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‘Answering For Success’ Program
By: Cg. Shaifur Azura Suhaimi
Date: 18 March 2009, Friday Place: Ridzuan Resort, Melawi, Bachok, Kelantan
Objectives: Solve the problems involving all Form 4 Chapters
Reference: SPM Paper 1 questions, 2008
Question
1.Diagram 1 shows the graph of the
function f ( x)  2 x  1 , for the
domains 0  x  5 . State the:
a) the value of t,
b) the range of f(x)
corresponding to the given
domain
[3marks]
Strategies
1. What is the meaning of
‘modulus’ or . The negative
value becomes positive value.
2. What is the x-intercept and yintercept
3. Given domain for x, means the
image of x can be found.
y
1
o
t
5
x
Diagram 1
2.
Given the functions
g:x→ 5x + 2 and
h : x → x2 – 4x + 3, find
a) g -1(6)
b) hg(x)
[4marks]
AFS Program @ S2fo2r
a) Find the inverse functions.
Use easy math method or
90’s method.
b) Find the composite functions
Solution
1. Find the x-intercept points, value of t can be
found.
a) x-intercept happened when y = 0,
2x  1  0
2x – 1 = 0
2x
=1
1
1
x =
, therefore t =
2
2
b) the range of f(x) :
when x = 0 , f(x) = 2(0)  1  1
x = 5 , f(x) = 2(5)  1  9
 the range of f(x) : 0  f ( x)  9
5x  2
4
 g-1(6) =
a) g ( x) 
5
1
x2
g-1(x) =
or Method 2:
5
x2
x2
=
g -1(x) =
5
5
6

2
6
2 4
g-1(6) =
g -1(6) =
=
5
5
5
Question
Strategies
f(x)
= 5 x+ 2
x - 2
f -1(x) =
5
3. Given the functions
a) f(5) means when x = 5, what
f(x) = x -1 and
is the value of f(x)?
g(x) = kx + 2,
find:
b) find composite function of
a) f (5)
gf(5), when gf(5) = 14, k = ?
b) the value of k such
that
gf (5) = 14
[3marks]
4. It is given that -1 is one Substitute x = -1 into quadratic
of the roots of the
equation, so the value of p can be
2
quadratic equation x – 4 found.
x – p = 0. Find the value
of p.
Solution
2 (b) hg(x) = h [g(x) ]
= h [5x + 2]
= [5x+2]2 – 4[5x+2] +3
= 25x2 +20x + 4 – 20x -8 + 3
= 25x2 -1
= (5x+1)(5x-1)
3 a)
b)
f(5) = 5 – 1
=4
gf(5) = g [f(5)]
= g [4]
= k(4) + 2
14 = 4k + 2
14 – 2 = 4k
4 k = 12
k = 12/4
k= 3
Method 1:
(-1)2 – 4 (-1) – p = 0
1+4–p=0
5–p=0
p=5
Method 2 :
Roots: α = -1 , β = p
SOR = -1 + p , POR = -p
Equations : x2 – (SOR)x + POR = 0
x2 – (p-1) x –p = 0
By comparison: x2 – 4 x – p = 0
x2 – (p-1) x –p = 0
AFS Program @ S2fo2r
Question
Strategies
Solution
By comparison: x2 – 4 x – p = 0
x2 – (p-1) x –p = 0
p–1=4
p = 4+1
p = 5
5. f (x) = p (x + q )2 + r, has
a minimum value of -4.
The equation of the axis
of symmetry is 3. State:
a) the range of the values
of p
b) the value of q
c) the value of r
[3marks]
6. Find the range of the
values of x for
(x – 3 )2 < 5 - x
p (x + q )2 + r
i) from the word minimum , we
know that p must be positive
numbers.
ii) p = coefficient
iii) x + q = 0 , equation of axis of
symmetry, given x = 3
iv) r = max or min point depends
on p
2
(x – 3 ) < 5 – x,
i) make sure inequalities in the form
of ax2 + bx + c = 0
ii) the words range & < symbol
means solution must sketch the
graph.
iii) use the concept of linear
inequalities.
