in 2019 using 2015 as the basis year.
S
Item Price index in 2019 Weightage (a) If the composite index for the cost of
based on 2015 making a tin of drink A in the year 2019
E 115 2 based on the year 2017 is 132, what is the
F 3 price index of raw material R?
G 150 p
H 1 (b) If the price of raw material R in the year
135 2017 was RM15, calculate the price of raw
material R in the year 2019.
140
10.2.1 10.2.2 203
Chapter 10 Index Numbers Additional Mathematics Fo4rm
(d) Solution:
2015 2017 2019
(a) For salt,
1.00
123.15 116 x = 0.80 × 100
I2019 = 116 × 123.15 = 125
2015 100
For sugar,
y
= 142.854 101.5 = 2.00 × 100
≈ 142.85 2.00
100
The corresponding cost of all the items in y = 101.5 ×
the year 2019 based on the year 2015 is
RM142.85. = 2.03
Try question 7 in Formative Zone 10.2 (b) ∑Iiwi = 1 28.5(60) + 125(25) + 101.5(15)
+ 114(20)
= 14 637.5
The index number of the T2 based on T0: ∑wi = 60 + 25 + 15 + 20
= 120
I T2 T0 T1 T2 I– = ∑Iiwi
= T1 × I T2 ∑wi
I T2 I T1
T0 100 T0 I T1 T1 = 14 637.5
T0 120
Example 15 = 121.9792
The table below shows the prices and the price
indices of four items used in the production of a ≈ 121.98
packet of cracker.
(c) Given the composite index for the cost of
making the crackers increased by 30% from
the year 2013 to the year 2019.
Price (RM) Price index in 130
per kg the year 2019
Item based on the 2013 2016 2019
2016 2019
Flour year 2016
Salt 2.00 2.57 121.98
Sugar 0.80 1.00 Increase 28.5%
Oil 2.00 (i) Price index in the year 2016 based on
5.04 y x the year 2013:
3.5
Increase 1.5% I2019
I2019 = 2016 ×
Increase 14% 2013 I2016
100 2013
Given the usage of item flour, salt, sugar and oil 130 = 121.98 × I2016 CHAP
are 60 g, 25 g, 15 g and 20 g respectively. 100 2013
(a) State the values of x and y. 10
(b) Calculate the composite index for the cost I2016 = 106.5748
2013
of making the crackers in the year 2019
based on 2016. ≈ 106.57
(c) It is given that the composite index for the (ii) The cost of making a crackers in the
cost of making the crackers increased by
30% from the year 2013 to the year 2019. year 2016:
(i) Calculate the composite index for the P520419 × 100 = 106.57
cost of making a packet of cracker in the P2016 = 57.5478
year 2016 based on the year 2013. ≈ 58 cent per packet
(ii) The cost of making a crackers is 54 Number of packet of cracker in 2019
cent per packet in the year 2013. Find = RM120 ÷ RM0.58
the maximum number of the packet = 206.90
of cracker that can produced using an ≈ 207 packets
allocation of RM120 in the year 2016.
Try question 8 in Formative Zone 10.2
180.2.12 205
Fo4rm Additional Mathematics Chapter 10 Index Numbers
For chicken, (ii) P2016 × 100 = 132.5
y = 12.00 × 100 550
8.00
= 150 P2016 = 132.5 × 550
100
For prawn,
z × 100 = 140 = 728.75
4.00
z = 140 × 4.00 Thus, the sale of food at Pak Abu’s
100 shop in the year 2016 is RM728.75.
= 5.60 (c) The production cost is expected to
decrease by k% from the year 2016 to
(b) (i) For crab, the year 2019.
The percentage of sale for the crab
106
= 360° − 90° − 120° − 80° 2014 2016 2019
= 70°
The ratio of the weightages 132.5 100 – k
= 90° : 70° : 80° : 120°
= 9 : 7 : 8 : 12 132.5 × 100 − k = 106
100
I– = 120(9) + 110(7) + 140(8) + 150(12)
36
100 − k = 106 × 100
= 4 770 132.5
36
100 − k = 80
= 132.5 k = 20
Thus, the composite index of the The sale of food is expected to decrease
sale in the year 2016 based on the by 20% from the year 2016 to the year
year 2014 is RM132.5. 2019.
Paper 2 (b) The price index of component L in the year
CHAP 2017 based on the year 2014 is 125. The
unit price of component L in the year 2017
10 1. The table below shows the unit price of four was RM30 more than its unit price in the
C4 components, K, L, M and N which is needed to year 2014. Calculate the values of q and r.
produce a type of laptop.
(c) The composite index for the cost to
Component Unit price (RM) in the year produce the laptop in the year 2017 based
2014 2017 on the year 2014 is 139. Calculate
(i) the price of the laptop in the year
K 40 p 2017 if the corresponding price in the
year 2014 was RM1 700,
L qr (ii) the value of m if the ratio of
M 40 70 components used to produce the
N 50 60 laptop is 2 : 1 : 3 : m.
(a) Given the price index of component K in
the year 2017 based on the year 2014 is
130. Find the value of p.
210
Chapter 10 Index Numbers Additional Mathematics Fo4rm
2. The table below shows the Sahara’s monthly (ii) The cost of making the health
C4 expenditure and weightage for the year 2013 supplement is RM35.00 per bottle in
the year 2011. Find the maximum
and 2018. number of bottle of health supplement
Expenditure Price that can be produced using an
Item (RM) index Weightage allocation of RM42 000 in the year
2013 2018 Clone 2019.
Transport 500 750 z 6 4. The table below shows the price indices and
SPM
SPMFood500 800 1609C5change in price indices of the expenditure of
vegetables plantation.
SeedsHouse
Insecticiderental360x1303 Change in
price index
FertiliserOthersy 405 1352ExpenditurePrice indexfrom 2017 to
Equipmentin 2017 based
maintenance(a) Find the values of x, y and z.2019
on 2013
(b) Find the composite index for the year 2018 Seeds 120 15% increase
based on the year 2013.
(c) Hence, find the average monthly Insecticide 110 5% decrease
expenditure for four item which are Fertiliser 135 10% increase
transport, food, house rental and others
in year 2013, if the average monthly Equipment
expenditure in year 2018 is RM651. maintenance 125 Unchange
Clone The table below shows the price and the price Bar chart below represents the mass of each
indices of four item A, B, C and D used in the expenditure used in vegetable plantation.
3. production of a type of health supplement.
C5
Item Price (RM) per Price Weightage Mass (kg)
bottle for the year index
35
2014 2019 30
25
A 40.00 60.00 150 30
15
B 18.00 21.60 120 50
C 80.00 90.00 112.5 12
D k 37.50 h 8 Expenditure
CHAP
(a) The price of item D is increased by 25%
from the year 2014 to the year 2019. 10
(i) State the value of h,
(ii) Find the value of k.
