CAPS
Mathematics
Learner’s Book
M. Bradley • J. Campbell • S. McPetrie
11
TOPIC
1
Algebra: Exponents and surds
Unit 1: Laws of exponents (Revision)
From earlier grades you know the definition of an exponent and the laws
of exponents.
Definition: an = a × a × a … to n factors where n is a natural number
KEY WORD
exponent (index) – a
number or variable that shows
how many times a number
(the base) is multiplied by
itself:
base → 34→index or exponent
=3×3×3×3
Laws where the exponents m and n are natural numbers:
1
am × an = am + n
2
am ÷ an = am − n
When multiplying, add the
When dividing, subtract the
exponents of like bases.
exponents of like bases.
Example:
Example:
2a3b2 × 5ab5 = 2 × 5× a3 + 1 × b2 + 5
26.36 ÷ 24.33 = 26 − 4.36 − 3
= 10a4b7
= 22.33
= 4 × 27
= 108
3
DID YOU KNOW?
1
2−2 = __
1
3−2 = __
2−1
1
3−1 = __
4
1
= __
2
9
3
20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
30 = 1
31 = 3
32 = 9
33 = 27
34 = 81
REMEMBER
2.4 means 2 × 4 whereas 2,4
4.
means 2___
10
(am)n = amn
When raising a number with an
exponent to another exponent,
multiply the exponents.
Example:
(a6)3 = a18
Using Law 1: (a6)3 = a6.a6.a6
= a18
WORKED EXAMPLE 1
WORKED EXAMPLE 2
Simplify: (–2a2b3)(4b4a)(–2a5b)2
14x y x
Simplify: ________
12 2
SOLUTION
SOLUTION
(–2a2b3)(4b4a)(–2a5b)2
(
(
)
7 5 6 4
21x y
)
7 5 6 4
14x y x
________
21x12y2
Apply Law 3 to raise last bracket
to exponent 2.
Apply Laws 1 and 2 before raising
to exponent 4.
= [–2a2b3)(4b4a)(–2)2a10b2]
2x y
= _____
Apply Law 1
= −2 × 4 × 4 × a2 + 1 + 10b3 + 4 + 2
= −32a13b9
4
(ab)n = anbn
When two or more numbers are
multiplied and raised to an
exponent, raise each one to the
exponent.
Example:
(2a4b2)3 = 21 × 3a4 × 3b2 × 3
= 23a12b6
= 8a12b6
4
(
)
1 3 4
3
24x4y12
3
= ______
4
16x4y12
81
= _______
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 4
2012/07/02 2:20 PM
WORKED EXAMPLE 3
x+1
WORKED EXAMPLE 4
x−1
4
.3
Simplify: _________
x−2
Simplify:
1
(3a3 – b)2 = 9a6 – 6a3b + b2
12
SOLUTION
2x + 2 – 2x
________
2
4x + 1.3x − 1
_________
12x − 2
3.2x
SOLUTION
Break down the numbers into prime
factor bases.
= ___________
2
x−2
Be careful how you apply the laws when
there are more than two terms.
1
(3a3 – b)2 = 9a6 – 6a3b + b2
Apply Laws 3 and 4.
2
(22)x + 1.3x − 1
(2 .3)
22x + 2.3x − 1
= __________
2x − 4 x − 2
2
.3
2x + 2 – 2x
________
3.2x
The ‘top’ has two terms, so factorise.
Apply Laws 1 and 2.
x
22x + 2 – (2x – 4).3x – 1 – (x – 2)
=
= 192
=
2
2 (2 – 1)
| 2x + 2 = 2x.22
= ________
x
26.31
3.2
2x(3)
= _____
3.2x
=1
EXERCISE 1
Simplify the expressions without using a calculator:
1
16 a3b2
1 a5b12 × ___
__
4
2
5
4
4
4 3
2(2a )
______
16a12
3a3 × 5a5
_________
(5a3)2
3
6a + 9a
________
4
5
px − 1px + 1
6
−2(– 4a2 + b3)2
7
164x ÷ 42x
8
12x + 1.27x − 2
___________
9
22 010 × 52 011
____________
10
3x + 3x + 1
________
5a2
1001 000
182x − 1
8
KEY WORD
negative exponent – a
number raised to a negative
exponent and the reciprocal of
that number with a positive
exponent:
1 or ___
1 = am
a−m = ___
m
−m
Negative exponents and the meaning of a0
because 5 ÷ 5 = 1 and 51 ÷ 51 = 50 = 1 | Using Law 2
For example: (2x)0 = 1; 2x0 = 2 × 1 = 2; (a + b)0 = 1
a0 = 1
1 because 1 ÷ 5 = __
1 and 50 ÷ 51 = 5−1 | Using Law 2
a−m = ___
m
a
Examples: 2−3
a
5
1 = __
1
= __
23 8
a –1 = __
or ( __
) ab
a
b
x10 = x10 − 12 = x−2 = __
1
___
12
2
x
a
__
b
( )
−2
x
a−2 = __
b2
= ___
−2
2
b
a
When you simplify an expression with negative exponents, use the exponent laws
first and write the final answer with positive exponents. This is easier than trying to
get rid of the negative exponents first.
Unit 1 Laws of exponents (Revision)
PLT MATHS LB 11 7th pgs (Real Book).indb 5
5
2012/07/02 2:20 PM
WORKED EXAMPLE 1
−3
2x
Simplify: _____
−3
(2x)
SOLUTION
2x−3 = _____
2x−3 = 2 × 23 = 16
_____
−3
−3 −3
(2x)
2 x
WORKED EXAMPLE 2
−3
2x
Simplify: ____
−2
2 x
SOLUTION
2x−3 = 21 + 2x−3 − 1 = 23x− 4 = __
8
____
4
−2
2 x
x
WORKED EXAMPLE 3
3x−1 + y
x y
| This problem has more than one term in the numerator.
Simplify: _______
−1
REMEMBER
SOLUTION
If the expression involves two
or more terms, you cannot
invert from the top to the
bottom of the fraction.
3
( __
3x−1 + y ______
3 + xy
3 + xy
x + y)
_______
=
= ______
× __xy = ______
y
x
y
__
x−1y
x
(
)
or
3x−1 + y
3x−1.x + y.x ______
3 + xy
_______
× __xx = __________
= y
−1
−1
x y
x y.x
EXERCISE 2
Simplify the expressions without using a calculator. Give your answer with
positive exponents.
6
1
12a0b3c−1
_________
4
6(x y) (x y)
____________
−2 5 −3
15a b c
2
3
−2
−1
3x−3y5
−2 −2
2
(3xy )
________
5
(2a )
______
−2
3x y
−1 2
2(a2)−1
−1 −2 2
3
9 x y
________
6
92x.3−2
______
(3xy−1)−3
81x − 1
7
x y
( _____
)
z
8
(2a )
______
9
a−2 + a−1 + a0
___________
10
3−2 + 30
_______
11
(2.3)−2 + 2.3−2
12
(2−1 − 5−1)2
13
xy
_______
14
(a3 + 3)−2
15
3.3x
________
16
(3x ) + (2x )
____________
17
(8.2
)(4.2
)
______________
18
12x + 1.27x − 2
___________
2 −3 −1
−2
3−1
x−2 + y−2
4 2
2 4
5.25.2−5.x10
−1 0
2(a0)−1
x+1
2x − 3
x−1
a
3x + 3x − 1
182x − 1
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 6
2012/07/02 2:20 PM
Exponential equations
Exponential equations have the unknown or variables in the exponent.
To solve equations with an unknown index, express each side with the same base and
| (a > 0, a ≠ 1)
use the property: if ax = ay then x = y
WORKED EXAMPLE 1
WORKED EXAMPLE 2
Solve for x: 2x = 4x + 1
Solve for x: 8x + 1 = 32 ÷ 4x
SOLUTION
SOLUTION
2x = 4x + 1
= (22)x + 1
= 22x + 2
x = 2x + 2
x − 2x = 2
−x = 2
x = −2
8x + 1 = 32 ÷ 4x
(2 )
= 25 ÷ 22x
23x + 3 = 25 − 2x
3x + 3 = 5 − 2x
5x = 2
2
x = __
KEY WORD
exponential equation – an
equation with unknowns or
variables in the exponent
3 x+1
5
EXERCISE 3
Solve for x:
1
73x + 1 = 49x – 1
2
32x + 1 = 27x – 2
3
4.82x + 1 = 32x + 2
4
5.9x + 1 = 15
5
2.32x + 1 = 54
6
( __12 )x – 5 = 64
WORKED EXAMPLES
These equations have more than one term on each side.
• Rearrange the equations so the unknown exponents are on one side.
• Factorise by taking out a common factor.
• Solve the remaining exponential equation.
WORKED EXAMPLE 1
WORKED EXAMPLE 2
Solve for x: 2x + 2x + 1 = 24
Solve for x: 3x + 1 = 8 + 3x − 1
SOLUTION
SOLUTION
2x + 2x + 1 = 24
2x + 2x.21 = 24
2x(1 + 2) = 24
2x(3) = 24
2x = 8
= 23
x=3
3x + 1 = 8 + 3x − 1
− 3x.3−1 = 8
3x(31 − 3−1) = 8
3x.31
(
)
( )
1 =8
3x 3 − __
3
x
__
3 8 =8
3
3x = 3
x=1
Unit 1 Laws of exponents (Revision)
PLT MATHS LB 11 7th pgs (Real Book).indb 7
7
2012/07/02 2:20 PM
EXERCISE 4
Solve for x:
1
3x − 3x − 2 = 24
2
2
3x + 1 = 3x + __
3
3x + 3x + 2 = 10
4
3 = 2.5x
5x + 1 − ___
5
2x + 2x + 2x = 96
6
12x + 12x = 288
7
2x + 2x – 2 = 5
8
22x + 4x + 1 = 80
3
25
EXERCISE 5
Solve for x:
8
1
1 = ___
1
8x. ____
x−1
4
32
2
3x + 1
1
= 9 . ___
3
3x
4
5x
5
22x.4x + 1 = 1
6
12x.12x = 144
7
2x .2x
x
1
= 9.___
x
2+2
2
27
27
1 = 25x
.___
−1
2
25
2−2
( )
+ 1 = 65
−x
8
1
4x + 1 × __
4
1
___________
= ___
2x + 1
9
1
81.9x − 1 = ___
10
102x .10x + 1 = 10 000
2
32
27
2
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 8
2012/07/02 2:20 PM
Unit 2: Simplify expressions with
rational exponents
KEY WORD
m
__
Rational exponents and the meaning of a n
1
__
1
__
__
1
__
__
__
32 × 32 = 31 32 = √3 because √3 .√3 = 3
1
__
1
__
1
__
__
1
__
__
__
__
and a3 × a3 × a3 = a1 a3 = 3√a because 3√ a .3√ a .3√ a = a
1
__
so an =
__
√a
n
m
__
1
__
___
and a n = √a where a > 0 and m and n are natural numbers
n
___
m
a n = ( am )n = n√am or a n = ( an ) = (n√a )m
1
__
m
__
• 83 is the exponential form
of the cube root of 8
__
3
• √ 8 is its radical form
__
1 m
__
m
__
rational exponents – an
exponent or index with
a rational number as the
exponent
For example:
3
__
The laws for natural number exponents apply to rational exponents.
• 54 is the exponential form
of the 4th root of 125
___
____
4
4
• √ 53 = √125 is its radical
form
WORKED EXAMPLE 1 AND SOLUTIONS
( 6√22 )6 = 22 | Because ( 226 ) = 22
___
1
____
1
__
√ −32 = −2
| Because (−25)5 = −2
√ 93 = 27
| Because 92 = [(32)3]2 = 33 = 27
5
2
1 6
__
__
3
3
__
1
__
Note: It is easier to write the number under the root as an exponent and to cancel
the exponents using Law 3 of the exponents.
WORKED EXAMPLE 2 AND SOLUTIONS
______
1
√64x16y8 = 4x8y4 | For square roots, divide the exponents by 2.
2
3
3
√ 64x10 + 36x10
_______
√27x9y12 = 3x3y4 | For cube roots, divide the exponents by 3.
___________
______
= √ 100x
10
| Add the like terms first. DO NOT square root each term.
5
= 10x
WORKED EXAMPLE 3
Simplify without using a calculator:
8 –
( ___
27 )
4
__
3
SOLUTION
8 – = __
( ___
( 23 )–
27 )
4
__
3
3
3
4
__
3
–4
4
3 = ___
81
2 = __
= ___
–4
4
3
2
16
WORKED EXAMPLE 4
Simplify:
2
__
___
3
a3. √a4
______
3 4 ___
__
a4. √a4
SOLUTION
2
__
___
4
__
6
__
a4.a4
a4
2
__
3
3 3
a3. √a4 _____
a3 = __
a2 = a
______
= a__3.a__1 = __
4
3
__
__
a
__
4
a4. √ a
Unit 2 Simplify expressions with rational exponents
PLT MATHS LB 11 7th pgs (Real Book).indb 9
9
2012/07/02 2:20 PM
WORKED EXAMPLE 5
Prove that:
___
√ ax
x
______
___ = ____
( 3√ ax )2
a
3
SOLUTION
___
√ ax
x
______
___ = ____
( 3√ ax )2
a
3
___
√ ax
x
___ | Multiply by 1.
___ × ____
LHS = ______
3
3
2
3
( √ ax )
√ ax
___
___
___
ax
x ax
ax
___ 3 = ______ = ____ = RHS
= x______
3
3
3
√
√
( √ ax )
ax
3
√
a
EXERCISE 6
1
Write in exponential form:
1.1
2
___
√ x2
3
____
√x5y6
1.2
4
2.2
6x4
___
1.3
5 √x15
2.3
( 3x2 )3
4
Write in radical form:
2.1
3
__
x5
1
__
2
__
EXERCISE 7
1
Evaluate without using a calculator:
3
√ 163
1.4
( 0,125 )
1.3
4
1.7
1.9
____
14
( 2___
25 )
1
– __
2
__
1.6
__
(–2)0 + 3√8 + ( √ 3 )2
2
–__
( 0,064 )
3
1.11 49–2( 1253 – 90 )
2
__
1
__
2
Simplify:
2.1
2.4
2.7
2.10
3
10
2.2
________
√ a8b16c20
2.5
3
__
–8
2.8
4
( )
√ b2 2
____
2.11
1
__
b– 3
1.10
3
__
81– 4
2
__
( 64a6 )3
____________
√ 125a12b15c21
3
(x )
__
√x 2
___
3
– __
___
8 .25
√______
4
_________
8
.2
√_________
16
n
3.3
5
n
n
4
– 92 )
3
__
x+2
x
______
1
__
3
__
2.3
1
__
√(a + b)3 × (a + b)2
3
__
( 81a16b8 )4
_____________
√ 15a15 + 17b15
2.6
5
2.9
3
2.12
3.2
2n
x+1
(
2
__
273
___
1
__
√ x5 .x3
________
___
√ x2 .3√ x4
3
______
3.1
1
– __
( 0,0625 )
2
___
3
11
( 1___
25 )
1
__
1
__
16
( ___
x )
3
1
–__
2
1.12 16–2 + 164 – 64 –3
( 27x27 )3
4
2
– __
1.8
Simplify:
3.5
____
√ 642
1.2
4
1.5
_______
√ 10 000
1.1
4
√ 81a–8b12
_________
–1
1 a2b –3
__
4
________
15 .3
√_________
9
.5
x
x+1
x
x–2
______
____________
3.4
√(a + b)2 + √a2 + 2ab + b2
3.6
(x2 – 6x + 9)
1
– __
2
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 10
2012/07/02 2:20 PM
Unit 3: Solve equations with
rational exponents
Equations with rational exponents have the unknown in the base.
To solve equations with an unknown base, raise both sides of the equation to
the exponent which makes the exponent of x equal to 1.
To simplify the other side, write the number as a base with an exponent.
WORKED EXAMPLE 1
WORKED EXAMPLE 2
5
__
3
__
Solve for x: x2 = 27
Solve for x: x3 = 32
SOLUTION
SOLUTION
3
__
5
__
x2 = 27
3
__
2
__
x3 = 32
2
__
5
__
x2 × 3 = (33)3
x = 32
=9
3
__
3
__
x3 × 5 = (25)5
x = 23
=8
Equations with rational exponents that
have two solutions
Remember that x2 = 9 has two solutions.
Method 2
x2 = 9
__
x = ±√ 9
= ±3
Method 1
x2 − 9 = 0
(x − 3)(x + 3) = 0
x = 3 or x = −3
Method 3
x2 = 32
1
1
__
__
(x2)2 = ±(32)2
x = ±3
An equation in which the top of the rational exponent is even will have
two solutions.
WORKED EXAMPLE 1
2
__
Solve for x: x−3 = 9
SOLUTION
Method 1
2
– __
x
3
Method 2
2
__
x− 3 = 9
–9=0
( x−__13 − 3 )( x− __13 + 3 ) = 0
1
− __
x
3
=3
or x
1 or
x = 3−3 = ___
27
2
__
3
__
3
−__
x−3 × −2 = ±(32)
1
− __
3
2
x = ±3−3
= −3
1
x = −3−3 = −___
27
1
= ±___
27
Unit 3 Solve equations with rational exponents
PLT MATHS LB 11 7th pgs (Real Book).indb 11
11
2012/07/02 2:20 PM
WORKED EXAMPLE 2
2
__
Solve for x: 5x− 5 = 20
SOLUTION
2
__
| The coefficient of x must be 1 before you can raise to an exponent.
5x− 5 = 20
x
2
−__
5
| Divide both sides by 5.
=4
5
__
2
__
x− 5 × − 2 =
5
−__
±(22) 2
x = ±2−5
1
= ± ___
5.
| Raise both sides to the exponent −__
2
| There are two answers because the top of the rational
32
exponent is even.
EXERCISE 8
Solve for x:
1
__
1
x4 = 3
3
7
x– 2 = __
5
x– 3 – 25 = 0
7
2x3 = 128
9
27x4 = 125
11
8x4 – 27 = 0
1
__
2
2
__
2
__
3
__
3
__
3
__
2
x– 2 = 8
4
x– 3 + 4 = 0
6
x3 – 16 = 0
8
125x3 – 1 = 0
10
49x–2 – 1 = 0
12
81x3 = 16
1
__
2
__
4
__
Always factorise equations that have more than one term on each side.
WORKED EXAMPLE 3
1
__
1
__
Solve for x: x2 + 3x4 − 18 = 0
SOLUTION
1
__
1
__
x2 + 3x4 − 18 = 0
( x4 + 6 )( x4 − 3 ) = 0
1
__
1
__
1
__
1
__
x4 = −6 and x4 = 3
__
| This is a trinomial with a 2:1 ratio of the exponents.
| Factorise the trinomial.
| Raise both sides to the exponent 4.
__
but 4√x = −6 has no solution because 4√ x > 0
1
__
x4 = 3
x = 34 = 81
12
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 12
2012/07/02 2:20 PM
EXERCISE 9
Solve for x:
1
__
1
__
2
__
1
__
1
x2 − 5x4 + 6 = 0
2
x3 − 7x3 + 10 = 0
3
x3 − x3 − 2 = 0
4
2x2 − x4 − 3 = 0
5
3x2 − 5x4 − 2 = 0
6
5x + 7√x − 6 = 0
7
x − x2 −12 = 0
8
x−1 − x−2 = 20
9
2x + 7√x −15 = 0
10
6
x−3 = √95
2
__
1
__
1
__
1
__
1
__
1
__
home work
__
1
__
1
__
__
5
__
___
Unit 3 Solve equations with rational exponents
PLT MATHS LB 11 7th pgs (Real Book).indb 13
13
2012/07/02 2:20 PM
KEY WORD
surds – irrational roots or
irrational numbers
The square root of a nonperfect square is an irrational
number and is called a surd.
For example:
__
1
__
3 is the surd form of 32
• √___
simplified
• √12 written in __
√3
surd
form
is
2
___
___
___
3
5
4
• √20 ; √30 or √25 are
surds of different radical
orders
Unit 4: Surds
In the previous unit we worked mostly with numbers in the exponential form.
In this unit we work with numbers in their surd form.
___
m
__
We can express all fractional exponents in surd form: a n = n√am
Multiplying and dividing surds
You can divide surds with the same roots.
You can multiply surds with the same
roots.
__
√a
n
__
n
n
× √b = √ab
___
___
__
√a
n
___
____
__
__
n
a
÷ √ b = n√__
b
___
Example: √2a × √5a2 = √ 10a3
__
___
__
√3
3
12 = 3√ 4
Example: √ 12 ÷ 3√ 3 = 3 ___
Reducing surds
You should always write a surd as the product of a rational and an irrational number.
___
__ __
__
√ a2b = √ a2 √ b = a√ b
____
____
___
__
__
Examples: √128 = √ 64.2 = √ 64 .√ 2 = 8√2
_____ 3 ___
__
__
___
√ 54 = 3√ 27.3 = √ 33 .3√ 3 = 3.3√ 3
3
Adding surds
You can only add like surds, so you should reduce surds first.
__
√a
__
√a
__
__
+ √ a = 2√a
__
__
__
__
____
+ √b = √a + √b
___
____
__
__
__
Example: √8 + √128 = √4.2 + √64.2 = 2√ 2 + 8√ 2 = 10√ 2
Rationalising denominators
When there is a surd in the bottom of a fraction, multiply the top and bottom by
the same surd to get rid of the surd at the bottom.
__
__
√b
a√b
a__ × ___
___
__ = ____
√b
√b
b
__
__
__
__
__
√5
5
√2
√2
√5
√5
3 + __
3√5 + 5
1__ × ___
__ = ___
__ = _______
Example: ___
or ______
× ___
√2
2
√2
__
__
√5
2 − √__
b × ______
1 − √__
b
Example: ______
1 + √b
=
__
1 − √b
2 − 3√b + b
__________
1−b
When there are two terms in the bottom of a fraction, multiply the top and bottom
by two terms that make the bottom the difference of two squares.
__
__
__
__
__
__
__
9 + 3√ 3 + 6√3 + 2(√3 )
√
3 + 2√__3
3 + √__
3 ____________________
9 + 9√3 + 6 ________
__
Example: _______
× ______
=
= __________
= 15 + 9 3
2
3 − √3
14
3 + √3
9 − (√ 3 )
2
9−3
6
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 14
2012/07/02 2:20 PM
WORKED EXAMPLE 1
___
__
___
Simplify: √ 18 – √8 + √50
SOLUTION
___
__
___
√ 18 – √ 8 + √ 50
___
___
____
= √9.2 − √4.2 + √25.2
__
__
__
= 3√2 − 2√ 2 + 5√2
__
= 6√2
•
•
Remember to reduce surds and add like terms or factorise if there are two
or more terms.
If there are brackets, multiply them out. Look for perfect squares and the
difference of two squares.
WORKED EXAMPLE 2
___
√
√ 80
20
___
Simplify: ____
SOLUTION
___
√ 20
____
___ =
√ 80
or
___
__
20 = __
1 = __
1
√___
80 √ 4 2
___
___
__
√ 80
√ 16.5
4√ 5
√ 20
√ 4.5
2√__
5 __
____
___ = ______
____ = ____
=1
2
WORKED EXAMPLE 3
___
√
√ 3 −1
27
__ − 3
Simplify: _______
SOLUTION
___
√ 27 − 3
_______
__
√ 3 −1
__
3√__
3−3
= _______
√ 3 −1
__
3(√3 −1)
__
= ________
(√3 −1)
=3
WORKED EXAMPLE 4
__
___
Simplify: (4 + √3 )2 − √48
SOLUTION
__
___
(4 + √3 )2 − √ 48
__
__
____
= 16 + 2(4)(√ 3 ) + (√3 )2 − √16.3
__
__
= 16 + 8√3 + 3 − 4√3
__
= 19 + 4√3
Unit 4 Surds
PLT MATHS LB 11 7th pgs (Real Book).indb 15
15
2012/07/02 2:20 PM
WORKED EXAMPLE 5
__
__
__
__
__
__
__
__
Simplify: (√ 2 + √ 8 + √ 7 )(√ 2 + √ 8 − √ 7 )
SOLUTION
__
__
__
__
(√2 + √8 + √7 )(√ 2 + √ 8 − √ 7 )
=
=
__
__
__
__
(√2 + 2√ 2 + √ 7 )(√ 2 +
__
__
__
(3√2 + √ 7 )(3√ 2 − 7)
__
__
2√ 2 − √ 7 )
| Reduce surds.
| Add like surds.
| Multiply out difference of two squares.
= 9(2) − 7
= 11
WORKED EXAMPLE 6
__
___24 ___ = 2√ 6 .
Show that __________
√ 96 − √ 24
SOLUTION
__
√
√6
__
√
__
6 _____
24__ × ___
____24 ___ = __________
__24
__ = ____
__
LHS = ____________
= 12 6 = 2√ 6 = RHS
√ 16.6 − √ 4.6
4√6 − 2√6
2√6
6
EXERCISE 10
1
Write in simplified surd form without using a calculator:
1.1
1.4
1.7
___
√ 32
____
√ 125
___
3
√ 16
______
1.2
1.5
1.8
1.10 √ 20a4b3
1.13
2
___
√ 72
___
√ 96
__
1.3
1.6
√ x3
1.9
______
1.12 √50x5y8
_____
3
1.15 √ 32 012
1.11 √217a14b13
_____
_____
√x11y7
3
___
√ 48
____
√ 128
___
3
√ 96
________
1.14 √22 011
Simplify:
__
__
√7 × √5
2.1
2.2
2.3
√ 72
____
__
2√5 × 3√ 5
2.4
2√3 × 3√ 2
2.5
2√12 × 3√ 8
2.7
2.9
___
√8
___
__
___
√ 45
____
___
√ 10
____
____
3
√ 12a × √ 9a5
3
___
__
√ 12 × √ 6
__
2.11 _________
√8
__
__
__
__
___
2.6
√ 98
____
__
2.8
2√2 × 3√ 3 × 6√ 6
2.10
5
√8
__
__
____
__
____
√ 8a3 × 5√ 8a7
___
___
√ 3a3 × √ 6a5
___
2.12 ___________
2
√ 2a
EXERCISE 11
1
16
Simplify without using a calculator:
___
__
___
√ 32 – √ 8 + 2√ 18
1.1
1.3
___
___
____
√ 50 + √ 98 – √ 128
1.2
1.4
___
___
___
√ 48 + 2√ 12 – √ 75
1.5
√ 48 – √ 12
__________
___
1.6
√ 72 + √ 8
_________
__
___
√ 75
___
____
___
____
___
5 √ 45 – √162 – √125 + √ 72
___
__
√2
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 16
2012/07/02 2:20 PM
__________
___ 5 ___
1.7
√ 20 + √ 45
___
___
√ 18 – √ 12
__________
___
___
√ 72 – √ 48
_____
_____
1.9
___
2 + √ 28
_______
__
1.8
1 + √7
___
√ 20 – 2
__
1.10 _______
√5 – 1
_____
_____
√ 32x10 – √ 18x10
____
1.12 ______________
8
√ 27m6 – √ 48m6
_____
1.11 ______________
6
√ 12m
√ 50x
EXERCISE 12
Expand and simplify where possible:
__
__
1
2√3 (3 – 2√3 )
3
__
__
2
(3 + √3 )(2 + √ 2 )
(2 + √2 )(2 – √2 )
4
(√ 5 – 2)2
5
(1 – √ 3 )2
6
(1 + 3√ 3 )2
7
(7 + 2√ 3 )(7 – 2√3 )
8
(√ 2 – √ 8 )2 + (0,5)–1
9
√4 – √12 × √4 + √12
10
__
__
__
__
_______
___
__
_______
___
__
__
__
__
__ 1
__
__ 1
__
√7
√7
× ( 2 + ___
( 2 – ___
2 )
2 )
2
2
EXERCISE 13
Rationalise the denominator:
1
1__
___
2
√3
4
6__
___
7
__
√
______
__ 3
√3 – 1
5
√2
8
3__
___
√3
__
√3
1 + __
______
√3
__
1 + √__
2
______
1 – √2
__
3
√2
___
__
6
√2
4 + __
______
9
1 + √__
3
______
√3
__
√2
__
3 – √2
EXERCISE 14
1
Show that:
1.1
__
√√37 + √5 .5√√37 – √5 = 2
1.4
√ 8 − √ 3 − √ 12
1
_______________
___
__
___ = __
1.6
√ 2 011
2
______________
_____2
_____ = __
√ 22 013 − √ 22 009 3
__________
___
__
_________
___
__
__
√ 80 − √ 5
__
___
√3 + 1
5 + √27
_______
__ = _______
2
4 − 2√3
1.7
__
13 __
1__ = ___
2
√__83 + 5√__23 − ___
2 √3
√6
_________
___15 __ = √ 5
1.5
__
1.2
5
1.3
__
8
______
= 16 + 8√2
1__
1 − ___
√2
__
___
√ 32 − 3√ 3 − √ 27
_____
___
√ 12
1.8
__
__
__
__
2
____
+ 2√ 3 + 31,5 = √147
_______
___
2
If m and n are rational numbers such that √ m + √n = √5 + √24 , calculate
a possible value for m2 + n2.
3
Simplify: (1 + √ 3x2 )2 − √12x2
4
If √ 2 = 1,414 show that 2,828 = 22.
___
__
_____
3
__
Now prove without a calculator or long division that:
1
_______
______
= 0,707
3
√ 2,828
Unit 4 Surds
PLT MATHS LB 11 7th pgs (Real Book).indb 17
17
2012/07/02 2:20 PM
REMEMBER
Solving equations with surds
To solve equations with surds,
arrange the equation with the
surd on one side of the equals
sign and the other terms on
the other side. Square both
sides of the equation and
solve for the unknown.
WORKED EXAMPLE 1
____
Solve for x: √x + 1 = 5
SOLUTION
____
√x + 1 = 5
____
(√ x + 1 )2 = 52
(x + 1) = 25
x = 24
Check the solutions:
for x = 24
______
LHS = √ 24 + 1 = 5 = RHS
REMEMBER
Always square sides and not
the terms:
(2 + 3) = 5
but
(2 + 3)2 = 25
22 + 32 ≠ 52
52 = 25
4 + 9 ≠ 25
Always check your solutions
when squaring both sides.
Squaring can create extra or
extraneous solutions.
The first statement is false but
after squaring the statement,
it is true.
__
√ x = –2
2 = −2
__
(√x )2 = (–2)2
(2)2 = (−2)2
22 = 4
x=4
WORKED EXAMPLE 2
____
Solve for x: 1 + √ x + 1 = x
SOLUTION
____
√x + 1 = x − 1
____
(√ x + 1 )2 = (x − 1)2
x + 1 = x2 − 2x + 1
x2 − 3x = 0
x(x − 3) = 0
x = 0 or x = 3
Check the solutions:
_____
for x = 0: LHS = √0 + 1 = 1; RHS = 0 − 1 = −1 x ≠ 0
_____
__
for x = 3: LHS = √3 + 1 = √ 4 = 2; RHS = 3 − 1 = 2 x = 3
EXERCISE 15
REMEMBER
To solve surd equations:
1 Isolate the surd.
2 Square both sides.
3 Solve the equation.
4 CHECK the solution!
Solve for x:
1
3
5
7
9
11
18
____
√x + 2 = 4
__
__
3√x = 2√3
_____
√ x2 + 7 − 4 = 0
____
√x − 2 = 4 − x
______
√ 3x + 13 − 1 = x
_____
√ 4 − 3x = 4 − 3x
4
____
√x − 3 + 3 = 0
__
__
√x
___
__ = 3√ 2
6
√ x2 − 9 = 4
2
8
10
12
√2
_____
____
√x + 5 − x = 3
____
x + √5 − x + 1 =
____
x + 2√x − 1 = 4
0
Topic 1 Algebra: Exponents and surds
PLT MATHS LB 11 7th pgs (Real Book).indb 18
2012/07/02 2:20 PM
Surds in other topics
You have worked with surds in Grade 10 in coordinate geometry and trigonometry.
Remember the distance between two points A(x1;y1) and B(x2;y2) is
________________
AB = √ (x1 − x2)2 + (y1 − y2)2
y
In the diagram, A(–1;6) and B(5;0), then:
________________
A(–1;6)
______
AB = √ (−1− 5)2 + (6 − (0))2 = √36 + 36
The answer is in simplified surd form.
Q(2;3)
To show that BC = 2PQ it is better to use simplified surd
form than round off to two decimal places.
P(–3;1)
B(5;0)
x
_________________
PQ = √ (2 − (−3))2 + (3 − (1))2
______
___
= √ 25 + 4 = √29
C(–5;–4)
__________________
BC = √ (5− (−5)) + (0 − (−4))
2
_______
2
____
= √ 100 + 16 = √116
______
___
= √ 4 × 29 = 2√29
y
You will work with the following types of surd equations when you make
lengths equal to each other.
P
Points A(1;5) and B(7;3) are given. Another point P(5;a) lies on the line x = 5.
Find the value of a if PA = PB.
______________
A(1;5)
B(7;3)
______________
√(5 − 1)2 + (a − 5)2 = √(5 − 7)2 + (a − 3)2
To solve this equation, square both sides.
O
____
Remember that: (√a + b )2 = a + b
x=5
(5 − 1)2 + (a − 5)2 = (5 − 7)2 + (a − 3)2
(4)2 + a2 −10a + 25 = (−2)2 + a2 − 6a + 9
16 −10a + 25 = 13 − 6a
−4a = −28
a=7
1
45°
30°
2
__
√3
Remember__the special angles in trigonometry.
√
3
1__
sin 60° = ___
and tan 30° = ___
2
Now rationalise the denominator:
__
√
√3
1
__
√2
45°
60°
√3
x
1
__
√
3 ___
1__ × ___
__
tan 30° = ___
= 3 which is the answer the calculator gives.
√3
3
1__
sin 45° = cos 45° = ___
√2
Rationalise the denominator:
__
√
√2
__
√
2 = ___
2
1__ × ___
__
sin 45° = cos 45° = ___
√2
2
Unit 4 Surds
PLT MATHS LB 11 7th pgs (Real Book).indb 19
19
2012/07/02 2:20 PM
Revision
1
2
Simplify using the laws of exponents:
Law 1: am × an = am + n
Law 2: am ÷ an = am – n
Law 3: (am)n = amn
Law 4: (ab)m = am × bm
Law 5: a0 = 1
1.1
15x4y6 × 2y2x0
1.2
3x3 × (3x3)3
____
9
1.3
4x3y5
______
12x9y0
1.4
213 × 214 − 20
________
1.5
2(x ) (x ) (−2x )
_______________
1.6
27x.9x + 1
________
1.8
(2a4 − 3b3)2
4 3
3 2
16(x4)5
(3x4)3
(2x3)4
1.7
+
____________
1.9
18x.24x + 1
_________
81.3−4x7
225
35x + 3
123x.4−x
[3 × 9]
1
Simplify leaving your answers with positive exponents by using the law: a−n = __
an
2.1
2−4 − 2−3 + 2−2
2.2
3x−4y
2.3
2
−1
9(a b)
_______
6ab−1
6(a−1b)2
2.4
2 a
( _____
3a b )
−2 5 −2
−1
2.5
_______
2.6
(3−1 − 2−1)−2
2.7
(2 + x−2)(2 − x−2)
2.8
(x + x−1)2 − (x − x−1)2
2.9
3
2 4
27x
4(ab)−2
y
________
x−2y + y−1
[4 × 9]
Without using a calculator, simplify
or evaluate the expressions with rational
___
m
__
n
m
√
exponents using the property a = a n and the exponential laws.
____
2
__
32−5
4
646 + √163
3.4
(0,16)−2
√27x6y12
3.6
√ 9x8 +16x8
(√125 )3
3.3
3.5
3
3.7
3
60 + √24 − √82 + (0,5)−2
3.9
3.11
2
__
3.2
3.1
5
__
____
_______
__
( )
1
−__
3
b ___
____
3
√ b2
___
1
__
−1
___
___
3
√ a2 × 3√ a5
__________
1
__
a3
1
__
5
( ________
4 −9 )
−1
________
3.10 x2( x2 − x−2 )
−2
−1
3.8
3
__
2
1
__
3
__
1
__
__
1
__
164.√3 .42
3.12 _________
2
1 __
__
27−2.83
[4 × 12]
20
PLT MATHS LB 11 7th pgs (Real Book).indb 20
2012/07/02 2:20 PM
4
5
Simplify by writing surds in their simplest surd form first.
Leave your answers with rational denominators.
__
___
___
√ 3 + √ 48 − √ 75
4.1
__
___
____
___
√ 3 − √ 32 + √ 108 − √ 72
4.2
4.3
___
__
√ 50 + √ 2
_________
4.4
√
3√18 −
___________
___ 50
6
___
We can reformat very large
or very small numbers using
exponents. This form is called
scientific notation.
2350000000 in scientific
notation is 2,35 × 109 and
0,00000000000761 is
7,61 × 10–12
We write only one significant
number in front of the
decimal. The exponent of the
base 10 indicates how many
times you have multiplied or
divided by 10, that is, moved
the decimal to the right or to
the left.
___
2√72
___
___
___
___
√
√5
4.5
60
__
(√15 × √20 ) − √ 75 − ____
4.6
___
__
__
14
__
√ 50 + (√ 3 × √ 6 ) − ___
____
√2
____
4.7
√ 48x6 − √ 18x6
_____________
1 4 ___
__
2 × 3 2 . √ x8
4.8
(√3 x − √2x )(√3 x + √2x )
4.9
√6 + 1
_______
___
4.10
___
2√___
11 – 3
________
√ 11 + 2
__
REMEMBER
___
__
___
__
√ 24
+2
[3 × 10]
Solve for x in equations with unknown exponents:
5.2
5.1
2x − 3 × 16 = 4x
4x
__
√
5
52x + 1 = ____
125
x2 − 1
−x
27
= ____
5.3
2x − 3 × 64 = __
5.4
3
5.5
32x + 1 − 32x = 18
5.6
3x + 2 = 42 − 5.3x
8
3
[3 × 6]
6
Solve for x in equations with rational exponents:
___
2
__
6.1
3
√ x2 = 25
6.2
3x3 − 12 = 0
6.3
(x
6.4
6 + 44√ x = 18
6.5
3x4 = 24x4
6.6
3x3 − 5x3 −12 = 0
+ 2 )( x2 − 8 ) = 0
3
__
3
__
1
__
__
2
__
1
__
[3 × 6]
7
Solve for x in equations with surds:
7.1
7.3
7.5
_____
√ 7x + 2 − 2 = 1
__
7.2
x − √x = 6
7.4
√ 2 − 7x + 2 = x
7.6
_____
_____
√ 3x − 2 − x = 0
____
√2 − x = x + 4
____
2√ x − 3 + 3 = x
[5 × 6]
21
PLT MATHS LB 11 7th pgs (Real Book).indb 21
2012/07/02 2:20 PM
TOPIC
2
Equations and inequalities
Unit 1: Completing the square
KEY WORDS
square – the product of two
identical factors
For example:
• 1=1×1
• 4=2×2
• 25 = 5 × 5
• x2 = x × x
• (x – 3)2 = (x – 3)(x – 3)
quadratic equation – an
equation in which the highest
index is squared
For example:
• x2 + 3x + 2 = 0
• x2 – x = 0
• 2x2 – 3x – 5 = 0
minimum – the smallest
possible value
maximum – the biggest
possible value
Completing the square allows learners to:
• solve quadratic equations, even if we cannot factorise them
• develop a quadratic formula which we can use to solve quadratic equations
• determine the turning point of a parabola graph
• determine the value of x for which an expression has a
minimum/maximum value
• determine the centre point of translated circles (Grade 12 Analytical geometry).
We can manipulate expressions so they take on the form of a perfect square. Although
the expressions will look different, they will not be different. Completing the square
enables us to predict the sign of an expression even if we do not know the specific
value(s) of x.
Unless we know the value of x, we do not know whether x is positive or negative.
Neither do we know whether –x is positive or negative.
y
x
A(0;–5)
P(2;–9)
y = x – 4x – 5
2
We know that:
• x2 is positive for all values of x ⇒ x2 ≥ 0
• –x2 is negative for all values of x ⇒ –x2 ≤ 0
If we write the expression x2 – 4x –5 as (x – 2)2 – 9 we can see that:
• (x – 2)2 ≥ 0 for all x ∊ ℝ and (x – 2)2 = 0 only when x = 2
• (x – 2)2 – 9 ≥ –9 for all x ∊ ℝ and (x – 2)2 – 9 = –9 only when x = 2
We say that this expression has a minimum value of –9 when x = 2.
In the same way, writing the expression –x2 + 4x + 5 as – (x – 2)2 + 9 we can see that:
• –(x – 2)2 ≤ 0 for all x ∊ ℝ and (x – 2)2 = 0 only when x = 2
• –(x – 2)2 + 9 ≤ 9 for all x ∊ ℝ and –(x – 2)2 + 9 = 9 only if x = 2
We say that this expression has a maximum value of 9 when x = 2.
22
Topic 2 Equations and inequalities
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WORKED EXAMPLE 1
Consider the expression: x2+ 6x + 9
1
Solve for x if x2 + 6x + 9 = 0.
State whether x2 + 6x + 9 has a minimum or a maximum value.
3
State the value of x for which we can find this minimum or maximum value.
4
State the minimum or maximum value of the expression.
SOLUTIONS
1
2
3
4
x2 + 6x + 9 = 0 ⇒ (x + 3)(x + 3) = 0 and so x = –3 or x = –3
The expression has a minimum value. | a = 1 > 0
x2 + 6x + 9 = (x + 3)2 and it has a minimum value when x = – 3.
(x + 3)2 ≥ 0, so its minimum value is 0.
WORKED EXAMPLE 2
Given the expression: –x2 + 6x – 9
1
Solve for x if –x2 + 6x – 9 = 0.
2
State whether the expression has a minimum or a maximum value.
3
State the value of x for which we can find this minimum or maximum value.
4
State the minimum or maximum value of the expression.
y
A(3;0)
A(0;–9)
x
C(6;–9)
y = –x2 + 6x – 9
SOLUTIONS
1
2
3
4
–x2 + 6x – 9 = 0 ⇒ x2 – 6x + 9 = 0
(x – 3)(x – 3) = 0 and x = 3
The expression has a maximum value. | a = –1 < 0
–x2 + 6x – 9 = – (x2 – 6x + 9) = –(x – 3)2 has a maximum when x = 3.
–(x – 3)2 ≤ 0, so its maximum value is 0.
REMEMBER
• 5x + 2 = 2x – 3
• x2 – 2x – 3 = 0
Inequalities have inequality
signs.
• x > 3 means the value of x
has to be greater than 3
• x ≥ –2 means that
minimum value of x is –2
• x < 7 means that the value
of x must be less than 7
• x ≤ 5 means that maximum
value of x is 5
Expressions have neither
equal signs nor inequality
signs.
• x2 – 3x – 4
• 2x – 5
Unit 1 Completing the square
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EXERCISE 1
KEY WORDS
coefficient – a number that
stands in front of a variable or
variables in a term
For example:
the coefficient of 3x2 is 3
the coefficient of x is 1
roots of an equation – the
solutions of the equation
real numbers – the set of
all rational and irrational
numbers
For example:
1 ; 2,5
Rational: –2; 0; __
__ 3
Irrational: √7 ; π
non-real numbers –
imaginary numbers or
complex numbers that are
not real
__
___
For example: √–2 , √–16
REMEMBER
LHS means Left Hand Side
RHS means Right Hand Side
1
Given the expression: x2 + 10x + 25
1.1
Solve for x if x2 + 10x + 25 = 0.
1.2
State whether the expression has a minimum or a maximum value.
1.3
State the value of x for which we can find this minimum or
maximum value.
1.4
State the minimum or maximum value of the expression.
2
Given the expression: x2 – 4x + 4
2.1
Solve for x if x2 – 4x + 4 = 0.
2.2
State whether the expression has a minimum or a maximum value.
2.3
State the value of x for which we can find this minimum or
maximum value.
2.4
State the minimum or maximum value of the expression.
3
Given the expression: –x2 – 12x – 36
3.1
Solve for x if –x2 – 12x – 36 = 0.
3.2
State whether the expression has a minimum or a maximum value.
3.3
State the value of x for which we can find this minimum or
maximum value.
3.4
State the minimum or maximum value of the expression.
4
Given the expression: –x2 + 8x – 16
4.1
Solve for x if –x2 + 8x – 16 = 0.
4.2
State whether the expression has a minimum or a maximum value.
4.3
State the value of x for which we can find this minimum or
maximum value.
4.4
State the minimum or maximum value of the expression.
There are many quadratic equations that we cannot solve by means of factorisation.
Completing the square enables us to solve equations which cannot be factorised.
Consider the equation: 2x2 – 12x + 3 = 0
We can break up the process of completing the square to solve an equation into steps:
• Divide both sides by the coefficient of x2 ⇒ x2 – 6x + 1,5 = 0 | Divide by 2.
• Add x and half the coefficient of x (with its sign) and square the result. Then
subtract the square of half the coefficient of x ⇒ (x – 3)2 – (–3)2 + 1,5 = 0
(x – 3)2 = 7,5
• Isolate the perfect square on the LHS and simplify the RHS ⇒___
indicating ± on the RHS ⇒ x – 3 = ±√7,5
• Square root both sides,___
Solve
for
x
⇒
x
=
3
±
7,5
= 5,74 or x = 0,26 | Correct to two decimal places.
√
•
The roots of an equation are non-real (imaginary) when a perfect square equals
a negative value.
The roots of (x – 2)2 = –9 are non-real or imaginary and the same applies to x2 = –5.
The roots of an equation are real and rational when a perfect square equals
a perfect square.
The roots of (x – 3)2 = 16 are real and rational. The same applies to (x – 4)2 = 1
and x2 = 9.
24
Topic 2 Equations and inequalities
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The roots are real and irrational when a perfect square equals a positive value
which is not a perfect square. If (x + 2)2 = 11 the roots are real and irrational.
The same applies if x2 = 7.
WORKED EXAMPLE 1
1
2
Solve for x by completing the square if (x – 3)2 – 5 = 0.
State whether the roots are real and rational, real and irrational or non-real.
SOLUTIONS
1
2
(x – 3)2 – 5 = 0 | Do not multiply out the brackets.
| Isolate the perfect square on the LHS.
(x – 3)2 = 5
__
√
|
x–3=± 5
Square root both sides; remember ± on the RHS.
__
| Solve for x, simplified surd form.
x = 3 ±√ 5
x = 5,24 or 0,76 | Correct to two decimal places.
These roots are real and irrational.
REMEMBER
For decimal rounding correct
to one decimal place, look at
the second decimal digit.
• if it is less than 5, leave the
first decimal unchanged.
For example: 2,74 ≈ 2,7
• if it is 5 or more, round up
the first decimal digit by 1.
For example: 2,76 ≈ 2,8
For decimal rounding correct
to two decimal places, look at
the third decimal digit.
For example: 5,321 ≈ 5,32
But: 5,326 ≈ 5,33
WORKED EXAMPLE 2
1
2
Solve for x by completing the square if x2 – 8x = 9.
State whether the roots are real and rational, real and irrational or non-real.
SOLUTIONS
1
2
x2 – 8x = 9
(x – 4)2 – 16 = 9 | (x + half the coefficient of x)2 – (half coefficient of x)2
= 25 | Isolate the perfect square on LHS and simplify RHS.
| Square root both sides; remember ± on the RHS.
x–4=±5
x = – 1 or x = 9 | Solve for x.
The roots are real and rational.
WORKED EXAMPLE 3
1
2
Solve for x by completing the square if 3x2 + 12x +15 = 0.
State whether the roots are real and rational, real and irrational or non-real.
SOLUTIONS
1
2
3x2 + 12x + 15 = 0
| Divide through by 3, the coefficient of x2.
x2 + 4x + 5 = 0
2
2
(x + 2) – 2 + 5 = 0 | (x + half the coefficient of x)2 – (half coefficient of x)2
| Isolate the perfect square on LHS and simplify RHS.
(x + 2)2 = –1
__
| Square root both sides; remember ± on RHS.
x + 2 = ±√–1
__
| Solve for x and note that √ –1 = i
x = – 2 ±i
The roots are non-real.
REMEMBER
ℝ
0
′
complex numbers
imaginary numbers
real numbers
rational numbers
integers
whole numbers
natural numbers
irrational numbers
0
ℝ
′
Unit 1 Completing the square
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The next worked example is a literal equation because it does not have numerical
coefficients. We follow the same steps in both numerical and literal equations.
WORKED EXAMPLE 4
Solve for x by completing the square if ax2 + bx + c = 0.
SOLUTION
ax2 + bx + c = 0
c =0
b x + __
x2 + __
| Divide through by a, the coefficient of x2.
c =0
b 2 – ___
b + __
( x + ___
a
2a )
4a
| (x + half the coefficient of x)2 – (half coefficient of x)2
a
a
2
2
b – 4ac
b 2 = _______
( x + ___
2a )
4a
2
2
_______
√
2
b – 4ac
b = ±________
x + ___
2a
2a
_______
√ 2
2a
–b± b – 4ac
x = ___________
| Isolate perfect square on LHS; simplify the RHS.
| Square root both sides; remember ± on RHS.
| You will use this quadratic formula in Unit 2, page 30.
EXERCISE 2
Solve for x by completing the square. Give your answers correct to two decimal places.
For Questions 1– 8 say whether the roots are real and rational, real and irrational or
non-real.
26
1
x2 – 6x –12 = 0
2
x2 + 8x + 5 = 0
3
2x2 + 4x – 8 = 0
4
5x2 – 20x + 20 = 0
5
2x2 + 5x – 12 = 0
6
3x2 – 24x = 27
7
2x2 – 12x + 12 = 0
8
x2 – 10x + 29 = 0
9
px2 + qx + r = 0
10
kx2 – mx – n = 0
11
2mx2 – mx + k = 0
12
3kx2 + 2nx – k = 0
Topic 2 Equations and inequalities
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Unit 2: Quadratic equations
You can solve equations by factorisation or by using the quadratic formula.
There are three ways to solve quadratic equations.
1
Factorisation – usually the quickest method
x2 – 4x – 12 = 0 ⇒ (x – 6)(x + 2) = 0 and x = 6 or x = –2
2
Quadratic formula – if you cannot find the factors.
(Proved in Worked example 2 on page 26 in Unit 1)
x2 – 4x – 12 = 0 ⇒ a = 1, b = – 4 and c = –12
_______
√ 2
2a _____________
–(–4)
± √ (–4)2 – 4(1)(–12)
= ______________________
2(1)
–b ± b – 4ac
x = ____________
___
√
4 ± 64
= _______
2
4 ± 8 = ___
12 or ___
–4
= _____
2
2
REMEMBER
Factors of 12 are numbers
which divide into 12 without
leaving a remainder:
• 1, 2, 3, 4, 6 and 12 are
factors of 12
• 1, 2, 3, 6, x and y are
factors of 6xy
• (x – 2) and (x – 3) are
factors of x2 – 5x + 6.
Factorising reminders: always
take out highest common
factor first, and then look for:
• a difference of squares
• a trinomial
• grouping in pairs.
4
= 6 or –1
KEY WORDS
3
Completing the square – if you are told to or prefer to. (Covered in Unit 1)
If x2 – 4x – 12 = 0, solve for x by completing the square.
x2 – 4x – 12 = 0 ⇒ (x – 2)2 – 16 = 0 and (x – 2)2 = 16
x – 2 = 4 or x – 2 = – 4 and x = 6 or x = –2
Check the degree of the equation and look for the same number of solutions as
the degree:
•
•
•
First degree equations are linear equations and usually have one solution.
Second degree equations are quadratic equations and usually have two solutions.
Third degree equations are cubic equations and have up to three solutions.
degree (in algebra) – the
highest index in an equation
or term
For example:
• 3x – 7 = 5 + x
is a first degree equation
• 5x2 – 2x – 3 = 0
is a second degree
equation
• x3 – 2x2 – 3x = 0
is a third degree equation
WORKED EXAMPLE 1
Solve for x if x2 – 5x = 0.
SOLUTION
| x is a common factor.
x2 – 5x = 0
x(x – 5) = 0 ⇒ x = 0 or x = 5 | If x is a common factor, x = 0 is a solution.
REMEMBER
• 0 × any real number = 0
• If A × B = 0 then A = 0 or
B=0
• If A + B = 0 and A = 0, then
B=0
• If x2 = 4, then x = ±2
WORKED EXAMPLE 2
Solve for x if 3x2 – 12x = 0.
SOLUTION
| The highest common factor is 3x.
3x2 – 12x = 0
3x(x – 4) = 0 ⇒ x = 0 or x = 4
Unit 2 Quadratic equations
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WORKED EXAMPLE 3
Solve for x if x2 – 9 = 0.
SOLUTION
x2 – 9 = 0 | This is the difference between two squares.
Method 1
(x – 3)(x + 3) = 0 and x = 3 or x = –3
Method 2
x2 – 9 and x = ±3
WORKED EXAMPLE 4
Solve for x if (x – 5)2 – 16 = 0.
SOLUTION
(x – 5)2 – 16 = 0 | This is a difference between squares.
Method 1
Method 2
[(x – 5) – 4][(x – 5) + 4] = 0
(x –5)2 = 16
(x – 9)(x – 1) = 0
x – 5 = ±4
x = 9 or x = 1
x = 5 + 4 or x = 5 – 4
x = 9 or x = 1
WORKED EXAMPLE 5
Solve for x if x2 – 5x − 6 = 0.
SOLUTION
x2 – 5x − 6 = 0 ⇒ (x – 6)(x + 1) = 0 and x = 6 or x = –1
WORKED EXAMPLE 6
Solve for x if x2 – x = 12.
SOLUTION
x2 – x – 12 = 0 ⇒ (x – 4)(x + 3) = 0 and x = 4 or x = –3
28
Topic 2 Equations and inequalities
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WORKED EXAMPLE 7
Solve for x if 10x2 –35x + 15 = 0.
SOLUTION
10x2 –35x + 15 = 0 | 5 is a common factor, so divide through by 5.
2x2 –7x + 3 = 0 ⇒ (2x – 1)(x – 3) = 0
1 or x = 3
2x – 1 = 0 or x – 3 = 0 ⇒ x = __
2
WORKED EXAMPLE 8
Solve for x if 6x3 –19x2 + 10x = 0.
SOLUTION
6x3 –19x2 + 10x = 0 | It is important to notice that x is a common factor.
x(6x2 – 19x + 10) = 0 | Do not divide by x, or you will lose the solution x = 0.
2 or x = 2,5
x(3x – 2)(2x – 5) = 0 ⇒ x = 0 or x = __
3
WORKED EXAMPLE 9
____
Solve for x if √x + 5 = x + 3.
SOLUTION
____
√x + 5 = x + 3
____
(√x + 5 )2 = (x +
| Square both sides to eliminate the square root.
3)2 | Remember to check your solutions at the end.
x + 5 = + 6x + 9 ⇒ x2 + 5x + 4 = 0
(x + 1)(x + 4) = 0 ⇒ x = –1 or x ≠ –4 | Eliminate the solution only after the check.
__
If x = –1, LHS = √4 = 2 and RHS = –1 + 3 = 2, so LHS = RHS
__
If x = –4, LHS = √1 = 1 but RHS = – 4 + 3= –1, so LHS ≠ RHS
x = –1 is the only solution | Checking is essential, not optional.
x2
Reminders about trinomial factorisation:
Both brackets have the same sign as the middle term if the trinomial ends with
a + sign.
x2 + 6x + 8 = (x + 2)(x + 4)
and x2 – 6x + 8 = (x – 2)(x – 4)
2
6x + 17x + 12 = (2x + 3)(3x + 4)
and 6x2 – 17x + 12 = (2x – 3)(3x – 4)
2x2 + 11x + 15 = (x + 3)(2x + 5)
and 2x2 – 11x + 15 = (x – 3)(2x – 5)
BUT
• The brackets have opposite signs if the trinomial ends with a – sign.
• The bigger value of the inner and outer products has the same sign as the middle
term of the trinomial, and the other value carries the opposite sign.
x2 + 5x – 6 = (x + 6)(x – 1)
and x2 – 5x – 6 = (x – 6)(x + 1)
3x2 + x – 14 = (x – 2)(3x + 7)
and 3x2 – x – 14 = (x + 2)(3x – 7)
2
6x + x – 12 = (3x – 4)(2x + 3)
and 6x2 – x – 12 = (3x + 4)(2x – 3)
•
Unit 2 Quadratic equations
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EXERCISE 3
Solve for x by means of factorisation:
1
3
5
7
9
11
13
15
x2 – 10x + 21 = 0
4x2 + 8x + 3 = 0
3x2 – 3(x – 1) = 4(2x + 1) + 3
x2 = x
x2 – 24 = 10x
(x + 2)(x – 5) + 6 = 0
4x2 + 11x + 6 = 0
6(x2 + 1) = 13x
2
4
6
8
10
12
14
16
2x2 – x – 6 = 0
3x2 + x – 2 = 0
x2 + 5x = 0
2x2 – 5x = 3
(x + 3)(x – 2) = 6
x2 – 6x = 5(1 – 2x)
(2x – 3)(4x – 3) = 9
(2x + 3)(2x – 5) = 9
17
3
19 x = __
x2 + ___
18
5 = 11
2x + __
x
20
3x – __
5
1 = ___
___
22
x + 1 = ___
1
______
24
4x2
26
2a2x2 + 5abx – 12b2 = 0
28
1 + √ 5x – 1 = 2x
12
2
2(2x – 1)
x(4x + 5) = ________
3
5 – 4x = ___
1
______
3
2x
5x(x – 3)
________
= 3(x – 5) + 6
2
19
21
23
6p2x2 – 19pqx + 10q2 = 0
25
____
√x – 1 + 3 = x
27
2
4
4x
14 – x
8x
9 = 13
+ __
2
x
_____
The quadratic formula enables us to solve quadratic equations that we
cannot factorise.
WORKED EXAMPLE
Solve for x if 3x2 + 5x – 1 = 0 | We cannot solve this equation by factorisation.
SOLUTION
_______
√
2
–b± b – 4ac
x = ___________
2a
__________
–5±√ 52 – 4(3)(–1)
= _______________ | a = 3; b = 5 and c = –1
2(3)
=
___
–5±√37
_______
6
| Simplified surd form.
= 0,18 or –1,85 | Correct to two decimal places.
•
•
•
30
Show only your substitution step and final answers.
Make sure you give your answer in surd form when you are asked to.
Pay attention to the number of decimal places you must round off to.
Topic 2 Equations and inequalities
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EXERCISE 4
Solve for x by using the quadratic formula, correct to two decimal places where
necessary.
1
2x2 – 7x + 3 = 0
2
7x2 + 4x – 12 = 0
3
5x2 + 9x + 3 = 0
4
3x(x – 3) = 4(1 – x)
2
5
5(2x – 1) – x = 7x(2 – x)
6
2(x – 1)(x + 1) = 5x
7
2 =2
x + ___
8
3x
x2 – 2px + 3p = 0
If you are given no specific instructions for solving an equation or determining
the roots:
• factorise if possible
• use the quadratic formula if you cannot find the factors
• complete the square if you prefer to.
EXERCISE 5
Determine the roots of the equations:
1
x2 – 5x – 6 = 0
3
(2x – 3)2 – 16 = 0
5
10x2 – 59x – 221 = 0
7
9
3
4 – y = __
____
y–1
2
p
3 = __
1
_____
– __
p–1 p 4
2
4
6
x2 – 10x – 200 = 0
2x2 – 5x – 6 = 0
5x(2x – 3) = 7
8
y2 – 6y + 7 = 0
10
2x – √ x – 1 = 3
____
A quadratic equation may have an unknown constant as well as one unknown root.
In this case you will be given one of the roots.
Substitute the root you know so that you can find the unknown constant. Then make
use of this result to determine the other root of the equation.
When you give your answer, state the other root.
WORKED EXAMPLE
REMEMBER
The root of an equation is
a value which solves the
equation.
A constant is a value which
does not change.
Given that x = 5 is a root of the equation px2 − 13x + 15 = 0, determine the value of
p. Then determine the value of the other root.
SOLUTION
Method 1
Substitute x = 5 into px2 − 13x + 15 = 0
p(5)2 − 13(5) + 15 = 0
⇒ 25p – 65 + 15 = 0
25p = 50 and p = 2
Now substitute p = 2 into the equation:
2x2 – 13x + 15 = 0 ⇒ (2x –3)(x – 5) = 0
3
The other root is x = __
Method 2
x – 5 is a factor of px2 − 13x + 15 = 0
Factorise by inspection:
(x – 5)(2x – 3) = 0 | The middle term
–13 = –3 – 10.
3
The other root is x = __
2
2
Unit 2 Quadratic equations
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EXERCISE 6
1
x = –3 is one of the roots of the equation given by kx2 + 3x – 9 = 0.
1.1
Determine the value of k.
1.2
Then determine the value of the other root.
2
x = 2 is one of the roots of the equation 3x2 + px – 10 = 0.
2.1
Determine the value of p.
2.2
Then determine the value of the other root.
3
x = – 1 is one of the roots of the equation 5x2 + 3x + a = 0.
3.1
Determine the value of a.
3.2
Then determine the value of the other root.
4
x = 3 is one of the roots of the equation 4x2 + tx –15 = 0.
4.1
Determine the value of t.
4.2
Then determine the value of the other root.
We can solve real life problems with quadratic equations.
• Decide what you have been asked to find.
• Let what you are finding be x.
• Translate the words into mathematical statements.
• Create an equation relating the statements.
• Solve the equation.
WORKED EXAMPLE 1
A group of Grade 11 learners goes out for lunch to celebrate their exam results.
They agree to share the expenses equally. The total bill is R1 200. Ten learners do
not have any money, so the others learners pay an extra R20 each. Determine the
number of learners in the group.
SOLUTION
Let the number of learners who go to lunch be x.
1 200 .
If all the learners pay, then each learner pays _____
x
1 200 .
Ten learners do not pay, so x – 10 learners each pay ______
x – 10
The learners who pay, pay R20 more, so the difference is R20.
1 200 – _____
1 200 = 20
______
x
| Extra amount to pay is R20.
x – 10
(x – 10)
x(x – 10)
1 200 × __
x – _____
1 200 × _______
______
= 20 × ________
x
x
x – 10
(x – 10)
x(x – 10)
1 200x – 1 200x + 12 000 = 20x2 – 200x ⇒ 20x2 – 200x –12 000 = 0
x2 – 10x – 600 = 0 ⇒ (x + 20)(x – 30) = 0
x ≠ 20, x = 30 only (x > 0)
There were 30 learners in the group.
Check the answer by comparing the amounts paid.
1 200 = R40, but if only 20 learners pay, each
If 30 learners pay, each learner pays _____
1 200 = R60.
learner pays _____
30
20
x = 30 is the correct solution because the payments differ by R20.
32
Topic 2 Equations and inequalities
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WORKED EXAMPLE 2
A man calculates that if his speed increases by 10 km/h, he will take 48 minutes
less to complete a journey of 240 km. Calculate his original speed.
SOLUTION
Let his original speed be x and the increased speed be x + 10.
240 .
Time taken to complete journey at original speed is ____
x
240 .
Time taken to complete journey at increased speed is ______
x + 10
48 = __
4 hours.
Difference in times is 48 minutes which converts into ___
60 5
240 – ______
240 = __
4
____
x
x + 10 5
240(5)(x + 10) – 240(5)x _________
4x(x + 10)
_____________________
=
5x(x + 10)
5x(x + 10)
1 200x + 12 000 – 1 200x = 4x2 + 40x
4x2 + 40x –12 000 = 0
x2 + 10x – 3 000 = 0 ⇒ (x + 60)(x – 50) = 0
x ≠ –60 or x = 50 (x > 0) | His original speed was 50km/h.
WORKED EXAMPLE 3
A bricklayer and his apprentice build a wall in 24 days. When each person works
separately, the apprentice takes 20 days longer than the bricklayer to complete
the job. Calculate the number of days each person takes to complete the job on
his own.
SOLUTION
Let the bricklayer take x days and the apprentice take (x + 20) days to build a wall.
1 and ______
1 of the wall each day.
The bricklayer and apprentice build __
x
x + 20
1 + ______
1 of the wall each day.
Working together, they build __
x
x + 20
1 of the wall.
Together they build the wall in 24 days, so in one day they build ___
24
1 + ______
1 = ___
1
⇒ __
x
x + 20
24
24(x + 20) + 24x __________
x(x + 20)
______________
=
24x(x + 20)
24x(x + 20)
24x + 480 + 24x = x2 + 20x
x2 – 28x – 480 = 0
(x – 40)(x + 12) = 0
x = 40 or x ≠ –12 (x > 0)
The bricklayer takes 40 days on his own and the apprentice 60 days on his own.
Unit 2 Quadratic equations
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EXERCISE 7
1
A piece of wire of length 300 millimetres is bent to form a rectangle with area
3 125 square millimetres. Calculate the dimensions of the rectangle.
2
The sum of the digits of a two-digit number is 13 and the product of the digits
is 36. Determine which two numbers fit this description.
3
Vula is a river guide on the Gariep River. Robert, a member of the group, is
injured. Vula paddles Robert to the nearest pick-up point, 12 kilometres away.
Vulu paddles back to his group. If the total paddling time for the return trip
(there and back) is five hours and the river flows at a constant speed of 1 km/h,
calculate the average speed that Vula paddles.
4
A motorcyclist travels from A to B at 40 km/h and from B to A at 60 km/h.
If the distance between A and B is x km, determine the average speed at which
the motorcyclist travels from A to B and back to A. (The average speed is not
50 km/h.)
5
Thabiso has a budget of R525 per month for petrol. The present price of petrol is
x cents per litre. If the price rises by 50 cents per litre, she can buy five litres less
petrol for R525. Calculate the present price of petrol.
6
6.1
6.2
Vusi sets out for a 40 kilometre run. After he runs for two hours he injures
his ankle. He takes an hour to walk the rest of the way. If he injured
himself after he had run 16 kilometres, he would have taken four hours to
complete the distance. Determine the average speed in kilometres per hour
at which Vusi ran.
Now determine the average speed in kilometres per hour at which
Vusi walked.
7
Two windmills work continuously and together can fill a reservoir in six days.
Working separately, one windmill takes nine days longer than the other to fill
the reservoir. Calculate how long it takes each windmill to fill the reservoir.
8
A reservoir is fed by two pipes of different diameter. The pipe with the larger
diameter takes three hours less than the smaller pipe to fill the reservoir. If both
pipes are opened simultaneously, the reservoir can fill in two hours. Calculate
how long it takes the pipe with the larger diameter to fill the reservoir on its
own.
9
A rectangular parking area has dimensions of 50 metres by 120 metres. If the
parking area is doubled by increasing both the length and the breadth by x
metres, determine the dimensions of the new parking area.
120 + x
120
new parking area
original parking
34
50 + x
50
Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 34
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Unit 3: Quadratic inequalities
Inequalities are often easier to understand if we use graphs to determine their solutions.
The equation x2 – 9 = 0 has one unknown and two solutions. These solutions
represent the x-intercepts of the graph y = x2 – 9.
When you solve an inequality you give the boundaries within which the solutions lie.
WORKED EXAMPLES AND SOLUTIONS
1
x2 – 9 > 0
(x – 3)(x + 3) > 0
+
–3
–
3
+
REMEMBER
To solve x2 – 9 > 0, look for
all possible values of x which
make the expression x2 – 9
positive.
but
To solve x2 – 9 < 0, look for
all possible values of x which
make the expression x2 – 9
negative.
| Number line solution
| Inequality notation
| Interval notation
x < –3 or x > 3
x ∊ (–∞;–3) ∪ (3;∞)
y
y = x2 – 9
+
+
+
+
(–3;0)
x
(3;0)
–
–
–
–
(0;–9)
x-axis divides the graph into + and – values
y-values on the x-axis are 0.
y-values above the x-axis are +.
y-values below the horizontal x-axis are –.
2
x2 – 9 < 0
(x – 3)(x + 3) < 0, x ∊ ℝ
+
–3
–
3
+
| Inequality notation
| Interval notation
–3 < x < 3
x ∊ (–3;3)
3
| Number line solution
x2 – 9 ≤ 0
(x – 3)(x + 3) ≤ 0, x ∊ ℝ
+
–3
–
–3 ≤ x ≤ 3
x ∊ [–3;3]
3
+
| Number line solution
| Inequality notation
| Interval notation
Unit 3 Quadratic inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 35
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EXERCISE 8
Solve for x and illustrate your answer on a number line.
Make a rough sketch of a parabola for each question to help you find the answer.
x2 – 16 < 0
4x2 – 9 < 0
1
3
2
4
x2 – 25 ≥ 0
x2 – 1 ≤ 0
EXERCISE 9
State the values of x for which the expressions are real:
______
√ x2 – 16
1
2
______
√ 4x2 – 9
Graphic interpretation of the solution
of inequalities in one variable
WORKED EXAMPLE AND SOLUTIONS
y
4
2
–2
y=x+2
y = –x2 + 4
2
x
Use the graphs to help you complete the table.
(Solutions are in the columns for Inequality notation and Interval notation.)
36
Statement
Inequality notation
Interval notation
1.1
–x2 + 4 ≥ 0
–2 ≤ x ≤ 2
x ∊ [–2;2]
1.2
x+2>–
x2
x < – 2 or x > 1
x ∊ (–∞;–2) ∪ (1;∞)
1.3
–x2
–2 < x < 1
x ∊ (–2;1)
1.4
–x2 + 4 ≤ 0
x ≤ – 2 or x ≥ 2
x ∊ (–∞;– 2] ∪ [2;∞)
1.5
–x2 + 4 ≤ x + 2
x ≤ – 2 or x ≥ 1
x ∊ (–∞;–2] ∪ [1;∞)
+4
+4>x+2
Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 36
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EXERCISE 10
1
Use the graphs to help you complete the table.
y
4
y = x2 + x – 6
2
–3
–1
–2
1
2
x
3
–2
–4
–6
y = –x2 + x + 2
Statement
2
Inequality notation
1.1
x2 + x – 6 > 0
1.2
–x2 + x + 2 > 0
1.3
–x2 + x + 2 ≥ x2 + x – 6
1.4
x2 + x – 6 ≤ 0
Interval notation
Use the graphs to help you complete the table.
y
y = –2x + 3
5
4
3
y = –x2 + 2x + 3
2
1
x
–2
–1
–1
1
2
3
4
–2
–3
–4
–5
Statement
2.1
–x2 + 2x + 3 > 0
2.2
–x2 + 2x + 3 > –2x + 3
2.3
–2x + 3 ≥ – x2 + 2x + 3
2.4
–2x + 3 ≤ 0
Inequality notation
Interval notation
Unit 3 Quadratic inequalities
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3
Use the graph to help you complete the table.
y
y = x 2 + 5x + 4
y = x + 4
4
x
–4
Statement
38
3.1
x2
3.2
x+4>0
3.3
x2 – 5x2 + 4 > x + 4
3.4
x2 + 5x + 4 < 0
–1
Inequality notation
Interval notation
– 5x + 4 ≥ 0
Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 38
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4
Use the graphs to help you answer the questions. Do not show calculations.
y
21
18
15
12
9
6
y=x+3
3
–4
–2
2
4
x
6
–3
–6
–9
y = –x2 – 2x –15
–12
–15
–18
4.1
4.2
4.3
4.4
Write down the value(s) of x for which x2 – 2x – 15 = 0.
For which values of x is x2 – 2x – 15 > 0?
For which values of x is x2 − 2x − 15 = x + 3?
For which values of x is x2 − 2x − 15 ≤ x + 3?
y
5
2
y = – __13x2 − __73x – 2
–7
–6
–5
–4
–3
1
–2
x
–1
–1
y = – __13x − 2
–2
5.1
1 x2 + __
7 x + 2 = 0.
Write down the value(s) of x for which __
5.2
1 x2 – __
7 x – 2 < 0?
For which values of x is – __
5.3
1 x − 2 = – __
1 x2 − __
7 x − 2?
For which values of x is – __
3
3
3
5.4
1 x + 2 − __
1 x2 + __
7 x + 2 > 0?
For which values of x is __
3
3
(3
3
3
) (3
3
)
Unit 3 Quadratic inequalities
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Unit 4: Equations in two unknowns
(simultaneous equations)
WORKED EXAMPLE AND SOLUTION
If y = –2x2 + 9x – 9 and y = –x – 1,
solve simultaneously for x and y.
Show all necessary calculations.
–2x2 + 9x – 9 = – x – 1
2x2 – 10x + 8 = 0
x2 – 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 4 or x = 1
If x = 4, y = – 4 – 1 = –5 ⇒ (4;–5)
If x = 1, y = –1 – 1 = –2 ⇒ (1;–2)
y
x
y = –x – 1
B(1;–2)
y = –2x2 + 9x – 9
A(4;–5)
Always substitute the value(s) you have found into the linear equation to find
the other value(s).
In Topic 5 you draw graphs of parabolas and straight lines. You then solve the
simultaneous equations graphically by determining the points of intersection of
the two graphs. A(4;–5) and B(1;–2) are the points of intersection of the graphs in
the Worked example.
EXERCISE 11
Solve simultaneously for x and y in each equation. Show all necessary calculations.
1
y = 3x2 – 2x – 8 and y = 5x – 2
2
y = –5x2 + 4x + 9 and y + 6x = 9
3
3 x2 + 12 and 3x + 2y = 12
y = −__
4
1 x2 – __
1 x + 6 and x + 2y = 8
y = __
5
1 (x – 1)(x + 6) and 2x + 5y = 5
y = – __
4
2
2
2
x2
6
y=
7
1 (x + 1)2 – 2
y = –x2 – 5x + 6 and y = __
8
1 (x – 5)2 + 6
y = (x – 2)2 – 3 and y = – __
9
y = –x2 – x + 4 and y = 2(x – 1)2 – 4
10
1 x2 + 4
y = (x – 3)2 – 5 and y = – __
– 5x + 4 and y – 2x + 6 = 0
2
2
2
In Topic 5 you draw the hyperbola and straight line graphs. You then determine the
point(s) of intersection of these graphs graphically. In the next exercise you determine
the point(s) of intersection between the hyperbola and the straight line graph
algebraically. You also determine the simultaneous solutions of other graphs.
40
Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 40
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WORKED EXAMPLE
y
4 , solve simultaneously
If y = 2x and y = _____
x–1
for x and y. Show all necessary calculations.
A
SOLUTION
4 ⇒ 2x(x – 1) = 4 | Multiply both
2x = _____
x–1
2
2x – 2x – 4 = 0
sides by (x – 1).
x2 – x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 or x = –1
If x = 2, y = 2(2) = 4 ⇒ A(2;4)
If x = –1, y = 2(–1) = –2 ⇒ B(–1;–2)
4
y = ____
x–1
x
B
y = 2x
It is not always possible to make either x or y the subject of an equation.
If x2 − 5xy + y2 = 7and x − 4y = 5, you cannot make either x or y the subject of the first
equation. In this case, use a substitution method and focus on the linear equation,
making x the subject to avoid fractions. Notice that x = 5 + 4y is fraction free, whereas
x − 5 contains a fraction.
y = _____
4
Start by substituting x = 5 + 4y into x2 − 5xy + y2 = 7 which gives
(5 + 4y)2 − 5y(5 + 4y + y)2 = 7.
Solve for y and then find the x value of each of the y values by substituting back into
the linear equation.
EXERCISE 12
Solve simultaneously for x and y in each equation. Show all necessary calculations.
1
2
3
4
3 −1
y = 3x + 5 and y = _____
x+2
1 +2
x + 4y = 9 and y = – _____
x−1
5 +1
5x + y + 14 = 0 and y = – _____
x+3
3 −2
x – 3y = 10 and y = _____
x−4
5
x2 – xy – 6y2 = 6 and x + 2y = 6
6
x2 – 3xy + 2y2 = 4 and 3x – 2y = 0
7
x2 – 5xy + y2 = 7 and x – 4y = 5
8
(x – 2y)(2x + y) = 12 and 7y – x + 9 = 0
9
4x2 – 7xy + 4y2 = 16 and y = 2x – 2
10
5x2 – 3xy + 3x – 2y2 + 4y = 5 and y + 2x = 1
11
4 | Factorise your equation by grouping.
y = (x + 2)2 – 5 and y = __
x
Unit 4 Equations in two unknowns (simultaneous equations)
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Unit 5: Nature of roots
To solve a quadratic equation you can:
• factorise
• use the quadratic formula
• complete the square.
It is sometimes useful to know whether an equation can be solved or not.
If you can solve an equation, you may be interested in the nature of its roots.
• Are the roots real or non-real (imaginary)?
• If the roots are real, are they rational or irrational?
• If the roots are rational, are they equal roots or are there two different roots?
_______
√ 2
2a
−b ± b − 4ac
Consider the quadratic formula: x = _____________
If you know the value of b2 − 4ac, then you know that:
Roots are real if b2 − 4ac ≥ 0, but roots are non-real (imaginary) if b2 − 4ac < 0.
You can further classify real roots as:
rational and unequal if b2 − 4ac = perfect square (not including 0)
rational and equal if b2 − 4ac = 0 | 0 is the smallest perfect square.
irrational and unequal if b2 − 4ac = positive number which is not a perfect square
Only an equation with rational roots can be solved by factorisation.
If you know the value of b2 − 4ac, do not work it out again when you use the
quadratic formula.
WORKED EXAMPLES
1
2
3
4
5
6
7
8
Without solving for x, fully classify the roots of x2 − 2x + 2 = 0.
Solve for x in as many different ways as possible if x ∊ ℝ.
Without solving for x, fully classify the roots of x2 − 6x + 9 = 0.
Solve for x in as many different ways as possible, if x ∊ ℝ.
Without solving for x, fully classify the roots of 2x2 − 5x − 7 = 0.
Solve for x in as many different ways as possible, if x ∊ ℝ.
Without solving for x, fully classify the roots of 3x2 + 5x − 3 = 0.
Solve for x in as many different ways as possible if x ∊ ℝ.
SOLUTIONS
1
2
3
42
If 1x2 − 2x + 2 = 0, then a = 1, b = −2, c = 2
∴ b2 − 4ac = (−2)2 − 4(1)(2) = 4 – 8 = −4
∴ b2 − 4ac < 0
∴ Roots are non-real.
No real solutions are possible.
If 1x2 − 6x + 9 = 0, then a = 1, b = −6, c = 9
∴ b2 − 4ac = (−6)2 − 4(1)(9) = 36 − 36 = 0
∴ b2 − 4ac = 0
∴ Roots are real (b2 − 4ac is positive), rational (b2 − 4ac is a perfect square) and
equal (b2 − 4ac = 0).
⇒ Roots are real, rational and equal.
Topic 2 Equations and inequalities
PLT MATHS LB 11 7th pgs (Real Book).indb 42
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4
Factors: (x − 3)(x − 3) = 0 ⇒ x = 3 or x = 3
_______
√ 2
2a
−(−6) ± 0
________
=
2(1)
6
__
=
2
−b ± b − 4ac
Quadratic formula: x = _____________
Note that these roots are
real, rational and equal.
=3
5
6
Completing the square: x2 − 6x + 9 = 0 ⇒ (x − 3)2 = 0 ⇒ x = 3
If 2x2 − 5x − 7 = 0, then a = 2, b = 5, c = −7
∴ b2 − 4ac = (−5)2 − 4(2)(−7) = 25 + 56 = 81
∴ Roots are real (b2 − 4ac is positive), rational (b2 − 4ac is a perfect square) and
unequal (b2 − 4ac ≠] 0).
Factors: 2x2 − 5x − 7 = 0 ⇒ (2x − 7)(x + 1) = 0 and x = 3,5 or x = −1
_______
√ 2
2a
−b ± b − 4ac ____
14 or ___
−4 = 3,5 or −1
Quadratic formula: x = _____________
= 5±9 = ___
4
4
_______
4
___
You already know √b2 − 4ac = √ 81 = 9
(
5 x = __
5
7 ⇒ x − __
Completing the square: x2 − __
)
2
25 = _______
56 + 25 = ___
81
7 + ___
= __
4
2
2
2 16
16
16
5
9
14
7
−4
__
__
___
__
___
x− =± ⇒x=
= or x =
= −1
4
4
4
4
2
7
8
x = 3,5 or –1
If 3x2 + 5x − 3 = 0, then a = 3, b = 5, c = −3
∴ b2 − 4ac = (5)2 − 4(3)(−3)
= 25 + 36
= 61
∴ Roots are real (b2 − 4ac is positive), irrational (b2 − 4ac is not a perfect
square) and unequal (b2 − 4ac ≠ 0).
___
√
Note that these roots are
real, rational and unequal.
___
√
−b ± 61 _______
Quadratic formula: x = ________
= −5± 61 = 0,47 or −2,14
2a
Completing the square:
x2
6
(
)
5 x = 1 ⇒ x + __
5 2 = 1 + ___
25 = ___
61
+ __
3
6
36
___
___ 36
±√61
−5±√61
5
__
_____
_______
x+ =
⇒x=
= 0,47 or − 2,14
6
6
6
Note that these roots
are real, irrational and
unequal.
EXERCISE 13
Fully classify the roots of each equation without solving the equation.
Solve each equation in as many different ways as possible if x ∊ ℝ.
1
3
5
7
5x2 – 2x + 1 = 0
7x2 – 3x + 4 = 0
4x2 – 12x + 9 = 0
25x2 + 20x + 4 = 0
2
4
6
8
7x2 – 3x – 4 = 0
2x2 – 3x + 5 = 0
3x2 + 5x + 1 = 0
x2 + 7x + 13 = 0
Unit 5 Nature of roots
PLT MATHS LB 11 7th pgs (Real Book).indb 43
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Revision
1
Solve for x in each equation, correct to two decimal places where necessary.
1.1
(x – 3)(2x – 5) = 0
1.2
(x – 5)(x + 2) = 6
1.3
(x – 3)(x + 2) = 6
1.4
(x – 2)(x – 3) = (2x – 5)(x + 3)
1.5
10x2 = 3
______
1.6
2x − 3 − __________
x−1
1
______
= __________
2
2
2
1.7
1.8
2
3
(6)
7x − 4
x −4
_____
2x − 5x + 2
√ 2x − 5 + 4 = x
_____
x − √2x − 3 = 3
2x + 3x − 2
| Remember to check your answers.
| Remember to check your answers.
Solve for x by completing the square:
2.1
3x2 – 12x = 6
2.1
(2x – 5)(3x – 2) + 7x = 22
Consider the equations and match each one to the statement
which best describes its roots.
3.1
x2 – 7x – 18 = 0
3.2
25x2 + 20x + 4 = 0
3.3
2x2 + 7x + 4 = 0
3.4
3x2 – 5x + 3 = 0
A
B
C
D
(2)
(5)
(5)
(7)
(8)
(6)
(6)
[45]
(5)
(6)
[11]
(4)
(4)
(4)
(4)
Real, rational and equal roots
Real, rational and unequal roots
Real, irrational and unequal roots
Non-real or imaginary roots
[16]
4
5
Consider the equation: 2x2 – 5x = 9
4.1
Without solving for x, discuss the nature of the roots.
4.2
Solve for x, correct to two decimal places, by:
4.2.1
completing the square
4.2.2
using the quadratic formula.
Solve for x and illustrate your answer on a number line.
5.1
x2 ≤ 64
5.2
81 – 4x2 ≥ 0
5.3
x2 – x – 12 > 0
5.4
x2 – 7x ≥ 0
(5)
(5)
(4)
[14]
(5)
(5)
(5)
(5)
[20]
44
PLT MATHS LB 11 7th pgs (Real Book).indb 44
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6
Refer to the graphs and complete the table.
y
2
y = x –4x + 3
y = 2x + 3
• (6;15)
15
12
9
6
3
•
•
–2 –1
– 1,5
3
•1
1
2
•
3
3
4
5
6
Statement
7
8
9
10
6.1
2x + 3 ≥ 0
6.2
(x – 1)(x – 3) ≤ 0
6.3
(x – 1)(x – 3) ≥ 2x + 3
6.4
(x – 1)(x – 3)(2x + 3) < 0
x
Inequality notation Interval notation
[20]
Solve for x and y if:
7.1
y = –x2 + 2x + 8 and 3x – y + 2 = 0
7.2
y – 4x + 14 = 0 and y = x2 – 3x – 4
(6)
(6)
7.3
4 and x + 2y + 4 = 0
y = _____
(7)
7.4
y = – 2x2 + 10x – 8 and y = x2 – 2x – 8
x–5
(7)
[26]
The price of petrol is Rx per litre. After an increase in price of R1 per litre,
you can buy seven litres less petrol for R504. Calculate the original petrol
price. Show all your calculations.
[6]
A group of learners raise money for a modelling and dance show in
a community hall. The total cost of producing the show is R6 000
and the learners plan to raise equal amounts of money. Eight learners
do not raise any money, so their friends raise an extra R25 each.
Determine the number of learners who participate in the show.
[7]
The sum of two numbers is 40 and the sum of their squares is 818.
Determine the numbers.
[7]
45
PLT MATHS LB 11 7th pgs (Real Book).indb 45
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TOPIC
3
Number patterns
Unit 1: Linear patterns
Consider these number patterns:
1
3; 5; 7; 9; … If the pattern continues in the same way, the next three numbers
will be 11; 13; 15.
2
2; 4; 8; 16; … If the pattern continues in the same, the next three numbers will
be 32; 64; 128.
3
1; 4; 9; 16; … If the pattern continues in the same, the next three numbers will
be 25; 36; 49.
It is often easy to continue a pattern without using a formula. In the patterns above:
1
You add 2 to each term to get the next term. This is a linear pattern.
2
You double each term to get the next term. This is an exponential pattern.
3
The difference increases by 2 each time, so you add 3, then 5, then 7, then 9,
then 11 and finally 13. This is a quadratic pattern.
The problem with this method is that finding the 1 000th term requires you to work
out 999 terms to work out the 1 000th term! To avoid working out terms you do not
need, you develop a formula based on the position of the term.
Consider the same patterns again, but pay attention to each number’s position in
the pattern.
1
T1 = 3; T2 = 5; T3 = 7
If you use the fact that each term increases by 2 and consider its position in the
pattern, then: T1 = 3 = 2 × 1 + 1; T2 = 2 × 2 + 1; T3 = 2 × 3 + 1; Tn = 2n + 1
2
T1 = 2; T2 = 4; T3 = 8; T4 = 16
You know that each term is doubling, but you need to link the terms to their
positions.
T1 = 2 = 21; T2 = 4 = 2 × 2 = 22; T3 = 8 = 2 × 2 × 2 = 23;
T4 = 16 = 2 × 2 × 2 × 2 = 24
The formula for the nth term, Tn, is Tn = 2n.
3
Although you can see that the difference increases by 2 each time, you need to
think about these numbers carefully and to look for a connection between the
value of the number and its position.
T1 = 1 = 11; T2 = 4 = 2 × 2 = 22; T3 = 9 = 3 × 3 = 32; T4 = 16 = 4× 4 = 42
The formula for the nth term, Tn, is Tn = n2.
KEY WORDS
n − the position of the term
nth term − the term in the
position n
The linear pattern formula for 3; 5; 7; 9; … is Tn = 2n + 1 where 2 is the common
difference between the numbers. In the same way that you can continue the pattern
forwards with 11; 13 and 15, you can continue the pattern backwards and work out
that the term before 3 is 1. ⇒ T0 = 2 × 0 + 1 = 1
The general formula for the value of the nth term is Tn = an + b where a is the common
difference, b is the value of T0 and n is the position of the term in the pattern.
46
Topic 3 Number patterns
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WORKED EXAMPLE 1
Consider the pattern: 3; 11; 19; 27; ….
1
Write the next three terms as if the pattern continues in the same way.
2
Write a formula for the nth term in the form Tn = … .
3
Determine the 200th term.
4
Which term has a value of 179?
REMEMBER
If T2 = 11
Tn = 11 and n = 2
SOLUTIONS
1
2
3
4
The terms increase by 8 each time, so the next three terms are 35; 43 and 51.
a = 8 and T0 = −5 ⇒ Tn = 8n − 5 | Check: T7 = 8(7) − 5 = 56 − 5 = 51
T200 = 8(200) − 5 = 1 600 −5 = 1 595
8n − 5 = 179 ⇒ 8n = 184 and n = 23, so T23 = 179
WORKED EXAMPLE 2
Consider the pattern: 17; 13; 9; 5; …
1
2
3
4
If the pattern continues in the same way, write down the next three terms.
Write down a formula for the nth term in the form Tn = … .
Determine the 163rd term.
Which term has a value of −255?
REMEMBER
If the value is −255
you know Tn = −255
and you need to find n.
Find the 163rd term tells you
n = 163, so find T163.
SOLUTIONS 2
1
2
3
4
The terms decrease by 4 each time, so the next three terms are 1; −3 and −7.
a = −4 and T0 = 21 ⇒ Tn = −4n + 21 | Check: T7 = −4(7) + 21 = −28 + 21 = −7
T163 = −4(163) + 21 = −631
−4n + 21 = −255 ⇒ −4n = −276 and n = 69, so T69 = −255
EXERCISE 1
1
Consider the pattern: 39; 34; 29; 24; 19; …
1.1
Write down the next three terms,
1.2
If the pattern continues in the same way, write a formula for the nth term
in the form Tn = … .
1.3
Determine the 22nd term.
1.4
Which term has a value of −251?
2
Consider the pattern: −11; −8; −5; −2; …
2.1
Write down the next three terms if the pattern continues in the same way.
2.2
If the pattern continues in the same way, write a formula for the nth term
in the form Tn = … .
2.3
Determine the 100th term.
2.4
Which term has a value of 178?
Unit 1 Linear patterns
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Unit 2: Quadratic patterns
You often refer to quadratic patterns as second difference patterns that have a
general formula: Tn = an2 + bn + c
A linear pattern has a constant first difference, but a quadratic pattern has a
constant second difference. Other patterns, such as exponential patterns, do not
have a constant difference.
WORKED EXAMPLE
The T0 method
Consider the pattern 1; 4; 9; 16; … and expand it as follows:
Step 1: Write T1 = 1; T2 = 4; T3 = 9; T4 = 16; T5 = 25 as indicated below.
