PremiumGeorge B. GCSE Maths tutor, A Level Maths tutor, A Level Further Math...
£24 - £28 /hr

George B.

Degree: Mathematics (Masters) - Warwick University

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About me

Hello! I’m George, a fourth-year undergraduate Maths student at the University of Warwick. I've been successfully tutoring on this site for over a year now with fantastic results and feedback from both students and parents. I am employed by the University of Warwick as a supervisor for first-year Maths students. This involves regular meetings where I support them with any lectured material that they're unsure about and help resolve any difficulties they're having with the course. My exam boards for GCSE and A-Level Maths were AQA and OCR (MEI) respectively. Don’t be deterred if these exam boards are different to yours! The vast majority of content between exam boards overlap so I’ll be able to cover anything you throw at me!

About my 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 our tutorials we could: - Go through certain sections of your course that you're struggling with. - Look at 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 booking a session it would be nice if you could send me a message telling me what you would like to cover along with your exam board. Doing so allows me to create a lesson plan so that I can come prepared with content specifically designed for you! Go ahead and send me a message if you have any unanswered questions. Additionally, feel free to book a 15 minute "Meet the Tutor Session" if you want to get to know me better!

Subjects offered

SubjectQualificationPrices
Further Mathematics A Level £26 /hr
Maths A Level £26 /hr
Maths GCSE £24 /hr
.STEP. Uni Admissions Test £28 /hr

Qualifications

SubjectQualificationLevelGrade
MathematicsDegree (Masters)FIRST
MathematicsA-levelA2A*
Further MathematicsA-levelA2A*
ChemistryA-levelA2A*
PhysicsA-levelA2A
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

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Ratings and reviews

4.9from 31 customer reviews

Zahid (Student) August 20 2017

Awesome teacher!

Daniel (Student) August 16 2017

Thank you so much for you help with my Further Maths problem. I can see you are very acknowledged in the topic of Maths and I would recommend You as a tutor to anyone! Thanks you!

Zahid (Student) August 12 2017

Really amazing teacher.

Oisin (Student) June 13 2017

Used lots of questions to help to understand ratio
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Questions George has answered

Given that a and b are distinct positive numbers, find a polynomial P(x) such that the derivative of f(x) = P(x)e^(−x²) is zero for x = 0, x = ±a and x = ±b, but for no other values of x.

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 A-level since we’re considerin...

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 A-level 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 non-zero 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. 

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2 years ago

676 views

How do I solve Hannah’s sweet question?

This 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 t...

This 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 n-1 sweets in the bag, 5 of which are orange (since she has eaten an orange sweet!). So the probability is 5/(n-1).

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/(n-1) = 30/n(n-1)

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(n-1) = 1/3

⇒ 90/n(n-1) = 1           (multiplying both sides by 3)

⇒ 90 = n(n-1)               (multiplying both sides by n(n-1))

⇒ n(n-1) – 90 = 0         (subtracting 90 from both sides)

⇒ n² – n – 90 = 0         (expanding the brackets)

Tah-dah. We’ve found the required equation and we’re done. This question was only worth three marks; a bit stingy in my opinion!

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2 years ago

1355 views

Let P(z) = z⁴ + az³ + bz² + cz + d be a quartic polynomial with real coefficients. Let two of the roots of P(z) = 0 be 2 – i and -1 + 2i. Find a, b, c and d.

This 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...

This 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.

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2 years ago

746 views
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