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Maths Ext Year Eleven
Topic 1: Methods in Algebra
Index Laws
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Binomial Products
In general,
Thus
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Difference of Two Squares:
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Algebraic Fractions
- Always factorise first.
- When adding/subtracting fractions, equate the denominator.
e.
Cubics
In general,
Thus
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In general,
And when is odd,
Thus
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Linear Equations
Make pronumeral the subject of formula:
Quadratic Equations
After factorising, if then or
Use quadratic formula:
To derive the quadratic formula: ,
- Complete the square: add to both sides of equation. Eg.
Simultaneous Equations
- Amount of equations required = number of pronumerals.
- Eliminate a variable:
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Substitute into
- Solving by substitution:
am × an = a ( m + n ) am ÷ an = a ( m − n ) ( am ) n = amn a − 1 = 1 a a 0 = 1 ( a b
) x = a
x bx ( a b )− 1 = b a x
a b = bxa OR( bx ) a
( a + b + c +... ) 2 = ( a 2 + b 2 + c 2 +... + 2 a b + 2 a c + 2 b c +... )
( a + b ) 2 = a 2 + 2 a b + b 2 ( a − b ) 2 = a 2 − 2 a b + b 2
( a + b )( a − b ) = a 2 − b 2
a b − 2 b 2 6 a 2 b
÷ a 2 − 4 a b + 4 b 2 3 a = b ( a − 2 b ) 6 a 2 b
× 3 a ( a − 2 b ) 2 = 1 2 a ( a − 2 b )
( a + b + c + d... ) 3 = a 3 + b 3 + c 3 +... + 3 a 2 b + 3 a 2 c + 3 a 2 d + 3 b 2 a +... + 6 a b c + 6 a b d +...
( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3
an − bn = ( a − b )( an − 1 + an − 2 b +... + a bn − 2 + bn − 1 ) n an + bn =( a + b )( an − 1 − an − 2 b +.. .− a bn − 2 + bn − 1 )
a 3 − b 3 = ( a − b )( a 2 + a b + b 2 ) a 3 + b 3 = ( a + b )( a 2 − a b + b 2 )
3 z + 2 = z − 9
z =− 11 2
a b = 0 a = 0 b = 0 x 2 + 9 x + 18 = 0 ( x + 6)( x + 3) = 0 ∴ x =−6,− 3
x =
− b ± b 2 − 4 a c 2 a 2 x 2 + 5 x −4 = 0 x =
− 5 ± 25 + 32 4 x = − 5 ± 57 4 a x 2 + b x + c = 0 x 2 + bax + ac = 0 x 2 + bax + ( 2 ba ) 2 −( 2 ba ) 2 + ca = 0
( x + 2 ba ) 2 − b
2 4 a 2
- ca = 0
( x + b 2 a ) 2 = b
2 − 4 a c 4 a 2 x + 2 ba =± b
2 − 4 a c 4 a 2
x = − b ± b 2 − 4 a c 2 a ( b 2
) 2 x 2 + 6 x −7 = 0 x 2 + 6 x + 9−7 = 9 ( x + 3) 2 = 16 x + 3 =± 4 x = 1,− 7
2 x + 3 y = 21→(1) 5 x + 2 y = 3→(2) (1)× 2 and(2)× 3 4 x + 6 y = 42→(3) 15 x + 6 y = 9→(4) (4)−(3) 11 x =− 33 x =− 3 x =− 3 (2) 3 y = 27 ∴ x =−3, y = 9
x = 7→(1), y + 1 x
= 9→(2)
y = 9− 1 7 = 62 7
Topic 2: Numbers and Surds
Real Numbers
All real numbers can be placed on the number line.
Subsets notated as , , , and.
All rational numbers expressed as , where.
Proof that is irrational by contradiction:
Assume is rational, thus where and are coprime.
is even. is even,.
is even. is even. If both are even this means they both have 2 as a factor, which contradicts the assumption at the beginning that they are coprime. Therefore, is not rational.
HCF and LCM
- For HCF, write both numbers in terms of prime factors, and then
multiply the common factors. and HCF =
- For LCM, write both numbers in terms of prime factors, and then
multiply the highest factors. and HCF =
Divisibility Tests
- All natural numbers.
- Even number.
- Digits sum to multiple of 3.
- Last two digits divisible by 4.
- Ends in 5 or 0.
- Divisible by 2 and 3.
- Double the last digit and subtract from the other digits, final digit is divisible by 7. Testing 2751:
is divisible by 7 8. Last three digits are divisible by 8. 9. Sum of digits is multiple of 9. 10. Ends in a 0. 11. Sum of even positioned digits = sum of odd positioned digits, or differ by a multiple of 11. Testing 4323: is a multiple of 11.
Recurring Decimals into Fractions
For , let.
Then multiply by , where is amount of repeating digits, two in this case.
Representing Real Numbers
All real numbers can be described:
- Geometrically:
- Algebraically:
- Interval Notation:
- Set Notation: Types of Intervals:
- Bounded - two endpoints (e. )
- Unbounded - one endpoint (e. )
- Closed - all endpoints included (e. )
- Open - all endpoints not included (e. )
- Degenerate - a single point (e. )
Significant Figures
- Starting from left, first non-zero digit is first sig fig.
- All non-zero digits are significant
- Zeroes at the end of whole number may or may not be significant, depending on rounding. e. 8200 has 2 significant figures if rounded to nearest 100, 3 if rounded to nearest 10, and 4 if rounded to nearest unit. If ambiguous, always say the largest amount (in this case 4)
- Zeroes at the end of number are significant if behind a decimal point.
- Zeroes between any non-zero digit are significant. e. 0 has 7 significant figures, 08200 has 9 significant figures.
Surd Operations
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Like surds can be added or subtracted.
Rationalising the Denominator
If denominator is pure surd, multiply fraction by :
If denominator is surd binomial, multiply fraction by where for the binomial the conjugate is.
ℝℚℤ𝕎 ℕ a b
a , b ∈ ℤ
2 2 2 = a b
a b
2 = a 2 b 2
→ 2 b 2 = a 2 → a 2 ∴ a a = 2 k
2 b 2 = (2 k ) 2 → b 2 = 2 k 2 → b 2 ∴ b a , b
2
1176 : (3× 23 × 72 ) 1260 : (2 2 × 32 × 5 ×7) 22 × 3 ×7 = 84
1176 : (3× 23 × 72 ) 1260 : (2 2 × 32 × 5 ×7) 23 × 32 × 5 × 72 = 17640
1 ×2 = 2→ 275 −2 = 273 3 ×2 = 6→ 27 −6 = 21 1 ×2 = 2→ 2 −2 = 0 ∴ 2751
3 + 3 = 4 + 2 ∴ 4323
0 = 0... x = 0 27 x = 0... x 10 a a
100 x = 32... 99 x = 32. x = 324 990
= 18 55 ∴0 = 18 55
x ≤ 2 (−∞, 2 ] { x = x ≤2}
4 < x < 6or− 2 ≤ x ≤ 4 x ≥ 2 or x < 3 x ≥ 2 or 4 ≥ x ≥ − 2 4 < x < 6or x < 3 x = 3
a × b = a b a b
= a b
( a ) 2 = a BUT a 2 =| a |
surd surd 3 2 5
= 3 2 5
× 5 5
=
3 5 10 conjugate conjugate a + b x a − b x 2 + 3 2 − 3
=
2 + 3 2 − 3
×
2 + 3 2 + 3
=
4 + 4 3 + 3 4 − 3 = 7 + 4 3
Rectangular Hyperbola and Exponential
Data that with inverse variation lies on rectangular hyperbola.