Step 2: Work out the first differences for 3; 5; 7 and 9 and record the ansyours in the
gaps as shown.
Step 3: Repeat the process, working out the second differences, which all equal 2.
Step 4: A common second difference confirms that this is a quadratic or second
difference pattern.
Work backwards one row at a time, filling in the number to the left of
each row.
These numbers have been enclosed in a square to show that they youre filled
in afterwards.
When you fill in the number under T0, you have the constant term.
T0
T1
T2
T3
T4
T5
0
1
4
9
16
25
d1
d2
1
3
5
2
7
2
2
9
2
d
2
Tn = an2 + bn + c with a = __2 ; T0 = c; T1 = a + b + c
You know that a = 1 and c = 0, but you still need to find b.
You know that T1 = 1 and by substitution into T1 = a + b + c you determine b.
T1 = a + b + c ⇒ 1 = 1 + b + 0 and so b = 0 and Tn = n2.
EXERCISE 2
For each number pattern, write down the values of the 6th and 7th terms. Use the
T0 method to determine the formula for the nth term, Tn. Work out the position
of the last value in each sequence.
1
3
5
7
9
48
0; 3; 10; 21; 36; … 17 020
−3; 1; 7; 15; 25; … 865
5; 3; −3; −13; −27; … −8 707
−1; −1; 0; 2; 5; … 1 769
0; −6; −17; −33; −54; … −873
2
4
6
8
10
1; 12; 29; 52; 81; … 9 857
2; 6; 12; 20; 30; … 992
1; 3; 6; 10; 15; … 1 540
1; 10; 24; 43; 67; … 439
0; −2; −2; 0; 4; … 4 690
Topic 3 Number patterns
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WORKED EXAMPLE
The second differences method
If we consider the quadratic formula Tn = an2 + bn + c and work out
the first six terms we get:
T1 = a(1)2 + b(1) + c = a + b + c
T2 = a(2)2 + b(2) + c = 4a + 2b + c
T3 = a(3)2 + b(3) + c = 9a + 3b + c
T4 = a(4)2 + b(4) + c = 16a + 4b + c
T5 = a(5)2 + b(5) + c = 25a + 5b + c
T6 = a(6)2 + b(6) + c = 36a + 6b + c
Now arrange these terms as shown:
T1
a+b+c
T2
T3
T4
T5
T6
4a + 2b + c
9a + 3b + c
16a + 4b + c
25a + 5b + c
36a + 6b + c
3a + b
5a + b
2a
7a + 2b
9a + b
2a
2a
11a + b
2a
Compare the result with the numbers for the pattern 1; 4; 9; 16; 25; …
T1
T2
T3
T4
T5
T6
1
4
9
16
25
36
5
3
2
7
2
9
2
11
2
Starting from the bottom row, write equations and solve each one as you work
your way to the top.
Step 1: 2a = 2 ⇒ a = 1
Step 2: 3a + b = 3, we know a = 1, so 3(1) + b = 3 ⇒ b = 0
Step 3: a + b + c = 3, we know that a = 1 and b = 0, so 1 + 0 + c = 1 and c = 0
The formula: Tn = n2
EXERCISE 3
Consider each number pattern and write down the next two terms. Using the second
differences method, determine the formula for each of the sequences below.
Now determine how many terms there are in each sequence if the value of the last
term is given.
1
3
5
1; 4; 11; 22; 37; … 667
−4; 0; 10; 26; 48; … 1 896
−9; −3; 10; 30; 57; … 3 432
2
4
6
8; 7; 4; −1; −8; … −617
7; 1; −8; −20; −35; … −2 099
19; 10; −7; −32; −65; … −1 085
Unit 2 Quadratic patterns
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REMEMBER
WORKED EXAMPLE
1
2
3
4
5
6
Pictures often contain patterns. You must be able to extract information from the
pictures in the same way as you do from numbers.
7
8
9
4, 5 and 6 are in the same
horizontal row.
A hexagon has six sides.
Figure 1
Figure 2
Figure 3
Figure 4
Consider the four figures and use them to help you complete the table. It would
take a long time to draw Figure 5, and this is not necessary.
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5 Figure n
No. of
horizontal
rows
1
3
5
Tn = 2n − 1
No. of
hexagons in
longest row
1
3
5
Tn = 2n − 1
No. of green
hexagons
1
1
No. of red
hexagons
0
6
No. of yellow
hexagons
0
n≥3
No. of blue
hexagons
0
n≥4
No. of
hexagons in
outer ring
0
Total no. of
hexagons
1
7
No. of
hexagons excl.
green
0
6
n≥2
Think carefully about your answers to Figures 1− 4 and then find the pattern. There
are often different patterns within one figure. Make use of each pattern as soon as
you discover it.
50
Topic 3 Number patterns
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SOLUTION
Fig. 1
Fig. 2
Fig. 3
Number of
horizontal rows
Fig. 4
Fig. 5
Fig. n
1
3
5
Tn = 2n − 1
No. of hexagons
in longest row
1
3
5
Tn = 2n − 1
No. of green
hexagons
1
1
1
1
1
No. of red
hexagons
0
6
6
6
6
6 (n ≥ 2)
No. of yellow
hexagons
0
0
12
12
12
12 (n ≥ 3)
No. of blue
hexagons
0
0
0
18
18
18 (n ≥ 4)
No. of hexagons
in outer ring
0
0
12
18
24
Tn = 6(n − 1)
Total no. of
hexagons
1
7
19
37
61
Tn = 3n2 − 3n + 1
No. of hexagons
excl. green
0
6
1
Tn = 3n2 − 3n
Outer rings: 0; 6; 12; 18 …
Linear pattern with common difference = 6 and T0 = − 6 ⇒ Tn = 6n − 6 = 6(n − 1)
Total number of hexagons:
T0
T1
T2
T3
T4
T5
1
1
7
19
37
61
d1
d2
6
0
6
12
6
18
6
24
6
Using the T0 method:
2a = 6 ⇒ a = 3 and c = T0 = 1. T1 = a + b + c, so 1 = 3 + b + 1 and b = −3
Tn = 3n2 − 3n + 1
Using the second differences method:
Step 1: 2a = 6 ⇒ a = 3
Step 2: 3a + b = 6, so 3(3) + b = 6 and b = − 3
Step 3: a + b + c = 1, so 3 + (− 3) + c = 1 and c = −1
Tn = 3n2 − 3n + 1
If the green hexagon is excluded, then the total number of hexagons is one less
than before.
Use the last result and subtract 1 from the formula.
Unit 2 Quadratic patterns
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EXERCISE 4
1
REMEMBER
A cube has 6 faces.
Face
You can see three faces in
this cube:
• one blue face
• two white faces
The stack of cubes has
three layers.
Figure 1
Figure 2
Figure 3
Consider the figures. The cubes are stacked against two walls so we can see
no faces on the other side. Copy and complete the table.
Fig. 1
Total number of blue faces
1
top layer
Total number of white faces
2
middle layer
Total number of faces visible
3
bottom layer
Number of layers
1
Number of cubes in bottom layer
1
Fig.2
Fig. 3
Fig. 4
Fig. 5
Fig. n
2
REMEMBER
The figure is a pentagon.
Figure 1
Each red dot is a bead.
Each line is a rod.
Figure 2
Figure 3
Figure 4
Figure 5
Figures 1−5 show the first five pentagonal numbers. A pentagon has five sides
and each polygon is pentagonal in shape. In each figure the red beads are joined
by black rods.
Complete the table.
Fig. 1
Fig. 2
Fig. 3
Number of red beads
1
5
12
Number of black rods
0
5
13
Number of beads and rods
1
10
25
Fig. 4
Fig. 5
Fig. n
Not all patterns are linear or quadratic. You can solve exponential and cubic patterns
by inspection, so you do not use special formulae or rules. In each number pattern
question, decide what type of pattern you have. Use the checklist to help you to
identify patterns:
• Numbers with a common difference form a linear pattern.
• Numbers with a common second difference form a quadratic pattern.
• Numbers with a common third difference form a cubic pattern.
52
Topic 3 Number patterns
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•
•
Numbers with no common differences can be broken up into prime factors.
If you cannot find a pattern, check whether adding or subtracting a constant
amount to each term results in a pattern of exponents or cubes.
WORKED EXAMPLES
Consider each number pattern and determine the nth term, Tn:
1
2; 4; 8; 16; 32; …
2
1; 8; 27; 64; 125; …
3
2; 8; 26; 80; 242; …
4
5; 10; 20; 40; 80; …
SOLUTIONS
1
2; 2 × 2; 2 × 2 × 2; …
21; 22; 23; 24; 25
Tn = 2n
2
1; 2 × 2 × 2; 3 × 3 × 3; 4 × 4 × 4; …
13; 23; 33; 43; …
Tn = n3
3
3 − 1; 9 − 1; 27 − 1; 81 − 1; 243 − 1, …
31 − 1; 32 − 1; 33 − 1; 34 − 1; 35 − 1; …
Tn = 3n − 1
Adding 1 to each given term gives you an exponent of 3, so express each term
as an exponent of 3 and then −1.
4
5; 5 × 2; 5 × 4; 5 × 8; 5 × 16
5 × 20; 5 × 21; 5 × 22; 5 × 23; 5 × 24
Tn = 5 × 2n − 1
Each index is 1 less than the position of the term. Instead of an index of n,
use n − 1.
The next exercise is a mixed exercise with linear and quadratic patterns as well as
unexpected patterns for which you find a rule by trial and error!
EXERCISE 5
If each pattern continues in the same way, write down the next two terms in
the pattern.
Determine a rule for the patterns. Then determine how many terms there are
in the sequence.
1
3
5
7
1; 5; 25; 125; 625; … 390 625
1; 8; 27; 64; 125; … 1 000
−2; −1; 4; 13; 26; … 18 334
2; 6; 18; 54; 162; … 13 122
2
4
6
8
14; 12; 4; −10; −30; … −2 146
6; 12; 24; 48; 96; … 3 072
13; 3; −15; −41; −75; … −635
0; 7; 26; 63; 124; … 9 999
Unit 2 Quadratic patterns
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Revision
1
2
3
4
5
Consider the pattern: 7; 3; −1; −5; −9; … −121
If the pattern continues in the same way:
1.1
Write down the next two numbers in the pattern.
1.2
Determine the formula for the nth term, Tn .
1.3
Determine the value of the 19th term.
1.4
Which term has a value of −121?
Consider the pattern: −14; −15; −12; −5; 6; … 4 080
If the pattern continues in the same way:
2.1
Write down the next two numbers in the pattern.
2.2
Determine the formula for the nth term, Tn .
2.3
Determine the value of the 25th term.
2.4
Which term has a value of 4 080?
Consider the pattern: 14; 14;12; 8; 2; … −96
If the pattern continues in the same way:
3.1
Write down the next two terms in the pattern.
3.2
Determine the formula for the nth term, Tn.
3.3
If the pattern is continued to include more terms, find the value
of the 51st term.
3.4
How many terms are there in the sequence?
Consider the quadratic pattern: −9; −6; 1; x; 27; …
4.1
Give the value of the second difference.
4.2
Determine an expression for the second difference in terms of x.
4.3
Determine the value of x.
4.4
Determine the 9th term in the sequence.
4.5
Which term in the sequence has a value of 397?
Consider the number patterns that follow. In each case state the value of
the unknown term and write the formula for the nth term, Tn .
5.1
7; 14; 21; 28; a; 42; …
5.2
−7; −2; 3; b; 13; …
5.3
14; 28; 56; c; 224; …
5.4
3; 21; 147; d; 7 203; …
5.5
6; 18; 36; e; 90; 126, …
(2)
(5)
(3)
(3)
[13]
(2)
(5)
(3)
(5)
[15]
(2)
(5)
(3)
(4)
[14]
(2)
(2)
(1)
(6)
(4)
[15]
(2)
(3)
(5)
(4)
(6)
[20]
54
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6
A children’s construction kit is designed so that flat, square tiles connect to
each other to create cubic storage boxes, which are open at the top. In the figures
each structure has been turned upside down so it is easier to count the tiles in
the base.
REMEMBER
flat tile
vertical
flat tile
Figure 1
Figure 2
The yellow surface is a flat tile.
The structure is upside down,
so the yellow tile is the base
tile. You can see two pink tiles.
Figure 3
Refer to the figures and complete the table.
Figure
1
2
3
Number of tiles in the base of
the structure
1
4
9
Number of vertical tiles in the
structure
4
12
Total number of tiles in
the structure
5
16
4
5
This cube has:
• four vertical pink tiles
• one yellow base tile
• there is no tile on the top.
Structure n
open top
33
56
[11]
7
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5
Figures 1− 5 show the first five heptagonal numbers. A heptagon has seven sides
and each polygon is heptagonal in shape. In each figure the red beads are joined
by blue rods. Complete the table.
Fig. 1
Fig. 2
Fig. 3
Number of red beads
1
7
18
Number of blue rods
0
7
19
Number of beads and rods
1
14
37
Fig. 4
Fig. 5
Fig. n
[15]
55
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TOPIC
4
Analytical geometry
KEY WORD
Cartesian plane – a twodimensional plane formed
by a horizontal number
line representing x-values
and a vertical number line
representing y-values, which
intersect at the point where
both values are 0; ordered
pairs or coordinates refer to
the position of points in the
plane
Unit 1: Equations of a line
From Grade 10 you know these formulae for working with points in the
Cartesian plane:
________________
• The distance between points A(x1;y1) and B(x2;y2) = √(x2 − x1)2 + (y2 − y1)2
•
•
(
x1 + x2 ______
y +y
The midpoint between points A(x1;y1) and B(x2;y2) = ______
; 1 2
Rise represents the change in
y values (the number of units
upwards), and run represents
the change in x values (the
number of units across) to get
from one point to another.
KEY WORD
)
2
(y
– y1)
2
The gradient of a line between points A(x1;y1) and B(x2;y2) = _______
(x2 – x1)
rise
(This is the same as considering ____
run from point A to point B, taking the direction of
the line into account to determine whether the gradient is positive or negative.)
= Positive gradient
REMEMBER
2
= Negative gradient
It is important to remember that:
• parallel lines have equal gradients
• perpendicular lines have gradients which are negative reciprocals.
(This implies that their product = –1).
3 and ___
3 × ___
– 4 are perpendicular because __
– 4 = –1
For example: gradients of __
4
3
4
3
You know that the standard form equation of a straight line is y = mx + c.
Another formula that we can also use for straight lines, particularly when we do not
know the y-intercept (the value of c) is:
reciprocal – the multiplication
inverse of a number is
obtained by interchanging
the numerator and the
denominator of that number
y – y1 = m(x – x1) where m represents the gradient of the line.
(x1;y1) represents the coordinates of any point on the graph.
WORKED EXAMPLE
A line passes through the points A(–1;3) and B(5;0).
1
Determine the equation of line AB.
2
Determine the equation of a line parallel to AB, and passing through
the point (2;–3).
3
Determine the equation of the line perpendicular to AB and passing
through the origin.
56
Topic 4 Analytical geometry
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SOLUTIONS
1
0 – 3 = ___
–3 = – __
1
m = _______
5 – (–1)
6
2
y – y1 = m(x – x1) using (x1;y1) = (5;0)
1 (x – 5)
y – 0 = – __
2
5 or 2y + x = 5
1 x + __
y = – __
2
2
2
1 | Parallel lines have equal gradients.
m = – __
2
y – y1 = m(x – x1) using (x1;y1) = (2;–3)
1 (x – 2)
y + 3 = – __
2
1x + 1 – 3
y = – __
2
The Worked example
would have been
simpler if you realised
that if the line passes
through the origin
c = 0, so y = mx + c
becomes y = 2x.
1 x – 2 or 2y + x = –4
y = – __
2
3
1 × 2 = –1
m = 2 | Perpendicular gradients; – __
2
y – y1 = m(x – x1) using (x1;y1) = (0;0)
y – 0 = 2(x – 0)
y = 2x
Important terminology
A
The median of a triangle is a line from the
vertex to the midpoint of the opposite side.
So, AD is a median of △ABC.
Note: D is perpendicular to BC only when
AB = AC.
B
D
C
KEY WORDS
A
The altitude (height) of a triangle is a line
from any vertex, perpendicular to the
opposite side.
So, AE is an altitude of △ABC.
Note: E will be the midpoint of BC only if
AB = AC.
B
A
The perpendicular bisector of a line passes
through the midpoint of the line, and is
perpendicular to the line.
F
So, FG is the perpendicular bisector of BC.
Note: FG will pass through point A only if AB = AC.
If AB = AC, then the altitude and
perpendicular bisectors are the same line.
C
E
vertex – a point on a triangle
where the sides meet, so a
triangle has three vertices
perpendicular – lines are
perpendicular when the angle
between the lines is 90°
equilateral triangle – a
triangle with all three sides
equal and all three angles
equal
B
C
G
In an equilateral triangle, this will be true for
all three possible medians, altitudes and perpendicular bisectors of the triangle.
Unit 1 Equations of a line
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WORKED EXAMPLE
Given points P(–2;6), Q(5;9) and R(3;–3).
1
Determine the equation of the
median of △PQR from point P.
2
Determine the equation of the
altitude of △PQR from point R.
3
Determine the equation of the
perpendicular bisector of QR.
4
Determine the point of intersection
of the median and the altitudes
found in Questions 1 and 2.
5
Determine the coordinates of point
S so that PQSR is a parallelogram.
y
Q(5;9)
P(–2;6)
x
R(3;–3)
SOLUTIONS
1
(
)
5 + 3 ;_____
9 – 3 = (4;3)
Midpoint of QR = _____
2
2
6 – 3 = ___
3 = – __
1
the gradient of the median is: m = ______
–2 – 4
–6
2
| Using point P(–2;6); you could also use midpoint (4;3).
1 (x + 2)
y – 6 = – __
2
1 x – 1 + 6 y = – __
1 x + 5 or 2y + x = 10
y = – __
2
2
2
3
Gradient of PQ = __
7
7
perpendicular gradient = – __
3
| Using point P(3;–3)
7 (x – 3)
y + 3 = – __
3
7 x + 7 – 3 y = – __
7 x + 4 or 3y + 7x = 12
y = – __
3
3
3
12 = 6
Gradient of QR = ___
2
1
perpendicular gradient = – __
6
| Using midpoint of QR, (4;3)
1 (x – 4)
y – 3 = – __
6
1 x + __
2 + 3 y = – __
1 x + ___
11 or 6y + x = 22
y = – __
6
4
3
6
3
The point of intersection will be the simultaneous solution
of the two equations.
1 x + 5 = – __
7x + 7
– __
2
3
–3x + 30 = –14x + 42
12
11x = 12 x = ___
11
Substitute this into either equation.
( )
3 + 5 = ___
–6 + 5 = ___
58
1 ___
y = – __
2 11
11
11
(
6 ;___
58
So the point of intersection is – ___
58
11 11
)
Topic 4 Analytical geometry
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5
We can use various methods to determine point S, but remember that
we must read points P, Q, S and R in this order for the vertices of the
parallelogram. If the parallelogram were formed with point S in the third
quadrant, it would be named PQRS.
Method 1
Find the equations of QS and RS. Then find the simultaneous solution to
these equations, as point S would be where those two lines intersect.
Gradient of QS = gradient of PR | Opposite sides of a parallelogram are parallel.
And gradient RS = gradient of PQ
3
and gradient of RS = __
9
gradient of QS = – __
5
7
3 (x – 3)
y + 3 = __
9 (x – 5)
y – 9 = – __
5
7
9 x + ___
45 + 9
= – __
3 x – __
9–3
= __
9 x + ___
90
= – __
3 x – ___
30
= __
5
5
5
7
5
7
3 x – ___
30
9 x + ___
90 = __
– __
5
5
7
7
7
3 (10) – ___
30 = 0
y = __
7
7
7
–63x + 630 = 15x – 150
–78x = –780
x = 10
So S = (10;0)
But we know the properties of a parallelogram, so we can determine point S
using quicker methods:
Method 2
The diagonals of a parallelogram bisect each other, so we can find the
coordinates of the midpoint of diagonal QR. This is also the midpoint of PS,
so we can find point S using the midpoint formula.
(
)
5 + 3 ;_____
9 – 3 = (4;3)
Midpoint of QR = _____
2
2
So, midpoint of PS is also (4;3).
6+y
–2 + x = 4 and _____
______
=3
2
2
–2 + x = 8 and 6 + y = 6
x = 10 and y = 0
S = (10;0)
Unit 1 Equations of a line
PLT MATHS LB 11 7th pgs (Real Book).indb 59
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Method 3
Because the opposite sides of
a parallelogram are parallel
and equal in length, the
quickest method to use is the
rise
____
run , which is the same for
opposite sides.
By inspection we can find the
coordinates of S:
From point P to point R there
is a rise of 9 and a run of 5, so
we determine 9 units down
and 5 units across from point
Q to find point S.
y
Q(5;9)
P(–2;6)
rise of 9
rise of 9
run of 5
S(10;0)
x
run of 5
R(3;–3)
Important pointers learnt from Worked examples
To find the equation of a median:
• Find the coordinates of the midpoint of the line through which the median
will pass.
• Find the gradient of the median, using the coordinates of the required vertex
and the midpoint.
• Use the formula y – y1 = m(x – x1) where (x1; y1) can be either the midpoint or
the relevant vertex.
To find the equation of an altitude:
• Find the gradient of the line on which the altitude stands.
• Then find the perpendicular gradient by inverting and changing to the
opposite sign, so that the gradients have a product of –1.
• Use the formula y – y1 = m(x – x1) and the coordinates of the relevant vertex
as (x1;y1).
To find the equation of a perpendicular bisector:
• Find the gradient of the line to be bisected perpendicularly, and then find
the perpendicular gradient.
• Find the coordinates of the midpoint of the bisected line.
• Use the formula y – y1 = m(x – x1) and the coordinates of the midpoint, (x1; y1).
To find the coordinates of the point of intersection of two lines:
• Find a simultaneous solution by equating the two equations.
• Solve for x.
• Substitute the value of x into either equation to solve for y.
To find the fourth coordinate of a parallelogram, given the other three points:
rise
• Use the method of inspection: determine the ____
run of one of the given sides of
the parallelogram.
rise
• Use the same ____
run on the opposite side.
60
Topic 4 Analytical geometry
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EXERCISE 1
1
In the figure K(x;y), L(–2;–1) and M (4;3) are the vertices of △KLM.
The equation of line KL is y = 5x + 9 and the equation of line KM is
5y + x – 19 = 0.
y
K(x;y)
M(4;3)
N
O
x
L(–2;–1)
1.1
1.2
Show that the coordinates of K are (–1;4).
Determine the equation of the median, KN, of triangle KLM, and express it
in the form y = mx + c.
1.3
Find the gradient of LM.
1.4
Then prove that KN is the perpendicular bisector of LM.
1.5
What can you deduce about triangle KLM without further calculations?
1.6
If L, M and the point J(7; p) are collinear, calculate the value of p.
1.7
Write down the coordinates of a point Q which will make KLQM a
parallelogram.
1.8
Determine the equation of a line parallel to KM and passing through
point N.
1.9
Determine the coordinates of P, the point of intersection of the line found
in Question 1.8, and line KL.
1.10 Prove that KM = 2 PN.
2
KEY WORD
collinear – points that lie on
the same line; the gradients
between any two of these
points will be the same
A(–3;3), B(3;–3) and C(6;12) are the vertices of triangle ABC and AD is a median
with D on BC. Determine:
2.1
the coordinates of D
2.2
the equation of AD
2.3
the equation of the altitude of triangle ABC, from point B
2.4
the equation of the perpendicular bisector of AC
2.5
the equation of line BC.
Unit 1 Equations of a line
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Unit 2: Inclination of a line
KEY WORDS
The angle of inclination of a line is the angle between that line and the positive
x-axis.
angle of inclination – the
angle that a line makes with
the positive x-axis
obtuse angle – any angle
between 90° and 180°
y
y
B
θ
y
A
B
θ
x
θ
x
x
B
A
A
Figure 1
Figure 2
Figure 3
y
The angle of inclination will always be between 0°
and 180°. We determine this angle using the formula:
tan θ = m
B
We can understand why when we consider the figure
on the right, which shows that:
θ
opposite
x
adjacent
opposite
adjacent
rise
tan θ = ________ = ____
run = m
A
Note: When the gradient is negative, the angle of inclination is obtuse (between 90°
and 180°) as shown in Figure 2.
To understand why, consider the graph of y = tan θ shown below.
Notice that the graph is below the x-axis (that is, has negative y values) when the
angle is between 90° and 180°.
An angle of inclination is between 0° and 180° so when we know that tan θ is
negative, θ would have to be an obtuse angle (any angle between 90° and 180°).
y
90°
62
180°
270°
360°
x
Topic 4 Analytical geometry
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To find the angle of inclination in the figure:
3=–1
tan θ = –__
3
reference angle (RA) = 45°
θ = 135°
The reference angle is
always the acute angle that
the line makes with the
x-axis. We can find it by
ignoring the negative sign,
and finding tan–1.
y
B
(–1;3)
RA
(2;0)
θ
x
A
WORKED EXAMPLES
Determine the angle of inclination of these lines, giving your answers correct
to two decimal places.
WORKED EXAMPLE 1
WORKED EXAMPLE 2
y
y
(1;5)
3
θ
α
x
x
(–4;–2)
–5
SOLUTION
SOLUTION
5 θ = 59,04°
tan θ = __
7 α = 54,46°
tan α = __
WORKED EXAMPLE 3
WORKED EXAMPLE 4
5
3
y
y
(–7;5)
2
β
3
θ
x
x
(0;–3)
SOLUTION
SOLUTION
2 reference angle = 33,69°
tan β = – __
8 reference angle = 48,81°
tan ϕ = – __
β = 146,31°
ϕ = 131,19°
3
7
Unit 2 Inclination of a line
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The angle between two lines
Because we can find the angle of inclination of lines, we can also find the angle
between two lines.
• Find the angle of inclination of both lines that form the required angle.
• Draw a line parallel to the x-axis, passing through the vertex of the required angle.
• Using Euclidean geometry to find equal angles, determine the required angle
through simple calculations.
WORKED EXAMPLES
y
A(–2;8)
Determine these angles, giving your answers
correct to one decimal place:
1
2
^C
AB
^B
AC
C(3;2)
x
B(–9;–10)
SOLUTIONS
1
First find the angles of inclination of lines
AB and BC:
y
A(–2;8)
18
tan θ = gradient AB = ___
∴ θ = 68,7°
7
θ
12 = 1
tan α = gradient BC = ___
∴ α = 45°
12
Now draw a line through point B,
parallel to the x-axis, and label the equal
corresponding angles θ and α.
It is now clear to see that the value of
^C = θ – α
AB
= 68,7° – 45°
= 23,7°
C(3;2)
x
B(–9;–10)
y
A(–2;8)
θ
α
B(–9;–10)
64
α
α
C(3;2)
x
θ
Topic 4 Analytical geometry
PLTMATHSLB11LB_04.indd 64
2012/07/14 2:29 PM
2
First find the angles of inclination
of lines AC and CB.
To visualise the angle of
inclination of line AC, we extend
line AC so that it it intersects the
x-axis.
6
tan β = gradient AC = – __
5
reference angle = 50,2°
β = 129,8°
We already know that α = 45°
y
A(–2;8)
C(3;2)
β
x
C
α β
x
α
B(–9;–10)
If we draw a line through C,
parallel to the x-axis, we can
find the equal alternate and
corresponding angles α and β.
^ B = 180° – β + α
So, AC
= 50,2° + 45°
= 95,2°
y
A(–2;8)
β
α
B(–9;–10)
In this example it may have been
easier to consider this method:
^ B is the exterior angle of
Since AC
△CMN,
^ B = α + CN
^M
AC
= α + 180° – β
= 45° + 180° – 129,8°
= 95,2°
y
A(–2;8)
C(3;2)
β
θ
M N
x
B(–9;–10)
Unit 2 Inclination of a line
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2012/07/02 2:20 PM
EXERCISE 2
1
2
3
Determine the angle of inclination of line AB where the gradient of AB is given:
1.1
m=2
1.2
m = –5
1.3
5
m = __
1.4
m = – 0,72
3
Determine the angle of inclination of these lines, correct to one decimal place:
2.1
1x – 8
y = __
2.2
5y + 2x = 10
2.3
y+x=3
2
Given points P(–1;–1), Q(4;2), R(7;–5) and S(–3;–7).
y
Q(4;2)
x
P(–1;–1)
R(7;–5)
S(–3;–7)
Determine (correct to one decimal place):
3.1
the acute angle between QR and the x-axis.
3.2
the angle of inclination of PQ
^R
3.3
PQ
3.4
the angle of inclination of SP
^R
3.5
SP
3.6
P^
SR
66
Topic 4 Analytical geometry
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Revision
1
2
3
4
5
6
A triangle with vertices A(–1;7), B(7;5) and C(5;–3) is given.
^ C = 90°
1.1
Show that AB
1.2
Prove that △ABC is isosceles.
1.3
Determine the area of △ABC.
(4)
(4)
(2)
[10]
P(–3;1), Q(2;5) and R(7;a) are three points in the Cartesian plane.
Find the value(s) of a in each case if:
2.1
P, Q and R are collinear
^ R = 90°
2.2
PQ
2.3
PR = QR.
(4)
(4)
(4)
[12]
P(3;5), Q(–1;–5), and R(4;1) are the vertices of △PQR.
3.1
Calculate the length of PQ (leave answer in simplest surd form).
3.2
Find the coordinates of M, the midpoint of PQ.
3.3
Find the equation of the perpendicular bisector of PQ.
3.4
If ME ∥ QR with E on PR, use analytical methods to calculate
the coordinates of E.
3.5
Find the equation of the median of △PQR drawn from Q.
^ R.
3.6
Calculate the size of PQ
3.7
Determine the coordinates of S, so that PQRS is a parallelogram.
(3)
(2)
(3)
(7)
(3)
(4)
(2)
[24]
Prove that the points A(1;6), B(1+ 3k;6 – k) and C(1 + 3n;6 – n) are
collinear for all real values of k and n.
[4]
Points A(–4;2), B(–2;–1) and C(1;1) are given.
5.1
Show that AB ⊥ BC.
5.2
Determine the area of △ABC.
5.3
Determine the coordinates of D if ABCD is a square.
5.4
Determine the equation of the median of △ABC from point A to BC.
5.5
Determine the equation of the perpendicular bisector of BC.
5.6
Determine the coordinates of the point of intersection of the
perpendicular bisector found in Question 5.5, and line AC.
^ C using analytical methods.
5.7
Determine BA
The figure shows a straight line
with equation 5y + 2x = 10, which
meets the x-axis at A and the y-axis
at B. The line BC is perpendicular
to AB, and meets the x-axis at C.
Showing all your working, find the
area of △ABC.
y
A(–4;2)
C(1;1)
x
B(–2;–1)
(3)
(4)
(2)
(4)
(4)
(7)
(6)
[30]
y
B
C
A
x
[8]
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TOPIC 1: REVISION CONTINUED
7
In the figure A(4;2), B(8;8), C and D are the vertices of a parallelogram
whose diagonals intersect at M(3;7).
y
C
B(8;8)
D
M(3;7)
A(4;2)
7.1
7.2
7.3
8
x
Determine the coordinates of C and D.
^ B = 90°.
Prove that AM
What kind of quadrilateral is ABCD? Give a reason for your answer.
(6)
(3)
(4)
[13]
Points P(–2;5), Q(5;9) and S(–1;–3) are given. R(m; n) is a fourth point in
the plane.
y
Q(5;9)
P(–2;5)
R
x
S(–1;–3)
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
9
Write down the values of m and n if PQRS is a parallelogram.
Show that in this case PQRS is a rhombus.
Write down the coordinates of the midpoint of QS.
Calculate the area of the rhombus PQRS.
^ R.
Calculate the size of PQ
Determine the equation of the line through S parallel to PR.
Determine the equation of the perpendicular bisector of PR.
Show that the perpendicular bisector in Question 8.7 passes
through S.
The coordinates of △ABC are A(1;7), B(3;7) and C(–5;–1). Calculate:
9.1
the perimeter of △ABC
9.2
the area of △ABC
^A
9.3
the magnitude of BC
9.4
the coordinates of D so that ABCD is a parallelogram
(2)
(3)
(1)
(4)
(2)
(4)
(4)
REMEMBER
A rhombus is a parallelogram
which has all sides equal, and
diagonals bisecting at right
angles.
(2)
[22]
(4)
(4)
(6)
(2)
[16]
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TOPIC 1: REVISION CONTINUED
10
11
12
13
Points A(–1;–3), B(6;5) and C are the vertices of a triangle.
The equation of AC is y = –3x – 6.
Determine, giving answers to two decimal places:
10.1 the angle of inclination of line AB
10.2 the angle of inclination of line AC
^ B.
10.3 CA
y
B(6;5)
C
Points P(3;7), R(1;1), S and T(9;10) are the vertices of
A(–1;–3)
a parallelogram PRST.
11.1 Write down the coordinates of S.
11.2 Find the equation of the line perpendicular to RS, passing through P.
11.3 Calculate the coordinates of the point of intersection of RS and the
line you found in Question 11.2.
A(–3;2), B(4;8), C and D(6;4) are the vertices of a parallelogram.
12.1 Find the coordinates of C.
12.2 Show that ABCD is a rhombus.
12.3 Write down the coordinates of M, the midpoint of AC.
12.4 Find the equation of BD.
12.5 Show that M lies on BD.
12.6 Describe in words the property of a parallelogram that you have
verified in your answers to Questions 12.3 to 12.5.
^ C.
12.7 Calculate the size of BA
x
(2)
(3)
(2)
[7]
(2)
(4)
(6)
[12]
(2)
(3)
(2)
(3)
(3)
(2)
(3)
[18]
In the figure, the equation of line AB is x – 3y + 6 = 0 and M is the
point (3;1).
y
A
B
M(3;1)
x
13.1 Determine the gradient of line AB.
13.2 Determine the equation of the straight line MN which is
perpendicular to AB and passes through point M.
13.3 Determine the equation of the line EF which is parallel to
AB and passes through point M.
13.4 Calculate the perpendicular distance between lines AB and EF.
(Leave your answer in surd form.)
(1)
(4)
(3)
(7)
[15]
69
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Formal assessment: Investigation: Ratios and the Golden Ratio
Task 1
If x2 – 3xy + 2y2 = 0
(x – 2y)(x – y) = 0
x = 2y or x = y
x = __
2
__
y
1
or __xy = __
1
1
A: For each equation:
1
2
3
•
•
•
x2 – 2xy + y2 = 0
2
(4)
2
(4)
5x – 3xy – 8y = 0
2
2
(4)
8x – 5xy – 13y = 0
Solve for x in terms of y.
Find the value of the ratio __xy .
Write the ratio __xy in decimal form, correct to three decimal places
where possible.
[12]
B: For each pair of simultaneous equations:
1
2
•
•
•
x + y = 13
and
5x – 8y = 0
(4)
x + y = 21
and
8x – 13y = 0
(4)
Solve for x and y.
Find the value of the ratio __xy .
Write the ratio __xy in decimal form correct to three decimal places where possible.
[8]
Task 2
You may have noticed that the numbers 1, 2, 5, 8 and 13 have appeared as solutions
in Task 1. These numbers form a sequence called the Fibonacci sequence.
In the Fibonacci sequence each number after the second term is the sum of the two
before it.
The table shows the first eight terms of the sequence (T1 = term 1):
70
T1
T2
T3
T4
T5
T6
T7
T8
1
1
2
3
5
8
13
21
Term 1 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 70
2012/07/02 2:21 PM
1
2
Write down the next three terms of the sequence
Find a new sequence by finding the ratios
(3)
T T T
T2 ___
T11
___
; 3 ; ___4 ; ___5 up to ___
.
T1
3
4
T2
T3
T4
(2)
T10
Find the ratios in Question 2 correct to three decimal places.
Describe what you notice.
(2)
(1)
[8]
Task 3
Fibonacci or ‘Leonard
of Pisa’ lived from
about 1170 to 1240
We can generate a sequence of numbers by continued square roots:
__
______
__
__________
______
__
_______________
____________
______
__
T1 = √x ; T2 = √1 + √x ; T3 = √ 1 + √1 + √ x ; T4 = √ 1 + √1 + √ 1 + √ x ;
___________________
_______________
____________
______
__
T5 = √1 + √1 + √1 + √1 + √x
1
Use a calculator (or Excel) to find the first ten terms of this sequence
for x = 1. Give your answers correct to four decimal places.
Calculator tip:
__
_______
For T1 press √1 = ; For T2 press √1 + Ans =
_______
For T3 press √ 1 + Ans = and so on, for each successive term.
2
3
Write down each term’s value in a table.
(2)
Find the first 11 terms of the same sequence for x = 2 and x = 3.
Record these values in the same table.
(4)
Describe what you notice about the values of the terms.
(2)
[8]
Task 4
We can generate a sequence of numbers by continued fractions:
1 ; T = 1 + _____
1 ; T = 1 + ________
1
1
1
T1 = 1 + __
; T4 = 1 + __________
; T5 = 1 + ____________
x
2
3
1
1
1
1
1 + __
x
1
1 + _____
1
1 + __
x
1 + ________
1
1 + _____
1
1 + __
x
1 + __________
1
1 + ________
1
1 + _____
1
1 + __
x
Use a calculator (or Excel) to find the first ten terms of this sequence for x = 1.
Give your answers correct to four decimal places.
Calculator tip:
1 = ; For T press 1 + ____
1 =
For T1 press 1 + __
2
1
Ans
1 = … and so on, for each successive term
For T3 pres 1 + ____
Ans
2
3
Write down each term’s value in a table.
(2)
Find the first ten terms of the same sequence for x = 2 and x = 3.
Record these values in the same table.
(4)
Describe what you notice about the values of the terms.
(2)
[8]
Investigation: Ratios and the Golden Ratio
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2012/07/02 2:21 PM
Investigation continued
Task 5
We have seen the number 1,618 (1,61803398 to be more accurate), in three
different situations.
• the ratio of the terms of the Fibonacci sequence
• continued square roots
• continued fractions.
This number is called ‘phi’ or φ. It is an important number and is also called
the Golden ratio. φ has many special properties. Let φ = 1,61803398
1
1
Write down the value of 1 + __
φ.
(1)
2
What do you notice?
3
1
Use the fact that φ = 1 + __
φ to solve for φ and show that φ = 1 +
4
5
__
√5
?
What kind of number is 1 + ___
Determine the value of 1 +
2
__
√5
___
2
__
√5
___
2
(2)
.
(3)
(1)
correct to nine decimal places.
(1)
[8]
Task 6
φ
The Golden rectangle is a rectangle in which the ratio of the sides is __
1
or approximately 1,618.
Throughout the ages this ratio has been recognised as aesthetically
pleasing. Many buildings and works of art have the Golden rectangle
in them. Two examples are the Parthenon in Greece and the United
Nations building in New York, where the ratio of the width of the
building compared with the height of every ten floors is the Golden ratio.
Rectangle 1 has sides x and y with the ratio of length to breadth
of x:y.
y
Rectangle 1
The Parthenon
x–y
Rectangle 2
Rectangle 2 is formed by creating a square of length y on the one side of
Rectangle 1. The remaining Rectangle 2 has sides of length x – y and y
with the ratio of length to breadth of y:x –y.
If the two rectangles have the same ratio of length to breadth, then they
y
are Golden rectangles. In other words: __xy = ____
x–y
72
The United Nations
Head Quarters
Term 1 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 72
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1
Show that the equation on page 72 is the same as x2 − xy − y2 = 0.
(2)
2
If y = 1, solve for x in the equation. Leave your answer in surd form.
(Why can we discard one of the values of x?)
(3)
What do you notice about this number?
(1)
3
4
2
Use the equation x − xy −
y2
= 0 to show that __xy = φ, the Golden ratio.
(2)
[8]
Task 7
Draw a rectangle on A4 size paper with dimensions
16,2 cm by 10 cm. Notice the ratio of length:breadth is
1,62 which is nearly 1,618, the Golden ratio. You have
created a Golden rectangle.
B
F
C
A
E
D
Study the pictures and use your rectangle to create your own
version. Each new rectangle is formed by drawing a square on
one side on the larger rectangle. If you join the dotted lines,
the diagonals of the squares, by a smooth curve, they form
a spiral.
Make your own version of this spiral and colour your diagram
in creatively. Hand this in with your answers.
[5]
y
y
Task 8
A4
Most paper is cut to internationally agreed sizes: A0; A1; A2; A3; A4 and A5.
x
This paper has the property that A1 is half the size of A0, A4 is half the size
of A3 and A5 is half the size of A4, and so on.
Find the ratio of the sides of A series paper. Use the diagram to show that the
__
ratio of the sides of the paper is √ 2 :1
A3
A5
A5
[5]
Total: 70 marks
Investigation: Ratios and the Golden Ratio
PLT MATHS LB 11 7th pgs (Real Book).indb 73
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Formal assessment: Term 1 Test
1
Simplify each expression as far as possible, showing necessary workings.
Do not use a calculator.
1.1
2x + 3 × 4x
_________
1.2
5
( ________
4 −9 )
1.3
2
2 × 8x + 1
1
__
−1
−1
(4)
−1
(4)
2
___
___
___
___
( √18 – √72 +
___
√ 50 )2
___
1.4
( √3x − √2x )( √3x +
1.5
√ 75k8 − √ 12k8
______________
____
√ 27k6
____
____
___
√ 2x )
(4)
[19]
2.1
11x + 8 = (x + 2)(3x + 4)
(4)
2.2
3(2x2 − 5) = x
(4)
2.3
5 = __
3
2x − __
(5)
x
2
2
2.4
3x − 7 = x (correct to two decimal places)
(4)
2.5
3
x2
4 = _____
_________
+ _____
2
(6)
2.6
2x2 − x − 3 > 0
(4)
2.7
1
2 = ___
(2)
2.8
2.25x + 1 = 250
(4)
x − 2x − 8
x
x−4
x+2
32
x
12 =
5x
(1)
3
__
2.10 x2 = 27
2
− __
2.11 3x
2.12
3
(2)
= 75
______
√ 10x − 1
2.13 3 − 2x +
4
(3)
Solve for x in each equation:
2.9
3
(4)
(3)
=3
______
√ 10x − 1
(2)
=0
Solve for x and y:
3.1
x − 8 = 2y and (x − 3)(y − 2) = y2 − 10
3.2
10 and y = __
5x + 5
y = _____
3.3
9x + 2.27y + 1 = 31−x and x2 + y2 + xy = 7
4.1
4.2
Solve for x by completing the square if 3x2 + 12x − 21 = 0
State whether the roots in Question 4.1 are real and rational,
real and irrational or non-real.
Find the nature of the roots of the equation 2x2 + 10x + 8 = 0
without solving for x.
4.3
x−2
3
(6)
[47]
(7)
(6)
(7)
[20]
(4)
(2)
(3)
74
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4.4
Given: P =
4.4.1
4.4.2
5
6
7
____
3
5 − __
√_____
x+1 x
For which value(s) of x will P be undefined?
For which value(s) of x will P be non-real?
(2)
(2)
[13]
Given the sequence: 2; 11; 20; 29; …
5.1
Determine the sixth term, T6.
5.2
Write down a formula for the general term, Tn.
(1)
(2)
[3]
Given the sequence: 6; 12; 20; 30; …
6.1
Find the next two terms.
6.2
Find a formula in terms of n for the nth term.
6.3
Which term is equal to 506?
7.1
7.2
(2)
(4)
(3)
[9]
Copy and complete the fourth and fifth lines the pattern:
12 = 1
12 + 22 = 1 + 4 = 5
12 + 22 + 32 = 1 + 4 + 9 = 14
12 + 22 + 32 + … = 1 + 4 + 9 + … = …
12 + 22 + 32 + … + … = 1 + 4 + 9 + … + … = …
1 n3 + bn2 + cn.
Given that the last number in the nth row is __
6
Find the values of b and c.
(2)
(6)
[8]
8
Shape 1
8.1
8.2
8.3
8.4
8.5
8.6
8.7
Shape 2
Shape 3
Shape 4
Find the number of green dots in the fifth and sixth shapes
of the pattern.
Find the number of orange dots in the fifth and sixth shapes of the
pattern.
Find a formula for the nth term of green dots.
Find a formula for the nth term of the orange dots.
How many green dots will there be in the ninth pattern?
How many orange dots will there be in the eleventh pattern?
Which pattern number will have 132 orange dots?
(2)
(2)
(2)
(4)
(1)
(2)
(3)
[16]
75
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TOPIC 1: TEST CONTINUED
9
The figure represents △ABC with A(3;8), B(17;6)and C(5;–6).
The altitude AD cuts the median BM at P.
y
A(3;8)
B(17;6)
P
M
D
x
C(5;–6)
Determine:
9.1
the length of AC
9.2
the coordinates of point M, the midpoint of AC
9.3
the equation of median BM
9.4
the gradient of line BC
9.5
the equation of altitude AD
9.6
the angle of inclination of line BC
^ B.
9.7
the size of AC
10
11
A(a;b), B(–12;1) and C(8;1) are the vertices of △ABC. M is
the midpoint of BC and N(3;6) is the midpoint of AC.
10.1 Draw a rough diagram to illustrate the information you are given.
10.2 Calculate the coordinates of M.
10.3 Calculate the values of a and b.
^ C.
10.4 Calculate the size of NM
10.5 Calculate the length of MN in simplified surd form.
10.6 Prove that NM ∥ AB.
10.7 Without further calculation, write down the length of AB.
10.8 Write down the equation of the perpendicular bisector of BC.
Given P(3;5), Q(k;6), R(–3;7) and S(–2;2). Determine the value of k if:
11.1 PR ∥ QS
11.2 PQ ⊥ RS
11.3 QR = QS.
(2)
(2)
(3)
(2)
(4)
(2)
(3)
[18]
(1)
(2)
(2)
(3)
(2)
(2)
(2)
(2)
[16]
(3)
(3)
(3)
[9]
76
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12
In the figure, OABC is a parallelogram with points A(–3;3), B(4;4)
and O, the origin. The diagonals intersect at D.
y
B(4;4)
A(–3;3)
D
O
12.1
12.2
12.3
12.4
12.5
12.6
C
x
Determine the coordinates of D.
Show that the equation of AC is 5y + x – 12 = 0
Find the gradient of AB.
Find the equation of OC.
Find the coordinates of C.
Find the size of the angle which AB (produced) makes with the
positive x-axis.
(2)
(4)
(1)
(2)
(2)
(3)
[14]
Total: 192 marks
77
PLT MATHS LB 11 7th pgs (Real Book).indb 77
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Term 1 summary
Topic 1
Exponents and surds
Laws for exponents:
1 am
×
an
=
am + n
Rules for surds:
__
___
__
n
n
1 n√ a × √ b = √ ab | Multiplying
__
__
__
n
a | Dividing
÷ √b = n√ __
2 am ÷ an = am – n
2
3 (am)n = amn
3 √ a2b = √ a2 √ b = a√ b | Reducing
4 (ab)m = ambm
1
5 a–n = __
n
___ a __
m
n
m
√
6 a = an
4
√a
n
b
___
__
√a
__ __
__
__
__
__
__
+ √a = 2√a | Adding like surds
__
__
√b
a√ b or ______
2 − √__
b × ______
2 − √__
b
a__ × ___
__ = ____
5 ___
b
√b
2 + √b
2 − √b
__
√
4
−
4
b
+
b
= __________ | Rationalising the denominator
4−b
√b
7 a0 = 1
Solving equations with:
Unknown exponents:
If ax = ay then x = y | a > 0, a ≠ 1
Rational exponents:
a
b
__
__
If xb = y c then x = ( y c )a | x > 0, x ≠ 1
Surd equations:
____
If √ x + a = x + b then x + a = ( x + b )2
and solve the quadratic equation.
Topic 2
Equations and inequalities
Solve quadratic equations in standard form ax2 + bx + c = 0 by:
• factorising (always try this first)
• completing the square (only if asked to)
_______
−b±√b2 − 4ac
• using the formula x = ____________ (answers usually given correct to two decimal places).
2a
Write all quadratic equations with fractions in standard form by:
• factorising the denominator and numerator (if possible)
• multiplying both sides of the equation by the LCD to get rid of fractions.
Solve simultaneous equations by:
• substitution of one equation (usually the linear one) into the other (usually the quadratic equation)
• equating the two y values (usually done with graphs in y form).
Solve quadratic inequalities by:
y
• factorising
• using a number line or using a graph.
Statement
–x2
+4≥0
Inequality notation
Interval notation
−2 ≤ x ≤ 2
x ∊ [−2;2]
– –
–2
+ +
+ +
2
–
–
x
Determine the nature of roots by:
• looking at both roots (solutions to equation) to decide if they are real or non-real.
If real decide if they are rational or irrational and equal or unequal.
• looking at b2 − 4ac (part of the quadratic formula under the square root)
b2 − 4ac < 0 ⇒ roots will be non-real
b2 − 4ac > 0 ⇒roots will be real
b2 − 4ac = 0 ⇒roots will be equal (and real)
b2 − 4ac = perfect square ⇒ roots will be rational (and real)
78
Term 1 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 78
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Topic 3
Number patterns
Patterns can be given as a sequence of numbers or a set of
figures which change according to a rule.
Linear patterns
Linear patterns have a constant first difference. Tn = an + b
where a is the common difference, b the value of T0 and n the
position of the term in the pattern.
Quadratic patterns
Quadratic patterns have a constant second difference.
Tn = an2 + bn + c, 2a = common second difference,
3a + b = first difference = T2 – T1
a + b + c = T1 or c = T0
Topic 4
Analytical geometry
Given points A(x1;y1) and B(x2;y2) in the Cartesian plane:
• The distance between A and B =
________________
√( x2 – x1 )2 + ( y1 – y2 )2
(
x1 + x2 ______
y +y
• The midpoint, M, between points A and B is M = ______
; 1 2
2
2
y
−
y
2
1
• The gradient of a line between points A and B is m = ______
x2 − x1
)
• The equation of the straight line through A and B is y = mx + c or y − y1 = m( x − x1 )
• The angle of inclination that line AB makes with the positive x-axis is found by letting tan θ = m
Parallel lines have equal gradients: m1 = m2
The product of the gradients of perpendicular lines is equal to –1: m1 × m2 = –1
We can find equations of perpendicular bisectors, altitudes and medians using all the above formulae.
Term 1 summary
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Term
2
80
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TOPIC 5
Functions: Effects of parameters
Unit 1
The effects of the parameters a, p
and q on parabolas
The effects of the parameters a, p
and q on hyperbolas
The effects of the parameters a, p
and q on exponential graphs
Real life applications
The average gradient between two
points on a curve
Unit 2
Unit 3
Unit 4
Unit 5
Revision
TOPIC 6
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 1
Unit 2
Unit 3
Unit 4
92
101
108
111
114
Functions: Trigonometric graphs
Revision of basic trigonometric graphs
using point by point plotting
The effects of the parameter k on
some trigonometric functions (graphs
of the functions defined by y = sin kx,
y = cos kx and y = tan kx)
Horizontal shifts
Determine the equations of
trigonometric graphs
Sketch graphs which have a change
in period and a horizontal shift
Revision
TOPIC 7
82
116
123
125
130
132
136
Trigonometry
Revision of Grade 10 trigonometry
Identities
Reduction formulae
Trigonometric equations – specific
and general solutions
Revision
Formal assessment: Assignment: Term 2 test
Term summary
Formal assessment: Examination Practice Paper 1
Examination Practice Paper 2
138
142
146
155
163
166
168
170
172
81
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TOPIC
2
5
Functions: Effects of parameters
KEY WORDS
parameter – a variable that
restricts or gives a particular
form or shape to the equation
it characterises
function – a relationship
between two variables,
usually x and y, where for
every value of x there is one
corresponding value for y
increase – to grow bigger
decrease – to grow smaller
reflect – create a mirror
image or reflection
y-intercept – the point
where the graph cuts the
x-axis (where x = 0)
Unit 1: The effects of the parameters a, p
and q on parabolas
We define trigonometric graphs by parametric equations which group the graphs
into families. When we assign values to the parameters and substitute them into
the equation, it becomes a specific equation with a specific shape that is restricted
by these values. Parameters do not change the type of graph, but define the
characteristics of a particular graph within a family of graphs.
We will investigate the effect of the parameters a, p and q on the graph of the
function defined by y = f (x) = a(x + p)2 + q. But firstly, we will revise the effects of a
and q on the parabola given by y = f (x) = ax2 + q by comparing the following graphs:
1 x2 as well as y = −x2; y = −4x2 and y = −__
1 x2.
f(x) = x2; g(x) = 4x2; h(x) = __
4
f (1) =
12
= 1⇒(1;1)
4
2
2
g(2) = 2 = 4⇒(2;4)
g(3) = 3 = 9⇒(3;9)
g(x) = 4x2
g(1) = 4(1)2 ⇒(1;4)
g(2) = 4(2)2 = 16 ⇒ (2;16)
g(3) = 4(3)2 = 36 ⇒ (3;36)
1 x2
h(x) = __
4
1 (1)2 = __
1 (1;0,25)
h(1) = __
4
4
1 (2)2 = 1 (2;1)
h(2) = __
4
9 (3;2,25)
1 (3)2 = __
h(3) = __
4
4
We can conclude that:
• the greater the value of a, the steeper the curve
• when a is negative the graph reflects in the x-axis
• changing the value of a has no impact on the y-intercept.
y
Parabolas
6
y = 4x2
y = x2
2
3
5
y = __14x2
4
3
2
1
–4
–3
–2
–1
0
–1
4
x
–2
–3
–4
–5
–6
y = –4x2
y = –x2
y = – __14x2
Axis of symmetry
82
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 82
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WORKED EXAMPLE
KEY WORD
Consider the function: f (x) = 2x2
1
Sketch f, showing the coordinates of three points.
2
If f is shifted up 1 unit, the translated graph is g. Draw g on the same system
of axes.
3
State the equation of g.
4
Sketch h, the reflection of f in the x-axis, on the same system of axes.
5
State the equation of h.
6
Sketch k, the reflection of g in the x-axis, on the same system of axes.
7
State the equation of k.
reflection – a mirror image
about a line of symmetry
SOLUTIONS
1
2
3
4
5
6
7
f (0) = 0 ⇒ (0;0)
f(1) = 2(1)2 = 2 ⇒ (1;2)
f (–1) = 2(–1)2 = 2 ⇒ (–1;2)
Add 1 to each y-value.
(0;0) → (0;1)
(1;2) → (1;3)
(–1;2) → (–1;3)
g(x) = f (x) + 1 = 2x2 + 1
Leave the x-coordinates
unchanged, but change
the sign of each y-value:
(0;0) → (0;0)
(1;2) → (1;–2)
(–1;2) → (–1;–2)
h(x) = –f (x) = –2x2
Leave the x-coordinates
unchanged, but change
the sign of each y-value:
(0;1) → (0;–1) (1;3) → (1;–3)
k(x) = –g(x) = –2x2 – 1
y
(–1;3)
g
f
(1;3)
(–1;2)
REMEMBER
(1;2)
1
x
0
–1
(–1;–2)
(1;–2)
(–1;–3)
(1;–3)
k
h
(–1;3) → (–1;–3)
The vertical axis is the y-axis.
The equation of the y-axis is
given by x = 0.
The horizontal axis is the
x-axis.
The equation of the x-axis is
given by y = 0.
The axes cross at the origin,
(0;0)
Coordinates
Give both the x- and
y-coordinates in the form
(2;3).
If the first coordinate is 0,
then you have a y-intercept:
(0;3) cuts the y-axis at 3
If the second coordinate is 0,
then you have an x-intercept:
(2;0) cuts the x-axis at 2
Vertical shifts require movement up or down only and do not change the
x-coordinates.
Reflections in the x-axis do not change x-coordinates, but change the signs
of the y-coordinates.
• f (x) + 2 will increase each y-coordinate by 2 and the graph will shift up 2 units.
• f (x) – 3 will decrease each y-coordinate by 3 and the graph will shift down 3 units.
• –f (x) will reflect f in the x-axis.
Unit 1 The effects of the parameters a, p and q on parabolas
PLT MATHS LB 11 7th pgs (Real Book).indb 83
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EXERCISE 1
REMEMBER
On the ‘same set of axes’
means you will draw more
than one graph on the same
sketch.
1
1 x2
Consider the function: f (x) = – __
2
1.1
Sketch the graph of f, showing the coordinates of three points.
1.2
If g is obtained by shifting f up 2 units, sketch g on the same system
of axes as f.
1.3
State the equation of g.
1.4
Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
1.5
State the equation of h.
1.6
Sketch k, the reflection of g in the x-axis, on the same system of axes
as f, g and h.
1.7
State the equation of k.
2
Consider the function: f (x) = –8x2
2.1
Sketch the graph of f, showing the coordinates of three points.
2.2
If g is obtained by shifting f down 1 unit, sketch g on the same system
of axes as f.
2.3
State the equation of g.
2.4
Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
2.5
State the equation of h.
2.6
Sketch k, the reflection of g in the x-axis, on the same system of axes
as f, g and h.
2.7
State the equation of k.
3
1 x2
Consider the function: f (x) = __
3
3.1
Sketch the graph of f, showing the coordinates of three points.
3.2
If g is obtained by shifting f down three units, sketch g on the same
system of axes as f.
3.3
State the equation of g.
3.4
Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
3.5
State the equation of h.
3.6
Sketch k, the reflection of g in the x-axis, on the same system of axes
as f, g and h.
3.7
State the equation of k.
4
Consider the function: f (x) = –3x2
4.1
Sketch the graph of f, showing the coordinates of three points.
4.2
If g is obtained by shifting f down 2 units, sketch g on the same system
of axes as f.
4.3
State the equation of g.
4.4
Sketch h, the reflection of f in the x-axis, on the same system of axes
as f and g.
4.5
State the equation of h.
4.6
Sketch k, the reflection of g in the x-axis, on the same system of axes.
4.7
State the equation of k.
REMEMBER
To find an x-intercept,
substitute y = 0.
To find a y-intercept,
substitute
84
Topic 5 Functions: Effects of parameters
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Horizontal shifts of the parabola y = x2
–6
–5
–4
–3
–2
–1
0
1
2
3
f
2
y=x
36
25
16
9
4
1
0
1
4
9
None
g
y = (x + 2)2
16
9
4
1
0
1
4
9
16
25
2 left
h
y = (x – 1)2
49
36
25
16
9
4
1
0
1
4
1 right
k
2
4
1
0
1
4
9
16
25
36
49
4 left
x
•
•
•
y = (x + 4)
Shift
Adding to x increases its value so the pattern of answers shifts to the left by the
amount added.
Subtracting from x reduces its value so the pattern shifts to the right by the amount
subtracted.
This pattern of horizontal shifting applies to all graphs, not only to the parabola.
The graphs in the table are sketched in the figure below. Notice that they are all
identical in shape.
The x-intercept in each graph clearly indicates how the graph has shifted.
y = (x + 4)2
y = (x + 2)2
y
y = x2
y = (x – 1)2
KEY WORD
x-intercept – the point where
the graph cuts the x-axis
(where y = 0)
–5
–4
–3
–2
–1
1
2
x
Each graph has equal roots and touches the x-axis, without crossing it.
There are three forms of each graph:
• y = (x + 2)2 is the turning point form
• y = (x + 2)(x + 2) is the x-intercept form
• y = x2 + 4x + 4 is standard form.
EXERCISE 2
Write each equation in the turning point form. State the turning point and then
describe the shift of each parabola from the origin (0;0).
1
2
3
4
y = x2 + 6x + 9
y = x2 – 8x + 16
y = x2 – 4x + 4
y = –x2 – 2x – 1
Unit 1 The effects of the parameters a, p and q on parabolas
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Horizontal and vertical shifts
Consider the parabola equation y = (x + p)2 + q:
• changing p results in a horizontal shift
• changing q results in a vertical shift
• p and q do not influence each other.
x
f
g
y = x2
–6
–5
–4
–3
–2
–1
0
1
2
3
Shift
36
25
16
9
4
1
0
1
4
9
None
2
12
5
0
–3
–4
–3
0
5
12
21
2 left, 4 down
2
y = (x + 2) – 4
h
y = (x – 1) + 2
51
38
27
18
11
6
3
2
3
6
1 right, 2 up
k
y = (x + 4)2 – 1
3
0
–1
0
3
8
15
24
35
48
4 left, 1 down
The graphs in the figure are drawn from the table. Notice that all the graphs are
identical in shape.
y
k
g
f
h
6
5
4
3
2
(1;2)
1
KEY WORD
vertical shift – with straight
up or down, no sideways
movement
horizontal shift – with
sideways, with no up or down
movement
axis of symmetry – a line
that divides a graph into
two identical halves and
about which a function is
symmetrical
–6
–5
–4
–3
–2
(–4;–1)
–1
–1
1
2
3
x
–2
–3
(–2;–4)
•
•
•
•
–4
y = x2 has a minimum value of 0 at (0;0) and an axis of symmetry given by x = 0.
y = (x + 2)2 – 4 has a minimum value of – 4 at (–2;– 4) and an axis of symmetry
x = –2.
y = (x – 1)2 + 2 has a minimum value of 2 at (1;2) and an axis of symmetry x = 1.
y = (x + 4)2 – 1 has a minimum value of –1 at (– 4;–1) and an axis of symmetry
x = – 4.
EXERCISE 3
Write each equation in standard form and in the x-intercept form. State the turning
point of each graph and then describe the shift of each parabola from the origin (0;0).
1
2
3
4
86
y = (x – 4)2 – 9
y = 4(x + 2)2 – 1
y = –(x + 5)2 + 16
y = –(x – 3)2 + 4
Topic 5 Functions: Effects of parameters
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Sketching a parabola
Check the sign of a:
• If a > 0, then the parabola ‘smiles’  and the arms go up.
• If a < 0, then the parabola ‘frowns’  and the arms go down.
Find the y-intercept by making x = 0 and solving for y:
• If y = 2x2 – 3x – 5, then the constant will be the y-intercept, so y = –5.
• If y = (x – 5)2 – 4, then y = (0 – 5)2 – 4 = 25 – 4 = 21.
Find the x-intercepts, make y = 0 and then solve the equation.
• Most graphs will have rational roots and so we can find the roots by factorisation:
y = 2x2 – 3x – 5 ⇒ 0 = (2x – 5)(x + 1) and x = 2,5 or x = –1
x-intercepts (–1;0) and (2,5;0)
• Use the quadratic formula for graphs with irrational roots:
y = x2 – 2x – 4 ⇒ x2 – 2x – 4 = 0 cannot be solved by factors.
quadratic formula:
a = 1, b = –2 and c = – 4 ⇒ b2 – 4ac = (–2) – 4(1)(– 4) = 20
___
(
) √
– –2 ± 20
x = __________
= 3,24 or –1,24
( )
•
21
x-intercepts (–1,24;0) and (3,24;0)
Parabolas do not necessarily have x-intercepts and in this case we cannot find
roots:
y = x2 – x + 2 ⇒ x2 – x + 2 = 0 cannot be solved by factors.
Test b2 – 4ac from the quadratic formula: a = 1, b = –1 and
c = 2 ⇒ b2 – 4ac = (–1)2 – 4(1)(2) = –7
No x-intercepts because b2 – 4ac = 0
Find the turning point:
• Every parabola has a turning point.
REMEMBER
b.
The x-coordinate is the axis of symmetry, given by x = – ___
2a
The y-coordinate is the minimum value if a > 0, then the parabola ‘smiles’ .
The y-coordinate is the maximum value if a < 0, then the parabola ‘frowns’ .
Find the y-coordinate by substituting the x-coordinate.
x +x
2
–1 + 2,5 ___
1,5
_______
x-intercepts (–1;0) and (2,5;0) ⇒ axis of symmetry is x =
=
= 0,75
2
2
•
1
2
If you know the x-intercepts, then the axis of symmetry is ______
•
b
If the roots are irrational or if there are no roots, then use the formula x = – ___
a > 0 means that a is positive.
a < 0 means that a is
negative.
minimum – smallest possible
value
maximum – largest possible
value
axis of symmetry – divides
graph into two identical
halves
2a
–( –2 ) = 1
–b = _____
y = x2 – 2x – 4 has a = 1 and b = –2, so the axis of symmetry is x = ___
2a
2( 1 )
(
)
– –1 = __
1
y = x2 – x + 2 has a = 1 and b = –1, so the axis is symmetry is x = _____
( )
21
2
Unit 1 The effects of the parameters a, p and q on parabolas
PLT MATHS LB 11 7th pgs (Real Book).indb 87
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WORKED EXAMPLE
1
2
3
4
5
1 (x – 4)2 + 2 and h(x) = x2 – 4x + 6 on the same system of axes, clearly
Sketch f (x) = x2 – 5x – 6, g(x) = – __
2
indicating the intercepts with the axes and the coordinates of the turning points.
Use your graphs to determine two values of x for which f (x) = g(x).
Solve algebraically for x if f (x) = g(x).
Determine h(4) – g(4).
Use your graph to determine two values of x for which g(x) – f(x) = 0.
SOLUTIONS
1
1 (x − 4)2 + 2 
g(x) = −__
f(x) = x2 – 5x – 6 
y-intercept: (0; – 6)
x-intercepts: (6;0) and (–1;0)
x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or x = –1
2
y-intercept: (0;– 6)
1 (0 − 4)2 + 2 = −6
− __
2
x-intercepts: (0;0) and (4;0)
1 (x − 4)2 + 2 = 0
− __
2
Turning point: (2,5;–12,25)
(x − 4)2 = 4
x − 2 = 2 or x − 2 = −2
x = 4 or x = 0
–1 + 6 = __
5 = 2,5
x = ______
2
2
y = (2,5)2 – 5(2,5) – 6 = –12,25
h(x) = x2 – 4x + 6
y-intercept: (0;6)
x-intercepts: None
No factors ⇒ check b2 – 4ac
a = 1, b = – 4 and c = 6
b – 4ac = (– 4)2 – 4(1)(6) = –8
No roots
Turning point: (2;2)
(– 4)
2(1)
b = – ____ = 2
x = – ___
2a
y = (2)2 – 4(2) + 6 = 2
Turning point: (4;2)
y
h
(4;6)
6
(2;2)
–1
f
2
(4;2)
x
6
–6
(2,5;–12,25)
2
3
4
5
88
g
x = 0 or x = 6
1 (x – 4)2 + 2 ⇒ – 2x2 + 10x + 12 = x2 – 8x + 16 – 4
x2 – 5x – 6 = – __
2
2
3x – 18x = 0 ⇒ 3x(x – 6) = 0 and x = 0 or x = 6
h(4) – g(4) = 6 – 2 = 4 units
x = 0 or x = 6
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 88
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WORKED EXAMPLE
1
REMEMBER
f(x) = 2x2 – 4x – 6 and g(x) = 2x – 6
1.1
Draw neat sketch graphs of f and g on the same system of axes. Indicate
the intercepts with the axes, as well as the coordinates of any turning
points.
1.2
Use your graphs to answer the questions:
1.2.1
For which value(s) of x is f(x) = g(x)?
1.2.2
For which value(s) of x is f(x) < g(x)?
1.2.3
For which value(s) of x is f(x).g(x) ≥ 0?
SOLUTIONS
1
f(x) = 2x2 – 4x – 6 
y-intercept: (0;– 6)
x-intercepts: (–1;0) and (3;0)
1.1
2x2 – 4x – 6 = 0
⇒ x2 – 2x – 3 = 0
(x – 3)(x + 1) = 0,
so x = 3 or x = –1
f (x) = 2x2 – 4x – 6 y
Turning point: (1;–8)
3 + (–1)
2
2=1
x = _______ = __
1.2
–1
2
y = 2(1)2 – 4(1) – 6 = –8
g(x) = 2x – 6
y-intercept: (0;– 6)
x-intercept: (3;0)
2x – 6 = 0
2x = 6 ⇒ x = 3
1.2.1
x = 0 and x = 3
1.2.2
0<x<3
1.2.3
x ∊ [–1;∞)
3
x
REMEMBER
–6
g(x) = 2x – 6
If A × B > 0, then A and B
are both positive or both
negative.
If C × D < 0, then C and D
must be opposite in sign.
f (x) = g(x) means that the
y-value of f is equal to the
y-value of g.
f (x) < g(x) means that the
y-value of f is less than the
y-value of g.
f (x).g(x) ≥ 0 means that the
product of the y-values of f
and g is positive or 0.
If a > 0, the arms of the
parabola go up  and the
parabola has a minimum
y-value.
If a < 0, the arms of the
parabola go down  and
the parabola has a maximum
y-value.
(1;–8)
x-intercepts and roots
represent the point where the
graph cuts the x-axis (where
y = 0)
EXERCISE 4
Label each graph clearly and indicate the intercepts with the axes as well as any
turning point(s).
1
1.1
Sketch f (x) = –x2 + 4x + 12 and g(x) = 4x + 8 on the same system of axes.
1.2
For which value(s) of x is f (x) = g(x)?
1.3
For which value(s) of x is f (x) ≥ g(x)?
1 x2 – x – 4 and g(x) = x – 4 on the same system of axes.
2
2.1
Sketch f (x) = __
2
2.2
For which value(s) of x is g(x) – f (x) = 2?
2.3
For which value(s) of x is f (x) ≤ g(x) ?
3
3.1
Sketch f (x) = –(x + 1)2 + 9 and g(x) = –3x + 6 on the same system of axes.
3.2
For which value(s) of x is f (x). g(x) ≤ 0?
Unit 1 The effects of the parameters a, p and q on parabolas
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Finding the equation of a parabola
We can represent a parabola in three forms:
• y = ax2 + bx + c | Standard form
• y = a(x – x1)(x – x2) | x-intercept formula, with x1 and x2 being the x-intercepts
• y = a(x – p)2 + q | TP formula, with TP (p;q)
If the x-intercepts and a random point are given, use the x-intercept formula.
WORKED EXAMPLE 1
Determine the equation of a parabola which has points (2;0), (– 4;0) and (4;8).
SOLUTION
y = a(x – x1)(x – x2) where x1 = 2, x2 = – 4, and the substitution point is ( 4;8 )
Substitute x-intercepts: y = a(x – 2)[x – (– 4)] ⇒ y = a(x – 2)(x + 4)
1
Now substitute the point: 8 = a(4 – 2)(4 + 4) ⇒ 8 = 16a and a = __
2
1 (x – 2)(x + 4) in x-intercept form or y = __
1 x2 + x – 4 in standard
The equation is y = __
2
2
form.
If the turning point and a random point are given, use the turning point formula.
WORKED EXAMPLE 2
Determine the equation of the parabola if it has a turning point (1;2) and passes
through (2;–1).
SOLUTION
y = a(x – p)2 + q where the turning point is (1;2) and the substitution point
is (2;–1).
Substitute the turning point: (1;2): y = a(x – 1)2 + 2
Now substitute the point: (2;–1) ⇒ –1 = a(2 – 1)2 + 2 and a = –3
The equation is y = –3(x – 1)2 + 2 in turning point form or y = –3x2 + 6x – 1 in
standard form.
EXERCISE 5
1
90
Determine the equation of the parabola which passes through the points:
1.1
(– 4;0), (1;0) and (0;12)
1.2
1.3
( __13;0 ), (3;0) and (0;1)
1.4
( __12;0 ), (–2;0) and ( – __12;6 )
( __32;0 ), (– 4;0) and (2;–3)
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 90
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2
2.1
2.2
2.3
2.4
Determine the equation of the parabola which has a turning point (3;4)
and passes through the point (4;2).
Determine the equation of the parabola which has a turning point (1;–2)
and passes through the point (2;1).
Determine the equation of the parabola with a turning point (–2;4) and a
y-intercept of 2.
5 ;– __
9
Determine the equation of the parabola which has a turning point __
2 8
and passes through the point (5;2).
(
)
EXERCISE 6
Complete the table and then draw each graph on its own system of axes.
• State each function in the three ways: y = ax2 + bx + c, y = a(x – x1)(x – x2) and
y = a(x – p)2 + q
• p is associated with the horizontal shift and q indicates the vertical shift.
• The first two rows have been completed for you as Worked examples with the
given information high-lighted in blue.
• Roots (x-intercepts) tell us where a graph passes through the x-axis and y = 0.
• Not all parabolas have x-intercepts.
Function
1
Shift
y = x2 – 6x + 5
y = (x – 1)(x – 5)
y = (x – 3)2 – 4
2
y = 2x2 + 4x + 6
No factorised form
2
y = 2(x + 1) + 4
Turning
point
Min/Max
Axis of
symmetry
y-intercept
Roots
3 right
4 down
(3;–4)
Minimum
y = –4
x=3
(0;5)
(1;0)
(5;0)
1 left
4 up
(1;4)
Minimum
y=4
x = –1
(0;6)
None
3
(– 4;9)
(0;–7)
4
y = –(x + 5)2 + 16
5
y = 3(x – 4)2 – 12
6
3 right
8 down
(0;10)
Unit 1 The effects of the parameters a, p and q on parabolas
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Unit 2: The effects of the parameters a, p
and q on hyperbolas
KEY WORDS
asymptote – a line that a
graph gets close to, but never
touches
xy = 4 has two asymptotes,
x = 0 and y = 0
If x = 0 or y = 0, then xy = 4
is undefined.
We will investigate the effect of the parameters a, p and q on the graph of the function
a
defined by f (x) = _____
x + p + q.
The effect of a on the hyperbola is the same as the effect of a on the parabola.
•
•
The greater the value of a, the steeper the curve..
If a is negative the graph reflects in the x-axis.
a+q
Consider the hyperbola: y = f (x) = __
x
1 . Changing q causes a
If q = 0 and a = 1, we have the basic hyperbola graph y = __
x
1
__
vertical shift upwards if q > 0, for example, y = x + 2 shifts up 2 units, but a vertical
1 – 1, shifts down 1 unit.
shift downwards if q < 0, for example, y = __
x
1
Reminders about the hyperbola y = __
x
asymptotes
•
•
•
x = 0 (vertical asymptote)
y = 0 (horizontal asymptote)
4
lines of symmetry
3
y=x
2
y = –x
1
Domain: x ∊ ℝ, x ≠ 0
horizontal asymptote y = 0
Range: y ∊ ℝ; y ≠ 0
–8 –7 –6 –5 –4 –3 –2 –1
y
–1
hyperbola
y=x
y = __1x
symmetry line
1
2
3
4
5
6
7
8
x
–2
–3
–4
symmetry line
vertical
asymptote
y = –x
x=0
1+2
Reminders about the hyperbola y = __
x
asymptotes
•
•
•
x = 0 (vertical asymptote)
y = 2 (horizontal asymptote)
lines of symmetry
y=x+2
y = –x + 2
Domain: x ∊ ℝ, x ≠ 0
Range: y ∊ ℝ, y ≠ 2
y
5
4
horizontal
asymptote
3
hyperbola
2
y=2
1
x
–4
–1
–1
symmetry
line
–2
y=x+2
–3
92
y = __1x + 2
–3
–2
1
2
3
4
symmetry
line
vertical
asymptote
x=0
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 92
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1–1
Reminders about the hyperbola y = __
x
• asymptotes
•
•
x = 0 (vertical asymptote)
y = –1 (horizontal asymptote)
lines of symmetry
y=x–1
y = –x – 1
Domain: x ∊ ℝ, x ≠ 0
Range: y ∊ ℝ; y ≠ –1
y
4
y = –x –1
y = x –1
3
symmetry
line
2
–4 –3 –2 –1
–1
y = __1x – 1
symmetry
line
1
1
–2
–3
2
x
3
4
y = –1
horizontal
asymptote
–4
–5 vertical
asymptote
hyperbola –6
x=0
1
Horizontal shifts of the hyperbola y = __
x
x
–5
–4
–3
–2
–1
0
1
2
3
4
Shift
f
1
y = __
1
– __
1
– __
1
– __
1
– __
–1
Undefined
1
1
__
1
__
1
__
None
g
1
y = _____
1
– __
1
– __
–1
Undefined
1
1
__
1
__
1
__
1
__
1
__
2 left
h
1
y = _____
1
– __
1
– __
1
– __
1
– __
1
– __
–1
Undefined
1
1
__
1
__
1 right
k
1
y = _____
1
– __
–1
Undefined
1
1
__
1
__
1
__
1
__
1
__
1
__
3 left
•
•
•
•
x
x+2
x–1
x+3
5
3
6
4
2
2
5
3
2
4
3
2
2
2
3
Adding to x increases its value so the pattern of answers shifts to
the left by the amount added.
Subtracting from x reduces its value so the pattern shifts to the
right by the amount subtracted.
This pattern of horizontal shifting applies to all graphs.
Note that there is no vertical shift and the horizontal
asymptote is y = 0.
Vertical asymptotes
• vertical asymptote of f (x) = __1x is x = 0
1 is x = –2
• vertical asymptote of g(x) = _____
x+2
1 is x = 1
• vertical asymptote of h(x) = _____
x–1
1 is x = –3
• vertical asymptote of k(x) = _____
x+3
2
4
3
4
5
4
3
5
6
2
3
7
6
Hyperbolae and horizontal shifts
y
3
1
h(x) = ____
x–1
2
1
–5 –4 –3 –2 –1
–1
1
2
3
4
x
–2
1
k(x) = _____
x+3
–3
–4
1
1
__
g(x) = _____
x + 2 f (x) = x
Notice that:
f (1) = g(–1) = h(2) = k(–2) = 1
shift 2 left
shift 1 right
shift 3 left
Unit 2 The effects of the parameters a, p and q on hyperbolas
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Horizontal and vertical shifts of the hyperbola
1
y = _____
x–p+q
Changing p results in a horizontal shift, but changing q results in a vertical shift.
1
2
Symmetry lines using
k
y = _____
x–p+q
Shift
Asymptotes
4 +3
y = _____
2 right
3 up
x=2
y=3
Both symmetry lines will
pass through (2;3)
x–2
5 –2
y = – _____
x+1
1 left
2 down
x = –1
y = –2
Both symmetry lines will
pass through (–1;–2)
Intercepts with
the axes
shifts
substitution
y=x
becomes
y = (x – 2) + 3
=x+1
Substitute (2;3) into
y=x+c
3=2+c
c=1
y=x+1
If x = 0
y = –2 + 3 = 1
y = –x
becomes
y = –(x – 2) + 3
= –x + 5
Substitute (2;3) into
y = –x + c
3 = –2 + c
c=5
y = –x + 5
If y = 0
4 +3
0 = _____
y=x
becomes
y = (x + 1) – 2
=x–1
Substitute (–1;–2) into
y=x+c
–2 = –1 + c
c = –1
y=x–1
If x = 0
y = –2 + 3 = 1
y = –x
becomes
y = –(x + 1) – 2
= –x – 3
Substitute (–1;–2) into
y = –x + c
–2 = 1 + c
c = –3
y = –x – 3
If y = 0
5 –2
= – _____
x–2
= 4 + 3(x – 2)
= 4 + 3x – 6
x+1
= –5 – 2(x + 1)
= –5 – 2x – 2
2x = 7
x = –3,5
WORKED EXAMPLES
1
6 +3
If f (x) = _____
x+1
1.1
1.2
1.3
1.4
State the equations of the asymptotes of f.
Determine the equations of the axes of symmetry of f.
Determine the coordinates of the x-and y-intercepts of f.
Draw f(x) and indicate the intercepts, the asymptotes and the
symmetry lines.
2
If g(x) is the reflection of f(x) in the x-axis, state the equation of g(x).
Answer all the questions in 1. for g(x), replacing f with g.
3
If h(x) is the reflection of f(x) in the y-axis, state the equation of h(x).
Answer all the questions in 1. for h(x), replacing f with h.
SOLUTIONS
1
1.1
6 +3
f (x) = _____
x+1
Vertical asymptote when x + 1 = 0 ⇒ x = –1 and horizontal asymptote
when y = 3
94
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1.2
Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x + 1) + 3 = x + 4 ⇒ y = x + 4
y = – (x + 1) + 3 = –x + 2 ⇒ y = –x + 2
Using substitution of the point (–1;3):
y = x + c ⇒ 3 = –1 + c and c = 4 ⇒ y = x + 4
y = –x + c ⇒ 3 = –( – 1 ) + c and c = 2 ⇒ y = –x + 2
1.3
To find the x-intercept, substitute y = 0.
6 + 3 = 0 ⇒ 6 + 3(x + 1) = 0 | × (x + 1)
_____
x+1
6 + 3x + 3 = 0 ⇒ 3x = –9 and x = –3
⇒ x-intercept (–3;0)
To find the y-intercept, substitute x = 0.
6 +3⇒6+3=9
y = _____
0+1
⇒ y-intercept (0;9)
1.4
y
x = –1
12
11
10
9
8
7
6
6 +3
f (x) = _____
x+1
5
4
y=3
3
2
1
–8 –7 –6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
x
–2
–3
–4
y=x+4
2
y = –x + 2
A reflection in the x-axis leaves x as it is, but changes the sign of y.
(x + 1
)
6 + 3 = – _____
6 –3
g(x) = –f (x) = – _____
2.1
2.2
x+1
Vertical asymptote when x + 1 = 0 ⇒ x = –1 and horizontal asymptote
when y = –3.
Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x + 1) – 3 = x – 2 ⇒ y = x – 2
y = –(x + 1) – 3 = –x – 4 ⇒ y = –x – 4
Using substitution of the point (–1;–3):
y = x + c ⇒ – 3 = –1 + c and c = –2 ⇒ y = x – 2
y = –x + c ⇒ – 3 = –(–1 + c) and c = – 4 ⇒ y = –x – 4
Unit 2 The effects of the parameters a, p and q on hyperbolas
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2.3
To find the x-intercept, substitute y = 0:
6 – 3 = 0 ⇒ – 6 – 3(x + 1) = 0 | × (x + 1)
– _____
x+1
– 6 – 3x – 3 = 0 ⇒ 3x = –9 and x = –3
⇒ x-intercept (–3;0)
To find the y-intercept, substitute x = 0:
6 – 3 ⇒ – 6 – 3 = –9
y = – _____
0+1
⇒ y-intercept (0;–9)
y
2.4
y = –x – 4
x = –1
3
2
1
–8 –7 –6 –5 –4 –3 –2 –1
–1
1
2
–2
(–1;–3)
3
4
x
y = –3
–3
−4
−5
6 – 3
y = _____
x+1
−6
–7
–8
–9
–10
y=x–2
–11
–12
3
A reflection in the y-axis leaves y alone, but changes the sign of x.
6 + 3 = _____
–6 + 3
h(x) = f (–x) = ______
–x + 1
3.1
3.2
x–1
vertical asymptote: ⇒ x = 1
horizontal asymptote: y = 3
Axes of symmetry: y = x + c and y = –x + c
Using shifts:
y = (x – 1) + 3 = x + 2
y = –(–x – 1) + 3 = –x + 4
3.3
Using substitution of the point (1;3):
y = x + c ⇒ 3 = 1 + c and c = 2 ⇒ y = x + 2
y = –x + c ⇒ 3 = –1 + c and c + 4 ⇒ y = –x + 4
To find the x-intercept, substitute y = 0:
– 6 + 3 = 0 ⇒ – 6 + 3(x – 1) = 0
_____
x–1
| × (x – 1)
– 6 + 3x – 3 = 0 ⇒ 3x = 9 and x = 3
⇒ x-intercept (3;0)
To find the y-intercept, substitute x = 0:
–6 + 3 ⇒ 6 + 3 = 9
y = _____
0+1
⇒ y-intercept (0;9)
96
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y
3.4
12
y = –x + 4
x=1
11
y=x+2
10
9
8
7
6
5
4
y=3
(1;3)
3
2
1
–6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
7
8
x
–2
–3
–4
–6 + 3
y = ____
x–1
EXERCISE 7
Sketch each graph on a separate system of axes. In each case:
• State the equations of the asymptotes.
• Determine the equations of the axes of symmetry.
• Determine the coordinates of the x-intercept.
• Determine the coordinates of the y-intercept.
• Clearly indicate the intercepts with the axes, the asymptotes and the
symmetry lines.
1
1 –3
f (x) = _____
2
g(x) if g(x) is the reflection of f(x) in the x-axis
3
h(x) if h(x) is the reflection of f(x) in the y-axis
4
4 –1
r(x) = – _____
5
t(x) if t(x) is the reflection of r(x) in the x-axis
6
v(x) if v(x) is the reflection of r(x) in the y-axis
7
3 +2
w(x) = – _____
8
z(x) if z(x) is the reflection of w(x) in the x-axis
9
d(x) if d(x) is the reflection of w(x) in the y-axis
x+2
x–3
x–1
Unit 2 The effects of the parameters a, p and q on hyperbolas
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Finding the equation of a hyperbola
a
y = ____
x – p + q is the general equation of a hyperbola.
The asymptotes are x = p and y = q.
The symmetry lines are y = x + c and y = –x + c.
Assume that the domain is x ∊ ℝ, x ≠ p, unless you are told otherwise.
KEY WORDS
domain – the set of all the
x-values of the graph
• exclude any values of x
which make y undefined
• check to see whether the
graph is restricted or not
range – the set of all the
y-values of the graph
• exclude any values which
make y undefined
• check to see whether the
graph is restricted or not
If the domain is restricted, then state the restriction when you state the equation.
You may be given the equations of the asymptotes and a random point and asked to
find the equation of the hyperbola and the equations of the symmetry lines.
WORKED EXAMPLE 1
A hyperbola has two asymptotes, x = 2 and y = –1, and it passes through
the point (4;1).
Determine the equation of the hyperbola and state the equations of its
symmetry lines.
SOLUTION
a –1
Substitute the p and q values: y = _____
x–2
a – 1 ⇒ 2 = __
a and a = 4
Now substitute: (4;1): 1 = _____
4–2
2
4 –1
The equation is: y = _____
x–2
Both symmetry lines pass through (2;–1), the point of intersection of the
asymptotes:
Symmetry line 1: y = x + c ⇒ –1 = 2 + c and c = –3, so y = x – 3
Symmetry line 2: y = –x + c ⇒ –1 = –2 + c and c = 1, so y = –x + 1
WORKED EXAMPLE 2
y
The equations of the graphs alongside are:
a
f (x) = _____
x + p + q and g(x) = mx + c
1
2
3
98
Determine the values of a, p, q, m and c.
Briefly explain how you found each solution.
State the domain and the range of f.
State the equation of each graph.
1
x
–2
Topic 5 Functions: Effects of parameters
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SOLUTION
1
p = –1 because the vertical asymptote is x = 1.
q = 0 because the horizontal asymptote is the y-axis.
a ⇒ –2 = _____
a and so a = 2.
To find a substitute: (0;–2) into y = _____
x–1
0–1
2=2
m is the gradient of the straight line, so m = __
1
c = –2 because this is the y-intercept of the graph.
2
Domain of f: x < 1, x ∊ ℝ and range of f: y < 0, y ∊ ℝ
3
2 , x < 1 and g(x) = 2x – 2
f (x) = _____
x–1
EXERCISE 8
1
A hyperbola has two asymptotes, x = –1 and y = 4, and it passes through the
point (1;3). Determine the equation of the hyperbola and state the equations of
its symmetry lines.
2
A hyperbola has two asymptotes, x = 2 and y = 1, and it passes through the point
(5;2). Determine the equation of the hyperbola and state the equations of its
symmetry lines.
3
A hyperbola has two asymptotes, x = –3 and y = –2, and it passes through the
point (–2;–3). Determine the equation of the hyperbola and state the equations
of its symmetry lines.
4
A hyperbola has two asymptotes, x = 4 and y = –3, and it passes through the
point (–1;– 4). Determine the equation of the hyperbola and state the equations
of its symmetry lines.
5
k
2
f (x) = _____
x + p – q and g(x) = ax + bx + c. A is the turning point
of the parabola.
5.1
State the domain and range of f.
5.2
Determine the values of k, p and q.
5.3
State the equations of the symmetry lines of f.
5.4
State the coordinates of A, B and C.
5.5
Does g have a maximum or a minimum value?
Briefly explain your answer.
5.6
Determine the equation of g by making use of:
5.6.1
the turning point formula
5.6.2
x-intercept method.
5.7
For which value(s) of x is f (x).g(x) < 0?
5.8
State the value of f (0) – g(0).
5.9
g is translated in such a way that the turning point of
the translated graph is (0;6).
5.9.1
Describe the translation in words.
5.9.2
State the equation of the translated graph.
5.10 State the equation of the graph obtained by translating
f 1 unit to the left and 2 units down.
y
x= 1
6
f
y=2
A
B
x
C
g
Unit 2 The effects of the parameters a, p and q on hyperbolas
PLT MATHS LB 11 7th pgs (Real Book).indb 99
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EXERCISE 9
Complete the table and sketch the graphs, clearly indicating the asymptotes,
intercepts with the axes and symmetry lines. Plot all the points in the table on the
graphs as well.
k
y = _____
x–p+q
1
Shifts
Asymptotes
Intersection of
Asymptotes
Symmetry
Lines
Intercepts
with axes
2 –3
y = _____
Point on graph
x
3
x–1
x = –1
2
1
y=2
x=3
3
(5;0)
(2;1)
4
4 –1
y = – _____
6
1 +4
y = _____
3
–1
y = –2
5
y
3
(0;2)
–1
x+5
5
x–2
EXERCISE 10
1 x2 + __
1 x + 3 and g(x) = _____
2 +1
Given f (x) = – __
4
1
2
3
4
5
6
7
8
9
10
11
12
13
14
100
4
x+1
State the equations of the asymptotes and the symmetry lines of g.
Draw f and g on the same system of axes, indicating the x- and y-intercepts with
the axes, asymptotes and symmetry lines.
Determine the coordinates of the points of intersection of f and the symmetry
line of g which has a positive gradient.
State the domain of g and the range of f.
Solve algebraically for x and y if f (x) = g(x).
For which value(s) of x is f(x) ≥ g(x)?
State the equation of the graph obtained by translating g 2 units right and 1 unit
down. Write your answer in the form y = …
Write down the equation of g(2x) in the form y = …
Write g(–x) in the form y = … and describe this transformation in words.
