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OCR (not MEI) FP1 - 20/01/12

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Reply 40
Avatar for Genesis2703
Genesis2703
OP
5
Original post by CraigKirk
Yeah, same really. Think about when you're in uni though, doing mega-hard stuff. You'll not even care about UMS scores then, as long as you got the grades that got you in. No-one gets 100% at uni (not that I've heard of...), so it'd be best to drop some of the perfectionism ASAP. I have to do this too, so I have to adjust also. :smile:


I wish i could go to uni but A) my familly is poorish anyway and B) since I live off the mainland, uni fares are FAR higher.

Also Im scared about the mainland as it is really really big :colondollar:

I just hope I can get a nice finance job or something...
Reply 41
Avatar for dybydx
dybydx
My answers:
1 a=12,arg(z)=arctan(512)=22.6oa=12, arg(z)=arctan(\frac{5}{12})=22.6^o
2 p=3,q=−2p=3, q=-2
3 6+3i,−6−3i\sqrt{6}+\sqrt{3}i, -\sqrt{6}-\sqrt{3}i
4 14n(n+1)(n+3)(n−2)\frac{1}{4}n(n+1)(n+3)(n-2)
5 a) (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
b) i) stretch scale factor 4 parallel to y-axis
ii) det(C)=4det(C)=4, the area of any shape transformed by matrix C will increase by a scale factor of 4
6 Circle centre (3,i)(\sqrt{3},i) radius 2, and half-line from origin passing through centre of circle
7 i) show that
ii) Mn=(3n03n−11)M^n=\begin{pmatrix} 3^n & 0 \\ 3^n-1 &1\end{pmatrix}
iii) proof by induction
8 i) show that
ii) nn+1\frac{n}{n+1}
iii) 1n+1\frac{1}{n+1}
9 i) det(X)=10−9a−a2det(X)=10-9a-a^2
ii) singular ifa=1a=1 or a=−10a=-10
iii) X−1=110−9a−a2(−a26−9a5−a−918−3a−a2a2−4)X^{-1}=\frac{1}{10-9a-a^2}\begin{pmatrix} -a & 2 & 6-9a \\ 5 & -a-9 & 18-3a \\ -a & 2 & a^2-4 \end{pmatrix}
10 i) α+β+γ=3,αβ+βγ+αγ=2,αβγ=−23\alpha+\beta+\gamma=3, \alpha\beta+\beta\gamma+\alpha \gamma=2, \alpha\beta\gamma=\frac{-2}{3}
ii) show that
a=−5,b=8a=-5, b=8
(edited 12 years ago)
Reply 42
Avatar for Genesis2703
Genesis2703
OP
5
8iii) is 1 - n/n+1 not 1/n+1 surely?
Reply 43
Avatar for dybydx
dybydx
Original post by Genesis2703
8iii) is 1 - n/n+1 not 1/n+1 surely?


They're the same thing:
1-n/n+1 = n+1/n+1 - n/n+1 = n+1-n/n+1 = 1/n+1
Reply 44
Avatar for Genesis2703
Genesis2703
OP
5
Original post by dybydx
They're the same thing:
1-n/n+1 = n+1/n+1 - n/n+1 = n+1-n/n+1 = 1/n+1


Oh yes thats right, its night time, not thinking straight :redface:
Reply 45
Avatar for Mr M (jr)
Mr M (jr)
1
Original post by dybydx
My answers:
1 a=12,arg(z)=arctan(512)=22.6oa=12, arg(z)=arctan(\frac{5}{12})=22.6^o
2 p=3,q=−2p=3, q=-2
3 6+3i,−6−3i\sqrt{6}+\sqrt{3}i, -\sqrt{6}-\sqrt{3}i
4 14n(n+1)(n+3)(n−2)\frac{1}{4}n(n+1)(n+3)(n-2)
5 a) (0−1−10)\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
b) i) stretch scale factor 4 parallel to y-axis
ii) det(C)=4det(C)=4, 4 represents the scale factor of the enlargement of any shape transformed by matrix C
6 Circle centre (3,i)(\sqrt{3},i) radius 2, and half-line from origin passing through centre of circle
7 i) show that
ii) Mn=(3n03n−11)M^n=\begin{pmatrix} 3^n & 0 \\ 3^n-1 &1\end{pmatrix}
iii) proof by induction
8 i) show that
ii) nn+1\frac{n}{n+1}
iii) 1n+1\frac{1}{n+1}
9 i) det(X)=10−9a−a2det(X)=10-9a-a^2
ii) singular ifa=1a=1 or a=−10a=-10
iii) X−1=110−9a−a2(−a26−9a5−a−918−3a−a2a2−4)X^{-1}=\frac{1}{10-9a-a^2}\begin{pmatrix} -a & 2 & 6-9a \\ 5 & -a-9 & 18-3a \\ -a & 2 & a^2-4 \end{pmatrix}
10 i) α+β+γ=3,αβ+βγ+αγ=2,αβγ=−23\alpha+\beta+\gamma=3, \alpha\beta+\beta\gamma+\alpha \gamma=2, \alpha\beta\gamma=\frac{-2}{3}
ii) show that
a=−5,b=8a=-5, b=8


All spot on.

