Pure:
1. Algebraic expressions
1. 1. Proof by contradiction
Steps:
1. Assume statement is false
2. Prove contradiction
3. Proving it right contradicts step 1 so it would prove the original statement
4. Write conclusion
Important:
● Odd number: 2n+1
● Even number: 2n
● Prime number (p1 + p2 +p3) + 1
a
● Rational number: , where a and b are integers
b
Important questions:
1.
, 2.
1. 2. Algebraic fractions
Work in the same way as numeric fractions. You can cancel common factors with common
denominators.
To multiply fractions:
1. Cancel common factors
2. Multiply numerators
3. Multiply denominators
To divide fractions: multiply the 1st one by the denominator of the 2nd one.
To add/subtract fractions: find common denominators.
1. 3. Partial fractions
A single fraction with 2 different linear factors in the denominator can be split into 2 separate
fractions with linear denominators, these are called partial fractions.
Steps:
1. Split fraction into the 2 or more different linear factors
and add the constants to find (A and B)
2. Multiply constants by the other linear factors that aren’t
below it
A ( x−4)+B ( x+1)
3. Use a. Substitution or b. equating coefficients ¿
( x +1)( x−4)
a. Substitution:
1. Let x = a value so the linear factor = 0 and solve
, A( x−4)+ B(x +1) A (4−4)+ B(4+1)
E.g. x = 4→ ¿ →5B = 5 →B = 1
( x+1)( x−4) (x +1)( x−4 )
Then let x = the other value
A( x−4)+ B( x +1) A (−1−4)+ B(−1+1)
x = -1 → ¿ →-5A→A = 1
( x+1)( x−4) (x +1)( x−4 )
b. Equating coefficients:
1. Expand brackets
2. Form simultaneous equations
3. Solve
1. 4. Repeated factors
Single fractions with repeated linear factors can also be split into separate fractions.
In these cases, there are 3 linear factors in the denominator: (x-5), (x+3) and (x+3). So you need
to split the fraction into 3 partial fractions. Notice, the repeated factor has 1 denominator with
the power and the other without it.
They are solved in the same way as normal partial fractions.
1. 5. Algebraic divisions
Improper algebraic fractions have a numerator with a degree equal to or larger than the
denominator.
2 3
x +5 x +8 x +5 x−9
E.g. improper fractions: or
x−2 x −4 x 2 +7 x−3
3
An improper fraction must be turned into a mixed one to express it in partial fractions. To do
this, you can use 1. Algebraic division or 2. The relationship f(x) = q(x) x divisor + remainder
1. Algebraic division: use long division, using the denominator as the divisor
Example:
, 2. Relationship: multiply by (x-2) and compare coefficients
Example:
2. Functions and graphs
2. 1. The modulus function
The modulus of a, is written as |a|, and its non-negative numerical value, meaning that
whatever it is, it will always be positive.
In general, a modulus function is of type y = |f ( x)|
● When f(x)⪰0,|f ( x)| = f(x)
● When f(x)⪯0, |f ( x)| = -f(x)
To sketch a graph of y=|ax +b|, sketch y = ax + b and
reflect the section of the graph below the x-axis in the
x-axis
1. Algebraic expressions
1. 1. Proof by contradiction
Steps:
1. Assume statement is false
2. Prove contradiction
3. Proving it right contradicts step 1 so it would prove the original statement
4. Write conclusion
Important:
● Odd number: 2n+1
● Even number: 2n
● Prime number (p1 + p2 +p3) + 1
a
● Rational number: , where a and b are integers
b
Important questions:
1.
, 2.
1. 2. Algebraic fractions
Work in the same way as numeric fractions. You can cancel common factors with common
denominators.
To multiply fractions:
1. Cancel common factors
2. Multiply numerators
3. Multiply denominators
To divide fractions: multiply the 1st one by the denominator of the 2nd one.
To add/subtract fractions: find common denominators.
1. 3. Partial fractions
A single fraction with 2 different linear factors in the denominator can be split into 2 separate
fractions with linear denominators, these are called partial fractions.
Steps:
1. Split fraction into the 2 or more different linear factors
and add the constants to find (A and B)
2. Multiply constants by the other linear factors that aren’t
below it
A ( x−4)+B ( x+1)
3. Use a. Substitution or b. equating coefficients ¿
( x +1)( x−4)
a. Substitution:
1. Let x = a value so the linear factor = 0 and solve
, A( x−4)+ B(x +1) A (4−4)+ B(4+1)
E.g. x = 4→ ¿ →5B = 5 →B = 1
( x+1)( x−4) (x +1)( x−4 )
Then let x = the other value
A( x−4)+ B( x +1) A (−1−4)+ B(−1+1)
x = -1 → ¿ →-5A→A = 1
( x+1)( x−4) (x +1)( x−4 )
b. Equating coefficients:
1. Expand brackets
2. Form simultaneous equations
3. Solve
1. 4. Repeated factors
Single fractions with repeated linear factors can also be split into separate fractions.
In these cases, there are 3 linear factors in the denominator: (x-5), (x+3) and (x+3). So you need
to split the fraction into 3 partial fractions. Notice, the repeated factor has 1 denominator with
the power and the other without it.
They are solved in the same way as normal partial fractions.
1. 5. Algebraic divisions
Improper algebraic fractions have a numerator with a degree equal to or larger than the
denominator.
2 3
x +5 x +8 x +5 x−9
E.g. improper fractions: or
x−2 x −4 x 2 +7 x−3
3
An improper fraction must be turned into a mixed one to express it in partial fractions. To do
this, you can use 1. Algebraic division or 2. The relationship f(x) = q(x) x divisor + remainder
1. Algebraic division: use long division, using the denominator as the divisor
Example:
, 2. Relationship: multiply by (x-2) and compare coefficients
Example:
2. Functions and graphs
2. 1. The modulus function
The modulus of a, is written as |a|, and its non-negative numerical value, meaning that
whatever it is, it will always be positive.
In general, a modulus function is of type y = |f ( x)|
● When f(x)⪰0,|f ( x)| = f(x)
● When f(x)⪯0, |f ( x)| = -f(x)
To sketch a graph of y=|ax +b|, sketch y = ax + b and
reflect the section of the graph below the x-axis in the
x-axis