If f is translated so that its turning point lies on the intersection of the symmetry
lines of g, state the equation of the translated graph in the form y = …
How must g be translated so that the x and y axes become the asymptotes of g?
1 ) and state the value(s) of x for which it is
Determine an expression for g( __
x
undefined.
Determine an expression for g(x – 4) and state the value(s) of x for which it is
undefined.
1 ) = 4g(–x)g(x – 4)
Show algebraically that: g(x) + g( __
x
Topic 5 Functions: Effects of parameters
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Unit 3: The effects of the parameters a, p
and q on exponential graphs
You will investigate the effect of the parameters a, p
and q on the graph of the function defined by
y = f (x) = abx + p + q where b > 0, b ≠ 1.
y
Exponential graphs
y = 2x
6
Although the exponential graph y = abx is different from
a,
the parabola y = ax2 and the hyperbola y = __
x
the effects of a are exactly the same.
•
•
y = 4 × 2x
7
y = __14 × 2x
5
4
3
The greater the value of a, the steeper the curve.
If a is negative the graph reflects in the x-axis.
2
1 __
1
–3
–2
–1
y=0
4
– _1_
–1 4
1
2
3
x
4
horizontal
asymptote
–2
–3
–4
–5
y = – __14 × 2x
–6
–7
Reminders about the exponential
1 x
graphs y = 2x and y = __
2
• These graphs share the same horizontal asymptote,
( )
•
•
•
y = 0.
Exponential graphs do not have any lines of symmetry.
Domain: x ∊ ℝ and Range: y > 0, y ∊ ℝ
y = 2x is an increasing function because as x
increases, y increases
•
1
y = __
(2)
x
y = –4 × 2x
y
6
y = 2–x
(2)
x
4
•
1
y = 2x is the reflection of y = __
•
•
For a reflection in y-axis, leave y alone and change sign of x.
For a reflection in x-axis, leave x alone and change sign of y.
(2)
1 x = 2–x
because __
y = 2x
5
is a decreasing function because as
x increases, y decreases
y = –2x
in the y-axis
Notice in the figure on the right that:
If y = 2x is reflected in the x-axis, the reflected graph is y = –2x.
If y = 2x is reflected in the y-axis, the reflected graph is y = 2–x.
3
(–1;2)
(–1;0,5)
–3
2
(1;2)
1
–2 –1
(–1;–0,5) –1
–2
(1;0,5)
1
2
3
x
y=0
(1;–2)
–3
–4
–5
–6
y = –2x
Unit 3 The effects of the parameters a, p and q on exponential graphs
PLT MATHS LB 11 7th pgs (Real Book).indb 101
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WORKED EXAMPLE
1
2
3
4
State the asymptote and y-intercept of y = 3x.
Consider the exponential graph:
f (x) = 3x + 1
2.1
State the asymptote of f.
2.2
State the y-intercept of f and the coordinates of two other points.
If g(x) is the reflection of f in the x-axis:
3.1
state the equation of g
3.2
state the equation of the asymptote of g
3.3
state the y-intercept g and the coordinates of two other points.
On the same set of axes, sketch f and g and show the asymptotes, the
y-intercept and two other points on each graph.
SOLUTION
1
2
y
The asymptote is y = 0 and the y-intercept
is (0;1).
2.1
The asymptote of is y = 1.
2.2
The y-intercept is (0;2) and two other
(
6
5
)
4
1 and (1;4).
points are −1;1__
3
4
3.1
3.2
3.3
3
g(x) = −f (x) = −(3x + 1) = −3x − 1
The asymptote is y = −1.
The y-intercept is (0;−2) and
1
two other points are −1;−1__
3
and (1;−4)
(
)
y = 3x + 1
(1;4)
3
( –1;1__13 )
–2
2
1
–1
See the graph alongside.
1
2
–1
( −1;−1__13 )
3
x
4
y = –1
–2
–3
–4
–5
(1;–4)
y = –(3x) – 1
–6
There is a table function on most calculators which will generate points for you.
Most scientific calculators will work according to the detailed instructions below:
Consider f (x) = –3x – 1 and follow the instructions below to generate a table of values:
• Select MODE.
• Press the number next to TABLE and your calculator will display f(X) =
• Press ‘–3’.
• Now Press ‘x□’, then ‘ALPHA’ followed by ‘X’.
• Use the replay button to take the cursor down to its normal position.
• Press ‘–1’.
• Your display will now show f (X) = –3x – 1
102
Topic 5 Functions: Effects of parameters
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•
•
•
•
•
•
•
•
•
•
Press ‘=’ and your display will change to ‘Start?’.
Press ‘–3’ = (or any other start value of your choice).
Your display will now change to End?.
Press ‘3=‘ (or any other end value of your choice).
Your display will now show ‘Step?’.
Press ‘1=‘ and a table of values will appear.
Choose the points you want to plot and ignore the rest.
If you want to edit your equation, or change the start, end or step values press ‘AC’.
Return to normal mode by selecting ‘MODE’, then press the number next to
‘COMP’.
Do not depend on the table for the asymptotes or the x-intercept.
EXERCISE 11
1
If f (x) = 5×2x, sketch f (x), –f (x) and f (–x) on the same system of axes.
2
If g(x) = 5x, sketch g(x), –g(x) and g(–x) on the same system of axes.
3
1
If h(x) = __
4
If t(x) = 3 × 2x – 1, sketch t(x), –t(x) and t(–x) on the same system of axes.
(3)
x
+ 2, sketch h(x), −h(x) and h(–x) on the same system of axes.
Horizontal shifts of the exponential graph y = 2x
y
k
(–4;1)
(–2;1)
g
f
(0;1)
h
(1;1)
x
x
–5
–4
–3
–2
–1
0
1
2
3
4
Shift
f
y = 2x
1
___
1
___
1
__
1
__
1
__
1
2
4
8
16
None
g
y = 2x + 2
1
__
1
__
1
__
1
2
4
8
16
32
64
2 left
h
y = 2x –
1
1
___
1
___
1
___
1
__
1
__
1
__
1
2
4
8
1 right
k
y = 2x +
4
1
__
1
2
4
8
16
32
64
128
256
32
16
4
8
64
32
2
8
2
16
4
8
2
4
2
4 left
Unit 3 The effects of the parameters a, p and q on exponential graphs
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REMEMBER
If an unknown value appears
more than once in the same
equation, it has the same
value each time it appears.
x0 = 1 or all x ∊ ℝ , x ≠ 0
If y = abx + p + q, q is not the
y-intercept.
Find the y-intercept by letting
x = 0.
The asymptote is given by
y=q
You may need to use
simultaneous equations to
determine the values of a,
b, p and q. Refer back to
Topic 2 if you need to revise
Horizontal and vertical shifts of the exponential
graph y = abx + p + q
WORKED EXAMPLE
Calculations required for sketching the shifted exponential graph:
y = abx + p + q
Shift
1
y = 5.2x – 2 + 3
Asymptote
y-intercept (x = 0)
x-intercept (y = 0)
2 right
3 up
y=3
y = 5.2–2 + 3 = 4,25
(0;4,25)
No x-intercept
y – 3, y ≠ 0
2
y = – 4.3x + 2 – 1
2 left
1 down
y = –1
y = – 4.32 – 1 = –37
(0; – 37)
No x- intercept
– 4.3 x + 2 ≠ 1
y < –1, y ≠ 0
3
y = 2.5x + 1 – 2
1 left
2 down
y = –2
y = 2.5 – 2 = 8
(0;8)
5x + 1 = 1 = 50
x = –1 ⇒ (–1;0)
4
1
y = 3 __
1 up
y=1
y = 3(1) + 1 = 4
(0;4)
No x-intercept
y > 1, y ≠ 0
5
1
y = –2 __
1 right
6 up
y=6
y = –2(3) + 6
= 0 ⇒ (0;0)
x=0
(0;0)
(2)
x
+1
( 3 )x – 1 + 6
Finding the equation of an exponential graph
in the form y = abx + p + q
All exponential graphs have a horizontal asymptote and if y = abx + p + q, then y = q is
the equation of the horizontal asymptote.
WORKED EXAMPLE
h(x) =
abx + p
y
+ q and passes through C(3;4) and D(2;8).
h
D(2;8)
The asymptote of h is given by y = 2.
If a = q, determine the values of b, p and q.
C(3;4)
y=2
x
SOLUTION
Substitute the points (3;4), (2;8) and q = 2.
q = 2 ⇒ y = 2bx + p + 2
(3;4) ⇒ 4 = 2b3 + p + 2 and 2 = 2b3 + p, so 3 + p = 0 and p = –3
1
(2;8) ⇒ 8 = 2b2 – 3 + 2 and 6 = 2b–1, so 3 = b- 1 and b = __
(3)
3
1 x–3 + 2
⇒ h(x) = 2 __
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Topic 5 Functions: Effects of parameters
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EXERCISE 12
y
y
f
(1;5)
y=2
g
y=2
2
1
x
–1
l
y = –1
x
(–1;–3)
x=1
h
–8
1
1.1
1.2
1.3
1.4
If f (x) = abx + c, determine the values of a, b and c
If g(x) = –2dx + p + e, determine the values of d, e and p
If h(x) = jrx + t, determine the values of j, r and t
v
If l(x) = _____
x – w + z, determine the values of v, w, and z.
EXERCISE 13
Complete the table by filling in the missing values. Then use the information in the
table to draw each graph on a separate set of axes. Clearly indicate the intercepts with
the axes, the asymptotes and the coordinates of at least two points.
y = abx + p + q
Shift
Asymptote
Intercepts
Point 1
Point 2
x
x
y
y
1
y = 3.2x + 1 – 6
–1
6
2
y = –5x – 2 + 1
–1
–4
3
1
y = 2. __
1
4
4
y = 2–x – 1
–2
–0,5
5
y = –5.3x – 1 + 5
2
4
4__
6
1
y = 2 __
( 5 )x – 1 – 2
2
( 3 )x + 1 – 2
9
48
Unit 3 The effects of the parameters a, p and q on exponential graphs
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7
d
x
f (x) = ax2 + bx + c, g(x) = ____
x + e + j and h(x) = vk + q
y
h
A(2;8)
f
x = –3
x
–1
–1
y = –2
g
B(–4;–10)
7.1
7.2
7.3
8
Determine the values of a, b and c if the turning point of f is A(2;8) and f
passes through the point B(–4;–10).
Determine the values of d, e and j if g passes through B(–4;–10) and has
asymptotes x = –3 and y = –2.
Determine v, k and q if h passes through (–1;0) and (0;–1) and has any
asymptote given by y = –2.
Sketched in the figure are the graphs f (x) = ax2 + bx + c;
k
g(x) = ____
x – p + q and h(x) = mx + n.
y
f
g
A
h
–2
x = –2
–1
Q
B
R
4
x
P(–1;–5)
T
A and B are the x- and y-intercepts of g respectively.
Graphs f, g and h intersect at P(–1;–5) and graphs g and h intersect at Q.
106
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 106
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8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
State the equation of the axis of symmetry of f.
For which value(s) of x is f (x) an increasing function?
Determine the equations of the symmetry lines of g.
For which value(s) of x is f (x) ≤ h(x)?
Determine the equations of graphs f, g and h.
Determine the coordinates of A and B.
Determine the coordinates of Q.
For which value(s) of x is h(x) ≤ g(x)?
State the values of x for which f(x). g(x) ≤ 0.
RT is parallel to the y-axis, with R on h and T on f.
8.10.1 Determine an expression for the length of RT.
8.10.2 Determine the maximum length of RT.
8.11 For which values of x is g an increasing function?
4 −1
8.12 Consider the function: g( x ) = − _____
x+2
8.12.1 Determine the equation of v if v is the reflection of g in the y-axis.
8.12.2 Determine the symmetry lines of v.
8.12.3 Sketch v on a new set of axes, clearly indicating the intercepts with
the axes and at least one other point.
8.12.4 Sketch and label the symmetry lines of v.
8.13 Given that w is obtained by translating g 2 units to the right and 1 unit up.
8.13.1 State the equation of w.
8.13.2 State the asymptotes of w.
8.13.3 State the equations of the symmetry lines of w.
8.13.4 Determine the shortest distance between w and the origin.
Unit 3 The effects of the parameters a, p and q on exponential graphs
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2012/07/02 2:21 PM
Unit 4: Real life applications
WORKED EXAMPLE
Vincent has money to invest and considers two options:
Option 1: Simple interest at a rate of 13,5% using the formula A = P(1 + in)
Option 2: Compound interest at a rate of 7,5% p.a. compounded annually using
the formula A = P(1 + i)n
1
Vincent wants know how long it will take for the two options to
become equal.
1.1
Set up an equation which would help him to work out how long it
would take.
1.2
Explain why is makes no difference how much money he invests at
the start.
2
Consider the formula: A = 1 + in | Ignore P as it does not affect the calculation.
2.1
Rewrite the formula to express it in terms of x and y.
2.2
Is the graph a straight line graph or an exponential graph?
3
Consider the formula: A = (1 + i)n | Ignore P as it does not affect the
calculation.
3.1
Determine 1 + i and give your answer in its simplest form
3.2
Rewrite the formula to express it in terms of x and y.
3.3
Is the graph a straight line graph or an exponential graph?
4
Sketch both graphs on the same system of axes, clearly labelling the graphs
and indicating the points used to plot the graph.
5
Use your graph to estimate the number of years it will take for the
investments to be equal. Give your answer to the nearest year.
SOLUTIONS
108
1
1.1
1.2
P( 1 + in ) = P( 1 + i )n ⇒ 1 + in = ( 1 + i )n
The initial amount cancels out when the equation is set up to compare
both options.
2
2.1
2.2
y = 1 + 0,135x
Straight line
3
3.1
3.2
3.3
1,075
y = 1,075x
Exponential graph
Topic 5 Functions: Effects of parameters
PLT MATHS LB 11 7th pgs (Real Book).indb 108
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4
Simple interest and compound interest
y
4
(18;3,68)
3
(20;3,7)
(12;2,38)
(9;1,92)
(6;1,54)
(3;1,24)
5
5
y = 1 + 0,135x
(15;2,96)
2
1
y = 1,075x
10
15
x
20
It will take 16 years to the nearest year.
EXERCISE 14
1
South African death rates since 1997
Number of deaths
1 200 000
1 000 000
y = 39 378x + 281 235
800 000
600 000
y = –1 709x2 + 56 468x + 249 902
400 000
200 000
0
0
5
10
15
20
25
30
35
Number of years since 1996
In the graph, death rates have been plotted and two possible models suggested.
1.1
If the linear model proves to be correct, briefly discuss the long-term
implications.
1.2
After how many years would the maximum number of deaths be reached
if the parabolic model is correct? Give your answer correct to two decimal
places.
1.3
According to the parabolic model, what will the maximum number of
deaths be? (Remember to give your answer as a whole number.)
1.4
State the year in which the maximum number of deaths occurs.
1.5
After which year does the number of deaths begin to decrease?
2
Jack is in Grade 11 and takes a gap year after school. He plans to start a savings
account with R10 at the end of the first month. For the next 11 months, he will
save double the amount of each previous month.
2.1
Calculate how much money he will add to his account at the end of the
fourth month.
2.2
Draw a graph which shows the first five monthly deposits.
Unit 4 Real life applications
PLT MATHS LB 11 7th pgs (Real Book).indb 109
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2012/07/02 2:21 PM
2.3
2.4
2.5
2.6
2.7
3
Is the graph continuous or discrete? Justify your answer.
Does the graph start at 0 months or 1 month? Explain your answer.
Do you think Jack will be able to maintain his monthly payments for
12 months? Justify your answer.
Determine the equation for your graph.
Check your equation by making use of your calculator.
This question is designed to show you that you can set up your own model from
raw data.
The data in the table was used to generate the graphs you used in Question 2.
Year
Number of
years since
1996
Number of deaths
per year in South
Africa
1997
1
316 559
1998
2
365 109
1999
3
381 037
2000
4
414 768
2001
5
500 082
2002
6
554 199
2003
7
572 620
2004
8
593 337
2005
9
605 408
3.1
3.2
3.3
3.4
110
Using your calculator or Excel, generate each of these models:
f (X) = A + BX; g(X) = A + BX + CX2; h(X) = A.BX
Determine the number of years since 1996 if the year is 2020.
Now determine the anticipated number of deaths in 2020 for each
different model.
Briefly discuss your findings and comment on the different outcomes.
Topic 5 Functions: Effects of parameters
PLTMATHSLB11LB_05.indd 110
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Unit 5: The average gradient between
two points on a curve
EXERCISE 15
REMEMBER
Consider the sketch of the parabola f (x) = –x2 + 2x + 8 and eight different
straight line graphs. Each straight line passes through point A and one other
point on the parabola.
y
G
F
D
I
–4
–3
–2
Function notation:
y = f( x )
If x = 3, then
y = f( 3 ) = − ( 3 )2 + 2( 3 ) + 8
= −9 + 6 + 8
=5
(3;5) lies on f.
E
H
–5
Gradient of straight line:
y2 − y1
m = _______
x2 − x1
C
–1
1
2
3
4
A
5
6
7
x
B
1
Determine the coordinates of the points A, B, C, D, E, F, G, H and I.
2
Consider AB and take note that xB – xA = 5 – (–3) = 8 units:
y
x
A(–3;–7)
B(5;–7)
The shaded area enclosed by the parabola and the line passing through A and B
is a large area, as shown in the figure.
Now consider AE and notice that xE – xA = 2 – (–3) = 5 units:
y
E(2;8)
x
A(–3;–7)
The shaded area enclosed by the parabola and the line passing through A and
E is smaller than the shaded area enclosed by the parabola and the line passing
through A and B.
Unit 5 The average gradient between two points on a curve
PLTMATHSLB11LB_05.indd 111
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Complete the statements by filling in the missing words:
The … the points are …, the greater is the enclosed area.
The … the points are …, the … is the enclosed area.
We use the gradient of the straight line between two points on a curve to
determine the average gradient of the curve between those points.
3
y –y
2
1
Use the formula m = ______
x – x and determine the average gradient of the parabola
2
1
f (x) = –x2 + 2x + 8 between the points:
3.1
3.3
3.5
3.7
4
3.2
3.4
3.6
3.8
A and B
A and D
A and F
A and H
A and C
A and E
A and G
A and I
Which average gradient that you determined in Question 3 best approximates
the gradient of the parabola at point A? Briefly explain your answer.
EXERCISE 16
1
4 –1
Consider the hyperbola: g(x) = _____
x–3
1.1
Determine:
1.1.1
g(–1)
1.1.2
g(0)
1.1.3
g(1)
1.1.4
g(2)
1.2
Determine the average gradient of g between:
1.2.1
x = –1 and x = 0
1.2.2
x = –1 and x = 1
1.2.3
x = –1 and x = 2
4 – 1 and indicate your solutions to Questions 1.1 and
1.3
Sketch g(x) = _____
x–3
1.2 on your sketch.
1.4
2
112
Which average gradient in 1.2 best represents the gradient of g at x = −1?
Consider the exponential graph: h(x) = 2 – x
2.1
Determine:
2.1.1
h(–2)
2.1.2
h(–1)
2.1.3
h(0)
2.1.4
h(1)
2.2
Determine the average gradient of h between:
2.2.1
x = –2 and x = –1
2.2.2
x = –2 and x = 0
2.2.3
x = –2 and x = 1
2.3
Sketch h(x) = 2– x and indicate your solutions to Questions 2.1, and 2.2
on the sketch.
2.4
Which average gradient in 2.2 best represents the gradient of h at x = −2?
Topic 5 Functions: Effects of parameters
PLTMATHSLB11LB_05.indd 112
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3
Consider the parabola: k(x) = x2 – 4x – 12
3.1
Determine:
3.1.1
k(–1)
3.1.2
k(0)
3.1.3
k(2)
3.1.4
k(4)
3.2
Determine the average gradient of h between:
3.2.1
x = –1 and x = 0
3.2.2
x = –1 and x = 2
3.2.3
x = –1 and x = 4
3.3
Sketch k(x) = x2 – 4x – 12 and indicate your solutions to Questions 3.1
and 3.2 on the sketch.
3.4
Which average gradient in 3.2 best represents the gradient of k at x = −1?
4
Determine the average gradient of the curve:
2 + 5 between x = 1 and x = 2
4.1
p(x) = – _____
x–3
4.2
q(x) = 2 × 3x – 1 between x = –1 and x = 1
4.3
r(x) = –2x2 – 5x + 7 between x = –2 and x = 1
Unit 5 The average gradient between two points on a curve
PLTMATHSLB11LB_05.indd 113
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Revision
1
f
y
h
g
F(–2;3)
O
E
C
D
A
x
B
1 – 1; h(x) = __
1 (x – 2)2 – 2; and f (x) = ax + q.
Sketched in the figure are g( x ) = – _____
x–1
2
A and B are the points of intersection of g and h. D and E are the points of
intersection between the vertical asymptote of g and graphs h and f respectively.
O and C are the x-intercepts of h. E and F are points on f.
1.1
Determine the values of a and q.
(4)
1.2
State the coordinates of the turning point of h.
(2)
1.3
State the equation of the axis of symmetry of h.
(1)
1.4
Determine the length of OC.
(2)
1.5
If P(–4;2) lies on h, state the coordinates of Q
if Q is the reflection of P in the axis of symmetry of h.
(2)
1.6
State the equations of the asymptotes of g.
(2)
1.7
Determine the length of ED.
(3)
1.8
Determine the coordinates of B.
(7)
1.9
An axis of symmetry of the graph of g is a straight line defined as
y = mx + c where m > 0. Write down the equation of this line in the
form y = …
(2)
1 (x – 2)2 – 2
1.10 Consider the graph: h( x ) = __
2
1.10.1 Describe in words the translation which must take place
so that the turning point of graph h is shifted to the origin.
1.10.2 State the equation of the translated graph in the form y = …
2
2 – 3; g(x) = – __
1 (x + 6)(x – 4)
Consider the graphs defined by: f (x) = – _____
4
x+1
and h(x) = –2x – 5
2.1
State the equations of the asymptotes of f.
2.2
Determine the equations of the symmetry lines of f and g.
2.3
Draw f, g and h on the same system of axes. Clearly label all
intercepts with the axes as well as the coordinates of any turning
points. Draw and clearly label all symmetry lines and asymptotes.
2.4
Solve for x if f (x) = h(x).
(2)
(2)
[29]
(2)
(6)
(10)
(6)
2.5
1 (x + 6)(x – 4) > 0?
For which value(s) of x is – __
(2)
2.6
2.7
For which values of x is (x).h(x) < 0?
If f (x) is translated 2 units to the right and 1 unit up, state
the equation of the translated graph in the form y = …
(4)
4
(2)
114
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2012/07/02 2:21 PM
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
3.
If h(x) is reflected in the y-axis, state the equation of the reflected
graph in the form y = …
If g(x) is translated 3 units left and 2 units down, determine the
equations of the translated graph. State your answer in the
form y = ax2 + bx + c.
Determine g(–2) – f (–2).
Determine the average gradient of f(x) between:
2.11.1 x = –7 and x = –2
2.11.2 x = –5 and x = –2
2.11.3 x = –3 and x = –2
Consider the gradients you found in Question 2.11 and briefly
explain which gradient best represents the gradient of f at x = –2.
Solve for x, correct to two decimal places, if g(x) = h(x).
Determine the average gradient of g(x) between:
2.14.1 x = – 4 and x = 0
2.14.2 x = – 4 and x = –3
Which gradient from your solutions to Question 2.14 best
represents the gradient of g(x) at x = – 4? Briefly explain your answer.
(2)
(4)
(3)
(3)
(3)
(3)
(2)
(4)
(3)
(3)
(2)
[64]
Nyati makes up his own graph application question for a maths project.
He decides to compare the height of a stone that is thrown up into the air,
against the time taken by the stone to reach that height.
After several trials he records these results, with t being the time in seconds
and h the height of the stone in metres.
t
0
1
2
3
4
5
h
15
20
21
18
11
0
3.1
3.2
3.3
3.4
3.5
How long did it take for the stone to hit the ground?
Draw a graph to represent the information in the table.
Determine an equation which models the relationship and
state your answer in the form h = …
After how many seconds did the stone reach its maximum height?
What was the maximum height reached?
(1)
(4)
(5)
(2)
(1)
[13]
115
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TOPIC
2
6
Functions: Trigonometric graphs
KEY WORDS
period – the number of
degrees needed to complete
a wave pattern or cycle
amplitude – half the total
distance between the
minimum and maximum
values
• always positive
maximum – minimum
• __________________
2
Unit 1: Revision of trigonometric graphs
using point-by-point plotting
The three basic trigonometric graphs are:
• y = sin x, y = cos x and y = tan x
Trigonometric graphs covered in Grade 10 include:
• y = asin x + p, y = acos x + p and y = atan x + p
You achieve point by point plotting by substituting x-values into an equation to find
the y-values.
Use the table function on your calculator to generate the table of points which you
need to plot. There are detailed calculator instructions in Unit 5 of this topic.
WORKED EXAMPLE
1
Use a calculator to complete the table for y = sin x.
x
When you cover Topic 7,
you will learn how to work
out the values in tables
using special triangles.
y
x
y
x
y
x
y
x
y
x
–360°
–330°
–300°
–270°
–240°
–210°
–180°
–150°
–120°
–90°
–60°
–30°
0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
y
360°
The range is
y ∊ (minimum;maximum)
2
3
4
5
6
Give the range of y.
Sketch the graph y = sin x by plotting the points from the table.
Give the period of the graph.
Give the amplitude of the graph.
Describe how the graph y = sin x must be shifted to become y = cos x.
2
SOLUTIONS
1
2
x
y
x
y
x
–360°
0
–330°
1
__
–300°
__
√3
___
–150°
1
– __
2
–120° –
__
√3
___
–180°
0°
116
__
√3
___
≈ 0,87
1 = 0,5
__
0
0
180°
0
360°
0
30°
210°
2
1
__
2
1
– __
2
y
2
__
√3
___
60°
240°
2
2
–
__
√3
___
2
x
y
x
–270°
1
–240°
–90°
90°
270°
–1
1
–1
–60°
120°
300°
y
__
√3
___
2
–
__
√3
___
2
__
√3
___
2
–
__
√3
___
2
x
y
–210°
1
__
–30°
1
– __
150°
1
__
330°
1
– __
2
2
2
2
Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 116
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2
Range: y ∊ [–1;1]
y
3
y = sin x
(–270°;1)
–360°
–270°
–180°
(90°;1)
1
–90°
90°
180°
–1
(–90°;–1)
270°
360°
x
(270°;–1)
4
Period = 360°
5
Amplitude = 1
6
y = sin x must shift 90° to the left to become y = cos x.
EXERCISE 1
1
2
1.1
REMEMBER
1.2
1.3
1.4
1.5
1.6
Copy the table provided for the Worked example and complete it for
y = cos x.
Give the range of y.
Sketch the graph y = cos x by plotting the points from the table.
Give the period of the graph.
Give the amplitude of the graph.
Describe the shift required for the graph y = cos x to become y = sin x
2.1
Copy the table below and complete it for y = tan x.
x
y
x
y
x
y
x
–360°
–315°
–270°
–225°
–180°
–135°
–90°
–45°
0°
45°
90°
135°
180°
225°
270°
315°
An asymptote is a line which
cannot be touched or crossed
by the graph to which it is an
asymptote.
y
360°
2.2
2.3
2.4
2.5
2.6
For which value(s) of x is tan x undefined?
y = tan x is undefined at x = 90°, which is one of the vertical asymptotes
of the tan graph. State the equations of three other vertical asymptotes of
y = tan x for x ∊ [–360°;360°].
Sketch y = tan x for x ∊ [–360°;360°], indicating the intercepts with the axes
and the asymptotes.
State the range and the amplitude of y = tan x.
Give the period of y = tan x.
REMEMBER
Division by zero is undefined.
Unit 1 Revision of trigonometric graphs using point-by-point plotting
PLTMATHSLB11LB_06.indd 117
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Important facts about the three basic trigonometric graphs
x ∊ [360°;360°] y = sin x
y = cos x
y = tan x
Amplitude
1
1
undefined
Period (P)
360°
360°
180°
Asymptotes
none
none
x = ±90°
x = ±270°
Range
y ∊ [–1;1]
y ∊ [–1;1]
y ∊ (–∞;∞)
90°
45°
Interval spacing 90°
y
y = tan x
(–270°;1)
(45°;1)
1
(360°;1)
(90°;1)
y = sin x
y = cos x
–360° –315° –270° –225° –180° –135° –90°
–45°
(–180°;–1) (–90°;–1)
45°
90°
135° 180° 225° 270° 315° 360°
–1
x
(180°;–1) (270°;–1)
–2
x = –270°
x = –90°
x = 90°
x = 270°
y = sin x and y = cos x each have a period of 360° because one wave pattern
takes 360°.
• Both graphs have turning points and intercepts which are spaced at 90°
intervals.
• Intervals help us to set up the scale on the x-axis.
• If you generate a table of values using your calculator, select a step of 90°.
1.
Consider the graph given by y = –cos x + __
2
y
(–180°;1,5)
value
1,5 maximum
is 1,5 units
1
amplitude = 1
(–270°;0,5)
(180°;1,5)
(–90°;0,5)
0,5
(90°;0,5)
(270°;0,5)
amplitude = 1
–360°
–300°
–240°
–180°
–120°
–60°
The period is 360° because it takes 360°
to complete one wave pattern
(–360°;–0,5)
•
118
60°
120°
minimum value
is –0,5 units
–0,5
180°
240°
300°
360°
x
(360°;–0,5)
1 unit up
the amplitude is 1 and the vertical shift is __
2
Topic 6 Functions: Trigonometric graphs
PLTMATHSLB11LB_06.indd 118
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•
•
•
•
the period is 360° and the interval spacing is 90°
the vertical shift creates a need for smaller steps of 60° or 30°
the range is y ∊ [– 0,5;1,5]
the maximum value is y = 1,5 and the minimum value is y = – 0,5.
It may not always be possible to determine the x-intercepts without complicated
calculations. It is acceptable to give the coordinates of the shifted x-intercepts in
these cases, rather than the x-intercepts.
y = tan x has a period of 180° because one wave pattern takes 180°.
• The equations of the asymptotes are x = –270°; x = –90°; x = 90° and x = 270°.
• Each asymptote is positioned in the middle of the period.
• (45°;1) is halfway between the (0°;0) and the asymptote x = 90°.
• Important values occur every 45°, so the interval spacing for y = tan x is 45°.
• If you generate a table of values using your calculator, select a step of 45°.
Consider the influence of a and q on the graphs:
y = asin x + q
y = acos x + q
y = atan x + q
KEY WORD
interval spacing – the
number of degrees between
critical points
a influences the vertical stretch of these graphs whereas q influences the vertical shift.
Amplitude
y = sinx
1
None
y = −3sinx + 1
3
Up 1
1 sinx − 2
y = __
1
__
Down 2
y = − 5cosx + 2
5
Up 2
y = 2cosx − 1
2
Down 1
y = 3tanx − 1
Undefined
Down 1
2
2
REMEMBER
Vertical shift
Graph
y = tanx has a period of 180° because one wave pattern takes 180°.
• The equations of the asymptotes are x = −270°; x = − 90°; x = 90° and x = 270°
• Each asymptote is positioned in the middle of the period.
• (45°;1) is exactly halfway between the (0°;0) and the asymptote x = 90°.
• Important values occur every 45°, so the interval spacing for y = tanx is 45°
• If you generate a table of values using your calculator, select a step of 45°.
Amplitude = half the total
distance between the
minimum and maximum
values.
Amplitude is always positive,
even if a is negative.
tan graphs have neither a
minimum nor maximum
value, and therefore
amplitude is undefined.
tan 45° = 1, so when x = 45°,
y = atan x + q will result in the
coordinates (45°;a + q).
WORKED EXAMPLE
Given f (x) = −tanx and g(x) = tanx + 1 or x ∊ [−90°;270°].
1
State the equation(s) of the asymptotes of f and g.
2
Draw f and g on the same system of axes.
3
State the range of g.
4
State the period of f.
5
State the amplitude of g.
6
Show that f (−26,57°) = g(−26,57).
7
State one other value of x for which f (x) = g(x).
8
State two values of x for which g(x) − f (x) = 1.
9
State two values of x for which g(x) − f (x) = 3.
10 For which values of x is f (x).g(x) > 0?
Unit 1 Revision of trigonometric graphs using point-by-point plotting
PLTMATHSLB11LB_06.indd 119
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SOLUTIONS
1
2
Asymptotes of f and g: x = −90°, x = 90° and x = 270°
y
f(x) = – tan x
g(x) = tan x + 1
3
2
(– 45° ;1)
– 90°
– 45°
(45° ;2)
(225° ;2)
1
–1
–2
(135° ,1)
45°
90°
135°
(45° ;–1)
(180° ;1)
180°
225°
x
270°
(225° ;–1
–3
–4
x = – 90°
3
4
5
6
7
8
9
10
x = 90°
x = 270°
Range of g: y ∊ (−∞;∞)
Period of f is 180°.
Amplitude of g is undefined.
f(−26,57°) = −tan(−26,57°) = 0,50 and g(−26,57°) = tan(−26,57°) + 1 = 0,50
−26,57° + 180° = 153,43°
x = 0° and x = 180°
x = 45° and x = 225°
x ∊ (−45°;0°) ∪ (135°;180°)
WORKED EXAMPLE
Given f (x) = sin x and g(x) = 2sin x for x ∊ [–360°; 360°]
1
Draw f and g on the same system of axes.
2
State the amplitudes of f and g.
3
State the ranges of f and g.
4
State the periods of f and g.
5
What is the maximum value of g?
6
For which values of x is f (x) = g(x)?
7
For which values of x is f (x) ≥ g(x)?
8
For which value(s) of x is f (x) – g(x) = 1?
9
For which values of x is f (x) an increasing function?
120
Topic 6 Functions: Trigonometric graphs
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SOLUTIONS
y
1
(–270°;2)
2
(90°;2)
(90°;1)
(–270°;1)
g(x) = 2sin x
–360°
–180°
180°
(–90°;–1)
1 unit
(–90°;–2)
2
3
4
5
6
7
8
9
f (x) = sin x
–2
(270°;–1)
360°
x
1 unit
(270°;–2)
The amplitude of f is 1 unit and the amplitude of g is 2 units.
The range of f is y ∊ [–1;1] and the range of g is y ∊ [–2;2].
Both f and g have a period of 360°.
Maximum value of g is 2 units.
x = 0° or ± 180° or ±360°
Note: f (x) = g(x) where the graphs intersect, as indicated by green dots on
the x-axis.
x ∊ [–180°;0] ∪ [180°;360°] | The thick orange lines indicate where sin x ≥ 2sin x
and the thick blue lines on the x-axis indicate where the solution is read.
x = –90° or x = 270° | There are four x-values for which the graphs differ by
1 unit, but only two of the values occur when sin x ≥ 2sin x and are indicated by
thick green vertical lines.
x ∊ [–360°; – 270°] ∪ [ – 90°;90°] ∪ [270°;360°] | The black dashes on the
orange graph indicate where sin x is an increasing function.
EXERCISE 2
For each question, draw the graphs on the same system of axes. State the period,
amplitude and range for each graph. If there are asymptotes, sketch and label them.
1
h(x) = 2cos x and g(x) = cos x for x ∊ [–180°;360°]
1.1
For which value(s) of x is g(x) = 0?
1.2
For which values of x is h(x) > 0?
1.3
For which values of x is h(x) an increasing function?
2
f (x) = tan x and g(x) = 2tan x for x ∊ [–180°;180°]
2.1
For which value(s) of x is f (x) = g(x)?
2.2
For which value(s) of x is f (x) – g(x) = 1?
3
3 cos x for x ∊ [–90°;270°]
f (x) = –2sin x + 1 and g(x) = __
2
3.1
For which values of x is f (x) – g(x) = 3?
3.2
State one value of x for which g(x) – f (x) = 1.
3.3
For which values of x is g(x) ≥ 0?
3.4
If f is reflected in the x-axis, state the new equation in the form y = …
REMEMBER
In an increasing function,
the y-values increase as x
increases.
In a decreasing function,
the y-value decreases as x
increases.
To reflect a graph in the
x-axis, leave x and change
the sign of y.
To reflect a graph in the
y-axis, leave y and change
the sign of x.
Unit 1 Revision of trigonometric graphs using point-by-point plotting
PLTMATHSLB11LB_06.indd 121
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4
1 tan x for x ∊ [– 45°;135°]
Given the graphs f (x) = –tan x and g(x) = __
2
4.1
For which value(s) of x is g undefined?
4.2
Determine g(45°) – f(45°).
4.3
If f is reflected in the x-axis, state the new equation in the form y = …
4.4
If g is reflected in the y-axis, state the new equation in the form y = …
EXERCISE 3
Complete the table for the graphs y = asin x + q, y = acos x + q and y = atan x + q.
Draw each graph on a separate set of axes for the stated domain. Indicate all x-and
y-intercepts, the coordinates of all turning points and endpoints.
Equation
122
a
Amplitude
Range
q
Shift Domain
1
y = 2sin x + 1
x ∊ [−180°;180°]
2
3 cos x – 1
y = __
x ∊ [−180°;180°]
3
y = tan x + 2
x ∊ [−90°;90°]
4
1 sin x – 1
y = __
x ∊ [−90°;270°]
5
1 cos x + __
1
y = – __
x ∊ [−90°;270°]
6
y = –2tan x
x ∊ [−135°;135°]
7
y = –sin x + 2
x ∊ [–270°; 90°]
8
y = –3cos x – 1
x ∊ [−180°;180°]
2
2
3
2
End
points
Topic 6 Functions: Trigonometric graphs
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Unit 2: The effect of the parameter k on
some trigonometric functions
We will consider the effect of the parameter k on the functions defined by y = sin kx,
y = cos kx and y = tan kx.
WORKED EXAMPLE
1
REMEMBER
Complete the table. (Use the table function on your calculator if you
need help.)
–180°
x
–135°
–90°
–45°
0°
45°
90°
135°
180°
f (x) = sin 2x
g(x) = cos 2x
The amplitude is half the
total distance between the
minimum and maximum
values. It is:
• always positive
• the result of
maximum – minimum .
__________________
2
2
Sketch f and g on the same system of axes for x ∊ [180°;180°].
3
State the amplitudes of f and g.
4
State the periods of f and g.
5
State the ranges of f and g.
6
For which value(s) of x is f (x) – g(x) = 1?
7
If f (x) = sin 2x is moved up 1 unit, state the equation of the shifted graph
in the form y = …
In trigonometric graphs,
a period is the number of
degrees needed to complete
one cycle of a pattern.
The range is represented by
y ∊ [minimum;maximum].
An interval spacing is the
number of degrees between
critical points.
SOLUTIONS
1
–180°
–135°
–90°
–45°
0°
45°
90°
135°
180°
f (x) = sin 2x
0
1
0
–1
0
1
0
–1
0
g(x) = cos 2x
1
0
–1
0
1
0
–1
0
1
x
y
2
(–180°;1)
(–135°;1)
1
(45°;1)
g(x) = cos (2x)
–180°
–135°
–90°
(–90°;–1)
f(x) = sin (2x)
–45°
(–45°;–1)
45°
–1
90°
(90°;–1)
3
The amplitude is 1 for both f and g.
4
Both f and g have a period of 180°.
5
Range for both f and g is given by y ∊ [–1;1].
6
f (x) – g(x) = 1 when x = –135°, –90°, 45° or 90°.
7
y = f (x) + 1 = sin 2x + 1
Unit 2
PLTMATHSLB11LB_06.indd 123
(180°;1)
135°
180°
x
(135°;–1)
The effect of the parameter k on some trigonometric functions
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EXERCISE 4
1
1.1
Complete the table correct to one decimal place where necessary.
–360°
x
–270° –180°
–90°
0°
90°
180°
270°
360°
1x
f (x) = sin __
2
1x
g(x) = cos __
2
1.2
1.3
1.4
1.5
1.6
2
Sketch f and g on the same system of axes for x ∊ [–360°;360°].
State the amplitude, the period and the range for each graph.
State two values of x for which g(x) − f (x) = 1.
State three values of x for which f (x) − g(x) = 1.
State two values of x for which f (x) = g(x).
1 x for x ∊ [–180°;180°].
Consider the graphs of f (x) = tan2x and g(x) = tan __
2
2.1
Write down the period of f (x) = tan 2x and state the step size.
2.2
1 x and state the step size.
Write down the period of g(x) = tan __
2.3
State the equations of the asymptotes of f and g for x∊ [–180°;180°].
2.4
Sketch f and g on the same system of axes for x ∊ [–180°;180°].
2.5
State one value of x for which g(x) – f (x) = 1.
2.6
2.7
2.8
1 x = 1.
State one value of x for which tan 2x – tan __
2
State the range of f.
State the amplitude of g.
2
EXERCISE 5
REMEMBER
Period
360°
y = asin kx has P = ____
k
360°
y = acos kx has P = ____
Complete the table and then sketch each graph on a new set of axes.
• In the table, k is the coefficient of x in each graph equation.
• Show all x- and y-intercepts and the coordinates of all turning points and
endpoints.
If
• a graph has asymptotes, sketch and label them.
k
k
180°
y = atan kx has P = ____
k
Steps (interval spacing)
P
S = __
4
Asymptotes (tan graph only)
are in the middle of the
period.
If the period is 180°,
x = 90° there will be a vertical
asymptote.
124
Period
Step
Domain
1
y = cos 3x
x ∊ [–30°;180°]
2
1x
y = tan __
x ∊ [–360°;360°]
3
1x
y = sin __
x ∊ [–360°;360°]
4
y = cos2x
x ∊ [–270°;90°]
5
y = tan3x
x ∊ [–60°;90°]
6
y = sin 2x
x ∊ [–90°;210°]
7
1x
y = cos __
x ∊ [–360°;360°]
8
1x
y = tan __
x ∊ [–360°;360°]
4
3
2
3
End Points
Topic 6 Functions: Trigonometric graphs
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Unit 3: Horizontal shifts
Horizontal shifts of y = sin x
x
–90°
f (x) = sin x
–1
–
g(x) = sin (x + 90°)
0
h(x) = sin (x – 30°)
__
√3
___
k(x) = sin (x + 60°)
•
•
–
–60° –30°
2
1
– __
__
√3
___
1
– __
1
__
__
√3
___
2
2
2
√
__
3
– ___
–1
2
1
__
0
2
2
2
0°
0
1
1
– __
2
__
√3
___
2
30°
1
__
2
__
√3
___
2
0
1
60°
90°
__
√3
___
2
√
None
__
3
– ___
√3
___
1
__
√3
___
30° right
1
__
0
1
– __
60° left
0
2
2
1
__
2
1
– __
1
__
1
__
__
√3
___
1
2
120° 150° Shift
2
__
2
__
√3
___
2
2
90° left
2
2
2
Adding to x increases its value so the pattern of answers shifts to the left by the
amount added.
Subtracting from x reduces its value so the pattern shifts to the right by the amount
subtracted.
The graphs in the table are sketched in the figure. They all have the same period and
amplitude, but some have shifted horizontally.
y
(0°;1)
f (90°;1)
(30°;1)
(120°;1)
0,87
h
(150°;0,87)
k
(150°;0,5)
x
G
– 90 °
– 60°
– 30°
(– 90°;– 0,5)
(– 90°;– 0,87)
0°
30°
60°
90°
– 0,5
g
(– 90°;–1) (– 60°;–1)
120°
150°
(150°;– 0,5)
(150°;– 0,87)
It is easy to generate points for plotting shifted graphs point by point. Make sure
that you know the period of the graph and the step, or the interval spacing between
critical points.
For all trigonometric graphs remember to show:
• x- and y-intercepts
• coordinates of turning points and coordinates of endpoints.
The domain determines the x-coordinates of the endpoints. Substitute these
coordinates into the graph equation to determine the y-coordinates. If you are
(
__
√
3
sketching y = cos x for x ∊ [−30°;240°], the endpoint on the left will be −30°;___
because y = cos (−30°) =
__
√3
___
2
1
__
because y = cos 240° = − .
2
(
2
)
)
1 ,
≈ 0,87, and the endpoint on the right will be 240°;−__
2
Unit 3 Horizontal shifts
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EXERCISE 6
1
1.1
REMEMBER
Use your graphs to answer these questions:
1.2
State one value of x for which h(x) – f (x) = 1.
1.3
For which values of x do f, g and h all increase as x increases?
1.4
For which values of x is g(x).h(x) ≥ 0?
The amplitude is half the
total distance between the
minimum and maximum
values. It is:
• always positive
• the result of
maximum – minimum .
__________________
2
A period is the number of
degrees needed for a wave
pattern.
The range is represented by
y ∊ [minimum;maximum].
An interval spacing is the
number of degrees between
critical points.
2
f (x)
g(x)
1.5
For which values of x is ____ ≤ 0?
1.6
If t( x ) = –sin x is the graph obtained by shifting g(x), how must g(x) be
shifted?
2.1
Draw f (x) = tan x, g(x) = tan (x + 45°) and h(x) = tan (x – 15°) on the same
set of axes for x ∊ [–135°;90°].
Use your graphs to answer the questions below:
2.2
State one value of x for which g(x) – f (x) = 1.
2.3
For which values of x ∊ [–135°;90°] do f, g and h all increase as x increases.
2.4
For which values of x is g(x).h(x) ≥ 0?
3
For which values of x is ____ ≤ 0?
3.1
Sketch f (x) = tan (x – 90°), g(x) = –2cos (x + 45°) and h(x) = –sin (x – 45°) on
the same system of axes for x ∊ [–135°;225°].
State the amplitude of g.
State the period of f.
State the range of h.
For which values of x is f (x) = 0?
State one value of x for which f (x) = h(x).
State the maximum value of g(x) – h(x) and give the corresponding x-value.
For which values of x is f (x).g(x) ≤ 0?
Evaluate: h(15°) – g(15°).
State three values of x for which h(x) – f (x) = 1.
If h is reflected in the x-axis, state the equation of the reflected graph in
the form y = …
If g is shifted 45° to the left, state the equation of the shifted graph in the
form y = …
For which values of x is h(x) a decreasing function?
For which values of x is f (x) ≥ 0?
If h is shifted 45° to the right, state the equation of the shifted graph in the
form y = …
3.12
3.13
3.14
3.15
4
f (x)
g(x)
2.5
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
4.1
4.2
4.3
4.4
4.5
4.6
4.7
126
Draw f (x) = cos x, g(x) = cos (x + 60°) and h(x) = cos (x – 90°) on the same
set of axes for x ∊ [–90°;150°].
1 and g(x) = 2cos x on the same set of axes for
Sketch f (x) = sin(x – 30°) – __
2
x ∊ [–120°;120°].
What is the period of f ?
What is the amplitude of g?
What is the range of f?
State one value of x for which f (x) = g(x).
For which value of x is g(x) – f (x) = 3?
Determine g(–60°) – f (–60°).
Topic 6 Functions: Trigonometric graphs
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4.8
4.9
4.10
4.11
4.12
4.13
4.14
5
For which values of x do both f (x) and g(x) increase as x increases?
For which values of x is f (x). g(x) > 0?
If g is reflected in the x-axis, give the new equation in the form y = …
If f is shifted 1 unit up, give the new equation in the form y = …
If f is shifted 30° to the left, state the new equation in the form y = …
1 is obtained by shifting f (x), describe the shift.
If h( x ) = –cos x – __
2
If t( x ) = –2sin x is obtained by shifting g(x), describe the shift.
Consider the functions f (x) = cos (x – 60°) and g(x) = tan (x – 15°) for
x ∊ [–210°;150°]
5.1
Sketch the graphs of f and g for x ∊ [–210°;150°] on the same system
of axes.
5.2
State the amplitude of f.
5.3
What is the period of g?
5.4
For which values of x is f (x) an increasing function?
5.5
For which value(s) of x is g( x ) undefined?
5.6
For which value(s) of x is f (x).g(x) ≥ 0?
5.7
State one value of x for which f( x ) = g( x ).
5.8
For which values of x ∊ [–120°;150°], is f( x ) ≥ g(x)?
5.9
For which value of x is g(x) – f (x) = 2?
5.10 State two values of x for which f (x) – g(x) = 1.
5.11 If h( x ) = –tan x is obtained by shifting and reflecting g( x ),
5.11.1 fully describe the shift and reflection
5.11.2 state the equations of the asymptotes of h(x) or x ∊ [–210°; 150°].
5.12 Describe the shift if t(x) = –sin x is obtained by shifting f (x):
5.12.1 to the right
5.12.2 to the left.
REMEMBER
In an increasing function,
the y-values increase as x
increases.
In a decreasing function,
the y-value decreases as x
increases.
To reflect a graph in the
x-axis, leave x and change the
sign of y.
To reflect a graph in the
y-axis, leave y and change the
sign of x.
A summary of trigonometric graphs
y = asin kx
y = asin (x + p)
y = asin x + q
y = acos kx
y = acos (x + p)
y = acos x + q
y = atan kx
y = atan (x + p)
y = atan x + q
All sine and cosine graphs have exactly the same properties as each other:
• Amplitude = a if a > 0, but amplitude = –a if a < 0
•
360°
Period = ____
k
The tan graph has unique properties:
180°
• Amplitude is undefined for all tan graphs and period = ____
•
k
Asymptotes are always positioned midway through each wave pattern.
Properties common to the trigonometric graphs:
•
•
P , where P = period
Step or interval spacing = __
•
•
Vertical shift will be up if q > 0, but down if q < 0.
Horizontal shift will be left if you add to x, but right if you subtract from x.
4
If there is a vertical shift, you may need to adjust the interval spacing, so consider
steps of 15°, 30°, 45° or 60°
Unit 3 Horizontal shifts
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Graph
Amplitude
Period
Steps
Asymptote with smaller
position (x) value
Vertical shift
Horizontal shift
y = 2sin 3x
2
120°
30°
None
None
None
y = –3cos x + 2
3
360°
90°
None
2 up
None
y = tan 2x – 1
Undefined
90°
22,5°
x = 45°
1 down
None
y = sin (x – 30°)
1
360°
90°
None
None
30° right
y = 2cos (x + 60°)
2
360°
90°
None
None
60° left
Undefined
360°
90°
180°
None
None
y = –2sin (x + 45°)
2
360°
90°
None
None
45° left
y = cos (x – 90°) – 2
1
360°
90°
None
2 down
90° right
Undefined
180°
45°
x = 135°
None
45° right
3 cos (x + 15°)
y = __
2
3
__
360°
90°
None
None
15° left
y = sin (x – 75°) + 1
1
360°
90°
None
1 up
75° right
1x
y = 2tan __
2
y = tan (x – 45°)
2
EXERCISE 7
There are eleven graphs in the table above and five of them are sketched here.
y
f
g
2
h
1
–105 ° –90° –75° –60° –45° –30° –15°
–1
0
15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165°
x
k
–2
t
128
–3
Topic 6 Functions: Trigonometric graphs
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Complete the table.
Equation
Endpoints (correct to
two decimal places)
Turning points
Range
f
g
h
k
t
Unit 3 Horizontal shifts
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Unit 4: Determine the equations
of trigonometric graphs
EXERCISE 8
y
1
REMEMBER
h
Period
360°
y = asin kx has P = ____
(300°;1) (360°;1)
1
f
g
0,5
k
–360°
360°
y = acos kx has P = ____
–180°–150°
30°
k
k
maximum – minimum .
__________________
2
In trigonometric graphs,
a period is the number of
degrees needed to complete
one cycle of a pattern.
The range is represented by
y ∊ [minimum;maximum].
An interval spacing is the
number of degrees between
critical points.
–1
(–360°;–1)
Steps (interval spacing)
P
S = __
4
Asymptotes (tan graph only)
are in the middle of the
period.
If the period is 180°,
x = 90° there will be a vertical
asymptote.
The amplitude is half the
total distance between the
minimum and maximum
values. It is:
• always positive
• the result of
360°
x
–0,5
180°
y = atan kx has P = ____
REMEMBER
180°
(120°;–1)
1.1
1.2
1.3
1.4
1.5
If f (x) = asin (x + b), write down the values of a and b.
If f (x) = dcos (x + c), write down the values of c and d.
What is the period of g?
If g(x) = psin qx, write down the values of p and q.
If h(x) = vcos wx, write down the values of v and w.
1.6
1 (x + t), write down the values of r and t.
If h(x) = rsin __
1.7
1.8
State two value(s) of x for which f (x) = h(x)
State one value of x for which f (x) − h(x) = 1.
2
y
2
h
2
g
f
1
–360°
360°
x
–1
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
State the period of f.
State the period of g.
State the amplitude of f.
State the amplitude of g.
If g(x) = sin ax + b, determine the values of a and b.
If f (x) = dcos (x + p), determine the values of d and p.
If f (x) = msin nx, determine the values of m and n.
If h(x) = vtan wx, write down the values of v and w.
2.9
1 (x + c) + t, determine the values of c and t.
If g(x) = cos __
2
2.10 For which value of x is g(x) – f (x) = 2?
130
Topic 6 Functions: Trigonometric graphs
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y
3
g
3
2
f
h
1
–270°
–195°
0
75°
135°
255° 270°
x
–1
–2
–3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
State the period of f.
State the amplitude of f.
State the period of g.
State the amplitude of g.
What is the period of h?
If f (x) = asin (x + b), determine the values of a and b.
If f (x) = dcos (x + c), determine the values of c and d.
If g(x) = vtan wx, determine the values of v and w.
If h(x) = pcos qx, determine the values of p and q.
For which values of x will f (x).g(x) ≤ 0?
For which value of x is f(x) = g(x)?
State two values of x for which g(x) − h(x) = 1.
Period
360°
y = asin kx has P = ____
EXERCISE 9
1
REMEMBER
If f (x) = cos x for x ∊ [–360°;360°], is moved up 1 unit and 20° to the right:
1.1
state the equation of the shifted graph in the form y = …
1.2
give the range of the shifted graph.
2
If g(x) = sin x for x ∊ [–360°;360°], is moved 1 unit down and 90° to the left:
2.1
state the equation of the shifted graph in the form y = …
2.2
state the minimum value of the shifted graph.
3
h(x) = atan bx has a period of 360° and passes through the point (90°;3)
determine the values of a and b.
4
If f (x) = tan x is shifted 45° to the right and 2 units down, state the equation of
the shifted graph in the form y = …
5
f (x) = –sin ax + b and g(x) = ctan dx. Both f and g have a period of 720° and both
pass through the point (–180°;2). The range of f is y ∊ [0;2].
5.1
Determine the values of a, b, c and d.
5.2
Sketch f and g on the same set of axes for x ∊ [–360°;360°].
k
360°
y = acos kx has P = ____
k
180°
y = atan kx has P = ____
k
Steps (interval spacing)
P
S = __
4
Asymptotes (tan graph only)
are in the middle of the
period.
If the period is 180°,
x = 90° there will be a vertical
asymptote.
Unit 4 Determine the equations of trigonometric graphs
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Unit 5: Sketch graphs which have a change
in period and a horizontal shift
In the equations y = asin k(x + p), y = acos k(x + p) and y = atan k (x + p):
• p indicates the horizontal shift
• pk indicates a change in period.
If the equations are simplified as shown below, then kp does not indicate the
horizontal shift y = asin (kx + kp), y = acos (kx + kp) and y = atan (kx + kp)
Factorisation solves the problem:
y = cos (2x – 60°) ⇒ y = cos 2(x – 30°) and the horizontal shift is 30°.
(2
)
1 x – 25° ⇒ y = sin __
1 (x – 50°) and the horizontal shift is 50°.
y = sin __
REMEMBER
Horizontal shifts
• to shift right, subtract
from x
• to shift left, add to x
2
WORKED EXAMPLE
(See calculator instructions on the next page.)
1
Consider the graphs f (x) = cos 3(x – 15°) and g(x) = sin (2x – 90°)
1.1
Complete the table.
Graph
Period calculations
• both sine and cos graphs
360°
P = ____
k
180°
• tan graphs P = ____
Amplitude
Period
Steps
Horizontal shift
f (x) = cos 3(x – 15°)
g(x) = sin 2(x + 45°)
k
Draw the graphs on the same system of axes for x ∊ [– 45°;135°].
1.2
SOLUTIONS
(There are detailed calculator instructions on the next page.)
1.1
Graph
Amplitude
Period
Steps
Horizontal shift
f (x) = cos 3(x – 15°)
1
360° = 120°
____
120° = 30°
____
15° right
g(x) = sin 2(x + 45°)
2
360° = 180°
____
180° = 45°
____
45° left
1.2
3
2
4
y
g(x) = sin 2(x + 45°)
(15°;1)
1
(135°;1)
f(x) y = cos 3(x – 15°)
0,7
–45°
–30°
(–45°;–1)
132
4
–15°
15°
–1
30°
45°
60°
75°
90°
105°
120°
135°
x
(75°;–1)(90°;–1)
Topic 6 Functions: Trigonometric graphs
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General calculator instructions for most scientific calculators used in schools
• Select ‘MODE’ and then press the number which appears next to the word ‘TABLE’.
The display shows ‘F(X) = ‘
– Remember that F(X) = y
– Your equation must be in the form f (x) = … or g(x) = … or h(x) = … or y = …
the
variable is entered using ‘ALPHA’, followed by ‘X’
•
– The variable may be θ, α, β, x, or any other letter, except y
• Enter the equation by selecting the appropriate keys.
• Remember that sin, cos and tan automatically appear with an open bracket; close
the brackets at the end by pressing the key ‘)’.
• Once you have entered your equation, press ‘=’, and once only. If this does not
work for your calculator, press ‘=‘ a second time, ignoring ‘g(X) =‘. Then follow the
prompts to enter the start and end values, followed by the step size.
• To edit your equation or change the start, end or step values press ‘AC’, then ‘=’.
REMEMBER
Period
360°
y = asin kx has P = ____
k
360°
y = acos kx has P = ____
k
180°
y = atan kx has P = ____
k
Steps (interval spacing)
P
S = __
4
When you have finished with the table, return to normal mode by selecting ‘MODE’,
followed by the number which appears next to ‘COMP’.
Generate a table of values for f (x) = cos [3(x – 15°)] for x ∊ [– 45°;135°]
• Without adjusting the step size, the table below will be generated.
Calculator display
•
•
•
•
360° = 120°
P = ____
3
120 = 30°
S = ____
4
Points to plot
X
F (X)
(x;y)
1
–45
–1
(–45°;–1)
2
–15
0
(–15°;0)
3
15
1
(15°;1)
4
45
0
(45°;0)
5
75
–1
(75°;–1)
6
105
0
(105°;0)
7
135
1
(135°;1)
It is important to note we have not found the y-intercept.
Press ‘AC’, followed by ‘=’ and then repeat ‘=’ until you get back to Step.
Change the step size to 15 by pressing ‘15’ followed by ‘=’
An expanded version of the table will come up which now includes the row:
4
0
0,7071
(0°;0,71)
Unit 5 Sketch graphs which have a change in period and a horizontal shift
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Now generate a table for g(x) = sin (2x – 90°) for x ∊ [– 45°;135°]
Calculator display
X
F (X)
360° = 180°
P = ____
2
180 = 45°
S = ____
Points to plot
4
(x;y)
1
–45
0
(–45°;0)
2
0
–1
(0°;–1)
3
45
0
(45°;0)
4
90
1
(90°;1)
5
135
0
(135°;0)
This table contains all the critical points which you need.
EXERCISE 10
Complete the table and sketch each pair of graphs on the same set of axes.
Graph
1.1
Amplitude
1 (x + 45°)
f (x) = tan __
2
g(x) = sin 3(x + 15°)
3.1
f (x) = tan (2x + 60°)
g(x) = 2cos (x + 30°)
4.1
Asymptotes
Shift
Domain and endpoints
1.2
2.2
3.2
)
)
x ∊ [–45°;270°]
(–45°; )(270°;
(–45°; )(270°;
)
)
x ∊ [–120°;60°]
(–120°; )(60°;
(–120°; )(60°;
)
)
x ∊ [–210°;150°]
(–210°; )(150°;
(–210°; )(150°;
1 sin (x + 30°)
f (x) = __
2
g(x) = cos (2x – 120°)
134
Steps
x ∊ [210°;105°]
(–210°; )(105°;
(–210°; )(105°;
f (x) = cos 2(x + 30°)
g(x) = sin 2(x – 60°)
2.1
Period
)
)
Refer to your solution to 1.1 and answer the questions that follow.
1.2.1
State four value(s) of x for which f (x) – g(x) = 1.
1.2.2
For which values of x is f (x).g(x) < 0?
Refer to your solution to 2.1 and answer the questions:
2.2.1
State the value(s) of x for which f (x) – g(x) = 1.
2.2.2
For which values of x is f (x).g(x) > 0?
2.2.3
For which value(s) of x is f (x) undefined?
Refer to your solution to 3.1 and answer the questions:
3.2.1
State the value(s) of x for which f (x) = g(x)
3.2.2
State the value(s) of x for which g(x) – f (x) = 2?
Topic 6 Functions: Trigonometric graphs
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4.2
Refer to your solution to 4.1 and answer the questions:
4.2.1
State the value(s) of x for which g(x) – f (x) = 1,5.
4.2.2
State the value(s) of x for which g(x) – f (x) = 0,5.
4.2.3
State the value(s) of x for which f (x) – g(x) = 1.
4.2.4
Determine f (0°) – g(0°).
4.2.5
For which value(s) of x is f (x) > 0?
EXERCISE 11
1
1 cos (x + 30°) and g(x) = sin (2x – 120°).
Consider the graphs defined by f (x) = –__
2
1.1
Sketch f (x) and g( x ) on the same system of axes for x ∊ [–210°;150°].
1.2
Use your graphs to answer the questions:
1.2.1
State two values of x for which f (x) = g(x).
1.2.2
State two values of x for which f (x) – g(x) = 0,5.
1.2.3
For which values of x is f (x) ≤ 0?
2
1 x + 45° and g(x) = 2sin (x – 30°).
Consider the graphs defined by f (x) = tan __
2
2.1
Sketch f (x) and g(x) on the same system of axes for x ∊ [–180°;180°].
2.2
Use your graphs to answer the questions below:
2.2.1
State two values of x for which g(x) – f (x) = 2.
2.2.2
For which values of x ∊ [–180°;0°] is f (x).g(x) < 0?
Consider the functions f (x) = asin (x + p), g(x) = tan k(x + d), h(x) = c and j(x) = e
x ∊ [–240°;270°].
3
(
)
y
j
g
h
(210°;2)
T
f
Z
x
V
R
U
REMEMBER
Period
360°
y = asin kx has P = ____
k
360°
y = acos kx has P = ____
k
180°
y = atan kx has P = ____
k
Steps (interval spacing)
P
S = __
4
Asymptotes (tan graph only)
are in the middle of the
period.
If the period is 180°,
x = 90° there will be a vertical
asymptote.
W
(30°;–2)
3.1
3.2
3.3
3.4
3.5
3.6
3.7
Write down the coordinates of R, T, U, V, W and Z.
Write down the values of a, p, k, d, c and e.
State the length of TU.
If f (x) is shifted 30° to the left, state the equation of the shifted graph.
How must f (x) be shifted if the equation of the shifted graph is y = –asin x?
If g(x) is shifted so that the new graph equation is y = tan kx, describe
the shift.
For which value(s) of x is g(x) – f (x) = 3?
Unit 5 Sketch graphs which have a change in period and a horizontal shift
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Revision
1
2
Write down the period, amplitude and range and the coordinates of the
y-intercept of each graph.
1.1
f (x) = 2sin (x + 30°)
1.2
g(x) = −cos 2x
1.3
h(x) = tan 2x
1.4
k(x) = sin (2x – 20°) – 1
Period
360°
y = asin kx has P = ____
k
360°
y = acos kx has P = ____
k
180°
y = atan kx has P = ____
1.5
1x
p(x) = 3tan __
1.6
1 cos (x – 20°)
q(x) = __
2.1
If h(x) = cos x is translated 90° to the right, state the equation of the
translated graph. Simplify your answer.
If h(x) = cos x is translated 90° to the left, will you get the same result
as 2.1? Explain your answer.
2.2
REMEMBER
k
2
[24]
2
(3)
(3)
[6}
y
3
E
−180°
F
A
Steps (interval spacing)
P
S = __
4
Asymptotes (tan graph only)
are in the middle of the
period.
If the period is 180°,
x = 90° there will be a vertical
asymptote.
G
H
D
180°
x
B
C
1 (x + 30°) – 1, x ∊ [–180°;150°]
The graphs f (x) = 2cos (x + 30°) and g(x) = tan __
2
are sketched.
3.1
Write down the coordinates of A, B, C, D, E, F, G and H.
(16)
3.2
State the equation of the asymptote of g.
(1)
3.3
For which values of x is
3.3.1
f (x) ≥ g(x) ?
(2)
3.3.2
f (x).g(x) ≥ 0?
(2)
3.4
State the period and describe the shift of g.
(3)
3.5
If f (x) is translated 60° to the left, state the equation of the
translated graph. Simplify your answer.
(3)
3.6
If g(x) is reflected in the y-axis, state the equation of the
reflected graph.
(2)
3.7
For which values of x is f (x) ≤ g(x) ?
(3)
3.8
For which values of x is f (x).g(x) ≥ 0?
(3)
3.9
If f (x) is reflected in the x-axis, state the equation of the
reflected graph.
(2)
3.10 If g(x) is reflected in the x-axis, state the equation of the
reflected graph and simplify your answer.
(3)
[40]
136
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4
Consider the functions defined by:
(2
)
3 x – 90° for x ∊ [–180°;180°]
f (x) = –2cos (x + 30°) and g(x) = sin __
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
State the period, amplitude, range and endpoints of f.
(6)
State the period, amplitude, range and endpoints of g.
(6)
Sketch f and g on the same system of axes. Indicate the intercepts
with the axes and the coordinates of the turning points and endpoints. (6)
Determine g(–120°) – f (–120°).
(1)
State one value of x for which f (x) = g(x).
(1)
__
For which value(s) of x is f (x) – g(x) = √ 3 ?
(2)
For which values of x is g(x) ≤ 0?
(2)
Write g(x) as a cos graph in the form y = ...
(3)
[27]
5
y
(45°;3)
(–315°;3)
C
f
(360°;1)
(–360°;1)
–270°
B
(270°;0)
(285°;0)
A
g
x
(225°;–1)
–2 D
Consider the graphs in the figure and use them to answer the questions.
5.1
5.2
State the amplitudes and periods of both f and g.
If f (x) = asin (x + b) + c and g(x) = dcos ex, write down the values of
a, b, c, d and e.
5.3
Write down the coordinates of A and B.
5.4
Determine the length of CD.
5.5
If f (x) = pcos (x + q) + v, write down the values of p, q and v.
5.6
If g(x) = wsin k(x + z), write down the values of w, k and z.
5.7
For which values of x is f (x) an increasing function?
5.8
For which values of x is f (x).g(x) ≤ 0?
5.9
State the range of −f (x).
5.10 State the range of g(–x).
5.11 For which values of x is f (x) < 0?
(4)
(5)
(4)
(3)
(6)
(6)
(6)
(6)
(2)
(1)
(4)
[47]
137
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TOPIC
2
7
Trigonometry
Unit 1: Revision of Grade 10 Trigonometry
REMEMBER
hypotenuse
θ
We work in the Cartesian plane for angles bigger than 90°.
The positive angle θ always rotates anti-clockwise from the positive x-axis about the
origin.
opposite
If angle θ rotates clockwise from the positive x-axis about the origin, it is negative.
______
adjacent
OP = r = √x2 + y2 (radius of the circle, r > 0)
We define the three
trigonometric ratios for the
angle θ in a right-angled
triangle as:
We now define the trigonometric functions as:
y
90°
h
y
Quadrant 2
If 90° < θ < 180°, point P is in
quadrant 2 and x < 0, y > 0.
o , cos θ = __
a , and
sin θ = __
h
o
tan θ = __
a
y
sin θ = _r ; cos θ = __xr ; tan θ = __x
Quadrant 1
P(x;y)
y
–x = −,
So, sin θ = _r = + , cos θ = __
r
r
y
y
180°
and tan θ = __
–x = –.
x O
θ
x 0° or 360°
If 180° < θ < 270°, point P is in
quadrant 3 and x < 0, y < 0.
–y
–x
__
So, sin θ = __
r–y = –, cos θ = r = –
and tan θ = __
–x = +
Quadrant 3
Quadrant 4
270°
If 270° < θ < 360° point P is in quadrant 4 and x > 0, y < 0.
–y
–y
x
__
__
So, sin θ = __
r = –, cos θ = r = + and tan θ = x = –
You can see the ratios change signs
as θ moves from 0° to 360°:
sin θ is positive in Quadrants 1 and
2.
cos θ is positive in Quadrants 1 and
4.
tan θ is positive in Quadrants 1 and
3.
90°
Quadrant 2
y
θ
180°
y
y
–y
1
r
–x
x
r
3
138
r
r
1
Sin
4
P(x;y)
Quadrant 4
270°
2
1
S
All
x
A
x
Tan
–y
3
y
y
2
y
x 0° or 360°
r
Use these diagrams to help you
remember the signs of the trigonometric ratios:
2
x
O
Quadrant 3
y
Quadrant 1
x
T
Cos
4
3
C
CAST shows
you in which
quadrant the
ratios are
POSITIVE
4
Topic 7 Trigonometry
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EXERCISE 1
1
In each case say which quadrant the given angle is in.
1.1
223°
1.2
153°
1.3
292°
1.4
–130°
1.5
– 400°
1.6
–51°
1.7
–182°
1.8
225°
1.9
520°
1.10 730°
2
Work out the sign of each ratio. (First use the CAST diagram, then check
your answer with a calculator.)
2.1
cos 115°
2.2
tan 150°
2.3
cos (– 40°)
2.4
sin (–30°)
2.5
tan (–170°)
2.6
sin 200°
2.7
cos 330°
2.8
tan 190°
2.9
sin 300°
2.10 cos (–60°)
3
In each case say which quadrant θ lies in if:
3.1
sin θ > 0 and cos θ < 0
3.2
sin θ < 0 and cos θ > 0
3.3
sin θ > 0 and tan θ < 0
3.4
cos θ > 0 and tan θ > 0
3.5
cos θ > 0 and tan θ < 0
3.6
sin θ < 0 and tan θ > 0
y
WORKED EXAMPLE
Find the length OP and then write down
the values of sin θ, cos θ, tan θ.
SOLUTION
θ
O
x
P(–5;–12)
___________
OP = √(–5)2 + (–12)2 = 13 | Pythagoras’ Theorem
y
–5 , tan θ = ____
–12 , cos θ = ___
–12 = ___
12
sin θ = ____
–5
5
13
13
–5
θ
O
x
–12
13
P
Unit 1 Revision of Grade 10 Trigonometry
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EXERCISE 2
1
In each figure, find the length OP (in surd form if necessary). Then write down
the values of sin θ, cos θ, tan θ.
y
1.1
P(7;24)
P(–3;4)
θ
O
y
1.2
x
O
y
1.3
θ
x
y
1.4
x
O
θ
θ
O
x
P(–5;–10)
2
P(5;–3)
In each figure, find the length of r, x or y (in surd form if necessary). Then write
down the three trigonometric ratios for each given angle (for example: sin 600°,
tan 135°, cos 210° or sin 300°). Give your answers with rational denominators.
y
y
2.1
2.2
__
P(1;√ 3 )
135°
60°
x
O
O
2
y
2.4
300°
210°
–1
x
O
y
2.3
140
P(–2;2)
x
x
O
2
__
–√ 3
Topic 7 Trigonometry
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WORKED EXAMPLE
1
8 and sin θ < 0, find the value of:
If cos θ = ___
17
1.1
2
tan θ
1.2
sin θ
_____
1.3
sin2 θ + cos2 θ
cos θ
If 7sin θ + 3 = 0 and 90° < θ < 270°, find the values of:
2.1
cos θ
2.2
1 + tan2 θ
SOLUTIONS
You are required to draw a diagram. Make sure that it is in the correct quadrant,
and that the lengths you mark have a – sign where appropriate.
1
8 and sin θ < 0
cos θ = ___
y
17
θ lies in Quadrant 4
_______
θ
y = –√172 – 82 = –15
1.1
1.2
1.3
2
(
8
7
________
) ( )2
_____
–15
17
P
y
θ lies in Quadrant 3
–3
sin θ = ___
x
O
–15
tan θ = ____
8
sin θ = ____
–15 ÷ ___
8 = ____
–15
_____
cos θ
17
17
8
–15 2 + ___
8
sin2 θ + cos2 θ = ____
17
17
225 + 64 = 1
= ________
289
___
___
–√ 40
x = –√72 – (–3)2 = –√49 – 9 = –√40
θ
___
x
O
√
2.1
– 40
cos θ = _____
2.2
–3___
1 + tan2 θ = 1 + _____
7
( –√40 )
2
9 = ___
49
= 1 + ___
40
–3
40
7
EXERCISE 3
REMEMBER
1
If 5 sin θ + 4 = 0 and cos θ > 0 , find the value of tan θ.cos θ.
2
10 – _____
5 .
12 and 90° < x < 270°, find the value of _____
If tan θ = ___
cos θ
5
cos θ
3
sin θ + cos θ .
If 7tan θ = 3 and cos θ > 0, find the value of ___________
4
If 5cos x + 3 = 0 and 180° < x < 360°, find the value of 3tan x + 25 sin2 x.
5
7 and 180° < x < 360°, find the value of 25 cos x – _____
7 .
If sin x = ___
tan x
6
3 , tan y = __
5 , x < 90° and y < 90°, prove that 74sin2 y + ______
7 = 34.
If cos x = __
2
7
If sin θ = k and 90° < θ < 180°, determine the value of tan θ in terms of k.
2sin θ
25
4
7
tan x
Steps to solve ‘from one ratio
to another’ type questions:
1 Isolate the trigonometric
ratio.
2 Decide in which quadrant
the angle lies.
3 Draw the triangle in the
correct quadrant.
4 Use Pythagoras’ Theorem
to find the third side.
5 Answer the question.
Unit 1 Revision of Grade 10 Trigonometry
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KEY WORD
trigonometric identity – an
equality which is true for
all values of an unknown
variable, for which both sides
of the identity are defined (so
no zero denominators)
It differs from a trigonometric
equation which is true only
for certain values of the
unknown variable.
Unit 2: Identities
sin θ and sin2 θ + cos2 θ = 1 are two standard trigonometric identities that you
tan θ = _____
cos θ
need to prove and learn to use.
An identity consists of two sides: Left Hand Side (LHS) and Right Hand Side (RHS).
To prove an identity you must prove LHS = RHS.
To prove the standard identities, use the basic trigonometric definitions in the
Cartesian plane. You can prove other identities using the two standard identities
to simplify each side.
sin x
Proof of tan x = _____
cos x
Proof of sin2 θ + cos2 θ = 1
y
P(x;y)
r
O
θ
y
x
LHS = tan θ = __
sin θ
RHS = _____
cos θ
y
_
x
x
LHS = sin2 θ + cos2 θ
= ( _r ) + ( __xr )2
y
y2
r
2
2
x
= __2 + __
2
r
= __
x
__
r
y
= __ × __r
r
y
x
y
x
= __
r
y2 + x2
r
| x2 + y2 = r2 Pythagoras
= ______
2
2
r = 1 = RHS
= __
2
r
LHS = RHS
In the Worked examples and exercises that follow, you do not use the trigonometric
definitions to prove identities or simplify expressions.
To prove identities follow these steps:
•
•
•
•
•
•
•
142
cos θ so that all the ratios are in terms of sin and cos.
sin θ or _____
1 = _____
Make tan θ = _____
tan x sin θ
cos θ
Consider the LHS or RHS as an algebraic expression and use algebraic
manipulations to simpifly further.
If there are fractions, find the LCD and add.
If there are fractions over fractions, simplify as you would in algebra.
Factorise where possible.
Use the square identity where possible in any of the forms:
sin2 θ + cos2 θ = 1 or sin2 θ = 1 – cos2 θ or cos2 θ = 1 – sin2 θ
Simplify both sides of the identity as far as possible.
Topic 7 Trigonometry
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WORKED EXAMPLE
Simplify the expression using the identities proved before:
(1 + tan2 x)(1 – sin2 x)
SOLUTION
sin x (1 – sin
( 1 + _____
cos x )
cos x + sin x (cos
= ( ____________
)
cos x
2
2
2
2
2
sin x
| Identity: tan x = _____
cos x
2
x)
2
x) | Identity: 1 – sin2 x = cos2 x
1 × cos2 x
= _____
2
cos x
| Identity: sin2 x + cos2 x = 1
=1
EXERCISE 4
Simplify the expressions as far as possible.
1
sin x
_________
3
2
tan2 x(1 – sin2 x)
(1 – cos x)
__________
4
8sin2 θ + 8cos2 θ
5
1 – tan x.cos x.sin x
6
(3 – 3sin θ)(3 + 3sin θ)
7
cos x
1 – _____
_____
8
1 + tan x ( sin x.cos x )
( _____
)
tan x
9
sin x.cos x
_______________
10
1
tan2 x – _____
2
cos x.tan x
2
sin x
sin x
tan x
1 + cos2x – sin2 x
cos x
WORKED EXAMPLE 1
Prove the identity.
sin x4x – cos4 x = 1 – 2cos2 x
SOLUTION
LHS = (sin2 x – cos2 x)(sin2 x + cos2 x)
= (sin2 x – cos2 x)(1)
= (1 – cos2 x) – cos2 x
= 1 – 2cos2 x
= RHS
Unit 2 Identities
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WORKED EXAMPLE 2
Prove the identity.
tan x + sin x = sin x
___________
1
1 + _____
cos x
SOLUTION
sin x
_____
cos x + sin x
cos x
LHS = __________
× _____
cos x
1
1 + _____
cos x
sin x + sin x.cos x
= _______________
cos x + 1
sin x(1 + cos x)
= _____________
(1 + cos x)
= sin x
= RHS
EXERCISE 5
Prove the identities.
144
1
2
3
4
sin x.cos x.tan x = 1 – cos2 x
cos3 x + cos x.sin2 x = cos x
(sin x + cos x)2 + (sin x – cos x)2 = 2
(1 – sin2 x)(1 + tan2 x) = 1
5
cos x
1 – sin x = _____
_____
6
cos x – cos3 x
________________
= tan2 x
2
7
sin θ – √ 1 – sin2 θ = 0
_____
8
1
tan2 θ + 1 = _____
2
9
sin x + cos x = cos x
___________
10
1 = _________
1
tan θ + _____
11
tan2 x – sin2 x = sin2 x.tan2 x
12
cos x + tan x = _____
1
________
cos x
1 + sin x
13
tan θ – sin θ.cos θ = tan θ
_______________
2
14
sin x = 1
tan2 x.cos2 x + ______
2
15
1
1
2
________
+ ________
= _____
2
tan x
sin x
cos x – cos x.sin x
________
tan θ
cos θ
1 + tan x
tan θ
sin θ.cos θ
sin θ
2
tan x
1 – sin x
1 + sin x
cos x
Topic 7 Trigonometry
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More complex identities
WORKED EXAMPLE
1 + cos x = ________
sin x
Prove the identity: ________
sin x
1 – cos x
You are not able to simplify the LHS or RHS. Therefore multiply one side by
1 + cos x (=1) or ________
1 – cos x (=1) to create a workable expression, without changing
________
1 + cos x
1 – cos x
its value.
SOLUTION
1 + cos x × ________
1 – cos x | Notice you are multiplying by 1.
LHS = ________
sin x
1 – cos x
cos2
| Notice the difference of 2 squares.
1–
x
= _____________
sin x(1 – cos x)
2
| Notice the identity 1 – cos2 x = sin2 x.
sin x
= _____________
sin x(1 – cos x)
sin x = RHS
= ________
1 – cos x
EXERCISE 6
Prove the identities.
1
cos x = ________
1 + sin x
________
cos x
2
cos2 x – cos x – sin2 x = _____
1 – _____
1
__________________
3
2sin x.cos x
1
sin x – ___________
+ cos x = ___________
4
sin2
x + sin x –
x = ________
sin x + 1
__________________
cos x
5
1 – cos x – ________
sin x = 0
________
6
sin x + 1
1
2 = ________
________
– _____
2
7
tan θ
1 = _____
tan θ + _____
2
8
sin x
_________
=
(1 + cos x)
9
cos2θ + cos2 θ = 2
sin2 θ + tan2θ _____
2
10
(2sin x – cos x)(2sin x + cos x) = 5sin2 x – 1
1 – sin x
tan x
2sin x.cos x + sin x
sin x + cos x
sin x
sin x + cos x
cos2
2sin x.cos x – cos x
sin x
1 + cos x
sin x + 1
cos x
tan θ
sin x – 1
sin θ
_______
√
1 – cos x
________
1 + cos x
( sin θ )
Unit 2 Identities
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Unit 3: Reduction formulae
In Unit 1 you learnt that as the positive angle θ rotates anti-clockwise from the
positive x-axis about the origin, the signs of x and y change and therefore the
trigonometric ratios change sign. This is also true for negative angles rotating
clockwise from the positive x-axis about the origin.
If θ is an acute angle that lies between the rotating arm and the positive or negative
x-axis, we name the angles in the different quadrants as follows:
Quadrant 1
Quadrant 2
Quadrant 3
Quadrant 4
θ
180° – θ
180° + θ
360° – θ
y
y
KEY WORD
y
reduction formulae –
trigonometric identities that
express the trigonometric
ratios of an angle of any size
in terms of the trigonometric
ratios of an acute angle
Example:
sin (180° + θ) = –sin θ and
cos (360° – θ) = cos θ
r
–x
180° – θ
θ
r
θ
x
y
180° + θ
y
–x
x
–y
r
θ
360° – θ
r
θ
x
y
r
θ
θ
x
y
x
r
–y
x
These are the reduction formulae:
y
–y
–y
sin (180° – θ) = _r = +sin θ
sin (180° + θ) = __
r = –sin θ
sin (360° – θ) = __
r = –sin θ
–x = –cos θ
cos (180° – θ) = __
r
–x = –cos θ
cos (180° + θ) = __
r
cos (360° – θ) = __xr = +cos θ
tan (180° – θ) = __
–x = –tan θ
tan (180° + θ) = __
–x = +tan θ
tan (360° – θ) = __
x = –tan θ
y
–y
–y
y
2
Notice that the ratios stay the same but the signs change according to the CAST
diagram (see Unit 1). We always write ratios as a function of θ (an acute angle) only.
1
Sin
All
x
Tan
3
Cos
WORKED EXAMPLE
Simplify as far as possible.
4
1
3cos (180° + x)sin (360° – x)
________________________
sin (360° + x)cos (180° – x)
2
sin (180° + θ) + cos (180° – θ)
_________________________
2
(–sin θ) + (–cos θ)
________________
2
2
tan (180° – θ).cos (360° – θ)
SOLUTIONS
1
3cos (180° + x)sin (360° – x)
________________________
sin (360° + x)cos (180° – x)
3(–cos x)(–sin x)
= ______________
sin x (–cos x)
= –3
2
2
(–tan θ).(cos θ)
sin θ + cos θ
= ____________
– sin θ
2
2
______.(cos θ)
cos θ
1
= – _____
sin θ
146
Topic 7 Trigonometry
PLT MATHS LB 11 7th pgs (Real Book).indb 146
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EXERCISE 7
REMEMBER
Simplify the expressions as far as possible.
1
cos (180° + x)tan (180° – x)sin (180° + x)
__________________________________
2
2sin (180° – x)cos (360° – x)
________________________
3
sin (360° – x).tan (180° – x).cos (360° – x)
___________________________________
4
cos (180° − x).tan (360° + x)
________________________
5
1 – _______________
Follow these steps to simplify
algebraic expressions with
reduction formulae:
1 Decide in which quadrant
the angle lies.
3 Use the CAST diagram to
determine the sign of the
ratio.
4 Remember that a squared
ratio is always positive.
5 Rewrite the ratio as a ratio
of the acute angle (θ).
6 Simplify the remaining
expression.
7 Use identities where
appropriate.
sin (180° – x)sin x
sin (180° + x)cos (180° – x)
sin (180° – x).sin (180° + x)
2
2
sin2(180°
− x)
sin2(180° + x)
1 – cos (180° + x)
Negative angles
θ – 360°
–180° – θ
θ – 180°
y
y
x
θ
θ – 360°
–θ
θ
θ
x
x
–θ
θ – 180°
y
x
y
–180° – θ
sin (θ – 360°) = sin θ
sin (–180° – θ) = sin θ
sin (θ – 180°) = –sin θ
sin (–θ) = –sin θ
cos (θ – 360°) = cos θ
cos (–180° – θ) = –cos θ
cos (θ – 180°) = –cos θ
cos (–θ) = +cos θ
tan (θ – 360°) = tan θ
tan (–180° – θ) = –tan θ
tan (θ – 180°) = tan θ
tan (–θ) = –tan θ
EXERCISE 8
Simplify the expressions as far as possible.
1
2sin (180° – x).cos ( – x)
_________________________
2
sin ( 180° + x ) . cos ( – x )
_____________________
+ sin (x – 360°).sin (x – 180°)
3
sin 180° + x
1 – ______________________
(
)
(
)
4
1 + sin ( – x)
____________
5
tan (x – 180°) + sin (180° + x).cos ( – x)
_________________________________
sin ( – 180° – x).cos (x – 180°)
tan ( – x)
2(
)
cos x – 180° .cos x – 360°
cos2(x – 180°)
sin2(360° – x)
Unit 3 Reduction formulae
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REMEMBER
The co-ratios are 90° – θ and 90° + θ
Co-ratios are also called complementary ratios.
Complementary angles add
up to 90°. (40° + 50° = 90°
and 20° + 70° = 90° are two
pairs of complementary
angles.)
y
a
θ
b
x
90° – θ
REMEMBER
Any function of (180° ± θ) or
(360° ± θ) is numerically equal
to the SAME function of θ.
Any function of (90° ± θ) is
numerically equal to the
co-function of θ.
90° + θ
θ
θ
c
Look at the figure and you will see that:
c
b
__
sin (90° – θ) = __
a = cos θ and cos (90° – θ) = a = sin θ
The angle (90° + θ) lies in Quadrant 2.
sin (90° + θ) = cos θ and cos (90° + θ) = –sin θ | because cos θ is negative in Quadrant 2
WORKED EXAMPLES
Examples: Simplify the expressions as far as possible.
1
2cos (90° – x) – sin (360° – x)
_________________________
2
cos (90° – x) + sin (90° + x)
________________________
2sin (90° – x) – cos (180° + x)
2
2
tan (180° + x).sin (90° + x)
SOLUTIONS
1
2cos (90° – x) – sin (360° – x)
_________________________
2sin (90° – x) – cos (180° + x)
2sin x – (–sin x)
2cos x – (–cos x)
= ______________
3sin x
= ______
3cos x
= tan x
2
2
2
cos (90° – x) + sin (90° + x)
________________________
tan (180° + x).sin (90° + x)
2
2
sin x + cos x
= ____________
–tan x.cos x
1
–1
= ____________
= _____
–sin x
_____ × cos x
cos x
148
sin x
Topic 7 Trigonometry
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EXERCISE 9
Simplify the expressions as far as possible.
1
tan (180° – x).sin (360° – x).cos (90° – x)
__________________________________
2
sin (180° + θ).cos (360° – θ).tan (–θ)
______________________________
sin (180° + x).cos (90° + x).tan (180° + x)
2cos (90° + θ).sin (90° + θ)
3
2sin (90° – x) + cos (180° – x)
_________________________
sin (90° – x) – cos (x – 180°)
4
cos (90° – β) + cos (β + 90°) + cos (180° – β) – cos (β + 180°) + cos (–β)
[ cos( 90º + x ) cos (–x) tan (360° – x) ] – [cos(360° – x) sin(x – 90º)]
5
6
[sin (180º – x) + sin (90º – x)][cos(90° – x) + cos(–x)]
7
tan (180° + x).sin (90° – x) ____________
cos (180° – x)
______________________
–
8
cos (90° – θ).tan (180° + θ).cos (360° – θ) – cos (–θ).sin (θ – 90°)
cos (90° – x)
sin (90° + x)
Numerical reductions
You have applied the reduction formulae to algebraic expressions where the angle is
a variable, for example x or θ. Now in numerical examples you apply the reduction
formulae to reduce a trigonometric ratio with an angle greater than 90° to a ratio with
an acute angle.
The acute angle, called the central angle, is the angle (θ) that lies between the rotating
arm and the x-axis.
WORKED EXAMPLES
REMEMBER
Write each function as a ratio of acute angles:
1
sin 215°
2
tan 120°
central
angle
cos 320°
y
y
215°
35° O
3
x
central
angle 60°
y
320°
120°
O
x
O 40° central
angle
x
Follow these steps when
reducing trigonometric ratios
with numerical angles.
1 Decide on the quadrant
and hence the +/– sign
(use the CAST diagram).
2 Keep the function
(trigonometric ratio)
the same.
3 Replace the angle with the
‘central’ angle (acute angle
between arm and x-axis).
SOLUTIONS
1
sin 215° = –sin 35°
Explanation: sin 215° = sin (180° + 35°) = –sin 35° sin (180° + θ) = –sin θ
2
tan 120° = –tan 60°
Explanation: tan 120° = tan (180° – 60°) = –tan 60° tan (180° – θ) = –tan θ
3
cos 320° = cos 40°
Explanation: cos 320° = cos (360° – 40°) = cos 40° cos (360° – θ) = cos θ
Unit 3 Reduction formulae
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EXERCISE 10
REMEMBER
To find the central or acute
angle θ quickly:
Quadrant 2:
125° = 180° – θ
θ = 180° – 125° = 55°
Quadrant 3:
235° = 180° + θ
θ = 235° – 180° = 55°
Quadrant 4:
305° = 360° – θ
θ = 360° – 305° = 55°
1
Rewrite each ratio as a ratio of an acute angle:
1.1
sin 122º
1.2
1.3
tan 133º
1.4
1.5
tan 145º
1.6
1.7
sin 245°
1.8
1.9
tan 191°
1.10
1.11 cos 230º
1.12
1.13 cos 350°
1.14
1.15 tan 288º
1.16
1.17 sin 295°
1.18
cos 156º
sin 166°
cos 99º
cos 225°
tan 209°
sin 216°
cos 300°
sin 302º
tan 322°
Numerical reductions with negative angles
You know that sin (–θ) = –sin θ , cos (–θ) = cos θ and tan (–θ) = –tan θ.