Rectangular hyperbolas have two perpendicular asymptotes.
Represented as or.
The orientation of basic exponentials ( where ) is determined by the base.
- For , starts shallow and increases rapidly as
- For , starts steep and decreases less rapidly as
Direct and Inverse Variation
A variable varies directly with a variable if
, for some non-zero constant of proportionality.
The graph of as a function of is thus a line through the origin. A variable varies inversely with a variable if
, for some non-zero constant of proportionality.
The graphs of as a function of is a rectangular hyperbola with asymptotes as the x-axis and y-axis.
Types of Relationships
Functions:
- One-to-one (e. ), passes vertical & horizontal line tests.
- Many-to-one (e. ), passes vertical line test only. Relations:
- One-to-many (e. ), passes horizontal line test only.
- Many-to-many (e. ), fails both tests.
Topic 4: Probability
Basics
means amount of members of.
is event space (favourable), a subset of , a finite sample space.
if sample space is uniform.
For numerical 2-steps, create 2-way array, with 1st step on x-axis.
For other multistage experiments, use tree diagram.
Complementary Events
- Complement of an event is defined as
Thus,
- Common notations include
Sets
Empty set has no members, denoted as
Two sets are equal if they have exactly the same members.
A and B = = Intersection
A or B = = Union
means in only A.
A B means every member of A is in B (A is subset of B). Counting Rule:
When (disjoint sets),
Three-way counting rule:
The counting rule becomes the addition rule when talking about probability: - When are mutually exclusive,
Multi-Events / Multiplication Principle
- If there are outcomes for independent experiment , outcomes for experiment , ... , outcomes for experiment , then there are outcomes for composite experiment. Applicable also to probability.
- If the composite experiments of a larger experiment are done without replacement, then all is the same, except with each successive action, one choice is removed from sample space.
Conditional Probability
- OR
Independent Events
and are independent if. This also means.
- Proving independent to when is known independent to :
And since ,
.
- can be rearranged as and for independent events. Conversely, if then independent.
Topic 5: Combinatorics
Factorial Notation
Factorial definition:
This definitions can be used to create common denominators:
Permutations -
- A permutation of objects is an ordered arrangement of the objects. Therefore,
- When there are positions to be filled with objects: choices for 1st position choices for 2nd position... choice for th position
- This can be represented by the notation
- Means “from arrange ”
- Restrictions: Line 5 cats and 5 dogs up so that they alternate?
arrangements.
y = 1 x x y = 1 y = ax a ≠ 1 a a > 1 x → ∞ 0 < a < 1 x → ∞
y x y = k x k y x y x y = k x k y x
y = 2 x y = x 2
| y |= x x 2 + y 2 = 5
| H | H E S P ( E ) = | E | | S |
E | E |=| S |−| E | P ( E ) = | E | | S | = | S |−| E | | S | = | S | | S | − | E | | S | = 1− P ( E )
E , E ′ , Ec
∅
A ∩ B A ∪ B A ∩ B ⊂ | A ∪ B |=| A |+| B |−| A ∩ B | | A ∩ B |= 0 | A ∪ B |=| A |+| B |
n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C )− n ( A ∩ B ) − n ( B ∩ C )− n ( A ∩ C ) + n ( A ∩ B ∩ C )
P ( A ∪ B ) = P ( A ) + P ( B )− P ( A ∩ B ) A , B P ( A ∪ B ) = P ( A ) + P ( B )
n 1 E 1 n 2 E 2 nm Em n 1 × n 2 ×.. .× nm E 1 × E 2 ×.. .× Em
P ( A | B ) = | A ∩ B | | B | = | A ∩ B | | B | ÷ | B | | S | = P ( A ∩ B ) P ( B )
A B P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) B A A B P ( B | A ) = P ( B ∩ A ) P ( A ) = P ( A ∩ B ) P ( A ) × P ( B ) P ( B ) = P ( A ∩ B ) P ( B ) × P ( B ) P ( A ) = P ( A | B )× P ( B ) P ( A ) P ( A | B ) = P ( A ) = P ( A )× P ( B ) P ( A ) ∴= P ( B ) P ( A | B ) = P ( A ∩ B ) P ( B ) P ( A ∩ B ) = P ( A | B )× P ( B ) P ( A ∩ B ) = P ( A )× P ( B ) P ( A ∩ B ) = P ( A )× P ( B )
x! ={ 0! = 1, x ( x −1)! for x ≥ 1
1 ( n −1)! + 1 ( n + 1)! = 1 ( n −1)!
- 1 ( n + 1) n ( n −1)!
= ( n + 1) n + 1 ( n + 1) n ( n −1)! ∴= n
2 + n + 1 ( n + 1)!
nP
####### n
n n A B A ≠ A A B n n n n − 1 n −( n −1) = 1 n nPn = n! n n
= 2 either cat or dog in 1 st ×5! dog per mutations ×5! cat per mutations = 28800
- When lining up objects, if there are of alike objects, of another type of alike object, and of a final type of alike object:
Permutations -
- When arranging objects where , from possible objects:
choices for 1st position choices for 2nd position... choices for th position
Thus
Restrictions: 6 people in boat with 8 seats, 4 on each side. What is the probability Bill and Ted are on left side, and Greg is on right?
Probability
Combinations
- Total number of subsets of , a finite set with elements =.
- A combination of objects is an unordered arrangement of the
objects. Therefore,
The number of combinations of objects for r positions is:
Restrictions: How many possible “three of a kind” hands can you
be dealt with 5 cards from a standard 52 card deck?
hands.
Division in Groups
- Case 1: Dividing objects into group with , group with :
Number of selections =
This is the same even for three groups:
Case 2: Dividing objects into two groups each with objects
Number of selections = as we can sort the two
equal groups times. - Similarly for dividing into groups of , and.
- Note that if groups are going to be arranged afterwards, there is no need to divide by arrangements!
Insertion Method
- Used for separation questions. Letters of BETWEEN are arranged in line. How many arrangements if all E’s are separated?
- Display as B_T_W_N: there are possibilities.
Using Separators
- Useful when dividing larger groups into smaller groups.
- In how many ways can 10 identical coins be allocated to 4 different boxes? Assign coins letter , and create 3 separators that act as separators between 4 boxes. Thus:
Circle Arrangement
The number of ways to arrange objects along a fixed circle is
This is since we create a fixed point as there is no front and back.
Pigeonhole Principle (ew)
- If pigeons are placed into pigeonholes, then there must be at least one pigeonhole with at least pigeons in it.
- In general, if or more objects are placed into holes, then at least one of the holes must contain over objects.
- Example 1: 16 positive integers are written down. At least how many numbers will leave same remainder when divided by 5? There are five possible remainders: 0,1,2,3,4. Thus there are five holes, i. We know
Therefore according to general rule, at least numbers will leave same remainder, i.
- Example 2: A computer generates random 3-letter words. How many words need to be generated to ensure 8 of same word? There are possible words, i. And we know we need repetitions, therefore
Total objects needed
Topic 6: Transformations and Symmetry
Translating Curves
- For horizontal shift, replace with , which moves units to the right.
- For vertical shift, replace with , which moves
units up.
- OR
Reflecting Curves
Vertical reflection: For reflection across y-axis, replace with.
