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 GenÃ¨ve

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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.

A question that haunts every IB HL Math student day and night. Most students beleive that if they solve all past exams and practice book problems they should easily make a seven. That is absolutely not true. There are five main study techniques that should allow any dedicated higher level student to make a 7. These techniques worked with me and they should work with each and every single one of you.

First thing, any student must detect his strengths and weaknesses. The way to do that is to start by solving book problems after every section in the book, throughly reading and understaning the material in every section first, and later move to past exam questions on that specific topic. I make this point because many teachers either jump to past exams without issuing book problems to students or they solve book problems with a few past exam problems. The correct order must be respected; book first then past exams.

Second, getting the right answer on book problems is not enough to show that you have what it takes to start solving past exams. You should time your self and practice on every single question untill you get to the point were all problems could be answered in less than a single minute. You should look at the solution to every problem and count the number of steps you made and compare that with the number of steps that are shown in the answers. If the number of steps you made are more or less that what is there you are still not trained enough for past exams. This all helps you save a lot of time on the exams because it forces you to only include the required steps that award full credit for the question with nothing more and nothing less. Remeber you are nothing to the examiner but an answer sheet. After this skill is mastered the student can move on to solving past exams with the same exact strategy.

Second, every student must control his concentration curve for all three papers. What in the world is a concentration curve?! Your concentration curve simply shows how your concertation on a standerdised exam varies with resepect to time. Scientifically speaking the conetration curve for any human being is a bell curve, that is it follows the repetetive cycle of increasing from zero, reaching a maximium, and then falling back to zero.

What do I mean by controlling your concentration curve during an exam? I mean that you should train yourself such that the gap between the increasing and decreasing portion of the curve, throughout your two hour papers 1&2 or one hour paper 3, should be wide enough so that your concentration increases in the first set of 6-8 mark problems of the exams, plateau at maximium contrentration while solving the 10-25 mark questions, and then start to decrease only when your done and reviewing your answers. So that all the effort and concentration is dedicated to the questions with the most marks. If you are anywhere in the Ib program do not have high hopes for a seven unless your concentration curve resembles the one I just described.

Third, every student should practice to solve past exam questions in a time less than that assigned to it by the IB organization. In other words, the claim by many teachers in paper 1 and 2 is that you have a 120 mark exam with 120 minutes (2 hours) to finish so that you should dedicate 1 min to every mark. This notion is complpetely wrong. In fact a 6 mark problem should be solved in less than 6 minutes. Moreover, the typical 6-8 mark problems should be solved in 2-3 minutes, the typical 10-15 mark problems should be solved in 5-7 minutes, and the typical 16-25 mark problems should solved in 10-12 minutes. This is because you must have at least 5-10 minutes to review and solve any questions you missed before the end of the exam if your aiming for a 7.

Fourth, every student must memorize the most important formulas. Many students beleive that math should only involve reason and understanding and that is perfectly true. But with standardized tests like the IB, SAT, GCSE, and many others the aim is to solve the problems with the least amount of time and effort while collecting the most marks. Memorising formulas does not come by simply reciting them, a student must be able to dervie them and master using them to solve problems. Some of the formulas i beleive should be memorized are included in the data booklet, like the vector cross product formula, but since there is absolutely no time to check the formula booklet in the exam, students must memorise them. Other formulas are mentioned in the book and some are even in answers to book problems. I will point out those formulas that need to be memorised to students during lectures.

Finally, I believe that students should not start solving a full past exams except by the second semester of the second year of the IB diploma. Up to that time a student must answer only the past exams questions that are related to the topic covered; working his way topic by topic seperately. I highly recomend not to attempt a full exam untill 90% of the material is covered which as I mentioned is by the second semester of the second year of the IB diploma. Any attempt of a full exam before then can be desappointing to students which is unrecomended especially to IB HL Math students which go through an extremely instense course with lots of stress and anxiety accompaning them in the process. However when a past exam attempt is made any seven seeker should aim for full credit on all three papers.

By precisely following those five steps any IB HL Math student should make a 7.

Answered by Youssef M.

Studies Mech Eng at Birmingham

To solve this problem, the maximum and minimum points of equations can be deduced through the differentiation process. This looks at the gradient of the function and the maximum/minimum value occurs when the gradient is zero.

The differentiation process is as follows:

f(x)=Ax^{n}

df(x)/dx = nAx^{(n-1)}

The equation

y = −16x^{2} + 160x - 256

becomes

dy/dx= -32x+160

after differentiation and set dy/dx=0

0=-32x+160

x=5

and the corresponding value for y is:

y=-16(5^{2})+160(5)-256= 144

And so the coordinate of the maximum point is:

(5,144)

The key to solving this equation is realizing that sin(2x) can be written in terms of sin(x) and cos(x) using a **double-angle formula**. (With trigonometric problems similar to this one, you should always check if any trigonometric identites like the double-angle formulae can be used, as these can often help you.)

Using your IB formula booklet, you will see that the double-angle formula for sine is:

sin(2x) = 2sin(x)cos(x)

Therefore, we can rewrite the given equation from:

2cos(x) = sin(2x)

to:

2cos(x) = 2sin(x)cos(x)

Next, we notice that both sides of the equation are multiplied by 2, so we can divide both sides by 2. This yields the equation:

cos(x) = sin(x)cos(x)

We can now divide both sides of the equation by cos(x), which leaves us with:

sin(x) = 1

Finally, we must think about the angles at which sin(x) is equal to 1. You should realize, perhaps by imagining the unit circle, that sine is equal to 1 whenever x = π/2 + n2π, where n is any integer.

However, this is not the final answer, as the problem gave us a restricted domain for x. x must be in between 0 and 3π. Therefore, the only possible values for x are π/2 and 5π/2.

So, the answer is:

x = π/2 and 5π/2

If the details of the distribution - its mean and standard deviation are known, the following transformation applies Z = (X - mean) / stdev where Z is now a random variable distributed as normal(0, 1). Now, a lookup table for values of the normal distribution can be used to find the probability you are looking for.

From the rules of logarithms, we know that:

**log(A) - log(B) = log(A/B)**

and thus:

**log(1-x) - log(x) = log[(1-x)/x]**

Therefore from the question, we know:

**log[(1-x)/x] = 1**

If we then take both sides of the equation as a power of 10:

**(1-x)/x = 10^1**

and then multiply both sides through by x:

**1-x=10x**

Solving for x:

**1=11x**

**x=1/11**

We can check our answer by inserting it into the original equation:

**log(1-x) - log(x) = log[1-(1/11)] - log[1/11]**

and using the rule **log(A) - log(B) = log(A/B):**

**l****og[1-(1/11)] - log[1/11] = log(10/11)-log(1/11)**

**= log[(10/11)/(1/11)] **

**= log(10) **

**= 1**

Thus we know x=1/11

Answered by Joshua D.

Studies Materials Science at Oxford, Corpus Christi College

This question is an example of the chain rule for differentiating.

Firstly, identify the inner function. In this case, it is x^{2 }- 3x. This function must be differentiated first:

d/dx (x^{2} - 3x) = 2x - 3

Secondly, identify the outer function. In this case, it is e^{z}, where z = x^{2} - 3x. This function must be differentiated second:

d/dz (e^{z}) = e^{z }

The final differentiated result is the derivative of the inner function multiplied by the derivative of the outer function:

dy/dx = (2x - 3)e^{z }= (2x - 3)e^{x^2 - 3x}

Answered by Ellie S.

Studies Physics at Oxford, Lady Margaret Hall

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