For this to be true, θ does not have to be acute.
WORKED EXAMPLES
Write each ratio as a ratio of positive acute angles
1
2
sin (–200°)
cos (–120°)
SOLUTIONS
y
1
central
angle
20°
–200°
150
O
y
2
x
central 60°
angle
x
O
–120°
sin (–200°) = sin 20°
cos( –120°) = – cos60°
Explanation:
sin (–200°) = –sin 200°
= –sin (180° + 20°)
= –(–sin 20°) = sin 20°
or –200° lies in Quadrant 2 where
sin θ > 0 and the central angle is 20°
sin (–200°) = sin 20°
Explanation:
cos (–120°) = cos 120°
= cos (180° – 60°) = –(cos 60°)
or –120° lies in Quadrant 3
where cos θ < 0 and the central
angle is 60°
cos (–120°) = –cos 60°
Topic 7 Trigonometry
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EXERCISE 11
1
Rewrite each ratio as a ratio of a positive acute angle.
1.1
sin (–112°)
1.2
cos (–125º)
1.3
tan (–93º)
1.4
sin (–106°)
1.5
tan (–135º)
1.6
sin (–204°)
1.7
cos (–265°)
1.8
tan (–189°)
1.9
tan (–237°)
1.10 cos (–251º)
1.11 cos (–340°)
1.12 cos (–292°)
1.13 tan (–286º)
1.14 sin (– 327º)
1.15 sin (–355°)
Numerical reductions with algebraic ratios
WORKED EXAMPLES
REMEMBER
If cos 20° = p, find these ratios in terms of p:
1
2
3
4
70°
cos 160°
sin (–70°)
cos 200°
sin 340°
1
20°
SOLUTIONS
_____
√1 – p2
p
| 160° = 180° – 20°
1
cos 160° = –cos 20° = − p
2
p
sin (–70°) = –sin70° = − __
1
3
cos 200° = –cos 20° = −p
4
sin 340° = –sin 20° = – _______
_____
√1 – p2
1
1 Draw a diagram. The given
angle is acute, so you
do not use the Cartesian
plane.
2 Find the third side using
Pythagoras’ Theorem.
3 Reduce the ratios to ratios
with acute angles (check
your signs).
4 Use the diagram to help
you write down the
required ratio.
| 200° = 180° + 20°
| 340° = 360° – 20°
EXERCISE 12
1
2
3
If cos 52° = k, find each ratio in terms of k.
1.1
cos 128°
1.3
tan 218°
1.5
cos 232°
1.2
1.4
1.6
sin 38°
sin 308°
tan (–142°)
If 2tan 25º = p, find each ratio in terms of p.
2.1
tan 155º
2.3
cos 335º
2.5
cos (–115°)
2.2
2.4
2.6
sin 205º
tan 65º
sin 295º
If sin 43° = a, express each ratio in terms of a.
3.1
sin 223°
3.2
3.3
sin (–137°)
3.4
cos 317°
cos 133°
Hint:
Draw a
diagram.
Unit 3 Reduction formulae
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Numerical reductions with special angles
You can apply the reduction formulae to the special angles you learnt in Grade 10.
Quadrant
Special angles in all four quadrants
1: θ
30°
45°
60°
0°
90°
2: 180° – θ
150°
135°
120°
180°
90°
3: 180° + θ
210°
225°
240°
180°
270°
4: 360° – θ
330°
315°
300°
360°
270°
REMEMBER
The figures show the triangles with special angles. The table of values is useful, but remember,
and use, the triangles.
sin
cos
tan
REMEMBER
1 Reduce the ratios to ratios
with acute angles (check
your signs).
2 Use the special angle
triangles to write down the
required ratio.
3 Simplify and evaluate the
remaining expression.
30°
45°
60°
1
__
2
1__
___
2
2
2
__
√3
___
√2
__
√3
___
1__
___
1
√3
__
√3
___
0°
90°
0
1
2
1
__
1
0
__
√3
0
Undefined
2
60°
1
30°
__
√3
__
√2
45°
1
45°
1
WORKED EXAMPLES
Evaluate the expressions without using a calculator.
1
cos 240°.sin 315°.tan 120°
______________________
2
cos2 330° – sin2 210°
__________________
sin 150°.tan 210°.cos 225°
tan2 135°.sin 90°
SOLUTIONS
1
(–cos 60°).(–sin45°).(–tan60°)
_________________________
(sin 30°).(tan 30°).(–cos45°)
1__ ( –√ 3 )
( –__12 )( –___
__ √__
√2 )
3 =3
______________
√ 3 ×___
=
1__ –___
1__
1
( __12 )( ___
(
)
)
√2
√3
__
2
2
2
cos 30° – (–sin 30°)
__________________
(–tan
45°)2.sin
__
√3 2
___
( ) – ( __1 )
90°
2
2
2
3 – __
1 = __
1
= __________
= __
2
(–1) .(1)
152
4
4
2
You can ignore the negative
signs for the squared ratios:
tan2 135° = tan2 45° = 12 = 1
Topic 7 Trigonometry
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EXERCISE 13
Evaluate the expressions without using a calculator.
1
4tan 330° cos 135° sin 300°
2
3tan 315° sin 225° sin 120°
_______________________
3
sin 240° tan 135°.tan 120°
_______________________
4
sin 210° cos180°.cos2 45°
_____________________
5
sin 135°.cos 315° + cos 360° tan 225°.sin 90°
6
tan 120° sin 210°
sin 240° √3 – _______________
cos 315° cos 210°
sin 150° sin 210°.cos 330°
1 + tan2 330°
(
__
cos 180°
)
Numerical reductions with co-ratios
You know that sin (90° – θ) = cos θ and cos (90° – θ) = sin θ.
To apply these formulae to numerical examples, work with acute angles only:
sin 40° = cos 50° because sin (90° – 50°) = cos 50°
cos 10° = sin 80° because cos (90° – 80°) = sin 80°
If you are given angles bigger than 90°, reduce them to acute angles where the ratio
stays the same, and then use the co-ratio formulae:
sin 145° = sin 35° = cos 55°
cos 250° = –cos 70° = –sin 20°
sin 302° = –sin 58° = –cos 32°
WORKED EXAMPLES
REMEMBER
Simplify the expressions as far as possible.
1
tan 210°.sin 480°.sin 170°
______________________
2
cos 70°.sin 150° + cos 240° cos 250°
______________________________
cos 225°.sin 315°.cos 260°
sin 20°
SOLUTIONS
1
1 Reduce angles to acute
angles.
2 Look out for pairs of ratios
that have complementary
angles.
3 Use co-ratios to make the
angles the same.
tan 30°.sin 60°.sin 10°
________________________
(–cos 45°).(–sin 45°).cos 80°
__
√3
1__ .___
___
.sin 10°
2
√
3
= _________________
1
1
__ ).(– ___
__ )sin 10°
(– ___
√2
√2
1 ÷ __
1=1
= __
2
2
2
cos 70°.sin 30° + (–cos 60°)(–cos 70°)
_______________________________
sin 20°
cos 70°.sin 30° + sin 30° cos 70°
= ___________________________
cos 70°
(2)
2cos 70°.sin 30° = 2sin 30° = 2 __
1 =1
= ______________
cos 70°
Unit 3 Reduction formulae
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EXERCISE 14
1
2
154
Write each ratio as a ratio of 15° or 70°.
1.1
sin 20°
1.3
sin 160°
1.5
cos 105°
1.7
cos 340°
1.2
1.4
1.6
1.8
sin 290°
cos 75°
cos 285°
cos 430°
Simplify and evaluate the expressions. Where possible, do not use a calculator.
2.1
sin 340°
________
2.2
sin2 (100°) ÷ cos2 370°
2.3
4sin 248°.tan 158°
________________
2.4
2cos 143°.cos 217°
________________
2.5
sin 147°.tan 114°
_______________
2.6
cos 149°.sin 239° – cos 301° sin 211°
2.7
sin ( –75° ).tan 420°
____________________
2.8
sin 145° – sin (–235°).cos 325°
_________
cos 430°
sin 202°
sin 127°.sin (–53°)
3cos 57°.tan 246°
sin 60°.sin 195°.tan 75°
cos (–55°)
Topic 7 Trigonometry
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Unit 4: Trigonometric equations – specific
and general solutions
In Grade 10 you learnt to solve trigonometric equations for angles less than 90°.
In Grade 11 you learn to solve for angles greater than 90° and for negative angles.
Solving equations with one ratio for angles in
the interval [0º;360º]
WORKED EXAMPLES
REMEMBER
Solve for x:
1
2sin x = 1
2
3tan x + 4 = 0
y
y
2
1
30°
30°
1
53,13°
x
O
3
2
O
4
53,13°
3
4
SOLUTIONS
1
2sin x = 1
2
3tan x + 4 = 0
1
∴ sin x = __
2
4
∴ tan x = – __
3
Calculator angle = 30°
Calculator angle = 53,13°
Quadrant 1:
x = 30°
Quadrant 2:
x = 180° – 53,13° = 126,87°
Quadrant 2:
x = 180° – 30° = 150°
Quadrant 4:
x = 360° – 53,13° = 306,87°
x
Follow these steps for solving
trigonometric equations with
one ratio.
1 Solve for the given ratio.
2 Determine in which two
quadrants the solution lies.
Use the CAST diagram.
This will be where the
given ratio is either positive
or negative.
3 Draw a diagram in
the Cartesian plane to
represent this.
4 Find the central or
‘calculator’ angle by
pressing shift sin–1 or
cos–1 or tan–1 of the
POSITIVE ratio.
5 Solve for the unknown
angle:
Quadrant 1:
x = calculator angle
Quadrant 2:
x = 180° – calculator angle
Quadrant 3:
x = 180° + calculator angle
Quadrant 4:
x = 360° – calculator angle
EXERCISE 15
Solve for x where 0° ≤ x ≤ 360°:
1
3
5
7
sin x = 0,68
cos x = 0,454
tan x = 3,526
3sin x + 1 = –1
2
4
6
8
tan x = –2,41
sin x = – 0,237
cos x = – 0 ,813
5tan x – 7 = 2
9
4cos2 x – 3 = 0
10
tan2 x – 1 = 0
______
11
4sin2 x = 2
12
2sin x + √ 3 ≤ 0
3
__
Unit 4 Trigonometrtic equations – specific and general solutions
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Solving equations with one ratio for angles in
the interval [–360º;360º]
WORKED EXAMPLES
Solve for x:
__
1
2sin (x + 20°) = √ 3
2
3cos (x – 25°) + 2 = 0
SOLUTIONS
Notice that the rotating angle is x + 20° for Worked example 1 and x – 25° for
Worked example 2.
y
y
2
1
2
x + 20°
x + 20°
60°
1
x – 25°
60°
O
3
x – 25°
4
x
48,19°
48,19°
x
O
x – 25°
4
3
x + 20°
To solve the equation follow the steps given in the Remember box on the
previous page. __
3cos (x – 25°) + 2 = 0
2sin (x + 20°) = √ 3
__
√
3
sin (x + 20°) = ___
2
cos (x – 25°) = – __
Calculator angle = 60°
Quadrant 1 (positive angle):
x + 20° = 60°
x = 40°
Quadrant 1 (negative angle):
x + 20° = –360° + 60°
x = –320°
Quadrant 2 (positive angle):
x + 20° = 180° – 60°
x = 100°
Quadrant 2 (negative angle):
x + 20° = –180° – 60°
x = –260°
Calculator angle = 48,19°
Quadrant 2 (positive angle):
(x – 25°) = 180° – 48,19°
x = 131, 81° + 25° = 156,81°
Quadrant 2 (negative angle):
(x – 25°) = –180° – 48,19°
x = –228,19° + 25° = –203,19°
Quadrant 3 (positive angle):
(x – 25°) = 180° + 48,19°
x = 228,19° + 25° = 253,19°
Quadrant 3 (negative angle):
(x – 25°) = –180° + 48,19°
x = –131,81° + 25° = –106,81°
2
156
3
Topic 7 Trigonometry
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EXERCISE 16
Solve for x where – 360° ≤ x ≤ 360°.
√
__
3
cos (x + 15°) = ___
1
1
sin (x – 22º) = – __
2
3
tan (x + 40º) = –1
4
__
√ 5 cos
5
cos x + cos 71º = sin 17º
6
3 tan x + sin 27º = 3
8
2sin (x + 50º) + 4tan 20º = 0
2
2
(x – 15º) = 2
7
__
√ 2 sin
9
5cos (x + 15°) – 2 = 0
10
3 tan (x – 10°) + 10 = 3
11
3sin (x + 32º) = 2sin 32°
12
tan (x + 50°)
___________
+1=3
(x – 27º) – 1 = 0
3
The concept of a general solution
You need not restrict the solutions to trigonometric equations to an interval of
[0º;360º] or [–360º; 360º]’
1:
Look at the solutions to sin x = __
2
1 , but so does sin 390° = sin (30° + 360°) = sin 30° = __
1
sin 30° = __
2
2
1
sin 750° = sin (30° + 2 × 360°) = __
2
1
and sin 1 110° = sin (30° + 3 × 360°) = sin 30° = __
2
1 , but so does sin 510° = sin (150° + 360°) = sin 30° = __
1
sin 150° = __
2
2
1 and sin 1 230° = sin (150° + 3 × 360°) = sin 30° = __
1
sin 870° = sin (150° + 2 × 360°) = __
2
2
You can summarise the solutions to trigonometric equations as the general solution
1:
of the equation sin x = __
2
x = 30° + n.360°, n ∊ ℤ or x = (180° – 30°) + n.360°, n ∊ ℤ
1.
You can see the solution graphically by drawing the graphs of y = sin x and y = __
2
1 in the interval [– 450º;450º].
The graphs show that there are five solutions to sin x = __
2
y
1
0,5
E
–450°
–360°
D
–270°
A
–180°
–90°
B
90°
C
180°
270°
360°
450°
x
–0,5
–1
Unit 4 Trigonometrtic equations – specific and general solutions
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Finding the general solution to equations with
one ratio
WORKED EXAMPLE
KEY WORDS
Find the general solution to the equation sin 2x = – 0 ,757.
general solution – the
formula which lists all possible
solutions to a trigonometric
equation; takes into account
the period of the ratios so
the angle can be positive or
negative
solution – a value of the
angle which satisfies a given
trigonometric equation
SOLUTIONS
Method 1
Using the positive ratio and ‘calculator
angle’ is a positive acute angle.
Calculator angle = 49,2°
(key in positive ratio)
(sin < 0 in Quadrant 3 and Quadrant 4)
Quadrant 3:
2x = 180° + 49,2° + n.360°
= 229,2° + n.360°(÷ by 2)
x = 114,16° + n.180°
Quadrant 4:
2x = 360° – 49,2° + n.360°
= 310,8° + n.360°
x = 155,4° + n.180°, n ∊ ℤ
REMEMBER
For sin x = a the general
solution is:
x = sin–1(a) + n.360°, n ∊ ℤ or
x = 180° – sin – 1(a) + n.360°,
n∊ℤ
For cos x = b the general
solution is:
x = cos–1(b) + n.360°, n ∊ ℤ or
x = –cos – 1(b) + n.360°,
n∊ℤ
For tan x = c the general
solution is:
x = tan–1(c) + n.180°, n ∊ ℤ
158
Method 2
Using the given ratio (+ or –) and the
‘calculator angle’ is acute, obtuse or
negative.
Calculator angle = – 49,2°
(key in given, negative ratio)
Use 2x = calculator angle
or 2x = 180° – calculator ang
2x = – 49, 2° + n.360°(÷ by 2)
x = –24,6° + n.180°
2x = 180° – (– 49,2°) + n.360°
= 229,2° + n.360°
x = 114,6° + n.180°, n ∊ ℤ
Notice that x = 155,4° + (–1).180° = –24,6°, which is the same solution
for Method 2.
EXERCISE 17
Find the general solutions to these equations.
1
sin (x – 16º) = 0,616
2
cos 2x = 0,789
3
cos 3x = – 0,123
4
tan (x + 56°) = 7,56
5
x = –1,421
tan __
6
sin (2x + 44º) = – 0,708
7
3cos (x – 15°) + 1 = – 0,456
8
2tan ( 2x – 10°) = 10,67
2
Topic 7 Trigonometry
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Solving equations with more than one ratio
WORKED EXAMPLES
1
2
3
KEY WORDS
Solve for x if cos x – 4cos x sin x = 0 and –180° ≤ x ≤ 360°.
Solve for x if 3cos2 x – 5sin x = 1 and write down the general solution.
Find the general solution to the equation 3cos x – sin x = 0.
SOLUTIONS
1
2
3
cos x + 4cos xsin x = 0 | Factorise by taking out a common factor.
cos x(1 + 4sin x) = 0
1
cos x = 0 or sin x = –__
4
x = 90° + n.360°
or calculator angle = 14,5°
x = –90° + n.360°
or x = 180° + 14,5° + n.360°
or x = 360° – 14,5° + n.360° n ∊ ℤ
Specific solution: Start with n = 0, then n = ±1, n = ±2 …
Continue to work out specific values of x until your answers fall outside
– 180° ≤ x ≤ 360°.
Then x = 90°; 270°; –90°; 194,5°; –65,5°; 345,5°; –14,5°
3cos2 x – 5sin x = 1
3(1 – sin2 x) – 5sin x = 1 | cos2 x = 1 – sin2 x
3 – 3sin2 x – 5sin x – 1 = 0
| trinomial
3sin2 x + 5sin x – 2 = 0
(3sin x – 1)(sin x + 2) = 0
1 or sin x = –2
sin x = __
3
calculator angle = 19,5° or No solution
Quadrant 1: x = 19,5° + n.360°
Qudrant 2: x = 160,5° + n.360, n ∊ ℤ
3cos x – sin x = 0
3cos x = sin x | ÷ both sides by cos x
sin x
3 = _____
cos x tan x = 3
calculator angle = 71,57°
x = 71,57° + n.180°, n ∊ ℤ
EXERCISE 18
Solve the equations: (GS = general solution)
1
2
3
4
5
6
7
8
9
3sin x – 2cos x = 0, find GS
3sin2 x – 2sin x = 0 and x ∊ [–360°;360°]
3tan2 x + tan x = 2 – 180° ≤ x ≤ 360°
5sin2 x – 3sin x – 2 = 0, find GS
4cos2 x – sin x = 1 and x ∊ [–180°;360°]
3cos2 x = 2(sin x + 1), find GS
3sin2 x – 5sin xcos x = 0 and x ∊ [–180°;180°]
10sin2 x + cos x = 8, find GS
4tan2 x – 8tan x + 3 = 0, find GS
specific solutions – solutions
which satisfy a given
trigonometric equation in a
restricted interval, such as
– 360° ≤ x ≤ 360°
REMEMBER
1 Convert all ratios to sine
and cosine to only one
ratio if possible.
Use the identities
sin x
tan x = _____
cos x ,
cos2 x = 1 – sin2 x,
sin2 x = 1 – cos2 x.
2 Simplify and factorise
to change into simpler
equations.
3 Find the general solution
and solve for the given
interval.
REMEMBER
If you use Method 2:
Key in the given ratio either
positive or negative (what is
given). The calculator angle
could be acute, negative or
obtuse. Learn these patterns
and DO NOT change them
for any situation:
sin x = …
x = calc + n.360°
or 180° – calc + n.360°, n ∊ ℤ
cos x = …
x = calc + n.360°
or – calc + n.360°, n ∊ ℤ
tan x = …
x = calc + n.180°, n ∊ ℤ
Unit 4 Trigonometrtic equations – specific and general solutions
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REMEMBER
1 Convert all ratios to sine
and cosine or only one
ratio if possible.
Use the identities
10
11
12
13
14
15
1 – 5sin2 x = 3sin x + 1 and x ∊ [–180°;180°]
6sin2 x + 7sin x + 2 = 0 and x ∊ [–360°;360°]
__
3cos2 x + √ 3 sin x cos x = 0, find GS
8sin x cos x + 4cos x + 6 sin x + 3 = 0, find GS
4cos2 x – sin2 x = 3cos x + 1 and x ∊ [–180°;180°]
3sin2 x + 2sin x.cos x = 4cos x + 6 sin x and x ∊ [0°;360°]
sin x
tan x = _____
cos x ,
cos2 x = 1 – sin2 x,
sin2 x = 1 – cos2 x.
2 Simplify and factorise
to change into simpler
equations.
3 Find the general solution
and solve for the given
interval.
Solving equations without a calculator
Look at these examples for x ∊ [0°;360°]:
If sin x = sin 41°: calculator ang = 41° so x = 41° or x = 180° – 41° = 139°
If cos x = –cos 58°: calculator ang = 58° so x = 180° – 58° = 122° or x = 180° + 58° = 238°
If cos x = sin 24°, then cos x = sin 66° (co-ratio) so x = 66° or x = 360° – 66° = 294
We can extend the concept of solving without a calculator to equations with algebraic
expressions for the angles.
WORKED EXAMPLE 1
Solve for x if sin (3x – 20°) = sin (x + 10°) and –360° ≤ x ≤ 360°.
SOLUTIONS
REMEMBER
If you use Method 2:
Key in the given ratio –
positive or negative (what is
given). The calculator angle
could be acute, negative or
obtuse. Learn these patterns
and DO NOT change them
for any situation:
sin x = … x = calc + n.360°
or 180° – calc + n.360°, n ∊ ℤ
cos x = … x = calc + n.360°
or – calc + n.360°, n ∊ ℤ
tan x = … x = calc + n.180°,
n∊ℤ
Find the general solution first:
Quadrant 1
3x – 20° = x + 10° + n.360°
2x = 30° + n.360°
x = 15° + n.180°, n ∊ ℤ
or Quadrant 2
3x – 20° = 180° – (x + 10°) + n.360°
4x = 190° + n.360°
x = 47,5° + n.90°, n ∊ ℤ
For – 360° ≤ x ≤ 360° use integral values for n
x = 15°; 15° + (1)180° = 195°; 15° + (–1)180° = –165°; 15° – (2)180° = –375°
x = 47,5°; 47,5° + (1)90° = 137,5°; 47,5° + (2)90° = 227,5°; 47,5° + (3)90° = 317,5°
x = 47,5°; 47,5° – (1)90° = –42,5°; 47,5° – (2)90° = –132,5°; 47,5° – (3)90° = 222,5°;
47,5° – (4)90° = 312,5°
WORKED EXAMPLE 2
Find the general solution of cos (2x + 30°) = sin x.
SOLUTIONS
Use co-ratio: sin x = cos (90° – x) and find the general solution:
cos (2x + 30°) = cos (90° – x)
Quadrant 1
or Quadrant 4
2x + 30° = 90° – x + n.360°
2x + 30° = –(90° – x) + n.360°
3x = 60° + n.360°
2x + 30° = –90° + x + n.360°
x = 20° + n.120°, n ∊ ℤ
x = –120° + n.360°, n ∊ ℤ
For Quadrant 4 you may use:
2x + 30° = 360° – (90° – x) + n.360°
x = 240° + n.360°
(which is the same as –120°)
160
Topic 7 Trigonometry
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EXERCISE 19
Solve the equations without a calculator: (GS = general solution)
1
2
3
4
5
6
7
8
9
10
cos x = sin 32° and x ∊ [0°;360°]
tan2x = tan 128° and x ∊ [–360°;360°]
sin (x – 40°) = cos 50° and x ∊ [–360°;360°]
tan x = tan (2x – 25°), find GS
cos (2x – 25°) = cos (38° – x) and x ∊ [–90°;180°]
sin (x + 10º) = sin ( 50º + 2x ), find GS
sin (2x – 5º) = cos ( x – 35º ), find GS
cos x = – cos 42° and x ∊ [–180°;360°]
cos (2x – 10°) = sin (x – 40°) and x ∊ [–180°;360°]
sin (x + 10º) = –sin (30º – 2x), find GS
EXERCISE 20
Solve for x in each equation.
1
2
3cos x + 2 = 1 and x ∊ [0°;360°]
sin x = 0,112 and x ∊ [–360°;360°]
_____
REMEMBER
2
3
4
5
tan (x – 10°) + 1 = 3 and x ∊ [–180°;180°]
15sin (x + 14º) = 14 and x ∊ [0°;360°]
2cos (2x + 20°) = 1, find GS
6
x = √ 2 and x ∊ [–360°;360°]
2sin __
7
8
9
10
11
12
13
14
4tan (2x – 40°) + 13 = 3, find GS
cos (x – 12°) = sin 56° and x ∊ [0°;360°]
13cos x + 12sin x = 0, find GS
4cos2 xsin x – 3sin x = 0, find GS
3cos2 x + 11sin x + 1 = 0 and x ∊ [–360°;360°]
5tan2 x + 7tan x = 6 and x ∊ [–180°;360°]
cos (2x + 20°) = cos (30° – x), find GS
2sin xcos x – cos2 x – 2sin x + cos x = 0 and x ∊ [–180°;180°]
__
2
1 If sin A = sin B, then A = B
or A = 180° – B
2 If cos A = cos B, then A = B
or A = 360° – B (or – B)
3 If tan A = tan B then A = B
or A = 180° + B
4 If sin A = cos B then
sin A = sin (90° – B) and
A = (90° – B) or
A = 180° – (90° – B)
Finding the values for which identities are invalid
or undefined
We have worked through trigonometric equations, so now we consider the values
for which identities or invalid or undefined.
Remember from Unit 2 that an identity is an equation which is true for all values of
the unknown variable, for which both sides of the identity are defined.
•
•
•
An identity is undefined if the denominator is zero.
___
If an identity is in the form A = √f(x) , it is not valid for the values of x which
make f (x) < 0
If an identity involves tan x, it is not valid for x = 90° + n.180°
sin x
tan x = _____
cos x and cos x = 0 when x = 90° + n.180°
KEY WORD
trigonometric equation
– an equation involving
trigonometric ratios which is
true only for certain values of
the unknown variable
When proving identities, you may be asked to find the values of the variable (angle)
for which the identity is undefined or invalid.
Unit 4 Trigonometrtic equations – specific and general solutions
PLTMATHSLB11LB_07.indd 161
161
2012/07/02 3:07 PM
WORKED EXAMPLE
1
tan x + sin x = sin x undefined?
For which values of x is the identity ___________
1
2
sin θ – √ 1 – sin2 θ = 0 hold true?
For which values of x does the identity _____
1 + _____
cos x
tan θ
________
SOLUTIONS
1
1
Undefined when tan x is undefined or when 1 + _____
cos x = 0
1
x = 90° + n.180° or 1 + _____
cos x = 0 cos x = –1 x = 180° + n.360°
2
Undefined or invalid when:
2.1
tan θ = 0 | θ = 0° + n.180
2.2
tan θ is undefined when θ = 90° + n.180°
2.3
1 – ________
sin2 θ ≤ 0 sin2 θ ≥ 1 which is never true
√ 1 – sin2 θ is defined for x ∊ ℝ
EXERCISE 21
In Exercises 4 and 5, you proved identities but did not state restrictions. Now find
the values of the variable x or θ for which the following identities are not valid.
1
cos x
1 – sin x = _____
_____
2
cos x – cos3 x
________________
= tan2 x
2
3
1
tan2 θ + 1 = _____
2
4
cos x + tan x = _____
1
________
cos x
1 + sin x
5
1
1
2
________
+ ________
= _____
2
6
cos x = ________
1 + sin x
________
cos x
7
cos2 x – cos x – sin2 x = _____
1 – _____
1
__________________
8
2sin x.cos x
1
sin x – ___________
+ cos x = ___________
9
sin x + 1
1
2 = ________
________
– _____
2
10
162
tan x
sin x
cos x – cos x.sin x
cos θ
1 – sin x
1 + sin x
cos x
1 – sin x
tan x
2sin x.cos x + sin x
sin x + cos x
sin x + 1
cos x
sin x
_________
=
(1 + cos x)
sin x
sin x + cos x
sin x – 1
_______
1 – cos x
√________
1 + cos x
Topic 7 Trigonometry
PLT MATHS LB 11 7th pgs (Real Book).indb 162
2012/07/02 2:22 PM
Revision
1
y
P(–5;4)
θ
O
1.1
1.2
x
Use the diagram to find the length OP.
Now write down the value of:
1.2.1
sin θ
1.2.2
41cos2 θ + 1
1.2.3
tan (180° – θ)
(1)
(1)
(2)
(2)
[6]
y
2
θ
O
x
__
P(√ 3 ;–1)
2.1
2.2
3
4
Use the diagram to find the length OP.
Now write down the value of:
2.2.1
cos θ.sin θ
2.2.2
tan2 θ + 1
2.2.3
θ
(1)
(3)
(2)
(2)
[8]
If 6sin θ = 5 and cos θ < 0, find, without a calculator, the value of:
3.1
1 – 2cos2 θ
3.2
cos θ.tan θ
If 4cos θ = 3 and 7tan β = 5; 90° < θ < 360° and β > 90°,
7 + 74sin2 β = 34
prove without using a calculator that _____
2
[6]
tan θ
5
(4)
(3)
[7]
tanA.sinA
4 and A + B = 90°, use a sketch to find the value of _________
If cos A = __
7
cosB
[5]
163
PLTMATHSLB11LB_07.indd 163
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TOPIC 7: REVISION CONTINUED
6
Prove each identity.
6.1
sin x.cos x = 1
sin2 x + _________
tan x
(4)
6.2
1
cos θ + cos θ.tan2 θ = _____
(5)
6.3
7
8
cos θ
1
1
cos θ tan θ + _____
= _____
tan θ
sin θ
(
)
6.4
cos2 x + (sin x – tan x)(sin x + tan x) = 1 – tan2 x
6.5
1 [(sin x + cos x)2 – 1]
sin xcos x = __
2
(5)
(6)
(5)
[25]
Simplify each expression.
7.1
2cos (90° + x) – sin (360° – x)
_________________________
(6)
7.2
sin (180° + x)
______________________________
(5)
7.3
2cos (–180° – x).sin (–x)
_______________________
(5)
7.4
cos ( 180° + x )
_____________________
+1
(5)
7.5
[ sin ( 90º + x ) sin ( – x) tan (180° – x) ] – [ sin ( 90° – x ) cos ( x – 180º ) ]
3sin (90° – x) – cos (180° – x)
tan (180° – x) tan 45°.cos (360° – x)
sin (x – 180°).cos (x – 360°)
2
sin( 180° – x ).cos( x – 90° )
Draw a diagram for each question.
8.1
If tan 28° = p, write down the following in terms of p.
8.1.1
tan 152°
8.1.2
cos (–28°)
8.1.3
sin 62°
8.2
If 2sin 41° = a, write down the following in terms of a.
8.2.1
sin 319°
8.2.2
cos 49°
8.2.3
cos 139°
(5)
[26]
(2)
(2)
(2)
(2)
(2)
(2)
[12]
164
PLT MATHS LB 11 7th pgs (Real Book).indb 164
2012/07/02 2:22 PM
9
Simplify without the use of a calculator:
cos 250°
9.1
sin 330° + ________
(5)
sin 160°
10
11
9.2
sin 120°.cos 135°.tan 240°
______________________
9.3
sin 55°
1__ sin 120° + __
1 tan 45° cos2 225° – ________
___
(6)
9.4
sin2 300° + sin 240° cos 150°
________________________
(6)
9.5
sin 225°.cos 135° + 2sin 210°.sin 90°
(6)
cos 150°.tan 150°.sin 315°
2
√3
cos 145°
tan 225°.sin 270°
Solve these equations.
10.1 sin (θ + 10°) = 0,757 and 0° ≤ x ≤ 360°
__
10.2 √2 sin 3 θ + 1 = 0 and – 180° ≤ x ≤ 180°
10.3 3cos θ = 2sin2 θ – 3. Find the general solution.
10.4 tan (2 θ – 30°) = tan 42°; –360° ≤ θ ≤ 360°
10.5 sin (x + 20°) = cos 2x. Find the general solution.
(4)
[27]
(3)
(4)
(7)
(6)
(6)
[26]
Find the values of x and θ for which the identities in Question 6
are invalid or undefined.
sin x.cos x = 1
11.1 sin2 x + _________
tan x
(3)
1
11.2 cos θ + cos θ.tan2 θ = _____
(4)
11.3
cos θ
1
_____
_____
cos θ tan θ +
= 1
tan θ
sin θ
(
)
(4)
[11]
165
PLT MATHS LB 11 7th pgs (Real Book).indb 165
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Formal assessment: Term 2 assignment
1
( 2 )x – 1 – 3
1
f( x ) = 3 __
1.1
1.2
1.3
1.5
1.6
2
3
3
4 +1
h( x ) = – _____
x+3
State the equations of the asymptotes of f and h.
Draw f, g and h on the same set of axes. Indicate the x- and
y-intercepts and the asymptotes of both f and h.
Use your graphs to help you answer the questions.
1.3.1
State one value of x for which f(x) = g(x) = h(x)
1.3.2
State one value of x for which f(x) ≠ g(x) = h(x)
1.3.3
For which values of x is f(x) ≥ 0?
1.3.4
For which values of x > –3 is f(x) ≤ g(x)?
1.3.5
1.4
1 ( x + 2 )2 – 3
g( x ) = __
h(x)
g(x)
For which values of x is ____ ≥ 0?
If y = –x + c is one of the symmetry lines of h, determine the value
of c.
If g is reflected in the y-axis, the reflected graph is given by y = …
State h(–x) in the form y = … and describe the translation in words.
1–1
Given f(x) = x2 – 1, g(x) = –2x + 1 – 1 and h(x) = __
x
2.1
State the equations of the asymptotes of g and h.
2.2
State the equations of the symmetry lines of f and h.
2.3
Draw f, g and h on the same system of axes. Indicate all intercepts
with the axes, the asymptotes of both g and h and the symmetry
lines of f and h.
2.4
Determine f (1) – g(1).
2.5
Determine g(–1) – h(–1).
2.6
State the ranges of f, g and h and the domain of h.
2.7
State the values of x and y for which f(x) = g(x).
2.8
For which value(s) of x is f(x) ≥ g(x)?
2.9
Solve algebraically for x and y if f(x) = h(x).
2.10 For which values of x is f(x) ≤ h(x)?
2.11 Write down the equation of f (2x) in the form y = …
2.12 If h is translated 3 units left and 2 units up, the translated
graph is given by y = …
2.13 If g is translated 2 units right and 1 unit down, the translated
graph is given by y = …
2.14 If g is reflected in the x-axis, write the reflected equation in
the form y = …
Draw a sketch for y = ax2 + bx + c, where:
3.1
a > 0, b < 0 and c < 0
3.2
a < 0, b < 0 and c > 0
3.3
a < 0, b < 0, c < 0 and b2 − 4ac
3.4
a > 0, b > 0, c > 0 and b2 − 4ac = 0
3.5
a < 0, b < 0, c < 0 and b2 − 4ac > 0
3.6
a > 0, b = 0 and c > 0
(3)
(12)
(1)
(1)
(2)
(2)
(4)
(2)
(2)
(3)
(2)
(5)
(12)
(2)
(2)
(4)
(2)
(3)
(5)
(2)
(3)
(3)
(3)
(2)
REMEMBER
(4)
(4)
(6)
(6)
(6)
(4)
The quadratic
formula
_______
√
2
−b± b − 4ac
x = ____________
2a
b2 − 4ac < 0
⇒ non-real roots
b2 − 4ac > 0
⇒ real, unequal roots
b2 − 4ac = 0
⇒ real, equal roots
166
PLT MATHS LB 11 7th pgs (Real Book).indb 166
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4
Consider the functions g(x) = psin qx and h(x) = vcos wx for x ∊ [–120°;90°].
y
1
A
h
B
g
E
0,92
F
0,71
D
C
0,38
–120° –97,5° –75°
4.1
4.2
4.3
4.4
4.5
4.6
67,5° 90°
x
State the period of both g and h.
Determine the values of p and q in g(x) = psin qx.
Determine the values of v and w in h(x) = vcos wx.
State the coordinates of D, E and F.
For which values of x is h(x) ≥ g(x)?
For which values of x does h(x) increase as g(x) decreases?
5
y
(60°;2)
h
E
g
–30°
C
(2)
(2)
(2)
(6)
(5)
(3)
(30°;1)
120°
x
D
A
B
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
State the period, the amplitude and the range of h and g.
(8)
If h(x) = asin (x + b) and g(x) = pcos (x + q), determine the values
of a, b, p and q.
(6)
If h(x) = dcos (x + c) and g(x) = vsin (x + w), determine the values
of c, d, v and w.
(6)
State the coordinates of A, B, C, D and E.
(10)
For which values of x is g(x) ≥ 0?
(2)
For which values of x is h(x).g(x) > 0?
(5)
For which values of x is h(x) a decreasing function?
(3)
For which values of x is g(x) an increasing function?
(3)
Total: 175 marks
167
PLT MATHS LB 11 7th pgs (Real Book).indb 167
2012/07/02 2:22 PM
Term 2 summary
Topic 5
The parabola
Standard form: y = f(x) = ax2 + bx + c
x-intercept form: y = f(x) = a(x – xint1)(x = xint2)
Turning point form : y = f( x ) = a( x – p )2 + q
a
The hyperbola: y = f( x ) = ____
x–p+q
The exponential function: y = f (x) = abx – p + q where b > 0, b ≠ 1
The effect of a, p and q is the same for all three graphs.
The effect of a
• Increasing a (so that a > 1) results in a vertical stretch.
• Decreasing a (so that 0 < a < 1) results in a vertical shrink.
• Changing the sign of a (a + to –) results in a reflection in the x-axis.
The effect of p and q
• Changing p results in a horizontal shift.
If p > 0, (x – p) shifts the graph to the right and (x + p) shifts the graph to the left.
• Changing q results in a vertical shift.
• If q > 0, graph moves up.
• If q < 0, graph moves down.
• p and q do not influence each other.
Graph
Turning point
Asymptotes
f( x ) = ax2 + bx + c or
f( x ) = a( x – p )2 + q or
f(x) = a(x – x1)(x – x2)
(p;q)
None
a
y = f( x ) = ____
x–p+q
None
y=q
x=p
f( x ) = abx – p + q
None
y=q
Domain (x); Range (y)
Axis of symmetry
x∊ℝ
y ∊ ℝ, y ≥ p if a > 0
x=p
–b
x = ___
x ∊ ℝ, x ≠ p
y ∊ ℝ, y ≠ q
y = (x – p) + q
y = –(x – p) + q
x∊ℝ
y ∊ ℝ, y > q (if a > 0)
y ∊ ℝ, y < q (if a < 0)
2a
None
• The average gradient of the curve between the two points on a curve is the same as the gradient of the
y –y
2
1
straight line between two points on a curve: m = ______
x –x
2
168
1
Term 2 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 168
2012/07/02 2:22 PM
Topic 6
Trigonometric graphs
Graph x ∊ [–360°;360°]
Amplitude
Period (P)
Asymptotes
Range
Interval spacing
y = sin x
1
360°
none
y ∊ [–1;1]
90°
y = cos x
1
360°
none
y = tan x
Not applicable
180°
x = ±90°
x = ±270°
y ∊ (–∞;∞)
y = asin kx + q
a if a > 0
– a if a < 0
360°
____
none
P
y ∊ [–a + q;a + q] S = __
y = acos kx + q
a if a > 0
– a if a < 0
360°
____
none
P
y ∊ [–a + q;a + q] S = __
y = atan kx + q
Not applicable
180°
____
P + n.P
__
y ∊ (–∞;∞)
y = asin (x ± p)
y = acos (x ± p)
y = atan (x ± p)
p results in a horizontal shift
to the left if p > 0
to the right if p < 0
k
k
y ∊ [–1;1]
45°
4
4
2
k
90°
P
S = __
4
Topic 7
Trigonometry
Trigonometry
Signs of ratios in all four quadrants
y
y
2
1
S
3
y
2
A
x
T
θ is the acute angle between
the arm and the x-axis
C
CAST shows
you in which
quadrant the
ratios are
POSITIVE
4
y
–y
1
r
r
–x
x
r
3
r
180° – θ
y
θ
x
–y
x
180° + θ
360° – θ
4
Identities
sin θ and sin2 θ + cos2 θ = 1
tan θ = _____
cos θ
Prove other identities using the formula above and algebra to show LHS = RHS.
Reduction formulae
sin (180° – θ) = + sin θ
cos (180° – θ) = –cos θ
tan (180° – θ) = –tan θ
sin (360° – θ) = –sin θ
cos (360° – θ) = + cos θ
tan (360° – θ) = –tan θ
sin (180° + θ) = –sin θ
cos (180° + θ) = –cos θ
tan (180° + θ) = + tan θ
sin (–θ) = –sin θ
cos (–θ) = + cos θ
tan (–θ) = –tan θ
Co-ratios
sin (90° – θ) = cos θ; cos (90° – θ) = sin θ; sin (90° + θ) = cos θ; cos (90° + θ) = –sin θ
Trigonometric equations
Standard pattern for general solutions:
sin x = … x = ‘calc’ + n.360° or 180° – ‘calc’ + n.360°, n ∊ ℤ
cos x = … x = ‘calc’ + n.360° or – ‘calc’ + n.360°, n ∊ ℤ
tan x = … x = ‘calc’ + n.180°, n ∊ ℤ
'calc' is obtained by pressing
shift sin–1 (or cos–1 or tan–1)
of the given ratio.
Term 2 summary
PLT MATHS LB 11 7th pgs (Real Book).indb 169
169
2012/07/02 2:22 PM
Formal assessment
Examination practice Paper 1
Time 2 hours
Question 1
Solve for x:
1.1 3(x2−4) = 4(x−3)
1.2 2 − 4x − x2 = 0 (correct to two decimal places)
(4)
(4)
1.3 √ 1 + x + 5=x
(4)
1.4
(4)
____
3x2 − x ≤ 0
______
3
1.5 4
1.6
x2−6
32x
=
(5)
3
__
27
2x4 = ___
(3)
[24]
4
Question 2
2.1 Consider the equation: x2 + 4xy − 21y2 = 0
2.1.1 Calculate the value of the ratio __xy , y ≠ 0.
2.1.2 Now calculate the values of x and y if x + 3y = 3.
2.2 If f(x) =
(4)
(4)
____
7 − x determine:
√_____
x+3
2.2.1 f(−2)
2.2.2 for which values of x, f(x) = 0
2.2.3 for which values of x, f(x) is real.
(2)
(2)
(2)
1 ) + ___
1 + f(−x)
2.3 Given f(x) = 2x, find a simplified expression for: f(x) + f(__
x
f(x)
(4)
[18]
Question 3
Simplify each of the following expressions. Do not use a calculator in this
question. Leave your answers with positive exponents and rational denominators
where necessary.
3.1
( )
__
√ b3 −1
____
(3)
1
__
b−2
−1
x +y
3.2 ______
−1
(4)
___9
__
3.3 _________
(4)
y
+x
√ 48 − √ 3
____
____
√ 45x5 − √ 20x3
_____
3.4 _____________
3
(4)
[15]
√ 125x
Question 4
4.1 Tsepo is doing a Life Sciences project in which he observes the number of
cockroaches entering a drain pipe per day.
170
Day
Number of cockroaches
1
1
2
10
3
23
4
40
Exam practice paper 1
PLT MATHS LB 11 7th pgs (Real Book).indb 170
2012/07/02 2:22 PM
4.1.1 Determine how many cockroaches he can expect to find on
day 5 and day 6.
4.1.2 Determine an algebraic formula to calculate how many cockroaches
there are on the nth day.
4.2 Examine the tiling pattern shown in the diagrams.
Stage 1
Stage 2
(2)
(5)
Stage 3
Stage 1
Stage 2
Stage 3
Stage 4
A: Number of striped tiles
3
5
7
9
B: Number of black tiles
1
4
9
16
C: Number of white tiles
2
6
12
20
D: Total number of tiles
6
15
28
45
Stage n
4.2.1 Find the rule for stage n of patterns A, B and C.
4.2.2 Will any stage require 21 striped tiles? Justify your answer with
calculations.
4.2.3 At what stage will a total of 120 tiles be needed to complete
the pattern?
4.2.4 If the dimensions of a tile are 0,5 m × 0,5 m, write a formula
for the total area in m2 of the tiling pattern at any given stage n.
(4)
(2)
(3)
(2)
[18]
Question 5
The sketch shows the parabola g(x) = x2 − 2x − 3 and the straight line f(x) = mx + c.
f(x) and g(x) intersect B and C.
5.1
5.2
5.3
5.4
5.5
5.6
x2
2
Write g(x) = − 2x − 3 in the form g(x) = a(x − p) + q.
Determine the lengths of OC and AB.
Give the coordinates of D, the turning point of g(x).
Find the values of m and c.
State the values of x for which g(x) ≤ f(x).
Give the new equations of the graph of g(x) if:
5.6.1 g(x) is moved two units to the right and 1 unit down
5.6.2 g(x) is reflected in the x-axis.
y
(3)
(4)
(2)
(2)
(2)
f
Q
(3)
(2)
[18]
A
B
O
x
C
R
Question 6
−6 − 1
Consider the function: f(x) = _____
g
P
D
x−3
6.1 Sketch the graph f(x) showing clearly the asymptotes and intercepts
with the axes.
6.2 For which values of x is f(x) > 0?
(5)
(2)
[7]
Total: 100 marks
Exam practice paper 1
PLT MATHS LB 11 7th pgs (Real Book).indb 171
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2012/07/02 2:22 PM
Formal assessment
Examination practice Paper 2
Time: 2 hours
Question 1
In the figure, A(1;−1), B(0;2) and C(6;4) are the vertices of a triangle.
y
C(6;4)
B(0;2)
x
A(1;–1)
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
Determine the gradient of BC
Determine the equation of line BC.
^ =90°.
Prove that B
Write down the coordinates of D if ABCD is a rectangle.
Write down the coordinates of the midpoint of AC.
Determine the equation of the perpendicular bisector of line AC.
^ C.
Find the size of angle BA
Determine the area of the △ABC.
If P (a;−1), B and C are collinear, find the value of a.
___
If Q (−1;b) is a point such that QB = √26 , find the value of b, b > 0
(2)
(2)
(3)
(2)
(2)
(4)
(4)
(4)
(3)
(4)
[30]
Question 2
2 and cos θ > 0, determine without finding the value of θ
2.1 If sin θ = − __
3
and with the aid of a sketch, the value of:
2.1.1 cos θ
2.1.2 tan2 θ × 5sin θ
(3)
(3)
1 , use a sketch to find the following in terms of p:
2.2 If cos 20° = __
p
2.2.1 sin 70°
2.2.2 cos 200°
2.2.3 tan 160°
172
(2)
(2)
(2)
[12]
Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 172
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Question 3
sin 210°.sin 225°
3.1 Evaluate without the use of a calculator: _______________________
2
cos 120° tan 150°cos 315°
sin 130°.cos (360° − θ).cos (−θ)
3.2 Simplify: sin(180° + θ).cos(90° + θ) − __________________________
cos 220°
Question 4
4.1 Prove that: cos A + cos (180°− A).sin2 A = cos3 A
(7)
(7)
[14]
(4)
2
(sin x − cos x)
1−sin x
1 − 2tan x
4.2 Given: _____________
= _____
2
2
cos x
4.2.1 Prove the above identity.
4.2.2 For which values of x in the interval [0°;360°] is the above identity
undefined?
(5)
(3)
[12]
Question 5
You may use calculators in this question; give answers correct to TWO decimal places.
5.1 Solve for x if tan (x – 10°)
= sin 72° and x ∊ [0°; 360°].
(3)
___
√
15
5.2 Solve for x if sin 2x = ____
, –270° ≤ x ≤ 180°.
8
2
5.3 Determine the general solution of 8sin x + 6cos x = 9.
(5)
(7)
[15]
Question 6
6.1 Draw sketch graphs of f(x) = cos (x−30°) and g(x) = sin 3x for −180° ≤ x ≤ 180°.
Label all the turning points and intercepts with the axes.
(7)
6.2 Read from your graphs the solution to the equation cos (x−30°) = sin 3x.
(2)
6.3 Confirm these solutions by solving the equation in 6.2
(6)
6.4 Write down the values of x for which g(x) ≥ f(x).
(2)
[17]
Coordinate Geometry: 30;
Trigonometry: 70
Total: 100 marks
Exam practice paper 2
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2012/07/02 2:22 PM
Term
3
174
PLT MATHS LB 11 7th pgs (Real Book).indb 174
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TOPIC 8
Measurement
Unit 1
Revision
Revision of Grade 10 measurement
TOPIC 9
Euclidean geometry and
measurement
Unit 1
Unit 2
Unit 3
Unit 4
Revision
Geometry revision
Circles, perpendicular lines through
the centre, chords, midpoints
Angle at centre theorem and cyclic
quadrilaterals
Tangents
176
185
188
190
194
202
211
TOPIC 10 Trigonometry: sine, cosine and
area rules
Unit 1
Unit 2
Proof and application of sine, cosine
and area rule
Two-dimensional problems using
sine, cosine and area rules
Revision
214
223
232
TOPIC 11 Finance, growth and decay
Unit 1
Unit 2
Simple and compound decay
The effect of different periods of
compound growth and decay
Revision
236
242
250
TOPIC 12 Probability
Unit 1
Unit 2
Unit 3
Addition and complementary rules;
Dependent and independent events
Venn diagrams
Using tree diagrams to solve problems
for events not necessarily independent
Contingency tables
Unit 4
Revision
Formal assessment: Term 3 Test 1
Term 3 Test 2
Term Summary
252
256
262
267
273
276
278
282
175
PLT MATHS LB 11 7th pgs (Real Book).indb 175
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TOPIC
2
8
Measurement
KEY WORDS
congruent – exactly the same
perpendicular height –
perpendicular distance
between bases
Unit 1: Revision of Grade 10 measurement
This work was covered in Grade 10, but is also examined in Grades 11 and 12.
We have explained the important concepts and summarised the formulae.
Right prisms and right cylinders:
• always have a pair of congruent bases which are parallel to each other
• the bases are perpendicular to the lateral surfaces
• the base of a right prism is a polygon and the base of a right cylinder is a circle.
REMEMBER
Area rectangle
= length × breadth
Area triangle
1 base × perpendicular
= __
2
height
Area circle = πr2
Circumference circle = 2πr
Lateral surface area cylinder
= 2πrh
rectangular prism
triangular prism
right cylinder
In the figures:
• the shaded surfaces are the bases
• the unshaded surfaces are rectangles which make up the lateral surface area
• the height of a right prism is the perpendicular height or perpendicular distance
between the bases
• the lateral surface area of a right cylinder is a rolled-up rectangle with dimensions
2πr and h.
Total surface area = 2 × area of base + sum of areas of all lateral surfaces
REMEMBER
The dimensions of a rectangle
are its length and breadth.
The dimensions of a right
prism are its length, breadth
and height.
The dimensions of a right
cylinder are its radius and
height.
176
Surface area is two-dimensional and if:
• one dimension is multiplied by a constant factor of k, then surface area increases
6x²
by a factor of k.
x
if
both
dimensions
are
multiplied
by
a
constant
factor
of
k,
then
the surface area
2
x²
•
3x²
2
increases by a factor of k .
6x²
2x
x
3x
2x²
3x²
Volume = area of base × perpendicular height
Volume = 6x³
2x
3x
Volume is three-dimensional and if:
• one dimension is multiplied by a constant
Volume = 6x³
factor of k, then the volume increases by a
51x²
factor of k
3
x
two
dimensions
are
multiplied
by
a
constant
•
18x²
factor of k, then the volume increases by a
51x² 27x²
factor of k2
6x
3x
18x²
9x
• three dimensions are multiplied by a
27x²
constant factor of k, then the volume
Volume = 162x³
6x
increases by a factor of k3.
9x
Volume
= 162
x³ effect:
Each dimension has been multiplied by a constant factor of three
with
this
2
• each surface area has increased by a factor of 9, which is 3
• the volume has increased by a factor of 27, which is 33.
Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 176
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Spheres and hemispheres
•
•
•
•
•
•
sphere
KEY WORDS
4 πr 3.
Volume of sphere = __
3
Total surface area sphere = 4πr 2.
2 πr 3.
Volume of hemisphere = __
d
hemisphere – half a sphere
r
3
Total surface area of hemisphere = 3πr 2.
The area of the flat section = πr 2.
The area of the curved section = 2πr 2.
r radius
Right pyramids and right cones
•
•
•
•
•
•
•
•
Only one base
The perpendicular height measures the distance between the vertex of the pyramid
or cone and the centre of its base.
The slant height runs from the side of the base to the top of the pyramid or cone
and is used to determine surface area.
right cone
The base of a right pyramid is a polygon, the base of a
right cone is a circle.
s
1 area of base × perpendicular height
h
Volume = __
3
Total surface area = sum of areas of all surfaces
Area of pyramid = area of base + area of lateral surfaces
Area of cone = πr2 + πr
s
Pythagoras’ Theorem is used in right cones and
right pyramids
•
•
•
•
•
A right-angled triangle is always formed
between h, s and r or h, s and x.
h
s
h is the perpendicular height to the base.
s is the slant height.
r
r is the radius of the circular base of a
right cone.
x is the shortest distance between the
perpendicular height and the edge of the base of a pyramid.
s
h
x
Unit 1 Revision of Grade 10 measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 177
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Summary of formulae:
(h represents the perpendicular height and s the slant height)
Solid
Shape
Right prism
Right pyramid
s
h
Volume
Total surface area
Area of base × h
area of base + areas of lateral surfaces
1 (Area of base) ×
__
3
h
area of base + areas of all triangles
x
Right cylinder
Right cone
h
s
πr2h
2πr2 + 2πrh
1 πr2h
__
3
πr2 + πrs
4 πr3
__
4πr2
2 πr3
__
3πr2 (includes flat circular surface)
r
Sphere
Hemisphere
178
3
r radius
3
Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 178
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Pyramids, right prisms and hemispheres are often combined to form complex shapes.
WORKED EXAMPLE 1
1
2
REMEMBER
K
AB = 29 units, D is the midpoint
A
of CE, DL = 21 units and
29
AK = 90 units.
B
1.1
Determine the area of the
base of the solid in terms of π.
C
D
1.2
Determine the total surface
21
area of the figure.
L
1.3
Determine the volume of the figure.
90
H
Area rectangle = l × b
b⊥h
Area △ = _____
2
J
G
E
If the solid is enlarged so that AB = 58 units, DL = 42 units and
AK = 180 units:
2.1
state the constant factor by which each dimension is multiplied
2.2
state the factor by which each surface area increases
2.3
determine the total surface area of the enlarged solid
2.4
state the factor by which the volume increases
2.5
determine the volume of the enlarged solid.
SOLUTION
Area circle = πr2
πr2
Area semicircle = ___
2
Circumference circle = 2πr
Circumference semicircle:
• πr + 2r, including diameter
• πr, not including diameter
LSA is lateral surface area.
LSA = perimeter of base ×
⊥h
TSA is total surface area.
Effects of multiplying each
dimension by a constant
factor of k:
• TSA increases by a factor
of k2.
• Volume increases by a
factor of k3.
| CD = DL, radii
BG = CE
= 42 units
_______
____
2
2
√
AF = 29 −21 = √400 = 20 | Pythagoras' theorem
1
1.1
Calculations for the area of the base:
1 ( 42 )( 20 ) = 420
Area of triangle = __
2
Area rectangle = (20)(42) = 840
1 πr2 = __
1 π(21)2 = 220,5π
Area semi-circle = __
2
1.3
2.1
2.2
2.3
2.4
2.5
Each dimension is multiplied by a constant factor of 2.
Each surface area increases by a factor of 22 = 4.
TSA = 4(18 663,05) = 74 652,20 units2
The volume increases by a factor of 23 = 8.
The volume = 8(175 744,91) = 11 247 675,24 units3
1.2
2
2
Area of base = 420 + 840 + 220,5π = 1 260 + 220,5π
Area of base = 1 260 + 220,5π
Lateral surface area = 2(29)(90) + 2(20)(90) + π(21)(90)
= 8 820 + 1 890π
Total surface area = 2(1 260 + 220,5π) + 8 820 + 1 890π
= 11 340 + 2 331π
= 18 663,05 units2
Volume = (1 260 + 220,5π) × 90 = 175 744,91 units3
Unit 1 Revision of Grade 10 measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 179
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WORKED EXAMPLE 2
Consider the cylindrical flask in the figure with its
hemispherical top. The radius of the cylinder is 6 cm
and the height of the flask is 25 cm. Give all answers
correct to two decimal places.
1
2
3
6
25
cm3
Determine the volume of the flask in
and in litres.
Determine the total surface area of the flask.
If the dimensions are trebled:
3.1
state the factor of enlargement of the total surface
area and then determine the surface area
3.2
state the factor of enlargement of the volume and
then determine the volume.
SOLUTION
1
The height of the cylinder without the hemispherical top is 19 cm.
2 πr3 = π( 6 )2(19) + __
2 π( 6 )3
V = πr2h + __
3
2
3
3
= 684π + 144π = 828π = 2 601,24 cm3 = 2,60 ℓ
TSA = πr2 + 2πrh + 2πr2 = 3πr2 + 2πrh
= 3π( 36 ) + 2π( 6 )( 19 ) = 336π = 1 055,58 cm2
3.1
The factor of enlargement is 32 = 9 and
A = 9( 336π ) = 3 024π = 9 500,18 cm2.
3.2
The factor of enlargement is 33 = 27 and
V = 27( 828π ) = 22 356π = 70 233,45 cm3 = 70,23 ℓ
WORKED EXAMPLE 3
H = 49 cm, h = 12 cm and r = 9 cm
Give all answers correct to two decimal places.
1
Calculate the volume and the surface area
of the solid. Give the volume in cm3 and
in litres.
2
If the dimensions are increased by a factor
of 1,5 determine the volume and surface
area of the enlarged solid.
180
40 cm
s
h
r
90 cm
H
Topic 8 Measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 180
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SOLUTION
1
REMEMBER
Volume conversions:
1 litre = 1 dm3 = 1 000 cm3
If each of the dimensions is
increased by a factor of k:
• the total surface area
increases by a factor of k2
• the volume increases by a
factor of k3.
If each of the dimensions is
increased by a factor of 1,5:
• the total surface area
increases by a factor of 1,52
• the volume increases by a
factor of 1,53.
Add the volumes of each of the three shapes.
Vcone
1 πr 2h = __
1 π( 9 )2( 12 ) = 324π
= __
3
3
Vcylinder = πr 2h = π( 9 )2( 40 ) = 3 240π
2 πr 3 = __
2 π( 9 )3 = 486π
Vhemisphere = __
3
3
V = 4 050π = 12 723,45cm3 = 12,72 ℓ
2
_______
____
s = √122 + 92 = √225 = 15 cm | Pythagoras’ Theorem
TSA = 2πr2 + 2πrH + πrs
= 2π(9)2 + 2π(9)(40) + π(9)(15)
= 162π + 720π + 135π
= 1 017π cm2 = 3 195 cm2
Volume = ( 1,5 )3( 4 050π ) = 13 668,75π = 42 941,64 cm3 = 42,94 ℓ
TSA = ( 1,5 )2( 1 017π ) = 2 288,25π = 7 188,75 cm2
WORKED EXAMPLE 4
A rectangular pyramid is
combined with a rectangular
right prism. The pyramid has
I
a height of 15 m, which is
indicated by AD. The prism
H
has a height of 4 m and the
length and breadth of both
the prism and the pyramid
are 40 m and 16 m respectively.
A
15
D
B
40
J
G
C
E
4
F
16
1
1.1
1.2
1.3
1.4
1.5
1.6
2
If each dimension is multiplied by a constant factor of five:
2.1
determine the factor by which the volume is increased
2.2
determine the volume of the enlarged solid
2.3
determine the factor by which the total surface area is increased
2.4
determine the total surface area of the enlarged solid.
Determine the volume of the solid.
State the length of BD and then determine the slant height AB.
Determine the area of the triangular face AIJ.
State the length of DC and then determine the slant height AC.
Determine the area of the triangular face AJG.
Determine the total surface area of the solid.
Unit 1 Revision of Grade 10 measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 181
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SOLUTION
REMEMBER
Complex solids are formed by
joining two or more separate
solids. The surface area of a
complex solid is always less
than the total surface areas of
the separate solids.
1
1.1
Volume of solid = Volume of prism + volume of pyramid
1 ( 40 )( 16 )( 15 )
A
V = ( 40 )( 16 )( 4 ) + __
3
= 2 560 + 3 200
= 5 760 m3
1.2
17
1 JG = 8 m
BD = __
2
AB2 = 152 + 82 = 289 | Pythagoras’ Theorem I
____
AB = √ 289 = 17 m
1.3
b × h = _______
40 × 17 = 340 m2
Area △AIJ = _____
1.4
1 IJ = 20 m
DC = __
1.5
= 152 + 202 = 62 | Pythagoras’ Theorem
____
AB = √ 625 = 25 m
J
16
×
25
b
×
h
_______
2
_____
Area △AJG =
=
= 200 m
2
A
2
2
25
AC2
2
2
G
16
2
1.6
TSA =
+
+ 2( 16 )( 4 ) + 2( 340 ) + 2( 200 )
= 640 + 320 + 128 + 680 + 400
= 2 168 m2
2.1
2.2
2.3
2.4
The volume is increased by a factor of 53 = 125.
V = 125( 5 760 ) = 720 000 m3
The TSA is increased by a factor of 52 = 25.
TSA = 25( 2 168 ) = 54 200 m2
( 40 )( 16 )
J
40
2( 40 )( 4 )
EXERCISE 1
Determine the volume and the surface area for each figure.
1
2
5
29 cm
21 cm
12
8 cm
3
4
25 cm
7m
48 m
65 cm
182
48 m
Topic 8 Measurement
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5
6
13
REMEMBER
12
15 cm
39 cm
Vhemisphere = __23πr3
Vcylinder = πr2h
Vcone = __13πr2h
36
TSAhemisphere = 3πr 2
100 cm
39
TSAcylinder = 2πr 2 + 2πrh
TSAcone = πr 2 + πrs
7
T
8
25 m
5m
J
K
24 m
N
6m
O
M
24 m
6m
R
24 m
OM = 24 units
OL = 32 units
OJ = 7 units
LR = 100 units
L
9
10
77
r = 24
The solid above is a regular
octahedron. Each edge is 10 units.
11
15 cm
EXERCISE 2
Where necessary, round off answers to two decimal places.
1.
A right cone has a volume of 3 392,92 units3 and a radius of 9 units.
1.1
Write down the formula for the volume of a cone.
1.2
Determine the height of the cone.
1.3
Determine the slant height of the cone.
1.4
Determine the surface area of the cone.
Unit 1 Revision of Grade 10 measurement
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2012/07/02 2:22 PM
1.5
REMEMBER
Remember that the surface
area of any complex solid
is less than the sum of the
surface areas of the shapes
which combined to form it.
Do not add the areas of any
surfaces at which the two
solids are joined.
• A hemisphere and a cone
that are joined by their flat
circular surfaces will each
lose that area from the
total surface area of the
complex solid.
• A cylinder and a cone
that are joined by their
flat circular surfaces will
each lose that circular
surface area.
• A square-based pyramid
and a right prism will be
joined by the base of the
pyramid so they will each
lose the area of the square
surface which joins them.
184
1.6
1.7
If the radius doubles, but the height remains the same:
1.5.1
describe the effect this will have on the volume of the cone
1.5.2
determine the volume of the cone.
If the height trebles, but the radius remains the same:
1.6.1
describe the effect this will have on the volume of the cone
1.6.2
determine the volume of the cone.
If all the dimensions are halved, determine:
1.7.1
the volume of the cone
1.7.2
the total surface area of the cone.
2.
A cylinder has a volume of 33 238,05 units3 and a radius of 23 units.
2.1
Write down the formula for the volume of a cylinder.
2.2
Determine the height of the cylinder.
2.3
Determine the total surface area of the cylinder.
1 , but the
2.4
If the height of the cylinder is multiplied by __
4
radius remains the same:
2.4.1
say how this will affect the volume
2.4.2
determine the volume.
2.5
If the radius halves, but the height remains constant:
2.5.1
say how this will affect the volume
2.5.2
determine the volume.
1:
2.6
If all the dimensions are multiplied by __
3
2.6.1
say how this will effect the volume
2.6.2
determine the volume
2.6.3
say how this will effect the total surface area
2.6.4
determine the total surface area.
3
A cylinder has a total surface area of 2 777,17 units2 and a radius of 17 units.
3.1
Write down the formula for the total surface area of a cylinder.
3.2
Determine the height of the cylinder.
3.3
Determine the volume of the cylinder.
4.
A sphere has a volume of 333 038,14 units3.
4.1
Write down a formula for the volume of a sphere.
4.2
Determine the radius of the sphere.
4.3
Determine the total surface area of the sphere.
4.4
If the radius of the sphere is doubled:
4.4.1
state the factor by which the volume
is increased
4.4.2
determine the volume
4.4.3
state the factor by which the total surface area is increased
4.4.4
determine the total surface area.
Topic 8 Measurement
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Revision
1
The base of a triangular right prism is an equilateral triangle.
Each side of the equilateral triangle is 15 m and the height of
the prism is 25 m.
E
A
F
B
1.1
1.2
1.3
2
3
15 m
C
D
25 m
Determine the volume of the prism.
Determine the surface area of the prism.
If the dimensions are multiplied by a constant factor of 4:
1.3.1
determine the volume
1.3.2
determine the surface area.
Consider the figure which indicates a 10 m long section of a concrete pipe.
The outer radius is 100 cm and the inner radius is 90 cm.
2.1
Calculate how thick the walls of the pipe are.
2.2
Express both radii in metres.
2.3
Determine the volume of concrete in the pipe, correct to
three decimal places.
2.4
What is the total surface area of the pipe, correct to three
decimal places?
2.5
If the dimensions are doubled:
2.5.1
state the factor of enlargement of the volume
2.5.2
determine the volume
2.5.3
state the factor of enlargement of the total surface area
2.5.4
determine the total surface area.
Consider the figure of a right cone which has a height of 35 m and
a slant height of 37 m.
3.1
Determine the radius of the circle.
3.2
Determine the volume of the cone.
3.3
Determine the surface area of the cone.
3.4
If the height is multiplied by five but the radius remains the same,
state the factor by which the volume will be enlarged.
3.5
If r is multiplied by three but the height remains the same, state
the factor of enlargement of the volume.
(6)
(4)
(3)
(3)
[16]
(2)
(2)
(6)
(6)
(2)
(2)
(2)
(2)
[24]
(3)
(3)
(5)
(2)
(2)
[15]
185
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TOPIC 8: REVISION CONTINUED
4
The rectangular right pyramid has a base with dimensions 96 units by 42 units.
The height of the pyramid is 20 units.
REMEMBER
Vhemisphere = __23πr3
D
Vcylinder = πr2h
Vcone = __13πr2h
20
C
H
F
A
4.1
4.2
4.3
4.4
4.5
E
96
TSAhemisphere = 3πr 2
TSAcylinder = 2πr 2 + 2πrh
G 42
TSAcone = πr 2 + πrs
B
Determine the volume of the pyramid.
(4)
Determine the slant heights DE and DG.
(6)
Determine the lateral surface area of the pyramid.
(5)
Determine the total surface area.
(4)
Determine the factor of enlargement of the volume if the shorter side of
42 units is doubled, but the other dimensions remain constant.
(2)
[21]
5
Determine the volume and surface area of a sphere with a radius of 24 cm.
(6)
6
Determine the volume and surface area of a hemisphere with a diameter
of 40 cm.
(6)
7
Consider a sphere with a volume of 212 174,79 units3
7.1
Write down the formula for the volume of a sphere.
7.2
Use the formula to determine the radius.
7.3
Determine the surface area of the sphere.
7.4
Determine the volume of the sphere if the radius is divided by five.
(1)
(3)
(3)
(3)
[10]
8
120 m
28 m
The diameters of the outer and inner shells of a semi-circular right
prism are 30 m and 28 m respectively. The length of the prism is 120 m.
8.1
Determine the area of the base in terms of π.
8.2
Determine the volume of the prism.
8.3
Determine the total surface area of the prism.
8.4
If the dimensions are halved, determine the total surface area of
the prism.
(4)
(3)
(4)
(4)
[15]
186
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9
10
Consider the right cone which has a volume of 1 005,31 cm3 and
a height of 15 cm.
9.1
Write down the formula for the volume of the cone.
9.2
Determine the radius of the base.
9.3
Determine the slant height of the cone.
9.4
Determine the lateral surface area in terms of π.
9.5
Determine the total surface area.
9.6
Determine the volume of the cone if the dimensions are multiplied
by a constant factor of 2,5.
(3)
[17]
Consider the square-based right pyramid. The volume or the pyramid
is 4 320 units3 and the height is 40 units.
10.1
10.2
10.3
10.4
10.5
10.6
11
(1)
(3)
(3)
(3)
(4)
Write down the formula for the volume of a pyramid.
Determine the dimensions of the base.
Determine the slant height of the pyramid.
Determine the lateral surface area of the pyramid.
Determine the total surface area of the pyramid.
Determine the volume of the pyramid if its height is doubled.
(1)
(3)
(3)
(4)
(3)
(3)
[17]
A cylinder has a volume of 9 424,78 units3 and a height of 30 units.
11.1 Write down the formula for the volume of a cylinder.
11.2 Determine the radius of the cylinder.
11.3 Determine the total surface area of the cylinder.
(1)
(3)
(4)
2 , determine the volume of the cylinder.
11.4 If the radius is multiplied by __
(3)
3
11.5 If all the dimensions are halved, determine the total surface area.
(3)
[14]
187
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2
9
Euclidean geometry and measurement
Unit 1: Geometry revision
Summary
Q
KEY WORDS
P
congruent – identical in all
respects
included angle – angle
between two known sides
hypotenuse – the side
opposite the right angle in a
right-angled triangle
y
T
a b
d c
x
e f
h g
V
z
x + y + z = 180° | ∠ sum △
W
R
|
|
|
|
|
a=c
a + b = 180°
y
vertically opposite ∠s
straight ∠
a=e
corresponding ∠s, PT ∥ VW
d=f
alternate ∠s, PT ∥ VW
d + e = 180°
co-interior ∠s, PT ∥ VW
a + b + c + d = 360° | revolution
REMEMBER
• Supplementary angles all
add up to 180°.
• Complementary angles all
c
B
p
b
a
TOPIC
R
Q
C
c2 = a2 + b2 | Pythagoras’ Theorem
To prove two lines parallel,
show one of the following:
• alternate angles are equal
• corresponding angles are
equal
• co-interior angles are
supplementary.
x = y + z | exterior ∠ of △
A
add up to 90°
• In a revolution, angles add
up to 360°
REMEMBER
x
z
q
P
r
If p2 = q2 + r2 ⇒ P = 90°
| converse Pythagoras’ Theorem
A D
C
B
F
E
△ABC ≡ △DEF (SSS)
Congruent: SSS – three pairs equal sides
J L
M
△GHJ ≡ △KML (SAS)
Congruent: SAS – two pairs equal sides
and included angle
188
R S
T
x
x
H
P
K
G
Q
U
△PQR ≡ △TUS (RHS)
Congruent: RHS – right angle,
hypotenuse, side
B
x
A
D
y
y
x
C
F
E
△ABC ≡ △DEF (AAS)
Congruent: AAS – two sets equal angles
and a pair of corresponding sides equal
Topic 9 Euclidean geometry and measurement
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Quadrilateral properties
A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
•
•
•
•
•
diagonal – a straight line
joining opposite vertices
bisect – to cut or divide into
equal parts
vertex – point where two
straight lines meet to form
an angle
perpendicular – at right
angles or at 90°
symmetry line – a fold line
that makes the shapes the
same on both sides
both pairs of opposite sides are equal
one pair of opposite sides is both equal and parallel
two pairs of opposite angles are equal
both diagonals bisect each other
area = base × perpendicular height
A rectangle is a parallelogram with a right angle.
•
•
•
KEY WORDS
a parallelogram with a right angle
a parallelogram with equal diagonals
area = length × breadth
A rhombus is a parallelogram with a pair of adjacent sides equal.
•
•
•
•
all four sides are equal
diagonals bisect each other at 90°
both diagonals bisect the corner angles
d1 × d2
area = base × perpendicular height or area = _______
2
d1
d2
A square is a rhombus with a right angle.
•
•
•
•
a rectangle with a pair of adjacent sides equal
equal diagonals bisect each other at 90°
both diagonals bisect corner angles into 45° and 45°
d ×d
2
1
2
area = side × side or area = _______
b1
• A rectangle has all
A trapezium is a quadrilateral with one pair of opposite
sides parallel.
•
•
an isosceles trapezium has equal sides which are not parallel
1 h(b + b ) where h is the distance between the
area = __
2
1
2
parallel sides, b1 and b2
A kite is a quadrilateral with two pairs of adjacent sides equal.
•
•
•
one diagonal bisects the other diagonal at 90°
one of the diagonals is a symmetry line
REMEMBER
h
b2
the properties of a
parallelogram as well as its
own special features.
• A square has all
the properties of a
parallelogram, a rectangle
and a rhombus.
• A rhombus has all
the properties of a
parallelogram.
d ×d
2
1
2
area = _______
where d1 and d2 are diagonals
Unit 1 Geometry revision
PLT MATHS LB 11 7th pgs (Real Book).indb 189
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Unit 2: Circles, perpendicular lines through
the centre, chords and midpoints
KEY WORDS
theorem – formal proof of a
geometric statement
construction – a line which
needs to be added to a sketch
chord – line with both
endpoints on the circle
converse – reversing the logic
and proving a theorem in
reverse
⊥ – symbol for perpendicular
radius – line from centre of
circle to circumference
A theorem is a formal proof of a geometric fact. You do not need to prove a
theorem every time you use it, but you can be asked to prove a theorem as a theory
question. Rather than learning each theorem by heart, make sure that you know the
construction and the method.
Theorem 1
The line drawn from the centre of a circle,
perpendicular to a chord, bisects the chord.
Given: Circle with centre O and chord PQ.
OR ⊥ PQ, with R on PQ.
Required to prove: PR = RQ
Construction: OP and OQ
Proof:
In △OPR and △OQR:
^ =R
^ = 90°
| OR ⊥ PQ
1) R
1
2
| radii
2) OP = OQ
3) OR is common
△OPR ≡ △OQR | RHS
| △OPR ≡ △OQR
PR = RQ
⇒ OR bisects PQ.
O
P
1
2
Q
R
Most theorems have a converse. In a converse theorem, the logic is reversed.
Converse (Theorem 1)
The line drawn from the centre of a circle to the midpoint of a chord will be
perpendicular to the chord.
Although you apply the converse theorems in application questions, you will not
prove any of the converse theorems.
WORKED EXAMPLE
OE ⊥ AB, CF = FD, OE = 63 cm, FE = 3 cm, AB = (2x − 18) cm and CD = 2x cm.
Determine the length of the radius.
SOLUTION
OF ⊥ CD | line from centre to midpoint chord
| OE ⊥ AB
EB = x − 9
| CF = FD
FD = x
2
2
2
| Pythagoras in △OFD
OD = 60 + x
C
OB2 = 632 + (x − 9)2 | Pythagoras in △OEB
A
3 600 + x2 = 3 969 + x2 − 18x + 81 | Radii
______
18x = 450 ⇒ x = 25 and radius = √4 225 = 65 cm
190
O
F
E
D
B
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 190
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Theorem 2
The perpendicular bisector of a chord passes through the centre of the circle.
Given: A circle with chord AB.
T is the midpoint of AB and QT ⊥ AB.
Required to prove: QT passes through the centre of the
circle.
Construction: RA and RB where R is any point on QT.
Q
R
Proof:
In △RAT and △RBT
^ =T
^ = 90°
| QT ⊥ AB
1) T
1
2
| given
2) AT = BT
3) RT is common
⇒ △RAT ≡ △RBT | SAS
| △RAT ≡ △RBT
RA = RB
2
1
A
B
T
This proves that any point on QT is equidistant from A and B. The centre of the circle
is also equidistant from A and B, so QT passes through the centre of the circle.
A R
WORKED EXAMPLE
A circle with radius 50 units has chords AB = 60 units
and CD = 28 units. TU and RS are the perpendicular
bisectors of AB and CD respectively. Determine the
shortest distance of each chord from the centre of
the circle.
T
O
U
Q
C
SOLUTION
TU and RS intersect at O, so O is the centre of
the circle. Join OB and OD.
| OP ⊥ AB
| Pythagoras in △OPB
D
S
A R
30
T
P
O
| OQ ⊥ CD
| Pythagoras in △OQD
30
50
B
AB is 40 units from the centre and CD is
48 units from the centre of the circle.
equidistant – the same
distance
intersect – to meet or cross
at a point
P
B
AP = PB = 30
OP2 = 502 − 302 = 1 600
OP = 40 units
CQ = QD = 14
OQ2 = 502 − 142 = 2 304
OQ = 48 units
KEY WORDS
C
14
U
50
Q
14
S
D
Reminders about Pythagoras’ Theorem
In a right-angled triangle, the square on the hypotenuse is equal to the sum of the
squares on the other two sides. If you know all three sides of a triangle but do not
know whether the triangle is right-angled or not, use the converse of Pythagoras’
Theorem. If the square on the longer side is equal to the sum of the squares on the
two shorter sides, then the triangle is right-angled and the right angle will lie between
the two shorter sides.
Unit 2 Circles, perpendicular lines through the centre, chords and midpoints
PLT MATHS LB 11 7th pgs (Real Book).indb 191
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A triangle with sides 20, 21 and 29 is right-angled because 202 + 212 = 841 and
292 = 841, but a triangle with sides 6, 7 and 8 is not right-angled because
62 + 72 = 85 but 82 = 64 and 85 ≠ 64.
WORKED EXAMPLES
REMEMBER
To prove two lines parallel,
show one of the following:
• alternate angles are equal
• corresponding angles are
equal
• co-interior angles are
supplementary.
D
A, B, C and D are points on the circumference
of the circle with centre O.
AB = 40 units, OE = 15 units and OF = 7 units.
1
Determine the radius of the circle.
2
Determine the length of the chord DC.
F
A
2
AE = EB = 20 units
OA2 = 202 + 152 = 625
OA = 25 units
The radius is 25 units
DF2 = 252−72 = 576
DF = 24 units = FC
DC = 48 units
B
E
SOLUTIONS
1
C
O
D
| OE ⊥ AB
| Pythagoras in △AOE
F
C
O
| Pythagoras in ∠DOF
| OF ⊥ DC
A
B
E
EXERCISE 1
O is the centre of the circle. Sketches are not drawn to scale. Give reasons unless you
are told otherwise.
D
1
DC is perpendicular to AB and cuts AB at E.
DE = 18 units and AB = 24 units.
1.1
Determine, with reasons, the length of
AE.
O
1.2
If the radius of the circle is x units,
express OE
in terms of x.
B
A
E
1.3
Calculate, with reasons, the value of x.
C
2
2.1
2.2
192
If RS = 48 units, OP = 7 units and
TU = 40 units.
2.1.1
Determine the radius of the circle.
2.1.2
Determine the length of OV.
If OP = 16 units, OV = 25 units and the circle has a
radius of 65 units, show that the sum of the chords
is 246 units.
R
P
O
T
V
S
U
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 192
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3
OE ⊥ AB, EB = 7 units, OE = 24 units and
DF = FC = 15 units.
Determine, with reasons:
3.1
AO
3.2
FO
3.3
FG
F
O
5
6
C
15
24
A
4
G
D
E
A circle with radius 65 units has two chords, PQ
and RT. PQ is 104 units in length and RT is
P
120 units in length.
4.1
Copy the sketch and show how you can
locate the centre of the circle by adding two
construction lines, one passing through chord
PQ and the other through chord TR.
Q
Name the centre of the circle O.
4.2
Determine the distance between the centre of the circle and:
4.2.1
chord PQ
4.2.2
chord TR.
A
AE = x units, CF = x − 4 units, EO = 7 units and
EF = 22 units. EOF is a straight line.
5.1
Prove that AB ∥ CD.
5.2
State the length of OF.
E
O
5.3
Determine AO in terms of x.
5.4
Determine CO in terms of x.
5.5
Why is AO = CO?
5.6
Solve for x.
B
5.7
State the lengths of AB and CD.
5.8
Determine the area of ABDC.
O is the centre of two concentric circles. The radii of these
circles are 65 units and 34 units. OG ⊥ AB and CD = 60 units.
6.1
Determine the length of:
6.1.1
CG
6.1.2
OG
6.1.3
AB
6.1.4
AC
^ F = θ:
6.2
If OB
^ in terms of θ.
6.2.1
determine E
2
6.2.2
prove that EH ∥ FB.
6.3
Determine the length of FB, in simplified surd form.
A
6.4
Join AO and AF.
6.4.1
Prove that AF = BF
6.4.2
What type of quadrilateral is AFBO?
6.4.3
Justify your answer to 6.4.1 in two different ways.
6.4.4
Calculate the area of AFBO.
B
7
T
KEY WORDS
concentric – circles that share
the same centre
R
REMEMBER
C
b1
F
h
b2
D
Area of trapezium
1 h(b + b )
= __
1
2
2
O
3
C
G 2
H
1
D
0
B
E
F
Unit 2 Circles, perpendicular lines through the centre, chords and midpoints
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Unit 3: Angle at centre theorem
and cyclic quadrilaterals
Theorem 3
The angle subtended by an arc at the centre of the circle is double the size of the
angle subtended by the same arc at any point on the circumference of the circle.
KEY WORDS
C
subtended by an arc –
formed by arc
arc – a part of the
circumference of a circle
reflex angle – angle greater
than 180°
C
1 2
1
A
1 2
2
Figure 1
B
O
1 2
O
O
A
1 2
1
C
2
A
B
B
Figure 2
Figure 3
Given: A, B and C are three points on the circle with centre O.
Construction: CO
^ B = 2AC
^ B, but in Figure 2 prove reflex
Required to prove: In Figures 1 and 3 prove AO
^ B = 2AC
^ B.
AO
Proof:
Figure 1 and Figure 2:
^ = x and C
^ =y
Let C
1
2
^ = x and B
^=y
| radii
⇒A
^
| ∠ sum △AOC
AOC = 180° − 2x
^ C = 180° − 2y
| ∠ sum △BOC
BO
^
^
^
|
AOB + O1 + O2 = 360° sum of ∠s around a point
^ B + 180° −2x + 180° – 2y = 360°
∴ AO
^ B = 2(x + y) = 2AC
^B
| sum of ∠s around a point
∴ AO
^
^
Figure 3: Let C1 = x and C2 = y
^ = x and B
^=x+y
| radii
A
^
| ∠ sum of △AOC
AOC = 180° − 2x
^
| ∠ sum of △BOC
BOC = 180° − (2x + 2y)
^ B = AO
^ C − BO
^ C = 2y = 2AC
^B
AO
Note that in Figure
^ B refers to the
2 AO
reflex angle.
V
The angle subtended by an arc at the centre of the circle is double the size of the angle
subtended by the same arc at any point on the circumference of the circle.
This theorem does not have a converse.
O
R
Theorem 3 proves that the angle subtended by an arc at the centre of the circle
is twice the size of the angle subtended by the same arc at any point on the
circumference of the circle.
P
T
S
If the angle at the circumference of a circle is half the size of an angle subtended by
the same arc at a point inside the circle, it is not necessarily true that this point will be
the centre of the circle.
In the figure alongside, both O and P lie on the circumference of the smaller circle
194
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 194
2012/07/02 2:23 PM
which passes through the centre of the larger circle and intersects the larger circle at R
and S.
^ =P
^
O
| ∠s on chord RS of the smaller circle
1O
^ | ∠ at centre of larger circle
^ = __
V
2
^ =P
^)
|O
1P
^
V = __
2
But, P is not the centre of the circle.
Although it is always true that the angle subtended by an arc at the centre of a circle is
twice the angle subtended by the same arc on the circumference, it is not necessarily
true that the converse applies.
REMEMBER
≡ is the symbol for
congruent.
Congruent means identical in
all respects.
Theorem 4
In Theorem 3 we proved that the angle subtended by an arc at the centre of the
circle is double the size of the angle subtended by the same arc at any point on the
circumference of the circle.
Using this result we can prove that the angles subtended by a chord of the circle, on
G
the same side of the chord, are equal.
x
F
x
F
O
2x
A
x
G
B
O
x
2x
B
A
F
Given: A, F, G and B are points on the circle with centre O.
Construction: Join AO and OB.
^
^=G
Required to prove: F
Proof:
^ B = 2F
^ | ∠ at centre
AO
^ B = 2G
^ | ∠ at centre
AO
^
^
F=G
The corollary of Theorem 4 is that equal chords
subtend equal angles.
U
In △OPQ and △OSR:
1) OP = OS | radii
2) OQ = OR | radii
3) PQ = RS | given
⇒ △OPQ ≡ △OSR | SSS
^ Q = RO
^ S | △OPQ ≡ △OSR
PO
^ Q = 2U
^ S = 2T
^ and RO
^ | ∠ at centre
PO
^
^
U=T
x
KEY WORDS
O
2x 2x
P
T
corollary – deduction based
on the result of a theorem
x
S
Q
R
Unit 3 Angle at centre theorem and cyclic quadrilaterals
PLT MATHS LB 11 7th pgs (Real Book).indb 195
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WORKED EXAMPLE
O is the centre of the circle and A, B, C, D
and E lie on the circle.
^ D = 21° and AB
^ E = 60°.
BA
Determine, with reasons, v, w, x, y and z.
E
A
SOLUTION
v = 21° |
w = 42° |
x = 318° |
y = 159° |
z = 60° |
x
O
w
21°
∠s on chord BD
∠ at centre
revolution
∠ at centre
∠s on chord AE
v
60
O
z
y
B
D
C
EXERCISE 2
O is the centre of the circle in each figure. In each case determine the value of the
lower case letters a, b, c, … . Give reasons for each statement.
1
3
2
N
32°
a
h
J
4
V
g
c
b
Q
N
e
O
22°
70°
f
d
R
T
E
21° u
t
g
g
b
T
A
O
d
N
O
j i
cd
5
27°
e
R
a b
32° a
M
M
M
f
a
Q
f e
Ob
c
K
P
c
R
B
y
D
x v
40°
z
w
C
P
196
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 196
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Theorem 5
KEY WORDS
The opposite angles of a cyclic quadrilateral are supplementary.
F
F
D
E
G
2
O 1
cyclic quadrilateral – a
four-sided figure with all
four vertices lying on the
circumference of a circle
supplementary – add up to
180°
revolution – the sum of the
angles around a point
produce – lengthen a
straight line; for example, AB
produced to C means that
ABC is a straight line
exterior angle – outside
angle of a polygon formed
by a line which has been
extended
interior angle – inside angle
E
2
G
1
O
D
Figure 1
Figure 2
Given: D, E, F and G are 4 points on the circle with centre O.
Construction: EO and GO.
^ F = 180°
^ +F
^ = 180° and DE
^ F + DG
Required to prove: D
Proof: (The proof is the same for both figures.)
^ =x
Let D
^
| ∠ at centre
O2 = 2x
^ = 360° − 2x
| revolution
O
1
^
| ∠ at centre
F = 180° − x
^
^
D + F = 180°
^ F = 180° | ∠ sum quad
^ D + DG
FE
Converse (Theorem 5)
A quadrilateral is cyclic if its opposite angles are supplementary.
Corollary: The exterior angle of a cyclic quadrilateral is equal to the interior
opposite angle.
REMEMBER
WORKED EXAMPLE
A
A, B, C and D lie on the circle and BC is
^ E = 100°.
^ = 85° and DC
produced to E. B
Determine, with reasons, x, y and z.
D
z
• Angles in any triangle add
up to 180°.
x
• Angles in any quadrilateral
add up to 360°.
• Angles around any point
100°
85°
B
SOLUTIONS
x + 85° = 180°
⇒ x = 95°
y = 80°
z + 80° = 180°
⇒ z = 100°
or z = 100°
| opposite ∠s cyclic quadrilateral
| straight ∠
| opposite ∠s cyclic quadrilateral
y
C
E
add up to 360°.
• Exterior angle of triangle
= sum of two interior
opposite angles.
• Exterior angle of cyclic
quadrilateral = one interior
opposite angle.
| exterior ∠ cyclic quadrilateral
Unit 3 Angle at centre theorem and cyclic quadrilaterals
PLT MATHS LB 11 7th pgs (Real Book).indb 197
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2012/07/02 2:23 PM
D
y
There are three ways to prove that a quadrilateral is cyclic.
1
If a line subtends equal angles at two different points on the A
x
same side of the line, then the four points are cyclic.
If x = y, then ABCD is a cyclic quadrilateral.
C
B
2
T
x
If one pair of opposite angles of a quadrilateral is
supplementary, then the quadrilateral is cyclic.
If x + y = 180°, then TUVW is a cyclic quadrilateral.
U
3
W
y
V
If the exterior angle of a quadrilateral equals the interior
opposite angle, then the quadrilateral is cyclic.
If x = y, then PQRT is a cyclic quadrilateral.
T
y
P
x
V
Q
R
EXERCISE 3
1
REMEMBER
To prove two lines parallel,
show one of the following:
• alternate angles are equal
• corresponding angles are
equal
• co-interior angles are
supplementary
Four points are concyclic if
they lie on the same line.
