Honors maths marking

PrecariousNuts

Well I got my mocks paper back and I'm puzzled over the marking scheme. Basicly, on nearly half of the questions I did I got only the absolute bare minium attempt mark even though the answer was correct. This is because I approached the question is a different way to his exam marking booklet thing.

For example, a question asked you to show that

cos^2(A) = 1/2(1 + cos2A)

I did it like this:

cos^2(A) = 1/2(1 + cos^2(A) - sin^2(A))
cos^2(A) = 1/2(cos^2(A) + (1 - sin^2(A)))
cos^2(A) = 1/2(cos^2(A) + cos^2(A))
cos^2(A) = 1/2(2cos^2(A))

cos^2(A) = cos^2(A)

Now the teacher insists this is wrong as I didn't change the left hand side into the right hand side. I did it the wrong way round, is the leaving cert marked like this?

drane2

The marking scheme is up on the Departments website. But I think you would get that wrong in the LC. You need to changed one of the sides to the other side.

Stark

That's perfectly correct to a logical person, but to play it safe you're best doing it that way in your rough work (if you find that way easier) and then printing it out in reverse in your actual exam script. That's what I used do.

Crash

well in those questions the idea is that you manipulate the previous identities to get that identity. so you'd start off with cos(a+b) identity, replace b with a And manipulate it from there. I think.

PrecariousNuts

More like

cos^2(A) = 1 - sin^2(A)
cos^2(A) = 1 - 1/2(1 - cos2A)
cos^2(A) = 1 - 1/2 + 1/2cos2A
cos^2(A) = 1/2 + 1/2cos2A
cos^2(A) = 1/2(1 + cos2A)

then?
Maths is maths and you can't really argue with that, they specified no particular to solve the problem in the question so I suppose it just seemed to be the best way to do it on the day. I guess I'll just do it the joe soap way in future then.

Thanks for clearing that up.

Mutant_Fruit

No, ignore that teacher. You can start with either side and prove it's equal to the other side.

Also, if for instance you get stuck at this line: 1 - 1/2 + 1/2cos2A

Its perfectly ok to whip out your rought work and start working from the other side back!. I.e. cos^2(A) = 1 - sin^2(A), then write cos^2(A) = 1 - 1/2(1 - cos2A).

By then you should realise how to finish it, so go back to your non-roughwork sheet, and finish it off properly. This approach has saved me in a question or two.

Anyway, if you proved one side equals the other, then you get full marks, providing you didn't make up your answer. It doesn;t matter one bit which side you choose to set equal ot the other (i.e. changin the right hand side to equal the left is the same as taking the left hand side and proving it's equal to the right)

Crash

yeah but mutant fruit is does depend on the question. it may have stated to use the identities. Precarious nuts will you post up the question?

honours maths is a weird beast. for instance the transformations piece - where you're not actually allowed assume (rightly) that a line using those translations will ALWAYS go to a line, you instead have to go back and prove it in reverse. unless it actually states that it will be a line.

PrecariousNuts

Question was a part (a), a one liner.

Prove that cos^2(A) = 1/2(1 + cos2A)

I didn't think anything of it since it was an (a) part so I just flew through it without thinking, I got 1 out of 10 marks for my original answer.

Mutant_Fruit

Well, you should have gotten 10/10 for it. If i were you, i'd get my paper checked by someone else. Its blatantly mis-marked in my opinion.

for instance the transformations piece - where you're not actually allowed assume (rightly) that a line using those translations will ALWAYS go to a line

Not always... check out this transformation.


x'= 10(x^2)/(x^2 + 1)

y'= 10(y^2)/(y^2 + 1)

Then, taking the line x + y = 0

Put in for (0,0), (2,-2) and (-3,3). (all points on that line).

It gives you the points (0,0), (8,8) and (9.9). These points are distinct, and collinear. If you connect these points you get the equation of the image, which is: x - y = 0.

So, is the image a line? NO!

Put in some more points, you'll notice its impossible to get a minus number (i.e. (-3,-3) ). Therefore it cannot be a line, its a line segment. By putting in more points you'll see you can't get more than (10,10) and less than (0,0).

Put in for the point (20,20) and you'll get the image cordinates as x= sqrt(-2), which is impossible.