Post on 04-Apr-2018
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________Name:
Class: Date:
1. Given that set P = {1, 2, 3, 4} and set Q = {2, 4, 6, 8} and the relation is Q istwo times P. Represent the relation using;a) An arrow diagramb) Ordered pairsc) A graph
SOLUTION:
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________
2. The following arrow diagram shows a relation.
q
rn
pm
F G
i) Determine the domain, codomain and the range of relation.
ii) State the object of q.iii) State the image of m.
SOLUTION:
3. The following set of ordered pairs shows a relation.{(1, 1), (2, 4), (3, 9), (4, 16)}
i) Determine the domain, codomain and range of this relation.ii) State the object of 9.iii) State the image of 4.
SOLUTION:
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________
4. The following graph shows a relation.
32SetA
1
12
9
6
3
0
SetB
i) Determine the domain, codomain and range of this relation.ii) State the object of 12.iii) State the images of 1.
SOLUTION:
5. State the type of relation for the following arrow diagram.
SOLUTION:
3
6
9
b
c
a
A B
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________
6. State the type of relation for the following ordered pairs.{(2, 3), (2, 4), (2, 5)}
SOLUTION:
7. State the type of relation for the following graph.
SOLUTION:
18
17
16
4321
15
0
8. The relation between setXand set Yis Yis the square root ofX. Draw anarrow diagram using any suitable objects and images. State the relation forthe two sets.
SOLUTION:
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________
9. State the type of relation for the following cases. Draw the arrow diagram withany suitable objects and images.
P = {Age of students}Q = {Students in a class}
i) Set P is the domain and set Q is the range.ii) Set Q is the domain and set P is the range.
SOLUTION:
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________
1. Given that set P = {1, 2, 3, 4} and set Q = {2, 4, 6, 8} and the relation is Q istwo times P. Represent the relation using;a) An arrow diagram
b) Ordered pairsc) A graph
SOLUTION:
a)
2
4
684
3
2
1
P Q
b) {(1, 2), (2, 4), (3, 6), (4, 8)}
c)
42 3
Set P
1
2
4
6
8
Set Q
0
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________2. The following arrow diagram shows a relation.
q
rn
pm
F G
i) Determine the domain, codomain and the range of relation.ii) State the object of q.
iii) State the image of m.
SOLUTION:i) Domain = {m, n}
Codomain = {p, q, r}Range = {p, q, r}
ii) Object of qis n.iii) Image of mis p.
3. The following set of ordered pairs shows a relation.
{(1, 1), (2, 4), (3, 9), (4, 16)}
i) Determine the domain, codomain and range of this relation.ii) State the object of 9.iii) State the image of 4.
SOLUTION:i) Domain = {1, 2, 3, 4}
Codomain = {1, 4, 9, 16}Range = {1, 4, 9, 16}
ii) Object of 9 is 3.
iii) Image of 4 is 16.
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________4. The following graph shows a relation.
32SetA
1
12
9
6
3
0
SetB
i) Determine the domain, codomain and range of this relation.
ii) State the object of 12.iii) State the images of 1.
SOLUTION:i) Domain = {1, 2, 3}
Codomain = {3, 6, 9, 12}Range = {3, 6, 9, 12}
ii) Object of 12 is 3.iii) Images of 1 are 3 and 6.
5. State the type of relation for the following arrow diagram.
SOLUTION:one-to-one.
3
6
9
b
c
a
A B
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________6. State the type of relation for the following ordered pairs.
{(2, 3), (2, 4), (2, 5)}
SOLUTION:
one-to-many.
7. State the type of relation for the following graph.
SOLUTION:many-to-one.
18
17
16
4321
15
0
8. The relation between setXand set Yis Yis the square root ofX. Draw anarrow diagram using any suitable objects and images. State the relation for
the two sets.
SOLUTION:
The type of relation is one-to-many.
2
2
3
3
9
4
X Y
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Activity Sheet Lesson 1Introduction to Relations
________________________________________________________________9. State the type of relation for the following cases. Draw the arrow diagram with
any suitable objects and images.P = {Age of students}Q = {Students in a class}
i) Set P is the domain and set Q is the range.ii) Set Q is the domain and set P is the range.
SOLUTION:i) Set P is the domain and set Q is the range.
Ram
Chong
Diana15
John14
P Q
The type of relation is one-to-many.
ii) Set Q is the domain and set P is the range.
15
Chong
Diana
Ram
14John
P Q
The type of relation is many -to-one.
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Activity Sheet Lesson 2Concept of a Function
________________________________________________________________Name:
Class: Date:
1. Determine whether each of the following relations are functions. Give areason for your answer.
(a)
______________________________________________________
(b)
_____________________________________________________
2. Determine whether each of the following relations are functions. Give areason for your answer.
(a) Ordered pairs of functionf= {(2,3), (1,4), (0,4), (1,5)}.
________________________________________________
(b) Ordered pairs of functionf= {(a,a), (b,b), (c,c)}.
________________________________________________
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Activity Sheet Lesson 2Concept of a Function
________________________________________________________________
3. Determine whether each of the following relations are functions. Give areason for your answer.(e)
_____________________________________________________
(f)
__________________________________________________
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Activity Sheet Lesson 2Concept of a Function
________________________________________________________________
4.
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Activity Sheet Lesson 2Concept of a Function
________________________________________________________________1. Determine whether each of the following relations are functions. Give a
reason for your answer.
(a)
It is not a function because object b has two images, m and o.
(b)
It is a function because every object has only one image.
2. Determine whether each of the following relations are functions. Give areason for your answer.
(a) Ordered pairs of functionf= {(2,3), (1,4), (0,4), (1,5)}.It is a function because every object has only one image.
(b) Ordered pairs of functionf= {(a,a), (b,b), (c,c)}.It is a function because every object has only one image.
3. Determine whether each of the following relations are functions. Give areason for your answer.
(a)
It is not a function because object c has no image.
(b)
It is a function because every object has only one image.
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Activity Sheet Lesson 2Concept of a Function
________________________________________________________________
4. Complete the following function notations based on the respective
representations.
(a)
(i) d (3) = 9 (ii) d( 2 ) = 4
(iii) d( 2 ) = 4 (iv) d(x) = x2
(b) Ordered pairs of function g = {(4,2),(3,1.5),(2,1)}
(i) g (3) = 1.5 (ii) g ( 4 ) = 2
(iii) g (x) =2
x
(c) (i) (i) f(8) = 2
(ii) (ii) f( 0 )=0
(iii) (iii) f(x) = 3 x
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Activity Sheet Lesson 3Range of function
________________________________________________________________Name:
Class: Date:
1. Determine the domain, objects, images and range for each of the followingfunctions.
(a) h:x 2cosx
x 2cosx
90 -1
120 0
240 1
270
(b) Ordered pairs of functions
f : x -2
x = {( -1,21 ), (2, -1), (4, -2)}
(c) p(x) = -x
-2 -1 0 1 2-1
2
- 3
- 4
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Activity Sheet Lesson 3Range of function
________________________________________________________________SOLUTION:
2. Sketch the graph of function d(x) = 2x 4in the domain -3 0x .
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Activity Sheet Lesson 3Range of function
________________________________________________________________Then determine the range of the function.
SOLUTION:
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Activity Sheet Lesson 3Range of function
________________________________________________________________1. Determine the domain, objects, images and range for each of the following
functions.
(a) h:x 2cosx
x 2cosx
90 -1
120 0
240 1
270
(b) Ordered pairs of functions
f : x -2
x = {( -1,21 ), (2, -1), (4, -2)}
(c) p(x) = -x
p(x)
-2 -1 0 1 2-1
2
- 3
- 4
x
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Activity Sheet Lesson 3Range of function
________________________________________________________________SOLUTION:a) Domain = {90, 120, 240, 270}.
Objects are 90, 120, 240 and 270Images are 1 and 0
Range = {1, 0}
b) Domain = {1, 2, 4}.Objects are 1, 2 and 4.
Images are 2, 1 and2
1
Range = {2, 1 and2
1}
c) Domain = {2, 1, 0, 1, 2}.Objects are 2, 1, 0, 1 and 2Images are 4, 1 and 0Range = {4, 1 and 0}
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Activity Sheet Lesson 3Range of function
________________________________________________________________2. Sketch the graph of function d(x) = 2x 4in the domain -3 0x .
Then determine the range of the function.