- To reflect in the line , replace with
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- Horizontal reflection: Reflection across x-axis, replace with.
- OR
To reflect in the line , replace with
- Reflection about Origin:
- Same as a rotation of
- Caused by a reflection across x-axis and across y-axis.
- Same as a rotation of
n r 1 r 2 rk Number of Permutations= n! r 1! r 2!... rk!
nP
####### r
r r < n n n n − 1 n − r + 1 r nPr = n ( n −1)( n −2)... ( n − r + 1)
= n ( n −1)( n −2)... ( n − r + 1)× ( n − r )( n − r −1)... (2)(1) ( n − r )( n − r −1)... (2)(1) = n! ( n − r )!
Sample space= 8 P 6 = 20160
Ways= 4 P 2 ( B an d T on lef t )× 4 P 1 ( G on r ight )× 5 P 3 ( rem aining three )
= 2880 = 2880 20160 = 1 7
S n 2 n n n A B A = A A B n nC r =
nPr r! = n! r !( n − r )! =( n r )
Hands= 13 C 1 (3 of kin d )× 4 C 3 (3 of 4 suits )× 12 C 2 ( rem aining 2)× 4 × 4
= 54912 nCn − r = n! ( n − r )!( n −( n − r ))! = n! ( n − r )! r! = nCr
( m + n ) m n m + nCm =( m + n )! m! n!
m + n + pC m × n + pCn =
( m + n + p )! m !( n + p )! × ( n + p )! n! p! = ( m + n + p )! m! n! p! (2 m ) m 2 mCm 2! = (2 m )! m! m !2! 2! (2 m + n ) mm n 2 m + nCn × 2 mCm =(2 m + n )! n !(2 m )!
× (2 m )! m! m !2!
= (2 m + n )! m! m! n !2!
4!cons× 5 C 3 E
C S 13! 3!10! = 286
n n! n = ( n −1)!
n k n k k n + 1 n k
n = 5 k n + 1 = 16 ∴ 5 k + 1 = 16 k = 3 k + 1 3 + 1 = 4
263 n = 26 3 = 17576 8 k + 1 = 8 k = 7 = k n + 1 = 123032 + 1 = 123033
x x − h f ( x ) h y = f ( x − h ) y y − k f ( x ) k y − k = f ( x ) y = f ( x ) + k
x − x y = f (− x ) x = a x 2 a − x y = f (2 a − x ) y − y − y = f ( x ) y =− f ( x ) y = a y 2 a − y 2 a − y = f ( x )→ y = 2 a − f ( x ) y =− f (− x ) 180 o
Reciprocal Functions
The graph of can be sketched by first drawing
When , then is undefined (ie. vertical asymptote)
When , then (asymptotes become point discontinuity)
Identify where as their reciprocals are identical.
Where increases, the reciprocal decreases, vice versa.
Where is positive, is also positive, vice versa.
If a point in is , in reciprocal it will be.
- Horizontal asymptotes in of will shift to in
Graphing Addition of Ordinates
The y-coordinate of a point is called the ordinate.
The x-coordinate of a point is called the abscissa.
If , each y-value of both functions are added
for each corresponding x-value. - Where , then , thus a x-intercept. - If has 0 at , then. - If two curves meet at so that their ordinates are equal, then - To sketch from two functions, simply rule many vertical lines across the function and add for each line, connecting dots.
- For graphs with asymptotes e where
and , therefore
- Since y is undefined at , it is still undefined in.
- Again, draw vertical lines and add normally.
- Exclusions in domain of original functions remains in
- E. if , then retains the vertical asymptote.
- Like functions retain symmetry when added, ie. Odd function +
Odd function = Odd function and vice versa.
- can be graphed by first graphing and and then adding ordinates together.
Addition of Absolute Value Functions
- For example: , each absolute value
will be a critical point, which creates three regions. - , therefore critical point at. Testing regions:
- , therefore critical point at. Testing regions:
Adding Ordinates of Bounded Functions
- For example , formed from and
- Draw points at sin graph turning points and all x intercepts.
- You end up with sin-like curve tilted anticlockwise 45 degrees.
Multiplication of Graphs
- can be graphed by first ,
- Where there are x intercepts, the function will change sign.
- Multiply signs of functions to determine new functions’ sign.
- Where , must equal (or in case of reflect/equal) the other curve.
- If , then so will the new function.
- However, in the case of when finding , while , thus we reach a dilemma.
- Dominance: the graph that gets steeper or shallower the quickest prevails, and we graph the accordingly. Therefore, in this case, for.
- Any exclusions in domain of original functions remain in new one.
- Even function even function = Even. Odd function odd function = Even. Odd function even function = Odd.
y = 1 f ( x ) y = f ( x )
f ( x ) = 0 1 f ( x ) f ( x )→±∞ 1 f ( x ) → 0
f ( x ) = 1,− 1 f ( x ) f ( x ) 1 f ( x ) f ( x ) <| 1 | >| 1 | f ( x ) y <| 1 | y >| 1 | 1 f ( x )
s ( x ) = f ( x ) + g ( x )
f ( x ) =− g ( x ) y = 0 g ( x ) x = a s ( a ) = 0 + f ( a ) = f ( a ) x = a s ( a ) = 2 f ( a ) O R 2 g ( a )
s ( x ) = f ( x ) + g ( x ) f ( x ) = x g ( x ) = 1 x s ( x ) = 1 + 1 x x = 0 s ( x )
s ( x ) f ( x ) = x + 1 x , g ( x ) = 1− 1 x y = f ( x ) + g ( x ) x = 0
y = f ( x )− g ( x ) y = f ( x ) y =− g ( x )
f ( x ) =| x + 3|+| 1 − x | = 0
x + 3 = 0 x =− 3
x ≤ −3; y =−( x + 3) + (1− x ) =− 2 x − 2 1 − x = 0 x = 1 −3 < x < 1; y = ( x + 3) + (1− x ) = 4 x ≥1; y = ( x + 3)−(1− x ) = 2 x + 2
y = x −sin x y = x y =−sin x
s ( x ) = f ( x ) g ( x ) y = f ( x ) y = g ( x )
f ( x ) o r g ( x ) = 1 s ( x ) y =− 1 f ( x ) o r g ( x )→ 0 o r ±∞ x → −∞ s ( x ) = x ex y = x →−∞ y = ex → 0
s ( x ) s ( x )→ 0 x → −∞
× × ×
y = x ex
Graphs of Squared Functions
- can be graphed by first graphing.
- All roots will become double roots.
- All stationary (turning) points will remain stationary
- All discontinuities remain.
- Values of horizontal and oblique asymptotes squared.
- If then , i. above original.
- If then , i. below original.
- Whenever new curve also
Division of Graphs
- can be thought of as
and same procedures as multiplication can be
followed except; - x intercepts of become vertical asymptotes or point discontinuities. - Find horizontal/oblique asymptotes, look at dominance, and check if curve cuts the asymptote. - Curve sticks to asymptotes except for when it cuts horizontal/oblique asymptotes.
Below, graphed against and
Absolute Value Graphs
- , reflect in the x-axis.
- , symmetry in y-axis, 1st quadrant reflected in 2nd.
- , symmetry in x-axis, reflect in x-axis.
- , symmetry in both x and y axes, 1st quad into all 4
- , symmetry in y axis then reflection in x axis.
- , symmetry in x axis then reflection in x axis.
- , symmetry in y axis then reflection in x axis, then symmetry in x-axis, as shown below.
