Currently unavailable: for new students
Degree: Mathematics (Masters)  Warwick University
About me:
Hey! I’m George, a tutor for Maths on MyTutorWeb. I'm a third year undergraduate Mathematics student currently studying at the University of Warwick.
Throughout school and university I've enjoyed helping and guiding friends through mathematical problems, enabling them to confidently tackle similar questions on their own. I’ve also tutored my younger sister in Maths throughout her GCSEs and Alevels helping her achieve great results.
My exam board for GCSE was AQA and for Alevel it was OCR MEI. Don’t let that deter you if it differs to yours; the vast majority of content between exam boards overlap so I’ll be able to cover anything you throw at me!
Modules I covered at Alevel: Core 14, Further Pure 12, Statistics 12, Mechanics 12, Decision 1, Numerical Methods.
The sessions:
In my tutorials the student is in charge; you dictate what we cover and the depth you want to go into your topic. During tutorials we can:
• Go through particular sections of your course which you're struggling with
• Go through particular questions which you can’t get your head around
• Identify areas of difficulty by going through past paper questions and build on them
Before the session I ask that you send me a 'TutorWebMail' indicating what you wish to cover plus your exam board. Doing so allows me to create a lesson plan so that I can come prepared with content specifically designed for you; click here to view an exemplar lesson plan that I've made. Additionally, after the session I'll send you the lesson plan I've created plus a (noncompulsory!) worksheet based on our tutorial, just to test yourself on whether you've understood everything during the session. An example of an exercise sheet that I've made for one of my students can be found here.
Availability:
Currently unavailable to new students.
What now?
Go ahead and send me a 'TutorWebMail' if you have any unanswered questions. Additionally, feel free to book a 15 minute 'Meet the Tutor Session' on the top right of the page if you wish to get to know me better.
I look forward to hearing from you! ☺
CRB/DBS Standard 
No 

CRB/DBS Enhanced 
No 
Please get in touch for more detailed availability
M (Parent) June 26 2016
Inge (Parent) March 1 2016
Abi (Student) December 20 2015
Ami (Parent) December 13 2015
This is a question from a STEP II paper.
STEP questions always give you just enough information to solve the problem; we’re looking to use everything that we’re given. The first thing we should do is differentiate f(x) by the product rule. This is a step up from Alevel since we’re considering a general function P(x), but the product rule still works just as usual:
f’(x) = P’(x)e^(x²)  2xP(x)e^(−x²) = e^(−x²)[P’(x) – 2xP(x)]
Now we can set the derivative we’ve found equal to zero for x = 0, ±a and ±b.
• f’(0) = P’(0) = 0
• f’(a) = e^(a²)[P’(a) – 2aP(a)] = 0 ⇒ P’(a)  2aP(a) = 0
• f’(a) = e^(a²) [P’(a) + 2aP(a)] = 0 ⇒ P’(a) + 2aP(a) = 0
• f’(b) = e^(b²)[P’(b)– 2bP(b)] = 0 ⇒ P’(b)  2bP(b) = 0
• f’(b) = e^(b²)[P’(b) + 2bP(b)] = 0 ⇒ P’(b) + 2bP(b) = 0
The only information we’re given that’s left is that the derivative of f(x) isn’t zero for any other values of x. So e^(−x²)[P’(x) – 2xP(x)] ≠ 0 for any other values of x. Since e^(−x²) is always nonzero we can deduce that P’(x) – 2xP(x) ≠ 0 for any other values of x.
We can now combine everything we know together: P’(x)  2xP(x) = 0 ONLY for x = 0, ±a, ±b.
How do we proceed now? We’ve used all the information given in the question. Let’s look back at what we’re asked to do: we’re asked to find a polynomial P(x) which satisfies the above conditions. The trick here is to equate the polynomial P’(x) – 2xP(x) to a polynomial that we already know equals zero ONLY for x = 0, ±a, ±b. Then by comparing coefficients we can find coefficients for P(x). Let’s use x(x  a)(x + a)(x  b)(x + b) = x(x²  a²)(x²  b²) = x⁵  (a² + b²)x³ + a²b²x.