198
Complete the statements by filling in the missing words:
1.1
The line drawn from the centre of a circle, perpendicular to a chord, …
1.2
The line drawn from the centre of a circle to the midpoint of a chord …
1.3
If PQ is the perpendicular bisector of chord AB, then PQ passes through …
1.4
If PQ and JK are the perpendicular bisectors of any two non-parallel chords
on the same circle, then PQ and JK will intersect each other and the centre
of that circle will lie on their …
1.5
The angle subtended by a chord at the centre of the circle is …
1.6
The angles subtended by a chord in the same segment of the circle …
1.7
The angle subtended by a diameter on the circumference of the circle is
always equal to …
1.8
If a chord subtends a right angle on the circumference of a circle, then the
chord is …
1.9
The opposite angles of a cyclic quadrilateral …
1.10 If the opposite angles of a quadrilateral are supplementary, then …
1.11 If a line subtends equal angles at two points on the same side of itself,
then …
1.12 If a line subtends unequal angles at two points on the same side of itself,
then …
1.13 If the exterior angle of a quadrilateral is equal to the interior opposite
angle, then …
1.14 If the exterior angle of a quadrilateral is not equal to the interior opposite
angle, then …
1.15 Equal chords subtend …
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 198
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2
A, B, C, D, E and F all lie on the circle.
F
^ E = 17°, BC
^ E = 86° and AE
^ C = 57°
DC
u
2.1
Is AE ∥ BC? Justify your answer.
2.2
Complete the statements:
A x
2.2.1
t = …°
(∠s on chord ED)
^ C = FE
^ C (…)
2.2.2
FD
y
^
2.2.3
BAE = …
(∠s on chord BE)
^ F = ...
2.2.4
AE
(∠s on chord AF)
v
B
2.3
Determine u, v, w, x, y and z in that order.
1
2
E
57°
w
D
17°
86°
C
REMEMBER
WORKED EXAMPLE
E
A
CD is a diameter of the circle with centre O.
^ E and DG ⊥ AB.
FD bisects CD
F
1
2
1 3
3
4
5
Prove that DEFG is a cyclic quadrilateral.
^ F = x, determine, with reasons,
If ED
C
three other angles equal to x.
Prove that EFGD is a kite.
If AB = 30 cm and CG = 9 cm,
determine the length of the radius.
Determine the length of EC.
(Leave your answer in simplified
surd form).
Equal chords subtend equal
angles.
Conversely, equal angles are
subtended by equal chords.
2
1
3
12
2
G
4
x
O
1
D
B
SOLUTIONS
1
2
3
^ D = 90°
| ∠ on diameter
CE
^
| given
OGB = 90°
DEFG is a cyclic quadrilateral | exterior ∠ = int opp ∠
(Do not use the word ‘cyclic’ in your reason when proving that a quadrilateral
is cyclic.)
^ = x | FD bisects CD
^E
D
1
^ = x | ∠s on chord FG
E
1
^ = x | ∠s on chord EF
G
2
EF = FG
^ = 90° − x
E
2
^ = 90° − x
G
3
ED = EG
DEFG is a kite
|
|
|
|
|
^ F = FD
^ G and EF and FG are chords on the circle EFGD
ED
adjacent complementary ∠s
straight ∠
^ opposite ∠s △DEG
^E = G
2
3
two pairs of adjacent sides equal
Unit 3 Angle at centre theorem and cyclic quadrilaterals
PLT MATHS LB 11 7th pgs (Real Book).indb 199
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4
AG = GB = 15 cm | OG ⊥ AB
Let the radius be k, then OG = x − 9
k2 = 152 + (k − 9)2 | Pythagoras in △OGB
k2 = 225 + k2 − 18k + 81
18k = 306° ⇒ k = 17
The radius is 17 cm
5
| diameter
CD = 34cm
GD = 25cm = ED | proved in 3
| Pythagoras in △ECD
EC2 = 342−252
____
___
EC = √531 = 3√ 59 cm
EXERCISE 4
1
A, B, C, D, E, F, G and H all lie on the circumference of the circle. Determine,
with reasons, w, x, y and z.
H
E
A
x
w
z
88°
B
y
C
2
D
79°
G
84°
F
Two circles intersect at W and Z. U and V lie on the smaller circle, Y and X lie on
^ = 101° and Y
^ = 82°.
the larger circle. V
Y
Z
U
d
101°
V
a
f
c
e
W
2.1
2.2
2.3
2.4
200
82°
b
X
Determine, with reasons, a, b, c, d, e and f.
Is UY ∥ VX? Justify your answer.
Is UV ∥ YX? Justify your answer.
Is UVXY a cyclic quadrilateral? Justify your answer.
Topic 9 Euclidean geometry and measurement
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3
^ F. BD and AE
A, B, C, D and E lie on the circle. AB = AF and DE bisects CE
^ = x.
produced meet at F and F
A
E
3
2
1
1
x
F
2
D
K
1
2
B
C
3.1
3.2
3.3
3.4
3.5
3.6
3.7
4
Name two cyclic quadrilaterals.
Prove that DE = DF.
^ C.
Prove that BF bisects AB
^
^
Prove that ABC = C.
Is ABCF a cyclic quadrilateral? Justify your answer.
Is ABKE a cyclic quadrilateral? Justify your answer.
Is AF ∥ BC? Justify your answer.
A, B, C and D are points on the circumference of the circle. COE is a straight line
^ = x.
such that E lies on AB. DC = BC and C
1
D
1
A
1 F
E 2
1
2
2 3
1 2
O
x
2
1
C
B
4.1
4.2
4.3
4.4
4.5
^ in terms of x.
^ and D
Determine the size of B
2
2
^ D.
Show that EC bisects BC
Prove that DF = FB.
Show that AEOD is a cyclic quadrilateral.
Draw ED and prove that EBOD is a kite.
Unit 3 Angle at centre theorem and cyclic quadrilaterals
PLT MATHS LB 11 7th pgs (Real Book).indb 201
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Unit 4: Tangents
Theorem 6
Two tangents drawn to a circle from the same point outside the circle are equal
in length.
KEY WORD
tangent – a straight line
which touches a circle, but
does not pass through it
Given: Point R lies outside the circle.
RP and RQ are tangents to the circle
with centre O.
Required to prove: RP = RQ
Construction: OP, OQ and OR
P
O
R
Proof:
In △RPO and △RQO
| radii
1) OP = OQ
^
^
| radius ⊥ tangent
2) P = Q = 90°
Q
3) OR is common
△RPO ≡ △RQO | RHS
⇒ RP = RQ
Theorem 7
REMEMBER
A
•
•
•
•
•
B
The shortest distance
between a point and a
line is the perpendicular
distance.
A radius is always
perpendicular to a tangent
at the point of contact.
A chord divides a circle into
two segments.
In the figure above, the red
segment is the alternate
segment.
Complementary angles add
up to 90°.
The angle between the tangent to a circle and the chord drawn from the point of
contact is equal to the angle subtended by the chord in the alternate segment.
Given: A, B and C are points on the circle with
centre O. DA is a tangent to the circle at A.
^D = C
^ and in Figure
Required to prove: In Figure 1, BA
^D = C
^
2, reflex BA
C
0
Construction: OA and OB
2
Proof:
^D = C
^
Figure 1: BA
^ =x
Let C
| ∠ at centre
O = 2x
^ =B
^ = 90° − x | ∠ sum isosceles △AOB, radii
A
2
1
^ =x=C
^
| Radius ⊥ tangent
A
1
^D = C
^
Figure 2: BA
^
Let C = x
^ = 2x
|
Reflex O
1
^
|
O2 = 360° − 2x
^ =B
^ = x − 90° |
A
2
1
^ = 90°
|
A
1
^D = x = C
^
BA
1
1
D
A
Figure 1
B
1
O
2
1
∠ at centre
Sum of ∠s around a point
∠ sum isosceles △AOB, radii
Radius ⊥ tangent
B
1
D
C
2
A
Figure 2
Converse (Theorem 7)
If the angle between a line and a chord equals
the angle subtended by the chord in the alternate segment, then the line is a tangent
to the circle.
202
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 202
2012/07/02 2:23 PM
Circle geometry summary with reasons
O
A
O
B
K
A
AK = KB | OK ⊥ AB
OR ⊥ AB
| line from centre to midpoint chord
E
F
A
B
0
C
^=F
^ | Equal chords AB & CD
E
D
B
Q
0
D
C
^ C = 2A
^
| ∠ at centre
BO
^ =D
^
| ∠s on chord BC
A
^
^
Reflex O = 2R | ∠ at centre
D
G
O
R
P
D
A
E
B
R
E
^ = 90° | ∠ on diameter
D
G
F
^ = 180°
^+G
E
| opposite ∠s cyclic quadrilateral
D
G
E
F
O
H
^H = D
^ | exterior ∠ cyclic quadrilateral
GF
A
^ C = 90°
OB
B
C
| Radius ⊥ tangent
D
F
G
D
E
E
DEFG is a cyclic quadrilateral
| DE subtends equal ∠s at F & G
A
1
2
3
B
C
^ =E
^ | tan AB, chord BD
B
1
^ =D
^ | tan BC, chord BE
B
3
Unit 4 Tangents
PLT MATHS LB 11 7th pgs (Real Book).indb 203
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2012/07/02 2:23 PM
T
M
A
x
P
180°– x
N
B
V
PA = PB | tangents from common point P
MNVT is a cylic quadrilateral
| opposite ∠s supplementary
A
C
A
D
D
B
^ B = 90°
AB is the diameter | AC
B
C
E
ABCD is a cyclic quadrilateral
^ E = interior opposite A
^
| exterior DC
WORKED EXAMPLE
PQ and RQ are tangents to the circle at V and
^ = 69°
W respectively. T
1
2
3
Prove that OVQW is a kite.
Prove that OVQW is a cyclic quadrilateral.
Determine, with reasons, the sizes of a, b,
c, d, e, f, g and h.
P
R
T
69°
O
e a
f
SOLUTION
1
2
3
204
OV = OW | radii
V
VQ = WQ | tangents from common
point Q
OVQW is a kite | two pairs of adjacent sides equal
g
d
W
U
c
b
h
^ Q = 90° | radius ⊥ tangent
OW
^
OVQ = 90° | radius ⊥ tangent
^ Q + OV
^ Q = 180°
⇒ OW
OVQW is a cyclic quadrialteral | opposite ∠s supplementary
a = 138°
b = 42°
c = 69°
d = 69°
e = 69°
f = 90°
g = 21°
h = 21°
|
|
|
|
|
|
|
|
Q
∠ at centre
opposite ∠s cyclic quadrilateral OVQW
tangent VQ, chord VW
tangent WQ, chord VW or equal tangents from common point Q
∠s on chord VQ or OQ symmetry line of kite OVQW
diagonals kite OVQW
radius ⊥ tangent
∠s on chord VO
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 204
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EXERCISE 5
KEY WORDS
O is the centre of the circle in each figure. In each case determine the value of
a, b, c, ... . Clearly state your reasons.
1
2
D
F
u
O
27°
z
31°
v
3
P
52ºf
x C
w y
B
Q
d
g V
0 a
ee
c
t
P
S
x
z
T
Q
b
70 º
R
50º
A
T
E
opposite sides – AD and
BC are opposite sides (no
common vertex)
adjacent sides – AD and DC
are adjacent sides (common
vertex D)
^ and D
^
opposite angles – B
are opposite angles
symmetry line – fold line
A
y w
R
D
U
S
C
B
WORKED EXAMPLE
ABED is a cyclic quadrilateral. FDCG is
a tangent to the circle at D. BEC is a
^ E = x.
straight line. AD ∥ BC, AD = AE and AD
1
2
3
E
REMEMBER
A
Write down, with reasons, five
other angles equal to x.
Prove that ABCD is a parallelogram.
Prove that ABED is an isosceles trapezium.
SOLUTIONS
1
2
3
^ =x
E
2
^ =x
D
1
^ =x
C
1
^E = x
3
^D = x
BA
1
2
B
|
|
|
|
|
isosceles △AED, AE = AD
tangent FD, chord AD
corresponding ∠s, AD ∥ BC
alternate ∠s, AD ∥ BC
exterior ∠ cyclic quadrilateral ABED
To prove a cyclic quadrilateral,
show one of the following:
• opposite angles are
supplementary
• a line subtends equal
angles on the same side
• an exterior angle equals an
interior opposite angle.
1
x
1
2
E
3
2
D
1
2 C
G
^D = D
^ =x
| proved in 1
BA
1
| alternate ∠s equal
AB∥FG
ABCD is a parallelgram | AB ∥ DC and AD ∥ BC
^ =x
^ =C
|E
DC = DE
3
1
| opposite sides parallelogram
DC = AB
⇒ AB = DE
ABED is an isosceles trapezium | AD ∥ BE and AB = DE
Remember to think about what you have already done as you start each new
question. In the Worked example above, Questions 2 and 3 both used the solution
from the previous question.
Unit 4 Tangents
PLT MATHS LB 11 7th pgs (Real Book).indb 205
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2012/07/02 2:23 PM
EXERCISE 6
1
REMEMBER
Complete the statements by filling in the missing words:
1.1
Tangents drawn to a circle from a common point outside the circle are …
1.2
The radius of any circle always meets a tangent to the circle at … at the
point of contact.
1.3
The angle between a tangent and a chord is … to the angle subtended
by the same chord in the alternate segment.
1.4
If the angle between a line and a chord is equal to the angle subtended
by the chord in the alternate segment, then the line is …
2
A
To prove two lines parallel,
show one of the following:
• alternate angles are equal
• corresponding angles are
equal
• co-interior angles are
supplementary.
54°
v
B 80° y
G
E
u
x
F
t
z w
C
D
Two circles touch at C. DE is a tangent to both circles at C. A and B lie on
^ = 54°; B
^ = 80°.
the larger circle, F and G lie on the smaller circle. A
2.1
Determine, with reasons, t, u, v, w, x, y and z.
2.2
Is DE ∥ BA? Justify your answer.
2.3
Is BA ∥ FG? Justify your answer.
3
N
60°
3x
J
x
K
M
2x
L
^ N = 3x, NK
^ M = x, MK
^ L = 2x and N
^ = 60°. M, K and N are points on the circle.
JK
JKL is a straight line.
3.1
Determine the value of x.
3.2
Prove that:
3.2.1
NK is a diameter of the circle
3.2.2
JKL is a tangent to the circle at K.
206
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 206
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4
5
^ F = 35°, OH
^ G = 31° and
DF is a tangent to the circle at E. KE
EK = HK. O is the centre of the circle. Determine:
^
4.1
E
4
^
4.2
K
^H
4.3
DE
^
4.4
G
^
4.5
O
1
^D
4.6
OE
^
4.7
E1
4.8
What type of quadrilateral is EOHK? Justify your answer.
4.9
Prove that OK perpendicularly bisects EH.
4.10 If KO is produced, will it pass through G? Justify your answer.
F
E
D
1
35°
2 3
O
4
K
1
2
31°
H
G
PN is the diameter of the circle with NRM a tangent to the circle
^ = x. O is the centre of
at N. P, N, T and Q lie on the circle and M
the circle.
5.1
Name four angles equal to 90°.
5.2
Determine two other angles equal to x.
5.3
Prove that MQTR is a cyclic quadrilateral.
5.4
Is PNRQ a cyclic quadrilateral? Justify your answer.
M
x
Q
3
1
2
2
P
1 R
2
1
T
1
2
6
3
2
O
1
3 N
G
EF is a tangent to the circle at C. CDG is a straight line. A, B, C and D are points
^ E = 53°, CB
^ D = 60° and AD
^ G = 105°.
on the circle. BC
6.1
Is BD ∥ EF? Justify your answer.
6.2
Complete the following statements by filling in the
A
missing words:
w
6.2.1
The angle between a tangent and a chord is …
x
6.2.2
The opposite angles of a cyclic quadrilateral are …
B
60°
6.2.3
The exterior angle of a cyclic quadrilateral is …
6.3
Determine angles p, q, v, w, x and y in that order.
Give reasons for your answer.
105° D
y
p
v
q
53°
F
C
E
Unit 4 Tangents
PLT MATHS LB 11 7th pgs (Real Book).indb 207
207
2012/07/02 2:23 PM
7
8
A, B, C and D lie on the circle. PCQ is a tangent,
^ = 18° and C
^ = 43°.
E is the midpoint of AC, A
2
1
O is the centre of the circle.
7.1
Determine the size of:
^
7.1.1
A
3
^
7.1.2
C
3
^C
7.1.3
AO
^
7.1.4
D
^A
7.1.5
PC
^C
7.1.6
AB
^ =O
^
7.2
Prove that: O
1
2
^ .
7.3
Determine the size of C
5
AC and BC are tangents to the circle. A, B,
^ C = 30°.
P and Q lie on the circle and AP
O is the centre of the circle.
8.1
Prove that AOBC is a kite.
8.2
Prove that AOBC is a cyclic
quadrilateral.
^ Q, O
^ ,
8.3
Determine the size of PA
P
2
^ B and AQ
^ B.
AC
8.4
Prove that APBC is a rhombus.
A
1 23
B
O
E
1
2
2
D
5
P
4
3
2
1
Q
C
A
1
2 3
Q1
30°
1
1 2
3
O 4
1
2
1
C
2
4 3
2 3
B
9
S
O
x
2
1 2 3
T 1 3
4
V
1
2
1
P
2
3 4
Q
R
SV is a diameter of the circle, with centre O. SV is extended to P. PR is a tangent
to the circle at Q. RO intersects QS at T. VQ ∥ OR and ^S = x.
9.1
Write down, with reasons, three other angles equal to x.
9.2
Prove that QRSO is a cyclic quadrilateral.
^ S in terms of x.
9.3
Write down, with reasons, the size of PQ
9.4
Prove that T is the midpoint of QS.
208
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 208
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EXERCISE 7
This is a mixed exercise. O is the centre of each circle.
1
P
10°
O
T
d e
a
f
c
Q
V
b
g
S
R
P, Q, R, S and T lie on the circumference of the circle and OR ⊥ QS.
1.1
Prove that OQRS is a kite.
1.2
Determine the size of a, b, c, d, e, f and g.
1.3
If QS = 48 units and VR = 2 units, determine the length of OQ, the radius
of the circle.
2
U
P
i g
d
31°
Q
α
20°
mO
l
c
V x k
y z f
n
j
h b
R
T
e
w
S
^ U = 31° and
QR produced, meets tangent US at S. SU touches the circle at T. PT
^ T = 20°.
RQ
2.1
Determine the size of:
^T
2.1.1
QP
^T
2.1.2
QR
^
2.1.3
QTU
2.2
Determine the angles a, b, c, d, e, f, g, h, i, j, k, l, m, n, w, x, y and z
in that order.
Unit 4 Tangents
PLT MATHS LB 11 7th pgs (Real Book).indb 209
209
2012/07/02 2:23 PM
3
4
PR is a diameter of the circle with centre O.
PR and chord SQ intersect at T. SR, QR and
PS are chords on the circle. OQ is joined
and RT ⊥ SQ.
3.1
Prove that PQRS is a kite.
^ = 29°, calculate, with reasons,
3.2
If P
1
the size of:
^
3.2.1
R
1
3.2.2
Q2
^
3.2.3
O
1
S
1
P
ABC is a tangent to the circle at B. E is the
midpoint of DF.
4.1
Prove that BOEC is a cyclic quadrilateral.
4.2
Prove that OC2 = OD2 + EC2 − EF2
4.3
Calculate BC if DF = 80 units, OB = 50 units
and DC = 10 units.
T1 2
O
1
2
2
4 3
2 1
1
2
1
R
1 23
Q
A
C
B
D
E
O
F
5
210
O is the centre of two concentric circles.
The radii of these circles are 37 units and 13 units.
OH ⊥ DE and OH = 12 units. KE is a tangent to
the smaller circle and touches it at M.
5.1
How far apart are the circumferences of
the circles?
5.2
Determine the length of:
5.2.1
FH
D
F
5.2.2
DH
5.2.3
DF
5.3
State the lengths of OM and OE.
5.4
Determine the length of KE, in simplified surd form.
5.5
What type of quadrilateral is OHEM?
K
M
O
H
G
E
Topic 9 Euclidean geometry and measurement
PLT MATHS LB 11 7th pgs (Real Book).indb 210
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Revision
1
O
A
D
B
C
O is the centre of the circle.
1.1
If AB = 54 units and DC = 3 units, determine with reasons,
the length of the radius.
1.2
What type of quadrilateral is OBCA? Fully justify your answer.
^ B in terms of x.
^ C = x, determine with reasons, the size of AO
1.3
If AB
(5)
(3)
(4)
[12]
2
O
F
A
B
9
C
E
D
AB and CD are parallel chords of a circle with centre O. AB is 40 units,
CD is 14 units and the distance between AB and CD is 9 units. If OF is
x units, determine, with reasons:
^A
2.1
OF
2.2
an expression, in terms of x,
2.2.1
for the radius OA.
2.2.2
for the radius OC.
2.3
Use the results from 2.2 to determine the length of the radius.
3
O is the centre of the circle, Q, R, T and P lie
on the circle and TOP is a straight line.
^ Q = 2x + 8° and R
^ = 4x − 14°.
PO
Determine, with reasons, the value of x.
P
2x = 8°
(2)
(2)
(2)
(5)
[11]
O
T
4x – 14°
Q
R
[4]
211
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TOPIC 9: REVISION CONTINUED
4
P
REMEMBER
1
2
______
1 2
T
Q
1
2
_______
___
√ 32 + 42 = √ ( 9 + 16 ) = √ 25
=5
__
__
√ 32 + √ 42 = 3 + 4 = 7
______
__
__
√ 32 + 42 ≠ √ 32 + √ 42
O
S
1 2
R
P, Q, R and S are points on the circumference of the circle with
^ R = 24°.
centre O. SOQ and PTR are straight lines. PO ∥ SR and OP
Determine, with reasons:
^
4.1
R
1
^
4.2
R
2
^
4.3
O
2
^
4.4
P
2
^S
4.5
^
4.6
Q
2
5
(2)
(2)
(2)
(2)
(2)
(2)
[12]
A
F
1
2
2
1
x
C
3
E
1
2
1
2
2
1
D
G
A and B lie on the smaller circle, D and E lie on the bigger circle and
C and F are the points of intersection of the two circles. ABG, FCG,
^ = x.
EDG and AFE are straight lines. B
1
5.1
Determine, with reasons, two other angles equal to x.
5.2
Prove that BCDG is a cyclic quadrilateral.
5.3
Join BD and then prove that
^D = E
^.
5.3.1
GB
^
^
5.3.2
BDG = A
5.3.3
What type of quadrilateral is ABDE? Justify your answer.
(4)
(2)
(4)
(4)
(2)
[16]
212
PLT MATHS LB 11 7th pgs (Real Book).indb 212
2012/07/02 2:23 PM
6
F
A
1
3
2
x
1
O
2
B
1
1
2
2
3 2
1
5
3
E
2
4 G
1
1
2
C
D
The diameter AG is produced to C. EC is a tangent to the circle at D.
A, D, G and F lie on the circle and DC = BC. AFB is a straight line
^ = x.
and A
1
6.1
Prove that ADCB is a cyclic quadrilateral.
^ D.
6.2
Prove that AC bisects BA
6.3
Prove that DG = FG.
^
6.4
Determine, with reasons, the size of C
2
^
^
6.5
Prove that G1 = O2.
6.6
Prove that FBCG is a cyclic quadrilateral.
6.7
For what value of x will DC be a tangent to the circle through
F, B, C and G?
(4)
(2)
(2)
(3)
(6)
(3)
(3)
[23]
D
7
O
1 2
A
x
C
B
AB and AD are tangents to the circle from A. O is the centre of the
^ = x. Prove that O
^ − ^A = 180°.
circle which passes through B, C and D. C
2
[8]
213
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TOPIC
10 Trigonometry: Sine, cosine
2
and area rules
Unit 1: Proof and application of the sine,
cosine and area rules
KEY WORDS
solve a triangle – to calculate
the lengths and sizes of the
unknown sides and angles
All triangles have three angles and three sides. If you are given three of any of
them, as long as one is a length, you can solve the triangle (find all the other
sides and angles).
Solving right-angled triangles – Revision
of Grade 10
For a right-angled triangle use the three trigonometric definitions:
hypotenuse
θ
opposite
adjacent
o
o , cos θ = __
a and tan θ = __
sin θ = __
h
a
h
WORKED EXAMPLE
Find the size of the unknown sides and angles of both triangles.
A
1
50 cm
To find a side:
C
35°
a
known side
d
c
REMEMBER
unknown side = sin, cos or tan
____________
E
2
θ
F
B
32 cm
58 cm
D
of the known angle
SOLUTION
or
known side = sin, cos or tan
____________
unknown side
of the known angle
To find an angle:
known side 1 = sin, cos or tan
___________
known side 2
of the unknown angle
1
In △ABC:
c = sin 35°
___
50
c = 50sin 35°
In △DEF
32 = tan θ
___
58
c
| substitute c from above
a = _______
tan 35°
= 40,96 cm
( 58 )
32
θ = shift tan−1 ___
= 28,68 cm
c
__
a = tan 35°
214
2
= 28,89°
58 = cos 28, 89°
___
d
58
d = __________
cos 28,89°
= 66, 24 cm
| or use Pythagoras
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 214
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EXERCISE 1
Solve the triangles.
1
A
2
A
13 cm
24 cm
C
4
A
A
33 cm
12 cm
33°
C
60 cm
B
B
B
C
69°
3
50 cm
B
C
Solving triangles with no right angles
For triangles that have no right angle you cannot use the trigonometric
definitions. You therefore need to know and prove two new rules called
the sine and cosine rules.
REMEMBER
We label triangles with no
right angles in this way:
side AB = c; side AC = b and
side BC = a
The sine rule and its proof
a = _____
c or _____
sin A = _____
sin B = sin __
C
b = _____
In any △ABC: _____
a
c
sin B
sin A
sin C
b
A
Proof:
A
A
c
c
b
h
B
c
h
b
D
C
a
Acute-angled triangle
Let AD = h
= height of △ABC with base BC
h and sin C = __
h
sin B = __
c
b
h = csin B and h = bsin C
Equate h on both sides:
csin B = bsin C
Divide both sides by bc
C sin B = ______
b sin C
_______
bc
bc
sin B = _____
sin C
_____
c
b
Let CE = h
= height of △ABC with base AB
Repeat the steps above to get:
sin B = _____
sin A
_____
a
b
sin A = _____
sin B = _____
sin C
_____
a
c
b
B
B
b
C
a
D
C 180° – c
Obtuse-angled triangle
Let AD = h
= height of △ABC with base BC
h and sin (180° − C) = __
h
sin B = __
c
b
but sin C = sin (180° − C)
h = csin B and h = bsin C
Equate h on both sides:
csin B = bsin C
Divide both sides by bc
c sin B = ______
b sin C
______
bc
bc
sin
B
sin
_____
_____
= cC
b
Let AD = h
= height of △ABC with base AB
Repeat the steps above to get:
sin B = _____
sin A
_____
a
b
sin A = _____
sin B = _____
sin C which is the same as
_____
a
c
b
a = _____
c
b = _____
_____
sin A sin B sin C
Unit 1 Proof and application of the sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 215
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WORKED EXAMPLE
REMEMBER
Use the sine rule if you know
a side–angle pair (that is, an
angle and the side opposite
it) and one other given side
or angle.
A
Solve for x in each triangle.
1
x is the unknown side use:
x
x
15
______
= ______
sin 66° sin 55°
15sin 66°
x = ________
sin 55°
To find a side:
B
= 16, 73 cm
unknown side
_______________________
sine of the angle opposite it
2
known side
= _______________________
sine of the angle opposite it
To find an angle:
sine of the unknown angle
_________________________
sine of the known angle
known side opposite the angle
= _________________________
55°
66°
x is the unknown angle use:
A
sin 36°
sin x = ______
_____
25
19
25sin 36°
sin x = ________
19
x
C
19 cm
x = 50,66° or x = 129,34°
Notice there are two possibilities for x
x = 180° − 50,66° = 129,34°
known side opposite the angle
15 cm
B
36°
C
25 cm
This is called an ambiguous case. Always check for ambiguity when solving for
angles. In many cases the diagram will show that no ambiguity occurs as the sum
of two of the angles will be greater than 180°.
The sine rule may be ambiguous if you know two sides and a non-included angle.
(SSA)
EXERCISE 2
Find the value of x in each triangle. The diagrams are not drawn to scale. Give all
answers correct to two decimal places.
1
2
A
A
18 cm
x
33 cm
B
B
35°
3
54°
B
216
x
4
A
108°
68 cm
52°
71 cm
C
C
A
x
54°
28°
C
B
x
42°
163 cm
C
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 216
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5
6
A
15 cm
A
100°
80 cm
x
B
18 cm
C
B
x
67°
65 cm
C
The cosine rule and its proof
For triangles that do not have a known side-angle pair you cannot use the sine rule.
You therefore need to know and prove the cosine rule.
In any △ABC: a2 = b2 + c2−2bccos A or b2 = a2 + c2−2accos B or c2 = a2 + b2 − 2abcos C
A
A
c
B
b
h
a–x
D
a
x
C
B
C
Acute-angled triangle
a
b
180° – c
c x
(a + x)
h
D
Obtuse-angled triangle
Let AD = h
= height of △ABC with base BC
Let AD = h
= height of △ABC with base BC
Let DC = x BD = a − x
Let DC = x BD = a + x
Apply Pythagoras’ Theorem to △ABD to
find c2
c2 = h2 + (a − x)2
= h2 + a2−2ax + x2
= a2 + (h2 + x2) − ax
= a2 + b2 − 2ax
Apply Pythagoras’ Theorem to △ABD to
find c2
c2 = h2 + (a + x)2
= h2 + a2 + 2ax + x2
= a2 + (h2 + x2) + 2ax
= a2 + b2 + 2ax
b2 = h2 + x2
x = cos C
__
| Pythagoras in △ADC
b
x = bcos C
c2 = a2 + b2−2abcos C
b2 = h2 + x2
| Pythagorus in △ADC
x = cos (180°− C)
__
b
x = −bcos C | cos (180°− C) = −cos C
c2 = a2 + b2 − 2abcos C
Unit 1 Proof and application of the sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 217
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2012/07/02 2:23 PM
WORKED EXAMPLE
REMEMBER
A
Solve for x in each triangle.
A
1
To find a side:
(unknown side)2 = (known
side 1)2 + (known side 2)2
− 2(known side 1)(known side
2) cos (∠ between 2 known
sides)
47 cm
B
To find an angle:
Change the subject of the
formula of the cosine rule to
get:
b2 + c2 − a2
cos A = __________
2bc
2
36 cm
x
48°
C
55 cm
29 cm
x
B
25 cm
C
SOLUTIONS
a2 + c2 − b2
or cos B = __________
x is the unknown side:
x2 = 472 + 552 − 2(47)(55)cos 48°
a2 + b2 − c2
or cos C = __________
2ab
x = √ 1 774, 594765...
x = 42, 13 cm
_____________
2ac
2
Use the cosine rule if you
know two sides and the
included angle (SAS) or three
angles (AAA).
___
| Press √ Ans do not round off
x is the unknown angle:
362 = 252 + 292−2(25)(29)cos x
2
2
2
25 + 29 − 36
cos x = _____________
2(25)(29)
= 0,11724
x = 83,27°
| Do not round off; use shift cos−1(Ans)
EXERCISE 3
Find the value of x in each triangle. The diagrams are not drawn to scale. Give all
answers correct to two decimal places.
1
B
3
18 cm
x
44°
C
54 cm
54°
B
x
A
58 cm
42°
63 cm
14,2 cm
C
C
28 cm
4
A
B
A
x
33 cm
218
2
A
B
x
16,3 cm
10,8 cm
C
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 218
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5
6
A
x
15 cm
10 cm
B
A
60 cm
60 cm
C
20 cm
B
42°
x
C
The area rule and its proof
1 base × height. If you are not given
You have learnt that the area of a triangle is A = __
2
or cannot calculate the height, you need to know and use the area rule.
1 bcsin A or Area △ABC = __
1 acsin B or Area △ABC = __
1 absin C
Area △ABC = __
2
2
2
Proof:
Let AD = h = height of △ABC with base BC
1a × h
Area △ ABC = __
2
h = sin B or __
h = sin C
__
c
b
h = csin B or h = bsin C
Substitute h:
1 acsin B or A = __
1 absin C
Area = __
2
2
Let CE = h = height of △ABC with base AB
Repeat the steps above to get:
1 acsin B or A = __
1 acsin A
Area = __
2
2
A
b
c
B
h
D a
C
Acute-angled triangle
Use the area rule if you know two sides and the included angle (SAS).
Unit 1 Proof and application of the sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 219
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2012/07/02 2:23 PM
WORKED EXAMPLE
A
Find the area of △ABC.
SOLUTION
1 acsin B = __
1 (65)(33)sin B
Area △ABC = __
2
2
^ first by using the Sine Rule
^ , find A
To find B
65 cm
sin A = ______
sin 54°
_____
33
65
(33)sin 54°
__________
sin A =
65
54°
B
C
33 cm
^ = 24,25°
A
^ = (180° − 24,25° − 54°)
B
= 101,75°
1 (65)(33)sin 101,75°
Area △ABC = __
2
= 1 050,03 cm2
EXERCISE 4
Find the areas of each triangle. The diagrams are not drawn to scale.
Give answers correct to two decimal places.
1
2
A
A
B
18 cm
18 cm
40 cm
36°
B
C
44 cm
3
52°
4
A
C
A
85°
B
5
37°
B
B
47°
60 cm
6
A
47°
C
A
25 cm
18 cm
220
C
28 cm
105 cm
28 cm
C
B
30°
123 cm
C
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 220
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Application of the sine, cosine and area rules
WORKED EXAMPLE
C
Find the value of x in the figure alongside.
SOLUTION
15
In right-angled △PBC:
15 = sin 75°
___
PC
15 = 15,53 cm
PC = ______
sin 75°
A
^ P = 20° | ext angle △APC
In △APC: AC
55°
75°
x
B
P
x
PC
______
= ______
sin 20°
sin 55°
15,53sin 20°
x = ___________ = 6,48 cm
sin 55°
EXERCISE 5
1
The diagrams below are not drawn to scale. All units are in cm. Give the answers
correct to two decimal places.
1.1
1.2
Find the area and perimeter of
quadrilateral ABCD.
ABCD is a trapezium. Find the
length of DC and area of ABCD.
A
70°
15
A
4
B
110°
B
12
7
67°
D
1.4
ABCD is a rectangle. Find the
^ E.
size of AC
A
C
C
D
1.3
35°
3
E
B
7
D
ABCD is a rhombus. Find:
^
1.4.1 D
1.4.2 Area of ABCD
1.4.3 Length of BD
10
12
x
14
B
A
C
D
10
C
Unit 1 Proof and application of the sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 221
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1.5 Find:
^C
1.5.1 the size of DB
1.5.2 the length of AC
1.5.3 the area of ABDC
1.6
Find the values of x and y.
B
27°
y
D
24 mm
A
40 mm
C
x
D
48°
62°
11
C
B
40°
45 mm
A
2
In each case find the length of BC in terms of the given letters and symbols.
2.1
2.2
B
A
x
A
0
x
C
y
2.3
B
β
0
2.4
A
C
β
x
B
0
0
2.5
0
α
C
A
2.6
B
A
C
β
C
α
a
B
D
x
A
0
F
β
B
α E
a
E
222
C
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 222
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Unit 2: Two-dimensional problems using
sine, cosine and area rules
REMEMBER
The angles of elevation and
depression are the angles
between the line of sight and
In Grade 10 you learnt about the angles of elevation and depression in
two-dimensional problems. You will now apply the sine and cosine rules
to two-dimensional problems.
the horizontal.
Application of triangle rules using
numerical values
Angle of
elevation
Horizontal
Angle of
depression
WORKED EXAMPLE
In the diagram AB represents a vertical
cliff which is 120 m high. An observer
at A sees two boats on the sea below.
The angle of depression from A to boat
C is 72° and from A to boat D is 65°.
A
65°
Horizontal
x
Angle of
depression
72°
120 m
The boats are in the same horizontal
plane as B, the foot of the cliff. AX
is parallel to BCD.
B
Angle of
elevation
D
C
Horizontal
Calculate the horizontal distance between the two boats.
REMEMBER
SOLUTION
1 Use geometry to fill in as
many angles as you can.
2 Start with the triangle
which has a given length.
3 Find a side which links the
two triangles (keep your
answer in the calculator).
4 Use this side to solve sides
or angles in the other
triangle.
5 Repeat the process if
necessary.
^ D = 72° − 65° = 7°
CA
^ B = 72° | alt angles AX ∥ BD and AD
^ B = 65° | alt angles AX ∥ BD
AC
120 = sin 72° AC = ______
120 = 126,18 m
In right-angled △ABC: ____
AC
sin 72°
126,18.sin 7°
CD = ______
AC CD = ________
ACsin 7° = ____________
In △ACD: _____
= 16,97 m
sin 7°
sin 65°
sin 65°
sin 65°
EXERCISE 6
1
A man standing at a point C looks up at
A
an angle of 35° to the top (A) of the cliff
AB
150 m away from him. He turns around
to walk 300 m in the opposite direction
(away from AB) at an angle of inclination
of 15°
to point C. A, B, C and D are in the same B
vertical plane. Calculate, to the nearest
metre, the distance between A and D.
300 m
35°
150 m
D
15°
C
Unit 2 Two-dimensional problems using sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 223
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2
The oil rig in the photograph has a width BC of 50 m. If the angle of elevation
from B to A, the top of the rig, is 32° and from C to A is 55°, calculate the height
AD of the rig.
A
32°
B
3
4
D
50 m
55°
C
A straight road with a slope of 12° leads directly
to the foot C of a tree BC. A and T, two points
on the road, are 8 m apart. The angle of elevation
^ C = 65°,
from A to the top of the tree is 40°. If BT
calculate the height, BC, of the tree.
B
8m
40°
12°
A
A
AD is a telephone pole and AB and AC are two wires
connected to the top and bottom of stay CB, which
is inclined at 64° to the horizontal line DC. The angle
of depression of B from A is 33° and of C from A is 59°.
If the length of stay BC is 3 m, calculate the length
of the pole AD.
33°
224
The drivers of two trucks which are
parked 5 m apart both measure the
angle of elevation from where they
are parked, to the top of the same
building. The angle of elevation of A,
the top of the building, from the truck
at point C is 65° and from the truck at
point D is 52°. Calculate the height, AB,
of the building.
T
D
59°
B
3m
64°
D
5
C
65°
C
T
52°
D 5m
65°
C
B
Topic 10 Trigonometry: Sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 224
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Applications of triangle rules using symbols
to do proofs
WORKED EXAMPLE 1
A
^ C = θ and AC = k.
In △ADC, AD = DC, AD
^
In △ABD, ABD = β.
k
kcos __θ
2
Prove that AB = ______
sin β
β
B
D
θ
C
SOLUTION
Find other angles:
180° − θ = 90° − __
θ
^ D = DA
^ C = _______
In isosceles △ADC: AC
2
^ B = 180° − θ
In △ADB: AD
2
Now use the sine rule to find the linking side AD:
(
)
ksin 90°− __θ
kcos __θ
2
k AD = ___________
2
AD
______
___________
____
=
=
sin θ
sin θ
sin θ
sin 90° − __θ
2
(
)
In △ADB use sine rule to find AB:
ADsin θ
AB
AD AB = _______
____________
= _____
sin (180° − θ)
(
AB =
)
sin β
sin β
[sin (180° − θ) = sin θ]
kcos __θ
2 sin θ
______
sin θ
___________
| substitute AD from above
sin β
kcos __θ
2
AB = ______
sin β
WORKED EXAMPLE 2
^ C = θ, AC = DC = r,
In the diagram AD
BD = 2r, AC = k and AB = 2k
A
2k
1
Use △ADC to express cos θ in
terms of r and k.
2
Use △ABD to express cos θ in
terms of r and k.
3
1.
Use your answers to show cos θ = __
B
2r
k
r
D
θ
C
4
Unit 2 Two-dimensional problems using sine, cosine and area rules
PLT MATHS LB 11 7th pgs (Real Book).indb 225
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2012/07/02 2:23 PM
SOLUTIONS
1
In △ADC use the cosine rule:
2
2
2
2
2
r + r − k = _______
2r − k
k2 = r2 + r2 − 2r2cos θ cos θ = __________
2
2
2
2r
^ B = 180°− θ
In △ADB AD
2r
Now use the cosine rule:
(2k)2 = r 2 + (2r)2 − 2(2r) r cos (180°− θ)
4k2 = r 2 + 4r2 + 4r2 cos θ
2
2
2
2
2
4k − 5r [cos (180° – θ) = – cos θ]
4k − r − 4r = ________
cos θ = ____________
2
2
4r
3
4r
Make the two cos θs equal to each other:
4k2 − 5r 2
2r 2 − k2 = ________
_______
2
2
2r
| × both sides by 4r 2
4r
4r2 − 2k2 = 4k2 − 5r2
| solve for k2
9r2 = 6k2
2
3r
k2 = ___
2
2
2
2r − k
Now substitute k2 into cos θ = _______
2
2r
3r 2
1r 2
__
2x 2 − ___
2 = ___
2 = __
1
________
cos θ =
2
2
2r
4
2r
EXERCISE 7
1
Use the given information in the diagrams to show that:
ksin β
sin (α + β)tan θ
ksin 2θsin α
1.2 d = ___________
1.1 d = _____________
A
0
α
β
sin (α + θ)
A
k
k
B
8sin θ
1.3. d = _________
sin θsin β
C
B
0
A
α
α
C
60°
0
4
D
D
D
226
B
C
d
β
d
2
d
k
A field has the shape of a trapezium with
a right-angled triangle added onto the
one side.
^ C = DC
^ E = x, AC
^ D = y and DE = a.
^ C = 90°, BA
AB
2.1
^ C in terms of x and y.
Find AD
2.2
acos (y − x)
Prove that: AC = __________
sin x
E
a
A
D
x
y
B
x
C
Topic 10 Trigonometry: Sine, cosine and area rules
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3
A cableway AB connects the top of two
mountains across an intervening valley.
The length of AB is d. From A the angle of
depression of B is α and the angle of
depression of the bottom of the valley at C
is β. From B the angle of depression of the
bottom of the valley at C is θ .
^ B = 180°− (θ + β)
3.1
Show that: AC
4
B
0
(θ + α)
sin (θ + β)
C
^ P = AC
^ B = θ,
In the diagram, QC = CB = x CQ
^
ACP = β and PQ ∥ AC
Express AC in terms of x and θ.
Express PC in terms of x, θ and β.
Hence show that the area of
4.1
4.2
4.3
D
A
P
Q
0
x
β
x .tan θ
△APC = _______
2
2
5
0
x
C
△XYZ is an isosceles triangle with z = y.
y2
β
d
Prove that: AC = dsin _________
3.2
A
α
B
X
2(Area of △XYZ)
= _______________
5.1
Show that:
5.2
Now use the Cosine Rule and your answer
to Question 5.1 to show that:
sin X
y
z
2
x sin X
Area of △XYZ = ___________
4(1 − cos X)
Y
6
x
Z
^ C = y. CB = a and AB = b.
^ C. AB
^ D = DB
^ C = x and BA
In △ABC, BD bisects AB
C
a
D
A
y
x
x
b
B
6.1
asin x
Prove that: DC = _________
6.2
bsin x
Prove that:AD = _________
6.3
Find the ratio DC:AD.
sin (x + y)
sin (x + y)
Unit 2 Two-dimensional problems using sine, cosine and area rules
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Applications of triangle rules in navigation
You can use trigonometry to solve surveying and navigation problems.
REMEMBER
• In navigation, a bearing
WORKED EXAMPLE
is the direction from one
object to another.
• Bearings are given as
angles rotated clockwise
from the north.
• Angles may also be given
in terms of North, South,
1
2
East and West.
North
West
S
An aeroplane flies from Q for 300 km at a bearing
of 310°. It changes it course to a bearing of 200°.
How far is the plane from its starting point
after travelling 450 km on this new course?
What is the bearing of the plane from its
starting point (that is, what is the bearing
of Q from P?)
200°
450 km
300 km
Q
310°
P
SOLUTION
East
The diagram alongside shows the acute triangles in △PSQ.
South
1
N
A
2
PQ2 = 3002 + 4502 − 2(300)(450)cos 70°
___________
PQ = √ 200 154,5613 = 447,39 km
^
450 km
447,39
300sin 70° = 0,63
^ Q = _________
sin SP
50°
North
S 130°
sin SPQ ______
_______
= sin 70°
300
N2
N3
20° 50°
300 km
N1
50°
447,39
^ Q = 39,06° | Not ambiguous SP
^Q ≠ 140,94°
SP
^
SPN3 = 20°
P
bearing of Q from P is 59,06°
P
A has a bearing of
50° from P
A has a bearing
N 50’E
Q
310°
N
EXERCISE 8
A
50°
Note: The diagrams accompanying these questions are not drawn to scale.
310°
P
A has a bearing of
310° from P
A has a bearing
N 50’W
1
A ship travels 80 km on a bearing of 28°, and
then travels for 190 km on a bearing of 155°.
How far is ship from its starting point?
North
B 155°
North
80 km
18°
190 km
C
A
228
Topic 10 Trigonometry: Sine, cosine and area rules
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2
Lighthouse A is situated directly north
of lighthouse B. From lighthouse A the
bearing of a ship 1 745 m away is 125° 62".
From lighthouse B the bearing of the ship
is 32° 43". Find the distance between the
lighthouses in km.
North
A 125° 62"
1 745 m
S
32° 43"
B
3
T is a tree on one side of a field. Its bearing
from A, which is on the opposite side of the
field, is 42°. B is a point 85 m from A on the
same side of the field as A. The bearing of the
tree from B is 330°. Calculate the width of
the field.
T
North
w
A
B
85
Applications of triangle rules in circle geometry
You can use trigonometry to find the lengths of chords and angles in geometric
problems. Apply your knowledge of circle geometry and trigonometry in the exercises
that follow.
WORKED EXAMPLE 1
A
O is the centre of the circle. BC = 15 units and
^ C = 52°. Calculate the radius of the circle.
BA
52°
SOLUTION
^ C = 104° | angle at centre
BO
Let BO = CO = r
O
B
15
C
Use cosine rule in △BOC:
152 = r 2 + r 2 − 2(r)(r)cos 104°
225 = 2r 2(1 − cos 104°)
r=
___________
225
= 9,52 units
√_____________
2(1 − cos 104°)
Unit 2 Two-dimensional problems using sine, cosine and area rules
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WORKED EXAMPLE 2
AB is the diameter of the circle and C is a point on the circumference. DC is a
tangent to the circle at C and meets AB at D.
dsin θ
^ C = θ, prove that: BD = ______
If AB = d and BA
2
A
cos 2θ
0
d
SOLUTION
^ D = θ tan
BC
B
| chord theorem
^ A = 90°
BC
| angle in semicircle
D
^ A = 180° − (90° + θ + θ)
CD
C
= 90°− 2θ | sum angles in △ACD
BC = sin θ BC = dsin θ
___
d
BC
BD = ____________
____
sin θ
sin (90° − 2θ)
dsin θ.sin θ = ______
dsin θ
BC sin θ
BD = ____________
= __________
2
cos 2θ
sin (90° − 2θ)
cos 2θ
EXERCISE 9
1
If O is the centre of the circle in each case, find x.
A
1.1
1.2
1.3
B
x
O
8
B
A
x
118°
D
44°
230
A
16°
15
C
2
C
O
D
x
ABC is a secant of the circle, and CT is the
tangent at T. AT = 16 cm, AB = 12 cm and
^ = 41°.
A
2.1
Calculate the length TB correct to
two decimal places.
^ B.
2.2
Calculate the size of the angle AT
2.3
Calculate the length of secant ABC.
2.4
Calculate the area of △TBC.
10
3
D
C
9
B
A
41°
12
16
B
T
C
Topic 10 Trigonometry: Sine, cosine and area rules
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3
O is the centre of the circle, AOD is a diameter.
AB = 14 cm, BC = 17 cm and AC = 9 cm.
^ correct to one
3.1
Calculate the size of B
decimal place.
3.2
Calculate the radius of the circle, correct
to one decimal place.
B
O
A
D
C
4
5
6
AB is the diameter of the circle and C
is a point on the circumference. DC is
a tangent to the circle at C and meets
^ C = θ:
AB at D. If DB = BC = a and BA
4.1
Prove that: DC2 = 2a2(1 + cos 2θ)
___
4.2
If DC = √ 12 and a = 2, show
that θ = 30°.
A
0
B
D
C
AB is a tangent to circle O with radius r.
^C = θ
AO
5.1
Find the area of △AOB in terms
of r and tan θ.
5.2
Find the area of △AOC in terms
of r and sin θ.
5.3
Use your results to show that
tan θ > sin θ where θ is acute.
O
C
0
B
x
A
In the sketch ABC is a cyclic quadrilateral.
^D = x
AB = AD = a and BC = CD = b and BA
6.1
Prove that: BC2 = 2a2(1 − cos x)
6.2
Find BD2 in terms of b and
a trigonometric ratio of x.
6.3
a
A
x
a
a
a = ________
1 + cos x
Use your results to show that: __
b
sin x
D
B
b
b
C
Unit 2 Two-dimensional problems using sine, cosine and area rules
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Revision
None of the diagrams are drawn to scale.
1
Solve for x in each triangle.
1.1
1.2
A
A
x
x
21 cm
64°
38°
B
35 cm
1.3
C
18 cm
62°
B
28 cm
1.5
47°
B
24 cm
C
38 cm
1.6
A
A
21 cm
x = Area Δ
x = Area Δ
B
A
x
35 cm
C
C
28 cm
1.4
A
x
B
54°
B
C
28 cm
28°
42°
88 cm
C
[3 × 6 = 18]
2
^ S = 54°.
In the diagram, PQ = PS and QP ∥ RS. PS = 18 cm and RS = 15 cm. PQ
P
18 cm
Q
54°
S
15 cm
R
Calculate (to two decimal places if appropriate):
^S
2.1
the magnitude of QP
2.2
the length of QS
2.3
the area of quadrilateral PQRS.
(3)
(3)
(3)
[9]
232
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3
A, B, C, D lie on a circle. AB = 7 cm, BC = 6 cm, AC = 10 cm, AD = 4 cm.
A
7
B
4
10
D
6
C
Calculate, correct to two decimal places:
^
3.1
the size of B
^ D.
3.2
the size of AC
4
(3)
(3)
[6]
P
A farmer wants to calculate the height, PQ, of
his windmill from the ground to the centre, P,
of its wheel. He measures the angle of elevation
from point R to point P to be 65,22° and from
another point S, 2 metres from R, to point P to
be 52,43°. Calculate the height PQ of the windmill.
65,22°
Q
5
A sail boat has sails consisting of two triangles
joined together by mast AB,
as shown in the diagram.
5.1
Calculate the height of the mast AB.
5.2
Calculate the area of △AEB.
5.3
Find the length of side BE.
R
52,43°
2m
[4]
A
2,8 m 30 °
44 °
3,75 °
E
C
S
D
1,70 m
B
(3)
(3)
(3)
[9]
P
6
Use the diagrams below to prove that:
6.1
xsin θ tan α
PS = __________
sin 2θ
α
S
0
Q
x
R
(4)
233
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TOPIC 10: REVISION CONTINUED
6.2
2a sin θ
BD = _______
A
cos β
0
a
B
150°
C
ß
(4)
[8]
D
7
In the figure, ECD is the wall of a building
with a flagpole BC at an angle to it. The
angles of elevation of B from C and A are θ
and α respectively and the angle of elevation
of C from A is β. AD = m.
B
E
msin (α – β)
sin (α + θ) cos β
Prove that: BC = ______________
θ
C
α
A
8
β
m
D
[5]
O is the centre of the circle with diameter AC and B another point on the
circumference.
B
A
radius = a
O
θ
C
If the radius of the circle is a, show that:
8.1
BC = 2a cos θ
8.2
Area △ABC = 2a2 sin θ cos θ
(2)
(3)
[5]
234
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9
A ship, Q, sails from P, on a bearing of 43° for 205 km. Another ship S also sails
from P on a bearing of 142° for a distance of 148 km.
Q
N
205
P
148
S
9.1
9.2
10
Calculate the distance between the two ships S and Q
What is the bearing of Q from S?
(4)
(4)
[8]
pqsin x cos y
2sin (x + y)
Use the figure to prove that: Area = ___________
P
x
S
p
y
Q
q
R
[5]
235
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TOPIC
11
Finance, growth and decay
2
Unit 1: Simple and compound decay
What is meant by decay
KEY WORD
depreciate – to lose value
Certain items depreciate in value over time. Each item and also different brands of
the same type of item, depreciate in different ways and at different rates. There are
two types of depreciation, or decay:
• simple decay, or the straight line method of depreciation
• compound decay, or the reducing balance method of depreciation.
Simple decay
In simple decay, the loss of value on an item each year is always a percentage of the
original value.
REMEMBER
p.a. stands for ‘per annum’,
which means each year.
We can explain this by considering this situation:
Peter bought a new car for R250 000. If the car depreciates in value at 15% p.a. on
the straight line method, the table below shows the value of the car over a five-year
period. In this situation, we calculate 15% of R250 000, and that amount remains the
constant loss of value each year.
Value of car in Rand
0
250 000
1
212 500
250 000 – 37 500
2
175 000
212 500 – 37 500
3
137 500
175 000 – 37 500
4
100 000
137 000 – 37 500
5
62 500
100 000 – 37 500
REMEMBER
To calculate 15% of a value
we multiply that value by
15 = 0,15 .
0,15: 15% = ____
(
100
)
KEY WORD
scrap value – the value of
an item after it has passed its
useful life in its original form,
and certain parts can be sold
as scrap
236
0,15 × 250 000
= 37 500
Age of car in years
It is clear that, under this method, a time will come when the value of Peter’s car will
be R0. In reality this is not likely to happen as a car will always have some value, even
if it is scrap value. It is important to understand that this is only a model, and that in
reality there may well be other issues that affect these values.
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 236
2012/07/07 11:27 AM
If we plot a graph of the value of Peter’s car over time, we see that it results in a
straight line with a gradient of –37 500:
REMEMBER
rise
Gradient = ___
run and this
gradient is negative because
the function values are
decreasing.
250 000
Value of car
200 000
150 000
100 000
50 000
1
2
3
4
Age of car in years
5
6
7
x
KEY WORD
appreciation – growth
in value; the opposite of
depreciation
We can see that according to this model, the car will have reached R0 after
approximately 6,6 years.
In Grade 10 you studied simple appreciation, which showed a similar linear
situation. In that case, where the function values were increasing, the line had a
positive gradient.
Instead of completing a table or drawing a graph, we can determine the value over
time after simple decay by applying the formula: A = P(1 – in)
A = the final amount; P = the present or principal value, n = the number of years
and i = the rate of interest
REMEMBER
Function values are the y
values obtained from each
value of x.
So, we could determine the value of Peter’s car after 5 years using this formula:
A = 250 000(1 – 0,15 × 5) = R62 500
REMEMBER
The formula for simple
appreciation is: A = P(1 + in)
WORKED EXAMPLES
1
2
3
A piece of equipment is expected to depreciate at 9% p.a. on the straight line
method. If its value is R11 352, 50 after 3 years, calculate its original value.
Calculate how long it would take an item valued at R30 000 to depreciate to a
value of R10 000 at 8, 5% p.a. on the straight line method. Give your answer
to the nearest year.
The total number of Aids deaths in South Africa reached 314 196 in 2005.
Over the next 4 years there was a decline in the number of deaths, so that
in 2010 there were 270 107 reported Aids deaths in South Africa. Using the
straight line method of depreciation, determine the rate per annum at which
the Aids deaths had decreased over these 4 years, correct to one decimal place.
REMEMBER
If interest is 12%, then
12 = 0,12
i = ____
100
REMEMBER
Round off money to two
decimal places ONLY at the
end of the calculation. This
will make your answer more
accurate.
Unit 1 Simple and compound decay
PLTMATHSLB11LB_12.indd 237
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2012/07/07 11:27 AM
SOLUTIONS
1
9 = 0,09 and n = 3
A = P(1 – in) where the final amount (A) = 11 325,50; i = ____
100
11 325,50 = P(1 – 0,09 × 3)
11 325,50 = 0,73P
P = R15 551,37
2
A = P(1 – in) where the final amount (A) = 10 000, the original amount
8,5
100
P = 30 000 and i = ____ = 0,085
10 000 = 30 000(1 – 0,085 × n)
1 = 1 – 0,85n)
= __
3
2 = –0,85n
– __
3
n ≈ 8 years
3
A = P(1 – in) where the final amount (A) = 270 107, the original amount
P = 314 196 and n = 4
270 107 = 314 196(1 – i × 4)
( 314 196 )
270 107 – 1 ÷ (–4) = i
_______
i = 0, 03508 = 3,5%
Compound decay
In compound decay, the loss of value of an item each year is calculated as a
percentage of the value that it was the previous year. Under this method, the loss is
greatest in the first year, becoming less as time passes. So it will be impossible for the
value of the item to become zero.
Consider this situation:
Thabile bought a car for R250 000 at the same time as Peter. His car depreciated at
15% p.a. on the reducing balance method. The value of his car over the first 5 years is
shown in the table.
238
0,15 × 250 000
= 37 500
Age of car in years
Value of car in Rands
0
250 000
1
212 500
250 000 – 37 500
2
180 625
212 500 – 31 875
3
153 531,25
4
130 501,5625
5
110 926,33
0,15 × 212 500 = 31 875
180 625 – 27 093,75
153 531,25 – 23 029,6875
130 501,5625 – 19 575,23438
0,15 × 180 625 = 27 093,75
0,15 × 153 531,25 = 23 029,6875
0,15 × 130 501,5625
= 19 575,23438
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 238
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The calculations in the right-hand boxes show that each year the decay, or loss of
value, is 0,15 × the value of the previous year. We can clearly see that these decay
values get smaller each year, which means that the rate at which the car is decreasing
in value each year reduces over time. This is evident when we consider the graph of
this situation:
y
REMEMBER
250 000
The gradient is shown by
the steepness of the graph,
so when the line becomes
less steep, the gradient is
decreasing.
Value of car
200 000
compound decay
150 000
100 000
simple decay
50 000
1
2
3
4
Age of car (years)
5
6
x
7
If we compare the compound decay in Thabile’s case to the simple decay in Peter’s
case, we can see how the loss in value of Thabile’s car is the same as that of Peter’s
in the first year. Thereafter, as the gradient decreases, the loss in value of Thabile’s
car becomes increasingly less and less as time passes. Although decreasing instead
of increasing, this situation is similar to the increasing gradient that you studied in
compound appreciation in Grade 10, which resulted in exponential growth. We used
an exponential formula for calculations involving compound appreciation.
REMEMBER
For compound appreciation:
A = P(1 + i)n
The formula for finding the value of an item after compound decay is similar:
A = P( 1 − i )n
A = the final amount; P = the Present or Principal value, n = the number of years and
i = the rate of interest
So, we could determine the value of Thabile’s car after 5 years using this formula:
A = 250 000(1 − 0,15)5 = R110 926,33
Unit 1 Simple and compound decay
PLTMATHSLB11LB_12.indd 239
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2012/07/07 11:27 AM
WORKED EXAMPLES
1
2
Cheryl leaves a bowl of soup to cool. The temperature of the soup decreases
on a reducing balance method at a rate of 4% per minute, and it is now 65 °C.
Calculate the temperature of the soup when she served it 6 minutes ago.
An investor bought shares in a certain company, but found that his money
had become half of it’s original value after a period of 4 years. What was the
annual rate of decay, on the reducing balance method, for the shares that he
had bought? Give your answer to two decimal places.
SOLUTIONS
1
A = P( 1 − i )n
65 = P(1 – 0,04)6
65
_________
=P
6
(1 – 0,04)
P = 83 °C (to the nearest degree)
Note: i) Although 65 is the present temperature, we are looking for the
temperature before it had decreased to 65, so 65 is the value A and we
need P.
ii) As the interest rate and the time period are both in minutes, we can
substitute them into the formula with no adaptations.
2
As we do not know the original amount of money we call it x, so the final
amount after 4 years will be half of x, that is, 0,5x.
A = P( 1 − i )n
0,5x = x(1 – i)4
Divide
by x, 0, 5 = (1 – i)4
____
4
√0,5 –1 = –i
i = 0,1591 = 15,91% p.a.
EXERCISE 1
1
Determine the value of R13 000 after it has depreciated for 3 years at 6% p.a:
1.1
on the straight line method
1.2
on a reducing balance.
2
A farmer bought a tractor. Five years later it had a book value of R168 345,22.
Determine the original value of the tractor if the annual rate of depreciation was
14% p.a.
2.1
on the straight line method
2.2
on the reducing balance method.
3
A machine costs R48 000 and has a scrap value of R8 000 after 10 years. Find the
annual rate of depreciation, correct to two decimal places, if it is calculated on
the:
3.1
straight line method
3.2
reducing balance method
KEY WORD
book value – the depreciated
value of a vehicle at any
specified time
240
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 240
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4
How long will it take an item bought for R12 800 to depreciate to R8 600 if
depreciation is calculated at 11,5% p.a. on the straight line method? Give your
answer to the nearest year.
5
An item is currently valued at R21 500. What was its value 3 years ago if it has
depreciated at a rate of 16,5% p.a. simple decay over the past 3 years?
6
Calculate the original value of a motor car that has a book value of R151 471,31
at the end of 8 years if depreciation is calculated at 13,5% p.a. on the reducing
balance method.
7
Four friends, Tim, Thabo, Tracy and Thandi each received a lump sum of
R12 000. Tim invested his money in shares for a new company called Newco,
Tracy invested her money in shares for a company called Womco, and Thandi
bought shares in a company called Steepco. Thabo did not invest with the stock
market and invested his money in an investment that guaranteed a growth in his
money of 5,5% p.a. simple interest. After 3 years Newco shares had depreciated
at a rate of 2% p.a. on the straight line method, Womco shares had appreciated
at a rate of 4,8% p.a. compound interest, and Steepco shares had depreciated at
a rate of 2,8% p.a. on the reducing balance method.
Calculate the value of each of the four friends’ investments at the end of 3 years.
Then state who had received the best return on their investment, and who had
received the worst return.
8
Rhino poaching is a serious problem, and has resulted in rhinos becoming an
endangered species, particularly the black rhino. Statistics suggest that there were
60 000 black rhino in 1970, but that only 4 200 remained in 2011.
8.1
Determine the annual rate of depreciation that these statistics represent if
the rate is:
8.1.1
simple decay
8.1.2
compound decay.
8.2
Using your answer to 8.1.2, determine the number of black rhino there
would be in 2050 if this rate of compound decay continues.
A black rhino
Unit 1 Simple and compound decay
PLTMATHSLB11LB_12.indd 241
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Unit 2: The effect of different periods of
compound growth and decay
KEY WORD
compounded quarterly
– interest is applied to an
investment every 3 months,
1 of the quoted annual
and __
4
rate is applied to the total at
each stage
In reality, compound interest is often compounded more often than once a year. Let
us consider the implications of this. Suppose that R10 000 is invested at an interest
rate of 12% p.a. for one year, but the interest is to be compounded quarterly. This
1 of 12% (= 3%) of the total at that
means that each quarter year (every 3 months), __
4
stage will be added to the investment. We can follow the growth of the investment in
the table below:
Quarters passed
Value of investment
0
10 000
1
10 300
10 000 + 0,03(10 000) [or 10 000 × 1,03]
2
10 609
10 300 + 0,03(10 300) [or 10 300 × 1,03]
3
10 927,27
10 609 + 0,03(10 609) [or 10 609 × 1,03]
4
11 255,0881
REMEMBER
To find a percentage increase:
increase
_____________
× 100
original amount
Therefore the percentage
1 255,09
increase = ________ × 100
10 000
10 927,27 + 0,03(10 927,27)
[or 10 927,27 × 1,03]
So, at the end of the year, the investment is worth R11 255,09. This is an increase of
R1 255,09.
= 12,5509%
KEY WORDS
nominal interest rate – the
annual interest rate that is
quoted
effective interest rate – the
actual rate achieved per
annum
In effect, therefore, the investment has had a growth of 12,5509% p.a. instead of the
quoted 12%. We call the quoted rate of 12% the nominal interest rate per annum,
and the actual growth of 12,5509% the effective interest rate per annum. The more
often the interest is compounded, the more the investment will benefit from the fact
that compound interest is calculated by multiplying the growing total by the rate of
growth each time interest is added to the investment.
When interest is compounded more often than once a year, the effective interest rate
will be higher than the nominal interest rate due to the effect of compound interest
which adds interest to a growing total.
Notation
•
•
We use i(m) to represent the nominal interest rate where the value of m represents
how often interest is compounded each year.
We use ieff to represent the effective interest rate per year.
So, in the situation we have discussed, we can say i(4) = 12% p.a. and ieff = 12,5509%
We saw that R10 000 became R11 255 (to the nearest Rand) in one year based on the
interest rate of 12% p.a. compounded quarterly. We could have calculated this final
value using the formula for compound interest:
A = P(1 + i)n
A = 10 000(1 + 0,125509)1 = R11 255,09 | Using the effective annual interest rate for
one year
242
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 242
2012/07/07 11:27 AM
OR
A = P(1 + i)n
0,12
A = 10 000 1 + ____
(
4
REMEMBER
)
4
= R11 255,09 | Using the quarterly interest rate for four quarters
(
)
0,12 4
4
0,12 4
____
divide by 10 000 1 + 0,125509 = 1 +
where 0,125509 = ieff and 0,12 = i(4)
4
So, 10 000(1 + 0,125509) = 10 000 1 + ____
(
)
1+i )
We can generalise these calculations into the following formula: 1 + ieff = ( _____
m
m m
It is important to remember that this formula shows the comparison between the
nominal and effective interest rates per annum, so the length of time that the money
is invested is irrelevant.
Whenever we use this
formula, the value that the
interest is divided by will
always be the same as the
value of the exponent. This
formula always looks at the
comparison of interest rates
per year, regardless of how
long the investment may last.
We could have used this formula to find the effective interest rate in the above
situation, where interest was quoted as 12% compounded quarterly, that is,
12 = 0,12
i(4) = 12% = ____
100
(
1 + i4
1 + ieff = _____
(
4
0,12
) = ( 1 + ____
4 )
4
4
0,12
4
ieff = 1 + ____
)
4
– 1 = 0,12550881 ≈ 12,5509%
WORKED EXAMPLES
1
Determine the effective annual interest rate of 9,5% compounded monthly,
correct to two decimal places.
SOLUTION
1
9,5
100
m m
( 0,095
12 )
0,095
= ( 1 + _____ )
12
= 1 + _____
(
)
(12) 12
12
12
12
2
KEY WORDS
– 1 = 0,0992475 ≈ 9,92%
Determine the nominal interest rate, compounded semi-annually, which
results in an effective annual interest rate of 11,5% p.a. (correct to one
decimal place).
SOLUTION
2
The value in the bracket refers
to how many times per year
interest will be compounded.
i(12) = 9,5% = ____ = 0,095
i ) = 1 + ___
i
1 + ieff = ( 1 + __
m
11,5
100
ieff = 11,5% = ____ = 0,115
i )
1 + ieff = ( 1 + __
m
REMEMBER
per annum – for the year
compounded semi-annually
– semi-annually means every
half year (every six months),
which means that interest is
twice a year, so this is notated
as i(2)
Find i(2)
m m
(
_____
)
2 2
i
1 + 0,115 = 1 + __
2
( √1,115 – 1 ) × 2 = i 2
i 2 = 0,11187 ≈ 11,2% p.a.
Unit 2 The effect of different periods of compound growth and decay
PLTMATHSLB11LB_12.indd 243
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3
Khabela deposits R7 000 into an account offering an interest rate of 8,5% p.a.
compounded monthly. Calculate how much he will have in his account at
the end of 4 years.
SOLUTION
3
8,5
100
0,085
12
8,5% = ____ = 0,085, so 0,085 per year = _____ per month
A = P(1 + i)n; n = 4 years = 4 × 12 months = 48 months
(
When interest is compounded
m times per year:
i is divided by m and n is
multiplied by m.
0,085
12
A = 7 000 1 + _____
REMEMBER
)
48
= R9 822,85
It is important to remember that the value of n must be the same unit of frequency
as that of the compounding interest. So the annual interest rate is divided by the
frequency per year, and the number of years is multiplied by the frequency per
year.
4
Keenan invested R15 000, and after 8 years his investment was worth
R32743,12. Determine the interest rate that he had received if interest was
compounded semi-annually.
SOLUTION
4
As interest is compounded every six months we are looking for i(2) p.a.,
(2)
i . As interest is applied
so every half-year the interest applied will be ___
REMEMBER
2
half-yearly, the value of n must be in terms of half-years.
i(2) is a notation that shows
that the nominal interest rate
is compounded twice a year.
This does NOT mean that the
value of i is squared.
So n = 8 years = 16 half-years.
So A = P(1 + i)n where A = 32 743,12 and P = 15 000
_________
32 743,12
– 1 = ___
√_________
15 000
2
i(2)
16
REMEMBER
Every three months means
four times per year, so this is
i(4) not i(3)
REMEMBER
Interest earned refers to the
amount of money that was
gained, that is, how much
more you had at the end
compared with what you
started with. This is not the
same as the RATE of interest,
which is a percentage.
Interest earned = A – P
(
)
(2) 16
i
32 743,12 = 15 000 1 + ___
2
i 2 = 0,05 × 2 ≈ 0,1 = 10% p.a.
5
Kirsten wanted to buy herself a car, but she first had to save more money. She
deposited the money into an account offering 9% p.a. interest, compounded
every three months. After 2 years she had R102 582,31. Calculate how much
interest she earned on her investment.
SOLUTION
5
Firstly we calculate how much money Kirsten deposited into the account:
9 = 0,09
A = P(1 + i)n where A = 102 582,31; i (4) = ____
100
and n = 2 years = 8 quarter years
(
0,09
4
)
8
0,09
4
102 582,31 = P 1 + ____ ; interest is 0,09 per year so ____ per quarter year
102 582,31
__________
= P P = R85 855,07
0,09 8
( 1 + ____
4 )
Therefore the interest that Kirsten earned
= R102 582,31 – R85 855,07 = R16 727,24
244
Topic 11 Finance, growth and decay
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Time lines
The formulae we have studied so far have involved a single transaction at a time.
When more than one transaction occurs which are linked to each other, it is useful to
picture the process over time with the help of a time line. If we list each transaction
on a time line, we will ensure that no mistakes are made with regard to interest gained
or lost in the transactions.
WORKED EXAMPLE 1
Henry deposited R4 000 into an account. The interest rate for the first 2 years was
6,5% p.a. compounded quarterly, 7% p.a. compounded semi-annually for the next
3 years and 8,5% p.a. effective thereafter. Calculate how much he will have saved
after 9 years.
SOLUTION
REMEMBER
8,5% p.a. effective means
that interest is compounded
once a year.
After the first two years, the amount accumulated will be:
(
0,065
4
A = P( 1 + i )n = 4 000 1 + _____
)
8
0,065
If the rate is 0,065 per year, then it is _____
4
per quarter, and n = 2 years, which is
8 quarter years.
The answer to this becomes the present value for the next 3 years when the interest
rate has changed, so that the value at the end of these 3 years will be:
(
0,065
4
A = P(1 + i)n = 4 000 1 + _____
0,07
) ( 1 + ____
2 )
8
6
0,07
2
A rate of 0,07 per year = ____
per half-year, and n = 3 years
= 6 half-years
P
The answer to this becomes the present value for the next 4 years, when the
interest rate changes again, this time to 8,5% effective, which means that it is
compounded annually. So the value at the end of these 4 years will be:
(
0,065
4
A = P(1 + i)n = 4 000 1 + _____
0,07
) ( 1 + ____
2 )
8
6
(1 + 0,085)4
P
It is important that we do not do separate calculations along the way and round off
answers. This will lead to the final result being inaccurate. Always consider the full
situation and do one calculation, rounding off to two decimal places at the end.
As we can see from the calculations above, each time there is a change in interest
rate we multiply by another bracket. Each bracket shows the rate of interest and
time period for that stage of the investment.
Unit 2 The effect of different periods of compound growth and decay
PLTMATHSLB11LB_12.indd 245
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2012/07/07 11:27 AM
We can visualise and summarise the situation using data on a timeline:
i 2 = 0,7
T0
i = 0,065
4
(
0,065
4 000 1 + _____
4
T2
)
i = 0,085
8
T5
(
0,065
4
4 000 1 + _____
4 000
0,07
) ( 1 + ____
)
2
8
6
T9
Final amount
Notice how we can use a dotted line to signify a change in interest and write the
interest rate for each time period above the time line. When you draw your own
time line, you do not have to write the accumulated amounts along the way each
time. The final amount at the end of 9 years will be:
(
0,065
4
A = P(1 + i)n = 4 000 1 + _____
0,07
(1 + 0,085)
) ( 1 + ____
2 )
8
6
4
= R7 752,21
WORKED EXAMPLE 2
Sunette took out a loan and paid it back in full with three payments: She paid
R10 000 after 2 years, R8 000 one year later, and R49 353,56 five years after the
loan was granted. If interest was charged at 7% p.a. compounded monthly, what
was the original value of her loan?
SOLUTION
Represent this information on a time line:
T0 i(12) = 0,07
REMEMBER
(
0,07
12
10 000 1 + ____
REMEMBER
Interest earned refers to the
amount of money that was
gained, that is, how much
more you ended up with
than what you started with.
This is not the same as the
RATE of interest, which is a
percentage.
T3
T5
Loan = x
1 = a–m
___
m
a
T2
(
0,07
12
8 000 1 + ____
(
0,07
12
49 353,56 1 + ____
)
–24
)
–36
)
–60
10 000
8 000
49 353,56
It is important to understand that we can do calculations relating to money
transactions over time at any stage on the time line, as long as all of the
transactions are compared at the same moment in time. As the value that we are
looking for in this question is at T0, it makes sense to find the value of each future
deposit at T0.
Part of each deposit made covers a portion of the loan debt, as well as some of the
interest that has to be paid on the loan. By finding the value at T0 for each deposit,
we find the portion of each deposit that covers the loan only. Therefore the sum of
these values at T0 will be the value of the loan.
246
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 246
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When working back in time, we find the P value when the A value is known. So,
make P the subject of the formula:
REMEMBER
Whenever you are taking
values back in time, divide by
the interest bracket, so the
exponent will be negative.
A = P(1 + i)n
A
∴ ______
n = P
(1 + i)
∴ A(1 + i)–n = P
Note: When working back in time on a time line we are not depreciating, so the
formula is NOT A(1 – i)n. We are, however, finding the value BEFORE any interest
growth, so we DIVIDE the future value by the interest bracket, resulting in the
bracket being raised to a negative exponent.
Therefore, in the example, we calculate Sunette’s original loan:
(
0,07
12
x = 10 000 1 + ____
)
–24
(
0,07
12
+ 8 000 1 + ____
)
–36
(
0,07
12
+ 49 353,56 1 + ____
)
–60
REMEMBER
= R50 000
When money is withdrawn
from an account, the loss to
the account is not only the
value of the withdrawal, but
also the potential interest
that the money would have
earned if it had not been
withdrawn. So the value of
the withdrawal together with
its potential interest growth,
will have to be subtracted
from the total in the account.
0,07
Note: To simplify this calculation, first calculate 1 + ____, then enter:
12
10 000 ×
ANS–24
+ 8 000 ×
ANS–36
+ 49 353,56 × ANS–60 = R50 000
WORKED EXAMPLE 3
Mrs Naidoo opened a savings account and the following transactions occurred:
She deposited R15 000 immediately.
Three years later she withdrew Rx.
After a further four years she deposited R21 500.
Seven years later (fourteen years after the initial investment), she had R25 735,50
in her account.
Interest is calculated at 6% p.a. compounded annually for the first five years,
and 10% p.a. compounded quarterly for the next nine years. Draw a time line to
represent the above and calculate the value of x.
SOLUTION
T0 i = 0,06
T3
T5
T7 i 4 = 0,1
T14
(
0,1
4
15 000(1 + 0,06)5 1 + ___
15 000
(
0,1
4
–x(1 + 0,06)2 1 + ___
–x
21 500
(
0,1
4
21 500 1 + ___
)
)
36
36
)
28
25 735,50
Unit 2 The effect of different periods of compound growth and decay
PLTMATHSLB11LB_12.indd 247
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2012/07/14 2:54 PM
Although the more instinctive way of calculating this is to work out how much
the original deposit had accumulated after 3 years, and to then subtract the
withdrawal. Find how much this total had accumulated to be at year 7, add
the next deposit, and then allow this total to grow with interest to year 14, this
[stop-start’] approach can be very tedious, and the calculations can become rather
unwieldy. It is far better to consider each transaction as separate values, and to
create an equation with these separate amounts once you have taken them all to
the same moment in time.
The most obvious moment in time would be to take each transaction to the end,
that is, to term 14, and the equation would then be:
(
)
(
)
(
0,1 36
0,1 36
0,1
– x(1 + 0,06)2 1 + ___ + 21 500 1 + ___
4
4
4
0,1 36
0,1 28
5
___
___
15 000(1 + 0,06) 1 +
+ 21 500 1 +
– 25 735,50
4
4
36
0,1
= x(1 + 0,06)2 1 + ___
4
15 000(1 + 0,06)5 1 + ___
(
(
(
)
)
(
(1 + 0,06) ( 1 + ___ )
4
0,1
15 000(1 + 0,06)5 1 + ___
)
(
36
0,1
+ 21 500 1 + ___
)
)
28
)
28
= 25 735,50
– 25 735,50
4
4
__________________________________________________
=x
0,1 36
2
It would have been easier, however, to take all of the values to term 5 where the
interest changes, as this would require no changes of interest, and therefore no
double interest brackets.
T0 i = 0,06
T3
T5
T7
i(4) = 0,1
15 000(1 + 0,06)5
15 000
–x
(
–x(1 + 0,06)2
0,1
4
21 500 1 + ___
(
0,1
4
25 735,50 1 + ___
)
)
–8
21 500
–36
25 735,50
So, at year 5:
REMEMBER
When money is withdrawn
from an account, the loss to
the account is not only the
value of the withdrawal, but
also the potential interest
that the money would have
earned if it had not been
withdrawn. So we subtract
the value of the withdrawal,
together with its potential
interest growth, from the
total in the account.
248
T14
(
0,1
4
15 000(1 + 0,06)5 – x(1 + 0,06)2 + 21 500 1 + ___
(
0,1
4
15 000(1 + 0,06)5 + 21 500 1 + ___
(
0,1
15 000(1 + 0,06)5 + 21 500 1 + ___
)
–8
)
–8
)
–8
(
0,1
4
– 25 735,50 1 + ___
(
(
0,1
4
= 25 735,50 1 + ___
0,1
– 25 735,50 1 + ___
)
–36
)
–36
)
–36
= x(1 + 0,06)2
4
4
__________________________________________________
=x
2
(1 + 0,06)
x = R24 154,27
Always remember that as long as all of the values are taken to the same moment
in time, we can the calculate at the most convenient moment on the time line. If
there is a change in the rate of interest, then the moment when the interest rate
changes is usually the most effective moment to choose.
Topic 11 Finance, growth and decay
PLTMATHSLB11LB_12.indd 248
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EXERCISE 2
1
Determine the effective annual interest rate, to one decimal place, for these
interest rates:
1.1
10,5% p.a. compounded quarterly
1.2
6,5% p.a. compounded daily
2
If an effective annual interest rate is 8%, determine the nominal interest rate p.a.
(correct to two decimal places) if interest was compounded:
2.1
every six months
2.2
every three months.
3
Khadija invested R25 000 into an account offering interest at 10,3% p.a.
compounded quarterly.
3.1
Determine how much she has in her account after 6 years.
3.2
How much interest has she received after 6 years?
3.3
What is the effective annual interest rate that Khadija obtained (to one
decimal place)?
4
Wouter has R8 000 to invest over a period of 10 years. If he requires R15 000 at
the end of the ten-year period, what annual interest rate, compounded monthly,
will he need? Give your answer correct to one decimal place.
5
Vicky invests R10 000 for a period of 10 years. During the first 3 years, the
interest rate is 9% p.a. compounded monthly. Thereafter, interest changes
to 12% p.a. compounded semi-annually. Calculate the future value of the
investment after 10 years.
6
Mr Jacobs invested R60 000. Four years later he withdrew R5 000 from his
account. After a further two years he deposited R8 000. Interest was 10% p.a.
compounded half-yearly. Use a time line to determine how much he had in his
account after a total of 10 years.
7
Mrs Mohamed opens a savings account and these transactions take place:
• She deposits R8 000 immediately, and a further R6 000 five years later.
• Two years after the deposit of R6 000 she withdraws R10 000.
• Interest is calculated at 10% p.a. compounded annually for the first two years,
and 9,5% p.a. compounded quarterly thereafter.
Draw a time line to represent the above and calculate the amount of money that
she will have saved after 10 years.
8
Uthmaan purchased a car five years ago. After paying a deposit, he took out a
loan for the balance that he owed. He paid off the loan with two payments: R30
000 after 2 years and a final payment of R113 582,40 which he made 5 years after
taking out the loan. Interest on the loan was 10% p.a. compounded monthly
during the first 3 years, and 11,5% p.a. effective for the remaining 2 years. Draw
a time line and determine the original price of the car.
9
Dumisani borrows R150 000 and plans to repay the loan over a period of 5 years
as follows:
• R5 000 after one year, R10 000 two years later, and a final payment of Rx at
the end of the five-year period. Interest is charged as follows:
• 14% p.a. for the first 3 years and then 10,5% p.a. compounded quarterly for
the last 2 years.
Calculate the value of Dumisani’s final payment by drawing a time line.
Unit 2 The effect of different periods of compound growth and decay
PLTMATHSLB11LB_12.indd 249
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Revision
1
2
3.
4
Determine the rate of depreciation for an item to become half of its
original value after 5 years on the straight line method of depreciation.
[4]
Determine the length of time (to two decimal places), that it will take
for an investment of R450 000 to be depreciate to R220 000, if the rate
of simple decay is 12% p.a.
[4]
What nominal annual interest rate, compounded monthly, would give
the same return on your investment as 9% p.a. effective? Give the answer
to one decimal place.
[4]
Sarah will need R800 000 to buy a flat in 5 years time.
4.1
Calculate how much she must deposit now into an account
offering 10% p.a. compounded monthly, to have the necessary
funds in 5 years time.
4.2
What was the effective interest rate, to 2 decimals, that Sarah
received each year?
(4)
(3)
[7]
5
Fatima invested R10 000 into an investment offering 8% p.a. compounded
quarterly. At the same time Bongi invested R10 000 into shares which
depreciated at a rate of 4, 5% p.a. on the reducing balance method. Calculate
how much better off Fatima is than Bongi after 5 years.
[6]
6
Thomas deposits R12 500 into an account at an interest rate of 9, 5%
compounded quarterly. Two years later, he deposits another R10 000.
Three years after that, he withdraws R8 000. Calculate how much will
he have in the account at the end of 8 years.
7
8
9
10
Marc buys a fridge on hire purchase at an interest rate of 12 % p.a.
After paying a 10% deposit, he continues to pay monthly payments
of R300 for 3 years.
7.1
Calculate the cash price of the fridge if he had bought it immediately.
7.2
Calculate how much interest he paid.
[8]
REMEMBER
(5)
(2)
[7]
Calculate how long it would take for an investment to become one third
of its original value at a rate of simple decay of 12% p.a. Give your answer
correct to one decimal place.
[4]
Determine the compound interest rate, correct to two decimal places,
Emile would require to triple his investment over a period of 10 years?
[4]
I plan to purchase two airline tickets in two years time. At the moment
the tickets cost R15 200.
10.1 Calculate the cost of each ticket in two years’ time if an inflation
rate of 6,5% p.a. is taken into consideration.
10.2 Calculate how much I need to invest now at 7, 5% p.a. interest
compounded quarterly, to provide for the purchase of these two
airline tickets in two years’ time.
Inflation uses compound
interest.
(3)
(4)
[7]
250
PLTMATHSLB11LB_12.indd 250
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11
12
13
14
15
16
17
Nicola makes two deposits into an account: a first amount of R5 000
and R2 000 four years later. The interest was 5% p.a. compounded
annually for the first 5 years, and then increased to 6% p.a. compounded
half-yearly. Calculate the total value of these deposits after 7 years.
[8]
Zafar receives a loan, which he plans to pay back in three payments:
R4 000 in 2 years’ time, R8 000 in 4 years’ time and R6 000 in 5 years’
time. The interest rate is 10% p.a. compounded semi-annually.
Calculate the amount of money that he receives as a loan now.
[6]
Mr Sandler deposits a lump sum into an account now to allow for
withdrawals when each of his three children starts tertiary education.
He calculates that he will need R30 000 for the first withdrawal in
5 years’ time, R32 500 for the second withdrawal 2 years later, and
R36 000 for the final withdrawal 10 years from now. If the interest rate
is 12% p.a. compounded monthly, calculate how much he deposits now?
[8]
Mpho deposits R15 000 into a savings account to save for a holiday in
3 years time. The account offers her 10% p.a. compounded monthly.
14.1 How much will she have accumulated at the end of 3 years?
14.2 What was the effective interest rate that Mpho received each year
(to two decimal places)?
14.3 How much interest had Mpho received at the end of 3 years?
Marisha invested R2 500 in a savings account. Calculate how much she
will have in the account at the end of 5 years if the interest rate is 7,5%
compounded quarterly for the first year, 6% compounded monthly for
the second year and 8,3% compounded semi-annually thereafter.
Kevin has a scooter that cost him R8 000. He wishes to replace it with
the equivalent model in 5 years’ time. If depreciation on his scooter is
at 15% per annum on the reducing balance method, and inflation is at
5,3% p.a., calculate:
16.1 the expected cost of the equivalent model in 5 years’ time
16.2 the expected book value of his scooter in 5 years’ time
16.3 the amount of money Kevin is expecting to have to pay in if he
sells his old scooter to offset the cost of the new one.
At the beginning of 2005 Joseph invested R10 000 into an account with
an interest rate of 6,5% compounded every 3 months. At the beginning
of 2007 an extra R5 000 was added to the account. At the beginning of
2010 he had to withdraw a certain amount from the account.
17.1 How much did he withdraw if the final amount yielded at the
end of year 2011 was R20 169,13?
17.2 What was the effective annual interest rate, correct to two
decimal places?
(4)
(3)
(2)
[9]
REMEMBER
[7]
Hire purchase uses simple
interest.
(3)
(3)
(2)
[8]
(10)
(3)
[13]
251
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TOPIC
12
2
Probability
Unit 1: Addition and complementary rules;
Dependent and independent events
Revision of Grade 10 probability
The probability of an event taking place is always:
total number of ways an event can occur
_________________________________________
total number of possible outcomes for the event.
KEY WORDS
theoretical outcome – the
result, in theory, that is
expected for a certain event
to happen
experiment – a trial situation
involving chance
1 . There is
For example, the probability of getting a 4 when you throw a die is __
6
only one 4 on a die, so the number of ways of getting a 4 is 1. The total number of
possibilities is 6, as there are 6 numbers altogether on a die. This means that one in
every six throws is likely to be a 4. This is the theoretical outcome, and although it
may not always be the case in an experiment, the more throws in an experimental
situation, the closer the experimental outcome will be to the theoretical outcome.
The probability of an event taking place will range between being impossible,
(probability = 0) to certain (probability = 1, or 100%).
These are some notations that we use in probability questions:
• P(A) means the probability of event A happening.
• P(A′) means the probability of event A not happening.
• P(A ∩ B) means the probability of event A and event B happening.
• P(A ∪ B) means the probability of event A or event B happening.
• P(A | B) means the probability of event A happening, assuming that event B is
taken as given.
Addition rule
In general, the addition rule is true in all cases where we consider two events:
•
REMEMBER
Mutually exclusive events
are events that have no
intersection or overlap, so it is
impossible for them to both
occur.
P(A or B) = P(A) + P(B) – P(A and B)
If events A and B are mutually exclusive, it follows that P(A and B) = 0
Therefore, the addition rule for mutually exclusive events is:
•
P(A or B) = P(A) + P(B) if A and B are mutually exclusive
WORKED EXAMPLES
1
2
252
If a die is rolled, determine the probability of it being a six or a one.
If a card is drawn from a pack of playing cards, determine the probability of it
being a ten or a heart.
Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 252
2012/07/02 2:24 PM
SOLUTIONS
1
Call event A ‘rolling a 6’ and event B ‘rolling a 1’. These are mutually
exclusive so:
1 + __
1 = __
2 = __
1
P(rolling a 6 or rolling a 1) = P(A or B) = P(A) + P(B) = __
6
2
6
6
3
Call event A ‘drawing a ten’ and event B ‘drawing a heart’. We cannot use
the addition rule in this example, because events A and B are not mutually
exclusive as there is an overlap. It is possible for the card to be a ten as well as
a heart (there is a ten of hearts). Therefore we use the general rule:
1 + __
1 – ___
1 = ___
4
P(A or B) = P(A) + P(B) – P(A and B) = ___
13
4
52
13
Complementary rule
We say A and B are complementary events if both these statements are true:
• They are mutually exclusive.
• P(A) + P(B) = 1 (This implies that the two events are exhaustive, as a probability
of 1 is 100%.)
This would then mean that P(B) = P(A′) = 1 – P(A)
KEY WORD
exhaustive events – events
which use up the full sample
space, that is, all possible
outcomes
WORKED EXAMPLES
Given the set of numbers: {3; 4; 5; 6; 7; 8}
Are these events complementary:
1
2
3
‘choosing an even number’ and ‘choosing a prime number’
‘choosing an odd number’ and ‘choosing a factor of 8’
‘choosing an odd number’ and ‘choosing a factor of 24’?
SOLUTIONS
1
The even numbers in the given set are A = {4; 6; 8} and the prime numbers
are B = {3; 5; 7). So, these events are mutually exclusive.
3 = __
3 = __
1 and P(B) = __
1 P(A) + P(B) = __
1 + __
1=1
and P(A) = __
6
2
6
2
2
2
these events are complementary.
2
The odd numbers in the given set are A = {3; 5; 7} and the factors of 8 are
B = {4; 8}. So, these events are mutually exclusive.
3 = __
5≠1
1 and P(B) = __
2 = __
1 P(A) + P(B) = __
1 + __
1 = __
and P(A) = __
6
2
6
3
2
3
6
these events are not complementary.