(b) Calculate the composite index for the cost (a) The price of the seeds in 2017 is RM60.
of making the health supplement for the Find the corresponding price in 2013.
year 2019 based on the year 2014.
(b) Find the price index of all the four
(c) It is given that the composite index for expenditure in 2019 based on 2013.
making the health supplement increased by Hence,
54.2% from the year 2011 to the year 2019. (i) calculate the composite index for the
(i) Calculate the composite index for the year 2019 based on the year 2013,
cost of making the health supplement (ii) find the cost price (per kg) in 2013
in the year 2014 based on the year if the corresponding cost in 2019 is
2011. RM6.00 per kg.
211
Tg 4 CHAPTER Circular Measure
1
SMART Important Learning Standards Page
217
1.1 Radian • Relate of angle measurement in radian and degree. 218
220
1.2 Arc Length of a • Determine 220
Circle (i) arc length, 222
(ii) radius, and 223
1.3 Area of Sector of a (iii) angle subtended at the centre of a circle. 224
Circle 226
• Determine perimeter of segment of a circle.
1.4 Application of • Solve problems involving arc length. 215
Circular Measures
• Determine
(i) area of sector,
(ii) radius, and
(iii) angle subtended at the centre of a circle.
• Determine the area of segment of a circle.
• Solve the problems involving areas of sectors.
• Solve problems involving circular measure.
Words
• Angle / Sudut
• Degree / Darjah
• Arc length / Panjang lengkok
• Radius / Jejari
• Area of sector / Luas sektor
• Circle / Bulatan
• Centre of circle / Pusat bulatan
• Circular measure / Sukatan membulat
• Chord line / Garis perentas
• Circumference of circle / Lilitan bulatan
• Radian / Radian
• Diameter / Diameter
• Angle subtended at the centre of circle / Sudut yang tercangkum di pusat bulatan
Fo5rm Additional Mathematics Chapter 1 Circular Measure
CHAP. Concept
1
Circular Measure
Conversion of angle measurement
× 180°
π
Radian Degree
× π
180°
Arc length of a circle Area of sector of a circle
B s Q
A r
r
θ O θA s
r
Or P
s = qr, q in radian A= 1 qr2, q in radian
Sine rule and cosine rule 2
can be used if involving a
Heron’s formula can
triangle be used if involving a
triangle
Application
216
Chapter 1 Circular Measure Additional Mathematics Fo5rm
1.1 Radian Example 1 CHAP.
Relating angle measurement in radian and Convert the angle in the unit of radian to the 1
degree degree. [Use π = 3.142]
5π
1. In circular measures, the angle can be measured (a) 1.15 radian (b) 6 radian
in 2 units, which are
(a) degree (°) and minute (ʹ). Solution:
(b) unit of radian (in or not in the terms of π).
(a) π rad = 180° 180°
2. Radian involves the angle that related with the π
radius and circumference of a circle. 1.15 rad = 1.15 ×
3. The diagram below shows the angle in degree = 1.15 × 180°
and minutes and radian while the radius have 3.142
the same length. = 65.89°
(b) π rad = 180°
5π 5π 180°
6 rad = 6 × π
= 5 × 180°
6
= 150°
57° 17'
O 1 rad Alternative Method
O
Substitute π =180° into the expression,
5π 5(180°)
6 = 6 = 150°
In degree and minute In radian Try question 1 in Formative Zone 1.1
57° 17ʹ = 1 rad
4. 1 radian is a measurement of an angle subtended Example 2
about the centre of a cirlce such as the arc length
is the same as the length of radius of circle. Convert
(a) 30° into radian unit, in term of π.
A (b) 200° into radian unit.
[Use π = 3.142]
rr Solution:
O 1 radian B (a) 180° = π rad
r
30° = 30° × π
180°
π
5. Hence, angle subtended about the centre of a = 6 rad
circle, ˙AOB is 1 radian if the arc length of AB is
equal to the radius of the circle. Alternative Method
AB = OA = OB = r
Substitute 180° = π into the expression,
180°
30° = 6 = π rad
6
6. The relationship between the measurement of (b) 180° = π rad
angle in radian with degree is
2π rad = 360° 200° = 200° × π
π rad = 180° 180°
3.142
200° = 200° × 180°
7. The conversion of angle measurement in the degree = 3.49 rad
to the radian and vice versa are as follow:
Try question 2 and 3 in Formative Zone 1.1
× 180° Calculator
π
Recheck the answer in Example 2(b) by using
Radian Degree calculator,
1. Press 2 0 0 × SHIFT EXP ÷ 1 8 0 =
× π
180° 2. The screen will display 3.490658504
1.1.1 217
Fo5rm Additional Mathematics Chapter 2 Differentiation
y 6. In determining the turning point is whether
minimum point or maximum point, two method
—ddyx = 0 Maximum is used.
point
2CHAP. —ddyx > 0 —ddyx < 0 (a) Sketch of tangent method,
(b) Second order derivative.
y = f(x)
BRILLIANT Tips
0 x1 – ␦x x1 x1 + ␦x x • Sketching of tangent method is used to
determine the nature of stationary points.
Value for x x – dx x x + dx
(–) • Second order derivative is used to determine
Sign for dy (+) 0 the nature of turning points.
dx
A Sketching of tangent method
Sketch of Example 33
the tangent
Given the curve y = 5x 3 + 2x 2 – 3x.
B Minimum point (a) Find the coordinate of turning points for the
A stationary point is minimum when the gradient of
the curve is changes from negative to zero and then curve.
to positive. (b) Hence, determine whether the turning point is
y maximum point or minimum point.
y = f(x)
Solution:
(a) y = 5x 3 + 2x 2 – 3x
d y
dx = 15x 2 + 4x – 3
= (5x + 3)(3x – 1)
d y
—ddyx < 0 —ddyx > 0 Turning point, dx = 0
(5x + 3)(3x – 1) = 0
x = – 53 1
Minimum —ddxy = 0 and x = 3
point + ␦x x
( ) ( ) ( ) When x = – 53 , y = 5 – 35 3 + 2 – 53 2 – 3 – 53
0 x2 – ␦x x2 x2
Value for x x – dx x x + dx = 36
25
dy ( ) ( ) ( ) When x = 1 1 1 1
Sign for dx (–) 0 (+) 3 , y = 5 3 3+2 3 2–3 3
Sketch of = – 2176
the tangent ( ) ( ) – 53 36
Thus, the turning point is , 25 and
1 – 1267
C Point of inflection 3 , .
