Maths Ext Year Eleven
Topic 1: Methods in Algebra
Index Laws
- a m × a n = a (m+n)
- a m ÷ a n = a (m−n)
- (a m ) n = a mn
1
−1
- a =a
- a0 = 1
a
ax
a
b
x
- ( b ) = bx
−1
- (b) = a
a
- x b = b x a OR ( b x )a
Binomial Products
In general,
(a + b + c + . . . ) 2 = (a 2 + b 2 + c 2 + . . . + 2a b + 2a c + 2b c + . . . )
Thus
- (a + b)2 = a 2 + 2a b + b 2
- (a − b)2 = a 2 − 2a b + b 2
Difference of Two Squares:
- (a + b)(a − b) = a 2 − b 2
Algebraic Fractions
• Always factorise first.
• When adding/subtracting fractions, equate the denominator.
a b − 2b 2 a 2 − 4a b + 4b 2
e.g.
÷
3a
6a 2 b
b (a − 2b)
3a
=
×
6a 2 b
(a − 2b) 2
1
=
2a (a − 2b)
Quadratic Equations
• After factorising, if a b = 0 then a = 0 or b = 0
x 2 + 9x + 18 = 0
(x + 6)(x + 3) = 0
∴ x = − 6, − 3
• Use quadratic formula: x =
−b ±
b 2 − 4a c
2a
2 x 2 + 5x − 4 = 0
−5 ± 25 + 32
x =
4
−5 ± 57
x =
4
To derive the quadratic formula: a x 2 + b x + c = 0,
b
c
x + =0
a
a
b
b 2
b 2 c
x2 + x + (
) −(
) + =0
a
2a
2a
a
2
b 2
b
c
(x +
) −
+ =0
2a
a
4a 2
2
b 2
b − 4a c
(x +
) =
2a
4a 2
2
b
b − 4a c
x +
=±
2a
4a 2
x2 +
x =
−b ±
b 2 − 4a c
2a
b
2
• Complete the square: add ( 2 ) to both sides of equation.
Eg. x 2 + 6x − 7 = 0
x 2 + 6x + 9 − 7 = 9
(x + 3) 2 = 16
x +3= ±4
x = 1, − 7
Cubics
Simultaneous Equations
a 3 + b 3 + c 3 + . . . + 3a 2 b + 3a 2 c + 3a 2 d + 3b 2 a + . . . + 6a b c + 6a b d + . . .
• Amount of equations required = number of pronumerals.
• Eliminate a variable:
• 2 x + 3y = 21 → (1)
In general, (a + b + c + d . . . )3 =
Thus
- (a + b)3 = a 3 + 3a 2 b + 3a b 2 + b 3
- (a − b)3 = a 3 − 3a 2 b + 3a b 2 − b 3
In general,
a n − b n = (a − b)(a n−1 + a n−2 b + . . . + a b n−2 + b n−1)
And when n is odd,
a n + b n = (a + b)(a n−1 − a n−2 b + . . . − a b n−2 + b n−1)
Thus
- a 3 − b 3 = (a − b)(a 2 + a b + b 2 )
- a 3 + b 3 = (a + b)(a 2 − a b + b 2 )
Linear Equations
Make pronumeral the subject of formula:
3z + 2 = z − 9
11
z =−
2
5x + 2y = 3 → (2)
(1) × 2 and (2) × 3
4x + 6y = 42 → (3)
15x + 6y = 9 → (4)
(4) − (3)
11x = − 33
x =−3
Substitute x = − 3 into (2)
3y = 27
∴ x = − 3, y = 9
• Solving by substitution:
x = 7 → (1), y +
y = 9−
1
62
=
7
7
1
= 9 → (2)
x
1
Topic 2: Numbers and Surds
Real Numbers
- All real numbers can be placed on the number line.
- Subsets notated as ℝ, ℚ, ℤ, 𝕎 and ℕ.
a
- All rational numbers expressed as b , where a , b ∈ ℤ .
Recurring Decimals into Fractions
For 0.327 = 0.3272727..., let x = 0.327.
x = 0.3272727...
Then multiply x by 10 a, where a is amount of repeating digits, two in
this case.
100x = 32.727272...
99x = 32.4
324
18
x =
=
990
55
18
∴ 0.327 =
55
Representing Real Numbers
All real numbers can be described:
-
Geometrically:
Algebraically: x ≤ 2
Interval Notation: (−∞, 2 ]
Set Notation: {x = x ≤ 2}
Types of Intervals:
- Proof that 2 is irrational by contradiction:
Assume
2 is rational, thus
a2
a
2 = where a and b are coprime.
b
→ 2b 2 = a 2 → a 2 is even. ∴ a is even, a = 2k.
b2
2b 2 = (2k ) 2 → b 2 = 2k 2 → b 2 is even. ∴ b is even.
If both a , b are even this means they both have 2 as a factor, which
contradicts the assumption at the beginning that they are coprime.
Therefore, 2 is not rational.
2=
HCF and LCM
- For HCF, write both numbers in terms of prime factors, and then
multiply the common factors.
1176 : (3 × 23 × 72 ) and 1260 : (22 × 32 × 5 × 7)
HCF = 22 × 3 × 7 = 84
- For LCM, write both numbers in terms of prime factors, and then
multiply the highest factors.
1176 : (3 × 23 × 72 ) and 1260 : (22 × 32 × 5 × 7)
HCF = 23 × 32 × 5 × 72 = 17640
Bounded - two endpoints (e.g. 4 < x < 6 or − 2 ≤ x ≤ 4)
Unbounded - one endpoint (e.g. x ≥ 2 or x < 3)
Closed - all endpoints included (e.g. x ≥ 2 or 4 ≥ x ≥ − 2)
Open - all endpoints not included (e.g. 4 < x < 6 or x < 3)
Degenerate - a single point (e.g. x = 3)
•
•
•
•
•
Significant Figures
1.
2.
3.
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.g. 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)
4.
Zeroes at the end of number are significant if behind a decimal
point.
5.
Zeroes between any non-zero digit are significant.
e.g. 0.06700802 has 7 significant figures, 08200.02300 has 9 significant
figures.
Surd Operations
Divisibility Tests
-
1. All natural numbers.
2. Even number.
3. Digits sum to multiple of 3.
-
4. Last two digits divisible by 4.
5. Ends in 5 or 0.
6. Divisible by 2 and 3.
7. Double the last digit and subtract from the other digits, final
digit is divisible by 7.
Testing 2751: 1 × 2 = 2 → 275 − 2 = 273
3 × 2 = 6 → 27 − 6 = 21
1×2 = 2→2−2 = 0
∴ 2751 is divisible by 7
8. Last three digits are divisible by 8.
9. Sum of digits is multiple of 9.
differ by a multiple of 11.
Testing 4323: 3 + 3 = 4 + 2
∴ 4323 is a multiple of 11.
a
b =
=
b
ab
a
b
- ( a)2 = a BUT
a2 = | a |
• Like surds can be added or subtracted.
Rationalising the Denominator
surd
• If denominator is pure surd, multiply fraction by surd :
3
2 5
=
=
3
2 5
×
5
5
3 5
10
conjugate
• If denominator is surd binomial, multiply fraction by conjugate
10. Ends in a 0.
11. Sum of even positioned digits = sum of odd positioned digits, or
a×
where for the binomial a + b
2+
3
2−
3
=
=
2+
3
2−
3
×
2+
3
2+
3
x the conjugate is a − b
x.
4+4 3+3
4−3
=7+4 3
2
Topic 3: Function and Graphs
- If discriminant is not perfect square, roots are irrational.
- For f (x ) = a x 2 + b x + c, functions of a > 0 and a < 0 are
Functions
- For f (x ) = a x 2 + b x + c, functions of △ < 0 and △ > 0 are
Relation - a rule that maps between two sets of values (x 2 + y 2 = 5)
Function - relation, uniquely maps from a set to another. (x 2 = y)
Independent Variable - the “input” of the function.
Dependent Variable - the “output” of the function.
Natural Domain: all possible values of x that can be substituted.
Restricted Domain: domains restricted when some values redundant,
positive and negative functions, respectively.
indefinite and definite functions, respectively.
Cubic Function
The classic shape has a horizontal point of inflection and can be
factorised into perfect cube y = k (x − a)3.
Otherwise it is a continuous curve
with two turning points.
y = a x3 + b x2 + c x + d
e.g. t ≥ 0 as independent variable for negative time is not needed.
1
- Fractions influence the domain e.g. in y = x , x ≠ 0 as undefined.
- No negative roots, e.g. in y = x + 3 − 5 − x → x + 3 ≥ 0,
5 − x ≥ 0 → ∴ − 3 ≤ x ≤ 5 since roots cannot be negative.
Range - all possible values of dependent variable, obtained by
Polynomials
substituting every value in domain.
- Even powers and absolute values are always ≥ 0. y = x 2 thus
has a range of y ≥ 0 as x 2 cannot be negative.
Polynomials that can be
represented as y = k (x − a) n will
- When domain is restricted, sub in endpoints and centre of domain
to gauge range.
- If you have constant in fraction numerator, fraction ≠ 0.
• Functions usually notated as y = f (x ) where x is independent, y is
dependent.
• If a straight line is drawn parallel to y axis, it will only cross
function at most once.
Linear Functions
Data demonstrating direct variation will lie on straight line.
Its function is known as the linear function, which is expressed as
y = m x + b where m is slope and b is the y intercept.
- Lines parallel to x axis ( y = c).
- Lines parallel to y axis (x = k ).
The Quadratic Polynomial and Parabola
-
Quadratic polynomial: a x 2 + b x + c
Quadratic function: y = a x 2 + b x + c
Quadratic equation: a x 2 + b x + c = 0
Indeterminate (independent variable): x
Roots: solutions to the quadratic equation
Zeroes: x intercepts of quadratic function
• Graph of quadratic function y = a x 2 + b x + c is a parabola.
- If a > 0, concave up. If a < 0, concave down.
- c = y intercept.
−b
- Axis of symmetry: x = 2a
• AOS also average of the zeroes.
- Vertex (the maximum or minimum): x value as AOS, and y
value by substituting AOS into the function.
• For monic quadratics: (x + a)(x + b) = x 2 + (a + b)x + a b
The Quadratic Function
- Vertex form: y = a (x − h)2 + k, where a measures concavity and
-
vertex is at (h , k ).
Discriminant: △ = b 2 − 4a c. If △ < 0, no x intercepts. If
△ = 0, one x intercept. If △ > 0, two x intercepts.
−b − △
,
)
Vertex = (
2a 4a
−b ± △
Zeroes =
2a
If discriminant is perfect square, then roots are rational.
take on the common basic shape.