3
The odd numbers in the given set are A = {3; 5; 7} and the factors of 24 are
B = {3; 4; 6; 8}. These events are not mutually exclusive as 3 is an element
of both sets, so there is an overlap.
these events are not complementary.
Unit 1 Addition and complementary rules; Dependent and independent events
PLT MATHS LB 11 7th pgs (Real Book).indb 253
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Note:
• Every time that you toss a
coin, the result ‘heads’ or
‘tails’ is not dependent on
the result of the previous
throw.
• Every time a child is
born the possibility of it
being a boy or girl is not
dependent on the sex of
the previous child born.
Dependent and independent events and the
product rule
Two events are independent when the result of the first event does not affect the
result of the second event. Examples of such events are:
•
When drawing a card from one pack and a card from another pack, the two events
are independent of each other. The possibilities of drawing whichever card is
drawn from the first pack will not be affected by the card that is drawn from the
second pack.
If a card is drawn from one pack and replaced, and then a second card is drawn,
the possibilities of whichever card is drawn second will not be affected by the card
that was drawn first, so these two events are independent.
A marble is chosen from a bag that contains 2 green and 5 yellow marbles, and
then replaced. If a second marble is then chosen at random, the possibilities of its
colour does not depend on the colour of the first marble removed.
•
•
The following situations show events that are NOT independent:
•
If a bag contains 3 red and 2 blue marbles, the probability of drawing a red marble
3 . If the marble is not replaced, the probability of drawing a red marble on a
is __
5
3 if the first marble had been
2 = __
1 if the first marble was red, but __
second draw is __
4 2
4
blue.
So the probability of drawing a red marble a second time when the first marble is
not replaced, is dependent on the outcome of the first draw.
If you draw two cards from the same pack without replacement, the probability of
what you draw the second time will depend on what you drew the first time, so the
two draws are not independent of each other.
•
REMEMBER
P(B | A) means the
probability of B happening,
given that event A has already
happened.
The product rule for independent events is: P(A and B) = P(A) × P(B)
The product rule for dependent events is: P(A and B) = P(A) × P(B) | A
WORKED EXAMPLES
1
2
3
254
If you roll two dice, what is the probability of getting a six on both dice?
2 and P(A ∩ B) = ___
4.
A and B are two events such that P(A) = __
5
15
If it is given that A and B are independent events, find P(B).
A bag contains four R2 coins, six R5 coins and three R1 coins. If two coins are
chosen randomly and not replaced, what is the probability that both coins
are R5 coins?
Topic 12 Probability
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SOLUTIONS
1
2
3
As the result on each die is not affected by the result on the other die, these
are independent events. We want to ‘get a 6 on the first die’ (call that event
A) and ‘get a 6 on the second die’ (call that event B).
1 × __
1 = ___
1
So, P(A and B) = P(A) × P(B) = __
6
6 36
As these are independent events, P(A ∩ B) = P(A) × P(B)
4 = __
2 × P(B)
___
15 5
5 = __
4 ÷ __
2 = ___
4 × __
2
P(B) = ___
15 5 15
2 3
As the first coin is not replaced, the outcome of the second event depends
on what coin was drawn first, so these are dependent events. The probability
6 but if this is not replaced, there will
of choosing R5 in the first draw is ___
13
be only five R5 coins left, out of a total of 12 coins left in the bag, so the
5.
probability of choosing a second R5 coin will be ___
12
5 = ___
6 × __
5
Therefore P(A and B) = P(A) × P(B) | (A) = ___
12
13
26
We will discuss further applications of independent events in the context of Venn
diagrams, tree diagrams and contingency tables.
EXERCISE 1
1
A coin is tossed and a die is rolled. Determine the probability that the outcome
will be:
1.1
a head on the coin and a 5 on the die
1.2
a tail on the coin or a prime number on the die
1.3
a head on the coin and not a 6 on the die.
2
Are the following events complementary? Give a reason for each answer.
2.1
Scoring less than 4 when rolling a die, and scoring more than 4 when
rolling a die.
2.2
Choosing all consonants from the word MATHS, and choosing an A
from MATHS.
2.3
Drawing a red card and drawing a king from a pack of cards.
Are the following events independent or dependent?
3.1
Selecting a card and then choosing a second card without replacing the
first card
3.2
Rolling a die and tossing a coin
3
4
A bag consists of five green marbles and eight blue marbles. If one marble is
drawn, then replaced, and a second marble is drawn, determine the probability
that:
4.1
both marbles are blue
4.2
the first marble is blue and the second marble is green
4.3
the first marble is green or the second marble is green.
5
Repeat questions 4.1 and 4.2, assuming that the first marble is not replaced
before drawing the second marble.
Unit 1 Addition and complementary rules; Dependent and independent events
PLT MATHS LB 11 7th pgs (Real Book).indb 255
255
2012/07/02 2:24 PM
Unit 2: Venn diagrams
Outcomes that apply to
both A and B
A
Anything
written here
shows outcomes
that are part
of the sample
space, but are
not relevant to
either A or B
When more than one event takes
place, we can represent the outcomes
of these events in various ways.
You studied the first representation
in Grade 10, and this assists you
with solving problems. In this
representation, we use circles to
represent each event. If it is possible
for both events to take place, the
circles will intersect. The area
enclosed in the overlap will show
when both events happened. We
place the circles inside a rectangle,
where the rectangle represents the
sample space.
What is written
here shows
the total of the
sample space
B
A∩B
Outcomes that apply
to A only
Outcomes that apply
to B only
The sketches that follow show possible Venn diagrams. In each case the notation
below the diagram describes what the shaded region represents.
A
B
A
B
A∪B
A
B
A ∩ B′
B′
A
B
A
B
KEY WORD
C
sample space – the full set of
data values
C
A∩B∩C
A
B
C
A
B
C
C ∩ (B ∪ A)′
256
(A ∪ B ∪ C)′
A ∩ B ∩ C′
Topic 12 Probability
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WORKED EXAMPLE 1
The Venn diagram shows the results of asking learners whether they play the
guitar or the piano.
1
2
Determine the probability that a learner chosen at random from the group:
1.1
plays the piano, but not the guitar
1.2
does not play either instrument
1.3
plays the guitar
1.4
plays the piano, given that he/she plays the guitar?
Give reasons to explain why the events ‘playing piano’ and ‘playing
guitar’ are not mutually exclusive, exhaustive or independent events.
A
piano
B
guitar
10
7
15
18
SOLUTION
1
2
1.1
From the Venn diagram, there are10 learners who play the piano but not the guitar, so:
10 = __
1
P(piano ∩ not guitar) = P(A ∩ B′ ) = ___
50 5
1.2
There are 18 learners who do not play either instrument:
18 = ___
9
P(no instrument) = ___
50 25
1.3
There are 7 + 15 = 22 learners who play the guitar:
22 = ___
11
P(plays guitar) = P(B) = ___
50 25
1.4
From the 22 learners who play the guitar, we can see that there are seven who also play the piano:
7
P(plays piano given plays guitar) = P(A | B) = ___
22
32
___
P( A ∪ B ) =
50
39
17 + ___
22 = ___
P(A) + P(B) = ___
50
50
50
∴ P( A ∪ B ) ≠ P(A) + P(B)
i.e. P( A ∩ B ) ≠ 0
Therefore these events are not mutually exclusive.
39 ≠ 1
17 + ___
22 = ___
P( A ) + P( B ) = ___
50
i.e. P( A ∪ B )´ ≠ 0
50
50
Therefore these events are not exhaustive.
7 = 0,14
P( A ∩ B ) = ___
50
187 = 0,1496
17 × ___
22 = _____
P( A ) × P( B ) = ___
50
50
1 250
P( A ∩ B ) ≠ P( A ) × P( B )
Therefore these events are not independent.
WORKED EXAMPLE 2
Fifty learners were asked if they had ever broken an arm, a leg or their nose.
21 had broken a leg
8 had broken their noses
5 had broken an arm and their nose
10 had not broken any of these
1
28 had broken an arm
9 had broken an arm and a leg
6 had broken their nose and a leg
Display this information in a Venn diagram. Then determine how many
learners had broken an arm, a leg and their nose.
Unit 2 Venn diagrams
PLT MATHS LB 11 7th pgs (Real Book).indb 257
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2012/07/02 2:24 PM
2
3
Determine how many learners had only broken a leg.
A learner is randomly chosen from those surveyed. Find the probability that:
3.1
the learner had broken an arm only
3.2
the learner had broken an arm given that he or she had broken a leg
3.3
the learner had not broken their nose
3.4
the learner had broken a nose and a leg
3.5
the learner had broken an arm or a leg.
SOLUTION
1
Always start by filling in the central area, which
represents the number of learners who had broken all
three body parts. We do not know this value, so we call
it x. Then look for the given information that refers to
the number in each overlapped area. Nine learners had
broken an arm and leg, so there are 9 – x in the rest of
the overlap for a broken arm and leg. Similarly, fill in all
the other overlap areas.
Broken
leg
We know that 21 learners had broken a leg. So the
broken leg circle must contain 21 in total. So 21 – (9 –
x) – x – (6 – x) = 6 + x in the non-overlapping part of the
circle. Similarly, for the other circles we get 14 + x and
x – 3 for the non-overlapping part.
We place the ten learners who have not broken any of
the three body parts in the square, but outside the circles.
Broken
arm
9–x
6+x
50
14 + x
x
6–x
5–x
x–3
Broken
nose
10
Now solve for x, knowing that the total number of people is 50.
6 + x + 9 – x + 14 + x + 6 – x + x + 5 – x + x – 3 + 10 = 50
x=3
We can now complete the Venn diagram without any variables.
2
Broken
leg
Broken
arm
6
9
50
17
3
3
2
0
Broken
nose
10
Nine learners had broken a leg only.
3.1
258
17
P(broken arm only) = ___
50
Topic 12 Probability
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2012/07/02 2:24 PM
3.2
3
9 = __
P(broken arm | broken leg) = ___
21
REMEMBER
7
Using the broken leg circle as the total sample space, we see that nine
learners (6 + 3) have also broken an arm.
3.3
42 = ___
21 | There are 8 learners in the broken nose
P(Not broken nose) = ___
50 25
circle: 3 + 3 + 2, so there are 42 who have not broken a nose, out of the
total 50 learners
3.4
3 | There are 6 learners in the intersection between
6 = ___
P(Nose ∩ Leg) = ___
50 25
Nose and Leg
3.5
40 = __
4 | There are 40 learners in total in the Arm and Leg
P(Arm ∪ Leg) = ___
50 5
circles: 9 + 6 + 3 + 3 + 17 + 2
The sample space is all
possible outcomes.
WORKED EXAMPLE 3
Eighty percent of the learners at a local high school play sport and 62,5 percent belong to a society. Fifty percent of
the learners play a sport and belong to a society.
1
2
3
4
5
Draw a Venn diagram to represent this information.
What is the probability that a learner who does not belong to a society, plays sport?
What is the probability that a learner belongs to a society or plays a sport?
What is the probability that a learner belongs to no society nor plays a sport?
Determine whether playing sport and belonging to a society are independent events in this high school.
i.e. = 100%
SOLUTION
Notice that in this example, percentages instead of actual numbers of
learners are given. We can change these percentages to decimal values,
and enter the probability of an event into each area of the Venn diagram,
instead of the actual values.
Society
1
Fill in the intersection, which is 50%, or 0,5. Then, knowing that
62,5% (or 0,625) belong to a society, we can determine that 10% (or
0,1) belong to a society but do not play a sport. (62,5% – 50% = 12,5%).
Similarly, 80% – 50% = 30% (or 0,3).
So 0,3 play a sport but do not belong to a society.
2
We need to find ______________________________________________.
Sport
0,125
(or 12,5%)
0,5
(or
50%)
1
0,3
(or 30%)
0,075
the number of learners who play sport
the number of learners who do not belong to a society
This is P(sport | does not belong to a society) = 0,3
3
P(society ∪ sport) = = 0,125 + 0,5 + 0,3 = 0,925
4
P(society or sport)′ = 0,075
5
P(Society ∩ Sport) = 0,5
P(Society × P(Sport) = 0,625 × 0,8 = 0,5 = P(Society ∩ Sport)
∴ these are independent events
Unit 2 Venn diagrams
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WORKED EXAMPLE 4
We know the following facts about a group of 32 learners:
• 12 of them like hamburgers
• 16 of them like hotdogs
• 8 of them like chips, but not hamburgers and hotdogs
• 7 of them like chips and hamburgers
• 13 of them like chips and hotdogs
• 2 of the learners who like hotdogs and hamburgers also like chips
• All the learners like either hotdogs, hamburgers or chips
1
2
Draw a Venn diagram to represent this information, using A for hamburgers, B for hotdogs and C for chips.
Explain with reasons why:
2.1
A, B and C are exhaustive events
2.2
A, B and C are not complementary events
2.3
B and C are independent events
32
A
2.4
A and B are not independent events.
B
SOLUTION
1
2
The second last piece of information tells us that 2 learners like all three, so we fill in
that first at the intersection of all 3 circles. The third piece of information tells us that
there are 8 who like chips only, so we fill that into the chips only part of the diagram.
Next, because 7 like chips and hamburgers, 7 – 2 = 5 must like chips and
hamburgers but not hotdogs.
Similarly, as 13 like chips and hotdogs, 13 – 2 = 11 must like chips and hotdogs but
not hamburgers
Since we do not know how many learners like hamburgers and hotdogs but not
chips, we let that value be x, and complete the rest of the diagram:
Hamburgers only: 12 – 5 – 2 – x = 5 – x
Hotdogs only: 16 – 2 – 11 – x = 3 – x
We know from the last piece of information given that none of the
32 learners who fall outside of any of the 3 circles, we can say:
5 – x + x + 2 + 5 + 11 + 8 + 3 – x = 32
Therefore x = 2
The Venn diagram is now complete:
2.1 P( A ∪ B ∪ C )´ = 0 so A, B and C are exhaustive.
32 = 1
2.2 P( A ∪ B ∪ C ) = ___
32
16 + ___
26 = ___
54 ≠ P( A ∪ B ∪ C )
12 + ___
P( A ) + P( B ) + P( C ) = ___
32 32
2 ≠0
Also P( A ∩ B ∩ C ) = ___
32
32
32
8
32
A
B
5
16 + ___
26 = ___
54 ≠ 1
12 + ___
Also P( A ) + P( B ) + P( C ) = ___
32
32
32
32
2
11
8
C
32
A
x
5
2
B
11
8
C
32
A
2
3
5
So A, B and C are not mutually exclusive.
C
2
8
1
B
11
C
So A, B and C are not complementary.
260
Topic 12 Probability
PLTMATHSLB11LB_13.indd 260
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13
2.3 P( B ∩ C ) = ___
32
13
16 × ___
26 = ___
P( B ) × P( C ) = ___
32
32
32
∴ P( B ∩ C ) = P( B ) × P( C )
So B and C are independent events.
2 = ___
1
2.4 P( A ∩ B ) = ___
32
16
3
16 = ___
12 × ___
P( A ) × P( B ) = ___
32
32
16
∴ P( A ∩ B ) ≠ P( A ) × P( B )
So A and B are not independent events.
EXERCISE 2
1
A group of 120 people visiting a music shop were interviewed and asked if they
had listened to Freshly Ground or Steve Hofmeyer. The results show that 73
had listened to Freshly Ground, 57 had listened to Steve Hofmeyr and 28 had
listened to both.
Using F for Freshly Ground and S for Steve Hofmeyr, draw a Venn diagram to
represent this data. Use your Venn diagram when you answer the questions.
1.1
How many people interviewed had listened to neither Freshly Ground nor
Steve Hofmeyr?
1.2
Determine the probability that a person chosen at random from the group
interviewed had listened to:
1.2.1
Freshly Ground and not Steve Hofmeyr
1.2.2
either Freshly Ground or Steve Hofmeyr or both
1.2.3
both Freshly Ground and Steve Hofmeyr.
1.3
Determine:
1.3.1
P(F ∪ S)’
1.3.2
P(S ∩ F’)
1.3.3
P(F ∪ S’)
1.4
Explain why listening to Freshly Ground and listening to Steve Hofmeyer
are not mutually exclusive, exhaustive or independent events, based on
this survey.
2
Let U (the universal set) be all integer values from 1 to 30. Let A, B and C be
subsets of U, so that: A = {x: x is a factor of 30}; B = {x: x is an odd number} and
C = {x: x is a perfect square}
2.1
List the elements of sets A, B and C.
This means ‘the number
2.2
List the elements of A ∩ B.
of elements in the set that
2.3
List the elements of B ∩ C.
follows’.
2.4
2.5
2.6
2.7
Determine n(A ∩ B ∩ C).
Copy and complete the Venn diagram and indicate the number of
elements that would be included in each section of the Venn diagram. Use
the Venn diagram to answer the following questions.
What is the probability that a number chosen randomly from 1 to 30,
will be:
2.6.1
a factor of 30
2.6.2
an odd number that is not a perfect square
2.6.3
not a perfect square
2.6.4
not a factor of 30, an odd number or a perfect square?
Show that, while being a factor of 30 and being an odd number from 1 to
30 are not mutually exclusive events, they are independent events.
KEY WORD
universal set – the sample
space, which is the full set of
data values
A
B
C
Unit 2 Venn diagrams
PLT MATHS LB 11 7th pgs (Real Book).indb 261
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2012/07/02 2:24 PM
A
B
0,6
3
Fifty people were surveyed at a shopping mall and asked whether they like to
shop at ‘Perfect Purchases’ or at ‘Shopaholics’ supermarkets. The Venn diagram
shows the results of the survey. A represents ‘Perfect Purchases’ and B represents
‘Shopaholics’, showing the probability of each choice.
3.1
Suggest what is implied by the two circles that do not overlap.
3.2
How many people said that they shop at ‘Perfect Purchases’?
3.3
Determine the value of x.
3.4
Determine:
3.4.1
P(A ∩ B)
3.4.2
n(A ∪ B)′
3.4.3
P(A ∪ B ′)
3.5
Determine whether A and B are complementary events showing your
reasoning.
4
A group of 70 learners were asked about their subject choice. Their responses
showed that: 32 take Physical Sciences, 43 take Mathematics, 25 take Life
Sciences and 6 take none of these three subjects. Also, 18 take Physical Sciences
and Mathematics but not Life Sciences; 12 take Life Sciences only and 5 take
Physical Science and Life Sciences.
0,3
x
4.1
4.2
D
M
0,14
0,06
0,44
0,1
0,02
0,04
0,08
W
262
Draw a Venn diagram to represent this information. Use the letters S, M
and L to represent Physical Sciences, Mathematics and Life Sciences.
Determine:
4.2.1
the number of learners who take Physical Sciences, Mathematics
and Life Sciences
4.2.2
P(M ∩ L)′
4.2.3
P(S | M)
4.2.4
nP(S ∩ M ∩ L′)
4.2.5
the probability that a learner who takes Life Sciences does not take
Mathematics
4.2.6
the probability that a learner taking Life Sciences and Mathematics
does not take Physical Sciences.
5
A veterinarian surveyed 50 of his clients, and found that 26 have dogs, 24 have
cats and 12 have birds. Ten clients have dogs and cats, four have dogs and birds,
one has a cat and a bird, and two have dogs, cats and birds.
5.1
Draw a Venn diagram to represent this information using the letters D, C
and B to represent dogs, cats and birds.
5.2
Determine:
5.2.1
n(D ∪ C ∪ B) ′
5.2.2
n(D ∩ C ′)
5.2.3
P(B | C)
5.2.4
P'(D ∩ C ∩ B) ′.
6
A group of 50 people were asked whether they like dark chocolate, milk chocolate or
white chocolate. The probabilities of each are shown in the Venn diagram.
6.1
Determine the probability that a person did not like any of the three types
of chocolates.
6.2
Determine the number of people from those surveyed who:
6.2.1
like all three types of chocolate
6.2.2
do not like any of these types of chocolate
6.2.3
like dark and white chocolate but not milk chocolate
6.2.4
like milk chocolate only
6.2.5
like dark chocolate and milk chocolate.
Topic 12 Probability
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Unit 3: Using tree diagrams to solve problems
for events not necessarily independent
When more than one event takes place consecutively or simultaneously, it is useful to
represent them as a tree diagram. We represent each event by a column of branches,
and the number of branches is determined by the number of possible outcomes for
that event. For example, if a die is thrown, there are six possible outcomes, numbers
1 to 6, which we represent by six different lines (or ‘branches’) drawn from a starting
point. If a coin is then tossed (with two possible outcomes, heads or tails), we draw
the possible ‘branches’ of heads or tails after the previous lines. We draw the tree
diagram as shown below.
1
__
1
2
H
1
__
T
2
1
__
6
1
__
2
1
__
6
2
H
1
__
T
2
1
__
1
__
3
6
2
H
1
__
T
2
1
__
1
__
6
4
1
__
6
2
H
1
__
T
2
1
__
1
__
6
5
2
H
1
__
T
2
1
__
6
2
H
1
__
T
2
REMEMBER
Simultaneously means at the
same time.
Consecutively means one
after the other.
Notice that we write the probability of each ‘branch’ alongside that branch.
This representation helps us decide which values we multiply and which values we
add. We always multiply probabilities along the branches. If there is more than one
set of branches that we can follow to achieve the result, add the final answers for each
pathway. So, in the example, if we want the probability of getting a prime number
and tossing a head, we follow the pathways: 2 then H; or 3 then H; or 5 then H.
So the probability of getting a prime number and tossing a head is:
3 = __
1 × __
1 + __
1 × __
1 + __
1 × __
1 = ___
1 + ___
1 + ___
1 = ___
1
__
6
2
6
2
6
2
12
12
12
12
4
Note: When each branch of all the events have equal probabilities, as they do in this
case, we can see that 3 of the 12 possible final outcomes satisfy the conditions set, so
3 = __
1.
the probability of achieving the required outcome is ___
12
4
It is also useful to remember that the events at the ends of the branches are mutually
exclusive and exhaustive. Therefore the sum of the probabilities of each branch will
equal 1.
Unit 3 Using tree diagrams to solve problems for events not necessarily independent
PLTMATHSLB11LB_13.indd 263
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WORKED EXAMPLE 1
Ken and Claire plan to have three children.
1
2
3
4
5
Draw a tree diagram to represent the possible combinations of the sexes
of their children.
Determine the probability that they will have 3 girls.
Determine the probability that they will have 2 boys and 1 girl.
Determine the probability that they will have 2 girls, given that their
first child is a boy.
Determine the probability that they will have at least one boy.
SOLUTIONS
1
__
1
B
1
__
2
1
__
G
1
__
1
__
2
2
B
2
B
1
__
2
G
2
B
1
__
G
2
1
__
1
__
B
1
__
2
2
2
B
1
__
G
2
G
1
__
1
__
2
G
2
B
1
__
G
2
Each column of branches represents the
possible results of a new child that is born, so
the first child could be a boy or a girl (B or G).
If the first child is a boy, the second child could
be a boy or a girl, so the first B is followed by
the branches B or G. If the first child had been
a girl, there is still the option of the second
child being a B or G. The same options of B or
G apply for the third child.
2
1 × __
1 × __
1 = __
1
P(3 girls) = __
3
There are three different pathways that result in two boys and one girl:
BBG or BGB or GBB
(Shown on the tree diagram with dotted lines.)
2
2
2
8
1
__
1
__
2
B
2
2
B
1
__
G
1
__
1
__
1
__
2
2
B
G
B
1
__
G
1
__
1
__
2
B
2
G
1
__
2
2
B
1
__
G
2
1
__
G
2
B
1
__
G
2
So, P(2 boys and 1girl)
= BBG + BGB + GBB
1 × __
1 × __
1 + __
1 × __
1 × __
1 + __
1 × __
1 × __
1
= __
2
2
2
1
__
(There is only one pathway that has three girls: GGG.)
2
2
2
2
2
2
2
2
3
1 + __
1 + __
1 = __
= __
8
8
8
8
Or, because each branch has equal probabilities, three of the eight possible
outcomes satisfy the requirement of having 2 boys and 1 girl.
264
Topic 12 Probability
PLT MATHS LB 11 7th pgs (Real Book).indb 264
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1
__
4
1
__
2
B
B
1
__
G
2
1
__
1
__
2
2
G
2
B
1
__
G
2
If we are given that the first child is a boy,
we consider only the branches that follow
that outcome. So there are only four possible
further outcomes, with only one resulting in
2 girls.
1.
So the probability of 2 girls is __
4
Or, we can see that the only possible pathway to follow the first boy is GG,
1 × __
1 = __
1.
which is __
2
5
2
4
There are many possible outcomes that will result in at least one boy. In
such a situation, it is quicker to determine the answer by realising that the
probability of anything happening will always be 1 minus the probability of
the required outcome not happening.
(P(A) = 1 – P(not A))
So P(at least one boy) = 1 – P(no boy) where the only way of having no boy is
1 = __
7
GGG, so P(at least one boy) = 1 – __
8
8
REMEMBER
A probability of 1 represents
certainty, or 100%.
WORKED EXAMPLE 2
Simon either walks or cycles to school. When he cycles to school, the probability
that he is on time is 0,9 but if he walks to school the probability that he is on time
is 0,8. He cycles to school 60% of the time.
1
2
3
Draw a tree diagram to represent this information.
Determine the probability that he is late for school on any day chosen at
random.
Are ‘walking to school’ and ‘being late’ independent events for Simon?
SOLUTIONS
1
The first event that will affect the answer is whether Simon walks or cycles
on the day chosen at random. It he cycles to school 60% of the time, he must
walk to school for the other 40% of days. This means that the probability of
60 = 0,6 and the probability of him walking is ____
40 = 0,4. Each
him cycling is ____
100
100
option lead to the possibility of him arriving at school on time or late. Use
the given probabilities for him arriving on time, and subtract them from 1 to
get the probabilities of him being late. Now draw the tree diagram as shown
on the right.
2
The probability of Simon arriving late for school is shown by the dotted
pathways. We calculate the probability as follows:
P(Late) = 0,4 × 0,2 + 0,6 × 0,1 = 0,08 + 0,06 = 0,14
3
P (walking and being late) = 0,08
0,4
0,6
Walk
Cycle
0,8
On time
0,2
Late
0,9
On time
0,1
Late
P( walking ) × P( being late ) = 0,4 × 0,14 = 0,56
∴ P (walking and being late) ≠ P( walking ) × P( being late )
So walking to school and being late are not independent events for Simon.
Unit 3 Using tree diagrams to solve problems for events not necessarily independent
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WORKED EXAMPLE 3
Based on their past achievements, a hockey coach knows that his team has a 65%
chance of winning their matches if they play a home game, and a 25% chance of
a draw. If they play an away game, they have a 55% chance of winning and a 40%
chance of losing. They play 50% of their matches at home. Draw a tree diagram,
and then determine the probability that they will lose their next match if it is not
known whether it is a home or away match.
SOLUTION
The first event that will affect the answer is whether they play at home or away.
The options are that they could win, lose or draw. So the tree diagram will have
two columns of branches, one for each possible outcome of these events:
50%
50%
win
65%
25%
Home
draw
10%
lose
win
55%
5%
draw
Away
40%
lose
So, the pathways that result in a loss
will be Home Lose and Away Lose as
shown by the dotted lines.
Notice that we can write probabilities
as fractions or percentages.
1
P(Lose) = 50% × 10% + 50% × 40% = 0,5 × 0,1 + 0,5 × 0,4 = __
4
WORKED EXAMPLE 4
A bag contains 6 white balls, 4 black balls and 5 green balls. A ball is drawn at
random and not replaced. A second ball is drawn.
1
2
Show that drawing a white ball in the first draw, and drawing a white ball in
the second draw are not independent events.
Draw a tree diagram and use it to determine that:
2.1
both balls are green
2.2
the first ball is black and the second ball is green
2.3
the two balls are white and green in any order.
SOLUTIONS
1
266
6 , but once that white ball has been removed and not
P(white first) = ___
15
replaced, there will be only 5 white balls left, and only 14 balls in total.
5 if a white ball had been drawn first,or ___
6 if a
Therefore P(white second) = ___
14
14
different colour had been drawn first. Therefore the probability of the second
event is affected by the first, and therefore the events are not independent.
It is important to realise that these two events are not independent of each
other. As the first ball is not replaced after it has been drawn, there will be one
less ball of a certain colour for the second choice. So the result of the second
draw will be affected by the colour of the ball that was drawn first.
Topic 12 Probability
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5
___
14
14
W
5
___
15
6
___
14
4
___
15
5
___
6
___
15
14
G
W
14
The probabilities of these branches are based
on the assumption that the first ball drawn was
green. So there are only 4 green balls, 6 white
balls and 4 black balls left to draw. The total
number of balls left in the bag is 14.
4
___
14
4
___
B
G
14
5
___
The probabilities of these branches are based
on the assumption that the first ball drawn was
black. So there are only 3 black balls, 6 white
balls and 5 green balls left to draw. The total
number of balls left in the bag is 14.
W
3
___
14
B
B
G
14
6
___
The probabilities of these branches are based
on the assumption that the first ball drawn was
white. So there are only 5 white balls, 4 black
balls and 5 green balls left to draw. The total
number of balls left in the bag is 14.
W
4
___
B
G
2.1
5 × ___
4 = ___
2
P(both green) = ___
2.2
5 = ___
4 × ___
2
P(black then green) = ___
2.3
15
14
21
In any order, so WG
or GW.
P(WG) + P(GW)
15
14 21
6 × ___
5 + ___
5 × ___
6 = __
2
P(white and green) = ___
15
14 15
14 7
WORKED EXAMPLE 5
A company buys stationery from two suppliers, A and B, and places twice as many
orders with A as with B. The probability that either supplier A or B will deliver an
order on time is 0,8.
1
2
Draw a tree diagram to represent this information.
Determine whether ordering from supplier A and the order not being
delivered on time are independent events.
SOLUTIONS
1
–23
–13
2
0,8
on time
0,2
not on time
0,8
on time
0,2
not on time
A
B
2 × 0,2 = ___
2
P( A and not on time ) = __
3
(3
15
)
6 = ___
2 × __
2 × 0,2 + __
1 × 0,2 = ___
2
P( A ) × P( not on time ) = __
3
3
∴ P( A and not on time ) = P( A ) × P( not on time )
So these are independent events.
45
15
Unit 3 Using tree diagrams to solve problems for events not necessarily independent
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2012/07/02 2:24 PM
EXERCISE 3
268
1
An ice cream shop sells ‘combos’ which consist of a cone with one scoop of
ice cream and a topping. The cones are plain or chocolate, the ice cream flavours
are vanilla, chocolate or banana, and the toppings are nuts, chocolate sprinkles
or smarties. Draw a tree diagram to represent these choices, and then answer the
questions.
1.1
Determine how many different ‘combos’ have smarties as a topping.
1.2
Thandi does not like chocolate. How many ‘combos’ can she choose from?
1.3
If Nina is allergic to nuts, determine the probability that she will have an
ice cream with chocolate sprinkles, given that she will not choose nuts as
a topping.
2
A bag contains 6 blue, 5 red and 9 white marbles. A marble is drawn and not
replaced, and another marble is then drawn. Draw a tree diagram to represent
this information, and use it to answer the questions.
2.1
Explain why drawing a second blue marble is not independent on drawing
a blue marble first.
2.2
What are the probabilities that:
2.2.1
both marbles are white
2.2.2
both marbles are blue
2.2.3
one blue and one red marble is chosen
2.2.4
neither of the marbles chosen is red?
3
There are eight houses in a school, five for day scholars and three for boarders.
Each house has chosen a Head of House. From these Heads of House, the learners
must select a Head prefect and a Deputy Head prefect. Assume that each Head
of House has an equal chance of being elected. Draw a tree diagram showing the
probabilities of each position being filled by a boarder or a day scholar. Then
answer the questions.
3.1
Determine the probability that the Head prefect of the school will be
a day scholar.
3.2
Is it true to say that the election of the Deputy is independent of the
election of the Head prefect? Give a reason.
3.3
What is the probability that both positions are filled by boarders?
3.4
What is the probability that at least one of the positions is filled by
a boarder?
Topic 12 Probability
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Unit 4: Contingency tables
Contingency tables are statistical tables that represent the relationships between two
or more variables. We show the frequencies of each variable in rows and columns. So,
if a survey of 100 people was done to analyse the relationship between gender and
right- or left-handedness, the results could be:
This represents the total of the first row,
that is, the total number of males surveyed.
Right-handed
Left-handed
Totals
Male
41
12
53
Female
38
9
47
Totals
79
21
100
This represents the total of the
second column, that is, the total
number of left-handed people.
This represents the total of the
first column, that is, the total
number of right-handed people.
This represents the total
of the second row, that
is, the total number of
females surveyed.
This represents the
total number of people
surveyed, and is the total
of the right-hand column
and the bottom row.
According to this survey, the probability of a male chosen at random being left12 ≈ 0,23, as 12 of the 53 males were left-handed. The probability of a
handed is ___
53
9 ≈ 0,19.
female being left-handed is ___
47
This survey suggests that males have a higher probability of being left-handed
than females, but we cannot draw such conclusions from a single survey of only
100 people.
Also notice that the events listed in the rows and those listed in the columns must be
complementary events, otherwise the total in the bottom right-hand corner will not
be the total sample space.
WORKED EXAMPLE 1
REMEMBER
Complementary events are
mutually exclusive (have no
overlap), and exhaustive (use
up all possibilities).
A group of Mathematics and Physical Sciences learners at a college were
interviewed about the subjects they studied. The college offered two different
Mathematics courses and two different Physical Sciences courses. No learner was
enrolled in more than one Mathematics or Physical Sciences course at any one
time. The results of the survey are recorded in the table.
Mathematics A
Physical Sciences A
Physical Sciences B
Total
6
20
13
Total
1
2
Mathematics B
23
Copy and complete the table.
How many learners were interviewed?
Unit 4 Contingency tables
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3
Find the probability that a learner chosen at random from those interviewed:
3.1
was enrolled in Mathematics A
3.2
was enrolled in Physical Sciences B and Mathematics B
3.3
was enrolled in Physical Sciences B, given that they were enrolled in
Mathematics A.
3.4
Five years later, a similar survey was conducted on 100 learners. It was
found that the probability of a learner:
• enrolled in Mathematics A and Physical Sciences A was 0,3
• enrolled in Mathematics A was 0,8
• not enrolled in either Mathematics A or in Physical Sciences A
was 0,1.
3.4.1
Copy and complete the contingency table.
Mathematics A
Not Mathematics A
Total
Physical Sciences A
Not Physical Sciences A
Total
3.4.2
3.4.3
3.4.4
3.4.5
3.4.6
Determine the probability that a learner surveyed takes
Mathematics A but not Physical Sciences A.
Determine the probability that a learner surveyed takes
Physical Sciences A but not Mathematics A.
How many learners surveyed were enrolled in Physical
Sciences A at that time?
Determine the probability that a learner surveyed does either
Mathematics A or Physical Sciences A.
Compare the probability for Mathematics A and for Physical
Sciences A in each table. Discuss whether the popularity for
each of these subjects has changed or not over the five years.
SOLUTIONS
1
By adding across rows and down columns, we can fill in the missing values
as follows:
20 – 6 = 14
Mathematics A
Mathematics B
Total
14
6
20
Physical Sciences A
Physical Sciences B
13
17
30
Total
27
23
50
13 + 17 = 30
23 – 6 = 17
14 + 13 = 27
2
270
20 + 30 = 50
and
27 + 23 = 50
50 learners were interviewed.
Topic 12 Probability
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3
3.1
total in Maths A
27
P(Maths A) = ____________________
= ___
3.2
17
P(Science B and Maths B) = P(Science B ∩ Maths B) = ___
3.3
number doing Science B and Maths A 13
P(Science B | Maths A) = ________________________________ = ___
3.4
total number surveyed
50
50
27
number doing Maths A
Notice that this time the values in the table represent probabilities and
not an exact number of learners. So the value in the bottom right-hand
block represents the total probability, which is 1. First fill in the given
values and the 1 in the bottom right-hand block (shown in bold). Then
add the rows and columns, and fill in the values.
0,2 – 0,1 = 0,1
3.4.1
0,2 – 0,1 = 0,1
Maths A
Not Maths A
Total
Science A
0,3
0,1
0,4
Not Science A
0,5
0,1
0,6
Total
0,8
0,2
1
0,8 – 0,3 = 0,5
1 – 0,4 = 0,6
1 – 0,8 = 0,2
3.4.2
3.4.3
3.4.4
3.4.5
3.4.6
0,3 + 0,1 = 0,4
P(Maths A, not Science A) = 0,5. The values in the table are
probabilities, so we only give the probability in the block that
is relevant to the question.
P(Science A, not Maths A) = 0,1
P(Science A) = 0,4 0,4 × 100 = 40 learners
P(Maths A or Science A)
= P(Maths A ∪ Science A)
= P(Maths A) + P(Science A) – P(Maths A ∩ Science A)
= 0,8 + 0,4 – 0,3 = 0,9
Or realise that:
P(Maths A or Science A) = 1 – P(not Maths A, not Science A)
= 1 – 0,1 = 0,9
REMEMBER
The definition: P(A ∪ B) =
P(A) + P(B) – P(A ∩ B)
27 ≈ 0,54 and P(Science A)
Five years ago, P(Maths A) was ___
20 = 0,4
was ___
50
50
Now, P(Maths A) = 0,8 and P(Science A) = 0,4
So over the five years, Maths A has become more popular,
whereas Science A has retained the same popularity.
Unit 4 Contingency tables
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WORKED EXAMPLE 2
A group of people were asked whether they had watched Gcina Mhlope’s plays or
whether they had listened to her story-telling. Based on the results of the survey,
the contingency table was drawn up showing the probabilities of these events.
Watched plays
Not watched plays
Heard stories
0,6
Not heard stories
0,26
Total
REMEMBER
For independent events:
P(A and B) = P(A) × P(B)
1
2
Total
0,35
Copy and complete the table.
Show that watching her plays and hearing her story-telling are independent
events.
SOLUTIONS
1
2
Watched plays
Not watched plays
Total
Heard stories
0,21
0,39
0,6
Not heard stories
0,14
0,26
0,4
Total
0,35
0,65
1
P(Plays and Stories) = 0,21
P(Plays) × P(Stories) = 0,35 × 0,6 = 0,21 = P(Plays and Stories)
Therefore watching the plays and hearing the stories are independent events.
WORKED EXAMPLE 3
A study was conducted to investigate the relationship between eye colour and
hair colour. A group of 250 people were observed. The results are:
• 12 people had red hair, 64 had blonde hair and 82 had black hair
• 35 of the blondes had blue eyes, and 21 had green eyes
• 68 of the brunettes had brown eyes
• 38 of those with black hair had brown eyes, and 41 had green eyes
• 5 of the redheads had blue eyes
• 116 people had brown eyes and 86 had green eyes
1
Copy and complete the contingency table.
Blonde hair
Brunette
Black hair
Red hair
Total
Blue eyes
Brown eyes
Green eyes
Total
272
Topic 12 Probability
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2
Based on these results, determine the probability of a person having:
2.1
blonde hair and green eyes
2.2
black hair and brown eyes
2.3
blue eyes, given that they are brunette
2.4
red hair, given that they have green eyes.
SOLUTIONS
1
First fill in the given values (shown in bold). Then calculate the missing
values by adding each column and row to equal the total at the end of each
row and column.
Blonde hair
Brunette
Black hair
Red hair
Total
35
5
3
5
48
Brown eyes
8
68
38
2
116
Green eyes
21
19
41
5
86
Total
64
92
82
12
250
Blue eyes
2
2.1
21
P(Blonde and green eyes) = ____
2.2
38 = ____
19
P(Black hair and brown eyes) = ____
2.3
5
P(Blue eyes | Brunette) = ___
2.4
250
250
125
92
number of redheads with green eyes
5
P(Redhead | Green eyes) = _______________________________ = ___
86
total with green eyes
EXERCISE 4
1
A stationery distributor investigated the number of defective pens that
they obtain from two different suppliers. They recorded their findings for a
consignment of pens from both suppliers in a contingency table.
Supplier A
Supplier B
Defective
18
Not defective
Total
1.1
1.2
1.3
1.4
1.5
Total
144
350
500
Copy and complete the table.
Determine the probability that a pen chosen at random from the whole
consignment will be defective.
Determine the probability that a pen chosen from supplier A will be
defective.
Determine the probability that a pen chosen from supplier B will not be
defective.
Based on these findings, which supplier should the distributor order from
in the future?
Unit 4 Contingency tables
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2
A group of 50 learners discuss the chances of their first rugby and hockey teams
winning their respective matches the following day. The opinions of the learners
were as follows:
• 30 thought that their rugby team would win
• 35 thought that their hockey team would win
• 6 thought that both their hockey and rugby teams would lose
• nobody thought that there would be a draw in either game.
2.1
Copy and complete the contingency table.
Hockey will win
Hockey will lose
Total
Rugby will win
Rugby will lose
Total
2.2
2.3
274
Based on these opinions, determine the probability that:
2.2.1
both the hockey and rugby teams will win their matches.
2.2.2
the hockey team will win and the rugby team will lose.
Show that according to these opinions, winning hockey and winning
rugby are independent events.
Topic 12 Probability
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Revision
1
Consider this list of events:
a
Scoring an even number when rolling a die
b
Scoring more than 3 when rolling a die
c
Scoring less than 5 when rolling a die
d
Scoring a 1 when rolling a die
e
Scoring an odd number when rolling a die
Which pair of the events are:
1.1
mutually exclusive but not complementary
1.2
exhaustive but not complementary
1.3
complementary?
2
A group of 100 people visiting a bookshop were asked whether they had
read Coconut by Kopano Matlwa or Rainmaker by Don Pinnock.
• 67 had read Coconut
• 59 had read Rainmaker
• 22 had read neither.
2.1
2.2
2.3
2.4
2.5
3
(2)
(2)
(2)
[6]
Represent this data in a Venn diagram.
Determine how many of the people visiting the store had read
both books.
Determine the probability that a person had not read either of
the books.
Determine the probability that a person chosen at random had
read Rainmaker only.
Determine the probability that a person chosen at random had
read Coconut, given that they had read Rainmaker.
A card is drawn from a pack of playing cards, replaced, and a second card
is drawn.
3.1
Use a tree diagram to show the probabilities of drawing red or black
cards for both drawings.
3.2
Determine the probability that both of the cards drawn were red.
3.3
Determine the probability that the cards were different colours.
3.4
Determine the probability that the second card was black.
3.5
Determine P(both black).
3.6
Determine the probability that at least one card was red.
(5)
(3)
(2)
(2)
(2)
[14]
(4)
(2)
(4)
(4)
(2)
(3)
[19]
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TOPIC 12: REVISION CONTINUED
4
Grade 9 learners are considering their subjects for Grade 10. They have to
choose one of Physical Sciences or Geography, and one of Life Sciences or
Accountancy. Complete the contingency table which shows the results of
their choices.
Physical Sciences
Life Sciences
Geography
15
30
Accountancy
10
Total
4.1
4.2
4.3
4.4
5
Total
50
Copy and complete the table.
Determine the probability that a learner chosen at random will do
Accountancy and Physical Sciences.
Determine the probability that a learner will take Geography, given
that they take Life Sciences.
Show that taking Geography and taking Accountancy are
independent events.
(2)
(2)
(4)
[13]
2 . If the team play two
The probability of a team winning a match is __
3
matches, determine the probability that they lose both matches, assuming
that there is no draw.
Use a tree diagram to answer the question.
6
(5)
A shelf in a bookshop had books written by South African authors.
Sarah found four books written by Mark Mathabane, nine by
Dalene Matthee and ten by James McClure. She took a book off the
shelf at random, did not replace it, and then took another.
6.1
Represent this situation in a tree diagram, showing the probabilities
of each branch.
6.2
Determine the probability that:
6.2.1
the books that she chose were by Mark Mathabane and
Dalene Matthee in any order
6.2.2
the books that she chose were by the same author
6.2.3
neither of the books were by Dalene Matthee
6.2.4
at least one of the books was by James McClure.
[4]
(9)
(4)
(4)
(6)
(4)
[27]
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7
At a garden nursery, 60 people were asked whether their gardens contained
shrubs, climbers or bulbs. Forty-one people said that their garden contained
shrubs, 25 said that their gardens contained climbers, and 21 said that their
gardens contained bulbs. The Venn diagram shows the rest of the results of
the survey where S stands for shrubs, C for climbers and B for bulbs.
S
C
60
2x
8
x
5
B
7.1
7.2
8
5
Determine the value of x.
Determine:
7.2.1
nP(S ∩ C)
7.2.2
P(S ∩ C ∩ B)′
7.2.3
P(C ∪ B)
7.2.4
P(S ∩ C ′ )
7.2.5
P(B | S)
7.2.6
P(S ∪ C ∪ B)′
(6)
(3)
(3)
(3)
(3)
(2)
(2)
[22]
A group of learners were asked whether they travel to school by bus or
by train. Some learners use both train and bus to get to school, some use
either a bus or a train, and others use neither a bus nor a train. The
contingency table shows the probabilities for this group of learners.
Travel by bus
Travel by train
Not travel by bus
0,1
Total
0,6
Not travel by train
Total
8.1
8.2
8.3
8.4
8.5
0,7
Copy and complete the table.
Determine the probability that a learner chosen at random does
not travel by train.
Determine the probability that a learner chosen at random travels
by train and not by bus.
Determine whether travelling by train and travelling by bus are
independent events.
Determine the probability that a learner travels by bus or by train.
(6)
(1)
(1)
(4)
(4)
[16}
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Formal assessment: Term 3 Test 1
1
A
B
F
15
49
H
C
D
G
J
100
E
L
The solid drawn above has a height of 100 units. The length of AL is
49 units, BC is 15 units and AF = DL.
1.1
Determine the area of the base, in terms of π.
1.2
Determine the perimeter of the base, in terms of π.
1.3
Determine the volume of the solid.
1.4
Determine the total surface area.
2
(4)
(3)
(3)
(5)
[15]
G
E
H
A
B
K
D
C
A paper weight is made up of a square-based right pyramid and a squared-based
right prism. It has a slant height of 34 units. AF = 32 units and ED = 7 units.
2.1
Determine the volume of the paper weight.
(5)
2.2
Determine the total surface area of the paper weight.
(5)
[10]
3
CD = 60 units, AB = 28 units and OH is 8 units longer than OG. OG = x units.
B
H
A
O
C
3.1
3.2
3.3
G
D
Express, giving reasons, the lengths of these sides in terms of x:
3.1.1
OH
3.1.2
AO
3.1.3
CO.
Solve for x.
State the length of the radius.
(1)
(3)
(3)
(3)
(1)
[11]
278
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4
O
Q
1
5
4
3
2
P
2 1
x
1
2
3
R
3
2
T
1
U
2
1
S
SQ and SP are tangents to the circle at Q and P. PQ produced meets the line
^ =x
through S at R. SR is parallel to UQ. PU produced meets SR at T and P
1
4.1
Find five angles equal to x.
4.2
Prove that PQST is a cyclic quadrilateral.
^ T = 3TP
^ S.
4.3
Prove that PQ
4.4
Prove that RQ = PS.
^ P in terms of x.
4.5
Determine QO
4.6
Join OU and then prove that OU is the perpendicular bisector of QP.
4.7
Is OQUP a cyclic quadrilateral? Justify your answer.
4.8
Determine the value of x if OQUP is a rhombus.
(10)
(4)
(4)
(3)
(4)
(4)
(3)
(2)
[34]
Total: 70 marks
279
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Formal assessment: Term 3 Test 2
1
Determine the total surface area of the shape in the diagram
(correct to one decimal place).
35 mm
10 mm
2
[5]
The box shown in the diagram contains two wooden shapes with
dimensions as shown. Determine:
8 cm
30 cm
20 cm
12 cm
10 cm
15 cm
40 cm
50 cm
2.1
2.2
3
the volume of each wooden shape
the percentage of the volume of the box that is taken up by the
two wooden shapes (to two decimal places).
(4)
(4)
[8]
In the figure A, P, Q and R are points on the circumference of the circle,
centre O. AB is a tangent to the circle at A. OR and AQ intersect at T and
^ B = 56°.
QT = TA. QA
Q
R
O
B
T
56°
A
P
3.1
3.2
3.3
3.4
Give a reason for each statement:
^ B = 90°
3.1.1
OA
^
3.1.2
OTA = 90°
Determine (with reasons) the size of each angle:
^R
3.2.1
AO
^
3.2.2
AQR
^Q
3.2.3
AP
Is PQ ∥ AB? Justify your answer.
If AQ is 8 units and the OA = 5 units, determine the length of TR,
giving reasons.
280
PLT MATHS LB 11 7th pgs (Real Book).indb 280
(1)
(1)
(2)
(2)
(2)
(4)
(3)
[15]
2012/07/02 2:24 PM
4
The figure shows circle ABCF with DC a tangent to the circle at C.
ABCD is parallelogram with side DA meeting the circle at F. E is a
^ = x.
point on AC so that EF ∥ CD. D
B
A
2
1
1
1
2
3 2
1
E
F
2
x
C
4.1
4.2
4.3
5
D
Prove CF = CD.
^ in terms of x, giving reasons.
Determine A
1
Prove that ECDF is a cyclic quadrilateral.
(5)
(4)
(5)
[14]
A farmer wants to subdivide a paddock ABCD by erecting a fence from
B to E and from E to C. The dimensions of ABCD are:
^ E = 35°
AB = 28 m; ED = 14 m; and BA
The farmer plans to use 41 m of fencing at an angle of 25° to BC.
Let the point where the fencing reaches line AD be E.
C
B
28
A
5.1
5.2
5.3
5.4
35°
25°
41
x
E
14
D
If x represents the angle between the fence BE and the
^ A), determine the value of x, correct to the
boundary AE (that is, BE
nearest degree.
Explain why there is no ambiguity with regard to the answer for 5.1.
If the farmer erects a fence from E to C, so that CE ⊥ BE, determine
the length of fencing that he will require (that is, determine the
length of CE), correct to the nearest metre.
Determine the area of the paddock ABCD, to two decimal places.
(3)
(1)
(3)
(6)
[13]
281
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Test 2 continued
6
^ C = y.
^ C has ABC = x; AC = b and BA
In the diagram △AB
^ C = θ.
Line CD is drawn perpendicular to BC, so that CD = a and BD
sin x __
b
Prove that: _____
cos y = a
A
y
B
b
x
C
a
0
D
7.
7.1
7.2
8.
8.1
8.2
9
Tom hears that the value of his car depreciates at 13% p.a. on a
straight line basis. If after 4 years the value of his car is R35 640,
calculate the original price of his car.
If Tom’s car had depreciated on a reducing balance method,
calculate the rate of depreciation if his car had depreciated to
R35 640 after 4 years. (Use your answer to 7.1 as the original
price of the car.)
Uthmaan deposits R8 000 into a savings account. Four years later
he withdraws R5 500 from the account. Interest is calculated at
7,8% p.a. compounded quarterly for the first three years,
and 9% p.a. compounded annually thereafter. Calculate how much
money will be in the account at the end of 6 years.
What was the effective annual interest rate that Uthmaan received
over the first 3 years?
Forty-two percent of the students at a local college live in the college
residence and fifty-five percent work part time. Ten percent of the students
work part time and live in residence.
9.1
Draw a Venn diagram showing the relationships.
9.2
Determine the probability that a student who does not live in
residence works part-time.
9.3
Determine the probability that a student chosen at random does
not live in residence or work part-time.
[7]
(3)
(3)
[6]
(8)
(3)
[11]
(4)
(1)
(1)
[6]
282
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10
Conrad investigates the effect of using a car phone on the speed that a person
drives. He interviewed 750 people in his area, and the results of his survey are
shown in the contingency table.
Speeding violation
Car phone user
Not a car phone user
No speeding violation
25
Total
250
450
Total
750
10.1 Copy and complete the contingency table.
10.2 Calculate these probabilities using the table:
10.2.1 P(person had no speeding violation)
10.2.2 P(person is a car phone user or person had no
speeding violation)
10.2.3 P(person is a car phone user given person had a
speeding violation)
10.3 Prove that, according to this survey, being a car phone user and
getting a speeding violation are independent events.
(5)
(1)
(3)
(2)
(4)
[15]
Total: 100 marks
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Term 3 summary
Topic 8
Measurement
Summary of formulae h represents the perpendicular height and s the slant height.
Solid
Volume
Total surface area
area of base × h
area of base + areas of lateral surfaces
1 (area of base ) × h
__
area of base + areas of all triangles
π r2 h
2πr2 +2πrh
1 πr2 h
__
3
π r2 +π r s
4 πr3
__
3
4πr2
2 πr3
__
3
3πr2 (includes flat circular surface)
Right prism
Right pyramid
h
s
r
s
h
3
x
Right cylinder
Right cone
h
s
r
s
h
x
Sphere
sphere
d
r
Hemisphere
r
284
radius
Term 3 summary
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Topic 10
Euclidean geometry and measurement
E
F
O
O
A
A
B
K
OR ⊥ AB (line from centre to midpoint
chord)
AK = KB (OK ⊥ AB)
A
D
A
B
R
B C
^ = F^ (equal chords AB & CD)
E
R
P
O
Q
D
O
D
D
B
C
^ C = 2A
^
BO
(∠ at centre)
^ =D
^
A
(∠s on chord BC)
^
^
Reflex O = 2R (∠ at centre)
E
G
E
G
O
F
^ = 180° (opp ∠s cyclic quad)
^+G
E
^ = 90° (∠ on diameter)
D
F
D
G
O
G
E
D
A
F
H
^H = D
^ (ext ∠ cyclic quad)
GF
B
C
^ C = 90° (rad ⊥ tan)
OB
E
DEFG is a cyclic quad
(DE subtends equal ∠s at F & G)
D
E
A
A
T
M
1 2
P
3
x
B
^ =E
^ (tan AB, chord BD)
B
1
^ =D
^ (tan BC, chord BD)
B
3
B
180° = x
C
V
N
MNVT is a cylic quad (opp ∠’s suppl)
PA = PB
(tangents from common point P)
A
D
C
B
A
D
B
^ B = 90°)
AB is the diameter (AC
C
E
ABCD is a cyclic quad
^ E = int opp A
^)
(ext DC
Term 3 summary
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Term 3 summary continued
Topic 10
Trigonometry: Sine, cosine and area rules
When solving for lengths or angles of triangles, remember:
a ; tan θ = __
o ; cos θ = __
o
• If it is a right-angled triangle, use the trigonometric ratios: sin θ = __
a
h
h
There are two cases of non-right-angled triangles that require the cosine rule. These are:
• SAS (that is, two sides and the angle between them). In this case we use:
a2 = b2 + c2 – 2bc cos A or b2 = a2 + c2 – 2ac cos B or c2 = a2 + b2 – 2ab cos C
• SSS (that is, three sides and no angles). In this case we use:
2
2
2
2
2
2
2
2
2
b + c – a or cos B = _________
a + c – b or cos C = _________
a +b –c
cos A = _________
2ac
2bc
2ab
All other cases of non-right-angled triangles, use the sine rule:
a = ____
c
b = _____
• When looking for a side, use: ____
sin A sin B sin C
sin A = ____
sin B = _____
sin C
• When looking for an angle, use: ____
a
c
b
• When you are given two sides and a non-included angle, look out for the ambiguous case. This occurs
when the side opposite the given angle is the shorter of the two given sides.
To determine the area of a triangle we can use:
1 base × height for triangles where the perpendicular height is known or
• area = __
2
1 ab sin c or __
1 ac sin b or __
1 bc sin a
area △ABC = __
2
2
2
Topic 11
Finance, growth and decay
• Growth takes place with:
simple interest: A = P(1 + in) or
compound interest: A = P(1 + i)n
Decay takes place with:
simple decay (called the straight line method): A = P(1 – in)
compound decay (called the reducing balance method): A = P(1 – i)n
• If the annual interest that is quoted (the nominal interest rate) is compounded more frequently than
once a year, the effective interest rate will be higher than the nominal interest rate, and is determined
using the formula:
(
)
(m) m
i
1 + 1eff = 1 + ___
m
• To determine the amount accumulated after an investment has been growing with compound interest
that is compounded k times per year: divide the quoted interest rate by k and multiply the number of
years by k.
• When more than one transaction occurs, draw a time-line to visualise what has happened over time.
Remember to take all values to any ONE moment in time, before adding or subtracting values. Use the
logic that ‘total of money in = total of money out’.
• When taking values back in time, you are finding the P value for a known A value, so the formula
becomes A = P(1 + i)–n
286
Term 3 summary
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Topic 12
Probability
• The probability of an event taking place is always:
total number of ways an event can occur
_______________________________________
total number of possible outcomes for the event
Notation:
• P(A) means ‘the probability of event A happening’.
• P(A′) means ‘the probability of event A not happening’.
• P(A ∩ B) means ‘the probability of event A and event B happening’.
• P(A ∪ B) means ‘the probability of event A or event B happening’.
• P(A | B) means ‘the probability of event A happening, assuming that event B
is taken as given’.
• For all events A and B: P(A or B) = P(A) + P(B) – P(A and B)
• P(A or B) = P(A) + P(B) if A and B are mutually exclusive.
• For complementary events (mutually exclusive and exhaustive):
P(B) = P(A′) = 1 – P(A)
• For independent events: P(A and B) = P(A) × P(B)
• Use Venn diagram, tree diagrams and contingency tables to represent data
and to assist in determining the probabilities of various outcomes.
Term 3 summary
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Term
4
288
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TOPIC 13 Statistics
Unit 1 Histograms and frequency polygons
Unit 2 Ogive curves
Unit 3 Variance and standard deviation
of ungrouped data
Unit 4 Symmetric and skewed data
Unit 5 Identification of outliers
Revision
Formal assessment: Term 4 Test
Term Summary
Formal assessment: Examination practice Paper 1
Examination practice Paper 2
288
294
299
304
307
312
316
318
320
323
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TOPIC
13
Statistics
Unit 1: Histograms and frequency polygons
Histograms
KEY WORDS
histogram – a graph with
rectangles that show the
frequency distribution of
grouped data with no space
between the groups, so the
rectangles touch each other;
we try to keep the groups of
equal width so that the height
of the rectangle (frequency
distribution) is the same as
the frequency
bar graph – a graph with
rectangles that show the
frequency with which
different types of data values
occur; these data values need
not be in consecutive groups,
so the rectangles do not
touch each other
Histograms are similar to bar graphs, but have important differences:
• In histograms the bars are adjacent to each other with no gaps between the
rectangles whereas in bar graphs the bars are sometimes separate rectangles.
• Bar graphs always have bars of equal widths whereas histograms sometimes have
bars of different widths.
• The height of the bar in bar graphs represents the frequency of each category
whereas in histograms the height of the bar represents the relative frequency or
frequency density. This is relevant when the bars are different widths and the area
of each bar becomes important.
WORKED EXAMPLE
A teacher analyses the ages of a group of children in a playground and tabulates
the data. Draw a bar graph to represent the data in the table.
Ages (years)
Frequency
0<x≤2
1
2<x≤4
3
4<x≤5
6
5<x≤7
8
7 < x ≤ 11
2
The problem with drawing a bar graph from the data in the table is that the age
groupings are not equally spaced, so the widths of the bars will not be the same.
Most of the groupings have a width of 2, so we adapt the frequency of the groups
which do not have a width of 2 to compare the area of each bar correctly with one
another.
Age group 4–5 has a width of 1, so we double the frequency from 6 to 12. Age
group 7–11 has a width of 4, so we halve the frequency from 2 to 1.
We now have this table:
290
Ages (in years)
Frequency
Relative frequency (Area)
0<x≤2
1
1
2<x≤4
3
3
4<x≤5
6
12
5<x≤7
8
8
7 < x ≤ 11
2
1
Topic 13 Statistics
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When drawing a histogram, we place the groups on the x-axis and the relative
frequency on the y-axis:
Histogram showing ages of children
14
Relative frequency
12
10
8
6
4
2
0
0–2
2–4
4–5
5–7
6–11
Number of children
We work on this principle:
area of bar (that is, group width × relative frequency) = frequency × standard width
We chose 2 as the standard width (the most common grouping width), so the area
of each bar is its frequency × 2.
REMEMBER
Be aware that histograms that have different group widths occur, but we deal only
with histograms where the group widths are the same, and therefore the relative
frequency (value represented on the y-axis) will always be the same as the frequency.
Frequency polygons
A frequency polygon enables us to represent the information in a frequency table
by means of line graphs. If we draw a line from the mid-class value of each bar in
a histogram and a line to the mid-class of the groupings before and after the given
groupings, we create a frequency polygon.
WORKED EXAMPLE
The frequency table shows the Mathematics marks obtained by Grade 11
Mathematics learners in a test out of 50 marks.
Number of learners
10 ≤ x < 20
25
20 ≤ x < 30
74
30 ≤ x < 40
66
40 ≤ x < 50
35
KEY WORD
frequency polygon – a
polygon that demonstrates
the frequency of each group
in a set of grouped data, and
that displays the spread of
the data
Note:
• It is important to complete the polygon on each end.
• The area of the polygon will equal the total area of the histogram.
Mark obtained (out of 50)
The area of each bar of a
histogram is important, so it
is easier if the width of each
group is constant. We use this
principle when looking at the
area of a frequency polygon.
REMEMBER
A polygon is a closed shape
that is made up of three
or more lines. Examples of
polygons are triangles (three
lines), quadrilaterals (four
lines) and pentagons (five
lines).
Unit 1 Histograms and frequency polygons
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1
2
Draw a histogram to represent the results shown in the table.
Draw a frequency polygon on the same set of axes as the histogram.
SOLUTIONS
1
80
Histogram showing learners’ marks
Frequency
60
40
20
0
2
KEY WORD
discrete data – data with a
restricted number of values;
includes units of quantity (for
example, number of people,
number of cars, with only
whole number values)
292
10
Mark obtained
(out of 50)
20
30
Marks
Mid-class value
40
50
60
Number of learners
(frequency)
0 ≤ x < 10
4,5
0
10 ≤ x < 20
14,5
25
20 ≤ x < 30
24,5
74
30 ≤ x < 40
34,5
66
40 ≤ x < 50
44,5
35
50 ≤ x < 60
54,5
0
Notice that the mid-class value for the group 0 ≤ x < 10 is not 5. This is discrete
data so the highest possible score less than 10 for a test out of 50, is 9. So we must
find the mid-point from 0 to 9.
We can now plot a point for each mid-class value at the height of the relevant
frequency (which should be in the middle of the possible scores), and then join
these points with lines.
To complete the polygon, consider groups of the same width before and after the
given grouping. We recognise that the frequency of those groups is zero.
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80
Histogram showing learners’ marks
Frequency
60
40
20
0
10 14,5
4,5
20 24,5
30 34,5
Marks
40 44,5
50 54,5
60
To understand why the area of the polygon is the same as the total area of the bars,
consider the triangles that are not common to the polygon and the bars (see triangles
A–K in the graph).
80
Histogram showing learners’ marks
E
D
F
Frequency
60
40
G
H
C
B
20
0
J
K
A
4,5
10 14,5
20 24,5
30 34,5
Marks
40 44,5
50 54,5
60
△A is part of the polygon but not part of the histogram.
△B is part of the histogram but not part of the polygon.
But area △A = area △B
Similarly, △C = △D; △E = △F; △G = △H and △J = △K
So △B + △D + △E + △ G + △ J = △A + △C + △F + △H + △K
Therefore the total area of the histogram equals the area of the polygon.
It is not necessary to draw a histogram before we draw a frequency polygon. All we
require is the mid-class value of each group and the frequency of each group.
Unit 1 Histograms and frequency polygons
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WORKED EXAMPLE
A traffic official records the speed at which vehicles travel on a road on a certain
day in a frequency table:
Speed of vehicle (km/h)
Number of vehicles
60 ≤ x < 70
12
70 ≤ x < 80
18
80 ≤ x < 90
32
90 ≤ x < 100
47
100 ≤ x < 110
52
110 ≤ x < 120
68
120 ≤ x < 130
45
130 ≤ x < 140
16
Draw a frequency polygon to represent this data.
SOLUTION
Firstly we need to complete the given table to include the mid-class values
as shown in the next table.
Speed of vehicle (km/h)
Mid-class value
Number of vehicles
50 ≤ x < 60
55
0
60 ≤ x < 70
65
12
70 ≤ x < 80
75
18
80 ≤ x < 90
85
32
90 ≤ x < 100
95
47
100 ≤ x < 110
105
52
110 ≤ x < 120
115
68
120 ≤ x < 130
125
45
130 ≤ x < 140
135
16
140 ≤ x < 150
145
0
KEY WORD
continuous data – data
without breaks and that can
have any value; includes all
units of measurement (for
example, height, mass)
294
Note that the mid-class value for the group 50 ≤ x < 60 is 55 even though 60 is not
included in the group. This is continuous data, which means that the highest
possible score less than 60 can be so close to 60, that we choose the mid-point
between 50 and 60. We can now plot a point for each mid-class value at the height
of the relevant frequency, and join these points with lines.
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y
Frequency polygon showing vehicle speeds
70
60
Frequency
50
40
30
20
10
50
60
70
80
90 100 110 120 130 140 150
Speed (km/h)
x
Unit 1 Histograms and frequency polygons
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Unit 2: Ogive curves
KEY WORD
cumulative frequency
– a running total of the
frequencies
An ogive curve is also called a cumulative frequency curve. Remember that:
• Ogive curves involve cumulative frequency and not individual frequencies.
• The graph is a smooth curve, and not straight lines that are joined together.
• Unlike frequency polygons, which work with mid-class values, this curve uses the
end points of each group.
• While the curve is connected to the x-axis at the end point of the previous group
(the lower limit of the first group), the curve does not return to a point on the
x-axis like the frequency polygon does.
WORKED EXAMPLE
A teacher is concerned about the results obtained by her class in the examinations.
She feels that the learners are spending too much time watching television instead
of doing homework or studying. She conducts a survey to discover how many
hours of television the learners watch each week. The results of this survey are
shown in the table.
1
2
3
4
5
Number of hours of television
watched each week
Number of learners (f )
0≤x<5
30
5 ≤ x < 10
6
10 ≤ x < 15
29
15 ≤ x < 20
27
20 ≤ x < 25
26
25 ≤ x < 30
20
30 ≤ x < 35
18
35 ≤ x < 40
7
40 ≤ x < 45
2
Draw an ogive curve to represent the data in the table.
Use the ogive curve to determine approximate values for the median, as well
as the upper and lower quartiles of the data.
Determine the modal group and the approximate mean of the data by using
the frequency table.
Could the mode have been determined by looking at the ogive curve?
Determine an approximate value for the interquartile range.
SOLUTIONS
1
296
We first need to determine the cumulative frequency so we can draw an ogive
curve. Extend the given table to represent this data.
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Number of hours of television
watched each week
Number of
learners ( f )
Cumulative
frequency
0≤x<5
30
30
5 ≤ x < 10
6
36
30 + 6
10 ≤ x < 15
29
65
36 + 29
15 ≤ x < 20
27
92
65 + 27
20 ≤ x < 25
26
118
92 + 26
25 ≤ x < 30
20
138
118 + 20
30 ≤ x < 35
18
156
138 + 18
35 ≤ x < 40
7
163
156 + 7
40 ≤ x < 45
2
165
163 + 2
Total f = 165
Notice that the total frequency should always be the same as the final
cumulative frequency value.
Now plot the end points of each group with the cumulative frequency. Join
the points with a smooth curve to create an ogive curve. The curve starts at the
end of the preceding group, which will be the start of the first group given.
y
Ogive curve showing learners’
hours of television viewing
Cumulative frequency
150
100
50
10
20
30
40
Number of hours watching television
50
x
Unit 2 Ogive curves
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2
In Grade 10 you learnt that the median is the middle data value. This means
1 (n + 1)th value where n represents the
that the median, (Q2), will be the __
2
1 (165
total number of data values. So, in this case, the median will be the __
2
th
rd
+ 1) value, that is, the 83 value. We can use the ogive curve to find an
approximation of this value by:
• finding 83 on the y-axis (as close as possible)
• reading across to the curve, and then down to the x-axis to find the
approximate corresponding x value.
1 (n + 1)th data value, that is,
The lower quartile, (Q1), will be the __
1 (165 + 1)th = the 41,5th value.
the __
4
4
3 (n + 1)th data value, that is,
The upper quartile, (Q3), will be the __
3 (165 + 1)th = the 124,5th value.
the __
4
4
These values are difficult to read off accurately. Statistics often works with
estimates and is usually based on a sample of data values, which is part of a
greater number of data values. The emphasis is on determining trends rather
than a high level of accuracy.
To determine the approximate upper and lower quartile values, we read the
values from the ogive curve in a similar manner to how we read the median.
Ogive curve showing learners’
hours of television viewing
y
Cumulative frequency
150
124,5
100
83
50
41,5
10
Q1
20
Q2
30
40
50
x
Q3
Number of hours watching television
The approximate values read off from the graph are:
Q1(lower quartile) ≈ 11; Q2 (median) ≈ 18,5; Q3(upper quartile) ≈ 26,5
298
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3
The modal group is easy to see as it is the group with the highest frequency
(group 0 ≤ x < 5). So the modal time is 0 to 5 hours.
To determine the approximate mean, we find the mid-class values, multiply
each value by the frequency, and divide the total by the total frequency
(covered in Grade 10).
Number of hours of television
watched each week
Number of
learners (f )
Mid-class
value (m)
Mean
m×f
0≤x<5
30
2,5
75
5 ≤ x < 10
6
7,5
45
10 ≤ x < 15
29
12,5
362,5
15 ≤ x < 20
27
17,5
472,5
20 ≤ x < 25
26
22,5
585
25 ≤ x < 30
20
27,5
550
30 ≤ x < 35
18
32,5
585
35 ≤ x < 40
7
37,5
262,5
40 ≤ x < 45
2
42,5
Totals
165
85
3 022,5
3 022,5
165
Therefore the approximate mean = _______ = 18,32 (to two decimal places)
4
We cannot read the mode from the ogive curve, but we can observe that the
modal group is the first group. The gradient of the ogive is steepest at this
stage, which shows that the change in frequency is highest over that portion
of the graph.
5
The interquartile range = upper quartile – lower quartile
= 26,5 – 11
= 15,5
EXERCISE 1
1
The table shows the heights of a group of learners.
1.1
Height (in cm)
Frequency
140 ≤ x < 150
15
150 ≤ x < 160
27
160 ≤ x < 170
18
170 ≤ x < 180
10
Exercise 1 has
questions that
cover Units 1
and 2.
Copy and complete the frequency table on the next page. Use the
information in the table to answer the questions that follow.
Unit 2 Ogive curves
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Height (in cm)
Frequency (f )
140 ≤ x < 150
15
150 ≤ x < 160
27
160 ≤ x < 170
18
170 ≤ x < 180
10
Mid-class value (m)
Mean m × f
Totals
1.2
1.3
1.4
1.5
1.6
300
1.1.1
Determine the approximate mean height (to two decimal places).
1.1.2
What is the modal group?
1.1.3
Determine the range.
Draw a histogram to represent the data in the frequency table.
On the same set of axes, draw a frequency polygon to represent this data.
Draw an ogive curve on a new set of axes.
Use your ogive curve to determine an approximate value for:
1.5.1
the median
1.5.2
the lower quartile
1.5.3
the upper quartile
1.5.4
the interquartile range
1.5.5
the 90th percentile
Draw a box-and-whisker diagram to represent the spread of this data.
Topic 13 Statistics
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Unit 3: Variance and standard deviation
of ungrouped data
Statisticians have derived formulae to find the variance and standard deviation of
a set of data values. It is important to understand these formulae and to know how
to use them. However, we can also calculate these values using a calculator, which
is a quicker method. Unless a question requires the use of the formulae, choose the
calculator method.
Using formulae
KEY WORDS
variance – the average of the
squared differences from the
mean
standard deviation – the
square root of the variance
Standard deviation is a way of measuring how spread out a set of data values are.
It is often useful to know whether most of the data values lie close to the mean, or
whether they are widely spread out with many values much higher or much lower
than the mean.
The graph shows two sets of data both with a mean of 100. But the blue data values
have a far greater standard deviation than the red data values. It is clear that the
blue data values are far more spread out than the red data values. So, a low standard
deviation indicates that the data values tend to be very close to the mean, whereas a
high standard deviation indicates that the data values are spread out and have a large
range.
The matric results of two schools are represented on the frequency polygons shown
below. Although the two schools have the same mean (58%), the results of the
learners in the school shown in blue (standard deviation of 16) are far more widely
spread than those of the learners from the school shown in red (standard deviation
of 9).
y
Frequency polygon showing standard deviations of matric results
standard deviation of red = 9
standard deviation of blue = 16
Frequency
Mean = 58%
58
x
Marks obtained (%)
The crucial aspect that we are interested in when determining the standard deviation,
is to find out how far each data value falls from the mean. This process leads us to
determine the variance (the average of the squared differences from the mean). The
standard deviation is therefore the square root of the variance.
Unit 3 Variance and standard deviation of ungrouped data
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The method we use to find the variance is:
__
sum of data values
• Determine the mean value for the set of data: x = ___________________
•
•
(
__
number of data values
)
Subtract the mean from each data value: (xi − x) where xi represents each value of x
__
Square each of these differences: (xi − x)2. This will ensure that all values will be
positive, as we are not interested in whether the data values are higher or lower
than the mean, but only how far they are from the mean. Also, by squaring we
allow the bigger differences to stand out, as these have an impact on the standard
deviation.
n
∑( xi − __x )2
•
Add these squared differences:
•
Divide the total of the squared differences by the number of data values:
( xi − x2 )
i=1
This result will represent the average of the squared differences from
__________
n
the mean, that is, the variance.
∑( xi − __x )2
i=1
The formula to find the standard deviation is therefore: σ = __________
n
•
n
i=1
∑
√
WORKED EXAMPLE
A game ranger measured the heights of different buck and recorded the results.
The heights at the shoulders are: kudu 150 cm; waterbuck 130 cm; duiker 100 cm;
steenbok 50 cm and dik-dik 35 cm
Heights (cm)
Heights of different buck
150
125
100
75
50
25
Determine the variance and standard deviation for this set of data.
SOLUTION
First we need to find the mean:
1 500 + 1 300 + 1 000 + 500 + 350 = _____
4 650 = 930
Mean = _____________________________
5
5
Now we can determine the variance using a table:
__
( xi − x )
( xi − x )2
1 500
570
324 900
1 300
370
136 900
1 000
70
4 900
500
–430
184 900
350
–580
336 400
Total
302
__
Heights in mm ( xi )
988 000
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n
Therefore the variance =
σ2
=
988 000 = 197 600
∑(xi − __x)2 = _______
5
i=1
Notice: the symbol for standard deviation is σ. As the variance is the square of the
standard deviation, we can use the symbol σ2 for variance.
_______
Therefore the standard deviation = σ = √197 600 = 444,5 mm.
We can now determine which heights are within one standard deviation of the mean:
Add the standard deviation value of 444,5 to the mean value 930 to get the upper
limit (1 374,5).
• Subtract the standard deviation value of 444,5 from the mean 930 to get the lower
limit (485,5)
• So any height that falls between 485,5 mm and 1 374,5 mm lies within one
standard deviation from the mean.
•
Any height that does not fall within one standard deviation of the mean would be
regarded as being particularly tall (the 1 500 mm buck) or short (the 350 mm buck).
Using a calculator
When there is a large set of data values, use a calculator to get the variance and/or
standard deviation. It would be tedious to go through the process necessary to apply
the formulae. Using the example about the heights of five different buck, we can use
our calculators to determine these answers.
•
•
•
•
Go to MODE: choose the STAT option.
Choose the VAR option: You will now have a table on your screen.
Type in each of the data values, pressing = after each entry.
You should now have this information on your screen:
X
•
•
•
•
•
•
1
600
2
480
3
170
4
410
5
300
Press AC.
Go to SHIFT 1 (the other STAT option).
Choose the VAR option.
Choose the xσn option, followed by =
You should have the answer 147,7 on the screen, which is the standard deviation.
If you want the variance, then square the standard deviation.
If you have data that occurs more often than once, first set up your calculator to
include a frequency column.
Unit 3 Variance and standard deviation of ungrouped data
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WORKED EXAMPLE
The table shows the number of questions answered correctly by a class of learners
in a general knowledge test consisting of ten questions.
1
2
Number of questions
answered correctly
Number of learners
0
3
1
2
2
1
3
5
4
7
5
6
6
2
7
4
8
5
9
2
10
1
Determine the mean number of questions answered correctly.
Determine the standard deviation (correct to one decimal place).
SOLUTIONS
1
Using a calculator:
• Go to SET UP.
• Scroll down until you see the option STAT.
• Now choose frequency ON. (You will now have created a frequency
column when you get to your table.)
• Clear (AC).
• Go to MODE and choose STAT.
• Choose VAR.
• Insert the two columns of values into the table, using the top central key
to scroll up, down , left and right. Remember to press = after each entry.
• Clear (AC).
• Go to the other STAT option (Shift 1).
• Choose the __VAR option.
• Choose the x option followed by =
This will show that the mean is 4,9 questions answered correctly.
2
You have already set up the table of values, so:
• Clear (AC).
• Go to Shift STAT again and choose VAR again.
• Choose the xσn option, followed by =
This will show that the standard deviation is 2,6.
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EXERCISE 2
1
A teacher asked a group of learners how long in minutes it took them to
complete their mathematics homework. They gave these answers:
12; 19; 33; 40; 24; 25; 15; 38
1.1
Determine the mean number of minutes taken by the learners to complete
their homework.
1.2
Determine the variance and standard deviation to two decimal places by
_________
n
completing the table
below using the formula: σ =
√
∑(xi − __x)2
i=1
__________
n
__›
( xi − x )
Time taken in minutes
12
__›
( xi − x )2
15
19
24
25
33
38
40
1.3
2
How many data values fall within one standard deviation of the mean?
A learner does a survey on 23 cars travelling past him. He counts the number
of passengers in each car. His results are recorded in the table.
Number of passengers
Frequency
1
2
2
9
3
5
4
4
5
3
Using your calculator, determine (correct to two decimal places):
2.1
the standard deviation
2.2
the variance
2.3
the mean.
Unit 3 Variance and standard deviation of ungrouped data
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Unit 4: Symmetric and skewed data
We describe how the data values are distributed throughout the range, relative to
the median.
• Symmetrical data values are balanced on either side of the median (the middle
value). If we draw a histogram (or frequency polygon) to represent symmetrical
data, we obtain a symmetrical shape.
•
y
10
8
6
4
2
2
4 Mode 6
Median
Mean
8
10
x
A symmetrical frequency polygon
A symmetrical histogram
This histogram, while not perfectly symmetrical, is close to being an exact image
on either side. When we have perfect symmetry as in the frequency polygon
with the middle value having the highest frequency, the median (middle value)
and modal values will be the same. When we work out the mean, it is also the
same value.
If we represent this spread in a box-and-whisker diagram, we can clearly see
the symmetry:
Skewed data values are more
spread out on one side than
on the other. Due to the high
frequency of many of the
smaller values, the median
(middle value) is closer to the
left. The values are more spread
out on the right, so we say that
this set of data is skewed to the
right (or the data is positively
skewed). Notice that the high
data values on the right will
cause the mean to be higher
than the median.
y
20
15
Frequency
•
10
5
100
Median
306
200
300 400
Ratio
500
600
x
Mean
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A box-and-whisker diagram of data values that are skewed to the right would be:
A frequency polygon can also show data that is skewed to the right:
y
10
8
6
4
2
2
4
6
8
10
x
y
The fact that there are some very low values will cause the mean to be
lower than the median.
A box-and-whisker diagram of this situation would be:
We can conclude:
• If the mean = the median, then the data is symmetrical.
• If the mean > the median, then the data is skewed to the right.
• If the mean < the median, then the data is skewed to the left.
It is possible for three sets of data values to have the same range and the
same interquartile range when one set is symmetrical, one is skewed to
the right and one is skewed to the left. See the box-and-whisker plots:
Frequency
In the histogram on the right, the mode (most frequent value) is the same
as the median, but the data values are more spread out on the left. We say
that this set of data values is skewed to the left (or negatively skewed).
24
22
20
18
16
14
12
10
8
6
4
2
45 46 47 48 49 50 51 52 53 54 55
Age
Mean
x
Median
Skewed to the right
Symmetrical
Skewed to the left
Notice that the widths of the boxes show the interquartile range and the widths of
the whiskers show the range, which are the same. But the central line which shows
the median is different, showing that some data values are skewed.
Unit 4 Symmetric and skewed data
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KEY WORD
outlier – a data value that
does not follow the trend of
the rest of the data
Unit 5: Identification of outliers
An outlier is a data value that does not follow the pattern or trend of the rest of the
data. We can sometimes see this clearly when the data values have been plotted in a
scatter plot.
y
20
x
15
10
5
x x xx
0
100
x
x xx x xx x x
xxx xx
xx
x
x
200
300
400
500
In this scatter plot, the data points follow a similar pattern and create a linear shape,
except for the point at approximately (350;16). This point clearly lies outside the
pattern created by the rest of the data values and we call it an outlier. If we omit
this outlier, the rest of the data values will have a strong positive correlation, and all
strongly follow a similar pattern that creates a straight line with a positive gradient.
In Grade 12 we will sketch scatter plots and discuss further the various types of
correlation that may be present. For now we must be able to visualise outliers by
recognising that they are points on a scatter plot which do not follow the pattern
created by the rest of the data values.
It is possible to determine whether a data value is an outlier without drawing
a scatter plot.
Any value that is either:
less than Q1– 1,5 × IQR, or greater than Q3 + 1,5 × IQR is an outlier.
WORKED EXAMPLE
The table indicates the CPI (consumer price index) inflation rate in South Africa
between 2005 and 2010.
1
2
3
308
Year
CPI rate
2005
3,4
2006
4,7
2007
7,1
2008
11,5
2009
7,1
2010
4,3
Determine the five-number summary for this set of data.
Draw a box-and-whisker plot to represent the dispersion of this data.
Calculate whether there are any outliers.
Topic 13 Statistics
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SOLUTION
1
Arrange these values in ascending order: 3,4; 4,3; 4,7; 7,1; 7,1; 11,5
The median is therefore between 4,7 and 7,1 which is 5,9.
( 4 )th data value, which is 4,075.
21 th data value, which is 8,2.
Q3 = ( ___
4 )
7
Q1 = __
So: Minimum = 3,4; Q1 = 4,075; Q2 (median) = 5,9; Q3 = 8,2;
Maximum = 11,5
2
3,4
3
3
4,075
4
8,2
5
6
7
8
11,5
9
10
11
12
IQR = 8,2 – 4,075 = 4,125
1,5 × 4,125 = 6,1875
Q1 – 6,1875 = – 2,11 and Q3 + 6,1875 = 14,3875
Therefore, although the value of 11,5 in 2008 was higher than the others
(causing the data to be skewed to the right), it was not high enough to be
classified as an outlier.
If the CPI in 2008 had been 14,5, then that data value would have been an
outlier, and the box-and-whisker diagram would look like this:
3
4
5
6
7
8
9
10
11
12
13
14
Notice that a very long whisker is often an indication that there is an outlier.
This outlier causes the data to be skewed to the right, even though the
median falls almost in the middle of the interquartile range.
Unit 5 Identification of outliers
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EXERCISE 3
Exercise 3 has
questions that
cover Units 4
and 5.
DID YOU KNOW?
In-migration refers to people
who are not born in a
particular area, but move and
settle there.
1
Raeez studied the in-migration figures for some municipalities of South Africa as
shown in the table.
Municipality
Province
In-migration figures in 2006
Nkangala DM
Mpumalanga
1 452
Siyanda DM
Northern Cape
1 504
Ehlanzeni DM
Mpumalanga
2 465
Nelson Mandela MM
Eastern Cape
6 715
Waterberg DM
Limpopo
11 694
Umgungundlovu DM
KwaZulu-Natal
13 149
Overberg DM
Western Cape
14 965
Bojanala Platinum DM
North West
20 168
Ekurhuleni MM
Gauteng
140 252
Zimbabwe
Botswana
Mozambique
Limpopo
Namibia
Mpumalanga
North West
Gauteng
Swazi land
Free State
Northern Cape
KwaZuluNatal
Lesotho
Eastern Cape
Western Cape
A provincial map of South Africa
1.1
1.2
1.3
1.4
1.5
1.6
Determine the five-number summary for the set of data in the table.
Draw a box-and-whisker plot to demonstrate this dispersion.
Describe the dispersion of this set of data.
Which aspect of this box-and-whisker diagram suggests that there is
an outlier?
Prove that the value of 140 252 is an outlier.
Give a possible reason for why the Ekurhuleni MM municipality had an
in-migration rate much higher than any of the other municipalities.
Raeez found a list which included more municipalities. This list is shown in the
table on the next page, with the figures arranged in ascending order.
310
Topic 13 Statistics
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Municipality
Province
In-migration in 2006
Nkangala DM
Mpumalanga
1 452
Siyanda DM
Northern Cape
1 504
Ehlanzeni DM
Mpumalanga
2 465
Southern DM
North West
4 914
Nelson Mandela MM
Eastern Cape
Waterberg DM
Limpopo
11 694
6 715
Umgungundlovu DM
KwaZulu-Natal
13 149
Overberg DM
Western Cape
14 965
West Coast DM
Western Cape
17 211
Metsweding DM
Gauteng
18 560
Boland DM
Western Cape
18 770
Bojanala Platinum DM
North West
20 168
Eden DM
Western Cape
22 983
eThekwini MM
KwaZulu-Natal
27 277
West Rand DM
Gauteng
42 674
City of Johannesburg MM
Gauteng
120 330
City of Cape Town MM
Western Cape
129 400
City of Tshwane MM
Gauteng
137 685
Ekurhuleni MM
Gauteng
140 252
1.7
1.8
Determine whether Ekurhuleni’s data value is still an outlier. Explain why
your answer reflects the analysis of in-migration figures better than the
result in Question 1.5.
Complete the frequency table which divides the values from the table
into groups.
In-migration groups
Frequency
0–30 000
30 000–60 000
60 000–90 000
90 000–120 000
120 000–150 000
1.9
Draw a histogram to represent the grouped data in the table.
1.10 Give a possible reason for the four municipalities at the upper end of the
range all having in-migration figures that are far higher than the rest.
2
Many concerned South Africans who are aware of the problems of water shortage
hold this opinion: ‘Another South African town runs out of water as two of its
dams dry up. Urgent attention is needed to prevent this happening to more
towns and cities across the country. Our nation’s use of water will have to change
to prevent total water scarcity.’
Refer to the table on dam levels and answer these questions:
2.1
Determine the mean percentage of water in the dams listed in the table.
2.2
Determine the standard deviation for water in the dams.
2.3
How many data values are within one standard deviation of the mean?
2.4
Determine the interquartile range.
Unit 5 Identification of outliers
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2.5
2.6
Are there any outliers in this set of data? Justify your answer.
Compare the median and mean values. State whether the data is
symmetrical, skewed to the left or skewed to the right.
The table shows the percentage filled of some dams around the country in
March 2010.
312
Dam
River
% full
Middel-Metabo Dam
Middel-Metabo River
Berg River Dam
Berg River
83,8
Eikenhof Dam
Palmiet River
46,8
Misverstand
Berg River
100
Steenbras Dam lower
Steenbras River
46
Steenbras Dam Upper
Steenbras River
69,2
6,9
Voelvlei Dam
Voëlvlei River
Wemmershoek Dam
Wemmers River
52,2
65
Groendal Dam
Swartkops River
40,1
Ernest Robertson Dam
Groot Brak River
87,2
Garden Route Dam
Swart River
32,2
Hartebeestkuil Dam
Hartenbos River
46,8
Impofu Dam
Krom River
49,7
Kromrivier Dam
Krom River
12,6
Wolwedans Dam
Groot Brak River
30,6
Korentepoort Dam
Korinte River
26,9
Duiwelshok Dam
Duiwenhoks River
26,9
Kommandodrift Dam
Tarka River
39,0
Bridle Drift Dam
Buffalo River
32,8
Gariep Dam
Orange River
95,0
Elandskloof Dam
Elands River
36,4
Roode Els Berg Dam
Sanddriftskloof River
36,9
Theewaterskloof Dam
Riviersonderend
70,6
Gamka Dam
Gamka River
Kammanassie Dam
Kammanassie River
Leeugamka dam
Leeu River
12,1
Miertjieskraal Dam
Brand River
0,0
Stompdrift Dam
Olifants River
0,0
9,4
33,9
Beervlei Dam
Groot River
0,0
Haarlem Dam
Groot River
23,2
Kouga Dam
Kouga River
33,9
Loerie Dam
Loerie Spruit
40,9
Darlington Dam
Sondags River
42,8
Grassridge Dam
Groot Brak River
41,2
Katrivier Dam
Kat River
48,6
Katse Dam
Malibamatso River
98,2
Vanderkloof Dam
Orange River
98,6
Vaal Dam
Vaal River
99,8
Source: Water Rhapsody
Topic 13 Statistics
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Revision
1
Fifteen households were surveyed in suburb A to find out how much each one
spent on electricity for a ten-day period. The results in rand are:
90 102 50 125 141 220 196 78 137 142 123 157 118 165 121
1.1
Determine:
1.1.1
the median
1.1.2
the lower quartile
1.1.3
the upper quartile.
Draw a box-and-whisker diagram to illustrate this data.
Calculate the mean expenditure on electricity for the 15 households.
Determine the standard deviation for the data to two decimal places.
Calculate the percentage of households whose expenditure on
electricity falls within one standard deviation of the mean.
The results of the same survey conducted in suburb B are shown
in the box-and-whisker diagram.
1.2
1.3
1.4
1.5
1.6
20
40
(3)
80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420
Compare and comment on the dispersion of household expenditures
on electricity in suburbs A and B.
Which suburb is more likely to have an outlier in their data values?
Justify your answer.
1.7
2
60
(2)
(1)
(1)
(4)
(3)
(2)
(3)
(2)
[21]
The bar chart shows the distribution of the marks scored for a test by
a group of learners.
Frequency
Distribution of marks
2.1
2.2
16
14
12
10
8
6
4
2
0
1
2
3
Marks out of 5
4
5
Show clearly that the mean mark is 3.
