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.

Integration by parts is not only a very useful techinque for integrating, but it is also one of the techinques that appears more in the IB Mathematics exams. The most important step is to choose the part of the integrand that works better for u or f(x), as a right choice will save you a lot of time (and time is everything in the IB!). Whenever you are in doubt, choose u or f(x) to be the following:
1. The LOgarithm. If there is none,
2. The ROot. If there is none,
3. The POlynomial. If there is none,
4. The Trigonometric function. If there is none,
5. The Exponential function.
To remember this rule, think about the sentence "LOve ROasted POtatoes Till the End", and its initials will tell you the order of the functions you need to look for to be u or f(x)!!

Answered by Xell B.

Studies Mathematics and Biology at Edinburgh

If the two graphs intersect, it means that they will share the same y and x coordinates at one particular point. (I will draw diagram to show point).
Therefore, you can set f(x)=g(x) so that x^2 -ax +a -1 = x-5
Then, x^2 -x(a+1) +a + 4 = 0
If they only intersect at one particular point, this means that the previous quadratic equation has only one solution. This is translated into an equation in terms of the determinant so that the determinant must be 0.
(If necessary, I will explain the difference number of solutions that one gets for different determinants).
Then, one requires b^2 - 4ac = 0 , where b is the coefficient multiplying the x, a is the coefficient multiplying the x^2 and c is the coefficient with no x in the previous equation.
This leads to (a+1)^2-4(a+4)=0 which is a quadratic equation in a:
a^2 -2a -15 =0 , which, using the quadratic formula, has solutions a= 5 and a=-3.

In a real 2-Dimensional function f(x) on the X-Y plane, we have the following relations between these concepts:
i) f'(x) is continuous if and only f(x) is differentiable; in fact, the continuity of f'(x) ensures that there are no points where the derivative tends to infinity, or has a possible multiple value. (picture as additional explanation)
ii) f(x) differentiable does not imply f(x) continuous, since we may have a function that is shifted up at a certain point, so it keeps to be differentiable, since there is no double derivative at that point, but the limits of x that tends to that point are different. (picture that function using a grapher)
iii) f(x) continuous does not imply f(x) differentiable. In fact a simple counter example could be f(x)=|x|. At x=0, f(x) is continuous, checkable using the definition. But the derivative assumes a double value at x=0, f'(0)=1 and f'(0)=-1. Therefore we found a counter-example.

An implicity function is one that is not expressed in the form y = f(x) such as the equation in the question. Instead of rearranging the equation to make y the subject, the equation can be differentiated using a technique called implicity differentiation. This involves differentiating each term on both sides of the equation. Differentiating x^3 will give 3x^2 and differentiating 34 will give 0. However differentiating y^4 will give (4y^3) X (dy/dx). This is achieved by using the chaing rule whereby d(y^4)/dx = (d(y^4)/dy) X (dy/dx).

Answered by Olavo M.

Studies Chemical Engineering at Edinburgh

This question is asking to find the values for x, such that h(x) is strictly less than g(x). We can write this as 9^x + 9 < 10*3^x and solve for x as follows.
9^x + 9 < 10*3^x
=> 3^(2x) - 10*3^x + 9 < 0
We let t = 3^x :
=> t^2 - 10*t + 9 < 0
=> (t - 9)(t - 1) < 0
By either sketching the quadratic, or by a sign diagram we find the values of t that satisfy this inequality :
1 < t < 9.
By substituting t = 3^x again, we find :
1 < 3^x < 9
=> 3^0 < 3^x < 3^2
=> 0 < x < 2.

Consider the Left-Hand-Side (LHS) of the equation first. LHS: 1/2 + 1 + 2 + 2^2 + ... + 2^10. We identify this as a geometric series by noticing that dividing any term u_(n+1) by the preceding term n the result is 2, eg. 1/(1/2) = 2. We also note that there are 12 terms, and that the first term is 1/2. From the formula booklet, section 1.1, we can find an equation for the sum of a finite geometric series: S_n = u_1(1-r^n)/(1-r). Where u_1 is the first term, r is the ratio of successive terms (u_(k+1))/u_k and n is the number of terms. In our case these take the values: u_1 = 1/2 , r = 2 , and n = 12. Substituting these in the equation we have : S_12 = 1/2(1 - 2^12)/(1 - 2)
= 1/2(2^12 - 1)
= 1*2^11 - 1/2
We can compare this last result for S_12 with the RHS of the original equation RHS: a*2^b + c, to find a = 1, b = 11, and c = -1/2.

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