(a) Point of inflection is a stationary point that does
( )(b) At point – 53 , 36
not change in sign. 25
(b) This point is not included in turning points.
Value for x – 45 – 53 – 25
Positive Value for d y 17 0 – 151
gradient dx 5
Zero gradient
Positive Sign for d y + 0 –
gradient dx
Negative Zero gradient Sketch of
gradient Negative the tangent
gradient Sketch of
the graph
256 2.4.4
Chapter 3 Integration Additional Mathematics Fo5rm
Explaining relation between the limit of the y ∫A = b y dx
sum of areas of rectangles and the area y = f(x) a
under a curve
1. The area under curve y = f(x) can be determined A
by integration.
Oa b x
y
y = f(x)
A 2. For the value of the area bounded by the curve CHAP.
and the x-axis,
(a) if the region above the x-axis, then the 3
integral value is positive.
bx
Oa yy
2. Dividing the area under curve between x = a and
x = b into a few rectangles, dL.
y y = f(x) ␦x A A
O
y2 xO x
A2
A3 y3 y ␦A= y␦x
y1 A1 (b) if the region below the x-axis, then the
integral value is negative.
O ␦x
yy
For each of the following rectangles,
is dx = b – a
(a) the width rectangles. n , where n is the
number of xO x
OA A
(b) the height of the rectangle can be obtained
from the function of the curve, yi.
Area of each rectangle, dAi = Height × Width (c) if the region bounded by below the x-axis
≈ yyiid×x dx and also above the x-axis, then the area of
≈ the region should be determined separately.
Hence, the total area of the n triangles, y
≈ di ∑=nA11d+AidA2 + dA3 + … + dAn
≈
≈ i ∑=n 1yidx A x
Oa bB c
3. As the width of the strips of rectangle is become Area of shaded region
thinner, the width of each rectangle, dx becomes = Area of A + Area of B
narrower and approaching to zero dx ˜ 0.
∫ ∫ = b y dx + c y dx
Hence, ab
Area under the curve = lim ∑ ydx
dx ˜ 0
∫ = b y dx
a
Determining the area of a region BRILLIANT Tips
A Area of a region between the curve and the The negative sign only indicates that the region is
x-axis below of x-axis. Therefore, negative signs can be
eleminated by using modulus, |a|.
1. The area under the curve bounded by x-axis = a
and x-axis = b given by:
3.3.2 275
Fo5rm Additional Mathematics Chapter 3 Integration
Determining the area between the curve and Example 14
a straight line
Given the curve y = 16 is intersects the line y = 2x
1. at point P. x 2
CHAP.
The area of shaded region y y = 2x
3 y y = –1x–62
y = f(x)
P
y = g(x) A
ᕡ
ᕢ B O 5x
aO b
x Find the area of the shaded region.
Solution:
Area under the Area under y = 16 …1 y
straight line the curve x 2 y = –1x6–2
y y y = f(x) y = 2x …2
y = f(x) P y = 2x
4
y = g(x) A y = g(x) A 1 = 2: 16 = 2x
1 x 2
– 2x3 = 16
BB
2x x3 = 8
aO b x=2 x
aO b x When x = 2, y = 2(2) O 2 5
=4
Thus, P is (2, 4).
The shaded area = 1 – 2 Area of the shaded region
= Area of triangle + Area under the curve
∫ ∫ = b g(x) dx – b f(x) dx ( ) ∫=1 16
aa 2 × 2 × 4 + 5 x 2 dx
2
∫ = b [g(x) – f(x) dx] [ ]= 4 +
a – 1x6 5
2
[ ]= 4 +
2. – 1x6 5
2
The area of shaded region [ ( )]= 4 + – 156 – – 126
y 24
y = f(x) = 4 + 5
ᕡ
= 44 unit2
ᕢ y = g(x) 5
Oa bx Try question 6 in Formative Zone 3.3
Area under Area under the Example 15
the curve straight line
Given the curve y = x(4 – x) intersects the straight
y y line y = x at point A.
y = f(x)
y = f(x) – y
1 2 y = g(x)
y = g(x) 4 y=x
A
Oa bx Oa bx 2
The shaded area = 1 – 2 y = x(4 – x)
O 24
∫ ∫ = b f(x) dx – b g(x) dx Find
aa (a) the coordinate of point A.
(b) the area of the shaded region.
∫ = b [f(x) – g(x) dx]
a
278 3.3.3
Fo5rm Additional Mathematics Chapter 3 Integration
2. The generated volume of a solid is formed from a (c) Volume of each cylinder, dHVeiight of cylinder
revolved of x-axis is as follow: = Base area of cylinder ×
(a) Rotate an area of shaded region completely T==oππtaxxlii 22vd×oylduyme
through 360° about the x-axis until its of n cylinders
generate a solid, approximately a cylinder. (d)
= V1 + V2 + V3 + …Vn
yy
≈ i ∑=n 1dVi
CHAP. y = f(x)
3x
D EO x ≈ i ∑=n 1πxi 2dy
(b) Divide the solid into n vertical cylinders ␦y
with a thickness of dx. Radius of
y cylinder
Radius of
cylinder (e) When the number of cylinders is sufficiently
large, that is n ˜ ∞, then dy ˜ 0. Hence, the
␦x generated volume of the solid:
x
i ∑=n 1πxi 2dy
∫ lim = b πx 2dy
(c) Volume of each cylinder, dHVeiight
= Base area of cylinder × of cylinder dx ˜ 0 a
(d) =T=oππtayylii 22vd×oxlduxme of n cylinders Determining the generated volume of a
region revolved at the x-axis or y-axis
= V1 + V2 + V3 + …Vn A Generated volume, V through x-axis
≈ i ∑=n 1dVi y
≈ i ∑=n 1πyi 2dx y = f(x)
(e) When the number of cylinders is sufficiently a bO x
large, that is n ˜ ∞, then dx ˜ 0. Hence, the
generated volume of the solid: The generated volume of a region bounded by the
curve y = f(x) is revolved through 360° about the
∑=n 1πyi 2dx x-axis is given by:
∫ lim i = b πy 2dx
∫V = π b y 2 dx
dx ˜ 0 a a
3. The generated volume of a solid is formed from a Example 17
revolved of y-axis is as follow: Find the generated volume, in terms of π, when
(a) Rotate an area of shaded region completely the shaded region in each diagram is revolved
through 360° about the y-axis until its through 360° about the x-axis.
generate a solid, approximately a cylinder. (a)
yy y
y = —x2
D
y = f(x)
E
O x O x
x
(b) Divide the solid into n horizontal cylinders
with a thickness of dy. O1 3
280 3.3.4 3.3.5
Fo5rm Additional Mathematics Chapter 5 Probability Distribution
3. The diagram below shows the normal Determining and interpreting standard
distribution graph and standard normal score, Z
distribution graph.
