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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.

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.

In order to solve the quadratic, we need to factorise it. Consider the coefficient on the x^2, its only factors are 1 and 3 so this tells us that in our factorising we will have something that looks like (x...)(3x...). All the signs are positive so we know that it willÂ also be a case ofÂ positive numbers (x+...)(3x+...).
Consider the factors of 6, by looking i can see that a combination on 3 and 2 is going to add up to get to 11 as 3x*3=9 and x*2=2x which totals 11x. Hence we have (x+3)(3x+2)=0. From that, I can show that my solutions will beÂ x=-3, -2/3 after equating each bracket to 0. Â

Answered by Natalie F.

Studies Mathematics at Warwick

Solving quadratics meansÂ we find out at what values of x the equation is equal to 0. This one we can solve by factorising: First take out a common factor of 2, to get: 2(x^{2}Â - 4x + 3). Now you have to find two numbers that add up to (-4) and also multiply to get 3. In this case the numbers are (-1) and (-2). So now we can re-write it as 2(x - 1)(x - 2). You can check this by multplying out the brackets and you should return to the original equation.Â (Now, remembering that when we multiply anything by 0 it is equal to 0.) The equation will equal 0 when (x-1) = 0 or when (x-2) = 0. These are two very simple equations to rearrange for x. __ANSWER:Â x = 1 and x = 2__ You can check this by substituting these numbers back into the original equation.

Answered by Hilde M.

Studies Physics & Music at Edinburgh

The nth term of the sequence is a formula that lets us find any value in the sequence if we know its position. In the given sequence there is a difference of 8 between each term, 13-5 = 8, 21-13=8, 29-21=8. This tells us that the formula for the nth term is going to look like 8n + c for some number c, but we need to find what this number c is. The first term in the sequence (n=1) is 5, so if we plug this into our equation we get 8(1) + c = 5, rearranging this we get c = 5 - 8 = -3. So c is -3.We could have chosen any term to find c. If we had chose useÂ n=2, for example, Â 8(2) + c = 13, c = 13 - 16 = -3, which is the same as we got before. This is because our formula is correct for every value so it doesn't matter which one we pick but using the first term is usually the easiest. To check our answer, we can check another term, the fourthÂ term is 29 so putting n=4 into our equation we get, 8(4) - 3 = 32 -3 = 29, which is what we expected, so we can be confident that we have the right answer So the formula for the nth term is 8n - 3

Since we are looking for a stationary this means the derivative will be equal to 0, so we will have to differentiate the equation. When we differentiate ( y = 2x + 27/x^2 ) we get ( dy/dx = 2 - 54/x^3 ). Because we are looking for a stationary point we set this equal to 0 and solve ( 2 - 54/x^3 = 0), rearranging this we get ( 2 = 54/x^3 ), multipying both sides by x^3 we get ( 2x^3 = 54 ), dividing this by 2, ( x^3 = 27 ), and now taking the cube root of both sides we get ( x = 3 ). So we know that the x-coordinate of the stationary point is 3, we can plug this back into the original equation to find the y-coordinate. Doing this we get ( y = 2*3 + 27/9 = 6 + 3 = 9 ), so the y-coordinate is 9. Thus the coordinate of the stationary point is (3,9)

This comes up in C4 in A level maths and differentiating it could come up in C3. You can write a^x as exp(ln(a^x))=exp(xln(a)) then differentiating this, you get ln(a)*exp(xln(a))=ln(a)a^x. By differentiating you can recognise the integral will be (a^x)/ln(a) +c or you can perform a u substitution where u=a^x then du=ln(a)a^x*dx.
dx=1/ln(a) * 1/u * du. Therefore the integral is now u/(u*ln(a)) du = 1/ln(a) du = u/ln(a) +c = a^x/ln(a) +c.
I have picked this since it could come up in C3 and C4 and I have had the same question asked to me by my peers before. The working can be further expanded by explaining how a^x can be written in terms of e and the natural logarithm, with these being inverse functions of each other, a topic within C3.

Prove by contradiction:
Assume negation to be true i.e. âˆš2 is rational
Then âˆš2 can be written in the form a/b where a and b are integers with no common factor (the fraction cannot be simplified)
=> a/b = âˆš2
=> a = bâˆš2
=> a^2 = 2b^2
=> a^2 is even, so 2 is a factor of a.
Therefore let a = 2k, where k is a whole number greater than zero
=> (2k)^2 = 2b^2 from above
=> 4k^2 = 2b^2
=> b = 2k^2
=> b is even, so 2 is a factor of b.
Therefore a and b have a common factor of 2.
This contradicts our original assumption that a and b have no common factor.
Therefore our assumption that âˆš2 is rational is false
Therefore âˆš2 is irrational.

Answered by Paul M.

Studies Informatics at Edinburgh

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