What order does P(x) need to be? The order of x⁵  (a² + b²)x³ + a²b²x is 5, so in order for P’(x) – 2xP(x) to have order 5 as well P(x) needs to have order 4.
Note that x⁵  (a² + b²)x³ + a²b²x has no x⁴ or x² terms so our P(x) should have no x³ or x terms to avoid x⁴ or x² terms cropping up in P’(x) – 2xP(x).
Thus P(x) = αx⁴ + βx² + γ for some α, β, γ to be determined.
P’(x) – 2xP(x) = (4αx³ + 2βx) – (2αx⁵ + 2βx³ + 2γx) = –2αx⁵ + (4α  2β)x³ + (2β  2γ)x
And now we equate coefficients:
–2αx⁵ + (4α  2β)x³ + (2β  2γ)x ≡ x⁵  (a² + b²)x³ + a²b²x
• –2α = 1 ⇒ α = 0.5
• 4α  2β =  a²  b² ⇒ 2  2β =  a²  b² ⇒ β = (a² + b²  2)/2
• 2β  2γ = a²b² ⇒ a² + b²  2  2γ = a²b² ⇒ γ = (a² + b² a²b²  2)/2
Hence P(x) = 0.5x⁴ + (a² + b²  2)x²/2 + (a² + b² a²b²  2)/2 is a solution.
see moreThis refers to a question in the Edexcel GCSE paper this year which took students to Twitter venting their frustration. The question is as follows:
‘There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.
Hannah takes a random sweet from the bag. She eats the sweet.
Hannah then takes at random another sweet from the bag. She eats the sweet.
The probability that Hannah eats two orange sweets is 1/3.
Show that n² – n – 90 = 0’
Seemingly out of nowhere you’re asked to prove that a certain quadratic equation holds using the information provided. The first three lines set up the situation whilst the fourth line provides you with some extra information to use to obtain the answer. Intuition should tell you that you need to calculate the probability that Hannah eats two orange sweet using the first three lines and then apply what you’re given in the fourth line.
So let’s do that. What’s the probability that the first sweet she eats from the bag is orange? There are n sweets in the bag, 6 of which are orange. So the probability is 6/n.
What’s the probability that the second sweet she eats from the bag is orange? Now there are n1 sweets in the bag, 5 of which are orange (since she has eaten an orange sweet!). So the probability is 5/(n1).
These two events are separate from one another, so the probability that both happen (i.e. both the sweets are orange) are the two probabilities multiplied together: 6/n × 5/(n1) = 30/n(n1)
But you’re told that this probability is 1/3! So all you need to do is set the expression equal to 1/3, rearrange and (hopefully!) obtain the required quadratic equation.
30/n(n1) = 1/3
⇒ 90/n(n1) = 1 (multiplying both sides by 3)
⇒ 90 = n(n1) (multiplying both sides by n(n1))
⇒ n(n1) – 90 = 0 (subtracting 90 from both sides)
⇒ n² – n – 90 = 0 (expanding the brackets)
Tahdah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!
see moreThis is a question from a FP1 paper. Here i denotes √1.
Fact that should be burned into your soul: ‘Complex roots of a polynomial equation with real coefficients form conjugate pairs’. This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and 1 – 2i are also roots of P(z) = 0.
So we now have 4 roots of our quartic equation P(z) = 0, so that’s all of them! We can now employ the factor theorem that you (probably) met in C1. Remember, this states that if z = α is a root of a polynomial then (z – α) is a factor of that polynomial.
So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that
P(z) = (z – (2  i))(z – (2 + i))(z – (1 + 2i))(z – (1  2i))
So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.
⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)
⇒ P(z) = (z²  2z – zi – 2z + 4 + 2i + zi  2i  i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i  4i²)
⇒ P(z) = (z²  4z + 5)(z² + 2z + 5) [Remember: i² =  1 by definition]
⇒ P(z) = z⁴ + 2z³ + 5z²  4z³  8z²  20z + 5z² + 10z + 25
⇒ P(z) = z⁴  2z³ + 2z²  10z + 25
Thus a = 2, b = 2, c = 10, d = 25 and we’re done.
see more