Thank you SO MUCH for the important part you played in helping Adam get the A level grades he needed to get to the course of his choice in Newcastle. To anyone contemplating getting Adams support - just do it :-) - having a helpful, knowledgable, friendly, same sort of age chap to ask about topics you need a bit more time on to get clarity - is invaluable. And Adam is reliable - flexible - friendly and supportive. Thanks and best of luck. Brenda - Mum to the Adam :-) - this Adam tutored.

Brenda, Parent from berks

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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 lessons. In my tutorials, 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 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.

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Since we are dealing with complex numbers and taking its modulus, we can rewrite
(z-i)=((-1)(i-z))=(i-z) doing the same for (z-1)=(1-z) we get (i-z)+(z+i)+(1-z)+(z-1)=(i+i+z-z+1+1+z-z)
=(2i+2)=4 as we are taking its modulus.

Answered by Yubo Z.

Studies Economics and Mathematics at Edinburgh

This thing we have to solve is an inequality and the solution we are looking for is an entire range of real number, something like "every x between 1 and 2", for example. To do this we need to build a sign diagram and to build a sign diagram we need to find the root of the respective equation first. This is because the roots are when the right hand side (rhs) changes sign. In the intervals between two different roots the sign stay the same.
In this case, the roots are x = -3 and x = 2. To compute the sign of the rhs before -3, we can simply compute the results when we substitute any (really, any!) number lower then -3, such as -4. We have: (-4)^2 -4 -6 = 6>0. So, the sign is positive. Same procedure for the (-3,2) interval and (2,infinity). The former gives us a negative sign, the latter a positive one. We want the intervals when the rhs is positive. Therefore the solution is: x<-3 and x>2.

Answered by Stefania G.

Studies PhD in Neuroscience at Imperial College London

Using LATEX (Logarithms, Algebra, Trigonometry, Exponential and Complex numbers) to determine which variable is du and which is dv/dx. This is decided by using the above acronym. For example in this question 'x' is an algebraic variable and 'cos(x)' is a trigonometric variable, hence 'x' is du and cos(x) is dv/dx. To solve this question, we use integration by parts and use the following formula. du.dv- integral(dv.(du/dx)dx).
du = x hence du/dx = 1 (differentiate du)
dv/dx = cosx hence dv = sinx (integrate dv/dx)
Plug in the values in the above equation.
Ans = xsinx + cosx + c

Answered by Vishnu P.

Studies Civil and Structural Engineering at Sheffield

We first apply a simple addition to make the inequality 0 on one side. We subtract 10, giving x^2-4x-12<0. Now we factorise the equation in x, intuitively or using the quadratic formula: x=(-b+sqrt(b^2-4ac))/2a or X==(-b-sqrt(b^2-4ac))/2a to give 2 values for x. In this case we can use intuition to get (X-6)(X+2)<0. We draw a graph of the function and deduce which values of X satisfy the inequality. Here, if -2 < x < 6 the inequality is satisfied.

(In terms of formatting, we'll use y' to represent dy/dx, exp() to represent the expoential e^(), int() to represent integration with respect to x, and 1ODE to mean First-Order Ordinary Differential Equation).
1ODE to consider: "(sin(x))y' - (cos(x))y = sin(2x)sin(x)".
There are typically two methods of solving 1ODE's - seperation of variables, and using a Multiplier M(x) known as an integrating factor. The latter of these methods is what we'll use here.
Firstly, let's recap. The standard form of a 1ODE is "y' + P(x)y = Q(x)", where P and Q are simply functions of x. The integrating factor is defined as M(x) = exp(int(P(x))). In order to use the integrating factor, we must have our 1ODE in the standard form (i.e. y' needs to be on it's own!).
The key to solving 1ODE's is to recognise the LHS (left-hand side), after multiplication by a factor (i.e. M(x)), as the derivative of a product. We'll see how this works in the example above.
We need our 1ODE in the standard form, so we'll start by dividing through by sin(x) to get y' on it's own. This gives "y' - cot(x)y = sin(2x)", remembering that cot(x)=1/(tan(x))=(cos(x))/(sin(x)). Now we have a form where we can let P(x)=-cot(x) and Q(x)=sin(2x), using the labels from our standard form equation above.
Next, we need to calculate our integrating factor. Using our definition and our P(x), we get M(x)=exp(int(-cot(x)))=exp(-ln|sin(x)|)=1/(sin(x)). [Can you see how we get this? Hint: write cot(x) in terms of sin and cos, then use Integration by Substitution with u=sin(x) to get the result]. We can ignore the constant of integration here.
Next, we can multiply our standard form 1ODE by the integrating factor to give "(1/(sin(x)))y' - [(cos(x))/(sin(x)sin(x))]y = 2cos(x)". The RHS has become simply "2cos(x)" because sin(2x)=2sin(x)cos(x), and the sin(x) from sin(2x) will cancel with the sin(x) in the integrating factor to leave cos(x). [Can you see this? If not, write it out and simplify!].
I said earlier that the method of solving a 1ODE is to recognise the LHS as the derivative of a product. So, let's write our 1ODE in this form. We get "d(y/(sin(x)))/dx = 2cos(x)". This is because when we evaluate the LHS (i.e. differentiate y/(sin(x)) with respect to x), we get exactly the LHS of our earlier 1ODE. [Can you see this? Try differentiating y/(sin(x)) with respect to x using the product rule].
The next step is to integrate both sides with respect to x. The LHS becomes "int(d(y/(sin(x)))/dx)", which simplifies to "y/(sin(x))" because our integral and differential both cancel. [Why? Because integration is the opposite of differentiating]. The RHS is a simple integration scenario, resulting in "2sin(x) + c". Don't forget the constant of integration (+c).
So, we now have our 1ODE in the form "y/(sin(x)) = 2sin(x) + c". We can multiply both sides by sin(x) to get it in the form y=..., so we have:
"y = 2(sin(x))^(2) + csin(x)".
This is the general solution to our 1ODE.
It is easy to see why this is an 8 mark question. There are lots of places where carelessness can create mistakes and cost marks, such as:
- Not dividing the RHS by sin(x) in the first step.
- Miscalculating the integrating factor, particularly by forgetting the exp() or integrating cot(x) wrong.
- Forgetting the constant of integration in the general solution. (You can see how costly this is as we now have csin(x)!)

Newton-Raphson is used for finding a numerical solution to equations of the form f(x)=0. The question asks for the square root of 3, which algebraically looks like this:
x = sqrt(3)
This can be rewritten in the form f(x)=0, giving:
x^2 - 3 = 0
(Note that in doing this you also create the negative solution by squaring it but this can be ignored in this question)
You can now calculate the Newton function for f(x), which is:
xn+1 = xn - (xn^2 - 3)/(2xn)
Now choose a sensible x0, 1.5 is a good place to start as you know the answer will be in (1,2) as 1^2 = 1 and 2^2 = 4
So now you just apply the iteration until the first four decimal places do not change:
x0 = 1.5
x1 = 1.75
x2 = 1.73214...
x3 = 1.73205...
x4 = 1.73205...
So after 4 iterations the first 4 decimal places do not change. So the answer is 1.7321

Answered by William S.

Studies Natural Sciences at Nottingham

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