- As power gets bigger, curve is
flatter at base and steeper at
sides.
• A polynomial is an expression of
the form a n x n + a n−1 x n−1 + a n−1 x n−2 + . . . + a 0 x 0 where
- n is a positive whole number (including 0)
- a n , a n−1, . . . , ao are coefficients
- a 0 x 0 = a 0 is the constant term
- a n x n is the leading term
- Highest power is the degree of the polynomial.
• When drawing y = P (x )
- y intercept is the constant
- x intercepts are the roots, found using P (x ) = 0
- As x → ± ∞, P (x ) acts like function of leading term alone.
- Even powered roots are in shape of a parabola.
- Odd powered roots are in shape of a cubic.
Circles
- Equation is (x − h)2 + ( y − k )2 = r 2 where (h , k ) is centre.
- Values for x , y restricted so that:
• h −r ≤ x ≤ h +r
• k −r ≤ y ≤ k +r
- (x − h)2 + ( y − k )2 = r 2 made of two pairs of functions:
• y =k ±
r 2 − (x − h) 2 for top and bottom of semicircle.
• x =h±
r 2 − ( y − k ) 2 for left and right of semicircle.
Half Parabolae
•
•
•
•
+
means top/right half.
−
means bottom/left half.
y is subject, symmetry parallel to x-axis
x is subject, symmetry parallel y-axis.
Functions with
Asymptotes
- Vertical asymptotes occur if lim f (x ) = ± ∞
x→a
• Functions never touch/cut vertical asymptotes.
lim f (x ) = b
x→±∞
f (x ) = ± ∞
- Oblique asymptotes occur if lim
x→±∞
- Horizontal asymptotes occur if
3
Rectangular Hyperbola and Exponential
Data that with inverse variation lies on rectangular hyperbola.
- Rectangular hyperbolas have two perpendicular asymptotes.
1
• Represented as y = x or x y = 1.
The orientation of basic exponentials (y = a x where a ≠ 1) is
determined by the base a.
- For a > 1, starts shallow and increases rapidly as x → ∞
- For 0 < a < 1, starts steep and decreases less rapidly as x → ∞
Direct and Inverse Variation
Multi-Events / Multiplication Principle
- If there are n1 outcomes for independent experiment E1, n 2
outcomes for experiment E 2, … , n m outcomes for experiment
Em, then there are n1 × n 2 × . . . × n m outcomes for composite
experiment E1 × E 2 × . . . × Em. 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
|A ∩ B|
OR
|B |
|A ∩ B|
|B |
=
÷
|B |
|S |
P (A ∩ B )
=
P (B )
A variable y varies directly with a variable x if
• P (A | B ) =
A variable y varies inversely with a variable x if
k
- y = x , for some non-zero constant k of proportionality.
• The graphs of y as a function of x is a rectangular hyperbola with
Independent Events
- y = k x, for some non-zero constant k of proportionality.
• The graph of y as a function of x is thus a line through the origin.
asymptotes as the x-axis and y-axis.
Types of Relationships
Functions:
• One-to-one (e.g. y = 2 x), passes vertical & horizontal line tests.
• Many-to-one (e.g. y = x 2), passes vertical line test only.
Relations:
• One-to-many (e.g. | y | = x), passes horizontal line test only.
Many-to-many (e.g. x 2 + y 2 = 5), fails both tests.
•
Topic 4: Probability
Basics
- | H | means amount of members of H.
- E is event space (favourable), a subset of S, a finite sample space.
|E |
- P (E ) = | S | 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 E is defined as | E | = | S | − | E |
|E |
|S | − |E |
|S |
|E |
=
=
−
= 1 − P (E )
|S |
|S |
|S |
|S |
Common notations include E, E ′, E c
Thus, P (E ) =
•
Sets
-
Empty set has no members, denoted as ∅
Two sets are equal if they have exactly the same members.
A and B = A ∩ B = Intersection
A or B = A ∪ B = Union
• A ∩ B means in only A.
- A ⊂ B means every member of A is in B (A is subset of B).
Counting Rule: | A ∪ B | = | A | + | B | − | A ∩ B |
• When | A ∩ B | = 0 (disjoint sets), | A ∪ B | = | A | + | B |
- Three-way counting rule:
n (A ∪ B ∪ C ) = n (A ) + n (B ) + n (C ) − n (A ∩ B )
−n (B ∩ C ) − n (A ∩ C ) + n (A ∩ B ∩ C )
• The counting rule becomes the addition rule when talking about
probability: P (A ∪ B ) = P (A ) + P (B ) − P (A ∩ B )
- When A , B are mutually exclusive, P (A ∪ B ) = P (A ) + P (B )
A and B are independent if P (A | B ) = P (A ).
This also means P (B | A ) = P (B ).
• Proving B independent to A when A is known independent to B:
P (B ∩ A )
P (B | A ) =
P (A )
P (A ∩ B ) P (B )
P (A ∩ B ) P (B )
P (B )
=
×
=
×
= P (A | B ) ×
P (A )
P (B )
P (B )
P (A )
P (A )
P (B )
And since P (A | B ) = P (A ), = P (A ) ×
P (A )
∴ = P (B ).
• P (A | B ) =
P (A ∩ B )
can be rearranged as
P (B )
P (A ∩ B ) = P (A | B ) × P (B ) and
P (A ∩ B ) = P (A ) × P (B ) for independent events.
Conversely, if P (A ∩ B ) = P (A ) × P (B ) then independent.
Topic 5: Combinatorics
Factorial Notation
Factorial definition: x ! =
0! = 1,
{ x (x − 1)! for x ≥ 1
This definitions can be used to create common denominators:
1
1
• (n − 1)! + (n + 1)!
1
1
(n + 1)n + 1
=
+
=
(n − 1)!
(n + 1)n (n − 1)!
(n + 1)n (n − 1)!
n2 + n + 1
∴=
(n + 1)!
Permutations - nPn
- A permutation of n objects is an ordered arrangement of the n
objects. Therefore, A B A ≠ A A B
- When there are n positions to be filled with n objects:
n choices for 1st position
n − 1 choices for 2nd position…
n − (n − 1) = 1 choice for nth position
• This can be represented by the notation nPn = n !
- Means “from n arrange n”
- Restrictions: Line 5 cats and 5 dogs up so that they alternate?
= 2either cat or dog in 1st × 5!dog per mutations × 5!cat per mutations
= 28800 arrangements.
4
• When lining up n objects, if there are r1 of alike objects, r 2 of
another type of alike object, and r k of a final type of alike object:
n!
Number of Permutations =
r1 !r 2 ! . . . r k !
Permutations - nPr
- When arranging r objects where r < n, from n possible objects:
n choices for 1st position
n − 1 choices for 2nd position…
n − r + 1 choices for rth position
- Thus nPr = n (n − 1)(n − 2) . . . (n − r + 1)
(n − r )(n − r − 1) . . . (2)(1)
= n (n − 1)(n − 2) . . . (n − r + 1) ×
(n − r )(n − r − 1) . . . (2)(1)
n!
=
(n − r )!
- 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?
Sample space = 8 P6 = 20160
Ways = 4 P2 (B and T on lef t) × 4 P1 (G on right) × 5P3 (remaining three)
= 2880
Probability =
2880
1
=
20160
7
Combinations
- Total number of subsets of S, a finite set with n elements = 2n.
- A combination of n objects is an unordered arrangement of the n
objects. Therefore, A B A = A A B
• The number of combinations of n objects for r positions is:
n
n!
n
nC = Pr =
=
r
r!
r !(n − r )! ( r )
- Restrictions: How many possible “three of a kind” hands can you
be dealt with 5 cards from a standard 52 card deck?
Hands = 13C1 (3 of kind ) × 4 C3 (3 of 4 suits) × 12C2 (remaining 2) × 4 × 4
= 54912 hands.
n!
n
• Cn−r = (n − r )!(n − (n − r ))!
n!
= nCr
=
(n − r )!r !
Division in Groups
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 C, and create 3 separators S that act as
13!
separators between 4 boxes. Thus:
= 286
3!10!
Circle Arrangement
• The number of ways to arrange n objects along a fixed circle is
n!
= (n − 1)!
n
- This is since we create a fixed point as there is no front and back.
Pigeonhole Principle (ew)
- If n pigeons are placed into k pigeonholes, then there must be at
n
pigeons in it.
k
• In general, if k n + 1 or more objects are placed into n holes, then
at least one of the holes must contain over k 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.e. n = 5
We know k n + 1 = 16
∴ 5k + 1 = 16
k =3
Therefore according to general rule, at least k + 1 numbers will
leave same remainder, i.e. 3 + 1 = 4
- Example 2: A computer generates random 3-letter words. How
many words need to be generated to ensure 8 of same word?
There are 263 possible words, i.e. n = 263 = 17576
And we know we need 8 repetitions, therefore k + 1 = 8
k =7
Total objects needed = k n + 1 = 123032 + 1
= 123033
least one pigeonhole with at least
Topic 6:
Transformations and Symmetry
Translating Curves
- For horizontal shift, replace x with x − h, which moves f (x )
• Case 1: Dividing (m + n) objects into group with m, group with n:
(m + n)!
m+n
h units to the right.
Number of selections =
Cm =
m !n !
• y = f (x − h)
- This is the same even for three groups:
- For vertical shift, replace y with y − k, which moves f (x )
m+n+pC × n+pC = (m + n + p)! × (n + p)! = (m + n + p)!
m
n
k units up.
m !(n + p)!
n !p !
m !n !p !
• y − k = f (x ) OR y = f (x ) + k
• Case 2: Dividing (2m) objects into two groups each with m objects
2mC
(2m)!
m
=
Number of selections =
as we can sort the two
2!
m !m !2!
- Vertical reflection: For reflection across y-axis, replace x with −x.
equal groups 2! times.
• y = f (−x )
- Similarly for dividing (2m + n) into groups of m, m and n.
- To reflect in the line x = a, replace x with 2a − x
2m+nC ×2m C = (2m + n)! × (2m)! = (2m + n)!
n
m
• y = f (2a − x )
n !(2m)!
m !m !2!
m !m !n !2!
Reflecting Curves
• 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 4!cons × 5C3 E possibilities.
- Horizontal reflection: Reflection across x-axis, replace y with −y.
• −y = f (x ) OR y = − f (x )
- To reflect in the line y = a, replace y with 2a − y
• 2a − y = f (x ) → y = 2a − f (x )
- Reflection about Origin: y = − f (−x )
- Same as a rotation of 180 o
• Caused by a reflection across x-axis and across y-axis.
