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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.
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