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A Level Pure Maths Summary Notes
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A LEVEL PURE MATHS REVISON NOTES1 ALGEBRA AND FUNCTIONS
a) INDICES
Rules to learn :
𝑛 𝑛
1 𝑚 𝑚
𝑥 𝑎 × 𝑥 𝑏 = 𝑥 𝑎+𝑏 𝑥 𝑎 ÷ 𝑥 𝑏 = 𝑥 𝑎−𝑏 (𝑥 𝑎 )𝑏 = 𝑥 𝑎𝑏 𝑥 −𝑎 = 𝑥 𝑚 = √𝑥 𝑛 = ( √𝑥 )
𝑥𝑎
3 1
Simplify 2𝑥(𝑥 − 𝑦)2 + 3(𝑥 − 𝑦)2 Solve 32𝑥 × 25𝑥 = 15
1
= (𝑥 − 𝑦) (2𝑥(𝑥 − 𝑦) + 3))
2 (3 × 5)2𝑥 = 151
1
= (𝑥 − 𝑦) (2𝑥 2 − 2𝑥𝑦 + 3)
2 2𝑥 = 1
1
𝑥=
2
b) SURDS
• A root such as √3 that cannot be written as a fraction is IRRATIONAL
• An expression that involves irrational roots is in SURD FORM
• RATIONALISING THE DENOMINATOR is removing the surd from the denominator (multiply by the conjugate)
2
Simplify Rationalise the denominator 2−√3
The conjugate of the denominator
2 2+√3 2 - √3 is 2 + √3 so that
√75 − √12 = ×2+
2−√3 √3 (2 - √3)( 2 + √3)
= √5 × 5 × 3 − √2 × 2 × 3 = 22 - √32
= 5√3 − 2√3 = 4 + 2√3 =1
= 3√3
c) QUADRATIC EQUATIONS AND GRAPHS
Factorising – identifying the roots of the equation ax2 + bx + c = 0
• Look for the difference of 2 squares x2 – a2 = (x + a)(x – a) or (ax)2 - b2 = (ax + b)( ax – b)
• Look for the perfect square x2 + 2ax + a2 = (x + a)2
• Look out for equations which can be transformed into quadratic equations
12
Solve 𝑥 + 1 − 𝑥 = 0 Solve 6𝑥 4 − 7𝑥 2 + 2 = 0
𝑥 2 + 𝑥 − 12 = 0 Let z = x2 6𝑧 2 − 7𝑧 + 2 = 0
(𝑥 + 4)(𝑥 − 3) = 0 (2𝑧 − 1)(3𝑧 − 2) = 0
1 1 2 2
x = -4 x = 3 𝑧= 2
𝑥 = ±√2 𝑧= 3
𝑥 = ±√3
Completing the square – identifying the vertex and line of symmetry
y = (x + a)2 + b vertex at (-a , b) line of symmetry as equation x = -a
Line of symmetry
x=2
www.mathsbox.org.uk
,Quadratic formula (and the DISCRIMINANT)
−𝑏±√𝑏2 −4𝑎𝑐
𝑥= for solving ax2 + bx + c = 0
2𝑎
The DISCRIMINANT b2 – 4ac can be used to identify the number of roots
b2 – 4ac > 0 there are 2 real distinct roots (graph crosses the x-axis twice)
b2 – 4ac = 0 there is a single repeated root (the x-axis is a tangent)
b2 – 4ac < 0 there are no real roots (the graph does not touch the x-axis)
d) SIMULTANEOUS EQUATIONS
Solving by elimination
3x – 2y = 19 × 3 9x – 6y = 57
2x – 3y = 21 × 2 4x – 6y = 42
5x – 0y =15 x = 3 ( 9 – 2y = 19) y = -5
Solving by substitution
x + y = 1 (y = 1 – x)
x2 + y2 = 25 x2 + (1 – x)2 = 25
2x2 – 2x – 24 = 0
2(x – 4)(x + 3) = 0 x = 4 y = -3 x=-3 y=4
If you end up with a quadratic equation when solving simultaneously the discriminant can be used to determine the
relationship between the graphs
If b2 – 4ac > 0 the graphs intersect at 2 distinct points
b2 – 4ac = 0 the graphs intersect at 1 point (or tangent)
b2 – 4ac < 0 the graphs do not intersect
e) INQUALITIES
Linear Inequality - solve using the same method as solving a linear equation but remember to reverse the
inequality if you multiply or divide by a negative number
Quadratic Inequality – always a good idea to sketch a graph
≤ ≥ plot the graph as a solid line or curve
< > plot as a dotted/dashed line or curve
If you are unsure of which area to shade pick a point in one of the regions and check the inequalities using the
coordinates of the point
f) POLYNOMIALS
www.mathsbox.org.uk
, • A polynomial is an expression which can be written in the form axn + bxn-1 + cxn-2 + … where a,b, c are
constants and n is a positive integer.
• The order of the polynomial is the highest power of x in the polynomial
• Polynomials can be divided to give a Quotient and Remainder
• Factor Theorem – If (x – a) is a factor of f(x) then f(a) = 0 and is root of the equation f(x) = 0
Show that (x – 3) is a factor of x3 – 19x + 30 = 0
f(x) = x3 – 19x + 30
f(3) = 33 -19 × 3 + 20
=0
f(3) = 0 so x – 3 is a factor of f(x)
g) GRAPHS OF FUNCTIONS
Sketching Graphs
• Identify where the graph crossed the y-axis (x = 0)
• Identify where the graph crossed the x-axis (y = 0)
• Identify any asymptotes and plot with a dashed line
𝑎 𝑎
y=mx + c y = kx2 y=kx3 y= Asymptotes at
y= Asymptotes at
𝑥 x = 0 and y = 0 𝑥2
x = 0 and y = 0
y is proportional to x2 𝑘 𝑎
y is proportional to x2 y is proportional to y is proportional to
𝑥 𝑥2
Modulus Graphs
• |x| is the ‘modulus of x’ or the absolute value |2|=2 |-2|= 2
• To sketch the graph of y = |f(x)| sketch y = f(x) and take any part of the graph which is below the x-axis and
reflect it in the x-axis
Solve |2x - 4|<|x|
2x - 4 = x 2x – 4 = -x
x =4 3x = 4
4
x=3
4
3
<x<4
www.mathsbox.org.uk
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