Xa x Xb = Xa+b
Maths Xa / Xb = Xa-b
(Xa)b = Xab
Yr 1 Pure Ch 1 Summary
X2 - a2 = (X-a)(X+a)
Algebra X-a = 1/Xa
[7] X1/a = a√ X
X0 = 1
Maths The discriminant (b2 – 4ac) can be used to find the
number of intersections.
Yr 1 Pure Ch 3 Summary
For number lines o means > or <, means ≤∨≥
Equations & Equalities For graphs dotted line means > or <, uninterrupted
[3] line means ≤∨≥ .
Domain- the set of possible inputs
Maths Range- the set of possible outputs
Yr 1 Pure Ch 2 Summary b2 – 4ac is the discriminant.
Quadratics < 0 means no real roots
= 0 means 1 real root
[6] > 0 means 2 distinct real roots
k
y=
x
Maths
Yr 1 Pure Ch 4 Summary
k
Graphs y=
x2
a) x3 graph [1] Steeper
than
b) x2 graph [1]
c) x4 graph [2]
d) k/x graph [1]
y = ax4 + bx3 + cx2 + dx + e
e) k/x2 graph [1]
, a)
a
Maths Vector = right a, up b
b
Yr 1 Pure Ch 4 Summary f (x + a) + b = left a, up b
b)
Transformations -f(x) reflects in the x-axis
a) Translation [2] f(-x) reflects in the y-axis
b) Reflection [2] c)
x
c) Stretching [2] f(ax) = scale factor
a
af(x) = scale factor a
Equation of a circle is (x-a)2 + (y-b)2 = r2, where (a,b)
is the centre & r is the radius.
A tangent to the circle is perpendicular to the radius
@ the point of contact.
Maths The perpendicular bisector of a chord will go
Yr 1 Pure Ch 6 Summary through the circle centre.
The angle in a semicircle is always a right angle.
Circles
[4]
n(n−1)
To find the nth row of Pascal’s triangle: 1, n,
2!
n ( n−1 ) ( n−2)
, , etc or use nCr
3!
n
Maths n
Cr is aka ( )
r
Yr 1 Pure Ch 8 Summary n!
To find nCr , do
Binomial Expansion th
r ! ( n−r ) !
The (n+1) row of Pascal’s triangle gives the
[5] coefficients of the expansion of (a + b)n
n
The binomial expansion is: (a + b) n = an + ( ) an-1 b + (
1
n n-2 2 n
) a b + … ( ) an-r br + … b
2 r
Maths a)
0 30 45 60 90
Yr 1 Pure Ch 10 Summary
Sin 0 ½
√2 √3 1
Trig identities & ratios 2 2
a) Fill the grid [15] Cos 1
√3 √2 ½ 0
2 2
0 30 45 60 90
√3
Sin Tan 0
3
1 √3 n/a
Cos b)
sin x
Tan Tan x =
cos x
b) Other trig identities [2] Sin2x + Cos2x = 1
Maths Xa / Xb = Xa-b
(Xa)b = Xab
Yr 1 Pure Ch 1 Summary
X2 - a2 = (X-a)(X+a)
Algebra X-a = 1/Xa
[7] X1/a = a√ X
X0 = 1
Maths The discriminant (b2 – 4ac) can be used to find the
number of intersections.
Yr 1 Pure Ch 3 Summary
For number lines o means > or <, means ≤∨≥
Equations & Equalities For graphs dotted line means > or <, uninterrupted
[3] line means ≤∨≥ .
Domain- the set of possible inputs
Maths Range- the set of possible outputs
Yr 1 Pure Ch 2 Summary b2 – 4ac is the discriminant.
Quadratics < 0 means no real roots
= 0 means 1 real root
[6] > 0 means 2 distinct real roots
k
y=
x
Maths
Yr 1 Pure Ch 4 Summary
k
Graphs y=
x2
a) x3 graph [1] Steeper
than
b) x2 graph [1]
c) x4 graph [2]
d) k/x graph [1]
y = ax4 + bx3 + cx2 + dx + e
e) k/x2 graph [1]
, a)
a
Maths Vector = right a, up b
b
Yr 1 Pure Ch 4 Summary f (x + a) + b = left a, up b
b)
Transformations -f(x) reflects in the x-axis
a) Translation [2] f(-x) reflects in the y-axis
b) Reflection [2] c)
x
c) Stretching [2] f(ax) = scale factor
a
af(x) = scale factor a
Equation of a circle is (x-a)2 + (y-b)2 = r2, where (a,b)
is the centre & r is the radius.
A tangent to the circle is perpendicular to the radius
@ the point of contact.
Maths The perpendicular bisector of a chord will go
Yr 1 Pure Ch 6 Summary through the circle centre.
The angle in a semicircle is always a right angle.
Circles
[4]
n(n−1)
To find the nth row of Pascal’s triangle: 1, n,
2!
n ( n−1 ) ( n−2)
, , etc or use nCr
3!
n
Maths n
Cr is aka ( )
r
Yr 1 Pure Ch 8 Summary n!
To find nCr , do
Binomial Expansion th
r ! ( n−r ) !
The (n+1) row of Pascal’s triangle gives the
[5] coefficients of the expansion of (a + b)n
n
The binomial expansion is: (a + b) n = an + ( ) an-1 b + (
1
n n-2 2 n
) a b + … ( ) an-r br + … b
2 r
Maths a)
0 30 45 60 90
Yr 1 Pure Ch 10 Summary
Sin 0 ½
√2 √3 1
Trig identities & ratios 2 2
a) Fill the grid [15] Cos 1
√3 √2 ½ 0
2 2
0 30 45 60 90
√3
Sin Tan 0
3
1 √3 n/a
Cos b)
sin x
Tan Tan x =
cos x
b) Other trig identities [2] Sin2x + Cos2x = 1
