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thecaped
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#1
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#1
Who Done The Paper Today?
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hattori
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#2
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(Original post by thecaped)
Who Done The Paper Today?
me,

how did you think it went...

(apparently I can only post once every 30 seconds.. 29...28...27.....)
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mensandan
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#3
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Here are the answers as I remember them.

1) Circle x^2 + (y-6.5)^2 = 25/4

2) Binomial:
N= -2, a = 3

Coefficient x^3 = -108

Series valid for modx < 1/3

3) Implicit Differentiation

a)Gradient at point was 7/5

b) Equation of line (Normal)

y = (-5/7)x + 19/7

4) Remainder theorem

p = -7
q= -14

b) Factorised (2x-1)(3x+4)(x-2)

5)Area of integral
a) 45Pi^2
b) 240(pi -1)
c) Error = 13.6 %

6) Partial Fractrion

=4/(2x-3) - 3/(x+1)

b) Integrated in form y= f(x)

y= 108((2x-3)^2)/((x+1)^3)

7) Turning points at (-2,-0.25) , ( 2, 0.35)

b) The first was a minimum the second a maximum
c) Sketch (Like a simgle heart beat on a heart monitor)

8) L1 and l2 intersect when lamda = 1 and mew = 2 at the point

Q (2i + 5j + 4k)

To prove l1 and l2 perpendicular

(i + 2j -k).(2i+j+4k) = 2 + 2 -4 =0 So Cos theta = 0
Which implies theta = 90 QED

Finally the area of the triangle was

(3Root14)/2

Let me know if you got the same. Mensandan

Just changed the partial fraction integral question, forgot the constant. Doh! No doubt I forgot it in the exam also, c'est la vie.
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*dave*
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#4
(Original post by mensandan)
Here are the answers as I remember them.

1) Circle x^2 + (y-6.5)^2 = 25/4

2) Binomial:
N= -2, a = 3

Coefficient x^3 = -108

Series valid for modx < 1/3

3) Implicit Differentiation

a)Gradient at point was 7/5

b) Equation of line (Normal)

y = (-5/7)x + 19/7

4) Remainder theorem

p = -7
q= -14

b) Factorised (2x-1)(3x+4)(x-2)

5)Area of integral
a) 45Pi^2
b) 240(pi -1)
c) Error = 13.6 %

6) Partial Fractrion

=4/(2x-3) - 3/(x+1)

b) Integrated in form y= f(x)

y= ((2x-3)^2)/((x+1)^3)

7) Turning points at (-2,-0.25) , ( 2, 0.35)

b) The first was a minimum the second a maximum
c) Sketch (Like a simgle heart beat on a heart monitor)

8) L1 and l2 intersect when lamda = 1 and mew = 2 at the point

Q (2i + 5j + 4k)

To prove l1 and l2 perpendicular

(i + 2j -k).(2i+j+4k) = 2 + 2 -4 =0 So Cos theta = 0
Which implies theta = 90 QED

Finally the area of the triangle was

(3Root14)/2

Let me know if you got the same. Mensandan
HOW ON EARTH did u remember that?!?!?
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mensandan
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#5
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Because I was concentrating so hard in the exam the paper and solutions have etched themselves in my mind. Useful but annoying skill, certainly helps when revising, did you do the paper? Mensandan
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prince_capri
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#6
(Original post by mensandan)
Here are the answers as I remember them.

1) Circle x^2 + (y-6.5)^2 = 25/4

2) Binomial:
N= -2, a = 3

Coefficient x^3 = -108

Series valid for modx < 1/3

3) Implicit Differentiation

a)Gradient at point was 7/5

b) Equation of line (Normal)

y = (-5/7)x + 19/7

4) Remainder theorem

p = -7
q= -14

b) Factorised (2x-1)(3x+4)(x-2)

5)Area of integral
a) 45Pi^2
b) 240(pi -1)
c) Error = 13.6 %

6) Partial Fractrion

=4/(2x-3) - 3/(x+1)

b) Integrated in form y= f(x)

y= ((2x-3)^2)/((x+1)^3)

7) Turning points at (-2,-0.25) , ( 2, 0.35)

b) The first was a minimum the second a maximum
c) Sketch (Like a simgle heart beat on a heart monitor)

8) L1 and l2 intersect when lamda = 1 and mew = 2 at the point

Q (2i + 5j + 4k)

To prove l1 and l2 perpendicular

(i + 2j -k).(2i+j+4k) = 2 + 2 -4 =0 So Cos theta = 0
Which implies theta = 90 QED

Finally the area of the triangle was

(3Root14)/2

Let me know if you got the same. Mensandan

how did u 5...b)

i agree with all the others !!!
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mensandan
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I thought this was the worst question.

