Oxford MAT papers

Does anyone have the solution to question 4 from the 1996 paper?
Thanks.

Reply 23

Also, what does Q3c even mean from the 1996 paper? The rest of the question I understand, but what does it mean to have a perpendicular distance 1 from the origin. How can I work that out?

Reply 24

Not been through the question fully, but if you write the equation of a line as
ax + by = d
Where the vector (a,b) has unit magnitude, then d is the perpendicular distance from the origin. Like the equation of a plane.

Too answer the question without this info, you could calculate the two lines which pass through the point and are also tangent to the unit circle centered on the origin.

Original post by tfeerick

Also, what does Q3c even mean from the 1996 paper? The rest of the question I understand, but what does it mean to have a perpendicular distance 1 from the origin. How can I work that out?

(edited 4 years ago)

Reply 25

Thanks for your help, this all makes logic sense to me.

But, even with this information, how do I find the answer? Particularly, using the unit circle method, as it seems to me that there are many different solutions which could work?

Thanks again.

Original post by mqb2766

Not been through the question fully, but if you write the equation of a line as
ax + by = d
Where the vector (a,b) has unit magnitude, then d is the perpendicular distance from the origin. Like the equation of a plane.

Too answer the question without this info, you could calculate the two lines which pass through the point and are also tangent to the unit circle centered on the origin.

Reply 26

There are a few ways to solvr the problem.
* firstly, sketch the point and unit circle. You should see that there would be two lines of intetest.

You could differentiate the circle and find the gradient at the tangent point. Then work out which lines with the same grad pass through both the tangent point and (1,2).

Or you could do a bit of reasoning (geometry) about the right angled triangles which are created at the tangent points.

Or a bit of algebra (linear, quadratic equations) on the coefficients (a,b).

However, the numbers indicate you'd have to be careful about handling inf gradient lines.

Original post by tfeerick

Thanks for your help, this all makes logic sense to me.

But, even with this information, how do I find the answer? Particularly, using the unit circle method, as it seems to me that there are many different solutions which could work?

Thanks again.

(edited 4 years ago)

Reply 27

Did you get this sorted

Original post by tfeerick

Thanks for your help, this all makes logic sense to me.

But, even with this information, how do I find the answer? Particularly, using the unit circle method, as it seems to me that there are many different solutions which could work?

Thanks again.

Reply 28

Yes thank you for your help.

Do you have the solution to Q4(d) I have been able to solve the other parts of question 4, but I'm not sure how the graphs are related?

Original post by mqb2766

Did you get this sorted

Reply 29

From a very quick look, the two integrals should be the same because of the (even) x^2 term which is the same in both integrals.
The limits are symmetric about 0 and the (odd) linear and cubic terms would both integrate to 0, as they have equal positive and negative area either side of 0. Hence threy play no part in the answer, there is no need to integrate them.

This sounds sensible, but looking at the form of the integrands, its almost as though they want you to factoize. In this case, I don't think thats the aprozch

Reply 30

Original post by mqb2766

From a very quick look, the two integrals should be the same because of the (even) x^2 term which is the same in both integrals.
The limits are symmetric about 0 and the (odd) linear and cubic terms would both integrate to 0, as they have equal positive and negative area either side of 0. Hence threy play no part in the answer, there is no need to integrate them.

This sounds sensible, but looking at the form of the integrands, its almost as though they want you to factoize. In this case, I don't think thats the aprozch

That makes a lot of sense - I never would have got it on my own though!
Thank you :smile:

Reply 31

The mat is reasonably time intensive, but its often worthwhile just looking at a question and thinking about the structure, symmetries, different approaches etc. Siklos STEP book has a good introduction and for Q3 above, the simplest two solutions, geometry or algebra on a,b, are the "naively" less obvious ways to proceed.

This question (4) gives you a bit of a hint in that they ask you to sketch the problems, so again think about whether preceding question parts can be used.

Original post by tfeerick

That makes a lot of sense - I never would have got it on my own though!
Thank you :smile:

(edited 4 years ago)

Reply 32

Has anyone done in the 1998 paper?
If so, I'd be interested in comparing answers.

Reply 33

Original post by tfeerick

Has anyone done in the 1998 paper?
If so, I'd be interested in comparing answers.

Want to post your solutions?
Not done, but can check some.

Reply 34

What do you think the answer is for 1(d). I think it is (v) satisfy none of these
For Q2(c) I don't understand where the sigma notation comes from, or how I can derive it?
For Q5(c) I'm not certain of the answer. I understand 4(C) and are able to explain why, but I'm not sure for 5(c). Is it Wn=Rn-1+0.5Rn-2 because half of the combination which ends in black will also start in black? Or is it more complex that

Reply 35

RichE

15

Original post by tfeerick

What do you think the answer is for 1(d). I think it is (v) satisfy none of these
For Q2(c) I don't understand where the sigma notation comes from, or how I can derive it?
For Q5(c) I'm not certain of the answer. I understand 4(C) and are able to explain why, but I'm not sure for 5(c). Is it Wn=Rn-1+0.5Rn-2 because half of the combination which ends in black will also start in black? Or is it more complex that

Who are you replying to? And which year are you referring to?

Reply 36

For 1d) i think its iv). When a!=6, the lines intersect. When b=8, this is ok with a=6.

Its 1998.

Reply 37

RichE

15

Original post by mqb2766

For 1d) i think its iv). When a!=6, the lines intersect. When b=8, this is ok with a=6.

Its 1998.

I concur, the answer is (iv)

Reply 38