OCR MEI C1 18th May 2016

Reply 20

MadeInMelzo

Had to use the quadratic formula maybe 2 or 3 times which always makes me think I made a mistake

Reply 21

C-rated

2

Original post by bethygirl

absolutely kicking myself. Was relying on getting full marks in this resit to make up for the hell of core 3 and 4. But I missed out the last question not realising I was to use the discriminant which I thought was just for intersection with x axis :frown:

I also made a mistake in simplifying my values for the x coordinates for part ii of the last questions, I subbed everything correctly into the quadratic formula but I'll have lost 2 marks due to simplifying wrong hence getting incorrect Y values :frown:

. So that's 7 marks I've already lost :frown:

uh so mad, definitely harder than last year.

I can tell you for sure that the grade boundaries will be lower, MEI actually put the nearly exact same stupid ass question from last year in this paper

Reply 22

hunter0raf42

1

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 ?

Reply 23

DrKillerX

1

Original post by TeeEm

How hard was the MEI C1, please?

Fairly easy. But its an annoying paper with graphs to draw n estimate.

Reply 24

DrKillerX

1

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 I did get a=-2 c=7 dun rmb the last question tho

Reply 25

sqr00t

2

im absolutely kicking myself, i knew how to do the entire paper, had time to check answers for half an hour and i cant believe i blatantly applied the remainder theorem wrong. instead i long divided and equated the remainder expression to the given remainder, while substituting the x-value of the denominator expression. yes, A should be -2 and C=7, but instead i got A=12 and c=-21, which substituting into the f(2), gives 11 as well .-. so i thought it was right then realised it wasnt.

last question, there should be two methods to do it, the discriminant and comparing gradients.

the linear line y=x+k should have a overall equation that is parallel to a gradient at a point on the curve, so i differentiated the curve and equated the gradient function to the gradient of y=x+k (1). to find the value of x where their gradients are equal. then that is the point where they would just touch. (the curve function is a smiley curve). hence at that point of x, x+k=curve equation. rearrange to k= something, sub the value of x i found, to find k at that point. and then k must be smaller than that value. i got k<-9

Reply 26

sqr00t

2

i almost made a mistake with the quadratic formulas because i have problems reading my own writing, upon checking i realised i wrote a 3 somewhere instead of a supposed 2.

Reply 27

Halanh

Anyone got any answers for some of the questions? Found it harder than most papers I've done. The graph question was hard

Reply 28

BlakeToad

1

For the last question I got k= -60/8, anyone else get this? It was B^2-4ac < 0 as it has no real roots. I thought it went pretty well, just one question asked for the coordinates but I only did the x value in the quadratic formula, you was meant to complete the square so you can get the y value apparently.

Reply 29

Exams147

1

Anyone got any answers for some of the questions? Found it harder than most papers I've done. The graph question was the hardest for me.

Reply 30

BlakeToad

1

Original post by Exams147

Anyone got any answers for some of the questions? Found it harder than most papers I've done. The graph question was the hardest for me.

What questions? I found the graph one fairly easy.

Reply 31

BlakeToad

1

Original post by sqr00t

im absolutely kicking myself, i knew how to do the entire paper, had time to check answers for half an hour and i cant believe i blatantly applied the remainder theorem wrong. instead i long divided and equated the remainder expression to the given remainder, while substituting the x-value of the denominator expression. yes, A should be -2 and C=7, but instead i got A=12 and c=-21, which substituting into the f(2), gives 11 as well .-. so i thought it was right then realised it wasnt.

last question, there should be two methods to do it, the discriminant and comparing gradients.

the linear line y=x+k should have a overall equation that is parallel to a gradient at a point on the curve, so i differentiated the curve and equated the gradient function to the gradient of y=x+k (1). to find the value of x where their gradients are equal. then that is the point where they would just touch. (the curve function is a smiley curve). hence at that point of x, x+k=curve equation. rearrange to k= something, sub the value of x i found, to find k at that point. and then k must be smaller than that value. i got k<-9

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

Reply 32

h.xlx

4

What did everyone get for the equation of the circle and the equation of the tangent?

Posted from TSR Mobile

Reply 33

BlakeToad

1

Original post by h.xlx

What did everyone get for the equation of the circle and the equation of the tangent?

Posted from TSR Mobile

my radius was 5, so it was =25 on the end. I cant remember the whole equation though. What did you get?

Reply 34

BlakeToad

1

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 ?

i got that, but for the last one it asked for a 'set' of values, so i just put + or - -60/8 in a panic, i didnt really get the 'set' of values part.

Reply 35

jessyoung

1

I'm an idiot moved the graph to the left instead of the right, but got the equation of the translated curve right. Does anyone remember how many marks the translation was worth?

Reply 36

jessyoung

1

But guys just remember it's so different under pressure.

Reply 37

BlakeToad

1

Original post by jessyoung

I'm an idiot moved the graph to the left instead of the right, but got the equation of the translated curve right. Does anyone remember how many marks the translation was worth?