1. Any continuous random variable X from normal
Area is distribution is standardised by:
the same
Z= X–m
s
a μb X za = a–σ–––μ– 0 zb = b–σ–––μ– z and produces the standard normal distribution ,
Z ~ N(0, 1), where
Normal distribution Standard normal X is a continuous random variable,
graph distribution graph m is mean and
s is standard deviation.
CHAP. 4. A continuous random variable, X can be 2. This changing process is known as distribution
standardised.
5 standardised by changing it to another
continuous random variable, Z as follow: Example 21
X ~ N(m, s 2) Find z-score if the value is X = 1 and X ~ N(12, 3).
where m = mean and
s = standard deviation Solution:
Standardised Z = X–m
s
X–m 1 – 12
Z= s = 3
Z ~ N(0, 1) = –3.367
where 0 = mean and
1 = standard deviation Try question 1 and 2 in Formative Zone 5.3
Example 22
5. The diagram below shows the percentage of data Given the mass of an apple and orange is normally
distribution graph for each standard normal distributed. Mean and standard deviation of apple
distribution. are 100 g and 15 g2 respectively while mean and
standard deviation of orange are 140 g and 25 g2
68.3% respectively.
of the data (a) Find the z-score if the mass of an apple 110 g
95.5% is chosen randomly.
of the data (b) If the standard score of an orange is –0.16,
99.7% determine the mass of orange.
of the data
Solution:
–3SD –2SD –1SD Mean +1SD +2SD +3SD
(a) Z = 110 – 100
From the graph above, 15
(a) 68.3% of the data lies within the standard = 0.667
z-score of apple is positive, which is 0.667.
deviation ±1 from the mean. This shows that the apple has mass more than
(b) 95.5% of the data lies within the standard the average apple.
deviation ±2 from the mean. (b) X– m = –1.6
(c) 99.7% of the data lies within the standard s
deviation ±3 from the mean. X – 140 = –1.6
25
X – 140 = –40
X = 100
Try question 3 in Formative Zone 5.3
326 5.3.2 5.3.3
Chapter 5 Probability Distribution Additional Mathematics Fo5rm
Determining the probability of an event for normal distribution
1. Besides using a scientific calculator or calculation, the probability of z-score for standard normal distribution
can also be determined by the standard normal distribution table.
2. The standard normal distribution table uses the concept that the probability of a normal distribution is given
by the area under the graph with the total area under the graph being 1 unit2.
3. Since the graph is symmetrical, P(Z > 0) = 0.5, the numeric table only gives the value of the area to the right
starting with 0.5 which is for P(Z . 0) = 0.
4. The table below shows the standard normal distribution table.
Upper Tail Probability Q(z) of the Normal Distribution N(0, 1)
z0 1 2 3 4 5 6 7 8 9 123456789
Substract
0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 4 8 12 16 20 24 28 32 36 CHAP.
0.1 0.4602 0.4562 0.4522 0.4483 0.443 0.4404 0.4364 0.4325 0.4286 0.4247 4 8 12 16 20 24 28 32 36 5
0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 4 8 12 15 19 23 27 31 35
0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.557 0.3520 0.3483 4 7 11 15 19 22 26 30 34
0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 4 7 11 15 18 22 25 29 32
0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 3 7 10 14 17 20 24 27 31
0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.02546 0.2514 0.2483 0.2451 3 7 10 13 16 19 23 26 29
0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 3 6 9 12 15 18 21 24 27
0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 3 5 8 11 14 16 19 22 25
0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1634 0.1611 3 5 8 10 13 15 18 20 23
1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 2 5 7 9 12 14 16 19 21
1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 2 4 6 8 10 12 14 16 18
1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 2 4 6 7 9 11 13 15 17
1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 2 3 5 6 8 10 11 13 14
1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 1 3 4 6 7 8 10 11 13
1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 1 2 4 5 6 7 8 10 11
1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 1 2 3 4 5 6 7 8 9
1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 1 2 3 4 4 5 6 7 8
1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 1 1 2 3 4 4 5 6 6
1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 1 1 2 2 3 4 4 5 5
2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 0 1 1 2 2 3 3 4 4
2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 0 1 1 2 2 2 3 3 4
2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 0 1 1 1 2 2 2 3 3
2.3 0.0107 0.0104 0.0102 011112222
0.0099 0.00964 0.0093 0.00991 3 5 8 10 13 15 18 20 23
0 94
0.0088 0.0088 0.0084 2 5 7 9 12 14 16 16 21
9 66 2
2.4 0.0082 0.0079 0.0077 0.0075 0.0073 2 4 6 8 11 13 15 17 19
0865 4
0.0017 0.00695 0.006 0.0065 0.0063 2 4 6 7 9 11 13 5 17
4 76 7 9
2.5 0.0062 0.0060 0.0058 0.0057 0.0055 0.0053 0.00523 0.0050 0.0049 0.0048 2 3 5 5 8 9 11 12 14
1470 4 9 840
2.6 0.0046 0.0045 0.0040 0.0042 0.0041 0.0040 0.00391 0.0037 0.0036 0.0035 1 2 3 4 6 7 9 9 10
63 752 987
2.7 0.0034 0.0033 0.0032 0.0031 0.0013 0.0029 0.00289 0.0028 0.0272 0.0026 1 2 3 4 5 6 7 8 9
7 6 6 7 07 8 04
2.8 0.0025 0.0024 0.0024 0.0023 0.0022 0.0021 0.00212 0.0020 0.0019 0.0019 1 1 2 3 4 4 5 6 6
6803 6 9 593
2.9 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.00154 0.0014 0.0014 0.0013 0 1 1 2 2 3 3 4 4
7159 4 9 949
3.0 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.00111 0.0010 0.0010 0.0010 0 1 1 2 2 2 3 3 4
5162 8 4 740
5.3.4 327
Fo5rm Additional Mathematics Chapter 5 Probability Distribution
5. If Z is a standard normal distribution with mean 8. If the probability is between z-score, which is a
is 0 and standard deviation is 1, thus the value and b:
of z gives the probability of standard normal
distribution. f (z)
f (z)
Oa b z
Oa z P(a , Z , b) = P(Z . a) – P(Z . b)
Probability by using Standard Example 23
CHAP. Normal Distribution Table
Given Z is a continuous random variable with a
5 bit.ly/2X8SFlz standard normal distribution. Find
(a) P(Z . 0.254)
6. If given the probability is more than z-score and (b) P(Z > 1.056)
(a) the value of a is positive, P(Z . a). (c) P(Z , –1.386)
(d) P(Z . –2.337)
f (z) (e) P(Z , 2.337)
(f) P(0 , Z , 1.242)
Oa z (g) P(–2.46 , Z , 1.281)
(h) P(–2.149 , Z , –0.214)
(b) the value of a is negative, P(Z . –a). (i) P(1.331 , Z , 2.147)
(j) P(| Z | . 1.471)
f (z)
Solution:
–a O z (a)
P(Z . –a) = 1 – P(Z , –a) f (z)
= 1 – P(Z . a)
7. If given the probability is less than z-score and 0 0.254 z
(a) the value of a is negative, P(Z , –a)
From the standard normal distribution table,
f(z) f(z)
=
–a O z Oa z z 54
Substract
P(Z , –a) = P(Z . a)
0.2 0.4013 15
(b) the value of a is positive, P(Z , –a)
f (z) P(Z . 0.25) = 0.4013
P(Z . 0.254) = 0.4013 – 0.0015
Oa z
= 0.3998
P(Z , a) = 1 – P(Z . a) Thus, P(Z . 0.254) = 0.3998.
