MAT Prep Thread - 2nd November 2016

I did this paper a few days ago and although I couldn't find a mark scheme, for question 6 I got exactly the same answers as you with exactly the same reasoning. So we would have got the same score... which is hopefully 15/15. If you did the rest of the paper, do you mind checking to see if your answers were the same as mine? Question 5 was a 'show that', but for questions 1, 2 and 3, I'm not sure what, if anything, I got wrong.

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I did all the questions (1-7) - I may have made some mistakes but tried to write some full solutions that may be useful

Thank you; that's very helpful. It's reassuring to see we got the same answers for everything but 1 (h). I don't know if you're doing A levels but part (d) required C3 [a year 13 module] knowledge so it wouldn't come up on any of the MAT papers today.

With part (h), I'm guessing this is also something you haven't studied yet. The right hand side of the equation is a geometric series, for which there's a formula when summing to infinity. Your range of $sinx$ and $sin^2x$ is correct, but there are plenty of infinite sums which equal 2, for instance $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$. In this case, the sum $\frac{2}{3} + (\frac{2}{3})^2 + (\frac{2}{3})^3 + ... = 2$ is relevant, and $sinx = \frac{2}{3}$ has two solutions between 0 and 2π, so I think the answer is (c).

For everything else, it looks like we had the same method, with one exception:
For 3(ii), rather than using your algebraic method (which is ingenious, but I think my method is simpler), I used my sketch in part (i). You know that y = f(x) should intersect the line y = -k at two points, one with a negative x value and another with a positive x value. You should be able to see from your graph that y = -k must be a tangent to the minimum you found at x = 1 + 1/√3. The y co-ordinate will be f(1 + 1/√3) = (1 + 1/√3)(1/√3)(1/√3 -1) = 2/(3√3). Thus -k = 2/(3√3), so [after rationalising] k = (-2√3)/9.

So if you're in year 12, why have you been doing MAT papers? Are you planning on applying to Oxford next year? Is it for fun? Or maybe a bit of both?

Integrate both sides of the given identity from -1 to 1, remember that the existing integral in that identity is just a number, you might find it helpful to call it $u = \int_{-1}^1 f(t) \, \mathrm{d}t$ and then integrate both sides like so:

$\int_{-1}^1 6 \, \mathrm{d}x + \int_{-1}^{1} f(x) \, \mathrm{d}x = 2\int_{-1}^1 f(-x) \, \mathrm{d}x + u\int_{-1}^1 3x^2 \, \mathrm{d}x$

which then simplifies down to $12 + u = 2u + 2u$ where $u$ is the answer you want.

Part ii

I feel dumb, but could you explain how


or am I missing something in the simplification?

They're definite integrals so you don't care what the variable inside is called.

I feel dumb, but could you explain how


or am I missing something in the simplification?

okay, integrate x dx between -1 and 1, what number do you get?Integrate t dt between -1 and 1, what number do you get?

The t and x inside the integral don't matter, they're called dummy variables as long as the integral is a fefinite one with limits, then the letter doesn't matter since the integral just represents a number anyway.

It's a bit (actually, almost exactly) why in summation notation, it doesn't matter what letter you choose. The sum from k=1 to 100 of k is the same as the sum from m=1 to 100 of m. The letters are just placeholders.

This has tripped up many a student in the past.

I'm confused

I don't understand how we know that the integral of t w/respect to t is the same as the integral of x w/respect to x? Same limits, but aren't they completely different functions?

I'm confused

I don't understand how we know that the integral of t w/respect to t is the same as the integral of x w/respect to x? Same limits, but aren't they completely different functions?

...remember that if you have a definite integral, that's just a really long way of writing a NUMBER. It's a CONSTANT. No matter what variable you use inside the integral, the entire thing eventually evaluates to some number. This is in contrast to indefinite integrals where it evaluates to a FUNCTION.

No, the function is f in either case..

Say that f(x) = x^3 (just for an example); then f(t) = t^3... t is simply a different variable; the function f is simply performing operations on whatever variable is input. If you were asked to sketch both, you would get the same curve... so integrating between the same limits would yield the same result

It's worrying that this has tripped me up and the exam is in 6 days lol.

Thanks guys!

i found that the old ones are much easier than the new ones

Good luck to everyone taking mat.
Multiple choice is probably the hardest part of the test and why marks are low take ur time and it will be decent. not the hardest mathematically but people tend to rush it and it is worth a lot.
dont be scared of q5 as sometimes it is literally trivial,in other instances it isn't.

can someone do 2004 Q1F and tell me what they get?