iv) (x – 3 )2 = expand the bracket
a) p  1
b) (x + q ) = 0
x=-q
q=-x
q = - (3)
q=-3
c) r = -4 ( minimum value )
x2 – 6x + 9 < 5 – x
x2 – 6x + 9 -5 + x < 0
x2 - 5x + 4 < 0
(x – 1) (x - 4) < 0
x – 1= 0
, x–4=0
x =1 or x = 4
x= 0
1
 Range of x :
AFS Program @ S2fo2r
4
1< x < 4
Question
7. Solve the equation:
162x – 3 = 34x
Strategies
Look at the base, not in the same
base. Must use the logarithms on
both sides of equations.
Solution
Method 1:
162x – 3 = 34x
42(2x – 3) = 34x
log 44x – 6 = log 34x
(4x – 6) log 4 = 4x log 3
4 x  6 log 3

4x
log 4
4x – 6 = 4x (0.7925)
4x – 6 = 3.17x
4x – 3.17x = 6
0.83x = 6
6
x=
0.83
x = 7.2289
Method 2:
7. Solve the equation:
162x – 3 = 84x
AFS Program @ S2fo2r
Look at the base, not in the same
base. Must use the logarithms on
both sides of equations
( 2 4 ) 2 x 3  ( 2 3 ) 4 x
28x – 12 = 2 12x
8 x – 12
= 12 x
4x
= -12
x
= -3
Strategies
Solution
Question
8. Given that log 4 x = log 23,
find the value of x.
1. Look at the base, not in the
same base.
2. Changing the base is needed to
solve the question.
log 4 x = log 2 3
log 2 x
 log 2 3
log 2 4
log 2 x
 log 23
log 2 2 2
log 2x = 2 log 2 3
log 2x = log 2 32
= log 2 9
 x = 9
AFS Program @ S2fo2r
Question
Strategies
Solution
a)
y
y
a)
y-intercep t
T(0,4)
x y
 1
3 4
T(0,4)
x-intercep t
S(3,0)
o
S(3,0)
x
13.(a) Write down the
equation of the straight line
ST in the form
x y
 1
a b
o
x
x y
  1 ------------(1)
a b
i) Identify a = x-intercept, b = y-intercept
ii) substitute a & b into formula (1)
(4  y ) 2  (0  x) 2  (0  y ) 2  (3  x) 2
b)
b) A point P (x, y) moves
such that PS = PT. Find
the locus of P.
y
y2 - 8y +16 + x2 = y2 + x2 – 6x + 9
-8y +16 + 6x -9 = 0
-8y + 7 + 6x = 0
6x – 8y + 7 = 0
y-intercept
T(0,4)
x-intercept
P
S(3,0)
o
AFS Program @ S2fo2r
b) PT = PS
x
 equation of locus : 6x – 8y + 7 = 0 or
3
7
y= x
4
8
Question
Strategies
Solution
18. Radius = 10cm
O
OP = PQ and
OPR  90
Find:
a) QOR in radians
b) the area, in cm2, of
the coloured region
O
5
P
5
10cm
a)
cos POR 
5
10
P
R
R
= 0.5
POR = cos-1(0.5)
= 60
60x3.142
i) OPR  90 , OP = PQ = 5cm
 QOR =
rad
180
ii) Use the right angle triangle OQR
= 1.0472 rad
Q
iii) Use circular measures formulae
O
10cm
P
10cm
5
R
Q
[4marks]
AFS Program @ S2fo2r
b) Area of coloured region =
Area of sector QOR – Area of triangle POR
1
1
= OQxORx   OPxPR
2
2
1
1
= (10) 2 (1.0472)  (5) 10 2  5 2
2
2
= 52.36 – 21.651
= 30.709cm2
Question
19. Two variables, x and y,
Strategies
i)
approximate change in y = y = 16x -2
ii)
given x = 4, x  4  h  4
=h
iii)
use
are related by the equation
16
y 2.
x
Express, in terms of h, the
approximate change in y,
when x changes from 4 to
4 + h, where h is a small
value.