328 5.3.4
Tg 4 CHAPTER Trigonometric Functions
6
SMART Important Learning Standards Page
• Represent positive angles and negative angles in a Cartesian Plane. 343
6.1 Positive Angles and
Negative Angles • Relate secant, cosecant and cotangent with sine, cosine and tangent 345
of any angle in a Cartesian plane. 346
6.2 Trigonometric
Ratios of any Angle • Determine the values of trigonometric ratios for any angle. 351
6.3 Graphs of Sine, • Draw and sketch graphs of trigonometric functions: 355
Cosine and (i) y = a sin bx + c
Tangent Functions (ii) y = a cos bx + c 357
(iii) y = a tan bx + c 358
6.4 Basic Identities where a, b and c are constants and b . 0. 358
360
6.5 Addition Formulae • Solve trigonometric equations using graphical method. 362
and Double Angle
Formulae • Derive basic identities:
(i) sin2 A + cos2 A = 1
6.6 Application of (ii) 1 + tan2 A = sec2 A
Trigonometric (iii) 1 + cot2 A = cosec2 A
Functions
• Prove trigonometric identities using basic identities.
• Prove trigonometric identities using addition formulae for
sin (A ± B), cos (A ± B) and tan (A ± B).
• Derive double angle formulae for sin 2A, cos 2A and tan 2A.
• Prove trigonometric identities using double-angle formulae.
• Solve trigonometric equations. 364
• Solve problems involving trigonometric functions. 366
Words • Complementary angle / Sudut pelengkap
• Basic identities / Identiti asas
• Degree / Darjah • Addition formulae / Rumus sudut majmuk
• Radian / Radian • Double angle formulae / Rumus sudut berganda
• Trigonometric ratio / Nisbah trigonometri • Half angle formula / Rumus sudut separuh
• Quadrant / Sukuan
• Reference angle / Sudut rujukan
341
Fo5rm Additional Mathematics Chapter 6 Trigonometric Functions
Concept
Trigonometric Function
Positive Angle Negative Angle Basic Identities
• sin2 A + cos2 A = 1
y y • 1 + tan2 A = sec2 A
Positive • 1 + cot2 A = cosec2 A
angle
θx θx
Negative
angle Addition Formulae
• sin (A B) = sin A cos B cos A sin B
6 ( ) ( )CHAP. • cos (A B) = cos A cos B sin A sin B
x° = x × π rad, q rad = q× 180 ° tan A tan B
180 π 1 tan A tan B
• tan (A B) =
Trigonometric Ratios Graph of Trigonometric
1 Functions
• sin A = cosec A Double Angle Formulae
• Sine, y = sin x • sin 2A = 2 sin A cos B
• cosec A = 1 • cos 2A = cos2 A – sin2 A
sin A y
• cot A = 1 = 2 cos2 A – 1
tan 1 = 1 – 2 sin2 A
A
• tan 2A = 2 tan A
sin cos –2π –3 2–π –π – ––2π1 0 2–π π 3 –2π 2π x 1 – tan2 A
tan 1 cot • Cosine, y = cos x Half Angle Formulae
y !• sin q =± 1 – cos q
2 2
sec cosec 1
x !• cos q =± 1 + cos q
2 2
–2π –3 2–π –π – ––2π1 0 2–π π 3 –2π 2π sin q 1 – cos q
• tan q = 1 + cos q = sin q
Complementary Angles 2
• sin q = cos (90° – q) • Tangent, y = tan x
• cos q = sin (90° – q)
• tan q = cot (90° – q) y
• cot q = tan (90° – q)
• sec q = cosec (90° – q) 1 • a = amplitude
• cosec q = sec (90° – q) • b = number of cycle in the range
x 0 < x < 2π for sine and
–2π –3 2–π –π – –2π 0 –2π π 3 –2π 2π
–1 cosine graph whereas
0 < x < π for tangent graph.
( )• c = translation 0
graph. c from basic
Method to Find Trigonometric Ratios • Period: (i) 2π for sine and
• Use calculator b
• Use unit circle cosine graph.
• Use trigonometric ratio of
(ii) π for tangent graph.
corresponding reference angle b
• Use right-angled triangles
342
Chapter 6 Trigonometric Functions Additional Mathematics Fo5rm
A Using calculator The following is the trigonometric ratios by using
For the value of cosecant, secant and cotangent, the x-coordinate and y-coordinate in unit circle
value can be calculated by using the reciprocal of the
following ratios. sin q = coordinate-y
cos q = coordinate-x
y-coordinate
cosec q = 1 q, sec q = 1 q, cot q = 1 q tan q = x-coordinate
sin cos sin
Example 8 Example 10
Find the value of (b) sec 505° (–0.643, 0.766) y
(a) sin 235°
Solution: (0.643, 0.766)
(a) sin 235° = –0.8192 130°
Calculator 50° x
1. Press sin 2 3 5 =
2. The screen will display –0.819152044 Based on the unit circle above, state the following CHAP.
value.
(a) sin 50° (b) cos 130° 6
(b) sec 505° = 1
cos 505° (c) cosec 130° (d) cot 50°
= –1.221 Solution:
Try question 3 in Formative Zone 6.2 (a) sin 50° = y-coordinate
= 0.766
(b) cos 130° = x-coordinate
Example 9 = –0.643
Find the value of (c) cosec 130° = sin 1
(a) tan 2.4 rad 130°
Solution: (b) cosec 2 π rad 1
5 = 0.766
(a) tan 2.4 rad = –0.9160 = 1.305
x-coordinate
Calculator (d) cot 50° = y-coordinate
Make sure MODE of calculator is in radian. = 0.643
1. Press tan 2 . 4 0.766
= 0.839
2. Press = and the screen will display Try question 4 in Formative Zone 6.2
–0.916014289 C Using trigonometric ratio of corresponding
reference angle
(b) cosec 2 π rad = 1
5 2 1. The value of trigonometric ratio of any angle
sin 5 π rad can also be determined using the value of
the trigonometric ratio of the corresponding
= 1.0515 reference angle of the angle.
