The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

OCR MEI - FP1 - 14th May 2015

Scroll to see replies

Reply 200
Avatar for Aph
Aph
22
Original post by StrangeBanana
That 1 should be negative. :oops: Fixed it now.

You were given the polynomial
2x3+px2+qx+r2x^3 + px^2 + qx + r

And some information: the sum of the roots is 6, the product of the roots is -10, and one of the roots is 4. You had to find p, q and r.

How did you go about doing it??? I know I did it in the most convoluted way going.
Original post by Aph
How did you go about doing it??? I know I did it in the most convoluted way going.


You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that. :smile:
Original post by StrangeBanana
You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that. :smile:


I did the same as that except rather than factorising, i subbed in x=4 and made it equal to zero..

Also, 7ii is still incorrect as the second part of the inequality should be 0<=x<2 rather than 3<=x<2 !
Original post by jpetersgill
I did the same as that except rather than factorising, i subbed in x=4 and made it equal to zero..

Also, 7ii is still incorrect as the second part of the inequality should be 0<=x<2 rather than 3<=x<2 !


Sounds good. :biggrin:

:colondollar:
Reply 204
Avatar for Aph
Aph
22
Original post by StrangeBanana
You could get p and r immediately from the sum and product of the roots, and then I factorised the cubic (because (x - 4) was a factor) and deduced q from that. :smile:

I got those 2 yes but what I then did:s-smilie:

∑α+β+γ=6,αβγ=−10[br]⇒β=2−γ[br]⇒4γ(2−γ)=−10\sum\alpha+\beta+\gamma= 6, \alpha\beta\gamma= -10[br]\Rightarrow \beta=2-\gamma[br]\Rightarrow 4\gamma(2-\gamma)= -10 and went from there... Do you think that's alright?
(edited 8 years ago)
Original post by Aph
I got those 2 yes but what I then did:s-smilie:

∑α+β+γ=6,αβγ=−10[br]=>β=2−γ[br]=>4γ(2−γ)=−10\sum\alpha+\beta+\gamma= 6, \alpha\beta\gamma= -10[br]=> \beta=2-\gamma[br]=> 4\gamma(2-\gamma)= -10 and went from there... Do you think that's alright?


Well I'm not sure about those upside down question marks. :tongue:

So that gives you the third root, then what?
Anyone know how many marks each question was worth?


Posted from TSR Mobile
Reply 207
Avatar for Aph
Aph
22
Original post by StrangeBanana
Well I'm not sure about those upside down question marks. :tongue:

So that gives you the third root, then what?

Well it gives you the 2 roots then you add the 4 and multiply out. You have to put something in about beta and gamma being bi-variable and if one takes one value the other takes its complex congigate.
Original post by Aph
Well it gives you the 2 roots then you add the 4 and multiply out. You have to put something in about beta and gamma being bi-variable and if one takes one value the other takes its complex congigate.


Ah, fair enough. It's a valid method, so as long as you got the right answer I'm sure it'll be accepted!
(edited 8 years ago)
Reply 209
Avatar for Aph
Aph
22
Original post by StrangeBanana
Ah, fair enough. It's a valid method, so as long as you got the right answer I'm sure it'll be accepted!

Good, that was scary.the answer is 3 right?
Original post by Aph
Good, that was scary.the answer is 3 right?


Nope, p = -12, q = 11, r = 20. :s-smilie: Bad luck.
Reply 211
Avatar for Aph
Aph
22
Original post by StrangeBanana
Nope, p = -12, q = 11, r = 20. :s-smilie: Bad luck.

Oh... Well I got the other 2 correct:s-smilie:
Original post by henrygriff28
I did the exact same as you, do you think we'll still get the one mark? I think that would be all I dropped and I was hoping for 100 UMS so will be close.


Yeah can't lose 2 marks for getting it entirely right! Bullsh*t question, I didn't think I'd have to repeat the obvious lol
How many marks would one lose for not shading correctly in the Argand diagram? i.e didn't shade all the way down to minus one


Posted from TSR Mobile
Reply 214
Avatar for acelle
acelle
2
Original post by StrangeBanana
Nope, p = -12, q = 11, r = 20. :s-smilie: Bad luck.


NUUUUUUU I got r=123 T.T
Reply 215
Avatar for Aph
Aph
22
Original post by acelle
NUUUUUUU I got r=123 T.T

There were 2 r's in the paper.
Reply 216
Avatar for acelle
acelle
2
Original post by Aph
There were 2 r's in the paper.


O. Thanks that calmed my heart ..

Posted from TSR Mobile
Original post by acelle
NUUUUUUU I got r=123 T.T


:P yeah the other r was 123, don't worry
Original post by StrangeBanana
Unofficial Markscheme

1. x=518,y=127x = \dfrac{5}{18}, y = \dfrac{1}{27}


2. z1=2+3j,z2=2−3jz_1 = 2 + 3j, z_2 = 2 - 3j

∣z1∣=∣z2∣=13|z_1| = |z_2| = \sqrt{13}

arg(z1)=0.983,arg(z2)=−0.983arg(z_1) = 0.983, arg(z_2) = -0.983


3. p=−12,q=11,r=20p = -12, q = 11, r = 20


4.


5i. Use standard summations.

ii. k=13k = \dfrac{1}{3}


6. Run-of-the-mill proof by induction.


7i. Sketch:


ii. Values of x for which y≥3y \geq 3 are:

x<−1,0≤x<2x < -1, 0 \leq x < 2


8i. α2=9+40j,α3=−115+236j\alpha^2 = 9 + 40j, \alpha^3 = -115 + 236j

ii. q=−7,r=123q = -7, r = 123

iii. z=5+4j,5−4jz = 5 + 4j, 5 - 4j, or −3-3

iv. z=5+4j,5−4j,−3z = 5 + 4j, 5 - 4j, -3 or 11


9i. A'(0, 0), B'(-4, 0), C'(2, 12)

ii. Horizontal enlargement with s.f. 4, vertical enlargement with s.f. 2

iii. The matrix required was:
Unparseable latex formula:

\dfrac{1}{48}\begin{pmatrix}[br]0 & 8 \\[br]-6 & 12[br]\end{pmatrix}



iv. 192 sq. units


Please marry me

BUT will i lose some marks for the argand diagram question because for the part where you had do draw the argument boundries i used dotted lines instead of solid?
Reply 219
Avatar for Buses
Buses
9
Original post by Darcy1
Please marry me

BUT will i lose some marks for the argand diagram question because for the part where you had do draw the argument boundries i used dotted lines instead of solid?


probs 1 or 2 marks because it shows that the line wasn't inclusive of that inequality, and it clearly was. So it sort of shows you didn't understand the condition fully.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you