if correctly attached then nogot same answer but I did half question on answer sheet and half on extra sheet would it cause any problem?????
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if correctly attached then nogot same answer but I did half question on answer sheet and half on extra sheet would it cause any problem?????
then lets take satisfaction.....I wrote the same exact thing.
The question gives point R (-1.3) and line of 3y + 2x = 33 as the mirror. We are looking for the image of R, (let's name it as P)
Rearranging the equation of the mirror gives us y = -2x/3 + 11
Therefore gradient (m) of mirror is -2/3.
Since we understand that the line PR will be PERPENDICULAR to mirror (properties of reflection), therefore m of PR = 1.5.
Work out the equation of PR from the point (-1,3) and m of 1.5
y - 3 = 1.5 (x+1)
y = 1.5 x + 4.5
Another properties of mirror is that distance from object to mirror = distance from image to mirror. Therefore point of intersection of line PR and the mirror will be the MIDPOINT for line PR.
Working out simultaneous equation on PR and mirror,
y = 1.5 x + 4.5
y = -2x/3 + 11
1.5 x + 4.5 = -2x/3 + 11
9x + 27 = -4x + 66
13x = 39
x = 3
Subs x = 3 to any of the 2 equation above, we will get y = 9
Since we know R, and the midpoint, we can work out P.
(-1 + x)/2 = 3 for the x-coordinate and (3+y)/2 = 9 for the y-coordinate
Therefore x = 7 and y = 15
it will be less than 60Is there any hope that the Math P1 gt will be around 55/56 ?
I mean it can't be 60 or above right ?
This paper was pretty hard compared to Nov. 2012 and June 2012 don't you think ?
got same answer but I did half question on answer sheet and half on extra sheet would it cause any problem?????
it will be less than 60
insha-allah
Less than 60 ? u mean 59 ? !! yeah that is exactly what its gonna beIs there any hope that the Math P1 gt will be around 55/56 ?
I mean it can't be 60 or above right ?
This paper was pretty hard compared to Nov. 2012 and June 2012 don't you think ?
that will be after November 2013 .. y do u neet original gt ? u will get ur results b4 thatWhen do you think the original gt will be out?
the point R doesn't matter, once we get the answer The examiner will ignore that mistakeIn Q 7 Guys point R is the reflection of point (-1,3) on that line giving and we were supposed to find R not that pointtt.....cuz i saw u guys took point R as (-1,3) and ur finding its reflection
answer to the first part was 10Hi fellow members,
Just want to ask you guys what the answers/solutions are for the quadratic equation in cos^2x (or was it cos^2 theta??)
Thanks!
(Were there four solutions?)
it was -2.5 instead of -2OKEI GUYS !
for the domain range question
1- the function was increasing, bcuz the gradient is above zero ( dy/dc = positive value)
2-the range is y>=2
domain -2<=x<0
don't ask me how, cuz I asked the same thing to my teacher (HOW??) and his reply was 'don't ask me how'
do u remember how much u got for the answer?the point R doesn't matter, once we get the answer The examiner will ignore that mistake
What so u mean to say that the function was less than -2.5 but greater than What ???I wrote same but real ans was 0<x<-2.5
That was a trick
Thanks for the link!My fellow xtremers, I present to you - the end to all discussions and arguments:
A hand written mark scheme for Mathematics P12 May/June 2013
http://dynamic-marketing.blogspot.com/2013/05/cie-pure-maths-paper-1-mayjune-2013.html
Thank me later
My fellow xtremers, I present to you - the end to all discussions and arguments:
A hand written mark scheme for Mathematics P12 May/June 2013
http://dynamic-marketing.blogspot.com/2013/05/cie-pure-maths-paper-1-mayjune-2013.html
Thank me later
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