AQA A2 MFP3 Further Pure 3 – 18th May 2016 [Exam Discussion Thread]

I messed up my Cambridge application just by making silly mistakes, doesn't get worse than that lol.

I'd rather have messed up my application and not have gotten an offer than missing my offer from silly mistakes after having worked my arsenic off for UMS interview prep and A2 grades tbh.

Are you taking step?


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Thankfully not!

Hmm okay, keep focusing on getting high ums scores in your other exams then. You haven't thrown away your offer after one exam today, I have little doubt that you still have a good chance at an A* in FM.


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I'd rather have messed up my application and not have gotten an offer than missing my offer from silly mistakes after having worked my arsenic off for UMS interview prep and A2 grades tbh.

There's death, heard that's slightly worse #firstworldproblems


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Yeah, not too good with getting high ums in FM because I run out of time and leave my silly mistakes behind! But I'm counting on 600/600 in chemistry too lol

I will applaud you for that don't know how you do it! I am settling with an A this year in Chemistry, I don't see much point in wasting time working for the A*.


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I will applaud you for that don't know how you do it! I am settling with an A this year in Chemistry, I don't see much point in wasting time working for the A*.


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Well unit 5 will be a little difficult but fingers crossed! You're a mathmo, you don't need chemistry! Instead you've got that evil STEP

I don't think there's any point in any mathmos going for the third A* tbh.

STEP is a beautiful but excruciatingly painful exam which I am not very good at :getscoat:


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Can someone remember what Q5 was? And post the question here please? Cheers

Hi there, I can't remember the question exactly but I believe this is roughly what it was:

5)a) Express $\dfrac{1}{(x+1)(x+2)}$ in the form $\dfrac{A}{x+1} + \dfrac{B}{x+2}$. $\ \ [1]$

Solution

If we multiply both sides by $(x+1)(x+2)$, we get $1 = A(x+2) + B(x+1)$. By substituting in $x = -1$ and $x = -2$, we can find the values of the constants $A$ and $B$. You could also do this by inspection but it's always better to show a method of some sort.

The answer is $\dfrac{1}{(x+1)(x+2)} = \dfrac{1}{x+1} - \dfrac{1}{x+2}$


5)b) Using the substitution $u = \dfrac{dy}{dx}$, solve the differential equation $\dfrac{d^2y}{dx^2} + \dfrac{1}{(x+1)(x+2)}\dfrac{dy}{dx} = \dfrac{x+2}{x+1}$, where $\dfrac{dy}{dx}=4$ when $x = 0$, and $y = 1$ when $x = 0$. Give your answer in the form $y = f(x)$. $\ \ [11]$

Solution

Noting that $\dfrac{du}{dx} = \dfrac{d^2y}{dx^2}$, we can rewrite the differential equation as $\dfrac{du}{dx} + \dfrac{1}{(x+1)(x+2)}u = \dfrac{x+2}{x+1}$. The original second order differential equation has now been transformed into a first order differential equation of the form $\dfrac{du}{dx} + Pu = Q$, which can be solved analytically.

This differential equation still looks quite tricky to solve, but that is where our answer to part a) comes in. If we rewrite it again as $\dfrac{du}{dx} + \left(\dfrac{1}{x+1} - \dfrac{1}{x+2}\right)u = \dfrac{x+2}{x+1}$, we can now find an integrating factor very easily:

$I = e^{\displaystyle\int\left(\frac{1}{x+1} - \frac{1}{x+2}\right) dx}\ \ = e^{\displaystyle \bigg(\ln(x+1) -\ln(x+2)\bigg)} \ \ =e^{\displaystyle\ln \left(\frac {x+1}{x+2}\right)}$

$\therefore I = \dfrac {x+1}{x+2}$

Multiplying throughout by the integrating factor, our first order differential equation becomes:

$\dfrac{x+1}{x+2}\dfrac{du}{dx} + \dfrac{1}{(x+2)^2}u = 1 \ \ \Rightarrow \ \ \dfrac{d}{dx}\bigg(\dfrac{x+1}{x+2}u\bigg) = 1$

If we then integrate both sides of this expression with respect to $x$, we get $\dfrac{x+1}{x+2}u = x + C$ which can be rearranged as $u = \dfrac{(x + C)(x+2)}{x+1}$.

Now, going back to the information given in the question, we know that $u = \dfrac{dy}{dx}$, and $\dfrac{dy}{dx}=4$ when $x = 0$. Using these bits of information, and our expression for $u$, we can find the value of $C$. It turns out that $C=2$, so:

$\dfrac{dy}{dx}=\dfrac{(x + 2)^2}{x+1}$

To find an expression for $y$, we need to integrate again. In its current form, the RHS of the above equation cannot be integrated. Using algebraic long division, we can rewrite it in a form that can:

$\dfrac{dy}{dx}= x + 3 + \dfrac{1}{x+1}\ \ \Rightarrow \ \ y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + D$

Using the boundary conditions $y = 1$ when $x = 0$, it can be shown that $D=1$. The solution of the differential equation in the form $y = f(x)$, subject to the boundary conditions given, is therefore:

$y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + 1$

Hope that helps anyone who is unsure

That's not true! I've seen your scores in some of your mocks, impressive stuff .

In the last few step I integrated by substitution but then I end up with a different c but I think it simplifies down to the same amswear do you think I would still get full marks on that question

In the last few step I integrated by substitution but then I end up with a different c but I think it simplifies down to the same amswear do you think I would still get full marks on that question

I did the same thing, v=x+1 leads to $frac{(x+1)^2}{2} + 2(x+1) + ln(x+1) - frac{3}{2}$ which should get full marks still.