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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 a fairly typical example of a question from the Further Maths syllabus.

We wish to demonstrate that for all integers n greater than or equal to 4, n! > n^{2} .

We briefly recap that n! is the product of all numbers from 1 up to n. For example 4! = 4x3x2x1 = 24.

As with any induction question, there are four steps that we will work through in turn. They are the base case, the induction hypothesis, the inductive step, and finally the conclusion.

We begin with the base case. Here we select the minimum integer satisfying the conditions, in this case n=4, and demonstrate that the inequality holds true. Clearly 4! = 24 > 4^{2} = 16

The second step is to assume our induction hypothesis.

For this, we simply assume that the inequality holds for n=k, for any valid integer k.

That is, we assume k! > k^{2}.

We will now use this in step three to show that IF the inequality holds true for n=k, then it also holds for n=k+1.

For step three we consider the left hand side of the inequality, for n=k+1, and show that it is indeed larger than the right hand side for the same n:

(k+1)! = (k+1) x k x (k-1) x (k-2) x.....x 3 x 2 x 1 = (k+1) x k! This is a key observation!

(k+1)! = (k+1) x k!

(k+1) x k! > (k+1) x k^{2 } By our Induction Hypothesis

(k+1) x k^{2} = k^{3} + k^{2}

k^{3}+k^{2} > k^{2} + 2k + 1 Because k^{3} > 2k+1 for k>4

k^{3}+k^{2}> (k+1)^{2}

Now linking up the chain of inequalities from beginning to end yields (k+1)! > (k+1)^{2} Which is the original inequality with n=k+1.

Hence our inductive step holds. Practically this means that if we know the inequality holds for n=k, then it also holds for n=k+1

The final step is to note that since our base case (n=4) holds. And our inductive step (n=k+1) holds, it follows that n=5 holds. Since n=5 holds, and n=k+1 holds, sp does n = 6, we can repeat this indefinitely, showing that any n greater than or equal to 4 satisfies the given inequality.

We wish to demonstrate that for all integers n greater than or equal to 4, n! > n

We briefly recap that n! is the product of all numbers from 1 up to n. For example 4! = 4x3x2x1 = 24.

As with any induction question, there are four steps that we will work through in turn. They are the base case, the induction hypothesis, the inductive step, and finally the conclusion.

We begin with the base case. Here we select the minimum integer satisfying the conditions, in this case n=4, and demonstrate that the inequality holds true. Clearly 4! = 24 > 4

The second step is to assume our induction hypothesis.

For this, we simply assume that the inequality holds for n=k, for any valid integer k.

That is, we assume k! > k

We will now use this in step three to show that IF the inequality holds true for n=k, then it also holds for n=k+1.

For step three we consider the left hand side of the inequality, for n=k+1, and show that it is indeed larger than the right hand side for the same n:

(k+1)! = (k+1) x k x (k-1) x (k-2) x.....x 3 x 2 x 1 = (k+1) x k! This is a key observation!

(k+1)! = (k+1) x k!

(k+1) x k! > (k+1) x k

(k+1) x k

k

k

Now linking up the chain of inequalities from beginning to end yields (k+1)! > (k+1)

Hence our inductive step holds. Practically this means that if we know the inequality holds for n=k, then it also holds for n=k+1

The final step is to note that since our base case (n=4) holds. And our inductive step (n=k+1) holds, it follows that n=5 holds. Since n=5 holds, and n=k+1 holds, sp does n = 6, we can repeat this indefinitely, showing that any n greater than or equal to 4 satisfies the given inequality.

Use demoivre's theorem (cosx+isinx)^n=cosnx+isinnx, put n=5 gives (cosx+isinx)^5 =cos5x +isin5x. Expand the LHS to give (let c=cos(x) and s=sin(x)); c^5+5c^4 is +10c^3 (is)^2 + 10c^2 (is)^3 + 5c(is)^4 + (is)^5.
Now compare the real parts of each side; cos5x= c5 -10c^3 s^2 + 5c s^2. Put s^2=1-c^2; cos5x = c^5 - 10 c^3 [1 - c^2] + 5 c [1 - 2 c^2 + c^4)] . Simplify to give: cos 5x= 16 (cos x)^5 -20 (cos x)^3 + 5 cos x

To find the gradient of the tangent, we can differentiate to give dy/dx=3x^2+9. We can now put in x=2 to find the gradient at (2,1): 3(2)^2+9=21. Therefore the gradient is 21 at (2,1).

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.

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