5
- Absolute Value Inequations METHOD ONE:
Symmetry
- Odd function: True for (prove using):
• f (−x ) = − f (x )
- Odd functions have point symmetry about the origin.
- All polynomials with only odd powers are Odd Functions.
- Even function: True for (prove using):
• f (−x ) = f (x )
- Even functions have line symmetry about the y-axis
- All polynomials with only even powers are Even Functions.
- Most functions are neither even nor odd.
Absolute Value
- | a | = { − a for a < 0
• Absolute value is magnitude from 0 only, not direction.
a,
1.
2.
3.
4.
5.
6.
for a ≥ 0
| − x | = |x |
|x − y | = |y − x |
|x y | = |x ||y |
|x |
x
| |=
|y |
y
| x |2 = x 2
x2 = | x |
• For absolute value graphs, e.g. y = | m x + b | , the part of
y = m x + b below the x-axis is reflected above the x-axis.
- NOTE: the equation | x | = something with pronumerals may
produce an answer that is not a solution, which must be verified.
E.g. | 2 x + 6 | = 3x − 1
2 x + 6 = 3x − 1 OR −2 x − 6 = 3x − 1
x = 7 OR x = − 1
But, subbing x = − 1 into original equation, | 4 | = − 4 which is
obviously false, so only x = 7 is valid solution.
Composite Functions
- Composite functions are when two or more functions combine to
create a new function.
• g ( f (x )) means the value of f (x ) is substituted as x into g (x ).
- Usually f (g (x )) and g ( f (x )) are different functions.
- Represented as ( f ∘ g)(x ) and (g ∘ f )(x ) respectively.
- For a value of x to be in the domain of f (g (x )),
• x must be in domain of g (x ), g (x ) must be in domain of f (x ).
• Value of f (g (x )) is in range of composite function only if x in
domain of f (g (x )).
- If domain of f (g (x )) = ∅, range is also ∅ and is empty function.
Topic 7: Further Graphs
Inequations
- The inequality sign only changes when you:
1. Multiply or divide by a negative number.
2. Take the reciprocal of both
sides.
- Quadratic Inequation:
x 2 − 3x + 2 ≥ 0:
(x − 2)(x − 1) ≥ 0
With this we can graph y = (x − 2)(x − 1)
Then we can check when y ≥ 0
∴ x ≤ 1, x ≥ 2
| 3x + 2 | ≤ 3
3x + 2 ≤ 3 O R − (3x + 2) ≤ 3
5
1
x ≤ OR x ≥ −
3
3
5
1
∴− ≤ x ≤
3
3
- Absolute Value Inequations METHOD TWO:
| x + 5 | < 2 (only works when both sides definitely positive)
(x + 5) 2 < 4 square both sides, solve as quadratic.
x 2 + 10x + 25 < 4 → (x + 7)(x + 3) < 0 then solve graphically.
∴−7< x <−3
- Inequations with Pronumerals in the Denominator
2
< 5 we know straight away that x ≠ − 3
x +3
Multiply by denominator squared: 2(x + 3) < 5(x + 3) 2
5(x + 3) 2 − 2(x + 3) > 0 → (x + 3)(5x + 13) > 0
13
,x ≠−3
Graphically solve, ∴ x < − 3 O R x > −
5
- Finding Changes of Signs through Critical Points
• Functions only change signs at x-intercepts or discontinuities.
- Critical points of an inequation found by moving all terms to
one side and finding when the function created changes sign.
2
−13 − 5x
< 5, create one algebraic fraction:
For
<0
x +3
x +3
Since this fraction is negative, either top or bottom is negative.
• Find critical point of numerator by −13 − 5x = 0
13
→x =−
which is when the function is zero.
5
• Find critical point of denominator by x + 3 = 0 → x ≠ − 3
which is the discontinuity of the function, as it is undefined.
- Then test the three regions created:
13
13
x < − 3, − 3 < x < −
, x >−
5
5
13
We find x < − 3 O R x > −
5
Graphs with Three Kinds of Asymptotes
- Curves always bend towards asymptotes.
• Never cross vertical asymptotes.
• Approach horizontal and oblique asymptotes as x → ± ∞
P (x )
R (x )
• y = A (x ) = Q (x ) + A (x )
- Where Q (x ) is horizontal/oblique asymptote.
- Solve A (x ) = 0 to find vertical asymptotes.
- If R (x ) = 0 has solution, function cuts horizontal/oblique.
Example: Graph y =
- x intercepts at :
0=
(x − 2)(x − 1)(x + 1)
(x + 2)(x − 3)
(x − 2)(x − 1)(x + 1)
→ (x − 2)(x − 1)(x + 1) = 0
(x + 2)(x − 3)
∴ (−1,0), (1,0), (2,0)
−2 × −1 × 1
1
1
=−
→ (0, − )
2 × −3
3
3
(x − 2)(x − 1)(x + 1)
x3 − 2x2 − x + 2
y =
=
(x + 2)(x − 3)
x2 − x − 6
3
2
2
→ x − 2 x − x + 2 = (x − x − 6)(x − 1) + 4x − 4
4x − 4
∴ y = x −1+
(x + 2)(x − 3)
- y intercept at y =
- Vertical asymptotes at (x + 2)(x − 3) = 0 → x = − 2,3
- Oblique asymptote at y = x − 1
- Cuts oblique asymptote at 4x − 4 = 0 → x = 1
Graph is found on the next page:
6
x ≤ − 3; y = − (x + 3) + (1 − x ) = − 2 x − 2
• 1 − x = 0, therefore critical point at x = 1. Testing regions:
−3 < x < 1; y = (x + 3) + (1 − x ) = 4
x ≥ 1; y = (x + 3) − (1 − x ) = 2 x + 2
Reciprocal Functions
1
- The graph of y = f (x ) can be sketched by first drawing y = f (x )
1
is undefined (ie. vertical asymptote)
•
f (x )
1
±
• When f (x ) → ∞, then f (x ) → 0 (asymptotes become point
When f (x ) = 0, then
Adding Ordinates of Bounded Functions
- For example y = x − sin x, formed from y = x and y = − sin x
• Draw points at sin graph turning points and all x intercepts.
• You end up with sin-like curve tilted anticlockwise 45 degrees.
discontinuity)
• Identify where f (x ) = 1, − 1 as their reciprocals are identical.
• Where f (x ) increases, the reciprocal decreases, vice versa.
1
• Where f (x ) is positive, f (x ) is also positive, vice versa.
• If a point in f (x ) is < | 1 | , in reciprocal it will be > | 1 | .
- Horizontal asymptotes in f (x ) of y < | 1 | will shift to
y > | 1 | in
1
f (x )
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 s (x ) = f (x ) + g (x ), each y-value of both functions are added
for each corresponding x-value.
- Where f (x ) = − g (x ), then y = 0, thus a x-intercept.
- If g (x ) has 0 at x = a, then s (a) = 0 + f (a) = f (a).
- If two curves meet at x = a so that their ordinates are equal,
then s (a) = 2 f (a) O R 2g (a)
- 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.g s (x ) = f (x ) + g (x ) where
1
1
f (x ) = x and g (x ) = , therefore s (x ) = 1 +
x
x
• Since y is undefined at x = 0, it is still undefined in s (x ).
• Again, draw vertical lines and add normally.
• Exclusions in domain of original functions remains in s (x )
1
1
- E.g. if f (x ) = x + x , g (x ) = 1 − x , then
y = f (x ) + g (x ) retains the vertical asymptote x = 0.
Multiplication of Graphs
- s (x ) = f (x )g (x ) can be graphed by first y = f (x ), y = g (x )
• Where there are x intercepts, the function will change sign.
• Multiply signs of functions to determine new functions’ sign.
• Where f (x ) o r g (x ) = 1, s (x ) must equal (or in case of
y = − 1 reflect/equal) the other curve.
• If f (x ) o r g (x ) → 0 o r ± ∞, then so will the new function.
• However, in the case of x → − ∞ when finding s (x ) = x e x,
y = x →−∞ while y = e x → 0, thus we reach a dilemma.
- Dominance: the graph that gets steeper or shallower the
quickest prevails, and we graph the s (x ) accordingly.
Therefore, in this case, s (x ) → 0 for x → − ∞.
- 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.
- Like functions retain symmetry when added, ie. Odd function +
Odd function = Odd function and vice versa.
- y = f (x ) − g (x ) can be graphed by first graphing y = f (x ) and
y = − g (x ) and then adding ordinates together.
y = x ex
Addition of Absolute Value Functions
- For example: f (x ) = | x + 3 | + | 1 − x | , each absolute value = 0
will be a critical point, which creates three regions.
• x + 3 = 0, therefore critical point at x = − 3. Testing regions:
7
Graphs of Squared Functions
- y = [ f (x )]2 can be graphed by first graphing y = f (x ).
• All roots will become double roots.
• All stationary (turning)
Square Root Graphs
- y=
•
•
•
•
•
•
•
•
points will remain stationary
• All discontinuities remain.
• Values of horizontal and
oblique asymptotes squared.
• If | f (x ) | > 1 then
[ f (x )]2 > f (x ), i.e. above
original.
• If | f (x ) | < 1 then
[ f (x )]2 < f (x ), i.e. below
y = [ f (x )]2 (blue) where
3
(black)
f (x ) = 2 +
(x + 2)(x − 1)
new curve also = 1
Division of Graphs
f (x )
- y = g (x ) can be thought of as
f (x ) ×
f (x ) is only defined if f (x ) ≥ 0
f (x ) ≥ 0 for all x in the domain.
Stationary points must still be stationary points
All discontinuities will remain.
Horizontal/oblique asymptotes may change (value rooted)
f (x ) < f (x ) if f (x ) > 1 i.e. new curve below old curve.
f (x ) > f (x ) if f (x ) < 1 i.e. new curve above old curve.
All f (x ) = 1 remain at 1.
a
• x intercepts require close inspection: for y = x b ,
a
- if b < 1, curve is concave down in 1st quadrant,
original.
• Whenever f (x ) = | 1 | ,
f (x ) can be sketched by first drawing and seeing:
1
and same procedures as multiplication can be
g (x )
followed except;
• x intercepts of g (x ) 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.
sin x
graphed against y = sin x and y = x
x
Below, y =
with a vertical tangent at the x intercept.
a
- If b > 1, curve is concave up in 1st quadrant, with
a horizontal tangent at x intercept.
- y 2 = f (x ) is simply a reflection of y = f (x ) 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 y = x
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 f −1(x )
- To test if a relation has an inverse function:
• Passes both vertical and horizontal line test OR
• When x = f ( y) rewritten as y = f (x ), y = f (x ) only has one
value (unique).