You needed to get the integral from 0 to 3pi^2 y dx

In terms of t, this was the integral from pi/4 to pi/2 of y dx/dt dt

Now dx/dt was 32t So you have The integral from pi/4 to pi/2 of 960tsin2t

once you solve this you get what I put.

Basically you had to use the chain rule to change the variable and adapt the limits. Hardly anyone ever gets those questions right. I am still hoping that I did. Oh well I am sure I have passed and that is good enough for me. Mensandan
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prince_capri
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(Original post by mensandan)
I thought this was the worst question.

You needed to get the integral from 0 to 3pi^2 y dx

In terms of t, this was the integral from pi/4 to pi/2 of y dx/dt dt

Now dx/dt was 32t So you have The integral from pi/4 to pi/2 of 960tsin2t

once you solve this you get what I put.

Basically you had to use the chain rule to change the variable and adapt the limits. Hardly anyone ever gets those questions right. I am still hoping that I did. Oh well I am sure I have passed and that is good enough for me. Mensandan

ooooohhhhhhhhhhhhhhhhhhhhhhhhhhh .................
i was thinking of this !!!!
i missed this part....i repent it now !!
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mensandan
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If the rest is right it sounds like we both did well. Thank god for that, that question was worth a few marks but not loads. Hope you did well. Mensandan
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prince_capri
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(Original post by mensandan)
If the rest is right it sounds like we both did well. Thank god for that, that question was worth a few marks but not loads. Hope you did well. Mensandan

yea...it was pretty ok...thanx a lot....a hell lot !!!
u were of great help..
hope uu achieve all ur goals....and we do all the same correct exams !!!
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Darkness
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#11
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Can anyone scan this paper in?
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prince_capri
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#12
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#12
(Original post by Darkness)
Can anyone scan this paper in?
maybe no....why ???
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hattori
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#13
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(Original post by mensandan)
Here are the answers as I remember them.

8) L1 and l2 intersect when lamda = 1 and mew = 2 at the point

Q (2i + 5j + 4k)

To prove l1 and l2 perpendicular

(i + 2j -k).(2i+j+4k) = 2 + 2 -4 =0 So Cos theta = 0
Which implies theta = 90 QED

Finally the area of the triangle was

(3Root14)/2

Let me know if you got the same. Mensandan
The area one i can remember the answer?? but i can't remember if I put root 14 or root 5 as one length and root 9 as the other, so it kinda matches up but i can't remember exactly... root5 keeps popping up in my head.

if your answers are all correct i did pretty well and if they're not then i'm right and did better than you (like **** i did )
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mensandan
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#14
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Since it was a right angled triangle I used
(1/2)bh

b=root6
h=root21

So 0.5root126
= 0.5root9root 14
= 1.5root14

This is if the points were (3, 7, 3) and (4, 6, 8)

When given the 2 x values. Once you got these you got new values of mew and lamda. I think this gives what I got, not sure.
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prince_capri
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(Original post by mensandan)
Since it was a right angled triangle I used
(1/2)bh

b=root6
h=root21

So 0.5root126
= 0.5root9root 14
= 1.5root14

This is if the points were (3, 7, 3) and (4, 6, 8)

When given the 2 x values. Once you got these you got new values of mew and lamda. I think this gives what I got, not sure.
ae u doing all this for the first time ???
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Darkness
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(Original post by prince_capri)
maybe no....why ???
Because I'm bored and would like to have a go. Uni doesn't start till Monday.

You see I finished A-level maths last year and did a ***** of a P3 paper (June 03). Would like to see how much it's been tamed down since then.
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prince_capri
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(Original post by Darkness)
Because I'm bored and would like to have a go. Uni doesn't start till Monday.

You see I finished A-level maths last year and did a ***** of a P3 paper (June 03). Would like to see how much it's been tamed down since then.
ok......well sorry cannot help u in this regard....perhaps u may ask one of ur friend!!!
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mensandan
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This is my first P3 exam, I really enjoyed it. I had a go at the June 2003 paper, I thought was fairly straight forward. I heard a big fuss about it last June, but I must say it was no harder than all the other papers, just different. Mensandan
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prince_capri
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#19
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(Original post by mensandan)
This is my first P3 exam, I really enjoyed it. I had a go at the June 2003 paper, I thought was fairly straight forward. I heard a big fuss about it last June, but I must say it was no harder than all the other papers, just different. Mensandan
ok....by the way, where are u from ???i mean which country ???
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mensandan
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#20
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Birmingham is my home town. No more exams till Summer. Thank god. Mensandan
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