Edexcel FP3 23rd June 2014 - OFFICIAL THREAD

does anyone know how many marks question 9 was?!
I just realised I missed the entire question :'(

Reply 44

Original post by ermm

does anyone know how many marks question 9 was?!
I just realised I missed the entire question :'(

It was 8 marks; 5 for the reduction formula and 3 for evaluating

I_2

Reply 45

Original post by sjc1996

did anyone else get y=-9/4x for question 5??? no idea if it's right but it's my only hope as i ****ed up on 9a :frown:

Er, I had

y=-\frac{4}{9m} x

, is that what you meant?

(edited 9 years ago)

Reply 46

Original post by Ktulu666

It was 8 marks; 5 for the reduction formula and 3 for evaluating

I_2

Ahh thanks, don't know why Edexcel use 8 questions for years and years and then decide to chuck an extra one in :mad:

Reply 47

sjc1996

Original post by Ktulu666

Er, I had

y=-\frac{4}{9m} x

, is that what you meant?

I think you're right, I got the 4 and 9 the wrong way round somehow as i was in such a rush

Reply 48

lilac12

Original post by VannR

Yeah. I was almost there apart from a spurious '+2' on the denominator that stopped me from completing my answer. Checked for signing errors in my expansion, but I could not find one :rolleyes:

there was supposed to be a +2 on the denominator as sech^2(x)=4/(e^2x+e^-2x+2) Is that what you meant?

Reply 49

Okay, here's what I remember that I got:
Q1a Can't remember, I just had the direction vector being the normal to the plane and the position vector being that of the point it went through
Q1b Again, can't remember, something like 3, -1, 2
Q1c Distance of

/sqrt{6}

Q2a Area of \frac{1}{2} \sqrt{6}
Q2b Proved the triple scalar product was zero, QED
Q2c I THINK this shows they are coplanar vectors, since if they weren't, then the tetrahedron whose sides they form would have non-zero volume, but it does have zero volume so it must be a plane i.e: the three vectors are coplanar
Q3 (If this was the two integrals question), for part a I had

ln(\frac{\sqrt{6} + \sqrt{2}}{2})

and

\frac{1}{6}e^{3x} - \frac{1}{2}e^x + \frac{1}{3}

Q4 (If this was the hyp function question), for part a I had to work from both ends of the identity then rewrite the entire thing in one direction, was fairly tedious, for part b I got

x = ln7

only
Q5 Proved the derivative, QED
Q6 This was horrible. Like, tedious as f***. Thanks for a C3 question edexcel. Part a was a proof, for part b I had

(\frac{3}{2}(cos\alpha + cos\beta), sin\alpha + sin\beta)

, for part c I had

k = \frac{4}{9m}

Q7 (If this was the area of a sphere one) Part a, QED on the derivative, part b QED, part c I wrote

\frac{\pi}{2}

since the equation was of a quarter circle, radius 1.
Q8 (If this was the matrices question) QED on the eigenvalue 1 based upon solutions to the characteristic equation, I got the others as -1 and 5 I believe. For an eigenvector, doing

Mr = \lambda r

gives

x=x, y=0, z=0

, so you can choose any value for x except zero (I picked 1 because its the simplest), so my eigenvector was \begin{pmatrix} 1 \\0 \\0 \end{pmatrix}
Q9 QED on the reduction formula (I did it by parts on

u = (x^2 + 1)^{-n}, \frac{dv}{dx} = 1

and the rewrote the resulting integral of

x^2(x^2 + 1)^{-n-1}

as

(x^2 + 1)^{-n} - (x^2 + 1)^{-n-1}

(edited 9 years ago)

Reply 50

Original post by ermm

Ahh thanks, don't know why Edexcel use 8 questions for years and years and then decide to chuck an extra one in :mad:

This whole paper was fairly jammy tbh. There wasn't a single question on ellipses/hyperbolae at an FP3 level; I refuse to accept question 6 as anything more than particularly messy C3 trig identities and C1 co-ordinate geometry. You had to work out a line between two point (C1) using C3 geometry, work out the midpoint of that line (C1), then show that if the gradient of your line (C1) from part a was constant, the midpoint followed this locus (eh, loci are a bit advanced for C1, but not FP3) Like, that's a fair proportion of their own course ignored. Good job

Reply 51

FibonacciPigeon

Original post by Ktulu666

Okay, here's what I remember that I got:
Q1a Can't remember, I just had the direction vector being the normal to the plane and the position vector being that of the point it went through
Q1b Again, can't remember, something like 3, -1, 2
Q1c Area of

\frac{1}{2} \sqrt{6}

Q2 Can't remember this question
Q3 (If this was the two integrals question), for part a I had

ln(\frac{\sqrt{6} + \sqrt{2}}{2})

and

\frac{1}{6}e^{3x} - \frac{1}{2}e^x + \frac{1}{3}

Q4 (If this was the hyp function question), for part a I had to work from both ends of the identity then rewrite the entire thing in one direction, was fairly tedious, for part b I got

x = ln7

only
Q5 Proved the derivative, QED
Q6 This was horrible. Like, tedious as f***. Thanks for a C3 question edexcel. Part a was a proof, for part b I had