- Each composite of a function and its inverse sends every number
Graph should have point discontinuity at x = 0
Absolute Value Graphs
•
•
•
•
•
•
•
y = | f (x ) | , reflect f (x ) < 0 in the x-axis.
y = f ( | x | ), symmetry in y-axis, 1st quadrant reflected in 2nd.
| y | = f (x ), symmetry in x-axis, reflect f (x ) > 0 in x-axis.
| y | = f ( | x | ), symmetry in both x and y axes, 1st quad into all 4
y = | f ( | x | ) | , symmetry in y axis then reflection in x axis.
| y | = | f (x ) | , symmetry in x axis then reflection in x axis.
for which it is defined back to itself:
• f −1( f (x )) = x for all x in domain of f (x )
f ( f −1(x )) = x for all x in domain of f −1(x )
•
- An identity function is a function whose output is the same as
input: I (x ) = x for all x 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.g., if we restrict domain of f (x ) = x 2 to x ≥ 0 i.e.
•
g (x ) = x 2 where ≥ 0, then g (x ) has in inverse function.
- g −1(x ) = x
| y | = | f ( | x | ) | , symmetry in y axis then reflection in x axis, then
symmetry in x-axis, as shown below.
8
Parametric Equations
- Parametric coordinates are alternative way of describing graphs.
• Cartesian Form: Curve described by one equation, points
described by two numbers.
• Parametric Form: Curve described by two equations, points
• For angles in the 4th quadrant:
- cos(360 o − θ ) = cos θ
- sin(360 o − θ ) = − sin θ
- tan(360 o − θ ) = − tan θ
• Table of trigonometric values at boundary points:
described by one number (parameter)
- To change from parametric to Cartesian equations, eliminate the
parameter (t ) from two parametric equations x = f (t ), y = g (t ).
• For example, x = 2 + cos θ and y = 1 + sin θ where θ is
parameter (when parameter is angle, known as polar coords):
cos θ = x − 2, sin θ = y − 1 → (x − 2) 2 + ( y − 1) 2 = 1
Thus we deduce the that curve is circle with radius 1 and centre (2,1)
- The parametrisation of a circle x 2 + y 2 = r 2 is
x = r cos θ, y = r sin θ
1
- The parametrisation of hyperbola x y = 1 is x = t , y = t
- The parametrisation of parabola x 2 = 4y is x = 2t , y = t 2
• Caution: The domain and range of the cartesian equation is
restricted by any restrictions in the respective parametric
Trigonometric Graphs
- y = sin x and y = cos x defined for all real x.
• Has range of −1 ≤ y ≤ 1 (amplitude is 1) and period of 360 o
• y = sin x is odd function, i.e. sin(−x ) = − sin x.
• y = cos x is even function, i.e. cos(−x ) = cos x.
equations.
Topic 8: Trigonometry I
Trigonometric Ratios
- Reciprocal ratios used to avoid confusion
between sin−1 and sin2, for example.
O
H
sin θ =
/ csc θ =
H
O
A
H
cos θ =
/ sec θ =
H
A
A
O
tan θ =
/ cot θ =
A
O
- “Co” in name of means it is complementary to the other function,
i.e. cot θ + tan θ = 90 o, sec θ + csc θ = 90 o etc.
- Exact value special triangles and table:
Angles of Any Magnitude
• For a point P(x , y) on unit circle, can be defined as (cos θ, sin θ )
• For supplementary angles θ and 180 o − θ (2nd quadrant):
- cos(180 o − θ ) = − cos θ
- sin(180 o − θ ) = sin θ
- tan(180 o − θ ) = − tan θ
• For angles in the 3rd quadrant:
- cos(180 o + θ ) = − cos θ
- sin(180 o + θ ) = − sin θ
- tan(180 o + θ ) = tan θ
- For y = tan x, x ∈ ℝ\(90 + 180k), when k ∈ ℤ
- For y = cot x , x ∈ ℝ\180k , when k ∈ ℤ
• Range is all real y, period is 180 o
• Both odd functions, i.e. tan(−x ) = − tan x etc.
- For y = csc x, x ∈ ℝ\180k , when k ∈ ℤ
- For y = sec x , x ∈ ℝ\(90 + 180k ), when k ∈ ℤ
• Range is y ≥ 1, y ≤ − 1, period is 180 o
• y = csc x is odd function, while y = sec x is even function.
9
Trigonometric Identities
• For any angle θ, the ratio identities:
sin θ
Basic Formulae
cos θ
- General form: a x + b y + c = 0 where a , b , c are constants.
- cos θ = tan θ (provided that cos θ ≠ 0)
- sin θ = cot θ (provided that sin θ ≠ 0)
• For any angle θ, the pythagorean identities:
- sin2 θ + cos2 θ = 1
- tan2 θ + 1 = sec 2 θ (provided that cos θ ≠ 0)
- cot 2 θ + 1 = csc 2 θ (provided that sin θ ≠ 0)
• For any angle θ, the complementary angle identities:
- cos(90 o − θ ) = sin θ
- cot(90 o − θ ) = tan θ (provided that tan θ is defined)
- csc(90 o − θ ) = sec θ (provided that sec θ is defined)
Sine Rule and Area Formula
h
- In △ C B M, a = sin B
→ h = a sin B
h
- In △ C A M, b = sin A
→ h = b sin A
a
b
∴
=
sin A
sin B
a
b
c
• More generally, in any △ A B C: sin A = sin B = sin C
- 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 sin θ, sin(180 − θ ) 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 < 180 o, then the supplement is also valid.
1
• From triangle above, area = 2 c h, and we know h = b sin A.
1
- ∴ Area = 2 b c sin A
Cosine Rule
- In △ B P C,
a 2 = (b − x ) 2 + h 2
2
a = b2 − 2 x b + x 2 + h2
- In △ B PA ,
x 2 = c 2 − h2
x
cos A = → x = c cos A
c
a 2 = b 2 − 2b c cos A + c 2 − h 2 + h 2
∴ a 2 = b 2 + c 2 − 2b c cos A
• To find third side given two sides and included angle:
a 2 = b 2 + c 2 − 2b c cos A
•
Topic 9: Coordinate Geometry
b2 + c 2 − a2
To find angle given three sides: cos A =
2b c
Term One
Finished!
y −y
2
1
- To find gradient: m = x − x
2
1
- Gradient-intercept form: y = m x + b
- Point-slope form: y − y1 = m (x − x1)
•
y2 − y1
Eq. through (x1, y1), (x 2 , y2 ) is y − y1 =
(x − x1)
x 2 − x1
(x 2 − x1) 2 + ( y2 − y1) 2
x1 + x 2 y1 + y2
- Midpoint of a segment: M = ( 2 , 2 )
- For two parallel lines, m1 = m 2
- Length of a segment: d =
1
- To prove perpendicularity, m1 × m 2 = − 1 → m1 = − m
2
- To prove collinear points a , b , c: mab = mbc
- Perpendicular distance between a point (x1, y2 ) and line
a x + b y + c = 0: d =
| a x1 + b y1 + c |
a2 + b 2
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
• a1 x + b1 y + c1 + k (a 2 x + b 2 y + c2 ) = 0
- For example, 2 x + y + 1 = 0, 3x + 5y − 9 = 0, (1,2):
2 x + y + 1 + k (3x + 5y − 9) = 0, then subs. (1,2)
2 + 2 + 1 + k (3 + 10 − 9) = 0 → 4k = − 5 → k = −
- Substituting back into equation,
5
4
5
(3x + 5y − 9) = 0
4
7x + 21y − 49 = 0 → x + 3y − 7 = 0
2x + y + 1 −
10
Division of Line Segment in a Ratio
- A B is divided m : n as shown. To find P,
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
m
× A B away from A.
m +n
Thus,
m
x = x1 +
(x − x1)
m +n 2
x1(m + n) + m x 2 − m x1
=
m +n
m x 2 + n x1
=
m +n
Similarly, solving for y gives
m y2 + n y1
y =
m +n
x 2 m + x1n y2 m + y1n
,
)
Therefore, P = (
m +n
m +n
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.g. P (55 ≤ X ≤ 60), but this requires integration.
Random Experiments & Random Variables
- If experiment has more than one outcome - random experiment
• X is the random variable, the result of a random experiment.
• P (X = x ) OR P (x ) denotes when X results in a value of x.
• In discrete probability distributions, ∑ P (X = x ) = 1
- If experiment only has one outcome - deterministic experiment.
• Thus, one certain outcome (e.g. 4) only. P (X = 4) = 1
Expected Value
- Expected value calculated by weighting each value by their
probability.
• Thus it is calculated as E (X ) = ∑ x p (x )
• It is also alternatively notated as μ, meaning “mean”.
Laws of Expectation
1. If a and b are constants, E (a X + b) = a E (X ) + b
Proof: E (a X + b) =
(a x + b)p (x )
∑
b p (x )
a x p (x ) +
∑
∑
=a
x p (x ) + b
p (x )
∑
∑
Since
x p (x ) = E (X ) and
p (x ) = 1
∑
∑
= a E (X ) + b
2. E (X + Y ) = E (X ) + E (Y )
=
Variance
- Variance (Var(X )) is a measure of spread about the mean.
• Take deviation (x − μ) of each value and square: (x − μ)2
- 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
(x − μ) 2 p (x )
∑
- Thus, also expected value: Va r (X ) = E ((X − μ)2 )
- The units of variance are square of whatever units values have.
each value: Va r (X ) =
Variance - Alternative Formula
- To conjure up an alternative formula for easier calculations:
Va r (X ) =
(x − μ) 2 p (x )
∑
2
=
(x − 2μ x + μ 2 )p (x )
∑
=
x 2 p (x ) −
2μ x p (x ) +
μ 2 p (x )
∑
∑
∑
=
x 2 p (x ) − 2μ
x p (x ) + μ 2
p (x )
∑
∑
∑
p (x ) = 1 and
x p (x ) = μ,
Since
∑
∑
Va r (X ) =
x 2 p (x ) − 2μ 2 + μ 2 =
x 2 p (x ) − μ 2
∑
∑
2
And therefore Va r (X ) also equals E (X ) − μ 2.
Uniform Distribution
- For a uniform distribution X = 1,2,3,...,n, where all values have
equal probability of P (x ) =
n +1
2
n2 − 1
Va
r
(X
)
=
•
12
1
:
n
• E (X ) =
Standard Deviation
- Square root of the variance: σ = Va r (X ) or σ 2 = Va r (X ).
- Spreading out all values by factor of k, results in k × σ.
- Adding constant amount to each data value doesn’t affect spread.