Square Root Graphs
- can be sketched by first drawing and seeing:
- is only defined if
- for all in the domain.
- Stationary points must still be stationary points
- All discontinuities will remain.
- Horizontal/oblique asymptotes may change (value rooted)
- if i. new curve below old curve.
- if i. new curve above old curve.
- All remain at 1.
- intercepts require close inspection: for ,
- if , curve is concave down in 1st quadrant, with a vertical tangent at the x intercept.
- If curve is concave up in 1st quadrant, with a horizontal tangent at x intercept.
- is simply a reflection of over the x-axis.
Inverse Relations
- The inverse relation returns a number to where it came from.
- Found by swapping variables, therefore:
- Domain of relation is range of inverse relation
- Range of relation is domain of inverse relation.
- A relation and its inverse reflect each other in
- Found by swapping variables, therefore:
Inverse Functions
- If there exists a one-to-one relationship between the two sets, then both the relation and the inverse relation are functions. - Inverse relation here is called inverse function. - Notated as
- To test if a relation has an inverse function:
- Passes both vertical and horizontal line test OR
- When rewritten as , only has one value (unique).
- Each composite of a function and its inverse sends every number for which it is defined back to itself:
- for all in domain of
- for all in domain of
- An identity function is a function whose output is the same as input: for all in its domain.
- When a function is many-to-one, you can restrict the domain of the function so that it is able to have an inverse function. - E., if we restrict domain of to i. , then has in inverse function.
####### -
y = [ f ( x )] 2 y = f ( x )
| f ( x )|> 1 [ f ( x )] 2 > f ( x )
| f ( x )|< 1 [ f ( x )] 2 < f ( x )
f ( x ) =| 1 |, = 1
y = f ( x ) g ( x ) f ( x )× 1 g ( x )
g ( x )
y = sin x x y = sin x y = x
y =| f ( x )| f ( x ) < 0 y = f (| x |) | y |= f ( x ) f ( x ) > 0 | y |= f (| x |) y =| f (| x |)| | y |=| f ( x )| | y |=| f (| x |)|
y = f ( x ) f ( x ) f ( x )≥ 0 f ( x )≥ 0 x
f ( x ) < f ( x ) f ( x ) > 1 f ( x ) > f ( x ) f ( x ) < 1 f ( x ) = 1 x y = x
a b a b < 1
a b
> 1,
y 2 = f ( x ) y = f ( x )
y = x
f − 1 ( x )
x = f ( y ) y = f ( x ) y = f ( x )
f − 1 ( f ( x )) = x x f ( x ) f ( f − 1 ( x )) = x x f − 1 ( x )
I ( x ) = x x
f ( x ) = x 2 x ≥ 0 g ( x ) = x 2 where≥ 0 g ( x ) g − 1 ( x ) = x
(blue) where (black)
y = [ f ( x )] 2 f ( x ) = 2 + 3 ( x + 2)( x −1)
Graph should have point discontinuity at x = 0
Trigonometric Identities
For any angle , the ratio identities:
(provided that )
(provided that )
For any angle , the pythagorean identities:
####### -
- (provided that )
- (provided that )
- For any angle , the complementary angle identities:
####### -
- (provided that is defined)
- (provided that is defined)
Sine Rule and Area Formula
In
In ,
More generally, in any :
- Used to find side when two angles + one side are known, or to find angle when two sides + one angle are known.
Ambiguous case may arise when two sides + non-included angle:
- Since the supplement of , is also positive, it may be a plausible answer in some cases.
- To check if supplement is valid, add to the given angle, and if the result is , then the supplement is also valid.
From triangle above, area = , and we know.
Cosine Rule
In
In
To find third side given two sides and included angle:
To find angle given three sides:
Topic 9: Coordinate Geometry
Basic Formulae
General form: where are constants.
To find gradient:
Gradient-intercept form:
Point-slope form:
Eq. through is
Length of a segment:
Midpoint of a segment:
For two parallel lines,
To prove perpendicularity,
To prove collinear points :
Perpendicular distance between a point and line :
Sufficiency Conditions for Shapes
- Kite
- Two pairs of adjacent sides are equal.
- One diagonal bisects the other diagonal at right angles.
- Trapezium
- One pair of sides are parallel.
- Parallelogram
- Both pairs of opposite sides are equal
- Both pairs of opposite sides are parallel
- Diagonals bisect each other.
- One pair of opposite sides equal and parallel.
- Both pairs of opposite angles are equal.
- Rhombus
- Diagonals bisect at 90 degrees.
- All sides equal.
- Any condition for parallelogram + one pair of adjacent sides are equal.
- Rectangle
- Any condition for parallelogram + one angle is 90 degrees.
- Angles at vertices are 90 degrees.
- Any condition for parallelogram + diagonals are equal.
- Square
- 4 equal sides and one angle is 90 degrees.
- Diagonals equal and bisect at 90 degrees.
Equation of Line Through Point and
Intersection of Another Two Lines
For example, : , then subs.
Substituting back into equation,
θ sin θ cos θ
= tan θ cos θ ≠ 0 cos θ sin θ = cot θ sin θ ≠ 0 θ sin 2 θ + cos 2 θ = 1 tan 2 θ + 1 = sec 2 θ cos θ ≠ 0 cot 2 θ + 1 = csc 2 θ sin θ ≠ 0 θ cos(90 o − θ ) = sin θ cot(90 o − θ ) = tan θ tan θ csc(90 o − θ ) = sec θ sec θ
△ C B M , h a = sin B → h = a sin B △ C A M h b = sin A → h = b sin A ∴ a sin A = b sin B △ A B C a sin A = b sin B = c sin C
sin θ sin(180− θ )
< 180 o 1 2 c h h = b sin A
∴Area= 1 2 b c sin A
△ B P C , a 2 = ( b − x ) 2 + h 2 a 2 = b 2 − 2 x b + x 2 + h 2 △ B PA , x 2 = c 2 − h 2 cos A = x c → x = c cos A
a 2 = b 2 − 2 b c cos A + c 2 − h 2 + h 2 ∴ a 2 = b 2 + c 2 − 2 b c cos A
a 2 = b 2 + c 2 − 2 b c cos A cos A = b
2 + c 2 − a 2 2 b c
a x + b y + c = 0 a , b , c m =
y 2 − y 1 x 2 − x 1 y = m x + b y − y 1 = m ( x − x 1 ) ( x 1 , y 1 ), ( x 2 , y 2 ) y − y 1 =
y 2 − y 1 x 2 − x 1 ( x − x 1 )
d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 M = (
x 1 + x 2 2 ,
y 1 + y 2 2 ) m 1 = m 2 m 1 × m 2 =− 1 → m 1 =− 1 m 2 a , b , cmab = mbc ( x 1 , y 2 ) a x + b y + c = 0 d =
| a x 1 + b y 1 + c |
a 2 + b 2
a 1 x + b 1 y + c 1 + k ( a 2 x + b 2 y + c 2 ) = 0 2 x + y + 1 = 0, 3 x + 5 y −9 = 0, (1,2) 2 x + y + 1 + k (3 x + 5 y −9) = 0 (1,2) 2 + 2 + 1 + k (3 + 10−9) = 0→ 4 k =− 5 → k =− 5 4
2 x + y + 1− 5 4
(3 x + 5 y −9) = 0 7 x + 21 y −49 = 0→ x + 3 y −7 = 0
Term One
Finished!