SOLUTION:
x 3 2 1 0d(x) 10 8 6 4
d x
10
8
6
4
0123x
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Activity Sheet Lesson 4Determine the image of a function given the object and vice
versa________________________________________________________________Name:
Class: Date:
1. Given h (x) =x
12 2 forx 0. Find the images whenx = 1, 2 and 4
SOLUTION:
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Activity Sheet Lesson 4Determine the image of a function given the object and vice
versa________________________________________________________________2. Givenf(x) = 4 x. Find the object for the images 3 and 0
SOLUTION :
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Activity Sheet Lesson 4Determine the image of a function given the object and vice
versa________________________________________________________________3. Given g (x) = 2 cosx for 0 x 90. Find the object for the image 2
SOLUTION:
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Activity Sheet Lesson 4Determine the image of a function given the object and vice
versa________________________________________________________________4. Givenf(x) = 2x + 3. Find
a) the images of
i) x = 1ii) x = 3
b) the objects of 5
SOLUTION:
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Activity Sheet Lesson 4Determine the image of a function given the object and vice
versa________________________________________________________________
SOLUTION 3
g (x) = 2
2 = 2 cosx
cosx =2
2
cosx = 1x = cos (1)x = 0
4. Givenf(x) = -2x + 3. Find
a) the images of
i) x = 1ii) x = 3
b) the objects of 5
SOLUTION:
(a) i) x =1 ii) x =1
f(1) = 2(1) +3 f(3) = 2(3) +3= 2 + 3 = 6 + 3= 1 = 3
= 3
(b) f(x) = 2x +3 and 2x + 3 = 5 2x +3= 5 2x = 5 3
( 2x + 3) = 5 2x = 22x = 8 x = 1x = 4
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
Name:
Class: Date:
1. Givenf(x) = 3x 6 and g(x) = 6x + 1. Find
a) fgb) gfc) ffd) g2
e) fg(0)
f) fg(1)
g) gf(2
1)
h) g2(2)
i) x given gf(x) = 3
j) x given g2(x) = 2
Solution:
1
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
2
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
2. Givenf(x) =9
42 +xfor2 x 2 and g(x) =x + 2.
a) Find(i)fg(ii)gf
b) Find the range of
(i)fg(ii)gf
Solution:
3
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
1. Givenf(x) = 3x 6 and g(x) = 6x + 1. Find
k) fgl) gfm)ffn) g2
o) fg(0)
p) fg(1)
q) gf(2
1)
r) g2(2)
s) x given gf(x) = 3
t) x given g2(x) = 2
Solution:
a) fg(x) =f[g(x)]
=f(6x +1)
= 3(6x + 1) 6
= 18x + 3 6
= 18x 3
fg(x) = 18x 3
b) gf(x) = g[f(x)]
= g(3x 6)
= 6(3x 6) + 1
= 18x 36 + 1
= 18x 35
gf(x) = 18x 35
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
c) ff(x) =f[f(x)]
=f(3x 6)
= 3(3x 6) 6= 9x 18 6
= 9x 24
ff(x) = 9x 24
d) g2(x) = gg(x)
= g[g(x)]
= g(6x + 1)
= 6(6x+1) +1
= 36x + 6 +1
= 36x + 7
g2(x) = 36x + 7
e) fg(0) = 18x 3
= 18(0) 3
= 3
f) fg(1) = 18(x) 3
= 18(1) 3
= 18 3
= 21
g) gf(2
1) = 18x 35
= 18(2
1) 35
= 9 35
= 26
6
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
h) g2(2) = 36x + 7
= 36(2) + 7
= 72 + 7= 65
i) gf(x) = 3
18x 35 = 3
18x = 3 + 35
18x = 38
x =18
38
=9
12
j) g2(x) = 2
36x + 7 = 2
36x = 2 7
36x = 9
x =36
9
=4
1
7
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
2. Givenf(x) =9
42 +xfor2 x 2 and g(x) =x + 2.
b) Find(iii)fg
(iv)gfc) Find the range of
(iii)fg
(iv)gf
Solution:
a)(i) fg(x) =f(x + 2)
=9
4)2(2 ++x
=9
442 ++x
fg(x) =9
82 +x
(ii) gf(x) =f(9
42 +x)
=9
42 +x+ 2
=9
1842 ++x
gf(x) =9
222 +x
b)
(i) To find the range offg, substitutex = 2 andx = 2 intofg(x)
fg(2) =9
8)2(2 +
=9
84+
8
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Activity Sheet Lesson 5Composite Functions.
________________________________________________________________
=9
12
=34
fg(2) =9
8)2(2 +
=9
84 +
=9
4
= 9
4
The range offg is9
4fg(x)
3
4
(ii) To find the range ofgf, substitutex = 2 andx = 2 into gf(x)
gf(2) =9
22)2(2 +
=9
82
gf(2) =9
22)2(2 +
= 2
The range ofgfis 2 gf(x) 9
82
9
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Activity Sheet Lesson 6Image of a Composite Function
________________________________________________________________Name:
Class: Date:
1. Given gf:x2x2 + 1 and g :xx +1, determine the function off(x).
Solution:
2. Given gf:x8x 9 andf:xx + 4, determine the function ofg(x).
Solution:
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Activity Sheet Lesson 6Image of a Composite Function
________________________________________________________________
3. Iffg :x 18
7+x andf:x 1
8
1x , what is the function of g(x)?
Solution:
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Activity Sheet Lesson 6Image of a Composite Function
________________________________________________________________
1. Given gf:x2x2 + 1 and g :xx +1, determine the function off(x).
Solution:Given gf:x2x2 + 1 g :xx + 1
gf(x) = 2x2 + 1 g(x) =x + 1When we write the composite function in the form ofg, we will get
g[f(x)] = 2x2 + 1When we apply the rule of g
f(x) + 1 = 2x2 + 1We will get,
f(x) = 2x2
2. Given gf:x8x 9 andf:xx + 4, determine the function ofg(x).
Solution:Given gf:x8x 9 g :xx + 4
gf(x) = 8x 9 g(x) =x + 4When we write the composite function in the form ofg, we will get
g[f(x)] = 8x 9When we apply the rule off
g(x + 4) = 8x 9Lety =x + 4, thereforex =y 4
g(y) = 8x 9= 8(y 4) 9= 8y 32 9= 8y 41
So g(y) = 8y 41Replacingy withx,
g(x) = 8x 41
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Activity Sheet Lesson 6Image of a Composite Function
________________________________________________________________
3. Iffg :x 18
7+x andf:x 1
8
1x , what is the function of g(x)?
Solution:
Given fg :x 18
7+x f:x 1
8
1x
fg(x) = 18
7+x f(x) = 1
8
1x
When we write the composite function in the form off, we will get
f[g(x)] = 18
1x
When we apply the rule off
1)(8
1xg = 1
8
7+x
And make the necessary calculation,
1)(8
1xg = 1
8
7+x
11)(8
1+xg = 11
8
7++x
)(8
18 xg = 8)2
8
7( +x
We will get,
g(x) = 7x + 16
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
Name:
Class: Date:
1. Use the method of inverse mapping to find the corresponding objects ofimages, for each of the given functions belowa) f:x5x + 2
Images 6 and 4.
b) h :x 2x + 9Images 0 and 5.
SOLUTION:
1
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
2. Given d(x) =
x
x 2wherex 0. Find (i) d1(2) (ii) d1(3).
SOLUTION:
2
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
3. Given g(x) =1
12
+
x
xwherex1. Find (i) g1(1) (ii) g1(5)
SOLUTION:
3
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
1. Use the method of inverse mapping to find the corresponding objects ofimages, for each of the given functions belowa) f:x5x + 2
Images 6 and 4.
b) h :x 2x + 9Images 0 and 5.
SOLUTION:a) Image is 6
5x + 2 = 65x = 6 2 = 8
x = 5
8
x =5
8or
5
31
Therefore the object for6 is5
8or
5
31 .
Image is 45x + 2 = 45x = 4 2 = 2
x =5
2
Therefore the object for 4 is5
2 .
b) Image is 02x + 9 = 02x = 9
x =2
9
x =2
9 or
2
14
Therefore the object for 0 is 2
9
or 2
14
.Image is 52x + 9 = 52x = 5 9 = 4
x =2
4 = 2
Therefore the object for 5 is 2.
4
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
2. Given d(x) =
x
x 2wherex 0. Find (i) d1(2) (ii) d1(3).