Copy and complete the table using the data on the bar chart.
xi
f
__
xi – x
__
(xi – x)2
(2)
__
fi × (xi – x)2
1
2
3
4
5
Total
–
–
(4)
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TOPIC 13: REVISION CONTINUED
2.3
How calculate the standard deviation of the mark distribution,
correct to two decimal places.
The same group of learners wrote a second test. The standard
deviation for the second test was 2,12. Compare and discuss the
results of the two tests.
2.4
3
(2)
[11]
A squad of 20 rugby players had their masses recorded as a stem-and-leaf plot.
Stem
Leaf
6
5
7
7
8
8
9
8
0
0
1
9
0
5
9
10
5
8
8
11
4
8
8
12
9
3.1
3.2
3.3
4
(3)
8
9
Write down the five-number summary for the data.
(5)
Use your five-number summary to draw a box-and-whisker
diagram to represent the data.
(3)
Use your box-and-whisker diagram to help you decide whether
the statements are true or false. If false, write down the correct statement.
A: 25% of the players weigh more than 108 kg.
B: There are more players whose weight lies in the third quartile
than in the second quartile.
C: The data is skewed to the right.
(3)
[11]
A group of learners wrote a standardised Mathematics test that was scored
out of 60. The results were represented in a cumulative frequency graph or
ogive curve.
Cumulative frequency curve of Mathematics results
250
Cumulative frequency
200
150
100
50
0
0
5
10
15
20
25
30 35
Mark
40
45
50
55
60
65
314
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2012/07/02 2:25 PM
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Expenditure (in rand)
x < 50
(4)
(2)
[15]
Frequency Cumulative frequency
2
50 ≤ x < 100
22
100 ≤ x < 150
58
150 ≤ x < 200
14
200 ≤ x < 250
4
5.1
5.2
5.3
5.4
5.5
5.6
6
(1)
(1)
(1)
(2)
(1)
(3)
The table shows the results from a survey of cell phone expenditure for
100 learners.
Copy and complete the table.
Draw an ogive curve for this data.
Use your graph to estimate the median cell phone expenditure for
this group of learners, indicating where you read off your answer.
Draw a frequency polygon for this data.
Determine an approximate mean for this data.
Discuss the dispersion of this data.
Grade 11 learners conducted a survey on weather
patterns in their area. They measured the
average rainfall per month against the average
temperature per month over seven months.
They represented their results
on a scatter plot.
6.1
What can you see from the graph about the
general trend of rainfall
patterns as the temperature increases?
(1)
6.2
Which reading would you consider an
outlier?
(1)
[2]
y
100
(1)
(4)
(2)
(4)
(4)
(2)
[17]
Scatter plot showing rainfall and temperature
90
80
Rainfall in mm
5
How many learners wrote the test?
How many learners scored at most 20 out of 60?
What was the median test score?
Determine the interquartile range. (Show your working.)
Which was the modal group score?
Determine whether there were any outliers in this set of data.
Use your calculator to determine an approximate value for:
4.7.1
the mean (show some working)
4.7.2
the standard deviation.
70
60
50
40
30
20
10
0
10
20
30
40
50
60
70
80
x
Temperature in °C
315
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Formal assessment: Term 4 Test
1
The percentage marks for a class of 19 learners for a Mathematics test were:
81; 80; 74; 75; 57; 55; 91; 88; a; 76; 61; 60; 83; 89; b; 69; 80; 90; 80
The box-and-whisker diagram shows the five point summary:
b
55
1.1
1.2
2
c
88
a
If the range of the data is 40 and the inter-quartile range is 23,
determine the values of a, b and c.
Describe the skewness of the distribution.
(4)
(2)
[6]
Snowskiers pass a point on a ski slope at a speed in km/h.
Their speeds are in the table.
Speed (km/h)
Frequency ( f i)
0 ≤ x < 10
10
10 ≤ x < 20
20
20 ≤ x < 30
45
30 ≤ x < 40
71
40 ≤ x < 50
21
2.1
2.2
2.3
2.4
Cumulative frequency
Copy and complete the table.
Draw a cumulative frequency curve (ogive).
Indicate clearly on your cumulative frequency curve the lower
quartile (Q1), median (Q 2) and upper quartile (Q 3).
Use your graph to estimate the number of skiers who passed the
point with speeds greater than 35 km/h.
(2)
(4)
(3)
(2)
[11]
316
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Formal assessment: Term 4 Test
3
A cricket coach recorded the number of minutes the batsmen in his
team needed to reach 50 runs during the most recent cricket season.
His results are tabled below:
Minutes
needed
Number of
batsmen (xi )
50 < x ≤ 60
12
60 < x ≤ 70
18
70 < x ≤ 80
28
80 < x ≤ 90
37
90 < x ≤ 100
5
3.1
3.2
4
Midpoint of
interval ( fi )
fi.xi
__
__
(x – x)2
f.(x – x)2
Complete the columns for the midpoints and the f.x and then
calculate an estimate of the mean. Give your answer for the mean
correct to one decimal place.
Complete the remainder of the table to calculate the standard
deviation, correct to one decimal place.
Consider the given data which represents the individual distances travelled
on a particular day by a group of cyclists:
23; 32; 32; 23; 29; 32; 27; 27; 36; 28; 11; 27; 27; 40; 36
4.1
Determine the five-number summary, and hence draw
a box-and-whisker diagram to represent this data.
4.2
Use your answer in 4.1 to discuss the dispersion of this data.
4.3
Use your calculator to determine, correct to one decimal place:
4.3.1
the mean number of kilometres travelled
4.3.2
the standard deviation.
4.4
Determine the percentage of trips that is within one standard
deviation of the mean.
4.5
4.5.1
By grouping the given data, complete the table:
Kilometres travelled
4.5.2
Frequency
10 < x ≤ 20
1
20 < x ≤ 30
8
30 < x ≤ 40
6
(4)
(5)
[9]
(7)
(3)
(3)
(3)
(3)
Midpoint
(2)
Use the grouped data to sketch a frequency polygon.
(4)
[25]
Total: 51 marks
317
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Term 4 summary
Topic 13
Statistics
We can represent grouped data graphically in various ways including:
• Histograms
Masses of fish
Frequency
15
10
5
10–12 12–14 14–16 16–18
Mass (kg)
• A bar graph
– The area of each bar is relevant; your groups should be equal in width if possible.
– The bars must be joined.
– Frequency polygons (drawn with or without a histogram)
Vehicles in a particular route
Frequency
2,5
2,0
1,5
1
0,5
0
0–6
7–12 13–18 19–24
Hours
• A frequency polygon on a histogram
Frequency
20
D
B
10
C
E
A
F
M
0
32
37
N
42
47
52
57
62
67
72
Marks
– Plot the midpoint of each group against the frequency of the group.
– Join each point as a jagged line graph.
– Plot the midpoint of the group before and after each given group on the x-axis,
to secure the polygon to the x-axis.
– The area enclosed by the polygon will equal the total area of the histogram.
318
Term 4 summary
PLTMATHSLB11LB_14.indd 318
2012/07/07 11:40 AM
• Cumulative frequency curve (ogive curve)
40
35
Frequency
30
25
20
15
10
5
0
10,5
15,5
20,5
25,5
30,5
35,5
40,5
45,5
Length (mm)
– Plot the end of each group against the cumulative frequency at the end of each
group.
– Join each point as a curved graph.
– Plot the end point of the group before the given groups on the x-axis to secure
the beginning of the curve to the x-axis.
You can analyse and understand the spread or dispersion of data better by
considering:
Range:
• When there is a large range, the data is widely spread between the maximum and
minimum values
• When there is a small range, the data values are all closely bunched.
Symmetrical or skewed data:
• When the mean is greater than the median, the data is skewed to the right.
• When the mean is less than the median, the data is skewed to the left
• When the mean equals the median, the data is symmetrical.
Standard deviation:
• This value reflects how close most of the data is to the mean
_________
√
n
∑(x − _x)
i
2
i=1
__________
n
• We can find the standard deviation by using the formula:
or using a calculator.
Outliers: Any data value that is less than Q1 – 1,5 × IQR or greater than
Q1 + 1,5 × IQR is an outlier.
Term 4 summary
PLTMATHSLB11LB_14.indd 319
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Formal assessment
Examination practice Paper 1
Time: 3 hours
The following formulae are given for you to use:
_______
√
2
−b± b − 4ac
x = ____________
2a
y
–
y
2
1
m = ______
x2 – x1 ; y = mx + c; y – y1 = m(x – x1)
A = P(1 + in); A = P(1 – in); A = P(1 + i)n; A = P(1 – i)n
Question 1
1.1 Solve for x:
1.1.1 2x2 + 3 = 7x
1 – _________
x–2
7x + 10
+ ________
1.1.2 _____
x – 2 x2 + 2x + 4 2(8 – x3)
_____
1.1.3 8 – 2√2x + 5 = x
2
__
1.1.4 3x3 – 12 = 0
1.1.5 4x + 1 = 322x – 1
(4)
(6)
(6)
(3)
(4)
1.2 Solve simultaneously for x and y if x2 – 2xy – 3y2 = 0 and 3x + y – 2 = 0.
(8)
1.3 Consider the expression x2 – 4x – 12 and determine the value(s) of x for which:
1.3.1 x2 – 4x – 12 < 0
(4)
1
is undefined.
1.3.2 __________
2
x – 4x – 12
1.4 Determine the nature of the roots of 2x2 – 3x – 8 = 0.
(2)
(3)
1.5 Fully simplify:
1.5.1
________
2 +2
√_________
2
___
___
399
396
396
___
1.5.2 ( √ 5x – √3x )( √ 5x +
___
√ 3x )
(3)
(2)
[45]
Question 2
2.1 Consider the sequence 3; 8; 13; 18; 23; …
2.1.1 Write down the next two terms in the sequence if the pattern
continues in the same way.
2.1.2 Write down a formula for the nth term in the form Tn = …
(2)
(2)
2.2 Consider the sequence –2; 3; 12; 25; 42; …
2.2.1 Write down the next two terms in the sequence if the pattern
continues in the same way.
2.2.2 Determine a formula for the nth term in the form Tn = …
2.2.3 Which term in the sequence has a value of 558?
(2)
(5)
(5)
2.3 Determine the value of x if 3; 12; x; 54; ... is a quadratic sequence.
(5)
2.4 Consider the sequence 6; 18; 54; 162; 486 …
2.4.1 Write down the next term in the sequence if the pattern
continues in the same way.
2.4.2 Write down a formula for the nth term in the form Tn = …
2.4.3 Hence write down a formula for the nth term of the sequence
11; 23; 59; 167; 491; … if the pattern continues in the same way.
(1)
(3)
320
(1)
[26]
Exam practice paper 1
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Question 3
5.
4 + 2, g(x) = − __
1 (x + 1)2 + 8 and h(x) = __
1 x + __
Given f (x) = _____
x–1
2
2
2
3.1 State the range of g.
(2)
3.2 Determine g(−1) – h(−1).
(2)
3.3 Solve for x if f (x) = h(x).
(6)
3.4 State the equations of the asymptotes of f.
(2)
3.5 Determine the x- and y-intercepts of f.
(3)
3.6 Sketch f, g and h on the same system of axes. Show all the x-intercepts and
y-intercepts and the coordinates of any turning point(s).
3.7 State the equation of v(x) if v is the reflection of f in the x-axis.
3.8 w(x) = g(x – 5) – 4 , write down the equation of w(x).
(13)
(2)
(2)
[32]
Question 4
g
y
(–1;16)
f
(6;8)
4
y=1
–2
3
x
Consider the sketch which is not drawn to scale.
f (x) = ax2 + bx + c passes through (−2;0), (3;0) and (6;8).
y = 1 is the asymptote of g(x) = kpx + q which passes through (–1;16) and (0;4).
4.1 Determine the equation of f and write your answer in the form
f (x) = ax2 + bx + c.
(5)
4.2 Determine the equation of g.
(5)
4.3 For which value(s) of x is f (x) an increasing function?
(2)
4.4 For which value(s) of x is f (x).g(x) ≥ 0?
Question 5
5.1 If an investment doubles in value over 8 years, calculate the rate of interest
per annum (to two decimal places), if the interest was:
5.1.1 simple interest paid annually
5.1.2 compounded annually.
(2)
[14]
(3)
(3)
Exam practice paper 1
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2012/07/02 2:25 PM
5.2 A company buys a bus at a cost of R2 750 000. Calculate the book value
of the truck after 8 years if the rate of depreciation on a reducing
balance method is expected to be 9% p.a.
5.3 Mr Smith has two sons who are 5 years and 13 years respectively.
He wishes to give each son R21 000 on his 21st birthday. Calculate
how much he should invest now if interest is 11,5% p.a. compounded
semi-annually.
(3)
(7)
[16}
Question 6
Thirty people were asked whether they had travelled by train, car or bus that day.
Some people had used one form of transport, some two and one person all three
forms. Seven people had used no transport that day. The details of this survey are
recorded in the Venn diagram.
T
C
5
4
6
1
x
2
3
B
7
6.1 Find the value of x.
(2)
6.2 How many people used both bus and train?
(2)
6.3 How many people have not used a train as transport?
(2)
6.4 Determine:
6.4.1 P(T ∩ C)′
6.4.2 P(B | T)
(2)
(2)
6.5 Two taxi companies, Call-a-cab and Terrific Taxis operate in a certain
suburb. Tracy calls Terrific Taxis 70% of the time when she needs a taxi,
and Call-a-cab on every other occasion. Terrific Taxis arrives late 10% of
the time and Call-a-cab arrives late 15% of the time.
6.5.1 Draw a tree diagram to represent this information, showing the
probability of each branch.
6.5.2 Determine the probability that she calls Call-a-cab and it arrives
on time.
(4)
(3)
[17]
Total: 150 marks
322
Exam practice paper 1
PLT MATHS LB 11 7th pgs (Real Book).indb 322
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Formal assessment
Examination practice Paper 2
Time: 3 hours
The following formulae are given for you to use:
y –y
2
1
m = ______
x – x ; y = mx + c; y – y1 = m( x – x1 )
2
1
________________
(
)
x1 + x2 ______
y +y
; 1 2 ; m = tan θ
d = √ ( x2 – x1 )2 + ( y2 – y1 )2 ; M _______
2
2
n
∑x
x = _____;
__
n
∑fx
x = _____;
__
n
∑( xi − x__ )2
i=1
var = __________
; SD =
n
n(A)
n(s)
P(A) = ____
_________
n
___› 2
√∑(
xi − x )
_______
i=1
n
Question 1
The table shows the number of registered voters in each province for the 2011
Municipal elections in South Africa:
Province
Registered voters
Northern Cape
Rounded to the nearest hundred thousand
572 140
600 000
Free State
1 386 521
1 400 000
North West
1 576 898
1 600 000
Mpumalanga
1 718 309
1 700 000
Limpopo
2 340 799
2 300 000
Western Cape
2 706 736
2 700 000
Eastern Cape
3 111 535
3 100 000
KwaZulu-Natal
4 648 733
4 600 000
Gauteng
5 592 676
5 600 000
Source: Electoral commission
1.1 Use the values rounded to the nearest hundred thousand in the last column
to answer the questions.
1.1.1 Determine the median of the given data.
1.1.2 Determine the interquartile range for the data.
1.1.3 Draw a box-and-whisker diagram on a diagram sheet like the one
below to represent the data.
(1)
(3)
Cumulative frequency
y
x
Percentage of voter turnout
(3)
Exam practice paper 2
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1.1.4 Use your box-and-whisker diagram a diagram sheet comment on
the symmetry of the data.
1.1.5 Calculate the mean number of registered voters per province
(to two decimal places).
1.1.6 Calculate the standard deviation for this data.
1.1.7 Determine whether there are any outliers.
(1)
(2)
(2)
(2)
1.2 The percentage of voter turnout at these elections, as recorded in each
province, is shown in the frequency table:
Percentage of voter turnout
Number of provinces
50 < x ≤ 55
2
55< x ≤ 60
4
60 < x ≤ 65
3
1.2.1 Draw an ogive curve on the diagram sheet like the one below,
to represent the voter turnout.
600 000
1 600 000
2 600 000
3 600 000
1.2.2 Given that the mean voter turnout for these elections was 57,6%,
use your ogive to determine the number of provinces that had
a voter turnout higher than the mean. Show on your ogive how
you read off your answer.
(4)
4 600 000
5 600 000
(2)
[20]
Question 2
In the diagram below, △ABC is given with vertices A(6;1), B(3;–5) and C(–11;2).
y
C(–11;2)
A(6;1)
x
B(3;–5)
^ C = 90°
2.1 Prove: AB
(3)
2.2 Determine the equation of the median AE of △ABC where E lies on BC.
(5)
2.3 Determine the coordinates of D, so that ABCD is a parallelogram.
(2)
2.4 Prove that AC = BD.
(4)
2.5 What kind of parallelogram is ABCD? Give a reason for your answer.
(2)
2.6 Determine the coordinates of F, the point of intersection of AC and BD.
(2)
2.7 Determine the equation of the perpendicular bisector of BC.
(4)
324
Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 324
2012/07/02 2:25 PM
2.8 Show that the perpendicular bisector found in 2.6 passes through point F.
^ D (to one decimal place).
2.9 Determine the size of AB
(2)
(5)
[29]
Question 3
3__ where θ ∊ [90°; 270°]. Determine, using a diagram,
3.1 Given tan θ = ___
√7
the value of:
3.1.1 sin θ
3.1.2 4cos2 θ – sin(180° + θ)
(4)
(4)
tan 120°.cos180°
3.2 Simplify without using a calculator: _______________________________
sin 160°.cos 70° − cos 20°.sin (−110°)
(8)
3.3 3.3.1 State the values of θ ∊ [−180°;180°] for which the expression
cos θ + tan θ will be undefined.
________
1 + sin θ
cos θ + tan θ = _____
1
3.3.2 Prove that: ________
cos θ
1 + sin θ
3.4 Solve the equations, correct to two decimal places:
3.4.1 3tan(2x + 10°) + 5 = 0 for 0° ≤ 2x + 10° ≤ 360°
3.4.2 Find the general solution for 4sin x cos x − 3sin2 x = 0.
(3)
(5)
(5)
(5)
[34]
Question 4
Given the functions f(x) = sin 2x and g(x) = tan x − 1.
4.1 Sketch the graphs of f and g on the same system of axes for
x ∊ [−180°; 180°], clearly labelling endpoints, turning points,
intercepts with the axes and asymptotes.
(6)
4.2 Use your graphs to determine the value(s) of x for which:
4.2.1 sin 2x < 0
(2)
f(x)
4.2.2 ____ ≥ 0
(2)
g(x)
4.2.3 If h(x) = f(x) − 3, write down the range of h.
(2)
[12]
Question 5
ABC is a triangle and D is a point on BC. AD is joined. DC = m, BD = 2m,
^ D = 30°, AB
^ C = α and AD
^ C = θ.
BA
A
30°
B
α
2m
θ
D
m
C
5.1 Prove: Area △ADC = 2 m2 sin α sin θ
5.2 If it is now given that m = 12 cm and θ = 70°, determine the length of
AC to the nearest centimetre.
(4)
(3)
[7]
Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 325
325
2012/07/02 2:25 PM
Question 6
The diagram represents a pencil with a hemispherical eraser on one end and a conical
sharpened point on the other end.
The total length of the pencil is 13 cm, and the cylindrical portion of the pencil, with
a diameter of 1 cm, is 10 cm long.
10 cm
1 cm
13 cm
Determine, to two decimal places:
6.1 the slant height of the conical end
(4)
6.2 the total volume of the pencil
(3)
6.3 the surface area of the cylindrical section.
(5)
[12]
Question 7
Q
R
A
B
P
7.1 In the diagram, AB is a tangent to the circle at P. Q and R are points on
the circumference of the circle. PQ, QR and RP are joined.
^ P.
^ B = RQ
Prove the theorem that states RP
(5)
7.2 In the diagram AE is a diameter of the circle with centre O and ABC and
^ = 50°. FG is a tangent to the circle at A.
EDC are straight lines with C
DA and BE intersect at H. BO is joined.
C
50°
B
1
F
3
1
2
2
2
D 1
1
A
3
H
1
2
O
2 1
G
E
326
Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 326
2012/07/02 2:25 PM
7.2.1
7.2.2
7.2.3
7.2.4
7.2.5
^ E = 90°.
Give a reason why AB
Hence prove that BCDH is a cyclic quadrilateral.
^ and E giving reasons.
^ ,A
Determine the size of angles H
1
2
2
^
^
^
^
If A1 = x, determine E1, O2 and O2 in terms of x, giving reasons.
^ =B
^
Prove that: A
1
2
(1)
(3)
(6)
(7)
(3)
[25]
Question 8
AF is a diameter of circle O with chords AB and DF. M is a point on the circumference
so that OM bisects AB at C. OB is joined, with OB ∥ DF.
D
F
O 1
B
1 2
C
A
M
^ = 2B
^.
8.1 Prove that F
(6)
8.2 If it is given that CM = 8 mm and AC = 12 mm, determine the length
of OC, giving reasons.
(5)
[11]
Total: 150 marks
Exam practice paper 2
PLT MATHS LB 11 7th pgs (Real Book).indb 327
327
2012/07/02 2:25 PM
Optional assessment: Pattern investigation
Task 1 Triangular numbers
Note:
1.1 The diagram on the right represents the 3rd pattern of arranging squares in a
triangular shape. Copy and complete the table below:
Number of layers
1
2
3
Number of squares in bottom layer
1
2
3
Total number squares
1
3
6
4
5
This is an individual
investigation. Complete
all questions on a
separate answer sheet.
6
(2)
1.2 You know the bottom row of numbers in the table as the triangular numbers.
Use the method of differences to show that the formula for the nth triangular
1 n(n + 1).
number is Tn = __
(5)
2
1.3 Use this formula to find the number of squares needed for 10 layers.
(1)
1.4 How many layers would use a total of 91 squares?
(3)
[11]
Task 2 Square pyramid numbers
2.1 If the diagram on the right represents the 3rd pattern of stacking boxes (each box
is an identical cube) in a square-base pyramid form, copy and complete the table:
Number of layers
1
2
Number of boxes in base
1
4
Total number of boxes
1
5
3
4
5
6
14
(2)
2.2 How many boxes are there in a base that is n boxes wide (find Tn for row 2)? (1)
2.3 The formula for the nth square-based pyramid number (that is, find Tn for row 3)
n3 + __
n2 + __
n.
is __
3
2
6
Show how you could have used the numbers in the bottom row of the table to
establish that the formula is a 3rd degree expression.
(2)
[5]
Task 3 Another look at the square bases
Each small square in the diagram has side of length 1 unit. Within the larger squares
there are smaller squares of side 1 unit, 2 units or 3 units, and so on. So within a
3-unit square there are 91-unit squares as well as some 2-unit squares, and so on.
328
Term 4 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 328
2012/07/02 2:25 PM
3.1 By counting in the diagram, complete the table:
Size of big square →
Total width
1 unit
No. 1-unit squares
Total width
2 units
Total width
3 units
1
Total width 4
units
Total width
5 units
16
No. of 2-unit squares
No. of 3-unit squares
1
No. of 4-unit squares
No. of 5-unit squares
Total no. of squares
55
(5)
(1)
3.2 Do you notice anything familiar about the numbers in the bottom row?
3.3 On the basis of what the table seems to show, suggest how we might calculate
the total number of squares (of all sizes) that can be found in a 9-by-9 grid.
(2)
At the moment the only way we can do this calculation is with a calculator: do it
now, and see what you get for the 9-by-9 situation.
Hint:
Hint: look at the table in
Question 2.1.
(1)
If you look more closely at the calculation, you may find a formula.
3.4 Copy and complete the table:
1-by-1
From Total row in table above
Multiply by 6 (call this Mk)
Find value of Wk (in rational form)
1
6
M1 = 1 × 2 × W1; W1 =
2-by-2
M2 = 2 × 3 × W2; W2 =
3-by-3
M3 = 3 × 4 × W3; W3 =
4-by-4
M4 = 4 × 5 × W4; W4 =
5-by-5
55
330
M5 = 5 × 6 × W5; W5 = 11
…
9-by-9
M9 =
(8)
3.5 Decide how many squares (of all sizes) there would be in a 9-by-9 situation
(1)
3.6 Does this answer match the answer you obtained earlier?
(1)
3.7 Now try to write down a formula in terms of n for the total number of squares
that can you can find in the n-by-n situation.
(3)
3.8 Now show that this formula is the same as the formula given in task 2.2.
(2)
[24]
REMEMBER
Remember that the last
column of the table
above gives numbers
that are six times too big.
Pattern Investigation
PLT MATHS LB 11 7th pgs (Real Book).indb 329
329
2012/07/02 2:26 PM
Pattern Investigation continued
Task 4 Compare the triangular, square
and pyramid numbers
4.1 Summarise your findings of the previous tasks in the table. Copy and complete
more columns, and use them to answer the questions that follow.
Number of layers
1
2
3
Number of squares in a triangular shape
1
3
6
28
Number of squares in a square base
1
4
9
49
Total number of cubes in a square pyramid form
1
5
14
140
4
5
6
7
8
9
10
11
12
13
(6)
We can see that one square or box (of 1 unit) can be arranged as a triangular number,
a square number or a square pyramid number.
4.2 What is the next number (of squares or boxes) that can be arranged in a square
pyramid and also in a triangular shape?
(2)
4.3 What number (of boxes or squares) can be arranged as either a square base or a
triangular shape?
(2)
4.4 A large number of boxes is stacked in a square pyramid. The pyramid is
disassembled; some of the boxes are arranged in a square and the rest in a
triangle. How many boxes must there have been?
In other words, we want to find a square pyramid number that equals the sum of a
square number plus a triangular number. Find two such square pyramid numbers. (6)
4.5 If the boxes were built as towers as shown below, copy and complete the table for
the number of boxes used to make each tower.
Figure 1
Figure 2
Figure 3
Copy and complete the table.
Figure number
(6)
1
2
3
4
5
6
n
Number of boxes in base
Total number of boxes used
How many boxes will a tower with 1 029 boxes in the base use?
330
(3)
[25]
Term 4 Assessment
PLT MATHS LB 11 7th pgs (Real Book).indb 330
2012/07/02 2:26 PM
Answers
Note: Some answers are too complicated to provide
in this section. Please ask your teacher to check your
solutions after consulting the Teacher’s Guide.
TOPIC 1: EXERCISE 1
1
4 a8b14
__
2
1
4
3a2
___
5
p2x
6
–32a4 + 16a2b3 – 2b6
8
8
___
5
5
9
27
TOPIC 1: EXERCISE 6
3
2
_
3a2
__
7
21 010 × 51 011 10
1
4a c
_____
4
2x
___
7
10
13
16
5b2
11
y3
3
y
___
x2z2
10
__
3
x3y3
_____
2
y + x2
5
__
x2
____
__
3
2.3 √( 9x4 )
4
2.2 6√
x
TOPIC 1: EXERCISE 7
3x
__
2
1
1.1
10
1.2
16
1.3
8
1.4
4
2
y3
__
27
3
3x
__
y
1.5
5
_
1.6
5
_
5
2
6
9
1.7
6
1.8
2
8
1
_
9
1+a+a
_______
1.9
25
__
2
1.10 – _
11
1
_
12
49
___
24
1.11 __
14
1
_________
15
9
_
17
28 + x
18
8
___
2
4
a6 + 6a3 + 9
2
a3
4
2
27
x = –3
2
x=7
3
x=5
4
1
x = –_
5
x=1
6
x = –1
8
4
1.12 8
2.1
3x 9
2.2
16a4
2.3
27a12b6
2.4
a2b4c5
2.5
5a4b5c7
2.6
2a3
2.7
8x6
2.8
x4
2.9
x2
2.10 b2
3
2.12 _
2.11 x2
TOPIC 1: EXERCISE 4
3
1
x=3
2
x = –1
3
x=0
4
x = –2
5
x=5
6
x=2
7
x=2
8
x=2
2
4
x=4
x = 3 or x = –1
3
x = 0 or x = –3
5
1
x = –_
6
x = –2 or x = 1
7
x = 2 or x = –2
8
x = –3
9
–5
x = __
10
–3 or x = 1
x = __
2
2
4
3.1
50
3.2
5
__
3.3
2
3.4
2(a + b)
3.5
(a + b)2
3.6
1
_____
3
(x – 3)
TOPIC 1: EXERCISE 8
TOPIC 1: EXERCISE 5
x = –7
6
3
7
100
1
1
1.3 5x 4
212x
TOPIC 1: EXERCISE 3
2
15
__
1.2 x4y2
5
2.1 √x3
TOPIC 1: EXERCISE 2
2 2
3
5 _
_
1.1 x3
2
1
x = 81
2
1
x=_
3
4
x = __
1
x = ±___
125
1
x = __
5
6
x = ±64
9
625
x = ___
12
8
x = __
4
x = – 64
5
7
x = 16
8
10
x=7
11
4
81
x = = __
16
49
81
27
TOPIC 1: EXERCISE 9
1
x = 16 or x = 81
2
x = 8 or x = 125
3
x = 8 or x = –1
4
81 or x = 1
x = __
16
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 331
331
2012/07/02 2:26 PM
__
__
5
x = 16
6
9
x = __
5
4 – 2√3
6
10 + 6√3
7
x = 16
8
x = 25
7
37
8
4
9
9
x=_
10
1
x=_
9
2
10
3
_
4
25
3
TOPIC 1: EXERCISE 10
1
__
4√ 2
1.1
__
6√2
1.3
__
8√2
1.5
__
1.7
2.√2
1.9
x√x
1.11
28 7 6√
3
___
√3
___
1.4
__
5√5
4
__
3√ 2
1.6
__
4√6
7
3 + √3
______
1.8
___
3
2√12
9
3 + √2 + 3√ 3 + √6
_______________
a b
1.12 5x y
2b
2
__
__
5
__
√3 + 3
______
8
–3 – 2√ 2
3
__
__
__
__
3
√6
___
6
2√ 2 + 1
3
__
7
2x
1
1.1
__
√2
__
√ 35
2.2
30
2.3
3
2.4
6√6
2.5
24√6
2.6
7
_
2.7
3__
___
2.8
216
__
__
1.3
__
= 16 + 8√2 = RHS
2
1.2
__
√3
__
√3
8 +5
_
LHS =
3a3
___
1.2
4√2
1.4
3√3
1.5
2
_
1.6
8
1.7
1__
___
1.8
2
1.9
1
_
1.10 2
1.11
1
–_
__
5
√5
2
1.12
2
√3
1
___
√_23 – ____
√ 2.3
__
__
√3
√3
√3
__
__
√ 2. √ 3
√2
√3
2√ 3
__
√3
1 _
2 –_
2
=7 _
__
2
√_23 ( 7 – _12 )
__
13
2 __
= √_
3( 2 )
__
13 _
2 = RHS
= __
√
2 3
=
1.3
x
_
5
_________
___
__
__
__
6√3 – 12
2
6 + 3√2 + 2√3 + √6
3
2
4
9 – 4√5
_________
___
__
5
5
√ 37 + √ 5 .√ √ 37 – √ 5 = 2
√
LHS =
1
__
__
√2
2 – ____
__
=7 _
__
__
__
4.2 + 5
___
√2
1 __ × ___
2 +5 _
2 – ______
__
__
=2 _
=
TOPIC 1: EXERCISE 12
2
√6
__
√_23
13
1__ = __
2 – ___
_
__
__
7√5
__
8(2) + 8√ 2
2–1
= ________
5
2.10 a2 √
64
2.12
1.1
__
__
___
3
3a2.√
4
__
√2 + 1
8__√2 × ______
__
= _____
√2 – 1
√2 + 1
2.1
__
8√ 2
√
√2
__
1 – ___
___
√2
__
8
_____
= 16 + 8√ 2
1__
1 – ___
√2
__
2
8
__
LHS = _____
× ___
1
__
3
3670.√
9
TOPIC 1: EXERCISE 11
332
3
1.14 21 005√2
2.11 3
1
2
TOPIC 1: EXERCISE 14
___
2 4√
___
2.9
__
√3
1
1.10 2a2b√5b
3
1.13 x3y2.√
x2y
2
TOPIC__ 1: EXERCISE 13
___
__
1.15
1.2
__
4√3
2
___________________
___
___
__
__
5
(√ 37 + √ 5 )(√ 37 – √5 )
√
_____________
___
__
5
(√ 37 )2 – (√ 5 )2)
√
_____
5
37 – 5
=√
___
5
32
=√
= 2 = RHS
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 332
2012/07/02 2:26 PM
1.4
__
__
___
√ 8 – √ 3 – √ 12
1
______________
___
__
___ = _
2
2
√ 32 – 3√ 3 – √ 27
__
__
__
2√__2 – √3__– 2√3__
LHS = ______________
4√2 – 3√3 – 3√ 3
__
__
2√__
2 – 3√__
3
= _________
4√2 – 6√ 3
__
__
2√2__– 3√3__
= ___________
2(2√2 – 3√3 )
__
4√5 – √5
__
√
√5
__
15√5 = √ 5 = RHS
= _____
4
RHS = 22 = √ 23 = 2√ 2
= 2(1,414) = 2,828 = LHS
1
1__
______
_____
= ____
3
3
3
√ 2 011
2
____________
_____2
_____ = _
√ 22 013 – √ 22 009 3
1 = __
1
= ____
1
1
_
3 _
_
2 010
( 22 )3
.2
√2
__
√ 2 (2
1.7
__
2 = RHS
=_
3
– 1)
___
√3 + 1
5 + √27
______
__ = ______
2
4 – 2√ 3
LHS =
__
√3 + 1
______
__
4 – 2√3
__
__
√
4 + 2√ 3
__
4 + 2 __
3
× ______
__
2√3
2(√3 )2
4√3 +
+4+
__
= __________________
42 – (2√3 )2
__
__
√
√
6 3 + 6 + 4 = _______
6 3 + 10
= _________
4
x = 36
5
x = ±3
6
x = ±5
7
x=3
8
x = –1
9
x=4
10
x = –4
11
4 or x = 1 12
x=_
1.4
3
1.6
1
_
4a8 – 12a4b3 + 9b6
1.9
24
2.1
7
__
2.2
3y
__
2.3
3
___
2.4
144b
_____
x
__
LHS = RHS
1.7
43x5
1.8
___
√ 12
__
+ 5)
2
____
+ 2√ 3 + 31,5 = √147
__
__
3
_
LHS = 2√3 + 2√3 + 32
__
__
= 4√3 + √33
__
__
= 4√3 + 3√3
__
= 7√ 3
____
______
RHS = √147 = √49 × 3
LHS = RHS
2
=
__
7√3
x=2
3x3
1.5
1
(3√3 + 5)
2
1.8
3
1.2
RHS = ________
__
3
TOPIC 1: REVISION
y
___
4
__
(3√3
4
x = __
No solution 3
1.3
+ 5)
2
2
= ________ = ________
4
2
√2
x = 14
30x4y8
__
2(3√3
__
1
1.1
16 – 12
__
TOPIC 1: EXERCISE 15
2 __ 2
= ____________
1 004
2
2
22
1,414
√2
√2
1__ × ___
__ = ___
= ___
= _____ = 0,707
__
21__005√2
______________
__
21 006√2 – 21 004√2
1 005√
__
√2_2
2
.2
______
______
LHS = _______________
2 012
2 008
=
__
3
_
√2,828
______
.2 – √2
____
2 2
(1 + √3x___
) – √12x2 ___
= 1 + 2√ 3x2 + 3x2 – 2√ 3x2
_____
√
___
3
5
15__ × ___
__
= ____
√2
__
___
__15 __
LHS = ________
1.6
___
m + n + 2√mn = 5 + 2√6
m + n = 5 and mn = 6
(m + n)2 = 25
m2 + 2mn + n2 = 25
m2 + n2 = 25 – 2mn
m2 + n2 = 25 – 2(6)
m2 + n2 = 13
√ 80 – √ 5
3×5
_______
___
__
___
________
___15 __ = √ 5
__
__
m + 2√ mn + n = 5 + √ 24
2
3√5
_______
___
__
+ √ n = √ 5 + √ 24
(√ m + √ n )2 = (√5 + √ 24 )2
1 = RHS
=_
1.5
__
√m
2.5
2.7
2.9
5
3x6
6
2
16
2a3
3b4
___
2
2.6
1
4 – __
4
x
2.8
3
x4
2
a12
36
__
25
4
2 2
x y
_____
y2 + x2
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 333
333
2012/07/02 2:26 PM
3
3.1
5
3.2
1
__
3.3
40
3.4
125
____
3.5
3x2y4
3.7
3.9
1
x = 7,58 or x = –1,58; Real, irrational
8
2
x = – 0,68 or x = –7,32; Real, irrational
3.6
5x4
3
x = 1,24 or x = –3,24; Real, irrational
5
3.8
a2
4
x = 2; Real, rational
5
3 or x = – 4 Real, rational
x=_
b2
3.10 x – 1
6
x = 9 or x = –1; Real, rational
3.12 36
7
x = 4,73 or x = 1,27; Real, irrational
8
x = 5 ±√– 4 ; Non-real or imaginary roots
9
–q±√ q – 4pr
x = __________
6
3.11 __
5
4
5
6
7
4.1
TOPIC 2: EXERCISE 2
4
0
4.2
__
__
7√3
–
__
10√2
4.4
1
__
3√3
4.6
__
√2
4.7
1
4.8
x
4.9
1
_
2
4.10 4 – √11
5.1
x=1
5.2
7
x = –_
5.3
x = –6
5.4
x = 0 or x = –3
5.5
x=1
5.6
x=1
4.3
√2
4.5
__
3
10
11
___
4
2
3
4
334
______
2
2p
________
2 + 4kn
√
m±
m
x = __________
2k
________
m±√ m2 – 8mk
x = __________
4m
_______
√
–n± n2 + 3k2
x = __________
3k
1
x = 3 or x = 7
2
x = –1,5 or x = 2
3
x = –1,5 or x = – 0,5
4
2
x = –1 or x = __
5
1 or x = 4
x = – __
3
6
x = 0 or x = –5
7
x = 0 or x = 1
8
x = – 0,5 or x = 3
9
x = 12 or x = –2
10
x = – 4 or x = 3
11
x = 4 or x = –1
12
x = –5 or x = 1
13
3 or x = –2
x = – __
14
x = 0 or x = 2,25
3 or x = __
2
x = __
16
x = 3 or x = –2
18
x = 0,5 or x = 5
20
5 or x = 1
x = – __
22
7 or x = –2
x = __
x = ±125
6.2
x = ±8
6.3
x = –2 or x = 4
6.4
x = 81
6.5
x = 64 or x = 0
6.6
– 64 or x = 27
x = ___
7.1
x=1
7.2
x = 1 or x = 2
7.3
x=9
7.4
x = –2
7.5
Both solutions are invalid.
15
7.6
x = 3 or x = 7
17
27
___
TOPIC 2: EXERCISE 3
6.1
19
TOPIC 2: EXERCISE 1
1
12
2
4
2
3
9
2
x = –__ or x = __
4
3
1
2
__
x = – or x = – __
4
3
3 or x = __
1
x = __
4
2
6 or x = 3
x = __
5
3
6
1.1
x = –5
1.2
Minimum
21
1.3
x = –5
1.4
0
23
2.1
x=2
2.2
Minimum
24
x = 1,5 or x = –1,5 or x = 1 or x = –1
2.3
x=2
2.4
0
25
x = __ or x = __
26
3b or x = – __
4b
x = __
a
3.1
x = –6
3.2
Maximum
27
x = 5 only
28
x = 2 only
3.3
x = –6
3.4
0
4.1
x=4
4.2
Maximum
4.3
x=4
4.4
0
2q
3p
5q
2p
8
2a
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 334
2012/07/02 2:26 PM
TOPIC 2: EXERCISE 4
TOPIC 2: EXERCISE 9
1
1
3 or _
2
2
1,05 or –1,63
3
– 0,44 or –1,36
4
2,26 or – 0,59
5
1,31 or – 0,64
6
2,85 or – 0,35
7
1,58 or 0,42
8
p±√ p2 – 3p
______
1
x = 6 or x = –1
2
x = –10 or x = 20
3
x = 3,5 or x = – 0,5
4
3,39 or – 0,89
5
13
17 or __
__
5
2
6
1,87 or – 0,37
7
2,11 or –2,61
8
4,41 or 1,59
9
There is no real solution.
10
x = 2 only
TOPIC 2: EXERCISE 6
2
3
4
1.1
1.2
2.1
2.2
3.1
3.2
4.1
4.2
k=2
The other root is x = 1,5
p = –1
5
The other root is x = – __
3
a = –2
2
The other root is x = __
5
t = –7
The other root is x = –1,25
1
Inequality notation
Interval notation
1.1
x < −3 or x > 2, x ∊ ℝ
x ∊ (−∞;−3) ∪ (2;∞)
7
8
9
length 125 mm; breadth 25 mm
The numbers are 49 and 94.
5 km/h
48 km/h
x ≠ –750 or x = 700
6.1
16 km/h
6.2
8 km/h
9 days and 18 days respectively
larger pipe takes 3 hours
150 m by 80 m
TOPIC 2: EXERCISE 8
1
2
3
4
−1 < x < 2, x ∊ ℝ
x ∊ (−1;2)
−2 ≤ x ≤ 2, x ∊ ℝ
−3 ≤ x ≤ 2, x ∊ ℝ
x ∊ [−2;2]
1.4
2.
Inequality notation
Interval notation
2.1
−1 < x < 3, x ∊ ℝ
x ∊ (−1;3)
x ≤ 0 or x ≥ 4, x ∊ ℝ
x ∊ (−∞;0] ∪ [4;∞)
1.3
2.2
2.3
x ∊ [−3;2]
0 < x < 4, x ∊ ℝ
x ∊ (0;4)
2.4
x ≥ 1,5, x ∊ ℝ
3
Inequality notation
Interval notation
3.1
x ≤ –2 or –1 ≤ x ≤ 2, x ∊ ℝ
x ∊ (−∞;−2] ∪ [−1;2]
x < −2 or 0 < x < 1, x ∊ ℝ
x ∊ (−∞;−2) ∪ (0;1)
3.2
3.3
3.4
x ∊ [1,5;∞)
x > −2, x ∊ ℝ
–2 < x < −1 or x > 2, x ∊ ℝ
x ∊ (−2;∞)
x ∊ (−2;−1) ∪ (2;∞)
4
4.1
4.2
4.3
4.4
x = –3 or x = 5
x < –3 or x > 5, x ∊ ℝ; x ∊ (– ∞;–3) ∪ (5;∞)
x = –3 or x = 6
–3 ≤ x ≤ 6, x ∊ [–3;6]
5.
5.1
5.2
5.3
x = – 6 or x = –1
x < – 6 or x > –1, x ∊ ℝ; x ∊ (– ∞;– 6) ∪ (–1;∞)
x = –6 or x = 0
5.4
–6 < x > 0, x ∊ ℝ; x ∊ (–6;0)
TOPIC 2: EXERCISE 7
1
2
3
4
5
6.
x ≤ – 4 or x ≥ 4 or x ∊ (– ∞;– 4] ∪ [4;∞)
x ≤ –1,5 or x ≥ 1,5 or x ∊ (– ∞;–1,5] ∪ [1,5;∞)
TOPIC 2: EXERCISE 10
1.2
TOPIC 2: EXERCISE 5
1
1
2
TOPIC 2: EXERCISE 11
1
16 and (3;13)
( – _23;– __
3 )
2
(0;9) and (2;–3)
3
(4;0) and (–2;9)
4
(2;3) and (–2;4)
5
17 and (–5;3)
( _45;__
25 )
6
(5;4) and (2;–2)
7
(–5;6) and (1;0)
8
(5;6) and (1;–2)
9
(2;–2) and (–1;4)
10
(0;4) and (4;– 4)
– 4 < x < 4 or x ∊ (– 4;4)
x ≤ –5 or x ≥ 5 or x ∊ (– ∞;–5] ∪ [5;∞]
–1,5 < x < 1,5 or x ∊ (–1,5;1,5)
–1 ≤ x ≤ 1 or ∊ [–1;1]
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 335
335
2012/07/02 2:26 PM
TOPIC 2: EXERCISE 12
6
1
(–1;2) and (–3;– 4)
2
(3;1,5) and (–1;2,5)
3
(–2;– 4) and (– 4;6)
4
(7;–1) and (1;–3)
5
(4;1)
6
(2;3) and (–2;–3)
7
(29;6) and (1;–1)
8
(2;–1) and (–5;–2)
9
(0;–2) and (3;4)
10
(1;–1) and (–1;3)
11
(– 4;–1), (1;4) and (–1;– 4)
TOPIC 2: EXERCISE 13
1
2
3
4
5
6
7
8
The roots are non-real.
The roots are real, rational and unequal.
4 or x = 1
x = – __
7
The roots are non-real.
The roots are non-real.
The roots are real, rational and equal.
x = 1,5 only
The roots are real, irrational and unequal.
– 0,23 or –1,43
The roots are real, rational and equal.
– 0,4 only
The roots are non-real.
Inequality notation
Interval notation
6.1
x ≥ – 2, x ∊ ℝ
x ∊ [–2;∞)
6.2
x ≤ – 2 or 1≤ x ≤ 3, x ∊ ℝ
x ∊ (– ∞;–2] ∪ [1;3]
6.3
–2 ≤ x ≤ 0 or x ≥ 4, x ∊ ℝ
x ∊ [–2;0] ∪ [4;∞)
6.4
1 < x < 3, x ∊ ℝ
x ∊ (1;3)
7.
8
9
10
7.1
(–3;–7) and (2;8)
7.2
(5;6) or (2;– 6)
7.3
(4;– 4) and (–3;– 0,5)
7.4
(0;– 8) and (4;0)
The original price was R8 per litre.
48 learners participated
17 and 23
TOPIC 3: EXERCISE 1
1
2
1.1
14; 9; 4
1.2
Tn = –5n + 44
1.3
T22 = – 66
1.4
T59 = –251
2.1
4; 7; 10
2.2
Tn = 3n – 14
2.3
T100 = 286
2.4
T64 = 178
TOPIC 3: EXERCISE 2
TOPIC 2: REVISION
1
2
T6 = 55 and T7 = 78; Tn = 2n2 – 3n + 1; n = 93
T6 = 116 and T7 = 157; Tn = 3n2 + 2n – 4; n = 57
1
3
T6 = 37 and T7 = 51; Tn = n2 + n – 5; n = 29
4
T6 = 42 and T7 = 56; Tn = n2 + n; n = 31
5
T6 = – 45 and T7 = – 67; Tn = –2n2 + 4n + 3; n = 67
2
3
4
5
336
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
2.1
2.2
3.1
3.2
3.3
3.4
4.1
4.2
5.1
5.2
5.3
5.4
x = 3 or x = 2,5
x = 5,77 or –2,77
x = 4 or x = –3
x = 2,48 or – 8,48
The roots are non-real.
7 or x = 1
x=_
3
x = 7 only
x = 6 only
x = 4,45 or x = – 0,45
x = 2,73 or x = – 0,73
B Real, rational and unequal roots.
A Real, rational and equal roots.
C Real, irrational and unequal roots.
D Non-real or imaginary roots.
The roots are real, irrational and unequal.
4.2.1 3,71 or –1,21
4.2.2
3,71 or −1,21
– 8 ≤ x ≤ 8; x ∊ ℝ, x ∊ [– 8;8]
– 4,5 ≥ x ≥ 4,5; x ∊ ℝ, x ∊ [– 4,5;4,5]
x < –3 or x > 4; x ∊ ℝ, x ∊ (– ∞;–3) ∪ (4;∞)
x < 0 or x > 7; x ∊ ℝ, x ∊ (– ∞;0) ∪ (7;∞)
6
7
8
1 n2 + _
1 n; n = 55
T6 = 21 and T7 = 28; Tn = _
2
2
3 n; n = 61
1 n2 – _
T6 = 9 and T7 = 14; Tn = _
2
2
3 n – 3; n = 13
5 n2 + _
T6 = 96 and T7 = 130; Tn = _
2
2
9
3 n + 1; n = 19
5 n2 + _
T6 = – 80 and T7 = –111; Tn = – _
10
T6 = – 80 and T7 = –111; Tn = n2 – 5n + 4; n = 71
2
2
TOPIC 3: EXERCISE 3
1
56; 79 Tn = 2n2 – 3n + 2; n = 19
2
–17; –28 Tn = –n2 + 2n + 7; n = 26
3
76; 110 Tn = –n2 + 2n + 7; n = 26
4
–53; –74 Tn = –1,5n2 – 1,5n + 10; n = 37
5
91; 132 Tn = 3,5n2 – 4,5n – 8; n = 32
6
–106; –155 Tn = – 4n2 + 3n + 20; n = 17
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 336
2012/07/02 2:26 PM
TOPIC 3: EXERCISE 4
1
TOPIC 3: REVISION
Fig 1
Fig 2
Fig 3
Fig 4
Fig 5
Total
number
of blue
faces
1
3
6
10
15
Total
number
of white
faces
2
Total
number
of faces
3
9
18
30
45
Number
of layers
1
2
3
4
5
Number
of cubes
in
bottom
layer
1
3
6
10
15
2
Number
of red
beads
6
12
20
30
Fig n
n
5
12
22
35
1,5n2 – 0,5n
1
10
25
46
73
2
1,5n + 0,5n – 2
3n2 – 2
TOPIC 3: EXERCISE 5
1
Tn = 5n – 1 There are nine terms in the sequence.
–3n2
3.1
3.2
3.3
3.4
T6 = –6 and T7 = –16
Tn = – n2 + 3n + 12
T51 = –( 51 )2 + 3( 51 ) + 12 = –2 436
n = 12
4
4.1
4.2
4.3
4.4
4.5
4
x – 8 or 28 – 2x
x = 12
Tn = 2n2 – 3n – 8
n = 15; T15 = 397
5
5.1
5.2
5.3
5.4
5.5
a = 35 and Tn = 7n
b = 8 and Tn = 5n – 12
c = 112 and Tn = 7 × 2n or Tn = 14 × 2n – 1
d = 1 029 and Tn = 3 × 7n – 1
e = 60; Tn = 3n2 + 3n
2
1
Number
of beads
and rods
3
n( n + 1 )
______
Fig n
38
T6 = 21 and T7 = 40
Tn = 2n2 – 7n – 9
T25 = 2( 25 )2 – 7( 25 ) – 9 = 1 066
n = 47
2
Fig
5
24
2.1
2.2
2.3
2.4
3n( n + 1 )
_______
Fig
4
13
2
n(n + 1)
Fig
3
5
T6 = –13 and T7 = –17
Tn = – 4n + 11
T19 = – 4( 19 ) + 11 = –65
n = 33
2
Fig
2
0
1.1
1.2
1.3
1.4
n( n + 1 )
______
Fig
1
Number
of black
rods
1
2
Tn =
3
T6 = 216 and T7 = 343 There are ten terms in
the sequence.
4
T6 = 192 and T7 = 384; Tn = 3 × 2n; n = 10
5
T6 = 43 and T7 = 64; Tn = 2n2 – 5n + 1; n = 97
6
T6 = –117 and T7 = –167; Tn = – 4n2 + 2n + 15;
n = 13
7
T6 = 486 and T7 = 1 458; Tn = 2 × 3n – 1; n = 9
8
T6 = 215 and T7 = 342; Tn = n3 – 1; n = 10
+ 7n + 10; n = 28
6
Structure
1
2
3
4
5
Structure n
Number of
tiles in the
base of the
structure
1
4
9
16
25
n2
Number
of vertical
tiles in the
structure
4
12
24
40
60
2n(n + 1)
Total number
of tiles in the
structure
5
16
33
56
85
3n2 + 2n
TOPIC 4: EXERCISE 1
1
1.1
Substitute y = 5x + 9 into
5y + x – 19 = 0
5(5x + 9) + x – 19 = 0
25x + 45 + x – 19 = 0
26x = –26
x = –1
y = 5(–1) + 9 = 4
K = (–1;4)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 337
337
2012/07/02 2:26 PM
1.2
3 x + __
5
y = – __
1.3
2
__
1.4
3 × __
2 = –1
– __
2
1.2
2
______________
______
1.6
p=5
1.7
Q = (3;–2)
1.8
1.9
y=
1x
– __
+
2
6
__
5
5
3 ;__
3
P = – __
22
(
)
_______________
___
1.10 KM = √ (–1 – 4)2 + (4 – 3)2 = √ 26
PN =
3
_______________
√( – __32 – 1 )2 + ( __32 – 1 )2
___
___
___
26 __
26 = ____
= √ ___
= 1 × √26
4
√
2
9 ;_
9
D= _
2.2
18
1 x + ___
y = __
2.3
y = –x
2.4
y = –x + 9
2.5
y = 5x – 18
5
4
2
3
2.1
a=9
2.3
51
a = – ___
4
8
___
3.2
(1;0)
3.3
5 x – __
5
y = __
3.4
7 ;3
E = __
3.6
18°
2
2
16 x – __
29
y = __
9
9
(2 )
(8;11)
6 – k – 6 = __
–k = __
–1
mAB = _______
3
1 + 3k – 1 3k
6
–
n
–
6
–n
–1
_______
__
__
=
=
=
1 + 3n – 1 3n
3
mAB = mAC A, B and C are collinear.
5
5.1
2
2
θ = 63,4°
1.2
θ = 101,3°
1.3
θ = 59,0°
1.4
θ = 144,2°
2.1
θ = 26,6°
2.2
θ = 158,2°
2.3
θ = 135°
3.1
66,8°
3.2
31,0°
3.3
82,2°
3.4
71,6°
3.5
81,8°
3.6
60,3°
8=4
2 = – __
1 and m = __
mAB = ___
BC
–8
3 and m = _
2
mAB = – _
BC
3
3×_
2 = –1
–_
1.1
1.1
5
a = – __
2√29
6
4
1 × 4 = –1 AB ⊥ BC
– __
4
^ C = 90°
AB
2
3
AB ⊥ BC
5.2
13 units
__
5.4
5.6
5.3
D = (–1;4)
4 x – __
2
y = – __
5.5
3 x – __
3
y = – __
( – __32;__32 )
5.7
45°
7.1
D = (–2;6)
2
7
7
29 units2
__
5
7.2
2
4
5 and m
1
mAM = – _
=_
MB
5
1
^ B = 90°
1 = –1 AM
–5 × _
5
7.3
^ B = 90°
3 DA
–2 and m = _
mAD = __
AB
3
2
ABCD is a square because it is a
parallelogram with diagonals bisecting
at right angles, and has right angles at its
vertices.
TOPIC 4: REVISION
1
2.2
3.1
mAC
5
TOPIC 4: EXERCISE 2
1
34 units2
3.7
(2 2)
2.1
1.3
3.5
2
KM = 2PN
2
___
= √ 4 + 64 = √ 68
AB = BC
△ABC isosceles
KN ⊥ LM and N is the midpoint of LM, so
KN is the perpendicular bisector of LM.
KL = KM
___
BC = √ (7 – 5)2 + (5 + 3)2
3
1.5
______
= √ 64 + 4 = √ 68
3
2
______________
AB = √ (7 + 1)2 + (5 – 7)2
8
8.1
8.2
R = (6;1)
Prove that the diagonals are perpendicular.
4 = –_
1 and m = __
12 = 2
mPR = – _
QS
8
2
6
1 × 2 = –1 PR ⊥ QS
–_
2
338
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 338
2012/07/02 2:26 PM
9
10
8.3
(2;3)
8.4
60 units2
TERM 1 INVESTIGATION
8.5
67,39°
8.6
1 x – __
7
y = –__
TASK 1A
x = 1 or _
x=1
_
1
y
y
8.7
y = 2x – 1
8.8
Subs x = –1 into y = 2x – 1
y = 2(–1) – 1 = – 3
the line passes through (–1;–3)
__
2
2
3
9.1
12 + 8√2
9.2
8 units2
9.3
8,13°
9.4
D = (–7;–1)
10.1 59,62°
10.2 108,43°
10.3 59,62°
11
11.1 S = (7;4)
11.2 y = –2x + 13
11.3 (5;3)
12
2
8 =_
1
mAC = __
16 2
1 = –1
–2 × _
2
So ABCD is a rhombus (parallelogram
with diagonals perpendicular must be a
rhombus or a square, which is a special case
rhombus).
12.4 y = – 2x + 16
12.3 (5;6)
12.5 Substitute x = 5 into the equation of BD
y = –2(5) + 16 = 6
So M(5;6) lies on DB
12.6 As we have shown that the midpoint of
diagonal AC lies on diagonal BD, we have
shown that diagonal AC is bisected by
diagonal BD.
^ C = 14°
12.7 BA
13.1 mAB
1
= __
3
1x
13.3 y = _
3
2
y = 21 – 8 = 13
13 = 1,625
x = ___
__
y
8
TASK 2
1
34; 55; 89
2
3; _
13 ; __
34 ; __
5; _
8 ; __
55 ; __
89
1; _
2; _
21 ; __
_
3
1; 2; 1,5; 1,667; 1,6; 1,625; 1,615; 1,619; 1,618;
1,618
4
The numbers in the sequence get closer and closer
to 1,618.
1
1
2
3
5
8
13
21
34
55
TASK 3
1 and 2
x
T1
T2
T3
T4
T5
1
1
1,41421
1,55377
1,59805
1,61185
2 1,41421 1,55377
1,59805
1,61185
1,61612
3 1,73205 1,65289
1,62877
1,62135
1,61906
T8
T9
T10
x
BD ⊥ AC
13
5
12.1 (13;10)
12.2 While we could show that the sides
are equal, it is quicker to show that the
diagonals are perpendicular.
4 = –2
mBD = – _
1,6 or _xy = 1
1,625 or _xy = 1
TASK 1B
x = __
8 = 1,6
__
1
y
3
2
T6
T7
1
1,61612 1,61744 1,617851
1,617978
1,618017
2
1,61744 1,61785 1,617978
1,618017
1,618029
3
1,61835 1,61813 1,618064
1,618043
1,618037
TASK 4
1 and 2
x
T1
T2
T3
T4
T5
1
2
1,5
1,66667
1,6
1,625
2
1,5
1,66667
1.6
1,625
1,61538
3
1,333333
1,75
1,57143
1,63636
1,61111
x
T6
T7
T8
T9
T10
1
1,61538
1,61905
1,6176471
1,61818
1,61798
2
1,61905
1,61765
1,6181818
1,61798
1,61806
3
1,62069
1,61702
1,6184211
1,61789
1,61809
13.2 y = – 3x + 10
___
3√10
13.4 _____
5
3
The terms of the sequence converge or get closer
and closer to 1,618
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 339
339
2012/07/02 2:26 PM
TASK 5
1
1,618033992
2
They are equal to one another
3
–(–1)±√ (–1)2 – 4 (1)( –1) 1±√5
x = __________________ = _____
2
2(1)
_____________
__
2
4
1 + √ 5 is an irrational number
______
5
1,618033989
3
4
T6 = 47
9n – 7
6
6.1
6.2
6.3
42; 56
Tn = n2 + 3n + 2 = (n + 1)(n + 2)
n = 21
7
7.1
12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30
12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25
= 55
7.2
1
b = 1; c = – __
8
8.1
8.2
8.3
8.4
8.5
8.6
8.7
T5 = 18 and T6 = 22
T5 = 20 and T6 = 30
Tn = 4n – 2
Tn = n2 – n = n(n – 1)
T9 = 34
T11 = 110
n = 12 or n = – 11
9
9.1
14,14
9.3
2
TASK 6
1
0
2
5.1
5.2
__
√
1± 5
But the ratio is positive φ = _____
__
5
__
1 ± √5
______
2
The number is the Golden Ratio φ.
_______
__
– b ± √ b2 – 4ac _
1 ± √5
x = ____________
_
; xy = ______
y
2a
2
TASK 8
__
√2
TERM 1: ASSESSMENT TEST
1
2
1.1
1
__
1.2
6
__
1.3
8
1.4
x
1.5
k
2.1
1
x = 0 or x = __
2
2.2
3
2.4
x = 1,70 or x = –1,37
2.5
x = –5
x = –5
340
9.5
y = –x + 11
9.6
θ = 45°
9.7
^ B = 53,13°
AC
10.1
A(a;b)
3
2
N(3;6)
2.6
3 or x < –1
x>_
2.8
1
x=_
2
B(–12;1)
2
M
C(8;1)
x
2.12 x = 1
3.1
(x;y) = (0;8) or (–1;6)
3.2
5 or (3;10)
(x;y) = ( – 4;– __
)
4.1
4.2
4.3
4.4
mBC = 1
13
3
5 or x = – __
x = __
2.13 x = 5
4
9.4
13
2.10 1
x=0
125
3.3
5 x – __
7
y = __
4
1
2.11 x = ___
3
M = (4;1)
5
– __
x = 2 or x =
2.9
9.2
y
10
2.3
2.7
5
6
^ C = 45°
10.4 NM
10.5 5√2
__
(x;y) = (–1;3) or (3;–1)
x=–2±
Roots are real and irrational.
Roots are real and rational.
4.4.1 x = – 1 or x = 0
4.4.2
x > –1
10.3 (3;6)
__
10.6 mMN = 1; mAB = 1
3
___
√ 11
10.2 (–2;1)
11
10.7 10√ 2
10.8 x = –2
11.1 k = –14
11.2 k = 8
11.3 k = 5
12
12.1 D = (2;2)
12.2 5y + x – 12 = 0
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 340
2012/07/02 2:26 PM
1
12.3 mAB = __
1x
12.4 y = _
12.5 C = (7;1)
12.6 θ = 8,13°
3
y
7
7
f
3
(–3;3)
g
(3;3)
TOPIC 5: EXERCISE 1
1
y
0
–3
h
3
x
k
(–2;2)
(2;2)
2
(–3;–3)
0
–2
2
–2
(–2;–2)
k
x
h
3.1
3.2
(2;–2)
3.3
3.4
g
3.5
f
1.1
Refer to graph.
3.6
1.2
Refer to graph.
1 x2 + 2
g(x) = – __
3.7
1.3
1.4
1.5
1.6
1.7
(3;–3)
–3
Refer to graph.
Refer to graph.
1 x2 – 3
g(x) = __
3
Refer to graph.
1 x2
h(x) = –f(x) = – __
3
Refer to graph.
1 x2 + 3
k(x) = –g(x) = – __
3
4
2
y
k
Refer to graph.
1 x2h(x) = –f(x) = 12x2
h(x) = –f(x) = __
h
(–1;5)
2
Refer to graph.
1 x2 – 2
k(x) = –g(x) = __
(1;5)
(–1;3)
(1;3)
2
2
2
x
y
–2
(–1;–3)
(1;–3)
k
(–1;9)
(1;9)
(–1;8)
h
(1;8)
(–1;–5)
(1;–5)
f
g
4.1
4.2
4.3
4.4
4.5
4.6
4.7
x
(–1;–8)
(1;–8)
(–1;–9)
(1;–9)
f
g
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Refer to graph.
Refer to graph.
g( x ) = – 8x2 – 1
Refer to graph.
h( x ) = –f( x ) = 8x2
Refer to graph.
k( x ) = –g( x ) = 8x2 + 1
Refer to graph.
Refer to graph.
g( x ) = –3x2 – 2
Refer to graph.
h( x ) = –f( x ) = 3x2
Refer to graph.
k( x ) = –g( x ) = 3x2 + 2
TOPIC 5: EXERCISE 2
1
2
3
4
y = ( x + 3 )2; TP ( –3;0 ); Shift 3 left
y = ( x – 4 )2; TP ( 4;0 ); Shift 4 right
y = ( x – 2 )2; TP ( 2;0 ); Shift 2 right
y = –( x + 1 )2; TP(–1;0); Shift 1 left
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 341
341
2012/07/02 2:26 PM
TOPIC 5: EXERCISE 3
1
2
3
4
3.2
y = – 8x + 7 = ( x – 7 )( x – 1 )
TP(4;–9); Shift 4 right, 9 down
y = 4x2 + 16x + 15 = ( 2x + 3 )( 2x + 5 )
TP (–2;–1); Shift 2 left, 1 down
y = –x2 – 10x – 9 = –( x + 9 )( x + 1 )
TP ( –5;16 ); Shift 5 left, 16 up
y = –x2 + 6x – 5 = –( x – 5 )( x – 1 )
TP(3;4); Shift 3 right, 4 up
x2
TOPIC 5: EXERCISE 5
1
2
TOPIC 5: EXERCISE 4
1
1.1
1.3
y = – 4x2 – 6x + 4
10 x + 1
y = x2 – ___
1.4
5x + 6
y = –x2 – __
2.1
y = – 2x2 + 12x – 14
2.2
y = 3x2 – 6x + 1
1 x2 – 2x + 2
y = – __
3
2
2
y=
1 x2
– __
2
5x + 2
– __
2
TOPIC 5: EXERCISE 6
1
84-
y = x² – 6x + 5
y
-
-
-
–2
-
5
-
-
1.2
2.4
f(x) = –x² + 4x + 12
12 -
-
y = –3x2 – 9x + 12
(2;16)
16 -
–4
1.1
2.3
g(x) = 4x + 8
y
x < – 4 and x = 2
2
4
6
x
1
2.1
(3;–4)
y
-
–4
-
-
-
f(x) =
–2
g(x) = x – 4
–4
2
1
x² – x – 4
2
-
4-
-
2
x = –2 and x = 6
x ∊ [ –2;2 ]
-
1.2
1.3
x
5
2
4
6
-
y = 2x² + 4x + 6
y
x
6
(–2;6)
(1;–4,5)
(–1;4)
2.2
2.3
3
x = 2 only
x ∊ [ 0;4 ]
x
3.1
g(x) = –3x + 6 y
(–1;9)
-
8
-
-
–2
-
-
–4
-
-
6-
2
4
x
f(x) = –(x + 1)² + 9
342
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 342
2012/07/02 2:26 PM
3
6
y
(–4;9)
y = –2x² – 12x + 10
y
10
–7
x
–1
x
5
1
–7
y = –x² – 8x – 7
(3;–8)
4
y
(–5;16)
TOPIC 5: EXERCISE 7
1
y
f
x = –2
–9
x
–1
–
–9
5
3
1
–5
x
-1
y = –x² – 10x – 9
y = –3
5
y
y = 3x² – 24x + 36
2,5
(–2;–3)
(–3;–4)
36
–5
y = –x – 5
y=x–1
6
2
x
2
y
g
y=x+5
(4;–12)
y = –x + 1
5
(–3;4)
y=3
(–2;3)
2,5
1
–5
–
5
3
1
x
x = –2
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 343
343
2012/07/02 2:26 PM
3
6
y
y
v
x = –3
h
y = –x – 1
x=2
(–1;1)
–4
5
3
1
1
3
x
1
x
5
(–3;–1)
–1
y = –1
(–5;–3)
–2,5
(2;–3)
–4
y = –3
(3;–4)
y=x+2
–5
y=x–5
7
y
x=1
4
y = –x – 4
y=x+1
5
y
r
x=3
y=2
w
2
3
2
–1
x
–1
(1;1)
1
3
2,5
x
4
(3;–1)
y=–1
y = –x + 3
(5;–3)
–4
8
y=x–4
y
y = –x + 2
5
y
t
x=3
y = –2
–1
–1
2,5
3
x
(1;–2)
4
–3
z
(5;3)
–5
y=1
(3;1)
x=1
–1
2
1
3
x
4
y = –x – 1
y=x–3
(1;–1)
–2
y=x–2
344
y = –x + 4
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 344
2012/07/02 2:26 PM
TOPIC 5: EXERCISE 9
1
y
–4
–2
y = –3
y = –x + 1
y=x–4
d
x=1
y=
-
x–3
-
-
-
-
-
-
4
y
543210–3 –2 –1 –1
–2 –3 –4 –5 y = –x – 5 –6
–7 –8 -
x
–2
-
3
-
k = – 4, p = 1 and q = 2
y = x + 1; y = –x + 3
A(1;2); B(0;0); C(2;0)
5.5
maximum; y ≤ 2 for all y ∊ ℝ
5.6
5.6.1 y = –2( x – 1 )2 + 2 = –2x2 + 4x
5.6.2
y = –2x( x – 2 ) = –2x2 + 4x
5.7
x ∊ (–∞ ; 0) ∪ (1; 2) ∪ (3; ∞)
5.8
6 units
5.9
5.9.1
1 left, 4 up
5.9.2
y = –2x2 + 6
–4
5.10 y = ___
x
(3;–2)
-
x ∊ ℝ , x ≠ 1; y ∊ ℝ , y ≠ 2
5.3
5.4
x
y = –x – 2
-
5.1
5.2
-
x–4
-
-
x+3
-
x–2
Symmetry lines: y = x – 7 and y = –x + 1
5
6
6542
+2
y=
x+1
3y=2
(1;3)
210–6 –5 –4 –3 –2 –1 - 1 2 3 4 5 6
–1
–2 –3 y = –x + 3 –4 –5 y = –x + 1
–6 -
-
4
Symmetry lines: y = x + 1 and y = –x – 5
5 –3
y = _____
4
x = –1
x+1
-
3
Symmetry lines: y = x – 1 and y = –x + 3
1 –2
y = – _____
2
y
-
2
Symmetry lines: y = x + 5 and y = –x + 3
3 +1
y = _____
5
3
2
-
2 +4
y = – _____
-
1
-
TOPIC 5: EXERCISE 8
-
-
x
1
-
–2,5
-
y=2
-
1
–3
–3
x–1
-
3
2
y=
43210–1 –2 –3 –4 –5 –6 –7 –8 -
5
-
5-
-
x = –1
-
y=x+3
-
y
-
9
1 2 3 4 5 6 7 8 9
3;–2) (7;1)
x
y = –2
y = –x + 1
x=3
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 345
345
2012/07/02 2:26 PM
TOPIC 5: EXERCISE 10
y
1
x=2
-
y
x = –1
-
-
-
-
-
Asymptotes: x = –1 and y = 1
Symmetry lines: y = x + 2 and y = –x
2
y=1
(2;1)
-
x
3 4 5 6 7 8 9
4- ( 1 ;3 1 )
2
16
y=x+2
3-
y = –x
f
2-
y=1
g
y=
2
x–2
-
0-
-
–1
-
–2
-
–3
-
-
–4
-
-
–5
-
-
y = –x + 3
-
1-
-
-
-
-
-
-
-
54 - (1;3)
3210–3 –2 –1 - 1 2
–1
–2 –3 –4 y = x – 1 –5
–6 –7 –8 -
-
4
1
2
3
4
5
6
7
x
–1-
+1
–2-
y
5
y=
4
x+5
–1
–34
–4–5-
y = –x – 6
x= 1
2
–6-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
–7-
x
–11 –10 –9 –8 –7 –6
–4 –3 –2 –1
( –5;1)
(–1;2)
y=–1
1
2
3
4
–1,3
–6
x=–5
y=
y
1
x–2
13
14
(3;5)
4-
(2;4)
–2 –3 -
4
2
4
1 unit to the right; 1 unit down.
1 = ______
g( __
) 3x + 1; undefined if x = –1 or 0
x
1+x
2 + 1; undefined if x = 3
g(x – 4) = _____
x–3
1 = 4g(–x)g( x – 4 )
g(x) + g( __
)
x
3x + 1 = _______________
2 + x + 1 + 3x + 1
2 + 1 + ______
LHS = _____
x+1
1+x
x+1
4( x + 1 )
4x + 4 = _______
= ______
=4
x+1
x+1
[
-
-
-
-
-
-
-
-
1
-
2x + 1
][ x – 3
2 + 1 _____
2 +1
= 4 – _____
–4 –3 –2 –1 –1
-
x–1
RHS = 4g(–x)g( x – 4 )
y=x+6
0-
-
Reflection in the y-axis.
3
1 x2 – __
1 x + __
y = – __
12
1-
9
-
2-
346
2 +1
y = ______
11
3-
y=x+2
8
8-
5-
]
7
10
+4
6-
y=4
16
–8
6
7
(
( 0;3 ); ( –3;0 ) or ( 3;1,5 )
x ∊ [–3;–1) ∪ [0;3] or –3 ≤ x < –1 or 0 ≤ x ≤ 3;
x∊ℝ
2
y = _____
5
6
–4
y=x+4
( – 4;–2 ); ( 1;3 )
g: x ∊ ℝ , x ≠ – 1; f: y ∊ (–∞;3,0625] or
1 ,y∊ℝ
y ≤ 3___
x
2
3
4
5
6
7
x–1
( x – 1 )( x – 3 )
x – 3 _____
x–1 =4
= 4( _____
x – 1 )( x – 3 )
]
2+x–3
–2 + x – 1 ________
= 4 _________
⇒ LHS = RHS
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 346
2012/07/02 2:26 PM
TOPIC 5: EXERCISE 11
1
4
y
t(x)
t(–x)
y
f(–x) 20 -
f(x)
(–1;5)
y=1
15 10 5-
–2
y = –1
–2
(1;2,5)
-
-
-
-
-
(–1;2,5)
x
(1;10)
–1
1
(1;–5)
-
(–1;10)
(1;5)
2
x
2
–t(x)
–5 (–1;–2,5)
–10 -
(1;–10)
TOPIC 5: EXERCISE 12
–15 -
12
–20 -
2
–f(x)
12.2 d = 5, e = 2 and p = 1
1 and t = 2
12.3 j = –1, r = __
5
y
g(–x)
g(x)
20 -
12.1 a = 3, b = 2 and c = –1
12.4 v = 2, w = 1 and z = 2
15 -
TOPIC 5: EXERCISE 13
10 (–1;5)
1
1
–5 -
(1;6)
-
-
-
-
–1
-
-
–2
y = 3 2x + 1 – 6
(1;0,2)
-
-
(–1;0,2)
–3
y
(1;5)
5-
2
(1;–5)
3
x
–10 -
x
0
–15 -
(–1;–3)
–20 -
3
–g(x)
h(x)
y = –6
h(–x)
y
2
(–1,5)
3
124
(1; 125
)
(1,5)
y=1
0
y=2
y = –2
(–1;–5)
y
x
x
–3
(1;– 37 )
(3;–4)
y = –5x – 2 + 1
–h(x)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 347
347
2012/07/02 2:26 PM
3
y=2
1 x+1
–
3
7
2
(–2;–4)
8
–1
–
x
(1;– 16
)
9
4
3
y = 2 –x –1
7.2
d = 8, e = 3 and j = −2
7.3
1 , q = −2 and v = 1
k = __
8.1
8.2
8.3
8.4
x=1
x > 1 or x ∊ (1;∞)
y=x+c⇒y=x+1
x ∊ [ –1;4 ] or –1 ≤ x ≤ 4, x ∊ ℝ
4 –1
f( x ) = x2 – 2x – 8; g( x ) = – _____
2
x+2
A( –6;0 ); B( 0;–3 )
Q( 2;–2 )
x ∊ ( –∞;–2 ) ∪ [–1;2]
x ∊ [ –∞;–6 ] ∪ [ 4;∞ )
8.10.1 RT = – x2 + 3x + 4
8.10.2 RT = 6,25
8.11 x ∊ ( –∞;–2 ) ∪ ( –2;∞ ) or x ∊ ℝ , x ≠ –2
4 – 2] – 1
8.12 8.12.1 [__
x
8.12.2 y = x – 3; y = –x + 1
8.12.3 Teacher to provide
8.12.4 Teacher to provide
4
8.13 8.13.1 w(x = –__
x
8.13.2 x = 0 and y = 0
8.13.3 y = x and y = –x
__
2
8.13.4 √
2
y
(1;–0,5)
x
y = –1
5
2
8.6
8.7
8.8
8.9
8.10
(–2;3)
y
y=5
TOPIC 5: EXERCISE 14
1
33
1
0
x
(2;–10)
y = –5 3
6
1 , b = 2 and c = 6
a = −__
8.5
y = –2
4
7.1
y
x–1
+5
y
1
1.1
The death rate will continue to increase at
a constant rate. In the long term this would
have a catastrophic impact on life in South
Africa and would ultimately mean that the
battle against HIV/AIDS has been lost.
1.2
1.3
1.4
1.5
16,52 years
Maximum = 716 349
during the year 2012
The number of deaths begins to decrease
after the end of 2012.
8
y = 2 ( 15 ) x – 1 – 2
x
0
(2;–1,6)
y = –2
348
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 348
2012/07/02 2:26 PM
2
2.1
2.2
200
-
180
-
160
-
TOPIC 5: EXERCISE 15
R80
1
2
(5;160)
140
-
120
-
100
-
80
-
3
(4;80)
60
-
40
-
4
A( –3;–7 ); B( 5;–7 ); C( 4;0 ); D( 3;5 ); E( 2;8 ); F( 1;9 );
G( 0;8 ); H( –1;5 ) and I( –2;0 )
The further apart the points are from each
other, the greater is the enclosed area. The closer
the points are to each other, the smaller is the
enclosed area.
3.1
mAB = 0
3.2
mAC = 1
3.3
mAD = 2
3.4
mAE = 3
3.5
mAF = 4
3.6
mAG = 5
3.7
mAH = 6
3.8
mAI = 7
mAI = 7 because the line through A and I, although
not a tangent, is much closer to the parabola than
the line between A and any of the other points.
(3;40)
20
TOPIC 5: EXERCISE 16
-
(2;20)
1
2.3
2.4
2.5
2.6
2.7
3
3.1
3.2
3.3
3.4
-
-
-
-
-
-
(1;10)
1
2
3
4
5
6
It is discrete. Each point represents the
month in which the deposit was made and
the amount of the deposit.
The graph starts when the first deposit
is made, which is at the end of the first
month.
No. His 12th payment will be R20 480.
y = 5.2x or y = 10.2x – 1
A = 5 and B = 2
f (x) = 281 234, 5833 + 39 377,95x
g(x) = 249 902,5 + 56 468,17727x
– 1 709,022727x2
h(x) = 303 433,3851 × 1,089710572x
24 years
f (24); 1 226 305 deaths
g(24); 620 742 deaths
h(24); 235,724 deaths
The linear model, f, shows that a steady
increase is anticipated in the long term.
The parabolic model, g, produces the
most encouraging result and indicates
that a steady decline in the death rate is
anticipated as the maximum would have
been reached in 2012.
The exponential model, h, shows that
an alarmingly rapid rate of increase is
anticipated in the death rate.
1.1
1.2
1.1.1
g(–1) = –2
1.1.2
7
g(0) = – __
1.1.3
g(1) = –3
1.1.4
g(2) = –5
1.2.1
1
m = – __
1.2.2
1
m = – __
1.2.3
m = –1
3
3
2
1.3
y
y=
4
x–3
–1
7
x
y = –1
1
(–1;–2)
–2 3
3
(2;–5)
3
2
1.4
1
m = – __
2.1
2.1.1
2.1.2
2.1.3
h( –2 ) = 4
h( –1 ) = 2
h( 0 ) = 1
2.1.4
1
h(1) = __
2.2.1
2.2.2
m = –2
3
m = – __
2.2.3
7
m = – __
2.2
3
2
2
6
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 349
349
2012/07/02 2:26 PM
2.3
2
y
y = 2 –x
(–2;4)
2.1
2.2
2.3
x = –1 and y = –3
f: y = x – 2 and y = –x – 4; g: x = –1
y
h(x) = –2x – 5
3
(–1;6,25)
6
1
2
x
m = –2
3.1
3.1.1
3.1.2
3.1.3
3.1.4
k(–1) = –7
k(0) = –12
k(2) = –16
k(4) = –12
3.2.1
3.2.2
3.2.3
–6
-
-
–2
-
-
–4
2
4
6
4
m = –5
4.1
m=1
4.2
8
m = __
4.3
m = –3
2.4
2.5
2.6
5
(4;–12)
2.7
2.8
2.9
f(x) = – 2 – 3
x+1
x = 0 or 2
x ∊ (–6;4) or –6 < x < 4, x ∊ ℝ
x ∊ (–∞;–6) ∪ (–2,5;4)
or x < –6 or –2,5 < x < 4
2 –2
y = – _____
x–1
y = 2x – 5
1 x2 – 2x – __
1
y = – __
4
4
2.10 g(–2) – f (–2) = 7
3
1
2.11 2.11.1 m = __
2.11.2
TOPIC 5: REVISION
1
(–3;–2)
y=x–2
(2;–16)
3.4
g(x) = – 14 (x + 6)(x+4)
x
x = –1
5
–12
4
–5
x
(–1;–7)
2
y = –3
m = –1
m = –1
m = –1
y
y = x² – 4x – 12
3.3
–1,6
–2,5
–4
(–2;–1)
(–7;–2,3) (–5;–2,5)
-
3.2
y = –x – 4
-
3
2.4
1.1
q = –1
1.2
A(2;2)
1.3
x=2
1.4
4 units
1.5
Q(8;2)
1.6
x = 1 and y = –1
1.7
ED = 1
1.8
B(3;–1,5)
1.9
y=x–2
3
1
m = __
2
2.11.3 m = 1
2.12 closest point to (–2;–1) is the point (–3;–2);
m = –1
2.13 x = 10,28 or – 4,28
1
2.14 2.14.1 m = __
2.14.2
2
5
m = __
4
2.15 (3;5,25) is much closer to (– 4;4) than ( 0;6 ) is
3
3.1
5 seconds
1.10 1.10.1 2 units left and 2 units up
1 x2
1.10.2 y = __
2
350
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 350
2012/07/02 2:26 PM
h
3.2
1.1
1.2
(2;21)
(1;20)
x = –90°, 90°or 270°
x ∊ (–90°;90°) ∪ (270°;360°)
–90° < x < 90° or 270° < x ≤ 360°
x ∊ (–180°;0°) ∪ (180°;360°)
or (–180°;0) ∪ (180°;360°)
(3;18)
1.3
(0;15)
2
(4;11)
period 180°; amplitude is undefined; y ∊ (–∞;∞) or
y ∊ ℝ; Asymptotes for both graphs: x = –90° and
x = 90°
y
2(45°;2)
g
1-
–90°
2
90°
180°
x
–1-
x = –90
–2-
x = 90
1.1
1.2
1.3
1.4
1.5
1.6
Teacher to provide
Range: y ∊ [ –1; 1 ]
Teacher to provide
Period is 360°
Amplitude = 1
90° to the right
2.1
2.2
2.3
2.4
2.5
Teacher to provide
tan x is undefined at ±90° and ±270°
x = –270°, x = –90° and x = 270°
Teacher to provide
Range y ∊ (–∞;∞), but the amplitude is
undefined.
Period 180°
2.6
-
-
–180°
TOPIC 6: EXERCISE 1
1
-
-
y = –2x2 + 7x + 15
t = 1,75 seconds
Maximum height = 21,125 m
3.3
3.4
3.5
-
(45°;1)
(5;0)
-
t
f
2.1
2.2
3
x = –180°, 0° or 180°
x = – 45° or x = 135°
period 360°; amplitude of f is 2; amplitude of
3 or 1,5.
g is __
2
Range of f: y ∊ [–1; 3]
[
3 ;__
5
Range of g: y ∊ –__
22
]
Range of g: y ∊ [ – 1,5; 1,5 ]
y
(–90°;3)
(270°;3)
3-
f
2(0;1,5)
–1 -
(360°;2)
2-
-
-
-
-
-
-
-
(360°;1)
1- g
-
180°
-
-
90°
270°
x
(90°;–1)
(180°;–1,5)
h
–180° –90°
-
–90°
-
360° ; Amplitude of h is 2; Amplitude of g is 1;
Range of h: y ∊ [–2;2]; Range of g: y ∊ [–1;1]
y
-
1
1-
-
TOPIC 6: EXERCISE 2
g
90° 180° 270° 360°
–1(–180°;1)
x
3.1
3.2
3.3
3.4
x = –90° or x = 270°
x = 90°
x ∊ [–90°;90°] or x = 270°
y = –f (x) = 2sin x – 1
(180°;–1)
–2-
(–180°;–2)
(180°;–2)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 351
351
2012/07/02 2:26 PM
4
3
period 180°; undefined amplitude.
Range y ∊ (–∞;∞) or y ∊ ℝ
Asymptote x = 90°
y
x = –90
2-
(–45;1)
1-
y
-
–90°
90°
g
x
–1-
2-
–2-
x = 90
1(135°;1)
f
4
y
-
(–45°;0,5)
–1 -
–1,5
(–90°;–1,5)
y
1-
2
(–90°;1)
0,5 -
y
–90°
2(–180°;1)
1-
(180°;1)
-
90°
180°
–1-
-
-
-
-
-
(–90°;–1)
(–135°;2)
90°
180°
x
(90°;–1)
x = –90
-
-
-
–135°–90°–45°
–1(–135°;2)
–2-
-
-
-
1-
0,5
–90°
(–45°;2)
2-
–1-
y
–180°
x
6
x
-
2
-
(–90°;–1)
180° 270°
y
-
-
-
–90°
90°
–0,5 -
g
–180°
1
6
-
3-
-
-
(90°;3)
(90°;12 )
-
TOPIC 6: EXERCISE 3
1
(180°; 56 )
-
2
5
x
(270°;–1,5)
-
x = 90°
1,5
y = –f (x) = tan x
1 tan (–x) = – __
1 tan x
y = g(–x) = __
270°
(180°;–1)
–1
x = 90
4.4
90°
180°
(90°;–0,5)
0,5
–2 -
4.1
4.2
4.3
-
–90°
-
-
90° (135°;–0,5)
-
x
-
-
-
(–45°;1)
45° 90° 135°
x
(45°;–2)
x = 90
–2(–180°;–2,5)
352
(180°;–2,5)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 352
2012/07/02 2:27 PM
7
TOPIC 6: EXERCISE 5
y
(–90°;3)
(–180°;2)
3
1
y
2
(120°;1)
1-
y = cos 3x
1
(90°;1)
-
–180°
–90°
-
-
–270°
-
-
90°
–30°
x
-
-
-
-
-
-
-
(–270°;1)
x
30° 60° 90° 120° 150°
–1 (60°;–1)
8
(180°;–1)
y
2
y
y = tan 14 x
90°
180°
(90°;–1)
x
(180°;1)
2.7
2.8
-
-
-
-
y = cos 2x
45°
90°
x
–1(–270°;–1)
(–90°;–1)
(90°;–1)
4
y = sin 2x
y
1-
–90°–45°
–1(–45°;–1)
(45°;1)
(210°;0,87)
-
f: x = –135°; x = – 45°; x = 45°; x = 135°
g: x = –180°; x = 180°
Teacher to provide
x = 90°
1 (90°) – tan 2(90°)
g(90°) – f (90°) = tan __
5
-
4
-
( __12 )
2
2.6
1-
–270°–225°–180°–135° –90° –45°
-
2.4
2.5
(–180°;1)
-
2.3
y
-
360° = 90°
180° = 360°; S = ____
P = ____
4
-
2.2
2
Teacher to provide
-
2.1
180° = 90°; S = ___
90° = 225°
P = ____
3
x = 360°
-
1.6
Teacher to provide
Teacher to provide
Amplitude of f and g: 1
x = –180° or 0°
Period of f and g: 720°
x = –360°, 180° or 360°
Range of f and g: y ∊ [–1;1]
x = –270°, 90°
1.5
2
x = –360°
-
1.1
1.2
1.3
1.4
x
180°
–180°
(180°;–1)
TOPIC 6: EXERCISE 4
1
-
-
–1
–2
–3
–4
-
–180°
–90°
(–90°;–1)
(180°;2)
2
1
-
(–180°;2)
45° 90° 135°180°
x
(135°;1)
= tan 45° – tan 180° = 1 – 0 = 1
x = ±90°
1 (–90°)
If x = –90°, then tan 2(–90°) – tan __
2
= tan (–180°) – tan (– 45°) = 0 – (–1) = 1
y ∊ (–∞;∞)
The amplitude is undefined.
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 353
353
2012/07/02 2:27 PM
(45°;1)
–90°–45°
(210°;0,87)
y = cos 12 x
-
-
–180°
x
180°
–1-
(–360°;–1)
(360°;–1)
Teacher to provide
TOPIC 6: EXERCISE 6
1
1.1
1.2
1.3
1.4
1.5
1.6
Teacher to provide
x = 90°
x ∊ (–90°;–60°)
x ∊ [0°;30°] or 0° ≤ x ≤ 30°
x = –90° or x ∊ [30°;90°]
30° to the left.
2
2.1
2.2
2.3
Teacher to provide
x = 0°
x ∊ (–135°;–90°) ∪ (–90°;–75°) ∪ (–75°;45°) ∪
(45°;90°)
x ∊ (–75°;– 4°] ∪ [15°;45°)
x ∊ (–135°;–90°) ∪ [– 45°;0°] ∪ (45°; 90°)
2.4
2.5
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Teacher to provide
g has an amplitude of 2.
f has a period of 180°.