Division of Line Segment in a Ratio
- is divided as shown. To find ,
We know that the red and green triangles are similar, as they have two corresponding angles equal, and therefore their corresponding sides are proportional. P can be defined as away from A.
Thus,
Similarly, solving for y gives
Therefore,
Topic 10: Discrete Probability Distributions
Probability Distributions
- A probability distribution is a set of all possible outcomes,
together with their corresponding probabilities. - Outcomes in probability distributions known as values.
- Probability distributions whose values are all counting numbers
and can be listed are known as discrete probability distribution. - Even though whole numbers and integers are technically countably infinite, they can still be “listed”.
- Continuous probability distributions include all real numbers as
values, and therefore have an infinite number of possible values. - Thus, the probability of a particular value occurring is zero. - Therefore, probability is recorded by measuring the probability that the random variable lies within an interval, e. , but this requires integration.
Random Experiments & Random Variables
If experiment has more than one outcome - random experiment
is the random variable, the result of a random experiment.
OR denotes when results in a value of.
In discrete probability distributions,
If experiment only has one outcome - deterministic experiment.
Thus, one certain outcome (e. ) only.
Expected Value
- Expected value calculated by weighting each value by their
probability. - Thus it is calculated as - It is also alternatively notated as μ, meaning “mean”.
Laws of Expectation
- If and are constants, Proof:
Since and
Variance
- Variance (Var ) is a measure of spread about the mean.
- Take deviation of each value and square:
- Squaring makes the deviation a positive number or zero.
- The square gets larger very quickly as deviation increases.
- Then take the weighted mean of the squared deviations for each value:
- Take deviation of each value and square:
- Thus, also expected value:
- The units of variance are square of whatever units values have.
Variance - Alternative Formula
- To conjure up an alternative formula for easier calculations:
Since and ,
And therefore also equals.
Uniform Distribution
For a uniform distribution , where all values have equal probability of :
Standard Deviation
- Square root of the variance: or.
- Spreading out all values by factor of , results in.
- Adding constant amount to each data value doesn’t affect spread.
- However, if each value of is divided by in , i ,
: spread is narrower by factor.
- If we have , does not affect the spread, and the spread must be positive, therefore
Sample Distribution
- A census performs an experiment on everything, while survey sample only some of the population.
- In simulations/trials, is replaced by (relative frequency).
The mean is denoted by , and corresponds to expected value:
Sample variance is and sample standard deviation is.
A B m : n P
m m + n
× A B
x = x 1 + m m + n ( x 2 − x 1 )
=
x 1 ( m + n ) + m x 2 − m x 1 m + n =
m x 2 + n x 1 m + n
y = m y 2 + n y 1 m + n P = (
x 2 m + x 1 n m + n ,
y 2 m + y 1 n m + n )
P (55≤ X ≤60)
X P ( X = x ) P ( x ) X x ∑ P ( X = x ) = 1
4 P ( X = 4) = 1
E ( X ) =∑ x p ( x )
a b E ( a X + b ) = a E ( X ) + b E ( a X + b ) =∑( a x + b ) p ( x ) =∑ a x p ( x ) +∑ b p ( x ) = a ∑ x p ( x ) + b ∑ p ( x )
∑ x p ( x ) = E ( X ) ∑ p ( x ) = 1 = a E ( X ) + b E ( X + Y ) = E ( X ) + E ( Y )
( X ) ( x − μ ) ( x − μ ) 2
Va r ( X ) =∑( x − μ ) 2 p ( x ) Va r ( X ) = E (( X − μ ) 2 )
Va r ( X ) =∑( x − μ ) 2 p ( x ) =∑( x 2 − 2 μx + μ 2 ) p ( x ) =∑ x 2 p ( x )−∑ 2 μx p ( x ) +∑ μ 2 p ( x ) =∑ x 2 p ( x )− 2 μ ∑ x p ( x ) + μ 2 ∑ p ( x )
∑ p ( x ) = 1 ∑ x p ( x ) = μ Va r ( X ) =∑ x 2 p ( x )− 2 μ 2 + μ 2 =∑ x 2 p ( x )− μ 2 Va r ( X ) E ( X 2 )− μ 2
X = 1,2,3,..., n P ( x ) = 1 n E ( X ) = n + 1 2 Va r ( X ) = n 2 − 1 12
σ = Va r ( X ) σ 2 = Va r ( X ) k k × σ
X k Z Z = X k Va r ( Z ) = Va r ( X k
)= E [( X k
) 2 ]−( μ k
) 2
= E ( X 2 ) k 2
− μ 2 k 2
= E ( X 2 )− μ 2 k 2 = Va r ( X ) k 2
= σ 2 k 2
= ( σ k ) 2 k
σ ( a X + b )+ b σ ( a X + b ) =| a | σ ( X )
p ( x ) fr x x =∑ x fr s 2 s s 2 =∑( x − x ) 2 fr =∑ x 2 fr − x 2
Differentiating Inverse Functions
- For inverse functions and ,.
E. differentiate by differentiating inverse function:
Solving for ,.
We want to find derivative of with respect to , so
, and therefore , thus.
Differentiating Non-Cardinal Powers
- We can prove the power rule works for a power of by
using first-principles differentiation. and
, which is indeed the same result as when power rule is used.
- Power rule also works for fractional indices.
- We can then prove that the power rule works on all negative
integer indices. We have , where is an integer.
Using the chain rule, where ,
as required.
Dividing Through by Denominator
- If the denominator of a function is a single term, first divide through by it, then differentiate.
However, it is just as practical to use to quotient rule, seen later.
Domain of
When is negative, cannot equal zero.
When is a fraction , where is an even number, cannot be negative.
When is irrational, cannot be negative.
Product Rule
- Product Rule: or
Proof: We have. If , then , , , i.
so as ,
as required.
Can also use when is square root, e.
Quotient Rule
- For a function where are both functions of ,
or more concisely,
Proof using product rule: let , let
Multiplying top and bottom of each expression by :
as required.
Reciprocal Rule
- For a function where is a constant, To differentiate
Rates of Change
At three arbitrary points on a function :
- If , then is increasing.
If , then is stationary, i. is a stationary point.
If , then is decreasing.
The average rate of change between two points and is the slope of the secant :
Displacement, Velocity and Acceleration
Displacement : Distance from a point, with direction.
Velocity : The rate of change of displacement with respect to time, i. speed with direction.
Acceleration ( ): The rate of change of velocity with respect to time, i. second derivative of.
Differentiability
A graph with no vertical asymptotes, point discontinuities, or any other gaps in the domain/range are said to be continuous.
To be smooth continuous, function cannot have cusps or corners.
A function is differentiable at a point if the curve is smooth continuous and the tangent is not vertical at the point.
For example, is not differentiable at as it is impossible to draw a tangent.
Similarly is not differentiable at as the tangent is vertical.
also not differentiable at as there is a cusp.
