modul tok jiring
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MODUL ADD MATH TOK JIRING 1 by CYY
KEM TEKNIK MENJAWAB
MATEMATIK TAMBAHAN
SMKA TOK JIRING
KUMPULAN HAK
(50 MARKAH)
2013
TOPICS
1. Functions
2. Quadratics Equation/Functions
3. Simultaneous Equation
4. Indices & Logaritme
5. Geometry Coordinate
6. Index Number
7. Solutions Of Triangles
8. Trigonometric Functions
DISEDIAKAN OLEH : CIKGU YUSRI BIN YAAHMAT
yusriyaahmat.blogspot.com
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MODUL ADD MATH TOK JIRING 2 by CYY
ALGEBRA
1. x = 2 4
2
b b ac
a
− ± −
2. am × a
n = a
m + n ----pan-pen
3. am ÷ a
n = a
m − n ----pan-pen
4. (am)n = a
m n
5. loga mn = loga m + loga n-----pan-pen
6. loga n
m = loga m − loga n -----pan-pen
7. loga mn = n loga m-----kuda pan
8. log
loglog
ca
c
bb
a= ---- kuda
9. ( 1)nT a n d= + − ------- kaki atok gbai
10. [2 ( 1) ]2
n
nS a n d= + −
11. 1n
nT ar
−=
12. nS = 1
)1(
−
−
r
ran
= r
ra n
−
−
1
)1(, 1≠r
13. ∞S = r
a
−1, | r | < 1
KALKULUS
1. y = uv, dx
dy = u
dx
dv + v
dx
du-----sida
2. y = v
u,
dx
dy =
2v
dx
dvu
dx
duv −
3. dx
dy =
du
dy ×
dx
du----3 p/u
4. Area under a curve
=
b
a
y dx∫ or = ∫b
a
dyx
5. Volume generated
= ∫b
a
dxy 2π or = ∫b
a
dyx2π
STATISTIK
1. x = N
x∑----tiada f
2. x = ∑
∑
f
fx
3. σ = N
xx∑ − 2)( =
22)(
xx
N
∑− tiada f
4. σ = ∑
∑ −
f
xxf 2)( =
22)(
fxx
f
∑−
∑
5. m = CLmf
FN
+
−21
6. I = 0
1
Q
Q × 100-2 thn
7. I = ∑∑
i
ii
W
IW----fungsi gubahan
8. rn P =
!)(
!
rn
n
−
9. rnC =
!!)(
!
rrn
n
−
10. P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
11. )( rXp = = rnrr
n qpC − , p + q = 1
12. Mean / Min = np
13. σ = npq
14. Z = σ
µ−X
=0 > 0
< 0 ≥ 0
Paksi-y
Paksi-x
1
1
4Q N=
3
3
4Q N=
DaTo TaBah
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MODUL ADD MATH TOK JIRING 3 by CYY
GEOMETRI
1. Distance = 221
221 )()( yyxx −+−
2. Midpoint
(x, y) =
++
2,
2
2121 yyxx
3. A point dividing a segment of a line
(x, y) = 1 2 1 2,nx mx ny my
m n m n
+ +
+ +
4. Area of triangle / Luas segi tiga
)()( 3123121332212
1yxyxyxyxyxyx ++−++
5. r = 22yx +
6. r̂ = 22
yx
yx
+
+ ji
TRIGONOMETRI
1. Arc length, s = rθ
2. Area of sector = 2
1 2r θ
3. AA 22 cossin + = 1
4. A2sec = A2tan1 +
5. A2cosec = A2cot1 +
6. sin 2A = 2 sinA cosA
7. cos 2A = cos2 A − sin
2 A
= 2 cos2 A − 1
= 1 − 2 sin2 A
8. )(sin BA ± = sinA cosB ± cosA sinB
9. )(cos BA ± = cosA cosB m sinA sinB
10. )(tan BA ± = BA
BA
tantan1
tantan
m
±
11. tan 2A = A
A2tan1
tan2
−
12. A
a
sin =
B
b
sin =
C
c
sin
13. a2 = b
2 + c
2 − 2bc cosA
14. Area of triangle = 2
1 ab sin C
FORMULA TAMBAHAN
1. x2 – (SOR)x + POR = 0 2 b c
xa a
− + = 0
2. f (x) = 2 2
2 4
b ba x c
a a
+ + −
3. 100
X YZ
×=
Z
Year C
X
Year A
Y
Year B
Sila t2
Vektor unit
SiLa
Anak
Panah
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MODUL ADD MATH TOK JIRING 4 by CYY
FUNCTIONS
1 :
Given that 43: −→ xxf and xxg 2: → ,
find fg(3).
