OCR MEI C1 18th May 2016

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Reply 60

sdeponte

https://docs.google.com/document/d/1GrOwL3hZzhJmn8rd9IyozRJX6e6-0fAlwbaEnJkJEnw/mobilebasic

unofficial mark scheme!!!

Reply 61

H3nry99

Thats not the paper I did?!

Reply 62

LXVI

was the turning point for the graph (2,-9) ? & Y intercept = -5 and X intercepts -1 and 5 ?

Reply 63

sqr00t

Original post by BlakeToad

Pretty sure thats not what you do for the last question, its just b^2-4ac < 0 since it has no real roots

this method actually works tho explanations are necessary (for both methods anyways). my problem was a silly arithmetic one. you can still learn this as a reference when you actually have to use gradients (in c2, c3, or c4).

differentiate curve with first principles.
this gradient function at a certain point would have the same gradient as y=x+k
so 1=4x-5, x=3/2. this is the boundary at which the line and the curve would JUST be touch.
sub x=3/2 into original curve equation to find a value for y.
sub y and x value into y=x+k to find k.
you can either sketch/ explain/ show with second differential, the nature of the quadratic, which is a smiley curve. therefore k must be smaller than -7.5 in order for them to not intersect or touch.

what i did wrong was i was thinking both methods in parallel and i equated x+k>curve, and subbed the x value into here.

double check with autograph or pen and paper. both methods would work and are mathematically valid, therefore marks should be allowed for either methods.

Reply 64

daniiaxox

I wonder if we'll get an unofficial mark scheme to check our answers and have a rough idea on how we did

Reply 65

Cybercelt

HI folks
I am just wondering how many people found this paper really hard?
Lots of surds and fractions in the working out which you couldn't solve without a calculator?
In all the past papers (I've done so many I have lost count) it was always easily worked numbers.
This paper was so out of my league.
I feel like the weeks I have spent revising for this hideous exam have been pointless> be lucky get a C at this rate!
My only hope is that I get marks for correct process.

(edited 7 years ago)

Reply 66

S199

Original post by hunter0raf42

Thought the paper was pretty good. The only tricky questions were whe you had to find A and C using the f(x) remainders. Did we all get A=-2 and C=7.

Also for the last quesitons: K < -60/8 ?

Yes.

Reply 67

Mo2351

It was a pretty good paper compared to past years .

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Reply 68

greenlimegaming

Ah everything seemed to be perfect until I rechecked Question 10. Appears I have done the circle where AB was the diameter meaning my midpoint and radius was incorrect but i still came out with an equation for the circle. In the subsequent question my tangent would also have been affected as the gradient would've have been the same when getting the perpendicular. How many marks would I have lost? Is there an error carried forward and for the first part where I did A and B midpoint and radius etc. gutted as I need to get near 100UMS in c2 in order to be averaging 80 going into A2 modules. (This was a resit for me)

Reply 69

suziew00zie

Original post by LXVI

was the turning point for the graph (2,-9) ? & Y intercept = -5 and X intercepts -1 and 5 ?

this is what I got, general consensus is yes :smile:

Reply 70

Mo2351

Do you think that the grade boundaries will be high like last year 63/64 as most people seemed to have found this paper quite easy ?

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Reply 71

sqr00t

Original post by ocnarfeam

For the last question. I realised that for the linear and quadratic to NOT intersect, the linear would have to intersect the y axis below the y value of the turning point on the quadratic. So, I differentiated the quadratic and found the coordinates for the turning point and then just said that for the linear(y=x+k) to NOT intersect, x+k<(y value of turning point)

Have I got anything correct there?

in order for them to not intersect, the linear line and curve must just not touch. and at the boundary of touching, their gradients would be the same at that point, hence an equivalent x value could be found, and y values and the k could be found.
I used this method, though you have to be careful calling it a turning point (as this only happens when the gradient function = 0), as long as you justify that the linear equation of the line must have k so that the its value of y is just below that of the curve its fine.

my problem was i accidentally subbed the x value for where the gradients are equal, into a separate equation as i was trying out the discriminant method at the same time.

Reply 72

S199

Original post by greenlimegaming

I'm pretty sure you should get error carried forward as long as your calculations ahead where all correct. I believe you would lose 1 mark for considering the centre of circle as mp of ab instead of ad which also affected your radius so they wouldn't penalize you for the wrong radius and wrong equation ahead. But i could be wrong

Reply 73

sqr00t

Original post by Jumbo97

11 ii) you can make x the subject of y=x+3 and therefore there's no need for the quadratic equation?

wrong. you need to find the x value first and then you can sub the x values into y=x+3. you definitely need the quadratic equation.

Reply 74

zebby1999

Original post by Mo2351

It was a pretty good paper compared to past years .

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at least it wasn't like the Jan 10 paper

Reply 75

sqr00t

Original post by Cybercelt

you're supposed to leave your answers in the most simplest rational form. ie. surds and fractions with whole numbers (ie. sqrt(3))

Reply 76

king9

Original post by TeeEm

How hard was the MEI C1, please?

Overall, it was pretty easy but the numbers used were very awkward which mean that final answers had to be left in surd form. Other questions weren't very explicit so took a little while to figure out what the question was asking.

Reply 77

sqr00t

Original post by king9

To whoever took the photo of, posted, and circulated the photos of the papers online. YOU ARE AT HIGH RISK OF DISQUALIFICATION. I'd still advise deleting them before it's too late.

Reply 78

Jumbo97

Original post by sqr00t

wrong. you need to find the x value first and then you can sub the x values into y=x+3. you definitely need the quadratic equation.

you make x=y-3 and y-3 for the x values in what ever the quadratic equation was to make it a y quadratic equation. You can work out Y and then sub your Y value into either y=x+3 or the the quadratic.

Reply 79

sqr00t

Original post by Jumbo97

you make x=y-3 and y-3 for the x values in what ever the quadratic equation was to make it a y quadratic equation. You can work out Y and then sub your Y value into either y=x+3 or the the quadratic.