Implicit Differentiation
- If you don't have a function, e. , use. , but cannot be differentiated in respect to
But implicit differentiation gives
. The derivative is in terms of as two
points would result if it was in terms of.
f ( x ) = y f ( y ) = x d y d x
× d x d y
= 1
y = 3 x xx = y 3 ∴ d x d y = 3 y 2
y x d y d x
= 1 3 y 2
y = 3 x y 2 = 3 x
2 d y d x = 1 33 x
2
n =− 1
f ( x ) = 1 x
f ( x + h ) = 1 x + h
f ′ ( x ) = lim
h → 0
1 x + h − 1 x h = lim h → 0
x − x − h h x ( x + h ) = lim h → 0
− 1 x ( x + h ) = − 1 x 2
f ( x ) = 1 xm m ≥ 2
y = 1 u , u = xm d y d x = d y d u × d u d x =− 1 x 2 m
× m xm − 1 =− m x − m − 1
f ( x ) = 5 x 3 + 2 x 2 + 4 3 x 2
= 5 3 x + 2 3 + 4 3 x − 2
∴ d y d x = 5 3 −2( 4 3 ) x − 3 = 5 3 − 8 3 x 3
= 5 x 3 − 8 3 x 3
xn
n x n m k k x
n x
d d x ( u v ) = u d v d x
- v d u d x
y ′ = u v ′ + v u ′
y = u v x → x + xδ y → y + yδ u → u + uδv → v + vδ ∴ y + yδ = ( u + uδ )( v + vδ ) y + yδ = u v + u vδ + u vδ + uδvδ → yδ = u vδ + v uδ + uδvδ yδ xδ
= uvδ xδ
vuδ xδ
uδ xδ
× vδ xδ
× xδ xδ → 0 d y d x = u d v d x
- v d u d x
- 0 u , v y = 2 x 2 x − 1 = 2 x (2 x −1)
1 2 → d y d x = (2 x )[ 1 2 (2 x −1)−
1 2 (2)] + (2 x −1)
1 2 (2)
= 2 x (2 x −1) − 12 + 2(2 x −1)
1 2 = 2(2 x −1)−
1 2 [ x + (2 x −1)] = 2(2 x −1) − 12 (3 x −1)→ ∴ d y d x = 2(3 x −1) 2 x − 1
y = u v
u , v x
d y d x =
vdud x − udvd x v 2
y ′ =
v u ′ − u v ′
v 2 y = u v − 1 U = u , V = v − 1 d y d x = V d U d x
- U d V d x = v − 1 d u d x − u v − 2 d v d x v 2 _d yd x_
vdu d x − udv d x v 2
f ( x ) = k v k d d x ( k v ) =
− k v ′
v 2 y = 6 4 x 2 + 3
→ d y d x = −6(8 x ) (4 x 2 + 3) 2
= − 48 x (4 x 2 + 3) 2
P ( x , f ( x )) f ( x ) d y d x
> 0 f ( x ) d y d x = 0 f ( x ) P d y d x < 0 f ( x ) P ( t 1 , Q 1 ) Q ( t 2 , Q 2 ) P Q = mPQ = Q 2 − Q 1 t 2 − t 1
( x ) ( v , x ·, d x d t )
a , v ·,·· x , d v d t , d 2 x d t 2 f ( x )
y =| x − 1 | x = 1
x = y 2 x = 0 y = ( x −2)
2 3 x = 2
x = y 2 d f d x = d f d y × d y d x d d x ( x ) = d d x ( y 2 ) y 2 x d d x ( y 2 ) = d d y ( y 2 )× d y d x ∴1 = 2 y d y d x → d y d x = 1 2 y y
x
- Find equation of tangent to at the point
Differentiate equation
. Both and are in the derivative as you need both
values to find a single point. Subbing point ,
Therefore
Thus, the equation of the tangent is
Topic 12: Polynomials
Polynomial Basics
- A real polynomial of degree is expressed as where and and is an integer.
- Degree (order): the highest index of the polynomial.
- Leading term: and leading coefficient:
- Monic polynomial: leading coefficient is equal to one.
- is polynomial equation.
- is polynomial function.
- Roots: solutions to
- Zeroes: intercepts of
Graphing Polynomials
- When drawing
- intercept is the constant
- intercepts are the roots, found using
- As , acts like function of leading term alone. Proof: Let
Then
Therefore, , showing that for large values of ,.
is the same sign as
- Even powered roots are in shape of a parabola.
- Odd powered roots are in shape of a cubic.
In this graph, is a root of multiplicity 4, is root of multiplicity 3, and is root of multiplicity 2.
Polynomial Division
- where is divisor, is quotient
- E. divided by First take the from and divide into the leading term , creating at the top. Now times by giving , and then subtract. Repeat until you cannot divide into the next term. , 5.
Remainder Theorem
- If a polynomial is divided by , Proof:. Let ,
. However, since is a divisor of degree 1, must be degree 0 or simply zero, so we can rewrite it as.
Factor Theorem
- If is a factor of , then. Converse also true. To factorise , the constant factors must be a factor of the constant (i. 36). These are thus , which can also be negative. Also possible to be fractional factors of the form. We test these, and we find , and thus is factor.
Polynomial Rules
- If has distinct real zeros, , is a factor of i. if and are zeros of then is factor.
- If has degree and has distinct real zeros then i. cannot have more linear factors than the degree of polynomial
- If has degree and has more than real zeros, then is the zero polynomial, i. for all
- If (identically equal), then the coefficients of each corresponding term must be equal.
- If degree polynomials for values of , then they are identically equal.
Sum and Product of Roots
- If and are the roots of , then:
and therefore:
and
For a cubic :
and and
For a quartic :
and and
and
Therefore, in general, for
(sum of roots, one at a time)
(sum of roots, two at a time)
(sum of roots, three at a time)
(sum of roots, four at a time)
Furthermore, due to the expansion of a perfect square,
x 2 + y 2 = 9 (1,2 2) → 2 x + 2 yd y d x
= 0→ 2 yd y d x
=− 2 x
∴ d y d x =− x y x y
(1,2 2) d y d x =− 1 2 2 y − 2 2 =− 1 2 2
( x −1)→ 2 2 y −8 =− x + 1
x + 2 2 y = 9 = 0
P ( x ) n P ( x ) = p 0 + p 1 x + p 2 x 2 +... + pn − 1 xn − 1 + pnxn pn ≠ 0 n ≥ 0
pnxn pn
P ( x ) = 0 y = P ( x ) P ( x ) = 0 x y = P ( x )
y = P ( x ) y x P ( x ) = 0 x →±∞ P ( x ) P ( x ) = anxn + an − 1 xn − 1 +... + a 1 x + a 0 P ( x ) xn = an +
an − 1 x +... + a xn − 1 +
a 0 xn lim x →±∞
P ( x ) xn = an x
P ( x ) an
x = 1 x =− 1 x =− 2
P ( x ) = A ( x ) Q ( x ) + R ( x ) A ( x ) Q ( x ) d e g ( R ( x )) < d e g ( A ( x )) x 3 + 2 x 2 − 4 x − 3 x x − 3 x 3 x 2 x − 3 x 2 x 3 − 3 x 2
Q ( x ) = x 2 + x + 3 R ( x ) =
P ( x ) ( x − a ) R ( x )= P ( a ) P ( x ) = A ( x ) Q ( x ) + R ( x ) A ( x ) = ( x − a ) P ( x ) = ( x − a ) Q ( x ) + R ( x )→ P ( a ) = ( a − a ) Q ( a ) + R ( a ) = R ( a ) A ( x ) R ( x ) r ∴ r = P ( a )
( x − a ) P ( x ) P ( a ) = 0 P ( x ) = 4 x 3 − 16 x 2 − 9 x + 36
1,2,3,4,6,9,12,18, factors of constant factors of leading coefficient P (4) = 0 ( x −4) P ( x ) = ( x −4)(4 x 2 −9) = ( x −4)(2 x + 3)(2 x −3)
P ( x ) k a 1 , a 2 , a 3 ,... , ak ( x − a 1 )( x − a 2 )( x − a 3 )... ( x − ak ) P ( x ) 1 − 2 P ( x ) ( x −1)( x + 2) P ( x ) n n a 1 , a 2 , a 3... , an P ( x ) = ( x − a 1 )( x − a 2 )... ( x − an )
P ( x ) n n P ( x ) P ( x ) = 0 x P ( x )≡ Q ( x )
n P ( x ) = Q ( x ) n + 1 x
α β a x 2 + b x + c = 0 a x 2 + b x + c = a ( x − α )( x − β ) = a ( x 2 − αx − βx + α β ) x 2 + b a x + c a = x 2 −( α + β ) x + α β
α + β =− b a
α β = c a a x 3 + b x 2 + c x + d = 0 α + β + γ =− b a α β + α γ + β γ = c a α β γ =− d a a x 4 + b x 3 + c x 2 + d x + e = 0 α + β + γ + δ =− b a α β + α γ + α δ + β γ + β δ + γ δ = c a α β γ + α β δ + α γ δ + β γ δ =− d a α β γ δ = e a a xn + b xn − 1 + c xn − 2... = 0
∑ a =−
b a ∑ α β =
c a
∑ α β γ =−
d a ∑ α β γ δ =
e a
∑ α
2 = ( ∑ α )
2 − 2 ∑ α β
y = ( x −1) 4 ( x + 1) 3 ( x + 2) 2
Topic 15: Trigonometry II
Radian Measure
radians, thus
Degrees to radians,
Radians to degrees,
Arcs and Sectors
- We know , to find arc ,
Since where in degrees
Thus, where is in radians.
