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William continues to provide, excellent support to my son - helping him to go beyond being able to answer a complex question with a by rote, step by step approach - to having a much fuller understanding of the concepts. This mean that he feels much better equipped to deal with what the exam might throw at him. Thanks very much Will : )

Susan, Parent from Hertfordshire

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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after Online Lessons. In my Online Lessons, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the Online Lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

This is an example of an inequalities question from FP2. For this, we will need to use the tools learned in this chapter. To start with, it may be tempting to multiply both sides of the inequality by (x+3) to get rid of the fraction, but doing this is wrong since in the case that (x+3) is negative (when x < -3), the direction of the inequality will not be preserved. Hence, we proceed by multiplying both sides by (x+3)^{2} (which is always non-negative). We then arrive at (x+3)^{2}(x+4) > 2(x+3). Using algebraic rearrangement and factorisation we can then get to (x+3)[(x+3)(x+4)-2] > 0. This is a good place to get to, since we can see that there is a quadratic (which we can factorise) in the second term. Expanding this out we reach (x+3)(x^{2}+7x+10) > 0. Now we can factorise the quadratic (we find 2 numbers 5 and 2 that add to 7 and multiply to 10) to get (x+3)(x+5)(x+2) > 0. We can clearly see this is a cubic expression on the left hand side. Now we can draw the graph y = (x+3)(x+5)(x+2), which must intersect the x axis at x = -5, -3 and -2 (since these value of x give a y value of 0). Now, looking at the annotated graph, we can see that the desired region (where y < 0) must be where x > -2 or -5 < x < -3. Note that we use strict inequality here and not equality aswell since if x were eqeal to these values, y would be equal to 0, which is outside of the constraint.

By definition z* = a - bj.
We can write z/z* = ((a+bj)/(a-bj))*(a+bj)/(a+bj).
We calculate this to be z/z* = (a^2-b^2)/(a^2+b^2) + j(2ab)/(a^2+b^2).
Therefore, Re(z/z*) = (a^2-b^2)/(a^2+b^2).
Im(z/z*) = (2ab)/(a^2+b^2).

Answered by Penelope J.

Studies Natural Sciences (Physics) at Cambridge

Multiply by complex conjugate
z = 50 / (3+4i) * (3-4i) / (3-4i)
Rationalise
z = 50 ( 3 - 4i) / 25 = 6 - 8i.

Let's say that T is a transformation of the two dimensional plane. Remember that we have the two standard unit vectors (1,0) and (0,1). These are, respectively, the unit vectors pointing in the positive direction on the x-axis and the y-axis. We first look at what the transformation does to these two vectors. This gives us two new vectors T(1,0) and T(0,1) which form the columns of the matrix corresponding to the transformation T.
For example, if T is the reflection in the y-axis we get the following. Since we reflect in the y-axis, all points on the y-axis stay fixed and so T(0,1) = (0,1). On the other hand, by reflection (1,0) in the y-axis we get the point (-1,0). Therfore, the matrix has columns (-1,0) and (0,1).

Firstly you should expand the brackets in this situation in order to collect the like terms, so get all the x's on one side and all the constants on the other side of the inequality. Expanding the bracket you get 1 - 2x + 6 > 4x. Now we try to collect the like terms so firstly I will add 2x to both sides to get 1 + 6 > 4x + 2x. Then we can simplify to get 7 > 6x. Finally to get x on it's own we divide through by 6 to get the final answer of x < 7/6. Note that if the 6 was negative then you would have to flip the sign of the inequality if you divided through by the -6.

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Our Further Maths A Level Tutors will give you that focused, individual attention which could make all the difference to your grade. Gaps in your learning, careless mistakes and ineffective time-management during the exams can all be overcome with the expert guidance of a Further Maths A-Level Tutor. Your Maths A Level Tutor will also ensure you have plenty of past papers to complete, as practice, repetition and excellent exam technique are key to Maths A Level success.