SOLUTION:(i) Find d1(2)
Let d1(2) =xd(x) = 2
x
x 2= 2
x 2 = 2xx + 2x = 2
3x = 2
x =3
2
d1(2) =
3
2
(ii) Find d1(3)
Let d1(3) =xd(x) = 3
x
x 2
= 3
x2 = 3x3xx = 2
2x = 2
x =2
2 = 1
d1(3) = 1
5
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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and
function________________________________________________________________
3. Given g(x) =1
12
+
x
xwherex1. Find (i) g1(1) (ii) g1(5)
SOLUTION:(i) Find g1(1)
Let g1(1) =xg(x) = 1
1
12
+
x
x= 1
2x 1 =x + 12xx = 1 + 1
x = 2g1(1) = 2
(ii) Find g1(5)Let g1(5) =x
g(x) = 5
1
12
+
x
x= 5
2x 1 = 5x + 52x 5x = 5 + 1
3x = 6
x =3
6 = 2
g1(5) = 2
6
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
Name:
Class: Date:
1. Find the inverse of the following functions.a) f(x) = 3xb) h(x) = 2x 5c) g(x) = 7x + 14
d) p(x) =3
102 +x
e) q(x) =2
23 x
f) f(x) = 43
7
+x wherex 3
4
g) h(x) =x2
5wherex 0
h) f(x) =14
3
xwherex
4
1
SOLUTION:
1
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
f(x) =2.5
3 x
f1(x) = 3 ______
OLUTION:
. f(x) = 2x 6
S
3
f1(x) =
_____
__ x __
OLUTION:S
2
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
4. f(x) =1
13
x
x
f1
(x) = _____
____ x
SOLUTION:
5. f(x) = 3x 11
f1(x) =
_____
____+x
SOLUTION:
3
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
1. Find the inverse of the following functions.i) f(x) = 3x
j) h(x) = 2x 5
k) g(x) = 7x + 14
l) p(x) =3
102 +x
m) q(x) =2
23 x
n) f(x) =43
7
+xwherex
3
4
o) h(x) =x2
5wherex 0
p) f(x) =14
3
xwherex
4
1
SOLUTION:a) letf(x) =y
y = 3x
x =3
y
f1(y) =
3
y
f1(x) =
3
x
b) let h(x) =yy = 2x 52x =y + 5
x =2
5+y
h1(y) =
2
5+y
h1(x) =
2
5+x
4
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
c) let g(x) =yy = 7x + 14
7x = 14 y
x =7
14 y
g1(y) =
7
14 y
g1(x) =
7
14 x
d) letp(x) =y
y =
3
102 +x
3y = 2x + 102x = 3y 10
x =2
103 y
p1(y) =
2
103 y
p1(x) =
2
103 x
e) let q(x) =y
y =2
23 x
2y = 3x 23x = 2y 2
x =3
22 y
q1(y) =
3
22 y
q1(x) =
3
22 x
5
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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra
________________________________________________________________
4. f(x) =1
13
x
x
f1
(x) = _____
____ x
SOLUTION:
f(x) =1
13
x
x
y =1
13
x
x
xyy = 3x1xy 3x =y1
x(y 3) =y1
x =3
1
y
y
f1(x) =
3
1
x
x
5. f(x) = 3x 11
f1(x) =
_____
____+x
SOLUTION:
f(x) = 3x 11y = 3x 113x =y + 11
x =3
11+y
f1(x) =
3
11+x
8
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Activity Sheet Lesson 9Determining the condition of an inverse function
________________________________________________________________Name:
Class: Date:
1. Determine whether the following functions have an inverse. Give a reasonfor your answer.(a)
0
f(x)
x
______________________________________________________
g(x)(b)
0
x
_____________________________________________________
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Activity Sheet Lesson 9Determining the condition of an inverse function
________________________________________________________________
(c)
a
b
c
u
v
BA
____________________________________________________________
(d)
w
x
y
p
r
q
QP
____________________________________________________________
(e) {(1, 3), (2, 3), (3, 7)}
____________________________________________________________
(f) {(6, 2), (7, 32), (8, 42)}
____________________________________________________________
(g)f:xtx2 + 3
____________________________________________________________
(h)p(x) =5
32 +x
____________________________________________________________
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Activity Sheet Lesson 9Determining the condition of an inverse function
________________________________________________________________Determine whether the following functions have an inverse. Give a reason for
your answer.(a)
0
f(x)
x
Has an inverse because it is a one-to-one function.
g(x)(b)
0
x
No inverse because it is a many-to-one function.
(c)
a
b
c
u
v
BA
No inverse because it is a many-to-one function.
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Activity Sheet Lesson 9Determining the condition of an inverse function
________________________________________________________________(d)
w
x y
p
r q
QP
Has an inverse because it is a one-to-one function.
(e) {(1, 3), (2, 3), (3, 7)}
No inverse because it is a many-to-one function.
(f) {(6, 2), (7, 32), (8, 42)}
Has an inverse because it is a one-to-one function.
(g)f:xtx2 + 3
No inverse because it is a many-to-one function.
(h)p(x) = 5 32 +
x
Has an inverse because it is a one-to-one function.
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Activity Sheet Lesson 10Recognising a Quadratic Equation
________________________________________________________________
Name:
Class: Date:
1. Which of the following is the general form of a quadratic equation?A) ax2 + bxy + c = 0, (x is a variable, a, b and c are real numbers and a 0)B) ax
2+ bx + c = 0, (x is a variable, a, b and c are real numbers and a 0)
C) ax3
+ bx2
+ cx + d= 0, (xis a variable, a, b, c and dare real numbers anda 0)
D) ax + b = 0, (xis a variable, a and b are real numbers and a 0)
SOLUTION:
2. Is 7
1x 10x2 = 4 a quadratic equation?
SOLUTION:
3. Is the following expression a quadratic equation?
x2 +
11
7x 18 = 5y
SOLUTION:
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Activity Sheet Lesson 10Recognising a Quadratic Equation
________________________________________________________________
4. Is the following expression a quadratic equation?3x2 x 18 < 0
SOLUTION:
1. Which of the following is the general form of a quadratic equation?
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Activity Sheet Lesson 10Recognising a Quadratic Equation
________________________________________________________________A) ax2 + bxy + c = 0, (x is a variable, a, b and c are real numbers and a 0)B) ax
2+ bx + c = 0, (x is a variable, a, b and c are real numbers and a 0)
C) ax3
+ bx2
+ cx + d= 0, (xis a variable, a, b, c and dare real numbers anda 0)
D) ax + b = 0, (xis a variable, a and b are real numbers and a 0)
SOLUTION:B) ax
2+ bx + c = 0, (x is a variable, a, b and c are real numbers and a 0)
2. Is 7
1x 10x2 = 4 a quadratic equation?
SOLUTION:First, we rewrite the expression as follows:
10x2 +
7
1x 4 = 0
We know the expression is an equation.It involves only one variable, which isx.Powers of the variable are 1 and 2, which are positive integers and thehighest power is 2.
Therefore,7
1x 10x2 = 4 is a quadratic equation.
3. Is the following expression a quadratic equation?
x2 +117 x 18 = 5y
SOLUTION:First, we rewrite the expression as follows:
x2 +
11
7x 18 5y = 0
We know the expression is an equation.It involves two variables,x andy.Therefore,x2 +
11
7x 18 = 5y is not a quadratic equation.
4. Is the following expression a quadratic equation?
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Activity Sheet Lesson 10Recognising a Quadratic Equation
________________________________________________________________3x2 x 18 < 0
SOLUTION:
Note that this is an inequality, therefore 3x2
x 18 < 0 is not a quadraticequation.
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Activity Sheet Lesson 11Determining the roots of a quadratic equation by substituting
________________________________________________________________
Name:
Class: Date:
1. Determine ifx= 2 is the root of this quadratic equation,2x2 + 16x+ 10 = 106.
SOLUTION:
2. Determine ifx
= 5 is the root of this quadratic equation,x
2
10x
+ 35 = 10.
SOLUTION:
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Activity Sheet Lesson 11Determining the roots of a quadratic equation by substituting
________________________________________________________________
1. Determine ifx= 2 is the root of this quadratic equation,2x2 + 16x+ 10 = 106.
SOLUTION:To determine:
1. Substitutex= 2 into the equation.
2. Determine whether the value on the LHS = value on the RHS.
Left-hand side = 2x2 + 16x+ 10= 2(2)2 + 16(2) + 10
= 2(4) + 32 + 10
= 8 + 42= 50
Right-hand side
Since LHS RHS, hence x= 2 is not a root of the equation.
2. Determine ifx= 5 is the root of this quadratic equation,x2 10x+ 35 = 10.SOLUTION:To determine:
1. Substitutex= 5 into the equation.
2. Determine whether the value on the LHS = value on the RHS.
Left-hand side =x2 10x+ 35
= (5)2 10(5) + 35
= 25 50 + 35
= 10
= Right-hand side
Since LHS = RHS, hencex= 5 is a root of the equation.