- Further, , segment :
where is in radians
- To find area of minor segment ,
Graphing Trig + Inverse Functions
- Refer to Topic 8 (convert to radians) and Topic 7 respectively.
Inverse Trig Functions
- For , As is not one-to-one, there is no inverse. But if we restrict domain , then we have
an inverse. Thus : - Domain: - Range:
- For ,
As is not one-to-one, there
is no inverse. But if we restrict domain
, then we have an inverse.
Thus :
- Domain:
- Range:
- For , As is not one-to-one, there is no inverse. But if we restrict domain (remember
),
then we have an inverse. Thus :
- Domain: all real
- Range:
As seen from the graphs:
(odd function)
(odd function shifted up )
(odd function)
( )
Inverse trig sometimes shown as etc.
: homogeneous of degree 2 in and
Manipulating Inverse Trig Graphs
Graph : Thus we establish
Domain:
Range: and graph the curve adhering to the new domain/range.
Graph : Thus we establish
Domain:
Range test endpoints: . Then test midpoint: . Therefore
Graph (as a relation ).
- Domain: all real
- Range: , therefore graph cannot simply be Instead, it zigzags between for all real.
Graph
- Domain: (domain of )
- Range (test endpoints):
, and then midpoint , therefore.
Proving Trig Symmetry and Identity
- To prove , for . We know that Therefore since. , as required.
- To prove for , It is evident that and are reflections of each other in the line , but algebraically, Let. for. Since , , because , and since
180 o = π 30 o = π 6 45 o = π 4 xo × π 180 xπ × 180 π
C = 2 πr l l = θ 2 π × 2 πr θ l = rθ θ A = πr 2 O A B AOA B = θ 2 π × πr 2 = 1 2 r 2 θ θ
A B 1 2 r 2 θ − 1 2 r 2 sin θ
y = sin− 1 x y = sin x
− π 2 ≤ x ≤ π 2 y = sin− 1 x − 1 ≤ x ≤ 1 − π 2 ≤ y ≤ π 2 y = cos− 1 x y = cos x
0 ≤ x ≤ π y = cos− 1 x − 1 ≤ x ≤ 1 0 ≤ y ≤ π y = tan− 1 x y = tan x
− π 2 < x < π 2 x ≠ π 2
- πk , k ∈ ℤ
y = tan− 1 x x − π 2 < y < π 2
sin− 1 (− x ) =−sin− 1 x cos− 1 (− x ) = π −cos− 1 x π 2 tan− 1 (− x ) =−tan− 1 x sin− 1 x + cos− 1 x = π 2 ∠sum△ arcsin x , arccos x , arctan x cos 2 x + sin 2 x = 1 sin x cos x
y = 5 sin− 1 x 3 − 1 ≤ x 3
≤ 1 → − 3 ≤ x ≤ 3
− π 2 ≤ y 5 ≤ π 2 → − 5 π 2 ≤ y ≤ 5 π 2
y = tan− 1 ( 3 − x 2 ) 3 − x 2 ≥ 0 → − 3 ≤ x ≤ 3 x = 3 → y = tan− 1 0 = 0 x =− 3 → y = tan− 1 0 = 0 x = 0→ y = tan− 1 3 = π 3 0 ≤ y ≤ π 3
y = sin− 1 sin x sin y = sin x − 1 ≤sin x ≤ 1 → x − π 2 ≤ y ≤ π 2 y = x − π 2 ≤ y ≤ π 2 x
y = sin sin− 1 x − 1 ≤ x ≤ 1 sin− 1 x
x = 1→ y = sin sin− 1 1 = 1 x =− 1 → y = sin sin− 1 (−1) =− 1 x = 0→ y = sin sin− 1 0 = 0 − 1 ≤ y ≤ 1
cos− 1 (− x ) = π −cos− 1 x y = cos− 1 (− x )→ − x = cos y 0 ≤ y ≤ π ∴ x =−cos y cos( π − y ) =−cos y cos( π − y ) = x → π − y = cos− 1 x 0 ≤ π − y ≤ π ∴ y = π −cos− 1 x sin− 1 x + cos− 1 x = π 2 − 1 ≤ x ≤ 1 y = cos− 1 x y = sin− 1 x y = π 4
a = cos− 1 x ∴cos a = x 0 ≤ a ≤ π sin( π 2 − a ) = cos a sin( π 2 − a ) = x ∴sin− 1 x = ( π 2 − a ) − π 2 ≤ π 2 − a ≤ π 2 sin− 1 x + a = π 2
a = cos− 1 x
sin− 1 x + cos− 1 x = π 2
Range should stop at − π 2 ≤ y ≤ π 2
Range should stop at 0 ≤ y ≤ π
Graph identical for y = cos cos− 1 x
Compound Angle Formulae
- If we wanted to find , refer to
the unit circle. We know: and Thus, using distance formula,
And, using the cosine rule:
Thus, equating the two:
- By replacing with , we can find ,
. Cos is even, and sin is odd, thus
- Using the identity , we can find ,
As just proven, we get
Replacing with , we get for :
To derive ,
We know
By dividing top and bottom by , we get:
Since we know tan is odd function, replacing with we get:
Double-Angle Formulae
By replacing both angles of compound angle formulae with ,
We can use the identity to create:
as well as,
The t-formulae
- To represent as algebraic functions,
Let. Thus using the double angle formula we get,
- , from which we draw a right angle triangle:
Therefore AC. We thus get:
Products to Sums
We know the four compound-angle formulae of sine and cosine:
Adding and subtracting the first two, and doing the same for the latter two yields the four product to sum identities: -
####### •
Sums to Products
Product to sum formulae, sub in and. For , we get: , rearranged as
This process is repeated for the other three products-to-sums, we get:
Topic 16: Binomials and Pascals
Binomial Expansions
- A binomial expression is one which contains two terms.