For the argand diagram in question 6 you would probably have needed to show that the circle passes through (0,0) to get full marks.
Reply 46
Avatar for TRLMaths
TRLMaths
Original post by Mr M (jr)
All spot on.

For the argand diagram in question 6 you would probably have needed to show that the circle passes through (0,0) to get full marks.


I agree. I got all those answers when i did the exam so it's looking good! Had to emphasise my circle (it was more elliptical due to my wobbly hand) going through 0,0 :')
Reply 47
Avatar for Ree69
Ree69
14
Does anyone have a copy of the question paper?
Reply 48
Avatar for JPM11
JPM11
I can't remember the question exactly but for 5 b) ii) weren't they asking about the the det of C which would be the area of the new shape, not the scale factor, as I seem to remember a similar question coming up on a past paper?
(edited 12 years ago)
Original post by JPM11

Original post by JPM11
I can't remember the question exactly but for 5 b) ii) weren't they asking about the the det of C which would be the area of the new shape, not the scale factor, as I seem to remember a similar question coming up on a past paper?


I also wrote an increase in the area by four times. This idea definitely came up in a past paper, and this was the answer.
Reply 50
Avatar for Ree69
Ree69
14
Original post by CraigKirk
I also wrote an increase in the area by four times. This idea definitely came up in a past paper, and this was the answer.


Surely the area increases from 1 to 2.5? Try sketching the transformed matrix...
Original post by Ree69

Original post by Ree69
Surely the area increases from 1 to 2.5? Try sketching the transformed matrix...


I haven't drawn it, but consider that the answer to the first part is a stretch SF4 in the y-direction. This is bound to increase the area by four times. Also, it's impossible for the same matrix create an image which is both a stretch and enlargement by SF4, because they aren't the same thing.
Reply 52
Avatar for Ree69
Ree69
14
Original post by CraigKirk
I haven't drawn it, but consider that the answer to the first part is a stretch SF4 in the y-direction. This is bound to increase the area by four times. Also, it's impossible for the same matrix create an image which is both a stretch and enlargement by SF4, because they aren't the same thing.


An enlargment is, by definition, a stretch in both directions. I'm sure this was just a stretch in the y direction (as you said, by SF 4).

So that's why I don't think it's the area increased by a scale factor of 4 :s-smilie:
Original post by Ree69

Original post by Ree69
An enlargment is, by definition, a stretch in both directions. I'm sure this was just a stretch in the y direction (as you said, by SF 4).

So that's why I don't think it's the area increased by a scale factor of 4 :s-smilie:


An enlargement by scale factor four would increase the area by scale factor of 16.

A stretch by scale factor 4 increases the area by a scale factor of 4. Take a 4x2 rectangle: I stretch it by four, making it 16x2 instead. Initial area=8, after stretch=32. 32=4x8; an increase in the area by 4 times.
Reply 54
Avatar for Ree69
Ree69
14
Original post by CraigKirk
An enlargement by scale factor four would increase the area by scale factor of 16.

A stretch by scale factor 4 increases the area by a scale factor of 4. Take a 4x2 rectangle: I stretch it by four, making it 16x2 instead. Initial area=8, after stretch=32. 32=4x8; an increase in the area by 4 times.


Ah, yeah, of course. I just realised the area did actually increase from 1 to 4. (not 1 to 2.5).

Oh that's painful. Another precious mark gone.
Hi everyone - I've been away from school for three days but, if anyone wants to send me a paper, I will try to resolve any debate.

That Mr M (Jr) character is going to convince people he is me!
Reply 56
Avatar for dybydx
dybydx
Original post by Mr M (jr)
All spot on.

For the argand diagram in question 6 you would probably have needed to show that the circle passes through (0,0) to get full marks.


I agree - it needs to start at (0,0), which I called the origin in my answers. Glad we agree.
Reply 57
Avatar for dybydx
dybydx
Thank you to the people who pointed out that I shouldn't have used the word 'enlargement'. I've amended my answers.
Reply 58
Avatar for dybydx
dybydx
A copy of the paper. Ignore the 001 on the end of the file name.
Reply 59
Avatar for rogersnm
rogersnm
4
Original post by CraigKirk
I know I got a=-5, but I know my value for b was wrong. I did a little work when I got home and figured b=10 in the end, which seems to be in disagreement with everyone here :tongue: This means it's probably b=-8 or something.


Either by substitution of x=ux = \sqrt{u} or by rearranging the α\alpha, β\beta, γ\gamma equations:
a=−5[br]b=8[br]c=−49a = -5[br]b = 8[br]c = \dfrac{-4}{9}

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