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Activity Sheet Lesson 12Determining roots of quadratic equations by inspection
________________________________________________________________
Name:
Class: Date:
1. From the list below choose the correct roots of the equation (x + 5)(x 2) = 0.
SOLUTION
a) For ,5=x 2
c) For =x
5=x
,
b) For ,d) For
2
5=x ,
2. Determine the roots of 0)3)(7( =++ xx by inspection.
SOLUTION:For ,7=x
For 3=x ,
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Activity Sheet Lesson 12Determining roots of quadratic equations by inspect
__________________________________________ ion
______________________
2
=x
00
0)7)(0(
0)25)(55(
0)2)(5(
=
=
=+
1. From the list below choose the correct roots of the equation (x + 5)(x 2) = 0.
SOLUTIONA) For ,5
=+ xx
So it is a root.
C) For 2=x ,
012
0)4)(3(
0)22)(52(
0)2)(5(
=
=+
=+ xx
So it is not a root.
B) For ,5=x
030
0)3)(10(0)25)(55(
0)2)(5(
=
=+
=+ xx
So it is not a root.
D) For 2
5=x ,
04
15
0)2
1)(
2
15(
0)22
5)(5
2
5(
0)2)(5(
=
=+
=+ xx
So it is not a root.
2. Determine the roots of 0)3)(7( =++ xx by inspection.
SOLUTION:For ,7=x
0)4)(0(
0)37)(77(
0)3)(7(
=
=++
=++ xx
So is a root.7=x
For 3=x ,
0)0)(4(
0)33)(73(
0)3)(7(
=
=++
=++ xx
So 3=x is a root.
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Activity Sheet Lesson 13Determining roots of quadratic equations by trial and
improvement________________________________________________________________
Name:
Class: Date:
1. Find the roots of the following equations.a) x2 +x 6 = 0b) x2 16 = 0c) 6x2 + 42x 60 = 0
Solution:
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Activity Sheet Lesson 13Determining roots of quadratic equations by trial and
improvement________________________________________________________________
1. Find the roots of the following equations.a) x2 +x 6 = 0b) x2
16 = 0
c) 6x2 + 42x 60 = 0
Solution:a) x= 3
x2 +x 6 = (3)2 + (3) 6
= 9 3 6= 9 9= 0
x= 2x2 + x
6 = (2)2 + (2)
6
= 4 + 2 6= 6 6= 0
The roots ofx2 +x 6 are 3 and 2.
b) x= 4x
2 16 = (4)2 16
= 16 16= 0
x= 4x2
16 = (
4)2
16
= 16 16= 0
The roots ofx2 16 are 4 and 4.
c) x= 56x2 + 42x 60 = 6(5)2 + 42(5) 60
= 6(25) + 210 60= 150 + 150= 0
x= 2
6x2 + 42x
60 =
6(2)2 + 42(2)
60
= 6(4) + 84 60= 24 + 24= 0
The roots of6x2 + 42x 60 are 5 and 2.
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Activity Sheet Lesson 14Determining the roots of quadratic equations by factorization
________________________________________________________________Name:
Class: Date:
1. Find the roots of the quadratic equation.x(6x+ 5) = 4?
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Activity Sheet Lesson 14Determining the roots of quadratic equations by factorization
________________________________________________________________
1. Find the roots of the quadratic equation.x(6x+ 5) = 4?
SOLUTION:x(6x+ 5) = 46x2 + 5x = 4
6x2 + 5x 4 = 0(2x 1)(3x+ 4) = 0
2x 1 = 0 or 3x+ 4 = 0
x=21 or x =
34
Hence,x=21andx=
34
are the roots of the equation.
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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the
square________________________________________________________________
Name:
Class: Date:
QUESTION 1(a)x2 2x 1 = 0
SOLUTION:
(b)x2 10x + 5 = 0
SOLUTION:
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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the
square________________________________________________________________
(c)x2 = 3(x + 2)
SOLUTION:
(d) 2p2 + 5p 6 = 0
SOLUTION:
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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the
square________________________________________________________________
(e) 6x2 + 7x = 3
SOLUTION:
(f) 8x = 12x2 5
SOLUTION:
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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the
square________________________________________________________________
QUESTION 1(a)x2 2x 1 = 0
SOLUTION:x
2 2x = 1
x2 2x + (
22 )
2 = 1 + (
22 )
2
(x 1)2 = 1 + 1(x 1)2 = 2
x 1 = 2
x 1 = 1.4142x = 1 + 1.4142 or x = 1 1.4142
= 2.4142 = 0.4142
The solutions are 2.4142 and 0.4142.
(f) x2 10x + 5 = 0
SOLUTION:x
2 10x + 5 = 0x
2 10x =5
x2 10x + (
210
)2 = 5 + (2
10 )2
(x 5)2 = 5 + (5)2
(x 5)2 = 20
x 5 =
20x 5 = 4.4721
x = 5 + 4.4721 or x = 5 4.4721= 9.4721 = 0.5279
The solutions are 9.4721 and 0.5279.
(g)x2 = 3(x + 2)
SOLUTION:x
2 = 3(x + 2)x
2 = 3x + 6
x
2
3x 6 = 0x
2 3x = 6
x2 3x + (
23 )2 = 6 + (
23 )2
(x 23 )2 = 6 + 4
9
(x23 )2 =
433
x23
= 433
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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the
square________________________________________________________________
x +127 =
1211
x +127 =
1211
or x +127 =
1211
x =127
+1211
x =127
1211
= 0.3333 = 1218
= 1.5The solutions are 0.3333 and 1.5.
(f) 8x = 12x2 5
SOLUTION:
8x = 12x2 512x
2 8x 5 = 0
x2
128 x
125 = 0
x2
32x
125 = 0
x2
32x =
125
x2
32x+ ( 2
32
)2 =125 + ( 2
32
)2
(x 31 )2 =
3619
x 31 = 36
19
x 31 = 0.7265
x = 0.3333 + 0.7265 or x = 0.3333 0.7265= 1.0598 = 0.3932
The solutions are 1.0598 and 0.3932.
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Activity Sheet Lesson 16Determining roots of quadratic equations by using a formula
________________________________________________________________
Name:
Class: Date:
1. Solve the following equation by using the formula.(x 2)(x + 1) = 3x 4
Solution:
2. Ifx = 2 is a root of the quadratic equation below, find the value oftand theother root using the formula.x
2 + tx 6t= 0
Solution:
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Activity Sheet Lesson 16Determining roots of quadratic equations by using a formula
________________________________________________________________
1. Solve the following equation by using the formula.(x 2)(x + 1) = 3x 4
Solution:(x 2)(x + 1) = 3x 4
x2x 2 = 3x 4
x2 4x + 2 = 0
a
acbbx
2
42
=
)1(2
)2)(1(4)4()4(2
=x
2
8164 =x
2
84 =x
2
224=x
22=x
x = 2 + 1.414 orx = 2 1.414x = 3.414 or x = 0.586
2. Ifx = 2 is a root of the quadratic equation below, find the value oftandthe other root using the formula.
x2 + tx 6t= 0
Solution:Sincex = 2 is a root, then it satisfies the equation, substitutex = 2 into theequation.
x2 + tx 6t= 0
(2)2 + t(2) 6t= 0
4
2t
6t= 04 8t = 08t= 4
t=2
1
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Activity Sheet Lesson 16Determining roots of quadratic equations by using a formula
________________________________________________________________
Therefore,x2 +2
1x 6(
2
1) = 0
2x2 +x 6 = 0
To find the other root,a = 2, b = 1 and c = 6.
a
acbbx
2
42
=
)2(2
)6)(2(4)1()1(2
=x
4
)48(11 =x
4
4811 +
=x
4
491=x
4
71=x
4
71+=x or
4
71=x
4
6=x or
4
8=x
211=x or 2=x
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Activity Sheet Lesson 17Determine roots of a quadratic equation
________________________________________________________________
Name:
Class: Date:
1. If 2 and 31 are the roots of a quadratic equation, what is the quadratic
equation?
SOLUTION:
2. Letp and q be the roots of the quadratic equation 3x2 + 12x 6 = 0. Form aquadratic equation with roots (p + 1) and (q + 1).
SOLUTION:
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Activity Sheet Lesson 17Determine roots of a quadratic equation
________________________________________________________________
1. If 2 and 31 are the roots of a quadratic equation, what is the quadratic
equation?
SOLUTION:1st way:First, we form the factorised form of the quadratic equation by using the roots.
x = 2 andx = 31 are the roots.
(x 2)(x (31 )) = (x 2)(x +
31 )
Then, we multiply the factors.
(x 2)(x +31 ) =x2 +
31x 2x
32
=x2 35x 3
2
Quadratic equation: x2
35x 32 = 03x2 5x 2 = 0
2nd way:x
2 (sum of roots)x + (product of roots) = 0sum of roots =p + q
= 2 31
=35
product of roots =pq
= 2 (31 )
= 32
x2 (sum of roots)x + (product of roots) = 0
x2 ( 3
5 )x + (32 ) = 0
x2
35x
32 = 0
3x2 5x 2 = 0
2. Letp and q be the roots of the quadratic equation 3x2
+ 12x 6 = 0. Form aquadratic equation with roots (p + 1) and (q + 1).
