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OCR (not MEI) C1 13/05/13

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Reply 240

ChestnutHero

I got k = -5,
A = -2, -27
and PQ as 4.5 or 9/2

Reply 241

roar96

How did you do question 2? The hidden quadratic?

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Reply 242

username1102419

Ok, I thought this was a really easy exam, but that's my own opinion- don't bite me for it!

I brought my answers out of the exam on scrap paper, so I'll do my best to translate. I'm pretty confident they're all right, if I'm wrong correct me, but I went through with teachers and other students afterwards and these seem to be correct answers. If anybody did get similar, shout up! :tongue:

1)i) 12 rt.5
ii) 4 rt.5
iii) 5 rt.5

2) x^3 = 1/8 and -1 ... therefore x = 1/2 and -1

3) f'(x) = -(12x^-3) +2
f''(x) = 36x^-4

4)i) 3(x+3/2)^2 + 13/4 [Because]
ii) Vertex is (-3/2, 13/4)
iii) b^2 - 4ac = 9^2 - 120 = -39

5)i) Basically the same graph as 1/x^2... Curves symmetrical and never touch any of the axes
ii) Stretch in Scale Factor 1/2 in y (vertical) direction

6)i) x^2 + (y+4)^2 = 40 Centre is (0, -4) Radius is rt.40 = 2 rt.10
ii) B co-ordinate is (-2, -10)

7)i) x < -1/8
ii) 0 </= x </= 5 ( </= means greater than or equal to! solved this one by drawing out the curve.. )

8) m of perpendicular is 1/3, so m of line is -3.. put into formula to get 3x + y -1/2 = 0 .. x2 to get into integers, 6x + 2y - 1 = 0

9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
ii) Vertex at x=1/4 so function is decreasing for x < 1/4
iii) Submit into formula for curve, use quadratic formula to solve, find that points P and Q are (-2,4) and (2.5,4) so distance is 4.5

10)i) Solve to find that k = -5
dy/dx = -3x^2 - 6x + 4 - k
0 = -3(-3)^2 - 6(-3) + 4 - k
k = -27 + 18 + 4 = -5

ii) d2y/dx2 = -6x - 6 ... submit -3 in to get 12, 12>0 therefore it's a minimum point
iii) This one was really long winded for 5 marks, basically you had to put the formulas together, solve the quadratic to find the two points on the cubic curve that satisfied a gradient of 9, you then put those two points (0 and -2) into the y=9x-9, and see which works. 0 doesn't work, but -2 does, so A = (-2, -27)

Hope this helps! Please point out any mistakes you think I've made, would be awesome if someone could scan a copy of the question sheet too :smile:

(edited 10 years ago)

Reply 243

a10

Original post by eggfriedrice

Same, I was tempted to ask for scrap paper to do the test again and see if I get the same answers, but instead I just decided to check through 3 times. Luckily I spotted and corrected a few mistakes.
Oh wow, their application thing opened early then. I haven't even touched my application in a few weeks lol, I literally have half a box to fill and send off.

yours sounds long hahaha, for mine I had two huge engineering questions and you pick one or both n u had to write a whole page for and the rest of it was on your university choices and grades and references.

Reply 244

John Dogg

Original post by roar96

How did you do question 2? The hidden quadratic?

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substitute y for x^3 then solve as normal i think you got -1/8 and -1 then take the 3rd root of these

Reply 245

Dugald

Original post by Anubis2

'Fraid you're wrong, the last is (-3/2, 13/4).

yeah I know, I remember now I did put (-3/2,13/4)

for y it was something like 10-17/4 that's where I got the 17/4 bit from in my head.

Reply 246

metabalo

Original post by Anubis2

You do not have to write the answer in any specific part of the answer booklet, only next to the correct question number. Congrats for getting (-2,-27) as that is right.

Thanks, but what I meant was that did the last question to find the tangent consist of about a third of the last inner page and the entire back outside page in the answer booklet? Because I only used the bottom space in the back inner page for the tangent question and I got a feeling I put the answer in the wrong question box. Sorry, I phrased the question badly.

Reply 247

a10

Original post by roar96

How did you do question 2? The hidden quadratic?

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let y = x^3 therefore y^2 = x^6 then substitute and factorise.

Reply 248

Big-Daddy

Original post by eggfriedrice

Could have sworn on my paper it said "coordinates". My eyes must be playing tricks on me. -_-

Original post by a10

it did say co-ordinates but only one of the pair satisfies the equation..

co-ordinates usually mean 2 points (x and y) if they wanted one value its usually referred to as ordinate.

Did the exam go well for you? :smile:

Original post by Zakee

It did say coordinates. However, remember a coordinate is simply one value. (i.e) find out the x coordinate in the equation (4,11), the coordinate is 4. The coordinates however are (4,11).

To be precise it specified "the coordinates of the point". Point, singular. Hence only one answer.

The coordinates weren't (4,11) - they were (-2,-27). That was an example, right?

Reply 249

roar96

Original post by John Dogg

substitute y for x^3 then solve as normal i think you got -1/8 and -1 then take the 3rd root of these

But i swear that gave you a cubic to solve...

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Reply 250

Majeue

Hi guys,
I accidentally put k=5 for 10i even though I did the right workings (can't add up). I also then used this for part iii with the tangent 9x-9 but It wouldn't factories as I got the wrong quadratic. How many marks will I lose?
thanks

Reply 251

queketth

Original post by roar96

But i swear that gave you a cubic to solve...

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No It didn't I am afraid that is the correct method!

Reply 252

Alex_Clarky

Original post by Dugald

what are the rules for A* at A2. say I happen to get a low A in this paper, will I still be able to get an A*?

to get an A* you have to have 80% of all ums and average 90% in C3 and C4 :smile:

Reply 253

niaphonic

Original post by roar96

But i swear that gave you a cubic to solve...

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the original equation was something like 8x⁶ + 7x³.. so you would end up with a quadratic when you substitute y=x³

Reply 254

BankOfPigs

I don't think they can ask you to solve cubics in core 1, thats restricted to c2 with factor theorem. Unless they already give you the factors.

For those questions they're always in a form that can easily turn into a quadratic.

Reply 255

Big-Daddy

Given that Mr M is posting answers at some point tonight, discussing solutions seems a little fruitless.

However, if anyone wants to post a question fully I can post my answer to it. Does anyone remember the two inequalities (questions not answers), or the 7-mark Question 8?

Reply 256

username1102419

Original post by metabalo

If you put the answer in the wrong box, don't worry. My teacher's an exam marker and said that it's just to help set it out. You'll get the marks for the working, even if it's in another question's box :smile:

Reply 257

TwlilightLoz

what did guys put for the question where you had to state what happens to graph as it transforms from 2/x^2 to 1/x^2. I wrote that it stretches parallel in the x axis by 0.5