X
- However, if each value of X is divided by k in Z, i.e Z = k ,
X
X
μ
) = E [( ) 2 ] − ( ) 2
k
k
k
E (X 2 ) μ 2
E (X 2 ) − μ 2
=
−
=
k2
k2
k2
2
σ
Va r (X )
σ
= ( ) 2 : spread is narrower by k factor.
=
=
k
k2
k2
Va r (Z ) = Va r (
• If we have σ (a X + b), +b does not affect the spread, and the
spread must be positive, therefore σ (a X + b) = | a | σ (X )
Sample Distribution
- A census performs an experiment on everything, while survey
sample only some of the population.
- In simulations/trials, p (x ) is replaced by fr (relative frequency).
- The mean is denoted by x, and corresponds to expected value:
• x = ∑ x fr
- Sample variance is s 2 and sample standard deviation is s.
2
2
2
2
• s = ∑ (x − x ) fr = ∑ x fr − x
11
Z-scores
Slope of Tangent to a Curve
- Z-scores are a common standardisation of data, which represent
• To find the tangent on point P (x , f (x )) on y = f (x ), we can find
the amount of standard deviations above or below mean.
x −μ
• z = σ , which is good because:
E (X − μ)
X−μ
μ −μ
=0
E (Z ) = E (
)=
=
σ
σ
σ
Va r (X − μ)
X−μ
σ2
Va r (Z ) = Va r (
)=
=
=1
2
σ
σ2
σ
∴ μz = 0 and σz = 1, mean is 0 and standard deviation is 1.
Topic 11: Differentiation
Gradient Function
- The gradient function (derivative) y = f ′(x ) measures how steep
the graph at the point (i.e. gradient).
• For a horizontal line f (x ) = c, f ′(x ) = 0.
• For an oblique line f (x ) = m x + b, f ′(x ) = m.
Derivative of a Semicircle
- Slope of a curve is defined as slope of tangent to curve at any
particular point.
r 2 − x 2 , the tangent on point A
f (x )
(x , f (x )) is perpendicular to the radius OA: mOA =
x
−x
m
Thus,
’s
negative
reciprocal
is
m
=
tangent
OA
f (x )
−x
2
2
∴ for a semicircle f (x ) = r − x , f ′(x ) =
•
r 2 − x2
• For a semicircle f (x ) =
Quadratic Derivative with Secants
1. Choose two points symmetrically on either side of point (x , x 2 )
2. Calculate the slope of secant joining the two points.
3. Let the two points be A (x − h , (x − h) 2 ), B (x + h , (x + h) 2 ).
4x h
(x + h) 2 − (x − h) 2
= 2x
4. Thus, mAB =
=
x + h − (x − h)
2h
5. Therefore for f (x ) = x 2 , f ′(x ) = 2 x is correct.
Finding Limits
1. Direct Substitution
the slope of the secant connecting P and Q (x + h , f (x + h))
elsewhere on the slope.
- The closer Q is to P, the more accurate the gradient will be.
f (x + h) − f (x )
f (x + h) − f (x )
, and thus we want lim
,
mPQ =
x +h −x
h
h→0
which is the slope of the tangent, known as the derivative of y with
dy
d
f (x ).
respect to x, symbolised as
or y′ or f ′(x ) or
dx
dx
dy
yδ
lim
, where δ is an infinitesimal change.
• d x represents xδ→0
xδ
- This process is called "differentiating from first principles"
- The derivative is a measure of the rate of change.
f (x + h) − f (x )
h
E.g. Find the equation of the tangent to y = x 2 − 5x + 2 at (1, − 2)
First find derivative: f (x ) = x 2 − 5x + 2,
f (x + h) = x 2 + 2 x h + h 2 − 5x − 5h + 2
2 x h + h 2 − 5h
dy
f (x + h) − f (x )
= lim
= 2x − 5
= lim
d x h→0
h
h
h→0
2 x − 5 is the gradient function, into which we sub (1, − 2) for m.
dy
When x = 1,
= 2(1) − 5 = − 3
dx
Thus, y + 2 = − 3(x − 1) → y = − 3x + 1
lim
• f ′(x ) = h→0
Rules for Differentiation
• Power Rule: Where f (x ) = x n, f ′(x ) = n x n−1
Proof: We know a 2 − b 2 = (a − b)(a + b) and
a 3 − b 3 = (a − b)(a 2 + a b + b 2 ) and thus we can see the pattern:
a n − b n = (a − b)(a n−1 + a n−2 b + a n−3 b + . . . + a b n−2 + b n−1)
We have f (x ) = x n, thus f (x + h) = (x + h) n
(x + h) n − x n
f ′(x ) = lim
h
h→0
= lim
h→0
(x + h − x )[(x + h) n−1 + (x + h) n−2 x . . . + (x + h)x n−2 + x n−1]
h
= lim (x + h) n−1 + (x + h) n−2 x . . . + (x + h)x n−2 + x n−1
h→0
= lim x n−1 + x n−1 . . . . + x n−1 = n x n−1
h→0
- The normal at P is the line perpendicular to tangent at P.
• Sum rule: If f (x ) = u (x ) + v (x ), then f ′(x ) = u′(x ) + v′(x )
e.g. lim x + 7 = 5 + 7 = 12
x→5
Proof :
2. Factorise and Cancel
x2 − 9
= lim x + 3 = 6
e.g. lim
x→3 x − 3
x→3
1
3. Special Limit - lim
=0
x→∞ x
x 3 + 3x 2 + 2 x − 1
e.g1. lim
, divide everything by highest power.
x→∞
4x 3 − 1
3x 2
3
x3
2x
1
2
1
+
+
−
1+ x +
−
3
1
x
x3
x3
x3
x2
x3
= lim
=
= lim
1
3
4
x→∞
x→∞
4x
4−
− 1
x3
x3
x3
u (x + h) − u (x )
v (x + h) − v (x )
= u′(x ) + v′(x )
= lim
+ lim
h
h
h→0
h→0
• Coefficient rule: If f (x ) = a u (x ), then f ′(x ) = a u′(x )
Look for coefficients of the biggest power, as they must cancel out.
4x − x 2
x3 + 2
1
0
e.g2. lim
e.g2. lim
= =∞
= =0
0
1
x→∞ x 3 + 1
x→∞ x 2 − 1
(x + 3)(x − 2)
Also used to find horizontal asymptotes: y =
(x − 1)(x + 1)
2
x +x −6
1
Horizontal asymptote at lim =
= ∴ at y = 1.
1
x→∞
x2 − 1
f ′(x ) = lim
h→0
f (x + h) − f (x )
u (x + h) + v (x + h) − u (x ) − v (x )
= lim
h
h
h→0
Proof:
f ′(x ) = lim
h→0
= a u′(x )
a u (x + h) − a u (x )
u (x + h) − u (x )
= a lim
h
h
h→0
dy
dy
du
d
n
n−1 × u′,
• Chain rule: d x = d u × d x OR d x (u ) = n u
if y is a function of u where u is a function of x.
- For example, in y = (4x 3 + 3)6, y = u 6, u = 4x 3 + 3
dy
= 6(4x 3 + 3)5(12 x 2 ) = 72 x 2 (4x 3 + 3)5
dx
• Chain rule also used for parametric equation, e.g. x = t 2 , y = 2t.
dy
dy
dt
1
1
=
×
=2×
= .
dx
dt
dx
2t
t
1
x
2
∴ eq. of tang. at T (t ,2t ) is y − 2t = (x − t 2 ) →y = + t
t
t
∴
12
Differentiating Inverse Functions
dy
dx
• For inverse functions f (x ) = y and f ( y) = x, d x × d y = 1.
3
E.g. differentiate y = x by differentiating inverse function:
dx
Solving for x, x = y 3. ∴
= 3y 2
dy
dy
1
=
We want to find derivative of y with respect to x, so
dx
3y 2
dy
1
3 2
3
.
=
y = x , and therefore y 2 = x , thus
dx
3 2
3 x
Differentiating Non-Cardinal Powers
- We can prove the power rule works for a power of n = − 1 by
using first-principles differentiation.
1
1
f (x ) =
and f (x + h) =
x
x +h
1
1
− x
x −x −h
−1
x+h
f ′(x ) = lim
= lim
= lim
h
h→0
h→0 h x (x + h)
h→0 x (x + h)
−1
, which is indeed the same result as when power rule is used.
=
x2
• Power rule also works for fractional indices.
- We can then prove that the power rule works on all negative
1
integer indices. We have f (x ) =
, where m ≥ 2 is an integer.
xm
1
Using the chain rule, where y = , u = x m,
u
dy
dy
du
1
=
×
=−
× m x m−1 = − m x −m−1 as required.
dx
du d x
x 2m
Dividing Through by Denominator
- If the denominator of a function is a single term, first divide
Quotient Rule
u
• For a function y = v where u , v are both functions of x,
dv
du
v
−u
dy
v u′ − u v′
dx
dx
=
or more concisely, y′ =
2
dx
v2
v
−1
Proof using product rule: let y = u v , let U = u , V = v −1
dy
dU
dV
du
dv
=V
+U
= v −1
− u v −2
dx
dx
dx
dx
dx
Multiplying top and bottom of each expression by v 2:
dv
du
v
−u
dy
dx
dx
=
as required.
dx
v2
Reciprocal Rule
k
d
k
−k v′
• For a function f (x ) = v where k is a constant, d x ( v ) = v 2
To differentiate y =
6
4x 2 + 3
→
dy
−6(8x )
−48x
=
=
dx
(4x 2 + 3) 2
(4x 2 + 3) 2
Rates of Change
• At three arbitrary points P (x , f (x )) on a function f (x ):
dy
- If d x > 0, then f (x ) is increasing.
dy
- If d x = 0, then f (x ) is stationary, i.e. P is a stationary point.
dy
- If d x < 0, then f (x ) is decreasing.
• The average rate of change between two points P (t1, Q1) and
Q − Q1
Q (t 2 , Q2 ) is the slope of the secant P Q: = mPQ = 2
t 2 − t1
Displacement, Velocity and Acceleration
- Displacement (x ): Distance from a point, with direction.
dx
through by it, then differentiate.
5x 3 + 2 x 2 + 4
5
2
4
f (x ) =
= x + + x −2
2
3
3
3
3x
8
5x 3 − 8
dy
5
4 −3 5
=
∴
= − 2( )x = −
3
3 3x
dx
3
3
3x 3
However, it is just as practical to use to quotient rule, seen later.
- Velocity (v, x· , d t ): The rate of change of displacement with
Domain of x n
Differentiability
m
• When n is a fraction k , where k is an even number, x cannot be
negative.
• When n is irrational, x cannot be negative.