SOLUTION:First, rewrite the quadratic equation in the following format.x
2 (sum of roots)x + (product of roots) = 0
3x2 + 12x 6 = 0 x2 (4)x 2 = 0Then, sum of roots =p + q
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Activity Sheet Lesson 17Determine roots of a quadratic equation
________________________________________________________________
= 4
product of roots =pq
= 2
ets form a quadratic equation with roots (p + 1)(q + 1):
(sum of roots)x + (product of roots) = 0
x [(p + 1)(q + 1)]x + [(p + 1)(q + 1)] = 0
x (p + q + 2)x + (pq +p + q + 1) = 0
ubstitutep + q = 4 andpq = 2 in the above equation.
hen, we get:
x2 (4 + 2)x + (2 + 4 + 1) = 0
x + 2x 5 = 0
L
2x
2
2
ST
2
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Activity Sheet Lesson 18Discriminants of a quadratic equation:
________________________________________________________________
3. Determine the type of roots in this quadratic equation.4x
2 20x + 25 = 0
SOLUTION:
4. Determine the type of roots in this quadratic equation.
32x
2 +65x +
71 = 0
SOLUTION:
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Activity Sheet Lesson 18Discriminants of a quadratic equation:
________________________________________________________________
5. Determine the type of roots in this quadratic equation.
3x2+ 5x + 15 = 0
SOLUTION:
6. Determine the type of roots in this quadratic equation.ax
2 + (2a + 3)x + a + 3 = 0
SOLUTION:
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Activity Sheet Lesson 18Discriminants of a quadratic equation:
________________________________________________________________
1. Determine the type of roots in this quadratic equation.x
2 + 3x + 7 = 0
SOLUTION:x
2 + 3x + 7 = 0b
2 4ac = (3)
2 4(1)(7)
= 9 28
= 19
b2
4ac < 0no roots.
2. Determine the type of roots in this quadratic equation.2x
2 + 5x + 4 = 0
SOLUTION:
2x2 + 5x + 4 = 0b
2 4ac = (5)
2 4(2)(4)
= 25 + 32
= 57
b2
4ac > 0two different roots.
3. Determine the type of roots in this quadratic equation.4x
2 20x + 25 = 0
SOLUTION:
4x2 20x + 25 = 0b
2 4ac = (20)2 4(4)(25)
= 400 400
= 0
b2
4ac = 0two equal roots.
4. Determine the type of roots in this quadratic equation.
32x
2 +65x +
71 = 0
SOLUTION:
32x
2 +65x +
71 = 0
b2
4ac = (65 )
2 4(
32 )(
71 )
=3625 +
218
=25279
b2
4ac > 0two different roots.
5. Determine the type of roots in this quadratic equation.
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Activity Sheet Lesson 18Discriminants of a quadratic equation:
________________________________________________________________
3x2+ 5x + 15 = 0
SOLUTION:
3x2+ 5x + 15 = 0
b2
4ac = ( 5 )2
4( 3 )( 15 )
= 5 12 5
= 5 26.833
= 21.833
b2
4ac < 0no real roots.
6. Determine the type of roots in this quadratic equation.ax
2 + (2a + 3)x + a + 3 = 0
SOLUTION:ax
2 + (2a + 3)x + a + 3 = 0b
2 4ac = (2a + 3)2 4(a)( a + 3)
= 4a2 + 12a + 9 4a2 12a
= 9
b2
4ac > 0two different roots.
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
Name:
Class: Date:
1. (a) Given that the quadratic equationx2 4qx +p2 = 0 has two different roots.Show that 4q2 >p2.
SOLUTION:
(b) Given that the equation m2x2 + 3nx + 2 = 0 has no roots. Find the relation
between m and n.
SOLUTION:
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
(c) If the equation 2kx2 (4k+ 6)x + 2k= 0 has no roots, what is the range ofvalues of k?
SOLUTION:
(d) If the equation rx2 6x + 2 = 0 has no roots, what is the range for r?
SOLUTION:
(e) If the following equation has two different roots, what is the range for h?x
2 x h = 0
SOLUTION:
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
(f) If the following equation has two different roots, what is the range for s?
0322
2= sxx
SOLUTION:
(g) Find the values of tif the following equation has two equal roots.x
2 8x = t2SOLUTION:
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
1. (a) Given that the quadratic equationx2 4qx +p2 = 0 has two different roots.Show that 4q2 >p2.
SOLUTION:a = 1, b = 4q, c =p2
b2 4ac > 0
(4q)2 4(1)(p2) > 016q2 4p2 > 0
16q2 > 4p2
4q2 >p2(b) Given that the equation m2x2 + 3nx + 2 = 0 has no roots. Find the relation
between m and n.
SOLUTION:a = m2, b = 3n, c = 2
b2 4ac < 0
(3n)2 4(m2)(2) < 09n2 8m2 < 0
9n2 < 8m2
(c) If the equation 2kx2 (4k+ 6)x + 2k= 0 has no roots, what is the range ofvalues of k?
SOLUTION:
a = 2k, b = (4k+ 6), c = 2kSince this quadratic equation has no roots
b2 4ac < 0
((4k+ 6))2 4(2k)(2k) < 016k2 + 48k+ 36 16k2 < 0
48k+ 36 < 048k< 36
k< 4836
k< 43
(d) If the equation rx2
6x + 2 = 0 has no roots, what is the range for r?
SOLUTION:a = r, b = 6, c = 2Since this quadratic equation has no roots
b2 4ac < 0
(6)2 4(r)(2) < 036 8r < 0
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
8r< 368r > 36
r >8
36
r > 29
r > 4 21
(e) If the following equation has two different roots, what is the range for h?x
2 x h = 0
SOLUTION:
x2 x h = 0
a = 1, b = 1, c = hSince this quadratic equation has two different roots
b2 4ac > 0(1)2 4(1)(h) > 0
1 + 4h > 04h > 1
h > 41
(f) If the following equation has two different roots, what is the range for s?
0322
2= s
xx
SOLUTION:
A =21 , b =
21
, c = -3s
Since this quadratic equation has two different rootsb
2 4ac > 0
(21 )2 4(
21 )(3s) > 0
41 + 6s > 0
6s > 41
s > 241
(g) Find the values of tif the following equation has two equal roots.x
2 8x = t2
SOLUTION:
x2 8x + t2 = 0
a = 1, b = 8, c = t2
b2 4ac = 0
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Activity Sheet Lesson 19Solving problems involving the use of the discriminants
________________________________________________________________
(8)2 4(1)(t2) = 064 4(1)(t2) = 0
64 = 4t2
t2
= 16t = 16
t = 4 or t= 4
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Activity Sheet Lesson 20Recognising Quadratic functions
________________________________________________________________
Name:
Class: Date:
1. (a) Is f(x) = 4x2 + x + x5 a quadratic function?
SOLUTION:
(b) Is the following expression a quadratic function?x2 +
85 x 13 = f(x)
SOLUTION:
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________________________________________________________________
(c) Is the following expression a quadratic function?
7x 21
xf(x) = 0
SOLUTION:
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Activity Sheet Lesson 20Recognising Quadratic functions
________________________________________________________________
1. (a) Is f(x) = 4x2 + x + x5 a quadratic function?
SOLUTION:
The function involves only one variable, which is x. The powers of the variable in the terms are 1, 2 and 5 which are
positive integers but the highest power is 5.Therefore, f(x) = 4x2 + x + x5 is not a quadratic function.
(c) Is the following expression a quadratic function?
x2 + 85x 13 =f(x)
SOLUTION:
First, we rewrite the function as follows:
f(x) = x2 + 85x 13
The function involves only one variable, which is x. The powers of thevariable in the terms are 1 and 2 which are positive integers and thehighest power is 2.
Therefore, x2 + 85x 13 =f(x) is a quadratic function.
(c) Is the following expression a quadratic function?
7x 21
xf(x) = 0
SOLUTION: First, we rewrite the function as follows:
f(x) = 7x +x2
The function involves only one variable, which is x.
The powers of the variable in the terms are 1 and 2. Note that 2 isnot a positive number.
Therefore, 7x 21
xf(x) = 0 is not a quadratic function.