Try question 3 in Formative Zone 6.2 2. The reference angle, a, is the acute angle formed
by rotating the ray OP and the x-axis on the
B Using unit circle Cartesian plane.
y y
P (x, y) OP2 OP1
1 x αx
θy
Ox
OP3 OP4
6.2.2 347
Fo5rm Additional Mathematics Chapter 6 Trigonometric Functions
Example 19 Example 20
Given f(x) = 4 cos 2x for 0 < x < 2π. Sketch the graph of the following trigonometric
(a) State the period of the graph function y = f(x). functions in the given range.
(a) y = sin x + 1 for 0 < x < 2π
Hence, state the number of cycle of the graph (b) y = –2 cos x for 0 < x < 2π
in the given range.
(b) State the amplitude of the graph. (c) y = | tan x | for 0 < x < 2π
(c) Write the coordinates of the maximum and the (d) y = | cos 2x | + 1 for 0 < x < 2π
minimum points.
(d) Sketch the graph of y = f(x). Solution:
(e) Using the same axes, sketch the graph of
(a) y = sin x + 1 for 0 < x < 2π
function y = –|4 cos 2x| for 0 < x < 2π.
1 Sketch the basic graph, y = sin x.
Solution: 2 The graph moves 1 unit upward, such that
Compare f(x) = 4 cos 2x with the basic cosine ( ) 0
function, f(x) = a cos bx + c. 1
2π translation .
2
(a) Period = π or 180°. Number of cycle, b = 2.
(b) Amplitude, a = 4 y
(c) Maximum point: (0, 4), (π, 4) and (2π, 4). ᕢ
( ) ( )CHAP. 3π
Minimum point: π , – 4 and 2 , – 4 2 y = sin x + 1
6 (d) To sketch graph function y = 4 cos 2x:2
1
Number of class = 2 × 2 × 2 = 8 O x
2π π –1 2–π π –32–π ᕡ 2π
Size of class interval = 8 = 4 y = sin x
x0 π π 3π 2π –2
2 2
y 4 – 4 4 – 4 4 (b) y = –2 cos x for 0 < x < 2π
1 Sketch graph of y = cos x.
Thus, the graph function of y = 4 cos 2x: 2 Reflect the graph at 1 on x-axis to make
y the graph of y = – cos x.
4 y = 4 cos 2x y
2
O –2π π –32–π 2π x 2
–2
1 y = – cos x ᕡ
–4 O 2π x
–1 ᕢ
(e) Steps in sketching the graph of y = –|4 cos 2x| –2π π –32–π
y = cos x
1 y = |4 cos 2x| is a reflection of graph on
–2
negative side of x-axis.
3 The value of a is –2. The maximum value is
2 y = –|4 cos 2x| is a reflection of graph at 1 (π, 2) and the minimum value is (0, –2) and
on x-axis.
(2π, –2).
y
4 y
2
2 ᕣ y = –2 cos x
O 2–π π –32–π x 1 y = – cos x
–2 2π
O –2π π –32–π 2π x
–1 ᕢ
–4 y = –|4 cos 2x|
Try question 3 in Formative Zone 6.3 –2
354 6.3.1
Chapter 7 Linear Programming Additional Mathematics Fo5rm
7.1
1. Shade the region which represented by the (c) In a week, Jason makes x cupboard P and y
following linear inequality. C1 cupboard Q. He has a capital of RM2 000.
1 2 The cost of making a cupboard P is RM200
(a) y < 2 x + 3 (b) y > 5 x – 1 and a cupboard Q is RM100.
yy 4. Represent each of the following linear
inequalities graphically. C3
6 3 (a) x > 4
y,8
y = –12x + 3 4 y = –52x – 1 2 (b) x , 7
2 1 y>1
(c) x > –2
–6 –4 ––220 2 x –3 –2 ––110 1 x y > 4
(d) x + y > 2
(c) y , 8 x – 5 (d) y < – 94 x – 3 y,6
3 (e) x + y < 4
y x + y . 1
y (f) x > y
x . 1
2 –8 –6 –4 ––210 x (g) y < 2x
y > x + 2
––220 246 x (h) y > 2x
–4 y = –83x – 5 y < 3x
–6 y = – –94x – 3 –2 CHAP.
–3
7
–4
2. Determine the linear inequality which satisfies 5. Shade the region R which satisfied the
each of the following regions. C2 following inequalities. C2
(a) (b) x>0
y>x
y y y < 2x + 1
x + y < 8.
4 y = –32x + 2 –73x 6
2 x
y = + 5 4
–4 ––220 24 2
–4 x=0
–3 –2 ––120 x 8 y = 2x + 1
1 6 y=x
(c) (d)
y y 4
2 x+y=8
4 4 0 2 4 6 8 y=0
2
3 y = – –83x + 4
2 –2–20 246 x
–4 y = –25x – 4
1
–10 x
123
3. Write a mathematical model for each of the 6. Draw y > 2x + 1 and y , 2x – 3 on the
following situations. C1 same graph. Is there any solution for these
(a) The number of participants for course A inequalities? C2
is at most three times than the number of
participants for course B. 7. Is the point (2, 1) a solution of the inequalities
(b) The number of male workers exceed the x + y . 1 and 2x + y , 8? C2
number of female workers by at most 40.
383
Fo5rm Additional Mathematics Chapter 8 Kinematics of Linear Motion
2. The diagram below shows the movement of (c) Displacement is maximum when v = 0, which is
an object thrown by Ahmad and its value of t = 4 s.
displacement and velocity with respect to the time. When t = 4, s = 4 + 8(4) – (4)2
s = 4 + 32 – 16
s = 20 m
t =2s Maximum Thus, the maximum displacement of the
s = 20 m displacement particle is 20 m.
v =0 t =3s (d) Velocity decreases, v , 0
t =1s s = 15 m 8 – 2t , 0
s = 15 m v = –10 m/s 2t . 8
v = +10 m/s t.4
t =4s Thus, the range of value t is t . 4 s.
Initial velocity s =0 (e) When the particle moves to the right, v . 0
t =0 v = –20 m/s 8 – 2t . 0
s =0 2t , 8
v = +20 m/s
t,4
Thus, the time interval of the particle moves to
the right is t , 4 s.
