# OCR MEI C1 2017 Unofficial Mark Scheme

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#1
1) Draw the graph that goes through (0,1) with a gradient of -2. The line y=-2x+1 [2 marks]

2)
i) (1 7/9)^-1/2 = 3/4 [3 marks]
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 [2 marks]

3) 6-x > 5(x-3) = x < 21/6 [3 marks]

4) Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) [4 marks]

5)
i) Centre: (-2, 3) Radius: 5^1/2 [2 marks]
ii) Line parallel to 5x + y = 4 and touches centre: y = -5x -7 [2 marks]

6)
r = (V/(a+b))^1/2
r^2 = V/(a+b)
r^2(a+b) = V
ar^2 + br^2 = V
br^2 = V - ar^2
b = (V-ar^2)/r^2 [4 marks]

7)
i) (5-2root7)/3+root7 = (29 -11root7)/2 [3 marks]
ii) 12/root2 + root98 = 13root2 [2 marks]

8) (a+bx)^5. Constant is 32. x^3 coefficient is -1080. a = 2, b = -3 [5 marks]

9) (n+2)^2 - n^2
= n^2 + 4n + 4 - n^2
= 4n + 4
= 4(n + 1) [4 marks]

10)
i) AB = root(5^2 + 5^2)
= root50
BC = (7^2 + 1^2)^1/2
= root50 [2 marks]
ii) Line perpendicular to AC which passes through B.
Grad of AC is -2, perp gradient is 1/2. y = 1/2x-1 [4 marks]
iii) (10, 4) [2 marks]
iv) Equation of line AD 7y = x + 18
Sub in x coordinate of point E (8, 3.8)
7y = 8 + 18
7y = 26
y = 26/7
26/7 > 3.8
Lies outside [4 marks]

11)
i) Cubic graph, cutting x-axis at -5, 3/2, 2, and y-axis at 30 [3 marks]
ii) f(x) = (x-2)(2x-3)(x+5)
g(x) = f(x+3)
g(x) = ((x+3)-2)(2(x+3)-3)((x+3)+5)
= 2x^3 +21x^2 +43x +24 [3 marks]
iii) x = -2 is a root of g(x) = 6. Divide g(x) = 6 by (x+2) to get other roots: (-17+ root217)/4 or (-17- root217)/4 [6 marks]

12)
i) (x+1/2)^2 + 11/4. Min point (-1/2, 11/4). Min y value is 11/4 - Above x-axis [4 marks]
ii) Intersection of x^2+x+3 and 2x^2-3x-9: (6, 45) and (-2, 5) [4 marks]
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13 [4 marks]
2
3 years ago
#2
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3 years ago
#3
What about the rearranging the forumla one?
0
3 years ago
#4
1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)
2.
i) (1 7/9)^-1/2 = 3/4 (3 marks)
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)
3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)
4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)
5. (x+2)^2 + (y-3)^2 = 5
i) radius is root 5, centre is (-2,3) (2 marks)
ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)
6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)
7.
i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)
ii) 12/root2 + root 98 = 13root2 (2 marks)
8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)
9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)
i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)
ii)Line perpendicular to AC which passes by B
Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)
iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)
iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)
11. f(x) = (x-2)(2x-3)(x+5)
i) Sketch graph (3 marks)
ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5
x = -1, -3/2, -8 are new roots
factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)
iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)
12.
i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)
ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13
1
3 years ago
#5
Our version can be found here.
0
3 years ago
#6
(Original post by LokiiR)
1. Sketched y= -2x + 1 (0,1) and gradient was -2 (2 marks)
2.
i) (1 7/9)^-1/2 = 3/4 (3 marks)
ii) (6x^5y^2)^3/18y^10 = 12x^15/y^4 (2 marks)
3. 6-x > 5(x-3) = 6 - x > 5x - 15 = -6x > -21 = x< 21/6 (3 marks)
4. Intersection of 2x+5y=5 and x-2y = (10/3, -1/3) (4 marjs)
5. (x+2)^2 + (y-3)^2 = 5
i) radius is root 5, centre is (-2,3) (2 marks)
ii) Line parallel to 5x + y = 4 and touches centre is y= -5x-7 (2 marks)
6. r = root (v/a+b) - Make B the subject : Two possible answers are b = (v-ar^2)/r^2 or b = v/r^2 - a (4 marks)
7.
i) (5-2root7)/3+root7 = (29-11root7)/2 (3 marks)
ii) 12/root2 + root 98 = 13root2 (2 marks)
8. (a+bx)^5 x^3 coefficient is -1080, constant is 32. a^5 = 32, so a = 2. b= -3 (5 marks)
9. Three consecutive integers and n is the smallest (therefore n, n+1 and n+2) show difference between squares of smallest and largest is 4 * middle. (n+2)^2 - n^2 = 4n+4. 4n+4 = 4(n+1). (4 marks.

