MAT Prep Thread - 2nd November 2016 - Page 27

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How do you do this question? 2013 paper.

You would draw the graph of y=sinx, then draw the graph of y = sinx reflected in x= pi. Then reflect the resulting graph in y=2. It helps to draw out each stage instead of doing it all in one. The questions are always in radians as well so it is easy to reflect in pi.

Reply 521

ShatnersBassoon

Original post by 11234

Can u use c3 c4 techniques or will u get penalised?

I've seen a couple of questions from past papers where they say "some students may have done this using the quotient rule" or similar things - like in A levels, every mathematically rigorous technique is equally valid. You can use FP3 techniques if you think it's easier that way.

Reply 522

some-student

2011 paper Question 5(v).

How many marks do you think I would get if I put this as my solution:

Let S_n be the number of solutions in a semi-grid size n.

When n is odd, the number of solutions is the same as that for n - 1, as the semi-grid simply has a blank diagonal added, i.e. S_n = S_n-1

When n is even, S_n = S_n-2 + 2^(n-1), as a grid of even n has all the same solutions as that for n-2 apart from the main diagonal, of which there are 2^(n-1) solutions, as already found in (iii).

We can now redefine it for odd n, S_n = S_n-3+ 2^(n-2), by substituting in n-1 for n to the above formula, as we know n-1 is even

Therefore for n >= 3, S_n=
{
S_n-2 + 2^(n-1); n is even
S_n-3+ 2^(n-2); n is odd
}

S₀= 0, S₁ = 0, S₂ = 2

I should have noticed the geometric series, but I did this recursive way instead. Would I get any marks at all for this? It does work but isn't on the solutions at all. Thanks :smile:

(edited 7 years ago)

2007 Q5 part v?

Original post by Mystery.

2007 Q5 part v?

Okay, so you should have found that the recursion depth of g(2^k) is k + 1.

So for g(2^l + 2^k), as l is greater than k, the power of 2^k will "run out" before the power of 2^l. So you'll get an expression similar to (iv), where you can apply your answer to (iv) to get an expression. Some more detailed information:

Spoiler

Reply 525

Mystery.

Original post by lewman99

Spoiler

I get it till it becomes g(2^(l-k)) in the second line

Reply 526

lewman99

Original post by Mystery.

I get it till it becomes g(2^(l-k)) in the second line

Okay, so we know from the previous part that g(2^k) = k + 1, where k is an integer. Now, both l and k are integers, and it'll be positive because l > k, so l-k will be an integer, so we can substitute l-k into our formula.

There's also another way to think about it. Essentially, you've got 2 "types" of steps: dividing by 2 when it's even, and subtracting 1 when it's odd. Now, because l > k, if you keep dividing by 2, l will become 1 in more steps than k will. So for each of these steps, we add 1 to the value of the function. This step then repeats l times, because it divides by 2 each time. But we also have to consider the step of subtracting 1 when it's odd, this occurs twice - when 2^k reaches 1 after k steps, and when 2^l reaches 1 after l steps. So overall, the value of the function is l + 2.

Reply 527

gavinlowe

Original post by ShatnersBassoon

Thank you for this information. Do you have any idea how much emphasis is usually placed on explanations in MAT? I'm applying for Maths+CompSci so I'll be doing question 6, which seems like a particularly explanation-heavy question. If I've got the correct answer (e.g. "Alice and Bob are liars; Charlie told the truth" :wink:

, is this usually sufficient to gain all the marks for that part of the question, or could I be penalised if my explanation is a little ineloquent?

We do want to see explanation, to check that you've got to the right answer by a valid method, rather than just guessing well. The explanation should be clear; a good rule of thumb is that it should be enough to convince one of your colleagues. However, it doesn't need to be very eloquent.

Gavin

Reply 528

danielhx

can someone explain to me part iv of Q4 of 2015. i dont understand how the lopsidedness is minimised when it is split into 3 equal regions since they didn't provide a diagram so i'm having problems visualising it. Thanks!

Reply 529

some-student

Original post by danielhx

Lopsidedness is the biggest area of the circle. Every time we make one area smaller, another area will get bigger and hence the lopsidedness will get bigger. To minimise the lopsidedness, we must split it into 3 equal regions, as if we have any bigger or smaller regions, the lopsidedness will be bigger.

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Reply 530

Mystery.

Do you need to know about harmonic series?

Reply 531

LaserRanger

Original post by Mystery.

Do you need to know about harmonic series?

Only C1 and C2 knowledge will be required, so I don't think it would be necessary. But, I suggest you to know more about different things, coz MAT will be more difficult in the later years, as a trend from the past few papers.

Reply 532

rtc16

Is anybody aware of good resources where you can get a grip for how to tackle longer answer questions as a lot of the time i am struggling to see how to best answer these questions

Reply 533

Carman3

Do you get marks for workings or just answers?

Reply 534

alfmeister

Could someone help me with 1J from the 2014 paper please? https://www.maths.ox.ac.uk/system/files/attachments/test14.pdf

I am struggling to understand how I would solve this and particularly the integral is confusing me a little.

Reply 535

Zacken

Original post by alfmeister

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.

Reply 536

alfmeister

Original post by Zacken

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.

Thanks, that makes more sense now. Making the integral = u makes it a lot simpler. I'll have to try to find some more of them but I am not too sure where to look. Do you know questions similar to this?