Answer : f(x) = 3x - 4 , g(x) = 2x
g(3) = 2( )
= ( )
fg(3) = f [ g(3) ]
= f ( )
= 3 ( ) - 4
= ( )
2 :
Given that xxf 23: −→ and 2: xxg → , find
gf(4).
Answer : f(x) = 3 – 2x , g(x) = x2.
f( ) = 3 – 2( )
= ( )
gf(4) = g ( )
= ( )2
= ( )
3:
Given that 23: −→ xxf , find f 2(2).
Answer : f(x) = 3x - 2
f(2) = 3( ) – 2 = ( )
f2(2) = f [ f(2) ]
= f ( )
= 3 ( ) – 2
= ( )
4:
Given that xxg 43: −→ , evaluate gg(1).
Answer : g(x) = 3 – 4x
g(1) = 3 – 4( ) =
gg(1) = g [g(1)]
= g ( -1)
= 3 – 4 ( )
= ( )
5 :
Given that 43: −→ xxf and xxg 2: → ,
find fg(x).
Answer : f(x) = 3x - 4 , g(x) = 2x
fg(x) = f [ g(x) ]
= f ( )
= 3 ( ) - 4
= 6x – 4
6 :
Given that xxf 23: −→ and 2: xxg → , find the
composite fuction gf.
Answer: f(x) = 3 – 2x , g(x) = x2.
gf(x) = g[f(x)]
= g ( )
= (3- 2x)2
=
EXAMPLE 1 :
Given that f(x) = 4x – 6 , find
÷ x + ( )
∴ f –1
(x) = 4
6+x
EXAMPLE 2 :
Given that f(x) = 2x + 3 , find f –1
(x).
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MODUL ADD MATH TOK JIRING 5 by CYY
QUADRATIC EQUATIONS/FUNCTIONS
1
Given that the roots of the quadratic equation
2x2 + (p+1)x + q - 2 = 0 are -3 and ½ . Find the value
of p and q.
x = -3 , x = ½
x + 3 = 0 or 2x – 1 = 0
(x + 3) ( 2x – 1) = 0
2x2 + 5x – 3 = 0
Comparing with the original equation :
p + 1 = , q - 2 =
p = , q =
2
Given that the roots of the quadratic equation
3x2 + kx + p – 2 = 0 are 4 and
- ⅔. Find k and p.
(Ans : k = -10 , p = -6)
3
The roots of the quadratic equation
2x2 + px + q = 0 are - 6 and 3.
Find
(a) p and q,
(b) range of values of k such that
2x2 + px + q = k does not have real roots.
Answer :
(a) x = -6 , x = 3
(x + 6) (x – 3) = 0
x2 + 3x - 18 = 0
2x2 + 6x – 36 = 0
Comparing : p = , q =
(b) 2x2 + 6x – 36 – k = 0
a = 2, b = 6, c = -36 - k
62 – 4(2)(-36 – k) < 0
k < – 40.5
4
The roots of the quadratic equation
2x2 + px + q = 0 are 2 and -3.
Find
(a) p and q,
(b) the range of values of k such that
2x2 + px + q = k does not have real roots.
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MODUL ADD MATH TOK JIRING 6 by CYY
Solve the inequality x2 + x - 6 ≥ 0
x2 + x - 6 ≥ 0
(x + 3) ( x – 2) ≥ 0
Consider f(x) = 0. Then x = -3 , x = 2
Range of x is : x ≤ -3 atau x ≥ 2
L4. Solve the inequality x2 + 3x - 10 ≥ 0.