- When the coefficients in the expansions of are arranged in a table, we get Pascal’s triangle. - Each row is symmetric/reversible. - Every number (except 1’s) is the sum of the two numbers above.
- To expand , there will be terms starting with , and gradually working through coefficients of th row until.
- If positive, all terms positive. If negative, terms alternate sign.
General Binomial Expansions
can be expanded with beginning, ending with , with coefficients of middle terms determined by Pascal’s triangle.
In each term, the sum of indices of and is.
is the coefficient of in
Thus,
From Pascal’s triangle, we also get:
where. Proving,
We see that and are the only terms with , and thus the coefficient will be.
cos( α − β )
A = (cos a , sin a ) B = (cos β , sin β )
A B 2 = (cos α −cos β ) 2 + (sin α −sin β ) 2
= cos 2 α −2 cos α cos β −cos 2 β + sin 2 α −2 sin α sin β + sin 2 β A B 2 = 2−2 cos α cos β −2 sin α sin β c 2 = a 2 + b 2 − 2 a b cos C A B 2 = 1 + 1−2 cos( α − β )
2 −2 cos α cos β −2 sin α sin β = 2−2 cos( α − β ) cos( α − β ) = cos α cos β + sin α sin β β − β cos( α + β ) = cos α cos(− β ) + sin α sin(− β ) cos( α + β ) = cos α cos β −sin α sin β sin θ = cos( π 2
− θ ) sin( α + β )
sin( α + β ) = cos( π 2 −( α + β )) = cos(( π 2 − α )− β ) cos( π 2 − α )cos β + sin( π 2 − α )sin β sin( α + β ) = sin α cos β + cos α sin β β − β sin( α − β ) sin( α − β ) = sin α cos β −cos α sin β tan( α + β ) tan( α + β ) = sin( α + β ) cos( α + β ) = sin α cos β + cos α sin β cos α cos β −sin α sin β cos α cos β tan( α + β ) = tan α + tan β 1 −tan α tan β β − β tan( α − β ) = tan α −tan β 1 + tan α tan β
θ sin( θ + θ ) = sin θ cos θ + cos θ sin θ sin 2 θ = 2 sin θ cos θ cos( θ + θ ) = cos θ cos θ −sin θ sin θ cos 2 θ = cos 2 θ −sin 2 θ sin 2 θ + cos 2 θ = 1 cos 2 θ = 2 cos 2 θ − 1 cos 2 θ = 1−2 sin 2 θ
tan( θ + θ ) = tan θ + tan θ 1 −tan θ tan θ tan 2 θ = 2 tan θ 1 −tan 2 θ
sin θ , cos θ , tan θ tan 1 2 θ = t
tan θ = 2 t 1 − t 2 h 2 = 4 t 2 + 1− 2 t 2 + t 4 = ( t 2 + 1) 2 = 1 + t 2 sin θ = 2 t 1 + t 2 cos θ = 1 − t
2 1 + t 2
sin( α + β ) = sin α cos β + cos α sin β sin( α − β ) = sin α cos β −cos α sin β cos( α − β ) = cos α cos β + sin α sin β cos( α + β ) = cos α cos β −sin α sin β
2 sin α cos β = sin( α + β ) + sin( α − β ) 2 cos α sin β = sin( α + β )−sin( α − β ) 2 cos α cos β = cos( α + β ) + cos( α − β ) −2 sin α sin β = cos( α + β )−cos( α − β )
A = α + β B = α − β 2 sin α cos β = sin( α + β ) + sin( α − β ) 2 sin 1 2 ( A + B )cos 1 2 ( A − B ) = sin A + sin B
sin A + sin B = 2 sin 1 2 ( A + B )cos 1 2 ( A − B )
sin A −sin B = 2 cos 1 2 ( A + B )sin 1 2 ( A − B )
cos A + cos B = 2 cos 1 2 ( A + B )cos 1 2 ( A − B )
cos A −cos B =−2 sin 1 2
( A + B )sin 1 2
( A − B )
(1 + x ) n
(1 + x ) n n + 1 1 n + 1 xn x x
( x + y ) n xn yn
x y n nC k =(
n k ) xk (1 + x ) n
(1 + x ) n =( n 0 ) +( n 1 ) x +( n 2 ) x 2 +... +( n n ) xn
( a + b ) n =( n 0 ) an +( n 1 ) an − 1 b +( n 2 ) an − 2 b 2 +... +( n n ) bn
nC k = n − 1 Ck − 1 + n − 1 Ck 1 ≤ k ≤ n − 1 (1 + x ) n = (1 + x )( n − 1 C 0 + n − 1 Ck − 1 xk − 1 + n − 1 Ckxk... n − 1 Cn − 1 xn − 1 ) (1)( n − 1 Ckxk ) ( x )( n − 1 Ck − 1 xk − 1 ) xk n − 1 Ck + n − 1 Ck − 1
- E. On an island, the population in 1960 was 1732 and in 1970 it
was 1260, find annual growth rate of nearest % assuming it is proportional to population. and , we know thus.
When , i.
Thus,
growth rate is %.
- In how many years will the population be half that in 1960?
Let as , thus.
, and we know
.
Thus, it takes 22 years to halve the population.
Modified Growth and Decay
- We must change the equation modelling growth and decay in
order to take into conditions such as temperature barriers.
, where is a fixed constant barrier.
The solution is thus where
- population at time
- = growth/decay constant
- initial population
- = time
Proof that given :
Let be difference between and.
Thus because is constant.
Therefore since
This is the same as the ordinary exponential model, so we can use i. and substitute in: as required.
Newton’s law of cooling states that the rate of change of temperature is proportional to , where is indoor temperature and is constant outdoor temperature, i
E. the outdoor temperature is and a heater malfunction has
caused the inside temperature to drop from to in half an hour. After how many hours is inside temperature but when and thus but we are told when i. , thus , and then log both sides:
We want to know what is when , thus.
We get thus but we know ,
= 2...
Therefore after about hours, the temperature dropped to.
d P d t
= k P P = A ek t A = 1732 P = 1732 ek t t = 10, P = 1260 1260 = 1732 e 10 k 10 k = log 1260 1732 → k = 1 10 log 1260 1732 =−0(4 d p )
∴ − 3
866 = 1732 ek t 1732 2
= 688 ek t = 1 2
k t = log 1 2 → t = 1 k log 1 2 k = 1 10 log 1260 1732 ∴ t = 1 1 10 log
1260 1732
×log 1 2
= 21.
d P d t
= k ( P − N ) N P = N + A ek t P = t k N + A = t P = N + A ek t d P d t = k ( P − N ) y = P − N P N d y d t = d P d t − 0 N d y d t = k ( P − N ) = k y y = P − N
P = A ek t y = y 0 ek t y = P − N
∴ P = N + A ek t
( T − A ) T A d T d t = k ( T − A ) 5 oC 20 oC 17 oC 10 oC? ∴ T = 5 + A ek t t = 0, T = 20 A = 15 T = 5 + 15 ek t t = 0, T = 17
17 = 5 + 15 e 0 k e 0 k = 12 15
0 k = log 12 15 → k = 2 log 12 15 t T = 10 10 = 5 + 15 ek t 1 3
= ek t log 1 3
= k t k = 2 log 12 15
∴ t =
log 13
2 log 1215
2 1 2 10 oC
nomnom...
Year 11 Maths for highschool studies
Subject: math advanced
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