Chris has worked throughly with my son on his IB past papers. I am extremely impressed with his professional approach to teaching, his quick response to my messages and his helpfulness in finding a solution with regard to timing, days, etc. He is very knowledgeable about the workings of MyTutorWeb, which put me at ease with the whole process. Thank you Chris! :o)

Sheila, Parent from Genève

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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.

First, we must evaluate what is given in the question. As it can be seen, the expression indicates that the problem consists of a first-order differential equation. We are also given the values of x and their respective y value. These indicate that the problem should be integrated and then solved to obtain the value for the integration constant. Finally, we must calculate the value of y for when x = 2.
Following these steps, the differential equation can be integrated to give y = 1/2*10*exp(2x) - 4*x + C. We are given that y = 6 when x = 0, thus the value of C is calculated as C = 6 - 5*exp(0) = 1. Thus the general expression of y is y = 5*exp(2x) - 4*x + 1. Substituting in the value of x = 2 gives y(2) = 5*exp(2*2) - 4*2 + 1 = 5*exp(4) - 7.

First we need to be clear how to construct a proof by induction. It has two main parts, the **base case** and the **inductive step**. Let *P(n) *denote that the statement is true for a fixed n. Our base case is to check that *P(0)* is true. Our inductive step is to show that if we assume *P(n) *is true, then we can derive that *P(n+1)* is true. This usually involves rearranging one side of *P(n+1)* untill we see one side of *P(n)* in some part of it then substituting the other side of *P(n)* and rearranging our way to the unseen side of *P(n+1)*. Finally we must remember to **conclude** the proof, drawing together what we have shown to show the statement is true in general. The conclusion is pretty much the same for any proof of this type.
The **base case** is simpler. Substitute *n=0 *into the LHS, then the RHS and check the inequality holds. To complete this part you need to make sure you understand the values of *2*^{0} and *0!*. For the **inductive step** assume *P(n)*. Write down this statement in full. Somewhere else write the statement for *P(n+1)* and remember it is our goal to derive this inquality. Starting with the LHS of *P(n+1)* argue your way to the RHS remembering to respect the inequality, so each new line should be greater or equal to the last. To complete this part it would be helpful to write out the meaning of some parts of the statement such as *(2(n+1))!* and *(n+1)! *^{2}. Our **conclusion** is that if *P(n)* is true then so is *P(n+1)*, and since *P(0) *is true, the statement is true for evey integer *n>=0*.

Answered by Connor F.

Studies Computer Science with Mathematics BSc at Manchester (Computer Science)

Unlike if we had just x in the numerature, here the chain rule will do us no good. Integration by parts in this case also is little help - we end up with expressions at least as difficult as we started with. For example:
Int x^{2}/(1+x^{2})dx = x^{2}tan^{-1}x - Int 2x tan^{-1}x dx
In this case, a very simple approach is to simply add and subtract 1 from the denominator:
Int x^{2}/(1+x^{2})dx = Int (x^{2}+1-1)/(1+x^{2})dx= Int (1+x^{2})/(1+x^{2}) - 1/(1+x^{2})dx = Int 1 - 1/(1+x^{2})dx
From here, do you know know you to solve the problem?

Euclid's algorithm is really useful to be able to, firstly, see if two numbers are co-prime, in other words to see if they share any common factors, but also to find solutions to equations. Say we have two integers that satisfy:
32x + 24y = 16
Then we use Euclid's algorithm to first calculate the greatest common divisor (gcd) of 32 and 24. Hopefully, the method of this is ok? So we get gcd(32,24) = 8. Now, we can reverse what we did to get our solutions to the equation above. But don't forget that we had the equation equal to 16, not 8. This is often used in exams to trip up students, so look out for that.

Answered by Abby R.

Studies Mathematics at Leeds

We start with the definitions of sine and cosine, which are, respectively: sinx = opposite/hypoteneuse and cosx = adjacent/hypoteneuse. We then square the analyzed expressions to get the following:

(opposite ^2)/(hypoteneuse ^2) + (adjacent ^2)/(hypoteneuse ^2) And since the denominators are the same, we can add the fractions to get: (opposite ^2) + (adjacent ^2) / (hypoteneuse ^2) But recall the Pythagorean Theorem, according to which: (opposite ^2) + (adjacent ^2) = (hypoteneuse ^2). So we get: [(hypoteneuse ^2)] / (hypoteneuse ^2) = 1. QED.

(opposite ^2)/(hypoteneuse ^2) + (adjacent ^2)/(hypoteneuse ^2) And since the denominators are the same, we can add the fractions to get: (opposite ^2) + (adjacent ^2) / (hypoteneuse ^2) But recall the Pythagorean Theorem, according to which: (opposite ^2) + (adjacent ^2) = (hypoteneuse ^2). So we get: [(hypoteneuse ^2)] / (hypoteneuse ^2) = 1. QED.

Answered by Eno A.

Studies MPhil at Cambridge alumni

Firstly look for key terms in the question, identifying that we are going to be finding the **sum** of n terms, and it is an **arithmetic series**. This allows us to know which equations to look for in our formula booklet, the following equations are relevant:
S_{n} = (n/2)(2u_{1} + d(n-1)) = n/2(u_{1} + u_{n})
u_{n} = u_{1} + d(n-1)
There are two ways of solving this problem, but first it is done by identifying the first term and the difference between the terms.
In this case u_{1} = 2, because it is the first term, and the difference between 2 and 4 is 2, which is the same difference between 4 and 6 which means that d = 2. Since we are trying to find the sum of all terms up to the 4th term then n=4. If we put these defined terms into the equation then we can derive the answer.
S_{n} = (n/2)(2u_{1} + d(n-1)) S_{4} = (4/2) (2 x 2 + 2(4-1)) S_{4} = 2 (4 + (2 x 3)) S_{4} = 20

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