• Particles stop, so velocity, v = 0 Try question 4 in Formative Zone 8.2
• For maximum or minimum displacement, Example 12
ds
dt = v = 0 A particle moves along a straight line with its
displacement, s meters, from a fixed point O given
• Particles return or through a fixed point O, by s = t 3 – 12t 2 + 36t + 10, where t is the time, in
s=0 seconds. [Assume motion to the right as positive]
CHAP. • Particles change their direction of movement, Find
8 v=0 (a) the initial velocity, in m s–1, of the particle,
(b) the values of t, in seconds, when the particle
Example 11
stops instantaneously,
A particle moves along a straight line with its (c) the range of value t when the velocity is
displacement, s meters, from a fixed point O given
by s = 4 + 8t – t 2, where t is the time, in seconds. negative.
[Assume motion to the right as positive]
Find Solution:
(a) the initial velocity, in m s–1, of the particle,
(b) the value of t, in seconds, when the particle Given s = t 3 – 12t 2 + 36t + 10.
ds
stops instantaneously, Velocity function, v = dt = 3t 2 – 24t + 36
(c) the maximum displacement of the particle,
(d) the range of values t values when the velocity (a) Initial velocity, t = 0 v = 3(0) 2 – 24(0) + 36
= 36
decreases, Thus, the initial velocity of the particle is 36 m s–1.
(e) the time interval when the particle moves to (b) Particle stops, v = 0
3t 2 – 24t + 36 = 0
the right. (t – 6)(t – 2) = 0
t = 6 or t = 2
Thus, the particle stops at t = 6 s and t = 2 s.
Solution: (c) Velocity is negative, v , 0
Given s = 4 + 8t – t 2. ds 3t 2 – 24t + 36 , 0
dt
Velocity function, v = = 8 – 2t (t – 6)(t – 2) , 0
Let (t – 6)(t – 2) = 0
(a) Initial velocity, t = 0 t = 6 or 2
When t = 0, v = 8 – 2(0)
= 8
Thus, the intial velocity of the particle is 8 m s–1.
(b) Particle stops, v = 0
8 – 2t = 0 26 t
2t = 8 Thus, the range of value t is 2 , t , 6 s.
t=4
Thus, the particle stops at t = 4 s. Try question 5 in Formative Zone 8.2
404 8.2.2
Fo5rm Additional Mathematics Chapter 8 Kinematics of Linear Motion
Example 20 When t = 5, s = 2(5) 3 – 6(5) 2 + 10(5)
3
A particle moves along a straight line through –16 32
a fixed point O with velocity 10 m s–1. The s = m
acceleration, a m s–2, is given by a = 4t – 12 where Number line,
t is the time, in seconds, after passing through
a fixed point O. [Assume motion to the right as t=0
positive] Find
(a) the time, in seconds, when the acceleration is t=1
zero, –16 –23 t = 5 0 4 2–3
(b) the minimum velocity, in m s–1, of the particle,
(c) the time, in seconds, when the particle stop Total distance travelled = 4 32 + 4 2 + 16 2
3 3
momentarily, = 26
(d) the total distance, in m, travelled by the particle
Thus, in the first 5th seconds, total distance
in the first 5 seconds. travelled of particle is 26 m.
Solution:
Alternative Method
Given the acceleration function, a = 4t – 12 (A) Using velocity-time graph
∫ ∫Velocity function, v = a dt = 4t – 12 dt Distance can be determine by finding the
area under the curve.
= 2t 2 – 12t + c v = 2t 2 – 12t + 10
When t = 0, v = 10
10 = 2(0)2 – 12(0) + c t 015
c = 10 v 10 0 0
So, v = 2t 2 – 12t + 10. Velocity-time graph for 0 < t < 5:
∫Displacement function, s = v dt v
= 2t 2 – 12t + 10 dt
∫CHAP.
8 = 2
When t = 0, s = 0 3 t 3 – 6t 2 + 10t + c 10
0 = 2(0) 3 – 6(0) 2 + 10(0) + c v = 2t2 – 12t + 10
3 A
c = 0 01 B 5
2
So, s = 3 t 3 – 6t 2 + 10t. t
(a) Acceleration is zero, a = 0
4t – 12 = 0
4t = 12 –8
t=3
Thus, the acceleration of the particle is zero at
t = 3 seconds. 1
dv
(b) Minimum velocity, dt = 0, a = 0 ∫Area A = 2t 2 – 12t + 10 dt
0
4t – 12 = 0 [ ] = 2 t 3 – 6t 2 + 10t 1
3 0
2
[ ] 3 – –0
t=3 = (1) 3 6(1) 2 + 10(1)
When t = 3, v = 2(3)2 – 12(3) + 10 = 4 32 m2
= –8
Thus, the minimum velocity of the particle is 5
–8 m s–1. ∫Area B = 2t 2 – 12t + 10 dt
(c) Particle stops, v = 0 1
2t 2 – 12t + 10 = 0
(t – 1)(t – 5) = 0 [ ]= 2 t 3 – 6t 2 + 10t 5
3 1
2 2
[ ] 3 – 3
t = 1 and t = 5 = (5) 3 – 6(5) 2 + 10(5) 4
Thus, the particle stops at t = 1 s and t = 5 s. = –21 1
2(0) 3 3
(d) When t = 0, s = 3 – 6(0) 2 + 10(0) = 21 31 m2
distance travelled
s=0m Total = Area A + Area B
2(1) 3
When t = 1, s = 3 – 6(1) 2 + 10(1) = 4 2 + 21 1
3 3
s = 4 2 m = 26 m
3
410 8.4.1
Chapter 8 Kinematics of Linear Motion Additional Mathematics Fo5rm
(B) Using displacement-time graph When t = 5, v = 6(5) – 18
2t 3 = 12
s = 3 – 6t 2 + 10t Thus, the acceleration when the particle stops
ds = 0, v = 0 is –12 m s–2 and 12 m s–2.
dt (b) The particle changes direction from point A to
t = 1 and t = 5 (Turning point)
2 B, v = 0
When t = 1, s = 4 3 From (a), it is known the particle changes the
s =
When t = 5, –16 2 direction at t = 1 and t = 5. So, when t = 1,
When s = 0, 3 the particle at point A whereas when t = 5, the
particle at point B.
( ) 23t 3 – 6t 2 + 10t = 0 t2t 2=0 Velocity-time graph:
3 – 6t + 10
v
t = 0, t = 2.21, t = 6.79 15
Displacement-time graph:
v = 3t2 – 18t + 15
s
1 4 3–2 2 5 6.79 t 01 5t
0 1 2.21 Point A Point B
–12
3 Distance AB = Area under the curve
5
–16 –32
∫= (3t 2 – 18t + 15) dt
1
Total distance travelled = 1 + 2 + 3 [ ] = t 3 – 9t 2 + 15t 5 CHAP.