[3 marks ]
AFS Program @ S2fo2r
Solution
y
dy y

dx x
dy
 y  xx
dx
dy
 32 x  3
dx
=  32
x3
 y  dy xx
dx
=  32
xh
x3
 32h
y =
x3
 32h
y 
( 4) 3
1
y   h
2
Question
Strategies
Solution
14. The points (0, 3), (2, t ) and
(-2, -1) are the vertices of a i) Find the area of a triangle. Use the
area formula, but remember to
triangle. Given that the area
of the triangle is 4 unit2, find eliminate the modulus.
Eg :
the values of t.
(0, 3 )
(2, t )
2x  2  5
2x - 2 =  5
(-2, -1)
Anti clock-wise reading
Area =
1
0 -2 2 0
2
3 -1 t
=4
3
1
(2t  6  6  2)  4
2
14 -2t =  8
14 -2t = 8
2t=6
t=3
AFS Program @ S2fo2r
or
or
14 -2t = - 8
2 t = 22
t = 11
Question
20. The normal to the
curve y = x – 5x at
point P is parallel to
the straight line
y = - x + 12. Find the
equation of normal
to the curve at point
P.
Strategies
Solution
i) use the Easy math method
P(3, -6 )
2
dy
y
y= x2-5x
dx
dy
= 2x - 5
dx
m normal=m straight-line
=-1
equation of normal:
m tan = 1
y - (-6) = -1 ( x - 3 )
y + 6 = -x + 3
y = -x - 3
dy
dx
=1
2x - 5 = 1
2x = 6
x= 3
when x = 3, y = (3)2-5(3)
= -6
AFS Program @ S2fo2r
Question
22. A set of seven numbers
has a mean of 9.
a) find  x
b) when a number k is
added to this set, the new
mean is 8.5. Find the value
of k.
Strategies
i)
ii)
Solution
use formula mean of
ungrouped data.
x
x
N
a) x 
x
9
x
N
7
 x  9x7
= 63
b) 8.5 =
xk
7 1
8.5 (8) = 63 + k
k = 68 – 63
k=5
AFS Program @ S2fo2r
FORM 5 TOPICS: REFER TO SPM EXAM 2008
USING THE SCAN/STRATEGY/FORMULAE/OPERATION/OUTPUT/RE-CHECK (S2FO2R ) CONCEPT
TO SOLVE PROBLEMS
Q
9.
QUESTION
It is given that the first four terms
of a G.P are 3, -6, 12 and x. Find
the value of x.
SCAN- STRATEGY / FORMULA
1. G. P – automatically have a
and r
2. Find r use T1 , T2 , T3
r=
10. The first three terms of an A.P are
46, 43 and 40. The n th term of
this progression is negative. Find
the least value of n.
T2 T3

T1 T2
1. A. P- automatically have a
and d
2. Find d use T1 , T2 , T3
d = T3 – T2 = T2 – T1
3. Tn = a + (n-1) d
11. In a G.P, the first term is 4 and the 1. G. P – automatically have a
common ratio is r. Given that the
sum to infinity of this progression
is 16, find the value of r.
AFS Program @ S2fo2r
and r
2. a = 4 , d = r , S = 16
OPERATION / OUTPUT / RE-CHECK
6
x

3
12
x  6(4)
x  24
a = 46
d = T3 – T2 = T2 – T1
d = 43 – 46
= -3
Tn = a + (n-1) d
a + (n-1) d < 0
46 + (n-1) -3 < 0
46 -3n + 3 < 0
-3n < -49
49
n>
3
1
n > 16
3
 n = 17
a
1
S 
r = 11 r
4
4
3
 r=
16 
1 r
4
a
1
1- r =
1 r
4
SCAN- STRATEGY / FORMULA
OPERATION / OUTPUT / RE-CHECK
1. Look at the x-axis and y- y  k
5x
axis
k
2. y-axis = log10y, means
log y  log x
5
log must be inserted on
log y  log k  log 5 x
both side of the given
3. Use S 
Q
QUESTION
12. The variables of x and y are
k
related by the equation y  x ,
5
where k is a constant. Diagram 12
shows the straight line graph
obtained by plotting log10y against
x.
k
a) Express the equation y  x in
5
its linear form used to obtain the
straight line graph shown in
diagram 12.
b) Find the value of k.
log10 y
o
(0, -2)
Diagram 12
AFS Program @ S2fo2r
x
equation.