- To be smooth continuous, function cannot have cusps or corners.
- A function is differentiable at a point if the curve is smooth
• When n is negative, x cannot equal zero.
Product Rule
d
respect to time, i.e. speed with direction.
2
· ·· d v d x
- Acceleration (a , v, x , d t , 2 ): The rate of change of velocity
dt
with respect to time, i.e. second derivative of f (x ).
- A graph with no vertical asymptotes, point discontinuities, or any
other gaps in the domain/range are said to be continuous.
continuous and the tangent is not vertical at the point.
• For example, y = | x − 1 | is not differentiable at x = 1 as it is
dv
du
• Product Rule: d x (u v) = u d x + v d x or y′ = u v′ + v u′
Proof: We have y = u v. If x → x + x δ, then y → y + y δ,
u → u + u δ, v → v + v δ, i.e. ∴ y + y δ = (u + u δ )(v + v δ )
y + yδ = uv + uvδ + uvδ + uδvδ → yδ = uvδ + vuδ + uδvδ
yδ
vδ
uδ
uδ
vδ
=u
+v
+
×
× x δ so as x δ → 0,
xδ
xδ
xδ
xδ
xδ
dy
dv
du
=u
+v
+ 0 as required.
dx
dx
dx
Can also use when u , v is square root, e.g. y = 2 x 2 x − 1
1
1
dy
1
−1
= 2 x (2 x − 1) 2 →
= (2 x )[ (2 x − 1) 2 (2)] + (2 x − 1) 2 (2)
dx
2
1
−1
−1
2
2
= 2 x (2 x − 1)
+ 2(2 x − 1) = 2(2 x − 1) 2 [x + (2 x − 1)]
dy
2(3x − 1)
−1
= 2(2 x − 1) 2 (3x − 1) → ∴
=
dx
2x − 1
impossible to draw a tangent.
• Similarly x = y 2 is not differentiable at
x = 0 as the tangent is vertical.
2
• y = (x − 2) 3 also not differentiable at
x = 2 as there is a cusp.
Implicit Differentiation
df
df
dy
2
• If you don't have a function, e.g. x = y , use d x = d y × d x .
d
d
(x ) =
( y 2 ), but y 2 cannot be differentiated in respect to x
dx
dx
d
d
dy
But implicit differentiation gives
( y2) =
( y2) ×
dx
dy
dx
dy
dy
1
∴ 1 = 2y
→
=
. The derivative is in terms of y as two
dx
dx
2y
points would result if it was in terms of x.
13
- Find equation of tangent to x 2 + y 2 = 9 at the point (1,2 2)
Differentiate equation → 2 x + 2y
dy
dy
= 0 → 2y
= − 2x
dx
dx
dy
x
= − . Both x and y are in the derivative as you need both
dx
y
dy
1
values to find a single point. Subbing point (1,2 2),
=−
dx
2 2
∴
Therefore y − 2 2 = −
1
2 2
(x − 1) → 2 2y − 8 = − x + 1
Thus, the equation of the tangent is x + 2 2y = 9 = 0
Topic 12: Polynomials
Polynomial Basics
• A real polynomial P (x ) of degree n is expressed as
-
Remainder Theorem
• If a polynomial P (x ) is divided by (x − a), R (x ) = P (a)
Proof: P (x ) = A (x )Q (x ) + R (x ). Let A (x ) = (x − a),
P (x ) = (x − a)Q (x ) + R (x ) → P (a) = (a − a)Q (a) + R (a)
= R (a).
However, since A (x ) is a divisor of degree 1, R (x ) must be degree 0
or simply zero, so we can rewrite it as r. ∴ r = P (a)
Factor Theorem
• If (x − a) is a factor of P (x ), then P (a) = 0. Converse also true.
To factorise P (x ) = 4x 3 − 16x 2 − 9x + 36, the constant factors
must be a factor of the constant (i.e. 36). These are thus
1,2,3,4,6,9,12,18,36, which can also be negative. Also possible to be
factors of constant
fractional factors of the form
.
factors of leading coefficient
P (x ) = p 0 + p1 x + p 2 x 2 + . . . + pn−1 x n−1 + pn x n where
We test these, and we find P (4) = 0, and thus (x − 4) is factor.
P (x ) = (x − 4)(4x 2 − 9) = (x − 4)(2 x + 3)(2 x − 3)
Degree (order): the highest index of the polynomial.
Leading term: pn x n and leading coefficient: pn
Polynomial Rules
pn ≠ 0 and n ≥ 0 and is an integer.
Monic polynomial: leading coefficient is equal to one.
P (x ) = 0 is polynomial equation.
1. If P (x ) has k distinct real zeros, a1, a 2 , a3, . . . , ak,
Roots: solutions to P (x ) = 0
2. If P (x ) has degree n and has n distinct real zeros
y = P (x ) is polynomial function.
Zeroes: x intercepts of y = P (x )
Graphing Polynomials
• When drawing y = P (x )
- y intercept is the constant
- x intercepts are the roots, found using P (x ) = 0
• As x → ± ∞, P (x ) acts like function of leading term alone.
Proof: Let P (x ) = a n x n + a n−1 x n−1 + . . . + a1 x + a 0
a0
a
P (x )
a
= a n + n−1 + . . . +
+
Then
xn
x
x n−1
xn
P (x )
= a n, showing that for large values of x,.
Therefore, lim
x→±∞ x n
P (x ) is the same sign as a n
- Even powered roots are in shape of a parabola.
- Odd powered roots are in shape of a cubic.
y = (x − 1)4 (x + 1) 3(x + 2) 2
In this graph, x = 1 is a root of multiplicity 4, x = − 1 is root of
multiplicity 3, and x = − 2 is root of multiplicity 2.
Polynomial Division
• P (x ) = A (x )Q (x ) + R (x ) where A (x ) is divisor, Q (x ) is quotient
• d e g (R (x )) < d e g (A (x ))
E.g. x 3 + 2 x 2 − 4 divided by x − 3
First take the x from x − 3 and divide into the
leading term x 3, creating x 2 at the top.
Now times x − 3 by x 2 giving x 3 − 3x 2, and then
subtract. Repeat until you cannot divide into the
next term. Q (x ) = x 2 + x + 3, R (x ) = 5.
(x − a1)(x − a 2 )(x − a3) . . . (x − ak ) is a factor of P (x )
i.e. if 1 and −2 are zeros of P (x ) then (x − 1)(x + 2) is factor.
a1, a 2 , a3 . . . , a n then P (x ) = (x − a1)(x − a 2 ) . . . (x − a n )
i.e. cannot have more linear factors than the degree of polynomial
3. If P (x ) has degree n and has more than n real zeros, then P (x )
is the zero polynomial, i.e. P (x ) = 0 for all x
4. If P (x ) ≡ Q (x ) (identically equal), then the coefficients of each
corresponding term must be equal.
- If n degree polynomials P (x ) = Q (x ) for n + 1 values of x,
then they are identically equal.
Sum and Product of Roots
- If α and β are the roots of a x 2 + b x + c = 0, then:
a x 2 + b x + c = a (x − α )(x − β )
= a (x 2 − α x − β x + α β )
b
c
x 2 + x + = x 2 − (α + β )x + α β and therefore:
a
a
b
c
• α + β = − a and α β = a
- For a cubic a x 3 + b x 2 + c x + d = 0:
b
d
c
• α + β + γ = − a and α β + α γ + β γ = a and α β γ = − a
- For a quartic a x 4 + b x 3 + c x 2 + d x + e = 0:
b
• α + β + γ + δ = − a and
c
α β + α γ + α δ + β γ + β δ + γ δ = and
a
d
e
α β γ + α β δ + α γ δ + β γ δ = − and α β γ δ =
a
a
- Therefore, in general, for a x n + b x n−1 + c x n−2 . . . = 0
b
• ∑ a = − a (sum of roots, one at a time)
c
• ∑ α β = a (sum of roots, two at a time)
d
• ∑ α β γ = − a (sum of roots, three at a time)
e
• ∑ α β γ δ = a (sum of roots, four at a time)
• Furthermore, due to the expansion of a perfect square,
α2 = (
α )2 − 2
αβ
∑
∑
∑
14
Multiple Roots
- A value x = a is a zero of multiplicity m of the polynomial P (x )
if P (x ) = (x − a) m Q (x ), where Q (a) ≠ 0.
- If x = a is a zero of multiplicity m ≥ 1 of P (x ), then x = a is a
zero of multiplicity m − 1 of the derivative P′(x )
Proof: since P (x ) = (x − a) m Q (x ) where Q (a) ≠ 0,
P′(x ) = m (x − a) m−1Q (x ) + (x − a) m Q′(x )
= (x − a) m−1(m Q (x ) + (x − a)Q′(x ))
= (x − a) m−1 R (x ) where R (x ) = m Q (x ) + (x − a)Q′(x )
But we know Q (a) ≠ 0, so substituting x = a into R (x ),
R (a) = m Q (a) + 0 which ≠ 0 so therefore:
x = a is a zero of multiplicity m − 1 in P′(x )
• If x = a has even multiplicity, then curve is tangent to x-axis
at x = a, and does not cross x-axis, i.e. (a ,0) is a turning point
• If x = a has odd multiplicity of ≥ 3, then is tangent to x-axis
at x = a, crossing x-axis, i.e. (a ,0) is a horizontal inflection.
- Once a zero of P (x ) is found, it should immediately be checked
whether it is double zero by subbing into P′(x ) and triple zero by
subbing into P′′(x ).
Topic 13:
Exponentials and Logs
Logarithms and Log Laws
• For x = a y, we know y = loga x, where a > 0, a ≠ 1.
- Logs and exponentials are inverse functions.
1. loga a x = x for all real x
2. a loga x = x for all real x > 0
3. loga 1 = 0 (as a 0 = 1)
4. loga a = 1 (as a 1 = a)
5. loga x + loga y = loga x y (as a x × a y = a x+y)
x
6. loga x − loga y = loga (as a x ÷ a y = a x−y)
y
7. loga x p = p loga x (as (a x ) p = a px)
logb x
8. loga x =
logb a
Proof: we know y = loga x and x = a y.
Thus, taking logs of base b on both sides: logb x = logb a y
logb x
logb x = y logb a → y =
as required.
logb a
Expo/Logs Inequations and Graphs
- Since for bases of a > 1, exponential function y = a x and log
function y = loga x are continually increasing, inequalities are
preserved, i.e. logging both sides doesn’t change sign.
• For example, in the inequation log3 x > 2, we can exponentialise
both sides according to the base: 3log3 x > 32 → ∴ x > 9
- The graphs of y = 2 x and y = log2 x are inverse functions and
are reflected in the line y = x.