1
= [(53 – 9(5)2 + 15(5)] – 8
2 2 –16 2 [(13 – 9(1)2 + 15(1)]
= 4 3 + 4 3 + 3
= 26 m = –25 – 7
= –32
Try question 3 and 4 in Formative Zone 8.4 = 32 m
Example 21 ∫(c) Displacement function, s = v dt
A particle moves along a straight line through a ∫= (3t 2 – 18t + 15) dt
fixed point P. The velocity of the particle, v m s–1,
is given by v = 3t 2 – 18t + 15 where t is the time, in = 3t 3 – 18t 2 + 15t + c
seconds, after passing through a fixed point P. The 3 2
particle begins to change direction of movement
at points A and B. [Assume motion to the right as = t 3 – 9t 2 + 15t + c
positive]. Find Initial displacement, v = 0 and t = 0
(a) the acceleration, in m s–2, when the particle 0 = (0)3 – 9(0)2 + 15(0) + c
c=0
stops, So, s = t 3 – 9t 2 + 15t.
(b) the distance between point A and B, Velocity decreases, a , 0
(c) the distance when a particle moves with 6t – 18 , 0
6t , 18
decreasing velocity. t,3
Solution: v
(a) Given velocity function, v = 3t 2 – 18t + 15
Particle stops, v = 0
3t 2 – 18t + 15 = 0 15
(t – 1)(t – 5) = 0 t
t = 1 and t = 5 O1 3 5
dv v = 3t 2 – 18t – 15
Acceleration function, a = dt
= d (3t 2 – 18t + 15) t = 1, s = (1)3 – 9(1)2 + 15(1) = 7
dt t = 3, s = (3)3 – 9(3)2 + 15(3) = –9
The distance travelled = 7 + 7 + 9 = 23 m.
= 6t – 18
When t = 1, a = 6(1) – 18 Try question 5 and 6 in Formative Zone 8.4
= –12
8.4.1 411
SPM MODEL PAPER
Paper 1
Time: 2 hours
Section A
(64 marks)
Instruction: Answer all questions
1. (a) Diagram 1 shows a part of graph for a (ii)
function of y = f(x).
y
y = f(x)
02 x
–3
SPM MODEL PAPER Diagram 1
State whether the function of f
(i) is a discrete or continuous,
(ii) has an inverse function or not. 2. (a) Diagram 2 shows a graph of function f for
[2 marks] domain 0 < x < 4 and its inverse function
a f –1.
(b) A function f is derived by f : x ˜ x , x ≠ 0
y
such as a is a constant. Given f –1( 2) = 2 , find A (4, 12)
(i) the value of a,
(ii) f 17(8). [3 marks]
f
Answer: f–1 B
(a) (i)
0x
(ii) –4
Diagram 2
Based on the graph, determine
(i) the domain of f –1,
(b) (i) (ii) the coordinate of point B on the graph
of f –1 that corresponding with the point
A on the graph of f. [2 marks]
Answer:
(i)
(ii)
418
(b) On the answer space below, sketch the graph SPM Model Paper Additional Mathematics
of y = |2x – 5| for 0 < x < 6. Hence, find the
range of value x such as y < 3. [3 marks] (b) The height, h m, of a hit golf ball from a
y surface of field after t seconds is given by the
function h(t) = 20t – 5t2. Find
(i) the time taken, t when the golf ball
touches the surface of the field again.
(ii) the maximum height, h reached by the
golf ball.
Hence, in the answer space below,
sketch the graph of h against t for the
movement of the golf ball. [3 marks]
Answer:
(b) (i)
0 x
3. (a) If a and b are the roots for quadratic (ii) SPM MODEL PAPER
equations x 2 + px + q = 0, such as p and q
are constant, express a 2b + b 2a in terms of
p and q. [2 marks]
Answer:
(a)
h (meter)
0 t (seconds)
419
SPM Model Paper Additional Mathematics
Section B
(16 marks)
Instruction: Answer any two questions only
13. (a) Given that y = (2x + 5)! 4x – 5 .
dy = ax such as a is a constant and state the value of a.
dx ! 4x – 5
∫(i) Show that
(ii) Hence, find the value of x dx. [4 marks]
! 4x – 5
(b) Diagram 8 shows a part of curve y = 4 , line x = 1, x = 2 and y = 4.
x 2
y
5 x=1 y=4
4
3Q
2 x=2
1 P y = —x42 x
0 12345
SPM MODEL PAPER
Diagram 8
(i) Find the area of region P, in unit2, bounded by the curve y = 4 , x-axis and line x =1 and x = 2.
(ii) Find the area, in unit2, of region Q x 2 4 and lines x= 2 and y = 4.
that bounded by the curve of y = x 2
[4 marks]
Answer: (ii)
(a) (i)
(b) (i) (ii)
425
Additional Mathematics SPM Model Paper
Paper 2
Time: 2 hours and 30 minutes
Section A
(50 marks)
Instruction: Answer all the questions
1. A married couple has three sons.At one time, the eldest child is 4 years older than the total age of his younger
bNreoxtthseervs.eTnwyoeayresalrastearg,oa,gteheofagtheroefetohfetyhoeumngises5t1cyheiladrswoalsd.41 of the age difference between his older brothers.
Determine the age of each son of the couple.
[6 marks]
2. Diagram 2 shows the mapping x to y which drerived by the function of f : x ˜ 9x – a and mapping of y to z
is derived by the function of g : y ˜ b y , y ≠ 12.
12 –
xy z
SPM MODEL PAPER 33
4
Diagram 2 [6 marks]
(a) Find the values of a and of b.
(b) Express the function that maps the element x to the element z in a similar form.
(c) Determine the element x that does not change when mapped to the z.
3. (a) The roots of equation 3x 2 – 2kx + k + 4 = 0 are a and b. If a2 + b 2 = 16 , find the possible value of k.
9 [3 marks]
(b) Diagram 3 shows a part of graph function f(x) = (x – 3)2 – 2 with minimum point A(h, 2k) and intersect
the f(x)-axis at 7. This graph is moved by 3 units to the right and 4 units upwards with new minimum
point, B as shown in the Diagram 3.
f(x)
7 [4 marks]
f(x) = (x – 3)2 – 2
B
0
A (h, 2k)
Diagram 3
Determine
(i) the value of h and of k,
(ii) the coordinate of minimum point, B,
(iii) the new equation of f(x) in the vertex form.
428
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