Section B

10. A(3,3) B (-2,-2) C (5,-1)
i) Show AB is BC. AB = ROOT 50, BC = ROOT 50 (2 marks)
ii)Line perpendicular to AC which passes by B
Grad of AC is -2, perp gradient is 1/2. Line is y= 1/2x-1 (4 marks)
iii) D is (10,4) because difference between B and C is 7, 1. Add 7 to x of A and 1 to y of A to get 10,4 (2 marks)
iv) 3.8 < 26/7. Find equation of lines and prove 8, 3.8 does not belong in them therefore it is outside rhombus (4 marks)
11. f(x) = (x-2)(2x-3)(x+5)
i) Sketch graph (3 marks)
ii) Show when (-3 0) occurs that g(x) is 2x^3 + 21x^2 + 43x + 24. You use f(x+3) and from that x+3 = 2, 3/2, -5
x = -1, -3/2, -8 are new roots
factors are (x+1)(2x+3)(x+8), multiply out to get the answer (3 marks)
iii)x = -2 is root of g(x)=6, prove this and find other roots. Other roots = (-17+-root217)/4 - Not sure of this (6 marks)
12.
i) Show y= x^2 + x + 3 as (x+a)^2 + b and it doesnt touch the x axis. Completed square is (x+1/2)^2 + 11/4 So minimum point is -1/2, 11/4 which is above x axis (4 marks)
ii) Intersection of x^2+x+3 and 2x^2-3x-9 are (6,45) and (-2,5) (4 marks)
iii) For what values of k is there no intersection of x^2 + x + k and 2x^2 - 3x- 9
Use b^2 - 4ac<0
k< -13
Do you think they'll allow both options for rearranging the formula?
0
3 years ago
#7
(Original post by RebeccaP10)
Do you think they'll allow both options for rearranging the formula?
they should, both are correct
1
3 years ago
#8
(Original post by RebeccaP10)
Do you think they'll allow both options for rearranging the formula?
For 11 part 3, the part inside of the root should say (17*17)-4(2)(18) which would be 289-144= 145, so in total it should read (if i'm correct) -17+-SQRT(145)/4
0
3 years ago
#9
(Original post by Libaan)
For 11 part 3, the part inside of the root should say (17*17)-4(2)(18) which would be 289-144= 145, so in total it should read (if i'm correct) -17+-SQRT(145)/4
Hey I think the equation for g(x) in q11)iii) has been written out wrong! Here is my working for the question... I hope this is right 1
3 years ago
#10
Ah, this is brill thank you!

I think/hope i got most of these.

Although i messed up the binomial expansion and the rhombus questions.

how many marks do you think i'd get for writing out the nCk x a^(n-k) x b^(k) for x^3 ?

Also, for the rhombus questions, i attempted to find the mid point for the 2 mark question. Then the 4 mark part, I didn't find an equation, i just said that point E was bigger than the max points of each coord so it's outside. How many marks for that?

ahh, I'm just hoping for at least 62!
0
#11
For the binomial Q, you might get method marks if you're lucky, but won't get the answer marks if you're final answer was wrong.

For the rhombus Q you might get some marks for the second part since you use you're answer from the last question (followed through) but probably not all of them.

I'm sure you did fine!

(Original post by gabriellenlewis)
Ah, this is brill thank you!

I think/hope i got most of these.

Although i messed up the binomial expansion and the rhombus questions.

how many marks do you think i'd get for writing out the nCk x a^(n-k) x b^(k) for x^3 ?

Also, for the rhombus questions, i attempted to find the mid point for the 2 mark question. Then the 4 mark part, I didn't find an equation, i just said that point E was bigger than the max points of each coord so it's outside. How many marks for that?

ahh, I'm just hoping for at least 62!
0
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