Pure mathematics 1(P1)
1.1 Quadratics
1.2 Functions
1.3 Coordinate geometry
1.4 Circular measure
1.5 Trigonometry
1.6 Series
1.7 Differentiation
1.8 Integration
, 1.1 Quadratics
the
completing Square
+
ax
+
bx c a(x p)+
q
=
+ +
Vertex: x =
-
p,y q =
·
y
X q
x
-
p
a tve,
=
minpt.
a - ve ,
= max pt.
Discriminant
b2-4ac
=
0 (one real roots) "pt of intersection, line is tangent to the curve
32-49s>0 (two real roofs-two distinct real roots t wo same
of real roots. realisti) (repeated
Find costs: (x -
1(0 0)
=
↓
bo-4aC30 (two distinct
real roots)
< X apt of intersection
b3-4ac<0 (no roots the line does intersect
not the
curve
Find roots (a,b) *
↳
a
*ab
Solve quadratic equation/inequality
(1 unknown (
(i) By factorisation i
x5 -
5x 6 +
(a
= -
5)(x 5) -
(ii)Byforniaa set
(iii) By competing the square
a(x +
p)c+ q
340 jx + +
q =
3(u- 2x+ 3) +
-
=
3/(x+1) 115]
3[(x 1) +2]
-
= +
3(x +1)2 b
=
+
To show that aquadratic function alwayshaveforall
is of
value c
always thecompeting the square >
always ve:
-
competing the square <0
E.g. aset
To show that isalways heforall of
value se
a
+
+x +1 = (x + t) +
Eacompeting the
square
(x + 1)">,8& more than 8
(a ) +1 -4
+
+ x
...
S o i salwaysthe for alla rathers
1.1 Quadratics
1.2 Functions
1.3 Coordinate geometry
1.4 Circular measure
1.5 Trigonometry
1.6 Series
1.7 Differentiation
1.8 Integration
, 1.1 Quadratics
the
completing Square
+
ax
+
bx c a(x p)+
q
=
+ +
Vertex: x =
-
p,y q =
·
y
X q
x
-
p
a tve,
=
minpt.
a - ve ,
= max pt.
Discriminant
b2-4ac
=
0 (one real roots) "pt of intersection, line is tangent to the curve
32-49s>0 (two real roofs-two distinct real roots t wo same
of real roots. realisti) (repeated
Find costs: (x -
1(0 0)
=
↓
bo-4aC30 (two distinct
real roots)
< X apt of intersection
b3-4ac<0 (no roots the line does intersect
not the
curve
Find roots (a,b) *
↳
a
*ab
Solve quadratic equation/inequality
(1 unknown (
(i) By factorisation i
x5 -
5x 6 +
(a
= -
5)(x 5) -
(ii)Byforniaa set
(iii) By competing the square
a(x +
p)c+ q
340 jx + +
q =
3(u- 2x+ 3) +
-
=
3/(x+1) 115]
3[(x 1) +2]
-
= +
3(x +1)2 b
=
+
To show that aquadratic function alwayshaveforall
is of
value c
always thecompeting the square >
always ve:
-
competing the square <0
E.g. aset
To show that isalways heforall of
value se
a
+
+x +1 = (x + t) +
Eacompeting the
square
(x + 1)">,8& more than 8
(a ) +1 -4
+
+ x
...
S o i salwaysthe for alla rathers