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Activity Sheet Lesson 20Recognising Quadratic functions
________________________________________________________________
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Activity Sheet Lesson 21Quadratic Functions
________________________________________________________________
2. Plot the graph of quadratic function based on the given tabulated values?
x 2 1 0 1 2 3
f(x) 16 6 0 2 0 6
SOLUTION
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Activity Sheet Lesson 21Quadratic Functions
________________________________________________________________
3. By using a suitable scale, plot the graph of the functiony =x2 + 2x + 2 for4 x 1.
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Activity Sheet Lesson 21Quadratic Functions
________________________________________________________________
5. By using a suitable scale, plot the graph of the functiony =x2 + 2x + 2 for4 x 1.To draw the graph for this function, we should follow the following steps.
i. Construct the table of values fory =x2
+ 2 x + 2.Whenx = 4, y = (4)2 + 2(4) + 2.
= 16 8 + 2= 10
Whenx = 3, y = (3)2 + 2(3) + 2.= 9 6 + 2= 5
Whenx = 2, y = (2)2 + 2(2) + 2.= 4 4 + 2= 2
Whenx = 1, y = (1)2 + 2(1) + 2.
= 1 2 + 2= 1
Whenx = 0, y = (0)2 + 2(0) + 2.= 0 + 2= 2
Whenx = 1,y = (1)2 + 2(1) + 2.= 1 + 2 + 2= 5
The table of values is
x 4 3 2 1 0 1
f(x) 10 5 2 1 2 5
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Activity Sheet Lesson 21Quadratic Functions
________________________________________________________________
8
6. By using a suitable scale, plot the graph of the functiony = 2x2 + 8 for
3 x 3.
Solution
To draw the graph for this function, we should follow the following steps.i. Construct the table of values fory = 2x2 + 8.Whenx = 3, y = 2(3)2 + 8
= 18 + 8= 10
Whenx = 2, y = 2(2)2 + 8= 8 + 8= 0
Whenx = 1, y = 2(1)2
+ 8= 2 + 8= 6
Whenx = 0, y = 2(0)2 + 8= 0 + 8= 8
Whenx = 1,y = 2(1)2 + 8= 2 + 8= 6
Whenx = 2,y = 2(2)2 + 8= 8 + 8
= 0Whenx = 3,y = 2(3)2 + 8
= 18 + 8= 10
The table of values is
x 3 2 1 0 1 2 3
f(x) 10 0 6 8 6 0 10
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Activity Sheet Lesson 21Quadratic Functions
________________________________________________________________
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________2. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is positive, a >
0 the parabola opens .
SOLUTION:
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________3. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative, a 0 the parabola opens .
SOLUTION:
If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is positive, a > 0
the parabola opens upward
f(x)
f(x) = ax
+ bx + ca > 0
x
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________
3. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative,a < 0 the parabola opens .
SOLUTION:If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative, a < 0
the parabola opens downward
f(x)
f(x) = ax+ bx + ca < 0
x
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________
4. If the absolute value of the coefficient ofx in a quadratic functionf(x) = ax+ bx + c increases, a , the parabola becomes .
SOLUTION:If the absolute value of the coefficient ofx in a quadratic function
f(x) = ax+ bx + c increases, a , the parabola becomes narrower
f(x)
f(x) = ax+ bx + cnarrow
a
x
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________
5. If the absolute value of the coefficient ofx in a quadratic functionf(x) = ax+ bx + c decreases, a , the parabola becomes .
SOLUTION:If the absolute value of the coefficient ofx in a quadratic function
f(x) = ax+ bx + c increases, a , the parabola becomes wider
f(x) = ax+ bx + cwider
a
f(x)
x
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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions
________________________________________________________________
6. Which function does the graph below represent?
f(x)
x
f(x) =x+ bx + c f(x) =x+ bx + c
f(x) = x+ bx + f(x) = bx + cc
SOLUTION:
The graph is a parabola, therefore, it is a quadratic function:f(x) = ax+ bx + c
Since the graph opens downward, therefore a is negative.
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________Name:
Class: Date:
1. Determine the shape and position of the graphs of the following functions withrespect to thexaxis by considering the value ofa and b 4ac.
(a) f(x) = x+ 5x + 3(b) f(x) = x 4x + 4(c) f(x) = x+ 4(d) f(x) = 3xx(e) f(x) = 1 +2xx
(a)
(b)
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________(c)
(d)
(e)
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________QUESTION 1
Determine the shape and position of the graphs of the following functions withrespect to thexaxis by considering the value ofa and b 4ac.
(f) f(x) = x+ 5x + 3(g) f(x) = x 4x + 4(h) f(x) = x+ 4(i) f(x) = 3xx(j) f(x) = 1 +2xx
SOLUTION 1
(a) For the functionf(x) = x + 5x + 3, a = 1, b = 5, c = 3b 4ac = (5) 5(1)(3) = 10
Since a > 0 and b 4ac > 0, the graph off(x) has the shape of and
intersects thexaxis at two points as shown below
x
b) For the functionf(x) =x 4x + 4, a = 1, b = 4, c = 4b 4ac = ( 4) 4(1)(4) = 0
Since a > 0 and b 4ac = 0, the graph off(x) has the shape of and
touches thexaxis at one point as shown below
x
(c) For the functionf(x) =x + 4, a = 1, b = 0, c = 4b 4ac = (0) 4(1)(4) = 16
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________Since a > 0 and b 4ac < 0, the graph off(x) has the shape of and
does not intersect thexaxis as shown below
x
(d) For the functionf(x) = 3xx,= x + 3x
a = 1, b = 3, c = 0b 4ac = (3) 4(1)(0) = 9
Since a < 0 and b 4ac > 0, the graph off(x) has the shape of and
intersects thexaxis at two points, as shown below
x
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________
(e) For the functionf(x) = 1 +2xx= x + 2x1
a =
1, b = 2, c =
1b 4ac = (2) 4(1)(1) = 0
Since a < 0 and b 4ac = 0, the graph off(x) has the shape of and
touches thexaxis at only point, as shown below.
x
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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function
________________________________________________________________
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Activity Sheet Lesson 24Maximum and minimum values of Quadratic Functions.
________________________________________________________________Name:
Class: Date:
1. Find the minimum value of the following quadratic function.f(x) = 4x 24x 13
2. Find the maximum value of the following quadratic function.f(x) = 2x+ 10x 1
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Activity Sheet Lesson 24Maximum and minimum values of Quadratic Functions.
________________________________________________________________
1.Find the minimum value of the following quadratic function.
f(x) = 4x 24x 13
SOLUTION:f(x) = 4x 24x 13
= 4(x 6x) 13= 4(x 6x + 9 9) 13= 4(x 6x + 9) 36 13= 4(x 3) 46
49 is the minimum value of functionf.
2. Find the maximum value of the following quadratic function.
f(x) =
2x+ 10x
1
SOLUTION:
f(x) = 2x+ 10x 1= 2(x 5x) 1
= 2(x 5x +4
25
4
25) 1
= 2(x 5x +4
25) +
2
23
=
2(x
25 ) +
223
2
23is the maximum value of functionf.
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Activity Sheet Lesson 25Sketching graphs of quadratic functions
________________________________________________________________Name:
Class: Date:
1. Sketch the graph of the quadratic function.f(x) = (x 1) 4
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Activity Sheet Lesson 25Sketching graphs of quadratic functions
________________________________________________________________f(x) =x+ 2x 8
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Activity Sheet Lesson 25Sketching graphs of quadratic functions
________________________________________________________________1. Sketch the graph of the quadratic function.
f(x) = (x 1) 4
SOLUTION:
f(x) = (x
1)
4= x 2x + 1 4= x 2x 3
a =1
a >0, therefore
f(x) = (x 1) 4
Turning point = (1, 4)
The dicriminant,
b 4ac = (2) 4(1)(3)= 4 + 12= 16
b 4ac > 0
The graph intersects thexaxis at two points.
x 2x 3 = 0(x 3) (x + 1) = 0
f(x)
x-1
-3
(1, -4)
3
x 3 = 0 or x + 1 = 0x = 3 x = 1
thexintercepts are (3, 0) and (1, 0).
When x = 0,f(0) = 0 2(0) 3
= 3Theyintercept is (0, 3)
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Activity Sheet Lesson 25Sketching graphs of quadratic functions
________________________________________________________________2. Sketch the graph of the quadratic function:.
f(x) = (x 2)+ 1
SOLUTION:
f(x) = (x 2)+ 1= (x 4x + 4) + 1= x+ 4x 4 + 1= x+ 4x 3
a =1
a 0
The graph intersects thexaxis at two points.
(x
2)
+ 1= 0(x + 3) (x 1) = 0
x + 3 = 0 or x = 1x = 3
The graph intersects thexaxis at (3, 0)
f(x)
x-3
(2, 1)
3
and (1, 0).