- This means that for y = a x and y = loga x,
loga (a x ) = x and a loga x = x
Topic 14: Extending Calculus
Differentiating Exponentials
- Let f (x ) = a x. Differentiating using first principles:
a x+h − a x
a x (a h − 1)
ah − 1
f ′(x ) = lim
→ lim
→ a x lim
h
h
h
h→0
h→0
h→0
As this is impossible to solve algebraically, we can notice that
ah − 1
a 0+h − a 0
f (0 + h) − f (0)
a x lim
= a x lim
= a x lim
h
h
h
h→0
h→0
h→0
The latter part of which is just f ′(0)’s first principles representation.
Therefore f ′(x ) = a x × f ′(0), f ′(0) is the slope of tangent at x = 0
Euler’s Number (e)
1
lim (1 + ) n ≈ 2.72
• e is an irrational number, defined as e = n→∞
n
- The exponential function f (x ) = e x is defined so that at x = 0 it
has a slope of 1, i.e. f ′(0) = 1.
• This means that for f (x ) = e x , f ′(x ) = e x or f ′(x ) = y.
Differentiating Exponential Functions
- If we have y = e f (x), let f (x ) = u so that y = e u, so chain rule:
dy
dy
d (e u ) d u
, which simplifies to
= e u × f ′(x ).
=
×
dx
dx
du
dx
f (x) to find derivative: d y = f ′(x )e f (x)
• Thus, for y = e
dx
d a x+b
a
x+b
= ae
- Essentially, d x e
- If we wish to differentiate non base-e exponential, e.g. y = a f (x)
dy
f (x)
• d x = f ′(x )(ln a)a , the proof of which is too hard right now.
- E.g. find tangent to y = e 2 x + 1 at point (1,e 2 + 1).
dy
dy
= 2e 2 x, when x = 1,
= 2e 2. Using point-slope formula:
dx
dx
y − (e 2 + 1) = 2e 2 (x − 1) → y = 2e 2 x − e 2 + 1
Natural Logarithm
• loge x usually written as ln x or log x.
- Inverse function to y = e x is
y = ln x
- For the graph y = log x,
• Domain is x > 0, range is all real y.
• x = 0 is vertical asymptote (since
•
•
•
•
ln 0 is undefined)
As x → 0+, y → − ∞.
As x → ∞, y → ∞.
The curve in its basic form is always concave down.
The curve has gradient 1 at x − intercept (1,0).
• To undo exponential, natural log both sides, e x = y → x = ln y
Rates of Change and Exponentials
- For growth and decay, we assume that growth and decay is
dP
= k P.
dt
• The solution of this differential equation is P = A e kt
- e.g. price $P of an item is inflating according to P = 150e 0.04t,
dP
where t is time in years. Find the rate
of inflation.
dt
d
P
If P = 150e 0.04t, the derivative would be
= (0.04)150e 0.04t
dt
= 6e 0.04t
proportional to the population:
15
Manipulating Inverse Trig Graphs
Topic 15: Trigonometry II
x
−1
- Graph y = 5 sin 3 : Thus we establish
Radian Measure
x
• Domain: −1 ≤ 3 ≤ 1 → − 3 ≤ x ≤ 3
• 180 o = π radians, thus
π
o
- 30 = 6
π
y
π
5π
5π
• Range: − 2 ≤ 5 ≤ 2 → − 2 ≤ y ≤ 2
π
o
- 45 = 4
and graph the curve adhering to the new domain/range.
π
o
- Degrees to radians, x × 180
180
- Radians to degrees, x π × π
- Graph y = tan−1( 3 − x 2 ): Thus we establish
• Domain: 3 − x 2 ≥ 0 → − 3 ≤ x ≤ 3
• Range test endpoints: x = 3 → y = tan−1 0 = 0
3 → y = tan−1 0 = 0. Then test midpoint:
π
π
x = 0 → y = tan−1 3 = . Therefore 0 ≤ y ≤
3
3
Arcs and Sectors
x =−
- We know C = 2π r, to find arc l,
θ
× 2π r where θ in degrees
2π
Thus, l = r θ where θ is in radians.
- Further, A = π r 2, segment O A B:
θ
1
AOAB =
× π r 2 = r 2 θ where θ is in radians
2π
2
1
1
To
find
area
of
minor
segment
A B, r 2 θ − r 2 sin θ
•
2
2
Since l =
- Graph y = sin−1 sin x (as a relation sin y = sin x).
• Domain: −1 ≤ sin x ≤ 1 → all real x
Graphing Trig + Inverse Functions
π
π
• Range: − 2 ≤ y ≤ 2 , therefore graph cannot simply be y = x
• Refer to Topic 8 (convert to radians) and Topic 7 respectively.
Instead, it zigzags between −
Inverse Trig Functions
π
π
≤ y ≤ for all real x.
2
2
- For y = sin−1 x,
As y = sin x is not one-to-one, there
is no inverse. But if we restrict
π
π
domain − ≤ x ≤ , then we have
2
2
an inverse. Thus y = sin−1 x:
• Domain: −1 ≤ x ≤ 1
π
π
• Range: − 2 ≤ y ≤ 2
- For y = cos−1 x,
Range should stop at − π
2
≤ y ≤
π
2
As y = cos x is not one-to-one, there
• Domain: −1 ≤ x ≤ 1
• Range: 0 ≤ y ≤ π
Range should stop at 0 ≤ y ≤ π
As y = tan x is not one-to-one, there
π
π
• Range: − 2 < y < 2
As seen from the graphs:
sin−1(−x ) = − sin−1 x (odd function)
•
π
−1
−1
• cos (−x ) = π − cos x (odd function shifted up 2 )
• tan−1(−x ) = − tan−1 x (odd function)
π
Proving Trig Symmetry and Identity
- To prove cos−1(−x ) = π − cos−1 x,
is no inverse. But if we restrict
π
π
domain − < x < (remember
2
2
π
x ≠ + π k , k ∈ ℤ),
2
then we have an inverse. Thus
y = tan−1 x:
• Domain: all real x
Graph identical for y = cos cos−1 x
x = 1 → y = sin sin−1 1 = 1
x = − 1 → y = sin sin−1(−1) = − 1, and then
midpoint x = 0 → y = sin sin−1 0 = 0,
therefore −1 ≤ y ≤ 1.
is no inverse. But if we restrict domain
0 ≤ x ≤ π, then we have an inverse.
Thus y = cos−1 x:
- For y = tan−1 x,
- Graph y = sin sin−1 x
• Domain: −1 ≤ x ≤ 1 (domain of sin−1 x)
• Range (test endpoints):
−1
−1
• sin x + cos x = 2 (∠sum △ )
- Inverse trig sometimes shown as arcsin x , arccos x , arctan x etc.
- cos2 x + sin2 x = 1 : homogeneous of degree 2 in sin x and cos x
y = cos−1(−x ) → − x = cos y for 0 ≤ y ≤ π
∴ x = − cos y. We know that cos(π − y) = − cos y
Therefore cos(π − y) = x → π − y = cos−1 x since 0 ≤ π − y ≤ π.
∴ y = π − cos−1 x, as required.
π
−1
−1
- To prove sin x + cos x = 2 for −1 ≤ x ≤ 1,
It is evident that y = cos−1 x and y = sin−1 x are
π
reflections of each other in the line y = , but
4
algebraically,
Let a = cos−1 x.
∴ cos a = x for 0 ≤ a ≤ π.
π
π
Since sin( − a) = cos a, sin( − a) = x
2
2
π
π
π
π
∴ sin−1 x = ( − a), because − ≤ − a ≤
2
2
2
2
π
sin−1 x + a = , and since a = cos−1 x
2
π
sin−1 x + cos−1 x =
2
16
Compound Angle Formulae
Products to Sums
- If we wanted to find cos(α − β ), refer to
- We know the four compound-angle formulae of sine and cosine:
the unit circle. We know:
A = (cos a , sin a) and B = (cos β , sin β )
sin(α + β ) = sin α cos β + cos α sin β
sin(α − β ) = sin α cos β − cos α sin β
Thus, using distance formula,
A B 2 = (cos α − cos β ) 2 + (sin α − sin β ) 2
cos(α − β ) = cos α cos β + sin α sin β
= cos2 α − 2 cos α cos β − cos2 β + sin2 α − 2 sin α sin β + sin2 β
A B 2 = 2 − 2 cos α cos β − 2 sin α sin β
And, using the cosine rule: c 2 = a 2 + b 2 − 2a b cos C
A B 2 = 1 + 1 − 2 cos(α − β )
Thus, equating the two:
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
- By replacing β with −β, we can find cos(α + β ),
= cos α cos(−β ) + sin α sin(−β ). Cos is even, and sin is odd, thus
• cos(α + β ) = cos α cos β − sin α sin β
π
Using
the identity sin θ = cos( − θ ), we can find sin(α + β ),
2
π
π
sin(α + β ) = cos( − (α + β )) = cos(( − α ) − β )
2
2
π
π
As just proven, we get cos( − α )cos β + sin( − α )sin β
2
2
• sin(α + β ) = sin α cos β + cos α sin β
- Replacing β with −β, we get for sin(α − β ):
• sin(α − β ) = sin α cos β − cos α sin β
- To derive tan(α + β ),
We know tan(α + β ) =
sin(α + β )
sin α cos β + cos α sin β
=
cos(α + β )
cos α cos β − sin α sin β
By dividing top and bottom by cos α cos β, we get:
tan α + tan β
• tan(α + β ) = 1 − tan α tan β
- Since we know tan is odd function, replacing β with −β we get:
tan α − tan β
• tan(α − β ) = 1 + tan α tan β
cos(α + β ) = cos α cos β − sin α sin β
- Adding and subtracting the first two, and doing the same for the
latter two yields the four product to sum identities:
•
•
•
•
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 )
Sums to Products
- Product to sum formulae, sub in A = α + β and B = α − β.
For 2 sin α cos β = sin(α + β ) + sin(α − β ), we get:
1
1
2 sin (A + B )cos (A − B ) = sin A + sin B, rearranged as
2
2
1
1
• sin A + sin B = 2 sin 2 (A + B )cos 2 (A − B )
This process is repeated for the other three products-to-sums, we get:
1
1
• sin A − sin B = 2 cos 2 (A + B )sin 2 (A − B )
1
1
• cos A + cos B = 2 cos 2 (A + B )cos 2 (A − B )
1
1
• cos A − cos B = − 2 sin 2 (A + B )sin 2 (A − B )
Topic 16: Binomials and Pascals
Binomial Expansions
- A binomial expression is one which
contains two terms.