x ≤ -5 , x ≥ 2
SIMULTANEOUS EQUATIONS
1. Solve x + y = 3, xy = – 10 .
x + y = 3 ........ (1)
xy = – 10 ........ (2)
From (1), y = ( )......... (3)
Substitute (3) into (2),
x ( ) = – 10
3x – x2
= – 10
x2 – 3x – 10 = 0
(x ) (x ) = 0
x = ( ) atau x = ( )
From ( ), when x = ( ) , y = 3 – ( ) =
x = ( ) , y = 3 – ( ) =
2. Solve x + y = 5, xy = 4 .
(Ans : x = 1, y = 4 ; x = 4, y = 1)
3. Solve x + y = – 2 , xy = – 8 .
(Ans : x = – 4 , y = 2 ; x = 2, y = – 4 )
4. Solve 2x + y = 6, xy = – 20 .
(Ans : x = – 2 , y = 10 ; x = 5, y = – 4 )
x
-3 2
y=f(x)
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MODUL ADD MATH TOK JIRING 7 by CYY
INDICES N LOGARITME
Solve the equation log2 (x+1) = 3.
Answers: log2 (x+1) = 3
x + 1 = 23
x =
L1. Solve the equation log2 (x – 3 ) = 2.
Jawapan:
Ans : x = 7
Solve the equation log10 (3x – 2) = – 1 .
Jawapan: 3x – 2 = 10-1
3x – 2 = 0.1
3x = 2.1
x = 0.7
L2. Solve the equation log5 (4x – 1 ) = – 1 .
Ans : x = 0.3
Solve the equation log3 (2x – 1) + log2 4 = 5 .
Ans : x = 14
L6. Solve the equation
log4 (x – 2) + 3log2 8 = 10.
Ans : x = 6
Solve the equation
log2 (x + 5) = log2 (x – 2) + 3.
Ans : x = 3
L8. Solve the equation
log5 (4x – 7) = log5 (x – 2) + 1.
Ans : x = 3
Solve log3 3(2x + 3) = 4
Ans : x = 12
L10 . Solve log2 8(7 – 3x) = 5
Ans : x = 1
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MODUL ADD MATH TOK JIRING 8 by CYY
Given log 2 T - log4 V = 3, express T in terms of
V. [4]
(Ans: T = 8V ½
)
L8. Given log 4 T + log 2 V = 2, express
T in terms of V. [4]
(Ans: 16V-2
)
Solve 42x – 1
= 7x. [4]
( Ans: x = 1.677 )
L10. Solve 42x – 1
= 9x. [4]
( Ans: x = 2.409 )
Solve the equation 3 3log 9 log (2 1) 1x x− + = .
[3]
(Ans: x = 1 )
6. Given that log 2m p= and log 3m r= ,
express 227
log16
m
m
in terms of p and r. [4]
(Ans: 3r – 4p +2 )
Solve the equation 5 3 68 32x x− += .
[3]
(Ans : x = 3.9 )
8. Given that 5log 2 m= and 5log 3 p= ,
express 5log 2.7 in terms of m and p. [4]
(Ans: 3p – m – 1 )
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MODUL ADD MATH TOK JIRING 9 by CYY
GEOMETRY COORDINATE
1
Given two points A(2,3) and B(4,7)
Distance of AB = 2 2( ) ( )+
= 20 unit.
2.
P(4,5) and Q(3,2)
PQ =
[ 10 ]
3.
The point P internally divides the line segment
joining the point M(3,7) and N(6,2) in the ratio 2
: 1. Find the coordinates of point P.
P =
+
+
+
+
12
)2(2)7(1,
12
)6(2)3(1
4.
The point P internally divides the line segment
joining the point M (4,5) and N(-8,-5) in the ratio
1 : 3. Find the coordinates of point P.
51,
2
5
P(0, 1), Q(1, 3) and R(2,7)
Area of ∆ PQR = 2
1 0 1 2 0
1 3 7 1
= 2
1[ ( ) – ( ) ]
= 1 unit 2
6
P(2,3), Q(5,6) and R(-4,4)
Area of ∆ PQR =
17
2
unit2
7
. Find the equation of a straight line that passes
through the point (5,2) and has a gradient of -2.
y = -2x + 12
8
. Find the equation of a straight line that passes
through the point (-8,3) and has a gradient of 4
3.
4y = 3x + 36
●
1
2
N(6, 2)
M(3, 7)
P(x, y)
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MODUL ADD MATH TOK JIRING 10 by CYY
INDEX NUMBER
Notes :
a. Price Index, I = 1
0
100P
P× , P1 = price at a specific time, P0 = price at the base year.
b. Composite Index , IW
IW
=∑∑
, I = price index, W = weightage
1. Table below shows the price indices and percentage of usage of four main
ingredients ,P,Q,R and S, in the production of a type of cake.