When x = 0,f(0) = (0 2)+ 1
=
( 2)
+ 1= 4+ 1= 3
Theyintercept is (0, 3)
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Activity Sheet Lesson 25Sketching graphs of quadratic functions
________________________________________________________________
3. Sketch the graph of the quadratic function:.f(x) =x+ 2x 8
SOLUTION:
f(x) = x+ 2x 8
a = 1
a >0, therefore
f(x) = x+ 2x - 8
= x
+ 2x +
2
2
2
2
- 8
= x+ 2x + 1 1 8= (x + 1) 9
Turning point = ( 1, 9)The dicriminant,
b- 4ac = (2) 4(1)(8)= 4 + 32= 36
b- 4ac > 0
The graph intersects thexaxis at two points.
x+ 2x 8 = 0(x 2) (x + 4) = 0
x 2 = 0 or x + 4 = 0x = 2 x = 4
f(x)
x4
8
(1, 9)
2
The graph intersects thexaxis at (2, 0)and (
4, 0).
When x = 0,f(0) = (0)+ 2(0) 8
= 8
Theyintercept is (0, 8)
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Activity Sheet Lesson 26Determining the range of values of x that satisfies a quadratic inequality
__________________________________________________________
1. Find the range of values of x for which x(2 x) >21
x2.
SOLUTION:
x(2 x) > 21x2
0 < x 0
34
0
2x x2 >21x
2
4x 2x2 >x2
4x 3x2 > 0Letf(x) = 4x 3x2
Whenf(x) = 04x 3x2 = 0
Forf(x) > 0, the range of
values ofx is 0 < x 0, the range of
values of m is m 4 and m 4.
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Lesson 27
Activity Sheet
Solving Simultaneous Equations
Name:
Class Date:
1. Solve the following simultaneous equations.
i) 2 3 4 0x y + =
2
5 1 0x xy =
ii) 8 3 7y x= +
3 1xy =
iii) 3 2 5x y =
2 2 3 0x y y =
iv)2
5 2 2 9y x x xy y+ = + + =2
2. The line 0y x = intersects the curve 1xy = at the points p and q . Find the
length ofpq .
3. Find the coordinates of the points of intersection between the curve2 2
2 3 3 0x y x y+ + = and the line 1 0x = .
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Answer
1. i) 2.698, 0.4653x y= = ;
0.1588, 1.227x y= =
ii)8 1
,3 8
x y= = ;
1, 1
3x y= =
iii)4 1
1 ,5 5
x y= = ;
3, 2x y= =
iv) 8, 5;x y= =
2, 1x y= =
2. 2.828 units
3. ( ) ( )1,1 , 1, 4
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Lesson 28
Activity Sheet
Simultaneous Equations Involving Real-Life Situations
Name:
Class: Date:
Solve the following questions.
1. A wire of length 120 cm is cut into two pieces. Each piece of the wire is then
bent to form a square. If the total area of the two squares formed is 650cm 2 , find
the length of each piece of the wire.
2. Given that the sum of the circumferences of two circles is 30 cm and the areas
of the circles differ by 117 cm 2 , find the diameter of each circle.
3. Given that the sum of two positive numbers is 18 and the sum of the square ofthe numbers is 194. Find the two numbers.
4. Rosma bought x birds and y rabbits for RM208. Given that she bought a total
of 20 birds and rabbits altogether. A bird costs RMx while a rabbit costs RMy
where y x> . Find the values ofx and y .
5. Given that triangle ABC is right angled at B, where ( 3)AB x= cm,
( 9)AC y= + cm and 15BC = cm. If the perimeter of the triangle is 36 cm, find
the values ofx and y .
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Lesson 29
Activity Sheet
Finding the Value of Numbers Given In Index Form
Name:
Class: Date:
Find the values of
1.3
( 6)
2.
5
23
3.
3
215
4. ( )3
1.3
5. ( )0
7
6.
0
1
16
7.
3
1
4
8.
2
223
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Answer
1. 216
2.32
243
3.93
2
125
4. 2.197
5. 1
6. 1
7. 64
8.
9
64
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Lesson 30
Activity Sheet
Finding the Value of Numbers Given in Index FormThat are Multiplied Divided or Raised to a Power.
Name:
Class: Date:
Simplify each of the following.
(a)4 3
5 5
(b)5 2
3 3
(c) ( ) ( )5 4
0.3 0.3
(d)3 5
9 3
(e) ( ) ( )2 44
2 8
(f)
3 49 3
27
(g) ( )3
64
4 16
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Answer
(a) 5
(b) 2187
(c) 0.3
(d) 177 147
(e) 1 048 576
(f) 2187
(g)1
16
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Lesson 31
Activity Sheet
Use Laws of Indices to Simplify Algebraic Expressions
Name:
Class: Date:
1. Simplify each of the following.
i)
1
2 1 1
16
2 4
x
x x
+
+
ii)
3 2
4 2
n n
n
y y
y y
+
+
iii)1
3 3n n+
2. Solve each of the following
i)2 4 3 2 1
5 5125
x x+ =
ii)3 2
32 4x x
=
iii)
2
4 2739
y
y
y
+
=
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Answer
1. i) 8
ii)
1
y
iii) (2)3n
2. i) 1x =
ii) 5x =
iii) 2y =
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1.
1. 8
2.1
y
3. (2)3n
b.
1. 1x =
2. 5x =
3. 2y =
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Lesson 32
Activity Sheet
Converting Equations in Index Forms to Logarithmic Forms and Vice Versa
Name:
Class: Date:
1. Convert each of the following index form to the logarithmic form.
i)4
3 81=
ii)
2 1
5 25
=
iii)
1
21
366
=
iv)
0
11
10
=
v)
3
2
81 729=
2. Convert each of the following logarithmic form to the index form.
i) 161
log 42
=
ii)5
1log 2
25=
iii) 101
log 102
=
iv)4
3 log 64=
v)2
log 3x=
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Answer
1. i)3
log 81 4=
ii) 51
log 225 =
iii)36
1 1log
6 2=
iv) 110
log 1 0=
v) 813
log 7292
=
2. i)
1
216 4=
ii)2 15
25
=
iii)
1
210 10=
iv) 34 64=
v)32 x=
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Lesson 33
Activity Sheet
Finding Logarithm of a Number
Name:
Class: Date:
Evaluate each of the following.
1.10
log 1000
2.10
log 0.001
3.5
10log 10
4.
1
3
10log 10
510
log 5.22
6.5
10log 1.2
7.3
10log 8
8.3
10log 1.55
9.3
10log 0.123
10 .
2
10
2log
9
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Answer
1. 3
2. 3
3. 5
4.1
3
5. 0.7177
6. 0.3959
7. 2.709
8. 0.5710
9. 2.730
10. 1.306
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Lesson 34
Activity Sheet
Finding Logarithm of Numbers by Using Laws of Logarithms
Name:
Class: Date:
1. Evaluate each of the following.
i)6 6
log 4 log 9+
ii)5 5
log 75 log 3
2. Given that log 2 0.631x
= and log 3 1.465x
= , evaluate
i) log 18x
ii) log 36x
iii) log 0.75x
iv) log 6x
3. Evaluate each of the following.
i) 2log5 log4 log10+
ii)8 8 8
1log 24 log log 3
8
iii)10 10
2 1log 125 log 160
3 2+
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Answer
1. i) 2
ii) 2
2. i) 3.561
ii) 4.192
iii) 1.161
iv) 1.048
3. i) 1
ii) 2
iii) 3
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Lesson 35
Activity Sheet
Simplifying Logarithmic Expressions
Name:
Class: Date:
1. Simplify each of the following.i)
3 3log 6 log 3+
ii) 3 3log 18 log 2 iii)
5 5 5log 20 log 5 log 4x+
2. Evaluate each of the following.
i)10 10
log 20 log 50+
ii)3 3
log 1 log 27
iii)2 2 2
1log 4 log 12 log
3+ +
iv)2
8log logx x
x x
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Answer
1. i)3
log 18ii) 2iii)
52 log x+
2. i) 3
ii) 3
iii) 4
iv) 4
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Lesson 36
Activity Sheet
Changing the Base of Logarithms
Name:
Class: Date:
1. Evaluate each of the following.
i) 3log 7
ii) 5log 6
2. Given that log 3a x= and log 5a y= , express the following in terms ofxand/ory
i) 3log 5
ii) 3log 15
3. Given log 2p k= and log 5p m= , express each of following in terms ofk
and/ormi) 2log 20
ii) 5log 50
4. Given that 2 0.631logx
= and log 3 1.465x = , evaluate
i) 4log 12
ii) 6log 18
5. Without using a calculator, find the value of
2 9 6 7 18log 18 log 6 log 2 log 9 log 7
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Answer
1. i) 1.771
ii) 1.113
2. i)y
x
ii)y x
x
+
3. i)2k m
k
+
ii) 2m km
+
4. i) 2.161
ii) 1.699
5. 1
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Lesson 37
Activity Sheet
Solving Problems Involving Logarithms
Name:
Class: Date:
1. Given that log 3p
k= and log 5p
m= , express the following in terms of
kand m.i) log 45
p
ii)25
log 3p
2. Given that a = log 3 xand b= log 3 y, express the following in terms ofa and b.i) 3log 27
yx
ii) 3
3
1
log log 9yx
3. Given 2hA = and 4kB = , express the following in terms ofhand k.
i)
2
4log
A
B
ii)16
log AB
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Answer
1. i) 2k m+
ii) 12
k
m
+
2. i)3 3a
b
+
ii)6a
b
3. i) h k
ii)8 4
h k+
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Lesson 38
Activity Sheet
Equations involving indices
Name:
Class: Date:
1. Solve the following equations:
i) 4 16x =
ii) 19 3 243x x =
iii) 32 72 2 0x x+ + =
2. Solve the following equations:
i) 3 2 52 8x x+=
ii) 164 4x x+=
iii) 3 227 9x x+ =
iv) 1 3 13(9 ) 81x x+ +=
v) 4 127 1x+ =
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Answer
1.