Double-Angle Formulae
- When the coefficients in the
- By replacing both angles of compound angle formulae with θ,
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
(1 + x ) n, there will be n + 1 terms starting with 1, and
•
gradually working through coefficients of n + 1th row until x n.
sin(θ + θ ) = sin θ cos θ + cos θ sin θ
• sin 2θ = 2 sin θ cos θ
cos(θ + θ ) = cos θ cos θ − sin θ sin θ
• cos 2θ = cos2 θ − sin2 θ
We can use the identity sin2 θ + cos2 θ = 1 to create:
• cos 2θ = 2 cos2 θ − 1 as well as,
• cos 2θ = 1 − 2 sin2 θ
tan θ + tan θ
tan(θ + θ ) =
1 − tan θ tan θ
2 tan θ
• tan 2θ = 1 − tan2 θ
The t-formulae
- To represent sin θ, cos θ, tan θ as algebraic functions,
1
θ = t. Thus using the double angle formula we get,
2
2t
• tan θ = 1 − t 2 , from which we draw a right angle triangle:
h 2 = 4t 2 + 1 − 2t 2 + t 4 = (t 2 + 1) 2
Therefore AC = 1 + t 2. We thus get:
2t
• sin θ = 1 + t 2
1 − t2
• cos θ = 1 + t 2
Let tan
expansions of (1 + x ) n are arranged
• If x positive, all terms positive. If x negative, terms alternate sign.
General Binomial Expansions
- (x + y) n can be expanded with x n beginning, ending with y n, with
coefficients of middle terms determined by Pascal’s triangle.
- In each term, the sum of indices of x and y is n.
k
n
n
• Ck = ( k ) is the coefficient of x in (1 + x )
n
n
2
n
• Thus, (1 + x ) = ( 0 ) + ( 1 )x + ( 2 )x + . . . + ( n )x
n
n
n
n
n
n
n−1
n−2 b 2 + . . . +
bn
• (a + b) = ( 0 )a + ( 1 )a b + ( 2 )a
(n)
- From Pascal’s triangle, we also get:
n
n
n
n
• nCk = n−1Ck−1 + n−1Ck where 1 ≤ k ≤ n − 1. Proving,
(1 + x ) n = (1 + x )( n−1C0 + n−1 Ck−1 x k−1 + n−1 Ck x k . . .n−1 Cn−1 x n−1)
We see that (1)( n−1Ck x k ) and (x )( n−1Ck−1 x k−1) are the only
terms with x k, and thus the coefficient will be n−1Ck + n−1 Ck−1.
17
• nCk = nCn−k where 1 ≤ k ≤ n − 1 (triangle symmetrical)
• nC0 = nCn = 1 (first/last in every row always 1)
Binomial Theorem
n
2
n
• (1 + x ) = ( 0 ) + ( 1 )x + ( 2 )x + . . . + ( n )x
n
n
n
n
n
n
n k
nC x k =
=
x
∑ k
∑ (k)
k=0
k=0
n n
n n−1
n n−2 2
n n
n
b + ... +
b
• (a + b) = ( 0 )a + ( 1 )a b + ( 2 )a
(n)
n
n
n n−k k
nC a n−k b k =
=
a
b
∑ k
∑ (k)
k=0
k=0
3 11
- For example, find the 5th term in the expansion of (5a − b ) ,
3
We know Tk+1 = 11Ck (5a)11−k (− ) k. ∴ k = 4
b
3 4
7
T4 = 11C4 (5a) (− )
b
2088281250a 7
=
b4
Relations between Binomial Coefficients
n
n
k=0
k=0
n
k
n
n
- Since (1 + x ) = ∑ Ck x , if we wanted to find ∑ Ck,
simply make x = 1, and thus we get:
n
nC (i.e., sum all numbers in row (k − 1), get 2n)
2n =
k
∑
•
k=0
- We have 2n = nC0 + nC1 + nC2 + . . . as proven. Let this be (1)
n
n
k
n
- If we let x = − 1 in, (1 + x ) = ∑ Ck x , however, you will get
k=0
(1 − 1) n = nC0 − nC1 + nC2 − nC3 + . . . = 0, let this be (2).
∴ (1) − (2) : 2n = 2nC1 + 2nC3 + 2nC5 + . . . and thus
n−1 = n C + n C + n C + . . .
• 2
1
3
5
- However, (1) + (2) : 2n = 2nC0 + 2nC2 + 2nC4 + . . . yielding,
n−1 = n C + n C + n C + . . . = n C + n C + n C + . . .
• 2
0
2
4
1
3
5
n
n
n
n
k
n
- To find ∑ k Ck, we must differentiate (1 + x ) = ∑ Ck x
k=1
k=0
n
n−1
n
k−1
Diff. both sides, we get n (1 + x )
(1) =
k Ck x
∑
k=0
n
To get rid of x, let x = 1 : n (2) n−1 =
k nCk
∑
k=0
n
n
Since 0 C0 = 0, we can conclude
k nCk = n (2) n−1
∑
k=1
Comparing Coefficients
- By equating the coefficients of x n on both sides of the identity
n 2 (2n)!
=
∑ (k)
(n !) 2
k=0
n
n
n 2
n n
We know (1 + x ) n =
+
x+
x + ... +
x thus
(0) (1)
(2)
(n)
n
n
n 2
n n
(1 + x ) n (1 + x ) n = (
+
x+
x + ... +
x )×
(0) (1)
(2)
(n)
(1 + x ) n (1 + x ) n ≡ (1 + x ) 2n, show
n
n
n
n 2
n n
(
+
x+
x + ... +
x )
(0) (1)
(2)
(n)
Finding coefficient of terms that multiply together to get x n, we get
n
n
n
n
n
n
n
n
n
n
+
+
+ ... +
+
( 0 )( n ) ( 1 )( n − 1 ) ( 2 )( n − 2 )
( n − 1 )( 1 ) ( n )( 0 )
n
n
=
, thus we can rewrite the coefficient:
( k ) (n − k)
n
n 2
n 2
n 2
n 2
n 2
=
+
+
+ ... +
=
∑
(0)
(1)
(2)
(n)
(k)
k=0
n
2n
Now looking at coefficient of x in (1 + x ) :
But we know
2n
2n
2n n
2n 2n
+
x + ... +
x + ... +
x
(0) (1)
(n)
( 2n )
2n
Thus we see the coefficient of x n =
. And therefore we get:
(n)
n
n 2
2n
=
, and writing in factorial notation,
∑ (k)
(n)
k=0
n
n 2 (2n)!
as required.
=
∑ (k)
(n !) 2
k=0
(1 + x ) 2n =
Topic 17: Rates of Change
Related Rates
- Where more than one rate, we must differentiate using chain rule
with respect to t, creating a relation between two rates.
dy
dy
dx
• i.e. using d t = d x × d t
- E.g. an ice cube of side length x = 20 is melting so that its
dimensions decrease at 1cm/s. What rate is its volume decreasing
when each edge is 5cm long?
dV
dx
= − 1, and that
We are trying to find
. We know that
dt
dt
dV
= 3x 2. Through the chain rule, we deduce:
V = x 3 thus
dx
dV
dV dx
=
×
= 3x 2 × −1 = − 3x 2 and subbing x = 5, we get
dt
dx
dt
dV
= − 75, thus volume will be decreasing at 75cm3/s.
dt
dy
dx dy
dz
- When there are 3 rates, we can use d t = d t × d z × d x
For example, a spherical balloon is expanding so that its volume
expands at a constant rate of 70mm3/s. What is the rate of increase
in surface area when r = 10mm?
dS
dV
4
Here we want to find
, and we know
= 70 and V = π r 3,
3
dt
dt
dV
2
thus
= 4π r . However, it is difficult to link V and S using
dr
dS
dV dS
dr
regular chain rule so we can use
.
=
×
×
dt
dt
dr
dV
dS
dr
1
.
= 8π r, and
=
S = 4π r 2, ∴
dr
dV
4π r 2
dS
140
dS
. When r = 10,
∴
=
= 14, i.e. increasing at 14mm2/s.
dt
r
dt
Growth and Decay
dP
• Growth and decay is proportion to population, i.e. d t = k P
• Solution is P = A e kt, where
- P = population at time t
- k = growth/decay constant
- A = initial population
- t = time
- Proof:
P = A e kt
dP
= k A e kt
dt
dP
= kP
dt
18
- E.g. 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.
dP
= k P and P = A e kt, we know A = 1732 thus P = 1732e kt.
dt
When t = 10, P = 1260, i.e. 1260 = 1732e 10k
1260
1
1260
→k =
log
= − 0.0318(4d p)
Thus, 10k = log
1732
10
1732
∴ growth rate is −3%.
- In how many years will the population be half that in 1960?
1732
1
= 688, thus e kt = .
Let 866 = 1732e kt as
2
2
1
1260
1
1
1
k t = log → t = log , and we know k =
log
2
k
2
10
1732
1
1
∴t =
× log = 21.786.
1260
1
2
log
10
1732
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.
dP
• d t = k (P − N ), where N is a fixed constant barrier.
• The solution is thus P = N + A e kt where
- P = population at time t
- k = growth/decay constant
- N + A = initial population
- t = time
dP
kt
- Proof that P = N + A e given d t = k (P − N ):
Let y = P − N be difference between P and N.
dy
dP
Thus
=
− 0 because N is constant.
dt
dt
dy
Therefore
= k (P − N ) = k y since y = P − N
dt
This is the same as the ordinary exponential model, so we can use
P = A e kt i.e. y = y0 e kt and substitute y = P − N in:
∴ P = N + A e kt as required.
nom nom…
• Newton’s law of cooling states that the rate of change of
temperature is proportional to (T − A ), where T is indoor
temperature and A is constant outdoor temperature, i.e
dT
= k (T − A )
dt
- E.g. the outdoor temperature is 5oC and a heater malfunction has
caused the inside temperature to drop from 20 oC to 17oC in half
an hour. After how many hours is inside temperature 10 oC ?
∴ T = 5 + A e kt but when t = 0, T = 20 and thus A = 15
T = 5 + 15e kt but we are told when t = 0.5, T = 17 i.e.
12
, and then log both sides:
17 = 5 + 15e 0.5k, thus e 0.5k =
15
12
12
→ k = 2 log
0.5k = log
15
15
We want to know what t is when T = 10, thus 10 = 5 + 15e kt.
1
1
12
We get = e kt thus log = k t but we know k = 2 log
,
3
3
15
1
log
3
= 2.46167…
∴t =
12
2 log
15
1
Therefore after about 2 hours, the temperature dropped to 10 oC.
2
19