Are you confused about the new topics introduced in GCSE Mathematics (9 - 1) specification?
Here is the solution; look no further than this small book.
The book, GCSE Mathematics (9-1) New Topics
, shows you how to master the new challenging topics with worked examples and images. The contents are offered in such a way that even a student with a modest
knowledge in elementary mathematics can understand them with ease.
This is the third book in Maths in Pamphlets Series
by the publisher on Amazon.
Steps have been taken to cater for all examination boards in the UK.
Every effort has been made to ensure the contents are rolled out smoothly so that the young students can grasp the concepts in an enjoyable manner, while making an irreversible, linear progress.
The topics include:
- The gradient of a curve
- Areas under the curves
- Venn Diagrams and Probability
- Composite Functions and Inverse Functions
- Linear, Quadratic and Geometric Sequences
- Trigonometric Values of Major Angles
- Turning Point of a Quadratic Curve
- A Bonus Chapter - algebraic proof and capture-recapture method
Some of the contents take the following form:
Finding the turning point of a quadratic curve
x2 - 2x - 5 = 0
By completing the square, you turned the above into,
(x - 1)2 -6 = 0
This is a translation of the curve, y = x2, 1 in the x-direction and 6 in the -y-direction.
So, the coordinates of the minimum point are (1,-6)
Finding the inverse function of a function
Let f(x) = 2x - 3
y = 2x - 3
Now, swap x and y around
x = 2y - 3
Make y the subject
y = (x + 3)/2 - this gives the inverse function
f-1(x) = (x + 3)/2
Iteration - finding a solution of an equation
If f(x) = 0, then rearrange it in x = g(x) form, where both f(x) and g(x) are functions of x
Let f(x) = x3 - 5x2 - 5
x3 - 5x2 - 5 = 0
x3 = 5x2 + 5
Divide both sides by x2
x = 5 + 5/x2
The iterative formula becomes,
xn+1 = 5 + 5/x2n
So, if you know xn, xn+1 can be found, provided that the initial value, x0 is given.
If x0 = 2, the values progress as follows:
- x0 = 2
- x1 = 6.25
- x2 = 5.128
- x3 = 5.1901
- x4 = 5.1856
- x5 = 5.1859
- x6 = 5.1859
Since the values approach a constant, the solution is taken as x = 5.186(3 d.p.)
Finding the nth term of a quadratic sequence - a unique, simple approach
Find the nth term of the sequence, 2, 5, 10, 17, 26...
The differences between the consecutive terms of the above: 3, 5, 7, 9
The differences between the consecutive terms: 2, 2, 2 of the above
So, the sequence is quadratic.
Let the nth term, N = an2 + bn + c, where a, b and c are constants to be found.
Since there are three unknowns, we need to make three equations.
n = 1; N = 2 => 2 = a + b + c
n = 2; N = 5 => 5 = 4a + 2b + c
n = 3; N = 10 => 10 = 9a + 3b + c
From the first two, we get:
3 = 3a + b
From the last two, we get:
5 = 5a + b
By solving the simultaneous equations, we get,
a = 1 and b = 0; sub them in the first equation,
c = 1
So, N = n2 + 1.
There are many more of these in the book. Please read them, practise and excel in the forthcoming