Ingredients
Price index for the
year 2012
(2010=100)
Percentage of
usage
P m 20
Q 105 30
R 108 10
S 120 40
(a) Calculate
(i) the price of ingredient Q in the year 2010 if its price in the year 2012 is RM 50.00,
(ii) the price index of R in the year 2012, based on the year 2008, given that its price
index in the year 2010, based on the year 2008 is 110.
(b) The composite index number of the cost of production of this type of cake in the year
2012,based on the year 2010 is 112.8. Calculate.
(i) the value of m,
(ii) the cost of these ingredients for the production of this type of cake in
the year 2012 if the corresponding cost in year 2010 is RM60.00.
Guided Solutions
a (i)
,
(ii) ( ) ( )
( )1 0 0
×=
b (i)
(ii)
5 0( ) 1 0 5 , ( )× = =x
x
( )( ) 1 0 5 (3 0 ) 1 0 8 (1 0 ) 1 2 0 ( 4 0 )( ) , ( )
1 0 0
+ + += =
mm
2012100 ( ) , ( )2012
60
× = =Q
Q
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MODUL ADD MATH TOK JIRING 11 by CYY
SOLUTION OF TRIANGLES
(1) Diagram 1 shows the triangle ABC.
Calculate the length of BC.
Answer :
8 . 2
( ) ( )
B C=
( )
( )( )
BC = ×
Using the scientific calculator,
BC = ( )
(1) Diagram 1 shows the triangle ABC.
Find ∠ACB.
( ) ( )
( ) ( )=
(3)
Find the value of x.
2 2 2( ) ( ) 2( )( ) cos( )x = + −
x =
(4)
Find the value of x.
2 2 2( ) ( ) 2( )( ) cos( )x = + −
[ 7.475 ]
Diagram 1
A
B C
600
15 cm
Diagram 1
10 cm
12.3
P
16.4 cm
x cm
Q R 67
0
5cm
x cm
7
cm
P Q
R
750
5
cm
12.3
P
16.4 cm
x cm
Q R 67
0
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MODUL ADD MATH TOK JIRING 12 by CYY
(5)
Find BAC∠ .
2 2 215 ( ) ( ) 2( )( ) cos BAC= + − ∠
)14)(13(2
151413 cos
222 −+=∠BAC
(6)
Find BAC∠ .
2 2 2( ) ( ) ( ) 2( )( ) cos BAC= + − ∠
[ 83.17°]
(7)
Find area of PQR
Area = a b s i n c1
2
( ) ( ) s in ( )=1
2
=
(8)
Find area of PQR
C
A
B
13cm 14 cm
15 cm Diagram 1
C
A
B
11cm 13 cm
16 cm
Diagram 2
12.3
P
16.4 cm
x cm
Q R 67
0
12.3
P
16.4 cm
x cm
Q R 67
0
5cm
7 cm P Q
R
750
5 cm
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MODUL ADD MATH TOK JIRING 13 by CYY
TRIGONOMETRIC FUNCTIONS
GRAPHS OF TRIGONOMETRIC FUNCTIONS
Paper 2
1. (a) Prove that θθθ 2cos2cottan ec=+ . [ 4 marks ]
(b) (i) Sketch the graph xy2
3cos2= for π2≤x≤0 .
(ii) Find the equation of a suitable straight line for solving the equation 1 - 4
3
2
3cos xx
π= .
Hence, using the same axes, sketch the straight line and state the number of
solutions to the equation 1 - 4
3
2
3cos xx
π= for π2≤x≤0 .
[ 6 marks]
2. (a) Sketch the graph y = cos 2x for 00 180≤≤0 x . [ 3 marks ]
(b) Hence, by drawing a suitable straight line on the same axes, find the number of solutions
satisfying the equation 180
-2sin2 2 xx = for 0180≤≤0 xo .
[ 3 marks ]
( SPM P2 No. 3 )
3. (a) Prove that cosec 2 x – 2 sin
2 x – cot
2 x = cos 2x. [ 2 marks ]
(b) (i) Sketch that graph of y = cos 2x for π2≤x≤0 .
