MAT Prep Thread - 2nd November 2016

No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.

No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.

Yuck, no. Geometry is shite, unless you mean some of the actual geomety like algebraic geometry.

No, a typical MAT prep-er won't know enough content to attempt the vast majority of STEP questions and will struggle overly much on the few questions that are accessible. It's not worth it, MAT is easy enough - you don't need so much preparation.

Would you mind explaining 2013 q1 (I)?

Thanks!

Would someone please explain how you do 2013 Q5 iii and iv?

Thanks!

How do you draw the graph for Q4(i) in 2013 paper?

Thanks

How many marks would I get for the question 6 of 2006? http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202006.pdf

6(i) If the message on the silver box is true, then the message on the gold box is false and hence the prize would be in the gold box. If the message on the silver box is false, then the message on the gold box must be false (if it was true, exactly one would be true) and hence either way the prize is in the gold box.

6(ii) If the message on the lead box is false, at least two messages would be true, and then the prize would be in both the silver and gold box - this is a contradiction. Hence the message is true. Therefore the messages on both the gold and silver boxes must be false (as the lead message is true) and hence he should choose the lead box.

6(iii) If the message on the lead box is true, the dagger is in the silver box. He should choose either gold or lead.

If the message on the lead box is false we have that at most one message is false, and as the lead message is false - both gold and silver messages are true and hence he should choose lead or silver.

The only consistent option is lead and hence he should choose the lead box.

We work out the number of functions that have the value of -1.

Imagine it as a 'tree'

3

6 7

and so on

We find how many times this tree splits until we get a maximum of F(100).

On each row, the number doubles each time - and on the nth line, we have 2^(n - 1) numbers. Taking the rightmost part, we find the highest value is 63. This is on the fifth line, of which there are 2^4 = 16 numbers.

We need to also find the numbers that are less than or equal to fifty - the sixth line will contain numbers stemming from these. We find the numbers we want on the fifth line are 48, 49 and 50. These therefore produce five numbers (96 to 100).

Our total number is 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 5 = 1 + 2 + 4 + 8 + 16 + 5 = 36.

Therefore our answer is 100 - 2 * 36 = 28 (b)

How do you draw the graph for Q4(i) in 2013 paper?

Thanks

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.

Yuck, no. Geometry is shite, unless you mean some of the actual geomety like algebraic geometry.



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.