3. Refer the new equation
to find k.
log y   x log 5  log k
log y   log 5 x  log k
log k = -2
k = antilog 10 (-2)
k = 10-2
1
k=
= 0.01
100
Q
QUESTION
15. The vectors a and b are non-zero
and non-parallel. It is given that
(h + 3 ) a = ( k – 5 ) b , where h
and k are constants. Find the
value of
a) h
b) k
(2 marks )
16. Diagram 16 shows a triangle
PQR.
Q
6b
P
SCAN- STRATEGY / FORMULA
1. The words non-zero and
non-parallel means :
(h + 3 ) = 0 and ( k – 5 ) = 0
(h + 3 ) = 0 and ( k – 5 ) = 0
a)
h+3=0
h =-3
b)
k -5 = 0
k=5
1. Change the ratio
QT : TR = 3 : 1 to a
fraction, to know the length
of QT and TR.
QT 3

TR 1
QT
3

TR
1
T
4a
R
Q
Diagram 16
The point T lies on QR such that
QT : TR = 3 : 1. Express in terms
of a and b :
a)
QR
b)
PT
AFS Program @ S2fo2r
OPERATION / OUTPUT / RE-CHECK
a)
QT =
6b
P
 QT  3,TR  1, QR  4
3
4
 6b  4a  QR
QR
T
4a
Diagram 16
QP  PR  QR
b)
PQ  QT  PT
6b +
3
(6b  4a) =
4
PT
Q
QUESTION
SCAN- STRATEGY /
FORMULA
OPERATION / OUTPUT /
RE-CHECK
9
6b  ( b)  3a  PT
2
3
PT  3a  b
2
17.
Given that sin θ = p, where p 1. Second quartile for
the trigo function is
is a constant and 90    180 .
referred.
Find in terms of p:
2. Use the triangle to
a) cosec θ
solve problems.
b) sin2θ
3. change
the
given
(3 marks)
questions using trigo-
90
p
1

180
1-p2
formulae
1
,
sin 
sin2θ = 2sinθcosθ
4. cosec θ =
1
sin 
1
=
p
a) cosec θ =
b) sin2θ = 2sinθcosθ
= 2 p (  1  p2 )
= -2p 1  p 2 )
AFS Program @ S2fo2r
Q
21.
QUESTION
SCAN- STRATEGY /
FORMULA
Given that
 (6 x
 1)dx  px  x  c ,
where
p and c are constants, find
a) the value of p
b) the value of c if
2
 (6 x  1)dx =13
when x = 1
23.
2
3
Diagram 23 shows
numbered cards.
six
1. Integral f(x) with
respect to x.
2. Compare the answer
with given value.
5
6
7
8
9
Diagram 23
A four-digit number is to be
formed by using four of these
cards.
AFS Program @ S2fo2r
= 2x3 + x + c
2x3 + x + c = px3  x  c
p = 2
b) 2x3 + x + c = 13 when x = 1,
substitute x = 1 into the equation
2(1)3 + (1) + c = 13
3 + c = 13
c = 13 – 3
c = 10
The words ‘is to be
formed’ means use
n
3
OPERATION / OUTPUT /
RE-CHECK
6 x3
a)  (6 x 2  1)dx =
 xc
3
pr .
1. use the
permutation rule
a)
6
p4  360
Or using ‘fill in the box’ method
How many
a) different numbers can be
formed?
b) different odd numbers can
be formed?
* refer to the notes
3,5,6,7,8,9
6,7,8,9
6
7
7,8,9
3
5
3
picked number
6
number to be picked
5,6,7,8,9
4
5
3 x 4 x 5 x 6 = 360
a) Different ways of numbers can be
formed are 360.
b)
3, 5, 7, 9
5p3=60
4p14
5p3x 4p1=60x4 = 240
Different odd numbers can be formed =
240
AFS Program @ S2fo2r
Q
QUESTION
SCAN- STRATEGY /
FORMULA
24.
