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OCR Maths C1 16/05/12 (not MEI)

Just wanted to know who found that hard and who found it easy, and which questions did you find the hardest? ect ect

For me I am pritty confident that I did well :albertein: :bunny: :bigsmile: so I'm happy!!

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Reply 1
I thought it was pritty hard.

Then again, I can't spell
For the last question, the discriminant was -76 right, so therefore the line doesn't intersect the circle?
Reply 3
Original post by As_Dust_Dances_
For the last question, the discriminant was -76 right, so therefore the line doesn't intersect the circle?


Yup :biggrin:
Reply 4
Original post by As_Dust_Dances_
For the last question, the discriminant was -76 right, so therefore the line doesn't intersect the circle?


I did not use the discriminant as you were suppost to work it out algerbically, but I'm sure you worked it out in your own way :smile:
Reply 5
Original post by Oglogski
I thought it was pritty hard.

Then again, I can't spell


Neither can I
Original post by ForensicCat
I did not use the discriminant as you were suppost to work it out algerbically, but I'm sure you worked it out in your own way :smile:


I didn't know any other way, so the discriminant ftw!
Reply 7
Just wondering, which collage/sixth form do you go to?
Reply 8
Original post by As_Dust_Dances_
I didn't know any other way, so the discriminant ftw!


I am not to sure what i did :tongue: I only had abot 5 minutes left to do it in
Reply 9
Original post by ForensicCat
I did not use the discriminant as you were suppost to work it out algerbically, but I'm sure you worked it out in your own way :smile:


You were supposed to sub y=2x into the equation of the circle, then simplify. So you get an quadratic equation in x.

Then, you use the discriminant (b^2 -4ac) of the formula and sub the coefficients in. You get -78 (or whatever it was)

Then you state something like 'discriminant < 0, therefore there are no points of intersection/ no real roots)'

At least that's what I did. I hope I got full marks for it.
Reply 10
Original post by samjj8
You were supposed to sub y=2x into the equation of the circle, then simplify. So you get an quadratic equation in x.

Then, you use the discriminant (b^2 -4ac) of the formula and sub the coefficients in. You get -78 (or whatever it was)

Then you state something like 'discriminant < 0, therefore there are no points of intersection/ no real roots)'

At least that's what I did. I hope I got full marks for it.


well, I sub'ed it in :smile: and then did some other equations like a madman hehe oh dear
Original post by ForensicCat
I did not use the discriminant as you were suppost to work it out algerbically, but I'm sure you worked it out in your own way :smile:



using the discriminant after substituting in y is algebraic.
Reply 12
Original post by WilliamGCSE
using the discriminant after substituting in y is algebraic.


oohhhh, may have lost a few marks there :0
Reply 13
Copying this over from the other C1 thread

1 x3 -6x2 -11
2 i) k=1/4
ii) k= -3/2
iii) k= 24
3
4 i) 2(x-5)2 -1
ii) minimum point = (5, -1)
5 i) graph
ii) a translation of 4 units in the positive x direction
iii) y = square root (1/5 x)
6 Normal to the curve = 4x -6y -29 =0
7 16+6 root 7, 16- 6 root 7 (16+- 6 root 7)
8 i)stationary point is at (-2, -48)
ii) this is a minimum
iii) x3 + 4 increase as x increases when x> -2
9 i) 0 < x < 4 (length can't be negative)
ii) 1.4 < y < 4.8
10 i) centre (5, -2), Diameter = 10
ii) y = 2x -12
iii) length CP = root 20, P is inside the Circle
iv) y=2x doesn't meet the circle because b2-4ac=-76 < 0

For question 3... (10/3,-2) I think... don't quote me on that, it's off the top of my head

Am I actually duplicating posts from two nearly identical threads hahaha. Crazy crazy crazy world.

curtsey of cheesy-craig.
(edited 11 years ago)
Original post by samjj8
Copying this over from the other C1 thread

1 x3 -6x2 -11
2 i) k=1/4
ii) k= -3/2
iii) k= 24
3
4 i) 2(x-5)2 -1
ii) minimum point = (5, -1)
5 i) graph
ii) a translation of 4 units in the positive x direction
iii) y = square root (1/5 x)
6 Normal to the curve = 4x -6y -29 =0
7 16+6 root 7, 16- 6 root 7 (16+- 6 root 7)
8 i)stationary point is at (-2, -48)
ii) this is a minimum
iii) x3 + 4 increase as x increases when x> -2
9 i) 0 < x < 4 (length can't be negative)
ii) 1.4 < y < 4.8
10 i) centre (5, -2), Diameter = 10
ii) y = 2x -12
iii) length CP = root 20, P is inside the Circle
iv) y=2x doesn't meet the circle because b2-4ac=-76 < 0

For question 3... (10/3,-2) I think... don't quote me on that, it's off the top of my head

Am I actually duplicating posts from two nearly identical threads hahaha. Crazy crazy crazy world.


I got that for q3 so looks hopeful :smile:
Reply 15
IMG_0243.jpg
the line y=2x doesn't meet the circle, i drew it out to check
Reply 16
Does anyone remember the question for 9i)? And how many points was it worth?
Reply 17
Original post by samjj8
Does anyone remember the question for 9i)? And how many points was it worth?


rectangle length 4x and width x-3, the area is less than 112, find possible values for x
Reply 18
4x(x-3) < 112 - So first mark

x^2-3x-28< 0 - Second mark

(x-7)(x+4) - 3rd mark/ 4th Mark

so -4 < x < 7 - which is where I stopped

Since -4 can't be negative...

its 4<x<7

How many marks would I have lost for not considering the negative value?

Have I got this question completely wrong?
Reply 19
Original post by As_Dust_Dances_
I got that for q3 so looks hopeful :smile:


Same, So hopefully it is right :h:

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