OCR MEI - C1 - 13th May 2015

Original post by Eux

What was the question again?

Right angled triangle of side length h, 2x + 1 and hypotenuse 3x. Part one is to show that h^2 = 5x^2 - 4x - 1

Then the second part was given h = root 7, find x.

Reply 362

Academic18

Original post by Eux

What was the question again?

5x^2-4x-8=0 find the x coordinate or something like that I think.

Reply 363

Original post by Academic18

5x^2-4x-8=0 find the x coordinate or something like that I think.

I think i remember the question but dont know where you got coorindates from.

Reply 364

Original post by 16Characters....

Right angled triangle of side length h, 2x + 1 and hypotenuse 3x. Part one is to show that h^2 = 5x^2 - 4x - 1

Then the second part was given h = root 7, find x.

I see, in which case + would have been correct, but as no units were given the negative answer should also be correct but not essential- perhaps an isw comment after +

Reply 365

Velocity_

18

For the triangle question when h2 = 7 (i think? because h = sqroot7) so would it be 7= (the thing it gave you for part i).
is it 7 or did we have to use sqroot 7 = ....
Anyone remember the question?

Reply 366

BECKl

has anyone made a mark scheme yet? honestly i'm not sure how it went, i feel like i have gotten an A but it all depends on whether any silly mistakes add up - i know i did well on Section A because i find those sorts of questions easy, but i feel as if question 12 is really gonna damage my grade. hoping for good grade boundaries :redface:

Reply 367

Academic18

Original post by Eux

I think i remember the question but dont know where you got coorindates from.

Sorry I don't knw either :P I meant the value of x. If you don't simplify your answer do you still get the marks? I didn't simplify my perpendicular bisector equation. :angry:

Reply 368

Original post by Academic18

Sorry I don't knw either :P I meant the value of x. If you don't simplify your answer do you still get the marks? I didn't simplify my perpendicular bisector equation. :angry:

Normally get o.e next to a lot of answers in mark schemes, so yes.

Reply 369

Original post by Velocity_

For the triangle question when h2 = 7 (i think? because h = sqroot7) so would it be 7= (the thing it gave you for part i).
is it 7 or did we have to use sqroot 7 = ....
Anyone remember the question?

7 = 5x^2 - 4x - 1 ==> 5x^2 - 4x - 8 x = 0 ==> (4 + 4*root(176))/10 = (4 + 4*root(11))/10

Which then simplifies to (2 + 2*root(11))/5 but I forgot that bit in the exam. Plus that's just the positive root. Read the above discussion about whether the negative one was needed. I personally found both but then said to discount the negative one but don't know if that is correct.

(edited 8 years ago)

Reply 370

chelseachampions

messed up the logic of maths questions....

Reply 371

GO97

8

Original post by 16Characters....

Probably your accuracy mark for 12ii and I think you might get full marks on 12iii as an error carried forward, but I cannot vouch for it I'm just another student like you, never marked an exam in my life.

ye just really wanted opinions, thanks for answering

Reply 372

bethygirl

Could someone please explain to me what I needed to do for the discriminant question? I knew I needed to discriminant so I wrote down b^2-4ac>0 to get a method mark but couldn't go anything further

Reply 373

jakoblauritzen

Original post by TazLi

Remember + and -

It said for when r>0

Reply 374

Anyone who's wondering if they'll get the mark for +/- take a look at this C1 paper from 2011 Q9. http://www.mei.org.uk/files/papers/2011_Jan_c1.pdf , mark scheme also in doc.

Reply 375

Original post by bethygirl

Could someone please explain to me what I needed to do for the discriminant question? I knew I needed to discriminant so I wrote down b^2-4ac>0 to get a method mark but couldn't go anything further

3x^2 + 12x + 13 = 2x + k ==> 3x^2 + 10k + (13-k) = 0
For the curve and line to intersect twice the above quadratic must have two distinct real roots, i.e. its discriminant > 0:

10^2 - 4(3)(13 - k) > 0
==> 100 - 156 + 12k > 0
==> -56 + 12k > 0
==> 12k > 56
==> k > 56/12 = 14/3

Reply 376

Velocity_

18

Original post by 16Characters....

7 = 5x^2 - 4x - 1 ==> 5x^2 - 4x - 8 x = 0 ==> (4 + 4*root(176))/10 = (4 + 4*root(11))/10

Which then simplifies to (2 + 2*root(11))/5 but I forgot that bit in the exam. Plus that's just the positive root. Read the above discussion about whether the negative one was needed. I personally found both but then said to discount the negative one but don't know if that is correct.

o..k, good so i think i got it. I have a horrible inkling in the back of my head i made it 5x^2 - 4x +6 but i doubt it... i remember doing 5x6 somewhere... i think. I dont even know anymore. i think i did what u said but i didn't simplify it because it was only A 3 mark question... :/ wait, i did to that method but i didn't simplify it.
Also for the graph, i didn't put the correct y value in (stupid mistake) but i correctly wrote the x axis on for the first question on section b. 4 marks... would i lose 1 or more? :/
For the second last question on completing the square, i got minimum value -2,1 or something that was correct but i stated that since the min y value was above the x axis, the curve wouldn't pass it. would that get me the 5 marks? :smile:

overall, i thnk it was decent. i need 82 ums. (i think i have got 57-61. i varies)

Reply 377

bethygirl

Original post by 16Characters....

3x^2 + 12x + 13 = 2x + k ==> 3x^2 + 10k + (13-k) = 0
For the curve and line to intersect twice the above quadratic must have two distinct real roots, i.e. its discriminant > 0:

10^2 - 4(3)(13 - k) > 0
==> 100 - 156 + 12k > 0
==> -56 + 12k > 0
==> 12k > 56
==> k > 56/12 = 14/3

I made them equal to eachother at the beginning will that be 1 method mark do you think?

Reply 378

Original post by jakoblauritzen

It said for when r>0

Is this Q1?

Reply 379