i) 2x =
ii) 2x =
iii) 3x =
2.
i) 5x =
ii)1
2x =
iii) 9x =
iv)1
5x =
v) 3x =
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Lesson 39
Activity Sheet
Equations Involving Logarithms
Name:
Class: Date:
Answer
1. Solve the following equations.
i) 4 15x =
ii) 3(4 ) 5
x
=
iii)1
4 8x+
=
2. Solve each of the following equations.
i)2 3
4 10 0x
=
ii)
1
0.3 0.4
x x+
=
iii)1
5 2x x+
=
iv)1
2 (2 ) 7x x+
=
v)1
7 (2 ) 10x x+
=
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Answer
1. i) 1.953x =
ii) 0.3685x =
iii) 0.5x =
2. i) 2.330x =
ii) 3.185x =
iii) 1.756x =
iv) 0.9037x =
v) 0.1352x =
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Lesson 40
Activity Sheet
Equations Involving logarithms
Name:
Class: Date:
1. Find the distance between each of the following pairs of points.
i) ( 1, 4) ( 4,0)A B and
ii) (0,3) ( 4,1)A B and
iii) (8,3) ( 8, 6)A B and
iv) (6, 2) (7,3)A B and
2. Given that the distance between ( 3, )P m and (2,3)Q is 10 units, find the
value of m .
3. Show that A(1, 7), B( 3, 5) and C( 1, 1) are vertices of an isosceles
triangle.
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Answer
1.
i) 5 units
ii) 4.472 units
iii) 18.36 units
iv) 5.099 units
2. 5.660m = , 11.66m =
3. AB = 20 units
BC= 20 units
AC= 20 units
SinceAB=BC, therefore ABC is an isosceles triangle.
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Lesson 41
Activity Sheet
Midpoint of Two Given Points
Name:
Class: Date:
1. Find the coordinates of the midpoint of each of the following pairs of points.
i) ( 1, 4) and ( 4,0)A B
ii) (0,3) and ( 4,1)C D
iii) (8,3) and ( 8, 6)P Q
iv) (6, 2) and (7,3)R S
2. If ( 3, )P m is the midpoint of the straight line joining (2,3)Q andR( n,10),
find the values of m and n.
3. The coordinates of pointsA, B and Care (p, -7), (-3, q) and (-1, -1) respectively.IfCis the midpoint of the lineAB, find the values ofpand q.
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Answer
1. i)1
( 2 , 2)2
ii) ( 2,2)
iii)1
(0, 1 )2
iv)1 1
(6 , )
2 3
2.1
62
m = and 8n =
3. p= 2and q= 6
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Lesson 42
Activity Sheet
Finding the point that divides a line segment according to a given ratio
Name:
Class: Date:
1. For each of the following, find the coordinates ofQ for which Q divides PQ
according to the ratio as stated.
i) ( 1, 3), ( 4,0); : 1 :2P R PQ QR =
ii) (0,3) , ( 4,1); : 1 :3P Q PQ QR =
iii) (8,3) , ( 8, 6); : 2 :3P Q PQ QR =
iv) (6, 4) , Q(6,3); : 2 : 5P PQ QR =
2. Point B is a point lying on the straight line joining (2,1)A and ( 2,5)C so that
3AB BC= . Find the coordinates ofB .
3. Given thatA( 2, 7),B( 3, q) and C( 8, 1) lie on the straight line such that
: :AB BC m n= , find
i) the ratio :m n
ii) the value of q
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Answer
1.
i) ( 2, 2)Q
ii)1
( 1,2 )2
Q
iii)3 3
(1 , )5 5
Q
iv) (6, 2)Q
2. ( 1,4)B
3.
i) : 1:5m n =
ii) 6q =
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Lesson 43
Activity Sheet
Finding Areas of Triangles
Name:
Class: Date:
By determining the areas of specific geometrical shapes, find the area of each of the
following trianglesABC.
(i) A(2,0) ,B(2,10) , C(5,5)
(ii) A(1,3) ,B(5,4) , C(4,3)
(iii) A(-2,3) ,B(6,5) , C(2,12)
(iv) A(2,-3) ,B(-5,10) , C(-5,-5)
(v) A(-1,1) ,B(-5,0) , C(4,-3)
(vi)A(2,3) ,B(1,8) , C(-2,-5)
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Answer
(i)2
15 unit
(ii)2
1.5 unit
(iii)2
32 unit
(iv)2
52.5 unit
(v)2
8.5 unit
(vi)2
14 unit
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Lesson 44
Activity Sheet
Finding Areas of Triangles by Using the Formula
Name:
Class: Date:
1. For each of the following, find the area of the triangleABCwith the given vertices.
i) A(2, 0), B(2,0) , C(5, 5)
ii) A(
1,
3), B(5,
4), C(4, 3)
iii) A(2,3) ,B(6,15), C(4, 12)
iv) A(6,5) ,B(5,10), C(4, 9)
v) A(0, 1) ,B(5, 0), C(4 ,3)
vi) A(2, 3),B(1, 8), C(0, 0)
2. Show that pointsA(2, 3), B(1, 1) and C(2,5) are collinear.
3. Find the value ofm if the pointsA(
2, 3) ,B(6, 5) and C(2, m) are collinear.
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Answer
1.
i) 210 unit
ii)2
20.5 unit
iii) 212 unit
iv) 2102 unit
v) 212 unit
vi) 26.5 unit
2. The area ofABC= 0
3. m= 4
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Lesson 45
Activity Sheet
Finding Areas of Quadrilaterals
Name:
Class: Date:
1. For each of the following, find the area of quadrilateralABCDwith the given vertices.
i) (2,0), ( 2,0), ( 5,5), (8,8)A B C D
ii) ( 1, 3), (5, 4), (4,3), ( 7,3)A B C D
iii) (2, 3), (6,15), (4,12), (0,0)A B C D
iv) (6, 5), ( 5, 10), ( 4,9), (1,1)A B C D
v) (0,1), ( 5,0), (4, 3), (8,0)A B C D
2. Show that the pointsA(4,7) ,B(2,3) , C(1,1) and ( 2, 5)D are collinear.
3. Find the value ofm if the points ( 5, )A m , ( 1,3)B , C(3,2m) andD(7,5) are collinear.
4. Find the possible values ofpif the area of the quadrilateral with vertices
(3, 2)A ,B(6,p), C(4,5) and ( 2,4)D is1
29
2
unit 2 .
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Answer
1. i)2
53 unit
ii)2
53.5 unit
iii)2
30 unit
iv)2
97 unit
v)2
26 unit
2. Hint: The area ofABCD= 0
3. m = 2
4. p = 1 or p=119
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Lesson 46
Activity Sheet
Gradients of Straight Lines
Name:
Class: Date:
1. Find the gradient of the straight line that passes through each of the followingpairs of points.
i) A(2, 0) andB(2, 8)
ii) B(5, 4) and C(4, 3)
iii) A(2, 3) and D(0, 0)
iv)A( 6, 5) and C( 4, 9)
v) C(4, 3) andD(8, 0)
2. Given that the gradient of the straight line passing through P(1,m2) and
Q(4, 9) is1
2, find the value ofm.
3. Given that the gradient of the straight line passing throughA(1,m) and
B(4m, 9) is 2, find the value ofm.
4. Given that pointsA(1,t) andB(12t, 8). If the gradient of the lineAB is2
3 ,
find the value oft.
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Lesson 47
Activity Sheet
Intercepts and Finding the Gradient of a Straight Line Using the Intercepts
Name:
Class: Date:
Answer1. State thex-intercept andy-intercept of the stra