The probability of Sarah
being chosen as a prefect is 3
5
while the probability of Aini
being chosen is 7 . Find the
12
probability that
a) neither of them is
chosen as a school
prefect
b) only one of them is
chosen as a school
prefect (4 marks )
a) Use the Probability X= (Sarah or Aini is chosen as a prefect)
3
2
of Independent
Sarah: p = , q =
5
5
events. Event A
7
5
Aini: p = , q =
occurs doesn’t
12
12
depend on
a) P(Both of them are not chosen)
outcomes of event
2
5
=
X
A.
5 12
i) Refer to AND:
1
alphabet N = 
=
6
ii) D = Darab 2
events
iii) P(AB) =
P(A) x P(B)
b) Use the Probability
of Mutually
exclusive events.
Both events cannot
occur at the same
time.
AFS Program @ S2fo2r
OPERATION / OUTPUT /
RE-CHECK
i)
Refer to OR:
Alphabet OR = b) P ( One of them is chosen)
ATAU, U
3 5
7
2
)( X )
=( X
ii) T = Tambah 2
5 12
12 5
events
29
iii) P(AUB) =
=
P(A or B) =
60
P(A) + P(B)
Q
QUESTION
SCAN- STRATEGY /
FORMULA
25.
The mass of group of students
in a school have a normal
distribution with a mean of
40kg and a standard deviation
of
5kg.
Calculate
the
probability that a student
chosen at random from this
group has a mass of
1. Found the words,
‘mean’ & ‘standard
deviation’. We
must refer to
Probability
Distributions.
2. What is a discuss
subject?
X = mass of group
of students
and 3. Change X to Z.
a) more than 45kg
b) between
35kg
47.8kg
[4marks]
AFS Program @ S2fo2r
X 
Z=

OPERATION / OUTPUT /
RE-CHECK
a) μ = 40, σ = 5
45  40
)
5
= P ( Z > 1)
= 0.1587*
* use the Z-Value table or
Calculator
P(X > 45) = P ( Z 
b) P(35 < X < 47.8) =
35  40
47.8  40
P(
Z
)
5
5
= P( -1< Z < 1.56 )
Q
QUESTION
SCAN- STRATEGY /
FORMULA
OPERATION / OUTPUT /
RE-CHECK
f(z)
z
-1
1.56
= 1 – P(Z< -1) – P ( Z > 1.56 )*
= 0.7819
* Use calculator : follow all the symbol &
operations from left to right.
AFS Program @ S2fo2r
NOTES: PERMUTATIONS & COMBINATIONS
PERMUTATIONS &
COMBINATIONS
PERMUTATIONS.
COMBINATIONS
The order of the objects
in the chosen set is taken
into consideration.
The order of the objects
in the chosen set is NOT
taken into consideration.
AFS Program @ S2fo2r
Q1:
CHEMISTRY
a) no digits is repeated
9
x
8
x
7
6
x
5
x
4
x
3
x
2
x
1
x
number of ways = 9x8x7x6x5x4x3x2x1
= 362,880
b) all the letters if the first letter is a vowel
2
x
8
x
7
x
6
x
5
x
4
E,I
2P1=2
number of ways = 2x8x7x6x5x4x3x2x1
= 80640
OR
number of ways = 2P1 X 8P8
= 2 X 40320
= 80640
AFS Program @ S2fo2r
8P8=40320
x
3
x
2
x
1
Q2: 5 Digit Numbers Are To Be Formed Using 0, 1, 2, 3, 4
a)
no digits is repeated
4
x
4
x
3
2
x
x
2,3,4
3,4
1,2,3,4 0,2,3,4
cannot
choose'0'
number of ways = 4x4x3x2x1
= 96
4
or
4p1=4
x
4
x
3
4p4= 24
number of ways = 4p1 x 4p4
= 96
AFS Program @ S2fo2r
x
2
1
ways of arrangement
4
numbers to be chosen
x
1
b)
the digits are allowed to repeat
0,1,2,3,4
0,1,2,3,4
4
5
5
