MAT Prep Thread - 2nd November 2016

STEP II, 2004, Q5 and STEP II, 2000, Q5.

I have never understood the 2000 step 2 q5 I think i skipped over that q before. Where do i start because i did integral by parts and cos it was minimised equated it to 0. Am I off track?

Completely. Why would you need to do it by parts? Just expand it out.

Completely. Why would you need to do it by parts? Just expand it out.

Do you have any mat resources?

first page of this thread

Do you get marks for workings or just answers?

Help with 2006 part J pls
http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202006.pdf

I was able to get a start on the the 2004 paper on the bit in the middle but I am a bit stuck on the end bit when I need to find the value of A and B. Could you give me a little hint on how to progress? Here's what I got so far.

STEP II, 2004, Q5 and STEP II, 2000, Q5.

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.

Do you know of any good geometry questions?

Sorry, why do we give the integral of f(x) between 1 and -1 the same value as the integral of f(t) between 1 and -1? i.e. "u"
Are they the same?
I let the integral of f(x) between 1 and -1 equal A.
And I let the integral of f(t) between 1 and -1 equal U.
Hence my confusion

STEP II, 2004, Q5 and STEP II, 2000, Q5.

Do you have any other recommendations for step questions to help prepare for MAT?

This wasn't to help prepare for MAT, it was just vaguely related to a MAT question.

Oh right, but do you think there are any particular step questions that would be worth doing anyway?