Range of h: y ∊ [ –1;1 ]
x = –90° or 90°
x = – 45°
3 units at x = 135°
x ∊ [–90°;0°) ∪ [45°;90°] ∪ [180°;225°]
h(15°) – g(15°) = –sin (–30°) + 2cos 60°
1 + 2 __
1 = 1__
1
= __
2
3.10
3.11
3.12
3.13
3.14
354
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
y
1-
3
5
(135°;1)
7
8
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
x
45° 90° 135°180°
–1(–45°;–1)
4
-
-
-
-
-
-
1-
3.15 y = –sin [(x – 45°) – 45°]
= –sin (x – 90°) = cos x
y = sin 2x
y
-
6
(2)
2
x = –135°, 45° or 225°
y = sin(x – 45°)
y = –2cos (x + 90°) = 2sin x
x ∊ (– 45°;135°)
x ∊ [–90°;0°) ∪ [90°;180°)
Teacher to provide
The period of f is 360°.
g has an amplitude of 2.
y ∊ [–1,5;0,5]
–12°
x = 0°
2,5 units
x ∊ (–60°;0°)
x ∊ [–120°;–90°) ∪ (60°;90°)
y = –g( x ) = –2cos x
1
y = sin (x – 30°) + __
2
1
__
y = sin x –
2
60° to the right
90° to the left
Teacher to provide
f has an amplitude of 1.
g has a period of 180°.
x ∊ (–120°;60°)
x = –75° or x = 105°
x ∊ [–210°;–165°] ∪ (–75°;–30°] ∪ [15°;10°)
x = 60°
x ∊ (–75°;60°] ∪ (10°;150°]
x = 120° Check: g(120°) – f(120°) = tan
(–135°) – cos (–180°) = 2
5.10 x = –210° or – 30° or 150° (any two)
5.11 5.11.1 15° to the left; reflected in the x-axis
(order does not matter)
5.11.2 x = ± 90°
5.12 5.12.1 210° to the right
5.12.2 150° to the left
TOPIC 6: EXERCISE 7
Teacher to provide
TOPIC 6: EXERCISE 8
1
1.1
1.2
1.3
1.4
f (x) = –sin (x – 35°) ⇒ a = –1 and b = –35°
f (x) = cos (x + 60°) ⇒ c = 60° and d = 1
720°
1
p = 1; q = __
1.5
1 ; w = __
1
v = – __
1.6
1 ; t = –180°
r = __
1.7
x = ±360°
1.8
x = 0°
2
2
2
2
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 354
2012/07/02 2:27 PM
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
3
Period of f is 360°
Period of g is 720°
Amplitude of f is 1
Amplitude of g is 1
1 x + 1 ⇒ a = __
1 and b = 1
g(x) = sin__
2
3
3.1
3.2
Teacher to provide
3.2.1 x = –120°, 0° or 60°
3.2.2
x = –30°
4
4.1
4.2
Teacher to provide
4.2.1 x = –120°
4.2.2
x = 60°
4.2.3
x = –210°, –30° or 150°
4.2.4
f (0°) – g(0°) = 0,75
4.2.5
x ∊ (30°;150°)
2
f (x) = cos (x + 90°) ⇒ d = 1 and p = 90°
f (x) = –sin x ⇒ m = –1 and n = 1
2.8
1
v = –1 and w = __
2.9
e = –180° and t = 1
2
2.10 x = 180°
TOPIC 6: EXERCISE 11
3.1
3.2
3.3
3.4
3.5
3.6
Period of f is 360°
Amplitude of f is 2 units
Period of g is 540°
The amplitude of g is undefined.
540°
f (x) = 2sin (x + 1°) ⇒ a = 2 and b = 1°
1
1.1
1.2
Teacher to provide
1.2.1 x = 120° or 60°
1.2.2
x = –210° or 150°
1.2.3
x ∊ [–120°;60°]
2
2.1
2.2
3.7
f (x) = 2cos (x – 7°)3.8
x ⇒ v = 1 and w = __
1
g(x) = tan __
Teacher to provide
2.2.1 x = ±180°
2.2.2
x ∊ [–180°;150°) ∪ (–90°;0°]
3
3.1
R(–60°;0); T(150°;2); U(–150°;–1);
V(120°;0); W(210°;–1) and Z(30°;1)
3.9
3
3
3
3
2 ⇒ p = –1 and = __
2
h(x) = –cos__
3.10 x ∊ [– 270°; –195] ∪ [–1°; 0°] ∪ [165°; 270°)
3.11 x = 135°
3.12 x = 0° or x = 135°
3.2
3.3
3.4
3.5
TOPIC 6: EXERCISE 9
1
1.1
1.2
y = cos(x – 20°) + 1
Range y ∊ [0;2]
2
2.1
2.2
y = sin(x + 90°) – 1 = cos x – 1
Minimum is –2
3
1x
a = 3 ⇒ y = 3tan __
4
y = tan (x – 45°) – 2
5
5.1
5.2
2
1 , b = 1, c = –2 and d = __
1
a = __
4
2
Teacher to provide
TOPIC 6: EXERCISE 10
1
2.
1.1
1.2
2.1
2.2
Teacher to provide
1.2.1 x = –210°, –165°, –30° or 15°
1.2.2
x ∊ (–210°;–165°) ∪ (–120;–75°)
∪ (–30°;15°) ∪ (60°;105°)
Teacher to provide
2.2.1 x = ±45°
2.2.2
x ∊ (–15°;45°) ∪ (105°;15°)
∪ (165°;225°)
2.2.3
x = 13°
3.6
3.7
1 , d = 60°, c = 120°
a = –2, p = 60°, k = __
2
and e = –240°
TU is 3 units.
y = –2cos x or y = –2sin (x + 90°)
y = –2sin [(x – 240°) + 60°] = –2sin (x – 180°)
= 2sin x
If must be shifted 60° to the right.
x = 30°
TOPIC 6: REVISION
1
1.1
1.2
1.3
1.4
The period of f is 360°, the amplitude 2, the
range is given by y ∊ [–2;2] and y-intercept
(0;1)
The period of g is 180°, the amplitude 1, the
range is given by y ∊ [–1;1] and y-intercept
(0;–1)
The period of h is 90°, the amplitude
undefined, the range is given by y ∊ (–∞;∞)
and y-intercept (0°;0)
The period of k is 180°, the amplitude 1, the
range is given by y ∊ [–2;0] and y-intercept
(0°;–1,34)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 355
355
2012/07/02 2:27 PM
1.5
1.6
The period of p is 360°, the amplitude
undefined, the range is given by y ∊ (–∞;∞)
and y-intercept (0°;0)
1 , the
The period of q is 360°, the amplitude __
2
1
1
__
__
range is given by y ∊ – ; and
2 2
y-intercept (0°;0,47)
[
]
2
2.1
2.2
y = cos (x – 9°) = sin x
No, because y = cos (x + 9°) = –sin x
3
3.1
3.2
3.3
3.4
3.5
A(–180°;–1,73); B(150°;–2); C(–180°;–3,73);
D(60°;0); E(–30°;2); F(–120°;0); G(0°;1,73);
H(0°; 0,27)
x = 150°
x ∊ [–180°;60°]
The period 360°, the shift is 30° to the left.
f (x) = 2cos(x + 90°) = −2sin x
3.6
1 (−x + 30) − 1 = −tan__
1 (x − 30) − 1
g(−x) = tan__
3.7
3.8
3.9
3.10
x ∊ [30°;150°)
x ∊ [−180°;−120°] and x = 60°
−f (x) = −2cos(x + 30°)
1 (x + 30°) − 1
−g(x) = − tan__
2
1 (x + 30°) + 1
= −tan__
4.1
Period of f is 360°
Amplitude of f is 2
Range of f: y ∊ [– 2;2]
__
__
Endpoints: (–180°;√3 ) and (180°;√ 3 )
Period of g is 240°
Amplitude of g is 1
Range of g: y ∊ [–1;1]
(–180°;0) and (180°;0)
Teacher to provide
g(–120°) – f (–120°) = 1
x = 6°
x = ± 180°
x ∊ [–60°;60°] or – 60° ≤ x ≤ 60°
3 x – 9° = –sin 9° – __
3x
g( x ) = sin __
2
4.2
4.3
4.4
4.5
4.6
4.7
4.8
(2
3x
= –cos __
)
(
TOPIC 7: EXERCISE 1
1
2
2
3
2
4
5.10 y ∊ (−2;1)
5.11 x ∊ (−195°;−75°) ∪ (165°;285°)
2
1.1
Q3
1.2
Q2
1.3
Q4
1.4
Q3
1.5
Q4
1.6
Q4
1.7
Q2
1.8
Q3
1.9
Q2
1.10 Q1
2.1
negative
2.2
negative
2.3
positive
2.4
negative
2.5
positive
2.6
negative
2.7
positive
2.8
positive
2.9
negative
2.10 positive
3.1
Q2
3.2
Q4
3.3
Q2
3.4
Q1
3.5
Q4
3.6
Q3
1.2
4
sin θ = __
TOPIC 7: EXERCISE 2
1
1.1
1.3
)
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
356
f has an amplitude of 2 and a period of 360°
g has an amplitude of 2 and a period of
1 080°
1
a = 2, b = 45°, c = 1, d = –2 and e = __
3
A(–75°;0) and B(–135°;–1)
__
CD = 3 + √2 = 4,14
p = 2, q = −45° and q = 1
1 and z = 270°
w = −2, k = __
3
x ∊ [−360°;−315°) ∪ (−135°;45°) ∪ (225°;360°]
x ∊ [−270°;−195°] ∪ [−75°;165°] ∪ [270°;285°]
y ∊ (−3;1)
25
7
cos θ = ___
25
24
tan θ = ___
7
__
√
√5
5√ 5 15
__
√5
–5__ = ___
–1__ × ___
____
__
√5
5√ 5 √5
5
–3
cos θ = ___
5
___
tan θ = 4
–3
__
√
5 _____
–10__ = ___
–2 × ___
__
sin θ = ____
= –2 5
cos θ =
5
__
–√ 5
= ____
5
–10 = 2
tan θ = ____
2
5
24
sin θ = ___
–5
1.4
–3
___
sin θ = ____
√ 34
5___
cos θ = ____
√ 34
–3
tan θ = ___
5
2
2.1
__
√
3
sin 60° = ___
2
1
cos 60° = __
2
__
√
__
3 √
tan 60° = ___
= 3
1
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 356
2012/07/02 2:27 PM
__
√
√2
2√2 √ 2
__
√2
–2
1
__ × ___
__
__ = ___
= ____
√2
2√2 √ 2
__
√
2
2 = ___
2 __ = ___
1__ × ___
__
sin 135° = ____
2.2
cos 135°
=
3
cos3 x + cos x.sin2 x = cos x
LHS = cos x (cos2 x + sin2 x) = cos x (1) = RHS
Teacher to provide
4
(1 – sin2 x)(1 + tan2 x) = 1
2
2
__
–√2
____
2
2 = –1
tan 135° = ___
–2
2
sin x )
LHS = (cos2 x)(1 + ______
2
–1
sin 210° = ___
2.3
cos x
2
__
–√ 3
____
cos 210° =
2
cos2 x + sin2 x = 1 = RHS
–1__ ×
tan 210° = ____
–√ 3
__
√
__
√3
___
__
√3
=
5
__
√3
___
3
2
sin x
LHS = RHS
2
__
–√ 3
tan 300° = ____
1
6
2
cos x
LHS = RHS
4
– __
2
–13
3
5
__
4
20
5
0
6
3 , tan y = __
5 , x < 90° and y < 90° ,
If cos x = __
3
cos x – cos x.sin2 x
cos x (1 – cos2 x)
cos x (1 – sin x)
1
5
cos x – cos3x
________________
= tan2 x
sin x = tan2 x
LHS = _______________
= ______
2
2
TOPIC 7: EXERCISE 3
7
________
sin θ – √ 1 – sin2 θ = 0
_____
tan θ
______
cos x – √ cos2 θ
sin θ × _____
LHS = _____
sin x
1
7
7
______
2
LHS = 74sin y +
tan2 x
= cos x – cos x = 0 = RHS
4
(
8
) ( ___7 )
3
5___ 2+ _____
7__
= 74 ____
2
√ 74
( 74 )
θ
cos θ
cos θ
9
7
cos θ
sin x + cos x = cos x
___________
1 + tan x
sin x + cos x = ___________
sin x + cos x
LHS = ___________
sin x
cos x + sin x
1 + _____
cos x
k
________
_____
√
– 1 – k2
___________
cos x
cos x
= (sin x + cos x) × _____________
(sin x + cos x)
= cos x = RHS
TOPIC 7: EXERCISE 4
sin x
cos
sin2 θ + cos2 θ = ______
1 = RHS
θ + 1 = ______________
LHS = ______
2
2
2
√
= 25 + 9 = 34 = RHS
1
8
1
tan2 θ + 1 = ______
2
sin2
25 2+ 7 × __
9
=74 ___
1
4
7
10
2
2
sin x
sin x
cos x × _____
cos x = ______
cos2 x
RHS = _____
sin x
sin x
1
1
cos 300° = __
7
tan x
sin x
cos x
1 – sin x = ______
1 – sin x = ________
LHS = _____
– 3
sin 300° = ____
2.4
cos x
1 – sin x = _____
_____
2
5
8
sin2 x
cos2 x
1
–1
3
6
9
sin x
10
9cos2 θ
tan x
_____
1 = _________
1
tan θ + _____
tan θ
sin θ.cos θ
cos θ = ____________
sin2 θ + cos2 θ
sin θ + _____
LHS = _____
cos θ sin θ
1
= _________
sin θ.cos θ
2
sin θ.cos θ
= RHS
TOPIC 7: EXERCISE 5
1
sin x
2
LHS = sin x.cos x._____
cos x = sin x
RHS = 1 – cos2 x
= sin2 x
LHS = RHS
11
tan2 x – sin2 x = sin2 x.tan2 x
2
2
2
2
sin x – cos x.sin x
sin x – sin2 x = _________________
LHS = ______
2
2
cos x
cos x
2 x (1 – cos2 x)
2
si
n
sin x.sin2 x
= __________
= _______________
cos2 x
cos2 x
= sin2 x.tan2 x = RHS
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 357
357
2012/07/02 2:27 PM
12
cos x + tan x = _____
1
________
3
cos x
1 + sin x
cos2 x + sin x.(1 + sin x)
cos x + _____
sin x = _____________________
LHS = ________
cos
x
1 + sin x
(1 + sin x).cos x
2 x + sin x + sin2 x
(1 + sin x)
co
s
__________________
= ______________
=
(1 + sin x).cos x
(1 + sin x).cos x
1
= _____
cos x = RHS
13
sin2 θ
sin x (sin x + cos x) – 1 + cos x (sin x + cos x)
sin x + cos x
2
sin2 θ
2
sin
θsi
n
_________θ
cos θ
tan θ.sin2 θ
_____________
________
= cos2 θ = _________
=
sin2 θ
sin2 θ
sin θ
sin x + cos x
2sin x.cos x + (sin2 x + cos2 x) – 1
sin x + cos x
= ____________________________
2sin x.cos x = ________________
2sin x.cos x + 1 – 1 = RHS
= ___________
sin x + cos x
4
2
2sin x + sin x – 1 (factorise)
= ________________
(2sin x – 1)(sin x + 1)
cos x (2sin x – 1)
sin x + 1
= ___________________ = ________
cos x
tan x
2
cos x
5
sin x
x = 1 = RHS
1 + sin x
1 – sin x
(1 – cos2 x) – sin2 x
sin x (1 + cos x)
2
sin x (1 + cos x)
6
cos x = ________
1 + sin x
________
cos x
1 + sin x
tan x
cos2
(1 – sin2 x) – 2 (sin x + 1)
(sin x + 1) cos x
2
1 – sin x – 2sin x – 2
= __________________
2
(sin x + 1) cos x
sin x
– (sin2 x + 2sin x + 1)
(sin x + 1)(1 – sin x)
= ___________________
2
x – cos x – (1 –
x)
LHS = _______________________
2sin x.cos x + sin x
–(sin x + 1)(sin x + 1)
(sin x + 1)(1 – sin x)(1 + sin x)
= _________________________
2cos2 x – cos x – 1 (Factorise)
= ________________
2sin x.cos x + sin x
(cos x – 1)
cos x – _____
1 = __________
RHS = _____
= LHS
sin x
sin x
sin x
sin x – 1
= ______________________
2
cos2
(2cos x + 1)(cos x – 1)
(cos x – 1)
= ____________________ = __________
sin x
sin x (2cos x + 1)
cos2 x
x – 2 (sin x + 1)
(sin x + 1) cos x
cos2 x – cos x – sin2 x = _____
1 – _____
1
__________________
2sin x.cos x + sin x
sin x + 1
RHS = __________________
2
cos x (1 + sin x)
1 – sin x
cos x (1 + sin x)
1 + sin x = RHS
= ________
= ______________
cos x
cos2 x
2
sin x + 1
1
2 = ________
________
– ______
cos2
cos x × ________
1 + sin x = ______________
LHS = ________
2
1 – sin x
2
sin x – sin x = 0 = RHS
= ______________
cos x
TOPIC 7: EXERCISE 6
1 – sin x
1 + cos x
= _________________
1 + sin x + 1 – sin x = ________
2
2
LHS = __________________
= ______
2
2
1
sin x
(1 – cos x)(1 + cos x) – sin2 x
sin x (1 + cos x)
cos2 x
(1 – sin x)(1 + sin x)
1 – cos x – ________
sin x = 0
________
LHS = _________________________
1
1
2
________
+ ________
= ______
1 – sin x
cos x
2sin x.cos x – cos x
2sin x.cos x – cos x
cos x
sin x × cos2 x + sin2 x × ______
LHS = ______
2
2
= sin x +
sin x + sin x – cos x = ________
sin x + 1
__________________
sin x + sin x – (1 – sin2 x)
2sin x.cos x – cos x
2
cos2
sin x + cos x
2
LHS = _______________________
sin x = 1
tan2 x.cos2 x + ______
2
2
2
2
= tan θ = RHS
2
2
sin x + sin x.cos x – 1 + cos x.sin x + cos x
= ____________________________________
sin θ – sin
θ
________________
sin2 θ
sin
θ
(1
– cos2 θ)
______________
sin x + cos x
LHS = ______________________________________
θ.cos2
sin θ – sin θ.cos θ
_____
15
sin x + cos x
tan θ – sin θ.cos θ = tan θ
_______________
cos θ
cos θ
LHS = _______________
= ______________
14
2sin x.cos x
1
sin x – ___________
+ cos x = ___________
– (sin x + 1)
(1 – sin x)
sin x + 1 = RHS
= ___________ = ________
7
sin x – 1
tan θ
1 = _____
tan θ + _____
2
tan θ
sin θ
cos θ = ____________
sin θ + cos θ
sin θ + _____
LHS = _____
2
cos θ
sin θ
2
cos θ.sin θ
1
= _________
cos θ.sin θ
sin θ × _____
1 = _________
1
RHS = _____
= LHS
2
cos θ
358
sin θ
cos θ.sin θ
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 358
2012/07/02 2:27 PM
8
sin x
_________
=
(1 + cos x)
RHS =
=
_______
1 – cos x
√________
1 + cos x
TOPIC 7: EXERCISE 11
_______________
1
1 – cos x × ________
1 + cos x
√________
1 + cos x
1 + cos x
_________
_________
sin x
√ (1 + cos x) √___________
(1 + cos x)
1 – cos2 x =
__________
2
2
2
sin x
= __________
= LHS
(1 + cos x)
9
( sin θ )
cos2 θ + cos2 θ = 2
sin2 θ + tan2 θ ______
2
( cos θ )( sin θ )
sin2 θ ______
cos2 θ
LHS = (sin2 θ + cos2 θ) + ______
2
2
= 1 + 1 = 2 = RHS
10
(2sin x – cos x)(2sin x + cos x) = 5sin2 x – 1
LHS = 4sin2 x – cos2 x = 4sin2 x – (1 – sin2 x)
= 4sin2 x – 1 + sin2 x = 5sin2 x – 1 = RHS
TOPIC 7: EXERCISE 7
1
4
–1
1
2
5
2
3
1
cos x
TOPIC 7: EXERCISE 8
1
–2
2
1
4
1
________
5
tan x
1 + sin x
3
1
______
2
cos x
TOPIC 7: EXERCISE 9
1
1
4
6
7
5
cos β
1 + 2sin x.cos x
2
8
2
tan θ
– _____
3
2
1
__
2
1
–sin 68°
tan 87°
tan 45°
–cos 85°
–tan 57°
cos 20°
tan 74°
sin 5°
1.2
1.4
1.6
1.8
1.10
1.12
1.14
–cos 55°
–sin 74°
sin 24°
–tan 9°
–cos 71°
cos 68°
sin 33°
TOPIC 7: EXERCISE 12
2
–1
1.1
1.3
1.5
1.7
1.9
1.11
1.13
1.15
3
1.1
–k
1.2
k
1.3
k
_______
_____
1.4
–√1 – k2
1.5
–k
1.6
k
_______
_____
2.2
–
√ 1 – k2
p
–__
2
2.1
_____
√ 1 – k2
p
_______
_____
√4 + p2
2.3
4
_____
– _______
2
2.4
2
tan 65º = __
p
2.5
–
2.6
2
_____
– _______
2
3.1
–a
3.2
√ 1 – a2
________
3.3
–a
3.4
–a
√4 + p
p
_______
_____
√4 + p2
√4 + p
______
1
TOPIC 7: EXERCISE 13
__
__
1
– √2
2
–3
3
4√ 3
4
3
___
5
3
__
6
5
– __
16
2
4
1
TOPIC 7: EXERCISE 14
1
TOPIC 7: EXERCISE 10
1
1.1
cos 70°
1.2
–sin 70°
1.1
1.3
sin 58°
–tan 47°
1.2
1.4
–cos 24°
sin 14°
1.3
cos 70°
1.4
sin 15°
1.5
–tan 35°
1.6
–cos 81°
1.5
–sin15°
1.6
sin 15°
1.7
–sin 65°
1.8
–cos 45°
1.9
tan 11°
1.10 tan 29°
1.7
sin 70°
1.8
cos 70°
2.1
–1
2.2
1
2.3
–4
2.4
–2
2.5
1
– __
2.6
1
2.7
2
2.8
sin2 35°
1.11 –cos 50°
1.12 –sin 36°
1.13 cos 10°
1.14 cos 60°
1.15 –tan 72°
1.16 –sin 58°
1.17 –sin 65°
1.18 –tan 38°
2
3
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 359
359
2012/07/02 2:27 PM
TOPIC 7: EXERCISE 15
TOPIC 7: EXERCISE 18
1
137,16°
2
92,54°
3
297°
4
346,29°
5
254,17°
6
215,61°
7
318,19°
8
240,95°
9
x = 30°; 150°; 210°; 330°
10
x = 60°, 120°, 240°, 300°
11
x = 45°,135°,225°,315°
12
Special angle = 60°; Q3: x = 240°; Q4: x = 300°
TOPIC 7: EXERCISE 16
1
2
3
4
5
6
7
8
9
10
11
12
x = 232°; x = –128° x = 352°; x = –8°
x = 15°; x = –345°; x = 315°; x = – 45°
x = 95°; x = –265°; x = 275° x = –85°
x = 41,57°; x = –318,43°; x = 348,43°; x = –11,57°
x = 91,90°; x = –268,10°; x = 268,10°; x = –91,90°
x = 40,32°; x = –319,68°; x = 220,32°; x = –139,68°
x = 72°; x = 342°; x = 162°; x = –198°
x = 176,71°; x = –183,29° x = 263,29°; x = –90,71°
x = 51,42°; x = –308,58°; x = 278,58°; x = –81,42°
x = 123,20°; x = –236,80°; x = 303,2°; x = –76,80°
x = –11,31°; x = –371,31°; x = 127,31°; x = –232,69°
x = 30,54°; x = –329,46°; x = 210,54°; x = –149,46°
TOPIC 7: EXERCISE 17
1
2
3
4
5
6
7
8
360
x–16º = 38,02° + n.360°; x = 54,02° + n.360° n ∊ ℤ;
x = 157,98° + n.360° n ∊ ℤ
x = 18,95° + n.180°; x = –18,95° + n.180°
x = 32,36° + n.120°, n ∊ ℤ;
x = 87,64° + n.120°, n ∊ ℤ; x = 32,36° + n.120°, n ∊ ℤ
x = –32,36° + n.120°, n ∊ ℤ
x = 26,46° + n.180°, n ∊ ℤ
x = 250,28° + n.360°, n ∊ ℤ;
x = –109,73° + n.360°, n ∊ ℤ
x = 90,54° + n.180°, n ∊ ℤ
x = 135,46° + n.180°, n ∊ ℤ
x = 90,54° + n.180°, n ∊ ℤ
x = 134,03° + n.360°; x – 15° = 240,97° + n.360°
x = 255,97° + n.360°; x = 134,03° + n.360°;
x = –104,03° + n.360°
x = 44,69° + n.90°, n ∊ ℤ
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x = 33,69° + n.180°
x = 0° + n.360°; x = 180° + n.360°;
x = 0°; 180°; 360°; 41,81°; –318,19°;
138,19°; –221,81°
x ≤ 360° x = 33,69°; 213,69°; –146,31°; – 45°; 135°
x = –23,58° + n.360° or x = 90° + n.360°;
x = 203,58° + n.360°, n ∊ ℤ
x = 48,59°; 131,41°; 270°; –90°
x = 19,47° + n.360° or n = 270° + n.360°
x = 169,53° + n.360°, n ∊ ℤ
x = 0°; 180°; –180°; 59,03°; –120,97°
x = –113,58° + n.360° or
x = –60° + n.360°, n ∊ ℤ
x = 56,31° + n.180° or 26,57° + n.180°, n ∊ ℤ
x = 0°; 180°; –180°; –36,87°; –143,13°
x = – 41,81°; 318,19°; 221,81°; –138,19°; –30°; 330°;
210°; –150°
x = 90° + n.180° or x = –60° + n.360°
x = –30° + n.360° or x = 138,59° + n.360°, n ∊ ℤ
x = 210° + n.360° or x = –138,59° + n.360°, n ∊ ℤ
x = ± 113,58°; 0°
x = 146,31°; 326,31°
TOPIC 7: EXERCISE 19
1
2
3
4
5
6
7
8
9
10
x = 58°; 302°
x = 64°; 154°; 244°; 334°; –26°; –116°; –206°;
–296°; –386°
x = 80°; –280°; 180°; –180°
x = 25° + n.180°, n ∊ ℤ
x = 21°; 141°; –13°
x = 40° + n.120°, n ∊ ℤ
x = 60° + n.360°
x = 138°; 222°; –138°
x = 44,67°; 166,67°; 286,67°; –73,33°;
–120°; 240°
x = 40° + n.360°, n ∊ ℤ
TOPIC 7: EXERCISE 20
1
2
3
4
5
6
7
8
9
x = 109,47° or 250,53°
x = 12,94°; –347,06°; 167,06°; –192,94°
x = 73,43°; –106,57°
x ∊ [0°;360°] x = 54,96°; 97,07°
x = – 40° + n.180°, n ∊ ℤ
x ∊ [–360°;360°] x = 90°; 270°
x = –14,10° + n.90°, n ∊ ℤ
x = 46° or x = 338°
x = – 47,29° + n.180° n ∊ ℤ
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 360
2012/07/02 2:27 PM
10 x = ± 150° + n.360°, n ∊ ℤ
11 x = –19,47°; 340,53°; 199,47°; –160,53°
12 x = 30,96°; 210,96°; –149,04°; –63,43°; 116,57°
13 x = 0° + n.360°, n ∊ ℤ
14 x = 26,57°; –153,43°; 0°
TOPIC 7: EXERCISE 21
1
2
3
4
5
6
7
8
9
10
x = 0° + n.180° or x = 90° + n.180°
x = 90° + n.180°
x = 90° + n.180°
x = 90° + n.180° or x = 270° + n.360°
x = 90° + n.180°
x = 90° + n.180°
x = 0° + n.180° or x = ±120° + n.360°
or x = 90° + n.360°
6.3
1
OP =
1.2
1.2.1
1.2.2
(
1.2.3
7
8
2
3
2.1
2.2
√ 41
2.2.3
θ = 330°
7.1
7.3
7.5
1
8.1
8.1.1
–p
8.1.3
1
_______
_____
8.2.1
a
– __
8.2.3
–√ 4 – a2
________
2
6
3.1
7
1 – 2cos2 θ = ___
3.2
5
cos θ.tan θ = __
6
34
5
tan A.sin A = ___
33
__________
6
6.1
sin2
6.2
= sin2 x + cos2 x = 1 = RHS
1
cos θ + cos θ.tan2 θ = _____
4
7.2
7.4
1
1
______
sin2 x
8.1.2
1
_______
_____
8.2.2
a
__
√1 + p2
2
_____
9.1
3
– __
9.2
3
9.3
7
__
9.4
3
– __
9.5
1
– __
2
4
√1 + p2
2
2
2
11.1 x = 0° + n.180° or x = 90° + n.180°, n ∊ ℤ
11.2 x = 90° + n.180°, n ∊ ℤ
11.3 x = 0° + n.180° or x = 90° + n.180°, n ∊ ℤ
18
4
–2
sin θ
11
3
5 and is in Quad 2
sin θ = __
( cos θ.sin θ )
)
)
10.1 θ = 39,20° or θ = 120,80°
10.2 x = 75°; – 45°; –165°; –15°; 105°
10.3 θ = ± 120° + n.360°
or θ = ± 180° + n.360°; n ∊ ℤ
10.4 θ = 36°; 126°; 216°; 306°; –54°;
–144°; –234°; –324°
10.5 x = 23,33° + n.120°; n ∊ ℤ
x = –70° + n.360°; n ∊ ℤ
4
4
__
sin θ
10
OP = 2
__
√3
2.2.1 – ___
2.2.2
cos θ.sin θ
tan x
– _____
8.2
4___
sin θ = ____
(5)
( cos θ
sin θ
1 = RHS
1
= _____
= cos θ _________
9
= 26
4
– __
)
sin2 θ + cos2 θ
= cos θ ____________
TOPIC 7: REVISION
___
√ 41
tan θ
cos θ
sin θ + _____
LHS = cos θ _____
x = 135° + n.180°
x = 90° + n.180
x = ± 180 + n.360°
(
1
1
cos θ tan θ + _____
= _____
TERM 2: EXAMINATION PRACTICE
cos B
16
sin x.cos x = 1
x + _________
PAPER 1
1
1.1
tan x
sin x.cos x × _____
cos x
LHS = sin2 x + _________
sin x
1
cos θ
2
si
n
sin2 θ
______
LHS = cos θ + cos θ. 2 θ = cos θ + _____
cos θ
cos θ
2 θ + sin2 θ
1
cos
_____
____________
=
=
= RHS
cos θ
cos θ
2
4
x = 0 or x = __
3
1.2
1.3
1.4
x = 0,45 or x = –4,45
x=8
1
0 ≤ x ≤ __
1.5
3 or x = 4
x = – __
1.6
81
___
2.1
2.1.1
x = 7 or __
x=3
__
y
y
2.1.2
3 or y = __
1
y = ___
3
16
2
10
2
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 361
361
2012/07/02 2:27 PM
2.2
3
4
3
x=7
x≤7
2.3
1
__
3.1
b6
3.2
3.3
√3
3.4
4.1
4.1.1
4.1.2
4.2.1
4.2
5
2.2.1
2.2.1
2.2.2
__
4.2.4
4
4
3
3.1
3.2
4.
–3
1
4.1
cos A + cos (180° – A).sin2A = cos3 A
LHS = cos A – cos A sin2A
= cos A(1 – sin2A)
= cos A.cos2A
= cos3A = RHS
4.2
4.2.1
x
(3x – 2)
_______
5
3n + __
2n2 + ___
1
___
D = (1;–4)
5.4
m =1; c = −3; y = x − 3
5.5
0≤x≤3
5.6
5.6.1
y = x2 – 6x + 4
5.6.2
y = –x2 + 2x + 3
5.2
2
(sin x – cos x)
1 – 2tanx
_____________
= _____
2
cos x
1 – 2sin x.cos x
= _____________
2
cos x
sin2 x – 2sin x cos x + cos2 x
LHS = ________________________
cos2x
AB = 4; OC = 3
1 – 2sin x cos x
= _____________
2
cos x
LHS = RHS
4.2.2
6.1
Teacher to provide
6.2
–1 < x < 3
x = 90° or 270°
5
5.1
5.2
5.3
233,56°
104,48°
x = 75,52° + n.360° or
x = 60° + n.360° = –75,52° + n.360° or
x = –60° + n.360°; n ∈ ℤ
6
6.1
6.2
6.3
6.4
Teacher to provide
x = 30°; 120°; –60°; –150°
x = 30° + n.180°
–60° ≤ x ≤ 120°
1
__
3
1.2
1x + 2
y = __
1.3
1 × –3 = –1; BC ⊥ AB; B
^ = 90°
mBC × mAB__
TOPIC 8: EXERCISE 1
1.4
D = (7;1)
1
3
3
( __72;__32 )
2
y = –x + 5
1.7
63,43°
1.8
10 units2
1.9
a = –9
1.10 b = 7
2.1
2.2
5
cos θ = ___
2.1.2
–8
___
2.2.1
1
sin 70° = __
p
2.2.2
–1
___
2.2.3
√p – 1
_______
3
3
Total volume = 71 994,83 cm3
4
TSA = 9 032,08 cm2
Volume pyramid = 5,376 m2
TSA = 4 704 m2
5
Volume truncated cone = 11 623,89 units3
TSA = 3 392,92 units2
6
Volume = 50 186,94 units3
TSA = 7 869,69 units2
Volume = 4 416 m3
7
8
_____
2
Volume = 18 849,56 cm3
TSA = 4 084,07cm2
Volume cylinder = 942,48 units3
TSA = 534,07 units2
3
__
√
2.1.1
p
cos2x
1 – sin x
2sin x
1 – ______
RHS = _____
2
cos x
4
5.3
1.6
362
y
__
4.2.2
4.2.3
(x – 1)2 – 4
1.5
2
2x
day 5 = 61, day 6 = 86
Tn = 2n2 + 3n – 4 or Tn = 2n2 + 3n – 4
A : 2n + 1
B: n2
C: n(n + 1) = n2 + n
n = 10
n=7
5.1
PAPER 2
1
1.1
5
___
TSA = 1 776 m2
Volume = 93 600 units3
TSA = 14 892 units2
p
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 362
2012/07/02 2:27 PM
9
Volume = 471,40 units3
TSA = 346,41 units2
10
Volume = 124 859,46 units3
TSA = 13 420,88 units2
Volume of sphere = 1 767,15 cm3
11
2.2
2.3
2.4
0,9 cm
5,969 m3
119,18 m2
3
3.1
3.2
3.3
12
5 277,88 m2
1 847,26 units2
4
4.1
4.2
4.3
4.4
26 880 units2
DE = 29; DG = 52
4 968 units2
9 000 units2
5
Volumesphere = 57 905,84 units3
TSA = 7 238,23 units2
6
Volumehemisphere = 16 755,16 cm3
TSA = 3 769,91 cm2
7
7.1
7.2
7.3
4 πr3
__
8
8.1
8.2
8.3
14,5π
5 466,37 units3
11 263,85 m2
9
9.1
9.2
9.3
9.4
9.5
1 πr2h
V = __
3
8 cm
17
136π
628,32 cm2
10
1 are of base × ⊥
10.1 Volume of pyramid = __
3
height
10.2 18 × 18
10.3 41
10.4 1 476 units2
10.5 1 800 units2
11
11.1 Volume of cylinder = πr2h
11.2 10 units
11.3 2 513,27 units2
TSA = 706,86 units2
TOPIC 8: EXERCISE 2
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
2
2.1
2.2
2.3
2.4
1 πr2h
Volume of cone = __
3
40 units
s = 41
Surface area of cone = 1 413,72 units2
1.5.1 It increases by a factor of 4.
1.5.2
V = 13 571,68 units3
1.6.1 It will increase by a factor of 3.
1.6.2
1.7.1
1.7.2
3
V = 10 178,76 units
V = 424,115 ≈ 424,12 units3
TSA = 335,43 units2
Volume of cylinder = πr2h
20 units
TSA = 6 214,07 units2
1.
2.4.1 The volume will be multiplied by __
4
units3
2.6
2.4.2
2.5.1
2.5.2
2.6.1
2.6.2
2.6.3
2.6.4
3
3.1
3.2
3.3
Surface area of cylinder = 2πr2 + 2πrh
9 units
Volume = 8 171,28 units3
4
4.1
4.2
4.3
4.4
4 πr3
Volume of sphere = __
3
43 units
TSA = 23 235,22 units2
4.4.1 8
4.4.2
V = 2 664 305,12 units3
4.4.3
4
4.4.4
92 940,88 units2
2.5
V = 8 309,51
1.
The volume will be multiplied by __
4
V = 8 309,51 units3
1
The volume will be multiplied by ___
27
3
V = 1 231,04 units
1.
It will be multiplied by __
9
2
TSA ≈ 698,45 units
TOPIC 9: EXERCISE 1
1
1.1
1.2
1.3
12 units
OE = 18 − x units
13 units
2
2.1
2.2
2.1.1
OS = 25
2.1.2
OV = 15 units
TU = 120 units
3.1
3.2
3.3
AO = 25 units
OF = 20 units
FG = 5 units
TOPIC 8: REVISION
1
1.1
1.2
2 435,70 m2
1 319,86 units2
2
2.1
10 cm
3
37 units
17 203,36 units2
3
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 363
363
2012/07/02 2:27 PM
4
4.1
4.2
perpendicular bisectors of PQ and RT
4.2.1 39 units
4.2.2
25 units
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
EO ⊥ AB; OF ⊥ CD; AB∥CD
OF = 15______
units
√
AO = x2 + 49
___________
CO = √( x − 4 )2 + 225 5.5
radii
x = 24
AB = 48 units and CD = 40 units
968 units2
6.1
6.1.1
CG = GD = 30 units
6.1.2
OG = 16 units
6.1.3
AB = 126 units
6.1.4
AC = 33 units
^ =θ
6.2.1 E
2
6.2.2 corresponding angles equal
____
FB = 7√130 units
6.4.1 △AGF ≡ △BGF
6.4.2
kite
6.4.3
Two pairs adjacent sides equal
OF is the perpendicular bisector
of AB
6.4.4
4 095 units2
6
6.2
6.3
6.4
1.9
1.10
1.11
1.12
Supplementary
Concylic quadrilateral
Four points are cyclic
The four points do not form a cyclic
quadrilateral
1.13 The quadrilateral is a cyclic quadrilateral
1.14 The quadrilateral is not a cyclic quadrilateral
1.15 Equal angles on the circumference
2
2.3
3
4
5
a = 64°; b = 296°; c = 148°
a = 90° ; b = 32°; c = 58°; d = 58°; e = 29°;
f = 29°; g = 29° ; h = 29°; i = 122°; j = 122°
a = 35°; b = 35°; c = 70°; d = 35°; e = 35°; f = 110°
g = 22°
a = 27°; b = 63°; c = 27°; d = 54°; e = 90°; f = 36°
g = 63°
t = 40°; u = 40°; v = 80°; w = 21°; x = 42°; y = 29°
z = 29°
TOPIC 9: EXERCISE 3
1.
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
364
Bisects the chord
Perpendicular to the chord
Centre of the circle
Point of intersection
Twice the size of the angle subtended by the
same chord on the circumference of
the circle
Are equal
90°
Diameter
No, co-interior angles not supplementary.
2.2.1 t = 17°
^ C = FE
^C
2.2.2
FD
^
^
2.2.3
BAE = BFE
^ F = AB
^F
2.2.4
AE
u = 77°; v = 103°; w = 77° ; x = 104°; y = 20°;
z = 20°
TOPIC 9: EXERCISE 4
1
w = 101°; x = 96°; y = 92°; z = 60°
2
2.1
2.2
2.3
TOPIC 9: EXERCISE 2
1
2
2.1
2.2
2.4
a = 101°; b = 79°; c = 82°; d = 98°; e = 98°;
f = 79°
^+X
^ = 161° and so the
No, because Y
co-interior angles are not supplementary.
^ +X
^ = 180° and so the
Yes, because V
co-interior angles are supplementary.
No, it is NOT a cyclic quadrilateral because
opposite angles are not supplementary.
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
ABCE; ABCD
^ = x; E
^ = x; ED = FD
B
1
1
^ = x; B
^ = x; BF bisects AB
^C
E
2
2
^
^
ABC = C
Opposite ∠s are not supplementary
Opposite ∠s not supplementary
AF ∥ BC
4
4.1
4.2
4.3
4.4
4.5
^ = 90° − 2x
D
2
Teacher to provide
Teacher to provide
Teacher to provide
Teacher to provide
TOPIC 9: EXERCISE 5
1
2
3
t = 59°; u = 59°; v = 63°; w = 63°; x = 58°; y = 59°;
z = 62°
a = 104°; b = 52°; c = 90°; d = 38°; e = 19°; f = 71°;
g = 123°
w = 70° ; x = 50°; y = 60°; z = 120°
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 364
2012/07/02 2:27 PM
TOPIC 9: EXERCISE 6
TOPIC 9: EXERCISE 7
1
1
2
1.1
1.2
1.3
1.4
Equal in length.
90°
Equal
Tangent to the circle.
2.1
t = 54°; v = 126°; w = 80°; x = 80°; y = 100°;
2.2
2.3
No, alternate angles unequal
Yes, corresponding angles equal
3.1
3.2
x = 30°; straight ∠
^ K = 90°; ∠sum △NMK
3.2.1 NM
3.2.2 Teacher to provide
1.1
1.2
z = 46°
3
4 and 5
Teacher to provide
6
No, alternate angles unequal
6.2.1
equal to the angle subtended by
the chord in the alternate segment.
6.2.2
supplementary.
6.2.3
equal to the opposite interior angle
p = 53°; q = 60°; v = 67°; w = 113°; x = 45°;
y = 22°
6.1
6.3
7
7.1
7.2
7.3
9
9.1
9.4
2.1
2.2
^ + 18° = 43°; tangent CQ,
A
3
^ = 25°
chord CD; A
3
^ = 18°; isosceles △OAC, radii
7.1.2
C
3
^ C = 144° ; ∠ sum △AOC
7.1.3
AO
^
7.1.4
D = 72°; ∠ at centre
^ A = 72°; tangent PC, chord CA
7.1.5
PC
^ C = 108°; (opposite ∠s cyclic
7.1.6
AB
quadrilateral)
Teacher to provide
Teacher to provide
Teacher to provide
9.3
2
7.1.1
8
9.2
1.3
^ = x (tan PQ, chord VQ)
Q
4
^ = x (corresponding ∠s, OR ∥ VQ)
R
Q2 = x ( isosceles △OSQ, radii)
^=^
R
S = x (from 10.1)
OQ subtends equal angles at S and R; QRSO
is a cyclic quadrilateral.
^ V = 90° (∠ on diameter)
SQ
Q4 = x (proved in 10.1)
^ P = 90° + x
SQ
^ = 90° (corresponding ∠s, RO ∥ QV)
T
2
ST = TQ (OT ⊥ SQ)
QV = VS (OV ⊥ QS)
OQRS is a kite (OR bisects QS
perpendicularly)
QR = SR (adjacent sides kite)
a = 10° (equal chords QR and SR)
b = 10° (∠s on chord QR)
c = 10° (∠s on chord RS)
d = 20° (∠ at centre)
e = 20° (∠ at centre)
f = 70° (exterior ∠ of △OQV)
g = 70° (isosceles △OQS, radii)
OQ is 145 units.
^ T = 90° (∠ on diameter)
2.1.1
QP
^ T = 90° (∠on diameter)
2.1.2
QR
^
2.1.3
QTU = 90° (radius ⊥ tangent)
a = 31° (tangent UT, chord PT)
b = 31° (tangent UT, chord PT or ∠s on
chord PT)
c = 62° (∠ at centre)
d = 20° (∠s on chord RT)
e = 20° (tangent ST, chord RT)
f = 70° (∠ △sum QTR)
g = 39° (∠ sum isosceles △OPT, radii)
h = 39° (radii)
i = 31° (radii)
j = 20° (radii)
k = 59° (adjacent complementary)
l = 140° (∠ sum △OQR)
m = 118° ( ∠ sum △OQP)
n = 40° (straight ∠)
w = 70° (∠ sum △QTS)
x = 101° (∠ sum △PTV)
y = 101° (vertically opposite)
z = 79° (straight ∠)
3
3.1
3.2
PR bisects SQ perpendicularly
3.2.1 P^SR = 90° (∠ on diameter)
^ = 61° (∠ sum △PSR)
R
1
^ = 61° PR symmetry line of kite)
3.2.2
R
2
^ +Q
^ = 61° (∠ sum isosceles
Q
2
3
△OQR, radii)
^ = 29° (∠s on chord SR)
Q
3
^ = 32°
Q
2
^ = 58° (∠ sum △QOR)
3.2.3
O
1
4
4.1
4.2
Teacher to provide
Teacher to provide
4.3
BC = 30 units
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 365
365
2012/07/02 2:27 PM
5
5.1
5.2
5.3
They are 24 units apart.
5.2.1 FH = 5 units
5.2.2
DH = 35 units
5.2.3
DF = 30 units
OM = 13 units (radius of small circle)
OE = 37 units
(radius of larger circle)
TOPIC 10: EXERCISE 1
1
2
__
5.4
KE = 40√ 3 (OM ⊥ KE)
5.5
OHEM is a cyclic quadrilateral
3
TOPIC 9: REVISION
1
1.1
1.2
1.3
2
2.1
2.3
r = 123
OC perpendicularly bisects AB;
OACB is a kite
^ B = 4x
AO
^A = 90°
OF
_______
2 + 400
2.2.1
AO = √ x__________
2.2.2
OC = √( x + 9 )2 + 49 (4)
____
OA = √625 = 25 units
3
x = 36°
4
4.1
4.2
4.3
4.4
4.5
4.6
5
5.1
5.2
5.3
6
6.1
6.2
6.3
6.4
6.5
6.6
6.7
7
366
^ = 24° (alternate ∠s, PO ∥ SR)
R
1
^
R2 = 66° (on diameter)
^ = 132° (at centre)
O
2
^ = 24° (sum isosceles △POQ, PO = OQ)
P
2
^S = 48° (∠s on chord RQ)
^ = 42° (∠ sum △RSQ)
Q
2
^ = x (exterior ∠cyclic quadrilateral)
F
1
D1 = x (exterior ∠cyclic quadrilateral)
BCDG is a cyclic quadrilateral
(exterior ∠ = interior opposite ∠)
5.3.1 Teacher to provide
5.3.2
Teacher to provide
5.3.3
ABDE is a cyclic quadrilateral.
(Exterior ∠ = interior opposite ∠)
DC subtends equal angles at A and B
^ = x (equal chords DC and BC)
A
2
^ D (A
^ =A
^ )
AC bisects BA
1
2
^
^
DG = FG (A1 = A2)
D2 = 90°
(∠ on diameter)
^ = 90°(∠s on chord AB)
C
2
Teacher to provide
exterior ∠ = interior opposite ∠
x = 30°
Teacher to provide
CB = 9,21 cm
AC = 25,71 cm
A^ = 21°
4
^ = 67,38°
C
^ = 22,62°
A
CB = 5 cm
^ = 57°
C
AC = 21,43 cm
CB = 39,35 cm
^ = 50,19°
C
^ = 39,81°
A
AC = 78,11 cm
TOPIC 10: EXERCISE 2
1
2
3
4
5
6
23,40 cm
37,97 cm
x = 55,36 cm or 124,64°
85,69 cm
x = 55,15°
x = 64,59°
TOPIC 10: EXERCISE 3
1
2
3
4
5
6
x = 37,96 cm
x = 22,70 cm
x = 75,15° or x = 104,85° (ambiguous case)
x = 40,75°
x = 104,47°
x = 89,18 cm
TOPIC 10: EXERCISE 4
1
2
3
4
517,25 cm2
157,19 cm2
200,86 cm2
(if AB rounded off)
200,83 cm2
(if AB answer from calculator)
965,16 cm2 (if AB rounded off)
965,13 cm2 (if AB answer from calculator)
5
6
Area = 221,19 cm2
Area = 658,18 cm2
TOPIC 10: EXERCISE 5
1
1.1
1.2
1.3
1.4
1.5
1.6
Perimeter ABCD = 31,84 cm
Area ABCD = 53,49 cm2
DC = 26,11 cm
Area ABCD = 289,73 cm2
^ E = 5,91°
AC
BD = 16 cm
986,21 mm2
x = 13,10 cm
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 366
2012/07/02 2:27 PM
2
2.1
______________
BC = √ x2 + y2 − 2xycos θ
xsin ( θ + β )
sin θ
2.2
BC = __________
2.3
BC = 2xcos θ
2.4
BC = _______________
2.5
2.6
asin ( α + β )sin α
sin β
asin
α
_________
BC =
sin βsin θ
xsin β
BC = _________
cos θsin α
TOPIC 10: EXERCISE 6
1
2
3
4
5
BC = 7,64 m
AD = 5,82 m
AB = 15,87m
Teacher to provide
TOPIC 10: EXERCISE 8
155,58 km
2,74 km
57,51 m
TOPIC 10: EXERCISE 9
1
1.1
1.2
1.3
x = 14,13 units
20,04 units
^ = 79,33°
D
2.1
2.2
2.3
2.4
TB = 10,5 units
^ B = 48,59°
AT
21,07 units
47,60 units2
3
3.1
3.2
^ = 31,9°
B
r = 8,51 units
4
4.1
4.2
Teacher to provide
Teacher to provide
5
5.1
5.2
5.3
1 r2tan θ
Area △OAB = __
2
1 r2sin θ
Area △OAC = __
2
tan θ > sin θ
6.1
6.2
6.3
2a2 (1 − cos x)
2b2 (1 + cos x)
Teacher to provide
2
6
TOPIC 10: REVISION
1
1.1
1.2
1.3
1.4
1.5
32,16 cm
x = 77,20° or x = 102,8°
x = 25,19
x = 77,73°
Area △ABC = 237,85 cm2
PLT MATHS LB 11 7th pgs (Real Book).indb 367
Area △ABC = 1 294,40cm2
(or 1 294,26 if AC is rounded)
2
2.1
2.2
2.3
^ S = 72°
QP
QS2 = 21,16 cm
282,46 cm2
3
3.1
3.2
^ = 100,29°
B
^ D = 23,18°
AC
4
PQ = 6,5m
5
5.1
5.2
5.3
6
Teacher to provide
7
Teacher to provide
8
Teacher to provide
9
9.1
9.2
10
Teacher to provide
AD = 440,63 m
AD = 21,73 m
TOPIC 10: EXERCISE 7
1
2
3
1.6
AB = 4,12 m
Area △AEB = 5,37 m2
BE = 2,97 m
SQ = 270,96 km
The bearing of Q from S is 10,35°.
TOPIC 11: EXERCISE 1
1
1.1
1.2
R10 660
R10 797,59
2
2.1
2.2
R561 150,73
R357 856,19
3
3.1
3.2
8,33%
16,40%
4
3 years
5
R42 574,26
6
R483 283,00
7
For Tim: R11 294,30
For Thabo: R13 980
For Tracy: R11 019,96
For Thandi: R10 683,33
Thabo got the best return and Thandi got
the worst.
8
8.1
8.2
2,27%
6,28%
335 rhinos left in 2050.
8.1.1
8.1.2
TOPIC 11: EXERCISE 2
1
1.1
1.2
10,9%
6,7%
2
2.1
2.2
7,85%
7,77%
3.1
R46 018,94.
3
Answers
367
2012/07/02 2:27 PM
3.2
3.3
12
R21 018,94
10,7%
13
4
6,3% p.a. compounded monthly
5
R29 587,21
6
R162 038,22
7
T0
T2
ieff = 0,1
T2
T3
T5
ieff = 0,115
R3 648,87
16
16.1 R10 356,95
16.2 R3 549,64
16.3 R6 807,31
17
17.1 R961,52
17.2 6,67% p.a.
30 000
R92 348,44
T0
9
T1
Lening
= 150 000
T3
ieff = 0,14
5 000
i(4) = 0,105
10 000
T5
TOPIC 12: EXERCISE 1
1
1.1
1.2
x
1.3
R253 122,40
2
TOPIC 11: REVISION
10% p.a.
2
4,26 years
3
8,6%
4
4.1
4.2
5
R 6 915,88
6
R33 454,56
7
7.1
7.2
8
5,6 years
9
11,61% p.a.
10
10.1 R17 240,22 per ticket
10.2 R29 718,84
11
2.3
3
R486 230,87
10,47%
R8 823,53 = cash price
R2 858,82
T0
2.1
2.2
1
ieff = 0,05
T2
3.1
3.2
4
T5
i(2) = 0,06
T7
5
5 000
1
___
12
3
__
4
5
___
12
No, because the events are not exhaustive
(4 is not included in either event).
Yes, because the events are mutually
exclusive and exhaustive.
No, because the king of hearts belongs to
both events, which means that the events
are not mutually exclusive.
These are dependent events; without
replacement means that when selecting
the second card, there will be one less card
to choose from, and the probabilities will
depend on the type of card that was selected
in the first draw.
These are independent events; the outcome
of tossing the coin (heads or tails) will not
be affected by the number that was thrown
on the die.
4.1
64
____
4.2
40
____
4.3
105
____
5.1
P(blue and blue) = P(blue) × P(blue | blue
8 × ___
7 = ___
14
first) = ___
2 000
169
169
169
13
x
R9 545,90
–36 000
15
x
113 582,40
–32 500
14.1 R20 222,73
14.2 10,47% p.a.
14.3 R5 222,73
x
R16 856,70
5.2
12
39
P(blue and green) = P(blue) × P(green | blue
10
8 × ___
5 = ___
first) = ___
13
368
T10
–30 000
14
–10 000
i(12) = 0,1
T5 i(12) = 0,12 T7
R41 510,55
T10
T5 i(4) = 0,0951T7
6000
T0
T0
x
P1
P2
P3
8 000
8
R12 389
12
39
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 368
2012/07/02 2:27 PM
TOPIC 12: EXERCISE 2
1
3.2
3.4
120
F
S
45
29
28
4
3.5
4.1
0,6 × 50 = 30
3.3
x = 0,1
3.4.1 P(A ∩ B) = 0
3.4.2
P(A ∪ B)′ = P(not A or B) = 0,1
∴ n(A ∪ B)′ = 0,1 × 50 = 5
3.4.3
P(A ∪ B′) = P(A or not B) = 0,6 + 0,1
= 0,7
Teacher to provide
S
M
70
x
1.1
1.2
45 + 28 + 29 + x = 120
∴ x = 18
3
45 = __
1.2.1 ____
1.2.2
1.2.3
1.3
1.3.1
12
L
20
4.2
P( S ∩ F’ ) = P(S and not F)
120
M
18
9
P(F ∪ S’) = P(F or not S)
120
1.4
Teacher to provide
2.1
A = {1; 2; 3; 5; 6; 10; 15; 30}
B = {1; 3; 5; 7; 9; 11; 13; 15; 17; 19; 21; 23;
25; 27; 29}
C = {1; 4; 9; 16; 25}
A ∩ B = {1; 3; 5; 15}
B ∩ C = {1; 9; 25}
n(A ∩ B ∩ C) = 1
B
3
12
4.2.1
4.2.2
4.2.4
30
9
1
4.2.6
5 learners take Maths, Physical
Sciences and Life Sciences
23
57
___
4.2.3 ___
70
13
D
C
3
8
2.6.1
2.6.3
5 = __
5
1 – ___
15
30
6
50
13
1
2
9
8 = ___
4
___
25
2
C
30
43
12
___
4.2.5
18
8
___
2
2.6
6
14
2
0
8
L
5.1
4
12
0
120
A
70
5
45 + 28 + 18 = ____
91
= ___________
2.2
2.3
2.4
2.5
6
S
29
= ____
2.
8
5–x
P( F ∪ S )′ = P(not F or S)
120
1.3.3
17 – x
x
120 8
45 + 28 + 29 = ___
17
___________
120
20
28 = ___
7
____
120 30
3
18 = ___
= ____
1.3.2
18
9
2.6.2
6
12 = ___
___
2.6.4
3
9 = ___
___
30
15
30
10
2.7
Teacher to provide
3.1
None of the shoppers interviewed like to shop
at both ‘Perfect Purchases’ and ‘Shopaholics’.
7
B
3
5.2.1 n(D ∪ C ∪ B)´ = 3
5.2.2 n(D ∩ C´) = 14 + 2 = 16
3
5.2.3 P(B | C) = ___
50
24
48 = ___
5.2.4 P(D ∩ C ∩ B)´ = ___
50
25
Answers
PLTMATHSLB11LB_16.indd 369
369
2012/07/14 3:02 PM
6
6.1
6.2
As the total probability = 1, we can calculate:
1 − (0,14 + 0,06 + 0,1 + 0,02 + 0,04 + 0,44
+ 0,08) = 0,12
6.2.1 0,1 × 50 = 5
6.2.2
0,12 × 50 = 6
6.2.3
0,02 × 50 = 1
6.2.4
0,44 × 50 = 22
6.2.5
0,16 × 50 = 8
1.5
2
2.1
TOPIC 12: EXERCISE 3
1
Teacher to provide
1.1
6
1.2
2
1.3
2
2.1
2.3
6 = __
1
___
12
2
When drawing the first blue marble there
are 6 blue marbles out of a total of 20
6 . Because
marbles, so the probability is ___
20
the marble is not replaced, the probability
of drawing a blue marble with the second
drawing will depend on the colour of the
first marble drawn. If it was blue, then
there will be 5 out of 19 possibilities of
drawing another blue marble, but if the first
marble had not been blue, the probability
of drawing a blue marble with the second
6.
drawing would be ___
95
3
2.2.3 ___
19
5
__
3.1
8
3.3
Hockey
will
lose
Total
Rugby will win
21
9
30
Rugby will lose
14
6
20
Total
35
15
50
2.2.1
21
___
2.2.2
14 = ___
7
___
3
___
3.4
28
2.2.2
3
___
2.2.4
21
___
Therefore winning hockey and winning
rugby are independent events.
TOPIC 12: REVISION
1
1.1
1.2
1.3
2
2.1
a and d, or b and d
b and c
a and e
100
Coconut
Not defective
14
Total
1.2
1.4
370
18 = ____
9
____
500 250
144 = ___
24
____
150 25
144
350
1.3
150
6
12 = ____
____
350
175
59 – x
x
22
3
6
338
Rainmaker
67 – x
38
9
___
12
50
38
Supplier Supplier
A
B
Defective
25
rugby) = P(win hockey and win rugby)
No. Once the head prefect has been chosen
there will only be 7 learners left to choose
from for the deputy, and the number of
boarders and day scholars left to choose
from will depend on whether the head
prefect was a boarder or day scholar.
1.1
50
21 ; P(win
P(win hockey and win rugby) = ___
35 ; P(win rugby) = ___
30 50
hockey) = ___
2.2
x = 48
2.3
22 = ___
11
____
2.4
59 − x = ____
11
______
2.5
48
___
100
100
50
100
59
3.1
1
__
TOPIC 12: EXERCISE 4
1
50
50
18
2.2.1 ___
3.2
Hockey
will win
35 × ___
30 = ___
21 P(win hockey) × P(win
___
50
50 50
20
3
Supplier A has marginally less probability of
supplying defective pens.
1
__
2
Total
18
Red
482
Black
1
__
2
1
__
1
__
2
Red
2
Red
2
Black
Black
1
__
2
500
3.2
1
__
4
3.3
2 = __
1
__
4
2
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 370
2012/07/02 2:27 PM
4
3.4
2 = __
1
__
3.5
1
__
3.6
3
__
4.2
10 = __
1
___
4.3
15 = __
1
___
4.4
4
8
2
4
4
50
5
30
1
__
5
2
8.5
1
__
5
1
__
3
Win
3
3
Win
1
__
Lose
3
1
__
1
__
Lose
8.1
8.2
8.3
8.4
3
Win
1
__
Lose
3
TOPIC 13: EXERCISE 1
1
1.1
1.4
1.5
1 × __
1 = __
1
P(LL) = __
3
6
3
9
6.1
3
___
Mathabane
22 ___
9
22
10
___
4
___
23
23
McClure
Mathabane
22
7
7.1
7.2
__
__
Time taken
in minutes
(xi – x)
(xi – x)2
189,0625
15
–10,75
115,5625
McClure
19
–6,75
45,5625
24
–1,75
3,0625
25
–0,75
0,5625
33
7,25
52,5625
253
38
12,25
150,0625
175
____
40
14,25
203,0625
36
____
6.2.2
87
____
7.2.2
759,5
√_____
8
–13,75
6.2.1
x=4
7.2.1
Mathabane
_____
12
22
6.2.4
1.2
25,75
Matthee
9
___
22
9
___
6.2.3
1.1
McClure
4
___
McClure
6.2
1
22 Matthee
22
1.1.1
158,29 cm
1.1.2
150 ≤ x < 160
1.1.3
40 cm
Teacher to provide
1.5.1 157 cm
1.5.2
151 cm
1.5.3
166 cm
1.5.4
15 cm
1.5.5
174 cm
Teacher to provide
TOPIC 13: EXERCISE 2
4
___
10
___
23
Matthee
22
Matthee
10
___
1.6
Mathabane
8
___
22
9
___
Teacher to provide
0,4
0,5
P(P) = 0,6 P(P) = 0,3 and P(B and T) = 0,1
0,6 × 0,3 = 0,18 so P(T) × P(B) ≠ P(T and B)
so they are not independent events.
P(B or T) = P(B) + P(T) – P(B and T)
0,3 + 0,6 – 0,1 = 0,8
253
253
91
____
253
759,5
16
13
___
15
7.2.3
17
___
30
7.2.4
5
___
7.2.5
12
___
7.2.6
1
___
2
1.3
Standard deviation = 9,74; Variance = 94,94
four data values
2.1
2.2
2.3
1,19
1,42
2,87
12
41
12
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 371
371
2012/07/02 2:27 PM
TOPIC 13: EXERCISE 3
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.6
Minimum value = 1 452
Maximum value = 140 252
Lower quartile = 1 984,5
Median = 11 694
Upper quartile = 17 566,5
Teacher to provide
very wide range (138 800), and is skewed to
the right.
very long whisker on the right
IQR = 17 566,5 – 1 984,5 = 15 582
17 566,5 + 1,5(15 582) = 40 939,5
So Ekurhuleni (140 252) is an outlier.
many job opportunities
So the City of Johannesburg, the City
of Cape Town, the City of Tshwane and
Ekurhuleni are all outliers. This is a bigger
sample, including a wider variety of
municipalities, so it will reflect a better result.
1.8
1.7
2
2.2
2.1
1.1
1.2
1.3
1.4
1.5
372
__
__
(xi – x) (xi – x)2
__
xi × (xi – x)2
f
1
5
–2
4
20
2
10
–1
1
10
3
15
0
0
0
Frequency
4
10
1
1
10
0–30 000
14
5
5
2
4
20
30 000–60 000
0
Total
45
…
…
60
60 000–90 000
1
90 000–120 000
0
120 000–150 000
4
3
45,16
29,16
24 values
IQR = 39,15
No outliers.
median < mean, so the data is skewed to
the right
TOPIC 13: REVISION
1
45
In-migration
groups
Teacher to provide
1.10 all cities, many people migrate towards
them in search of employment.
2.1
2.2
2.3
2.4
2.5
2.6
1 × 5 + 2 × 10 + 3 × 15 + 4 × 10 + 5 × 5 = 3
___________________________________
xi
2.3
2.4
1,15
The results in the second test were very
widely spread, with a greater discrepancy
between results than in the first test.
3.1
Minimum value = 65;
Maximum value = 129
So LQ = 79,25; Median = 89,5; UQ = 108
3.2
Teacher to provide
A: True
B: False: Each quartile represents 25% of the
data. The values in the third quartile
are spread out more than in the second
quartile.
1.9
2
In suburb B, the data values are mostly
bunched on the left (mostly low values)
with a long whisker on the right, so the data
is skewed to the right, with a range of 370
and an IQR of 80. In suburb A, the range is
narrower (170), and so is the IQR (55).
The spread is closer to being symmetrical,
but the mean is slightly less than the
median, with the median closer to the lower
quartile than the upper quartile, so the data
is skewed to the right.
Suburb B is more likely to have an outlier
because of the long whisker to the right.
1.1.1
125
1.1.2
102
1.1.3
157
Teacher to provide
131
42,03
˙%
73,3
3.3
4
4.1
4.2
4.3
4.4
4.5
4.6
200
60
27
36 – 18 = 18
20 to 30
It is impossible to get less than –9 or greater
than 63 for a test out of 60, so there were no
outliers.)
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 372
2012/07/02 2:27 PM
4.7
5
5.1
Mid-class
value
Cumulative
Frequency
Frequency
5
20
20
15
60
40
25
120
60
35
170
50
45
190
20
55
200
10
4.7.1
27
4.7.2
12,88
Expenditure
Frequency
x < 50
Cumulative
frequency
2
2
50 ≤ x < 100
22
24
100 ≤ x < 150
58
82
150 ≤ x < 200
14
96
200 ≤ x < 250
4
100
6.1
6.2
1.1
(5 5)
1.3.1 x ∊ (−2;6)
1.3.2
x = −2 or x = 6
The roots are real, irrational and unequal.
1.5.1 3
1.5.2
2x
2.1
2.4
2.1.1
28; 33
2.1.2
Tn = 5n − 2
2.2.2 Tn = 2n2 − n − 3
2.2.3
n = 17 or −16,5
54 − x ; 66 − 2x
3x = 87 and x = 29
2.4.1 6; 18; 54; 162; 486; 1 458
2.4.2 Tn = 2 × 3n
2.4.3
3.1
3.2
3.3
3.4
3.5
3.6
Tn = 2 × 3n + 5
y ≤ 8, y ∊ ℝ or y ∊ (−∞;8)
g(−1) − h(−1) = 6
x = 1 and y = 2
x = 3 or –3
(−1;0)
h
2,5
y=2
x
–5
Median read off from the ogive is
approximately R122.
Teacher to provide
123
almost symmetrical
The rainfall decreases with increase in
temperature.
The reading 30 °C and 80 mm rainfall
1.1.1
x = 1,5 or x = 2
1.1.2
5
____________
2(x2 + 2x + 4)
–1
x
3
–2
x=1
g
TERM 4: EXAMINATION PAPER 1
1
8
(–1;8)
-
Cumulative frequency
6
7 = 0,875
x = __
f
100 150 200 250
median
Cell phone expenditure (R)
5.4
5.5
5.6
1.1.5
3
__
y
50
5.3
x = ± ( 22 )2 = ± 8
1.3
2.3
y
100 80 60 50,5 40 20 0-
1.1.4
3 ;__
1
(1;−1) and __
2.2
3
5.2
x = 2 only
1.2
1.4
1.5
2
1.1.3
4
3.7
4 −2
v( x ) = − _____
3.8
1 ( x − 4 )2 + 4
w( x ) = − __
4.1
1 x2 − __
1x − 2
y = __
4.2
1 x+1
g( x ) = 3 __
x−1
2
3
3
(5)
Answers
PLTMATHSLB11LB_16.indd 373
373
2012/07/14 3:04 PM
5.
6
[2 )
4.3
1 ;∞ or x ≥ __
1, x ∊ ℝ
x ∊ __
4.4
x ∊ [ −∞;−2 ] ∪ [ 3;∞ ) or x ≤ − 2
or x ≥ 3, x ∊ ℝ
5.1
5.2
5.3
5.1.1
i = 12,5%
5.1.2
i = 9,05%
A = R1 293 194,45
R12 094,38
6.1
6.2
6.3
6.4
x=2
3 people
17
5
6.4.1 P(T ∩ C)′ __
6.5
6.4.2
6.5.1
2
2
y
3 D
6
2.1
^ C = 90°
AB
2.2
1 x − __
1
y = __
2.3
D = (–8; 8)
2.4
AC = √ 290 ; BD = √290 = AC
4
2
____
____
0,15
arrives late
2.5
ABCD is a rectangle; parallelogram with
equal diagonals or parallelogram with
90° vertex
0,85
arrives on time
2.6
3
5 ;__
F = − __
0,1
arrives late
2.7
2.8
Calls Terrific Taxis
0,9
6.5.2
6
B(3;–5) 3
Calls Call a cab
0,7
A(6;1)
α
x
θ
E
6
3
P(B | T) = ___
13
0,3
F
C(–11;2)
arrives on time
3
51
P(Call a Cab and On Time) = ____
200
(
)
22
13
y = 2x + ___
2
3
y = __
2
2.9
^ D = 66,8°
AB
3.3
3.1.1
3
sin θ = − __
4
y
TERM 4: EXAMINATION PAPER 2
1
1.1
1.2
1.1.1
1.1.2
1.1.3
1.1.4
1.1.5
1.1.6
1.1.7
1.2.1
Median = 2 300 000
LQ = 1 500 000; UQ = 3 850 000 ;
IQR = 2 350 000
Teacher to provide
The data is skewed to the right.
Mean = 2 622 222,22
Standard deviation = 1 512 744,22
No outliers.
Teacher to provide
3__
tan θ = ___
√7
– 7
0
x
–3
y
3.1.2
Cumulative frequency
10
3.2
8
3.3
6
3.4
4
2
50
55
57,6
60
65
__
√3
___
4
1
2
3.3.1
3.3.2
3.4.1
3.4.2
1
_____
cos θ
RHS
x=0
x = n.180° or
x = 53,13° + n.360°, n ∊ ℕ
x
Percentage of voter turnout
1.2.2
374
There were four provinces below the
mean, so there were five provinces
above the mean.
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 374
2012/07/02 2:28 PM
4
4.1
Teacher to provide
4.2
4.2.1
4.2.2
4.2.3
−90° < x < 0° or 90° < x < 180°
−180° ≤ x < −135° or 0° ≤ x < 45°
or x = 180°
including –180°, 0° and 180°;
excluding –135° and 45°
–4y ≤ –2
5
5.1
5.2
area △ADC = 2(m)2 sin α) sin θ
AC = 29 cm
6
6.1
s = 2,55 cm
6.2
2 π( 0,5 )2 + π( 0,5 )2( 10 ) + __
1 ( π )( 0,5 )2( 2,5 )
V = __
6.3
SA = 31,42 cm2
3
3
23 π = 9,03 cm3
___
8
7
7.1
7.2
^P
^ B = RQ
RP
7.2.1 Angle in semi-circle
7.2.2
BCDH is a cyclic quadrilateral
C
B
F
1 3
2
2
D1
2 3
1 H
21
1
2
3
A
4
12
0
G
E
7.2.3
7.2.4
7.2.5
8
8.1
8.2
^ = 40°
A
^ = 50° − x
A
3
^ =B
A
1
2
^ = 2B
^
F
OC = x – 8 = 5 mm
Answers
PLT MATHS LB 11 7th pgs (Real Book).indb 375
375
2012/07/02 2:28 PM
Glossary
corollary a deduction based on the result of a theorem
A
amplitude half the total distance between the minimum
and maximum values; it is always positive
angle of inclination the angle that a line makes with the
positive direction of the x-axis
cumulative frequency a running total of the frequencies
cyclic quadrilateral a four-sided figure with all four points
lying on the circumference of a circle
appreciation growth in value; the opposite of depreciation
D
arc a part of the circumference of a circle
decrease to grow smaller
asymptote a line that cannot be crossed by a graph; it tends
towards, but never touches, a graph
degree (in algebra) the highest index in an equation or term
axis of symmetry a line that divides a graph into
two identical halves and about which a function is
symmetrical
diagonal a straight line joining opposite vertices
B
bar graph a graph with rectangles that show the frequency
with which different types of data values occur; data
values need not be in consecutive groups, so the bars do
not touch each other
bisect to cut or divide into two equal parts
book value the depreciated value of a vehicle at any
specified time
discrete data data with a restricted number of values;
includes units of quantity (for example, number of
people), with only whole number values
domain the set of all x-values of a graph or function;
excludes any values of x which make y undefined
E
effective interest rate the actual rate achieved per annum
equidistant the same distance
equilateral triangle a triangle with all three sides equal and
all three angles equal
C
Cartesian plane a two-dimensional plane formed by
a horizontal number line representing x-values and
a vertical number line representing y-values, which
intersect at the point where both values are 0; ordered
pairs or coordinates refer to the position of points in
the plane
chord a line with both endpoints on the circle
coefficient a number that stands in front of a variable or
variables in a term
collinear points that lie on the same line; the gradients
between any two of these points will be the same
compounded quarterly interest is applied to an investment
1
of the quoted annual rate is
every 3 months, and __
4
applied to the total at each stage
compounded semi-annually semi-annually means every
half year (every six months), which means that interest is
twice a year
concentric circles that share the same centre
congruent identical in all respects
construction a line which needs to be added to a sketch
continuous data data without breaks and that can have any
value; includes all units of measurement (for example,
height, mass)
converse reversing the logic and proving a theorem in
reverse
376
depreciate to lose value
exhaustive events events which use up the full sample
space, that is, all possible outcomes
experiment a trial situation involving chance
exponent (index) a number or variable that shows how
many times a number (the base) is multiplied by itself
exponential equation an equation with unknowns or
variables in the exponent
exterior angle outside angle of a polygon formed by one of
the sides which has been extended
F
frequency polygon a polygon that demonstrates the
frequency of each group in a set of grouped data, and
that displays the spread of the data
function a relationship between two variables, usually x and
y, where for every value of x there is one corresponding
value for y
G
general solution the formula which lists all possible
solutions to a trigonometric equation; takes into account
the period of the trigonometric function so an angle can
be positive or negative
Glossary
PLT MATHS LB 11 7th pgs (Real Book).indb 376
2012/07/02 2:28 PM
H
hemisphere half a sphere
hence sometimes used to show that you should make use of
the answer that you have found
histogram a graph with bars that show the frequency
distribution of grouped data with no space between
the groups, so the bars touch each other; groups are
usually equal in width so that the height of the rectangle
(frequency distribution) is the same as the frequency
horizontal sideways
horizontal shift when sideways, with no up or down
movement
hypotenuse the side opposite the right angle in a rightangled triangle
I
included angle the angle between two known sides
increase to grow bigger
interior angle inside angle
intersect to meet or cross at a point; to overlap (sets)
interval spacing the number of degrees between critical
points on a trigonometric graph
perpendicular height the perpendicular distance between
bases or between the vertex of a polygon and the base
opposite it
produce to lengthen a straight line; for example, AB
produced to C means that ABC is a straight line
Q
quadratic equation an equation in which the highest index
is squared
R
radius a line from the centre of circle to its circumference
range the set of all y-values of a graph; the y-values of the
graph which exclude any values which make y undefined
rational exponent an exponent or index with a rational
number as the exponent
real numbers the set of all rational and irrational numbers
reciprocal the multiplicative inverse of a number; obtained
by interchanging the numerator and the denominator of
that number
reduction formulae trigonometric identities that express
the trigonometric ratios of an angle of any size in terms of
the trigonometric ratios of an acute angle
reflect to create a mirror image or reflection
M
reflection a mirror image about a line of symmetry
maximum the largest possible value
reflex angle an angle greater than 180°
minimum the smallest possible value
revolution the sum of the angles around a point
N
roots (of an equation) the solutions of the equation
n the position of the term in a pattern
S
nth term the term in the position n
sample space the full set of data values
negative exponent a number raised to a negative exponent
and the reciprocal of that number with a positive
exponent
scientific notation the form in which you use exponents to
write very large or very small numbers
nominal interest rate the annual interest rate that is quoted
non-real numbers imaginary numbers or complex numbers
that are not real numbers
O
obtuse angle any angle between 90° and 180°
outlier a data value that does not follow the trend of the
rest of the data
scrap value the value of an item after it has passed its useful
life in its original form, and certain parts can be sold as
scrap
solution the value of x or any other variable which solves
an equation; a value of the angle which satisfies a given
trigonometric equation
solve a triangle to calculate the lengths and sizes of the
unknown sides and angles
specific solutions solutions which satisfy a given
trigonometric equation in a restricted interval
P
square the product of two identical factors
parameter a variable that restricts or gives a particular form
or shape to the equation it characterises
standard deviation the square root of the variance
subtended by an arc formed by an arc
per annum for the year
supplementary angles that add up to 180°
period the number of degrees needed to complete a wave
pattern or cycle
surds irrational roots or irrational numbers; the square root
of a non-perfect square is an irrational number
perpendicular at right angles or at 90°; lines are
perpendicular when the angle between the lines is 90°
symmetry line a fold line that makes the shapes the same
on both sides
Glossary
PLT MATHS LB 11 7th pgs (Real Book).indb 377
377
2012/07/02 2:28 PM
T
tangent a straight line which touches a circle, but does not
pass through it
theorem a formal proof of a geometric statement
theoretical outcome the result, in theory, which is expected
for a certain event to happen
trigonometric equation an equation involving
trigonometric ratios which is true only for certain values
of the unknown variable
trigonometric identity an equality which is true for all
values of an unknown variable, for which both sides of
the identity are defined (so no zero denominators)
U
universal set the sample space, which is the full set of data
values
V
variance the average of the squared differences from the
mean
vertex a point on a triangle where the sides meet, so a
triangle has three vertices; a point where two straight
lines meet to form an angle
vertical straight up or down
vertical shift straight up or down, with no sideways
movement
vertical shrink grows smaller without changing the
y-intercept
vertical stretch grows taller without changing the
y-intercept
X
x-intercept, root the point where the graph cuts the x-axis
(where y = 0)
Y
y-intercept the point where the graph cuts the y-axis
(where x = 0)
378
Glossary
PLT MATHS LB 11 7th pgs (Real Book).indb 378
2012/07/02 2:28 PM
Index
A
D
B
E
altitudes 57−8, 79
amplitude 116−7, 119, 121−4, 130−1, 135−7, 169, 376
angles
corresponding angles 64−5, 188, 192, 198, 206
acute angles
central 149−50, 155
co-interior 188, 192, 198, 206
complementary 148, 153, 188
equal 64, 188, 195, 198−9, 205
exterior 65, 188, 195, 198−199, 203, 205
negative 146−150, 155−6
opposite 189, 197−8, 205
positive 138, 146, 156
reference 63, 65
applications of triangle rules 225, 228−9
area rectangle 176, 179
area rules 175, 214, 220, 223, 286
asymptotes 92−4, 97−107, 124, 130−1, 134−6, 376
average gradient 81, 111−3, 115, 168
axes 83−4, 188−9, 91, 94−7, 100, 102−3, 105
axes of symmetry 94−6, 376
base 4, 7, 11, 21, 55, 176−9, 184−5
bearing 228−9, 235
box-and-whisker diagram 300, 306–310; 313–314, 316–317
298, 304−5, 308−9, 312, 316
C
calculator instructions 116, 132−3
CAST diagram 139, 146−7, 149, 155
circle 22, 138, 176, 177, 179, 185, 190−9
circumference 176, 179, 190, 192, 194−5, 197−9
concentric 193, 210
diameter 34, 179, 186, 198−200
geometry 3, 22, 56, 64, 79, 175, 188, 223, 229
reflex angle 194
circumference 176, 179, 190, 192, 194−5, 197−8, 200,
209−10, 212
coefficient 12, 24−6
common factor 7, 27, 29, 159
complementary rule 175, 252−3
compound decay 175, 236−9, 241, 286
compound growth, periods of 175, 242−5, 249
compound interest 108−9, 241−2, 250, 284
congruent 176, 188, 195
constant factor 176, 179, 181, 185, 187
contingency table 269−72, 273– 5, 283, 287
converse 188, 190−1, 195, 197, 202, 376
cos graphs 132
cos, cosine 19, 18, 116−119, 127, 159−60, 175, 214−5,
217−8, 221, 223, 226, 286
cosine and area rules 175, 214, 223, 284
cumulative frequency 296−8, 314, 316, 319
cyclic quadrilateral 175, 194, 197−9, 376
data values 256, 259−61, 288, 296, 299−301, 305, 307,
308, 311, 313, 319
decay 175, 236−40, 242, 250, 286
decimal 19, 21, 25, 30, 44, 250
decimal places 19, 24, 30, 44
degrees 116, 119, 123−4, 126, 130, 135−6, 376
depreciation 236−7, 240–1, 250, 282, 322, 376
annual rate of 240−2
straight line method of 236−7, 250
depression 223−4, 227
angle of 223−4, 227
difference 14−6, 27, 46, 48, 52, 79, 108, 288, 290, 301
common 7, 27, 29, 46, 48, 51−2, 79, 127, 159, 191,
205−6
first 48, 79
squared 301−302
dispersion 308, 310, 313
distance 19, 34, 56, 69, 79, 107, 116, 119, 123−4, 126,
130, 136, 176−7, 189, 191, 193, 202−3, 129–135, 317,
376
domain 92−3, 98−101, 122, 124−5, 134, 166
effects of parameters 81−2, 84, 86, 88, 90, 94, 96, 98,
elevation, angle of 223−4, 233−234
events 175, 252−6, 259, 262–7, 269−5, 277, 283, 287
complementary 253, 267, 285
exclusive 252
independent 175, 252−4, 257, 259–61, 270, 272, 274−5,
281, 285
exponential equations 7, 376
exponential graphs 81, 101, 103−7
shifted 104
exponents 3−14, 16, 18, 20−1, 53, 78, 85, 87, 89, 91, 170
negative 5−6
positive 5−6, 20, 170
F
factorisation 24, 27, 29−30, 42, 87
factors 4, 22, 27, 31, 42−3, 53, 87−8, 253
Fibonacci sequence 70, 72
finance 175, 236, 238, 240−8, 286
first degree equations 27
form
exponential 9−10, 14, 239
radical 9−10
standard 56, 78, 85−6, 90, 168
formula 22, 26−7, 30−1, 42−4, 46−9, 51−2, 54, 56, 59−60,
62, 75, 78−9, 81, 87, 90, 99, 108, 112, 146−9, 151−3,
158, 166, 169, 171, 176−8, 183−7, 218, 237, 239−43,
245, 284, 299−303, 319−23
general 46, 48
x-intercept 90
formulae 52, 56, 79, 81, 146−53, 69, 176, 178, 245, 182,
286, 301, 303, 305, 319
Index
PLTMATHSLB11LB_17.indd 379
379
2012/07/07 11:54 AM
frequency 244, 287−98, 303−4, 307, 311, 314−9, 323−4
relative 290−1
total 297, 299
frequency distribution 290, 377
frequency polygons 289−96, 300–1, 306−7, 315, 317−318,
376
frequency table 291, 294, 299−300, 311, 324
functions 81−2, 84, 86−112, 116−8, 120−8, 130−8, 167,
325, 376
trigonometric 81, 123, 138
G
gradient 56−69, 76−7, 79, 81, 99−100, 111−15, 168, 172,
237, 239, 299, 308
graph, amplitude, period, steps, asymptotes, shift, domain
134
graph, amplitude, period, steps, horizontal shift 132
graph, amplitude, vertical shift 119
graph equation 124−5, 135
graph, turning point, asymptotes, domain 168
graphs 35−8, 40, 45, 78, 81−2, 85−9, 93, 98, 100−1,
103−8, 111, 114, 116−22, 124−28, 130−7, 157, 166,
168−9, 173, 290−1, 325
bar 290
reflected 101, 115, 121−2, 126−7, 136, 166, 171
sketching 87, 104, 125, 133, 135
H
histograms 290−4, 300, 306, 311, 318, 377
horizontal 35, 50−1, 56, 81, 83−6, 91−6, 99, 101, 103−4,
125, 127, 129, 132−3, 168−9, 223−4, 377
horizontal asymptote 92−6, 99, 101, 104
horizontal shift 81, 85−6, 91, 93−4, 103, 125, 127, 129,
132−3, 135, 168−9, 377
hyperbola 40, 81, 92−5, 97−9, 101, 112, 168
I
inequalities 3, 22−4, 26, 28−42, 78
interest 108−9, 237, 239−51, 282, 286, 301−302, 321−2
changes 245, 248−9
interest rate 240, 242−6, 248−51, 282, 286
effective 242−3, 250−1, 286
effective annual 242−3, 249, 252, 282
nominal 242−3, 249, 286, 377
interior opposite angle 197−8, 205
interquartile range 294, 297−8, 307, 311, 315, 323
intersection 40, 58, 60−1, 67, 69, 98, 100, 114, 212, 252,
259−60, 324
L
laws 3−5, 7, 9, 20, 78, 85, 87−9, 91
of exponents 3−5, 7, 20, 85, 87, 89, 91
linear equations 27
linear patterns 3, 46−7, 79
literal equations 26
M
median 57−8, 60−1, 67, 76, 79, 294, 296, 298, 306−9, 312,
315, 316, 319, 323−4
380
midpoint 56−60, 67−9, 79, 172, 175, 179, 190−1, 193,
198, 203, 208, 210, 285, 317−8
chord 175, 190−3, 195−6, 198−9, 202−7, 210−11,
229−30, 285
formula 59
mode 102−3, 109, 115, 133, 136−7, 251, 296, 299, 303−4,
306−7
N
number 3−5, 9, 11, 14, 17, 21, 24−5, 30, 32−3, 34−6, 42,
44−56, 70−75, 78−9, 102−3, 108, 123, 133, 140, 171,
237, 239, 241, 244, 252−7
imaginary 24−5
irrational 14, 24−5
natural 4, 9, 25
rational 17, 25
real 24−5
number patterns 3, 46, 48, 50, 54, 79
O
obtuse 62, 158−60, 215, 217
obtuse-angled triangle 215, 217
ogive curve 287, 294−98, 314–315, 319, 324
opposite sides 59−60, 189, 205
parallel 56−61, 64−5, 67−9, 77, 79, 107, 176, 188, 189,
198, 205−6, 211, 223, 279
P
parabola 22, 36, 40, 81−3, 85−7, 89−92, 99, 101, 111−13,
168, 171
parallel 57−61, 64−5, 67−9, 79, 176, 188−9, 192, 198,
205−6, 211, 223, 279
parallelogram 58—61, 67−9, 77, 189, 205, 281, 324
parameters 81−4, 86−8, 92
pattern 3, 46−54, 75, 79, 85, 93, 116, 118−9, 123−7,
135−6, 159−60, 169, 171, 308, 315, 320
of answers shifts 85, 93, 125
cubic 27, 52, 55
exponential 48
second difference 7, 48−9, 51−2, 54, 79
perpendicular 56−8, 60−1, 67−9, 76, 79, 172, 175, 176−8,
189, 190−3, 198, 202, 207, 277, 279, 324−5, 377
bisector 57−8, 60−1, 67−8, 76, 79, 172, 191, 198, 279,
324
gradient 57, 58, 60
height 176−8, 189, 286, 288
lines 56, 79, 175, 190−1, 193
polygon 52, 55, 176−7, 197, 290−301, 306−7, 315, 317−8
positive ratio 155, 158
prime number 253−4, 263
probability 252−5, 257−75, 280, 287, 322
of event 252, 273, 287
Pythagoras’ Theorem 139, 141−2, 151, 177, 179, 181−2,
188, 190−2, 200, 214, 217
Q
quadrant 59, 138−9, 141, 146−50, 155−6, 158−61, 169,
377
quadratic equation
22, 24, 27, 30–2, 42, 77, 377
quadratic patterns 3, 48−53, 79
Index
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quadrilateral 68, 175, 189, 193−4, 197−201, 203−5, 207,
110−13, 221, 231−2, 279, 281, 291
axis of 82, 86−7, 91, 107, 114, 168
symmetry lines 94, 97−100, 107, 114, 166
opposite angles 189, 197−8, 205, 207
R
radius 138, 176−8, 180, 183−7, 190−3, 199−200, 202−4,
206, 209, 211, 229, 231, 234, 278, 284, 377
range 92−3, 98−101, 116−24, 126, 129−31, 135−7, 166−9,
252, 294, 197−8, 301, 306, 307, 309, 311, 315–6, 319,
T
tan graphs 119, 127, 132
tangents 175, 202, 204−9, 213, 279, 285, 378
theorem 139, 141, 151, 175, 177, 179, 181−2, 188, 190−1,
194−5, 197, 201−2, 217, 230, 326, 378
321, 323, 325, 377
rational exponents 3, 9, 11, 21, 78, 377
time line 245−49, 286
ratios 3, 70−1, 73, 138, 140, 142, 146, 148−9, 151−3,
tree diagram 175, 263−8, 275−6, 287, 322
158−61, 169, 286, 377
triangle rules 223, 225, 228−9
given 19, 23−4, 31−2, 53, 58, 66, 75−9, 82, 87, 100,
triangles 116, 152, 178, 214, 217, 223, 228, 233, 284, 286,
104, 106, 107, 119−23, 139−40, 151, 153−60, 166,
291, 293
170−1, 173, 190−1, 194−5, 197, 199, 202, 214
acute-angled 215, 217
equilateral 57, 185
required 151−2
reduction formulae 81, 146−7, 149−52, 169, 377
non-right-angled 286
reflection 82−4, 94−7, 101−2, 107, 114, 127, 168, 321, 377
right-angled 138, 177, 188, 191−2, 214, 221, 223, 226,
revolution 188, 196−7, 377
roots 3, 9, 14, 24−6, 31−2, 42−4, 71−2, 74, 78, 85, 87−9,
286
trigonometric definitions 142, 214−5
91, 166, 230, 377
trigonometric equations 81, 155, 157, 159, 161, 169, 378
cube 9, 52−3, 55
trigonometric graphs 81−2, 116−25, 127, 130, 169
equal 42, 44, 85, 166
equations of 81
irrational 14, 87,
trigonometric ratios 138, , 140, 146, 149, 161, 286
rational 42, 87
trinomial 12, 27, 29, 159
unequal 44, 166
turning points 88−9, 114, 118, 121−2, 124−5, 137, 173,
325
S
sample space 253, 256, 258, 260–1, 269, 376
U
sequence 48−9, 53−4, 70−2, 72, 79, 320
undefined 75, 92, 98, 100, 114−8, 120, 127, 134, 152,
of numbers 71, 79
161−2, 173, 320, 325
simultaneous equations 3, 40−1, 70, 78, 104
sine 132, 159, 160, 175, 214−23, 225, 286
using 175, 223
V
variance 289–302, 305, 378
sine rule 215−7, 220, 225, 286
Venn diagram 175, 256−61, 275, 277, 282, 287, 322
standard deviation 287, 301−305, 311, 313−6, 317, 319,
vertex, vertices 57–61, 64, 67−9, 76, 82, 172, 177, 189,
324, 377
of ungrouped data 289, 300
197, 205, 324
vertical asymptote 82, 92−6, 99, 114, 117, 224, 130−1,
statistics 237, 241, 289−90, 298, 318
surd equations 18−9, 78, 236
135−6
vertical shift 83, 86, 91−4, 104, 118−9, 227, 168, 378
surd form 14, 16, 19, 21, 25, 30, 67, 69, 73, 76, 140, 193,
199, 210
surds 3−4, 14−9, 21, 78, 377
X
x-intercepts 35, 87−91, 114, 119, 321, 378
surface area 176−87, 278, 280, 284, 326
lateral 176, 179, 186−7
symmetry 82−3, 86−7, 91−101, 107, 114, 166, 189, 204−5,
306, 324
Y
y-intercepts 82−4, 87−9, 94, 99, 100, 102, 104, 106, 122,
124−5, 166, 321, 378
Index
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