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Unofficial mark scheme ocr core 1 16/05/12

UNOFFICIAL MARK SCHEME

1 x3 -6x2 -11
2 i) k=1/4
ii) k=-3/2
iii) k= 24
3 i) gradient = 3/5
midpoint PQ (10/3, -2)
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+- 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) -7 < x < 4 (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 = 2 root 5, P is inside the Circle
iv) y=2x doesn't meet the circle

can anyopne help with the rest of the answers and i will update this post :smile:
(edited 11 years ago)

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Reply 1
Question 4: 2(x-5)^2 - 1
Therefore (+5, -1)
Original post by KoalaKim
UNOFFICIAL MARK SCHEME

1 x3 -6x2 -11
2 2(x-5)2 -1 (or this might have been another question?)
3
4
5 i) graph
ii) a translation of 4 units in the positive x direction
iii) y = square root (1/5 x)
6
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) -7 < x < 4
ii) 1.4 < y < 4.8
10 i) centre (5, -2), Diameter = 10
ii) y = 2x -12
iii) length CP = 2 root 5, P is inside the Circle
iv) y=2x doesn't meet the circle

can anyopne help with the rest of the answers and i will update this post :smile:

8(i) it was (-2,-48)
9(i) is 0<x<4 (how can you have a -ve length)
Reply 3
Original post by I_am_god_123
8(i) it was (-2,-48)
9(i) is 0<x<4 (how can you have a -ve length)


it can be negative because it gets squared to would go positive anyways
Reply 4
Original post by I_am_god_123
8(i) it was (-2,-48)
9(i) is 0<x<4 (how can you have a -ve length)


:O didn't think of that, depends whether length is a vector..., because if its in the opposite direction then it could be negative
Original post by max25
it can be negative because it gets squared to would go positive anyways


no, if the length is 3y then y cannot be a -ve value!?
Original post by KoalaKim
:O didn't think of that, depends whether length is a vector..., because if its in the opposite direction then it could be negative


length would not be a vector in the context of the question
Original post by KoalaKim
UNOFFICIAL MARK SCHEME

1 x3 -6x2 -11
2
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
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) -7 < x < 4
ii) 1.4 < y < 4.8
10 i) centre (5, -2), Diameter = 10
ii) y = 2x -12
iii) length CP = 2 root 5, P is inside the Circle
iv) y=2x doesn't meet the circle

can anyopne help with the rest of the answers and i will update this post :smile:


For question 7 i got 16+6root 7 and 2+6root 7? As you 3+or-root 7 from the quadratic right? And I let y=x^1/2 so you had to square each one. You get 9-7 for the second one.
Reply 8
thats what i did but the question asked for +- of the same surd so had to be 16 root 3 plus or minus
Reply 9
Original post by As_Dust_Dances_
For question 7 i got 16+6root 7 and 2+6root 7? As you 3+or-root 7 from the quadratic right? And I let y=x^1/2 so you had to square each one. You get 9-7 for the second one.


you get (3+ root7)2 and (3- root 7)2

The first gives 9 +6 root 7 +7

the second gives 9-6root 7 +7

because -root 7 * -root 7 = 7 because - x - =+
Original post by KoalaKim


you get (3+ root7)2 and (3- root 7)2

The first gives 9 +6 root 7 +7

the second gives 9-6root 7 +7

because -root 7 * -root 7 = 7 because - x - =+


oh well, hopefully only lost 1 mark..
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<0
(edited 11 years ago)
What do people think the grade boundaries will be like? Do you think 70/72 will be 100?
Reply 13
2i was 1/5 .. ii was -3/2 and iii was 24 ... Also for 6 i got 4x-6y+13 ?
Reply 14
What was question 6 again?
Reply 15
Original post by cheesy-craig
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<0


thanks
Reply 16
Original post by jameslad
2i was 1/5 .. ii was -3/2 and iii was 24 ... Also for 6 i got 4x-6y+13 ?


2)ii was 1/4

it was + 29 not 13
Reply 17
Original post by Oliver94
What was question 6 again?

it gave you the equation of a curve and asked you to fin the noraml at x=2

6/(x2) - 5

find the normal to the curve in the form ax +by +c =0
Reply 18
Is mr m going to post answers
Original post by benex381
2)ii was 1/4

it was + 29 not 13


actually it was -29, as you got 6y+21 = 4x-8
(edited 11 years ago)

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