(ii) Hence, using the same axes, draw a suitable straight line to find the number of
solutions to the equation 3( cosec2
x – 2 sin2 x – cot
2 x) = 1-
π
x for π2≤≤0 x .
State the number of solutions. [ 6 marks ]
(SPM P2 No.5)
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MODUL ADD MATH TOK JIRING 14 by CYY
INTEGRATION
1. 43x dx∫ 2. 52
3x dx∫
3. 6
2
3dx
x∫ 4.
4
7dx
x∫
5. 3(2 3)x dx−∫ 6. ( )42
5 43
x dx−∫
7.
2
1
8x dx =∫
=
[12]
8.
4
3
2
x dx =∫
=
[60]
9.
2
1
(2 1)(2 1)x x dx− +∫
=
[ 25
3]
10.
3
2
1
(3 2)x dx−∫
=
[38]
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MODUL ADD MATH TOK JIRING 15 by CYY
11. Given that 3
1( )g x dx∫ = 6, find
a) the value of 3
1
( )
2dx
g x∫ ,
b) the value of 1
35 ( ) dxg x∫ ,
c) the value of k such that 3
1[ ( ) ]g x k dx+∫ = 10.
Guided Solutions
a) 3
1
( )
2dx
g x∫
= 3
1( )
1
2g x dx∫
= 1
( )2
= ( )
c) 3 3
1 1( )g x dx k dx+∫ ∫ = 10
( ) + [ ]3
1kx
= 10
( ) − ( ) = 4
k = ( )
12. Given that 5
2( )f x dx∫ = 9, find
a) the value of 5
2
2 ( )
3dx
f x∫ ,
b) the value of 2
54 ( )f x dx∫ ,
c) the value of k such that 5
2[ ( ) ]f x kx dx+∫ = 30.
b) 1
35 ( )g x dx∫
5( ) = ( )
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MODUL ADD MATH TOK JIRING 16 by CYY
13. Diagram shows the curve y = 5x4 and the straight line x = p.
If the area of the shaded region is 32 unit2, find the value of p.
Guided Solutions
Area = b
ay dx∫ =
4
0
5p
x dx∫ = ( )
[ ]0
p = ( )
( ) – ( ) = 32
p = ( )
14. Diagram shows the shaded region bounded by y-axis, the curve y2 = 4x and
a straight line y = k.
Given that the area of the shaded region is 9
4 unit
2, find the value of k.
Answer :
y2 = 4x
y = k
0 x
y
y = 5x4
x = p
0 x
y
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MODUL ADD MATH TOK JIRING 17 by CYY
PAPER 2
15. Diagram shows the straight line PQ is normal to the curve 2
13
xy = + at M (3, 4). The straight line
MN is parallel to y-axis.
Find
a) the value of h,
b) the area of the shaded region,
c) the volume of revolution, in terms of π, when the region bounded by the curve, the
y-axis and straight line y = 4 is rotated through 360° about y-axis.
Guided Solutions
a) m1 = dy
dx =
2
3
x =
2( )
3 = ( )
m2 = 1
1
m
− = ( ) =
4 0
3 h
−
− (m2 is gradient of PQ)
h = ( )
b) Area of region A = 23
0
( 1)3
xdx+∫ = ( )
Area of region B = ( )
Hence, the area of the shaded region = A + B = ( )
c) The volume of revolution = π4
2
1
x dy∫ = π4
1
( ) dy∫ = ( )
.
B A
N Q (h, 0) O x
y
y = 2
13
x+
M (3, 4)
P ••••
•
•
•
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MODUL ADD MATH TOK JIRING 18 by CYY
A O
B
y
P
C
x k
Q
32 += xy
9=+ xy
16. Diagram shows the curve 32 += xy intersects the straight line AC at point B.
It is given that the equation of straight line AC is y + x = 9 and the gradient of the curve at point B is
4. Find
a) the value of k,
b) the area of the shaded region P,
c) the volume of revolution, in terms of π , when the shaded region Q is rotated through
360º about the x–axis .
9. Diagram shows the curve 22 1y x= + intersects the straight line 9 9y x= − at point (2, k).
Find,
a) the value of k,
b) the area of the shaded region,
c) the volume of revolution, in terms of π, when the region bounded by the curve, the
y-axis and straight line y = 9 is rotated through 360° about y-axis.
END OF MODULE
(2, k)
y = 2x2 + 1
y = 9x−9
x
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