Honours Maths Paper Error

analyse this

Was just wondering whether there was an error on the honours maths paper 2? In the last part of (c) in the vectors question it asks to prove r bisects angle. Should there be an arrow to imply that it is a vector?...or am i reading it wrong?

cocoa

obviously I can't say for sure but the I read it was this:

the point, r, is on the bisector of poq.

r is a point, or(arrow) is a vector and, confusingly, r (arrow) is it's abbreviation.

JC 2K3

r is a point.

r with an arrow is a vector.

The question was "Show that r is on the line that bisects the angle"

This could also be phrased as "Show that r with an arrow bisects the angle"

No misprint.

md99

speaking of maths misprints.....

In Paper I, the first diff. question. f(x)=, g(x)=, find f'(x) and g'(x) HENCE Prove f(x) + g(x) is a constant....

Was anyone else convinced this was a misprint?? My grinds teacher is convinced they meant add the diff's of each, which does give a constant (0).... That's what I did, although I strongly suspect a 0 for my efforts.... I had no idea that it was possible to add Tan-1 quantities...... But the word 'Hence', that's what made me think it.

seandoiler

md99 wrote:

speaking of maths misprints.....

In Paper I, the first diff. question. f(x)=, g(x)=, find f'(x) and g'(x) HENCE Prove f(x) + g(x) is a constant....

Was anyone else convinced this was a misprint?? My grinds teacher is convinced they meant add the diff's of each, which does give a constant (0).... That's what I did, although I strongly suspect a 0 for my efforts.... I had no idea that it was possible to add Tan-1 quantities...... But the word 'Hence', that's what made me think it.

if you differentiate a constant you get 0, the derivative of f(x) + g(x) is simply f'(x) + g'(x), now since this is zero, it implies that f(x) + g(x) has to be a constant....there are two ways to do the addition part, firstly you can put in any acceptable value of x and simply work out the value, as it is constant or you can use a triangle method...whether or which the answer is pi/2

JC 2K3

md99 wrote:

speaking of maths misprints.....

In Paper I, the first diff. question. f(x)=, g(x)=, find f'(x) and g'(x) HENCE Prove f(x) + g(x) is a constant....

Was anyone else convinced this was a misprint?? My grinds teacher is convinced they meant add the diff's of each, which does give a constant (0).... That's what I did, although I strongly suspect a 0 for my efforts.... I had no idea that it was possible to add Tan-1 quantities...... But the word 'Hence', that's what made me think it.

If that were the case then what would be the point of (c)(iii)?

You either add the diffs, get zero and therefore the sum of the two functions is a constant, or put f'(x) = -g'(x), integrate both sides and you get a constant of integration, so f(x) + g(x) = c.

I can't see why anyone would think it was a misprint. I originally added the diffs and got zero and said that was a constant, but then scribbled it out when I saw I had misread it.

Nehpets

I hope very little marks will go for that

md99

Nehpets wrote:

I hope very little marks will go for that

I too, brother

JC 2K3

It's prob 10 marks for part (i), 5 for part (ii) and 5 for part (iii).

Nehpets

Part one I'd prob get a blunder, or attempts and probably nothing for the other two