(\frac{3}{2}(cos\alpha + cos\beta), sin\alpha + sin\beta)

, for part c I had

k = \frac{4}{9m}

Q7 (If this was the area of a sphere one) Part a, QED on the derivative, part b QED, part c I wrote

\frac{\pi}{2}

since the equation was of a quarter circle, radius 1.
Q8 Can't remember this question
Q9 QED on the reduction formula (I did it by parts on

u = (x^2 + 1)^{-n}, \frac{dv}{dx} = 1

and the rewrote the resulting integral of

x^2(x^2 + 1)^{-n-1}

as

(x^2 + 1)^{-n} - (x^2 + 1)^{-n-1}

Did you get the sphere one? Couldn't prove it was 4 pi r^2 and ended up with something weird like 4 pi ^2 r... :/

Reply 52

MelonBath

Original post by Ktulu666

Okay, here's what I remember that I got:
Q1a Can't remember, I just had the direction vector being the normal to the plane and the position vector being that of the point it went through
Q1b Again, can't remember, something like 3, -1, 2
Q1c Area of

\frac{1}{2} \sqrt{6}

Q2 Can't remember this question
Q3 (If this was the two integrals question), for part a I had

ln(\frac{\sqrt{6} + \sqrt{2}}{2})

and

\frac{1}{6}e^{3x} - \frac{1}{2}e^x + \frac{1}{3}

Q4 (If this was the hyp function question), for part a I had to work from both ends of the identity then rewrite the entire thing in one direction, was fairly tedious, for part b I got

x = ln7

only
Q5 Proved the derivative, QED
Q6 This was horrible. Like, tedious as f***. Thanks for a C3 question edexcel. Part a was a proof, for part b I had

(\frac{3}{2}(cos\alpha + cos\beta), sin\alpha + sin\beta)

, for part c I had

k = \frac{4}{9m}

Q7 (If this was the area of a sphere one) Part a, QED on the derivative, part b QED, part c I wrote

\frac{\pi}{2}

since the equation was of a quarter circle, radius 1.
Q8 Can't remember this question
Q9 QED on the reduction formula (I did it by parts on

u = (x^2 + 1)^{-n}, \frac{dv}{dx} = 1

and the rewrote the resulting integral of

x^2(x^2 + 1)^{-n-1}

as

(x^2 + 1)^{-n} - (x^2 + 1)^{-n-1}

I got exactly the same as you for all, I assume your very good at maths so they are right :smile:

And also does Arsey post answers to this?

Reply 53

Original post by FibonacciPigeon

Did you get the sphere one? Couldn't prove it was 4 pi r^2 and ended up with something weird like 4 pi ^2 r... :/

Yeah that one was sneaky
You had to set the limits as r and -r rather than pi and 0...it then comes to 4pi r^2

Reply 54

leebf1

Original post by sjc1996

did anyone else get y=-9/4x for question 5??? no idea if it's right but it's my only hope as i ****ed up on 9a :frown:

yes i got this

Reply 55

Original post by Ktulu666

This whole paper was fairly jammy tbh. There wasn't a single question on ellipses/hyperbolae at an FP3 level; I refuse to accept question 6 as anything more than particularly messy C3 trig identities and C1 co-ordinate geometry. You had to work out a line between two point (C1) using C3 geometry, work out the midpoint of that line (C1), then show that if the gradient of your line (C1) from part a was constant, the midpoint followed this locus (eh, loci are a bit advanced for C1, but not FP3) Like, that's a fair proportion of their own course ignored. Good job

ahaha preach! Yeah Edexcel like ****ing with people

Reply 56

lilac12

Original post by FibonacciPigeon

Did you get the sphere one? Couldn't prove it was 4 pi r^2 and ended up with something weird like 4 pi ^2 r... :/

you had to use your part a 1+(dy/dx)^2 = r^2 / r^2 - x^2 , Area = 2pi * integral of [ y (1+ (dy/dx)^0.5 ] ... using x=r and x= -r as limits

simplified down to 2pi * [rx] = 2pi * 2r^2 = 4pi r^2

Reply 57

Original post by ermm

Yeah that one was sneaky
You had to set the limits as r and -r rather than pi and 0...it then comes to 4pi r^2

That's what I did for the sphere one, yeah. You can't use pi and 0 since you're not dealing with the parametric form of a circle, and since the question has already forced you into calculating

1 + (\frac{dy}{dx})^2

, they're wanting you to go down the integral-with-respect-to-x route.

Reply 58

Pascal678

Original post by lilac12

you had to use your part a 1+(dy/dx)^2 = r^2 / r^2 - x^2 , Area = 2pi * integral of [ y (1+ (dy/dx)^0.5 ] ... using x=r and x= -r as limits

simplified down to 2pi * [rx] = 2pi * 2r^2 = 4pi r^2

I cheated a bit and used the limits 2r and 0, is this acceptable?

Also k=4/9M guys?

Reply 59