5 x 5
x
x
x
0,1,2,3,4
1,2,3,4 0,1,2,3,4
cannot
choose'0'
number of ways = 4x5x5x5x5
= 2500
or
4
4p1=4
x
5
x
5
x
5
(5p1)4= 625
number of ways = 4p1 x (5p1)4
= 2500
AFS Program @ S2fo2r
x
5
numbers to be chosen
ways of arrangement
numbers to be chosen
c)
even numbers are obtained with no repeat of digits
Case 1: 0 is chosen to fill the last number
ways of arrangement
4
numbers to be chosen
1,2,3,4
x
3
x
2,3,4
2
x
1
x
even number will
end with 0,2,4:
now "0" is
selected
4
3,4
number of ways = 4x3x2x1x1
= 24
ways of arrangement
3
numbers to be chosen
1,3,4
x
3
0,1,3
x
2
x
1,3
1
3
number of ways = 3x3x2x1x2
= 36
Total number of ways = 24 + 36
= 60
AFS Program @ S2fo2r
1
x
2
even number will
end with 2,4:
now 2 or 4 is
selected(2p1=2)
n pr
Q3:
1, 2, 3, 4, 5, 6
3 digit even numbers formed, exceed 200
i)
2
3
4
x
3, 5
x
5, 1, 4, 6
2, 4, 6 - even
ways of
arrangement
number to be
chosen
ways of arrangement = 2 x 4 x 3
= 24
3
ii)
x
2, 4, 6
2
4
x
5, 1, 3, 6
4, 6 - even
ways of arrangement = 3 x 4 x 2
= 24
AFS Program @ S2fo2r
ways of
arrangement
number to be
chosen
Q4:
Words: FORMAT, n = 6
2
x
i)
4
5
O, A
x
x
F, R, M, T, A
3
R, M, T, A
M, T, A
ways of arrangement = 2 x 5 x 4x 3
= 120
or
x
x
x
O, A
2p1
5p3
ways of arrangement = 2p1 x 5p3
= 120
AFS Program @ S2fo2r
ways of
arrangement
an alphabet
to be chosen
Q5:
Word : VOWELS , n = 6
a) vowels are separated
C
C
V
C
C
V
V
C = CONSONANT
V
V = VOWEL
V
arranging vowels = 5p2 = 20
arranging consonant = 4! = 4p4=24
 total number of ways = 20 x 24
= 480
b) the consonant are next to each other
V
W
x
L
x
S
x
x
4!
3!
 total number of ways = 4! x 3!
= 144
AFS Program @ S2fo2r
x
Q5:
c) each arrangement starts and ends with a consonant
V
x
x
x
C
x
C
L, S, O, E
arranging of consonant = 4p2
= 12
arranging of others = 4p4 @ 4!
= 24
 total number of ways = 12 x 24
= 288
AFS Program @ S2fo2r
W
x
Q6: How many different odd numbers can be formed using digits
3, 4, 5, 6, 7? How many of these are more than 50,000?
a) the last digit ends with 3, 5, 7
x
x
x
x
4, 5, 6, 7
3, 5, 7
4! or 4p4 = 24
selected number
3p1=3
 no. of ways = 3 x 24
= 72
b) case 1: the first digit is '5' and '7'
2
2
x
5, 7
2p1
x
x
4, 6, 7
3p3
start with '5' or '7' = 2 x 3p3 x 2
= 24
AFS Program @ S2fo2r
> 50,000
x
3, 7
2p1
Q6: b)
case 2: the first digit is '6' and end with odd number
3
1
x
x
x
6
3p3 = 6
start with '6' = 1 x 3p3 x 3
=1x6x3
= 18
total number of ways = 24 + 18
= 42
AFS Program @ S2fo2r
x
3, 5, 7
3!
Q7: 7 students are lining up for a photo session. In how many ways
can they be arranged if:
a) the tallest is in the middle
5
6
x
x
1
4
3
x
2
x
x
1
x
T
no. of ways = 1 x 6p6
= 720
b) the two shortest are at the end of the line and the tallest is in the middle
2
x
x
s, s
2p1
no of ways = 2 x 4p4 x 1
= 48
AFS Program @ S2fo2r
x
1
T
x
x
x
1
s