# Maths A level Core 1 Last resit 2019 May 15th Unofficial Mark SchemeWatch

Announcements
#1
if anyone can help fill in the rest cheers

Question 1
A) 63
B) 5 + âˆš2 (i think)

Question 2
A) dy/dy
B) d2y/dx2
C) @x=9, d2y/dx2 = 44

Question 3
- equate curves and use b^2-4ac < 0 as proof they don't intersect

Question 4
A) x = -1/2 (i think)
B) -7<x<3/4 (or something like that)
C) -1/2<x<3/4

Question 5
A) differentiate
B) âˆš13
C) t = 11/2
D) area = 39/8

Question 6
A) f(x) = 12/x + 5, find eqn. of normal @(-2,1)
B) B(4,8) C(-4,2)

Question 7
A) Â£8000
B) d = 12 which follows u17 = 280 (i think)

Question 8
integration

Question 9
more intergration?

Question 10
A) stretch 2
B) translate 2 in negative x-axis
C) k = -4, a = 1
Last edited by Virolite; 5 days ago
0
5 days ago
#2
I donâ€™t agree with most of your questions
(Original post by Virolite)
if anyone can help fill in the rest cheers

Question 1
A) 63
B) 5 + âˆš2 (i think)

Question 2
A) dy/dy
B) d2y/dx2
C) @x=9, d2y/dx2 = 44

Question 3
- equate curves and use b^2-4ac < 0 as proof they don't intersect

Question 4
A) x = -1/2 (i think)
B) -7<x<3/4 (or something like that)
C) -1/2<x<3/4

Question 5
A) differentiate
B) âˆš13
C) t = 11/2
D) area = 39/8

Question 6
A) f(x) = 12/x + 5, find eqn. of normal @(-2,1)
B) B(4,8) C(-4,2)

Question 7
A) Â£8000
B) d = 12 which follows u17 = 280 (i think)

Question 8
integration

Question 9
more intergration?

Question 10
A) stretch 2
B) translate 2 in negative x-axis
C) k = -4, a = 1
0
#3
can you be more specific?
(Original post by ian.blanch)
I donâ€™t agree with most of your questions
0
5 days ago
#4
How many marks was question 3? I didnâ€™t use b^2-4ac instead I equated the equations? Will I not get any marks?
0
#5
4 marks i think, you'll probs get all the marks by solving for x and proving no solutions
(Original post by Maarlo)
How many marks was question 3? I didnâ€™t use b^2-4ac instead I equated the equations? Will I not get any marks?
0
5 days ago
#6
What was the base length of the triangle?
0
5 days ago
#7
I think it was difference in x coordinates of A and B coz they lie on the line L1. It was 3 or 2 i think.
0
#8
5-t
What was the base length of the triangle?
0
5 days ago
#9
I got Â£7800 for the total of the 25 years
(Original post by Virolite)
if anyone can help fill in the rest cheers

Question 1
A) 63
B) 5 + âˆš2 (i think)

Question 2
A) dy/dy
B) d2y/dx2
C) @x=9, d2y/dx2 = 44

Question 3
- equate curves and use b^2-4ac < 0 as proof they don't intersect

Question 4
A) x = -1/2 (i think)
B) -7<x<3/4 (or something like that)
C) -1/2<x<3/4

Question 5
A) differentiate
B) âˆš13
C) t = 11/2
D) area = 39/8

Question 6
A) f(x) = 12/x + 5, find eqn. of normal @(-2,1)
B) B(4,8) C(-4,2)

Question 7
A) Â£8000
B) d = 12 which follows u17 = 280 (i think)

Question 8
integration

Question 9
more intergration?

Question 10
A) stretch 2
B) translate 2 in negative x-axis
C) k = -4, a = 1
0
4 days ago
#10
what does everyone think of the paper? easier than last year or harder?
0
4 days ago
#11
100% easier
what does everyone think of the paper? easier than last year or harder?
0
4 days ago
#12
How do you find t? Do you have to find the length ac and the length bc? And equal ?
0
4 days ago
#13
So A was (5,8) and B was (3,11) and AC=BC and the point C is (t,8) so:
5-t = (square root) (11-8)(squared) (3-t)(squared)
5-t = (square root) (9) (9-6t t(squared))
5-t = (square root) (18 - 6t t(squared))
(5-t)(squared) = (18 -6t t(squared))
(25- 10t t(squared)) = (18 -6t t(squared))
Therefore 25-18 = -6t 10t
So 7 = 4t
t= 7/4

Hope that helps!
How do you find t? Do you have to find the length ac and the length bc? And equal ?
Last edited by Mackenziekav; 4 days ago
1
4 days ago
#14
I think for question 6b) the x coordinates were /-3 but I might be wrong
1
4 days ago
#15
Yes they were 3,9 and -3,1
(Original post by aslaterm32)
I think for question 6b) the x coordinates were /-3 but I might be wrong
1
#16
bro i did the whole thing with (5+t) instead of (5-t) how many marks do you reckon i dropped?

it was 3 marks for t and 2 for area
(Original post by Mackenziekav)
So A was (5,8) and B was (3,11) and AC=BC and the point C is (t,8) so:
5-t = (square root) (11-8)(squared) (3-t)(squared)
5-t = (square root) (9) (9-6t t(squared))
5-t = (square root) (18 - 6t t(squared))
(5-t)(squared) = (18 -6t t(squared))
(25- 10t t(squared)) = (18 -6t t(squared))
Therefore 25-18 = -6t 10t
So 7 = 4t
t= 7/4

Hope that helps!
0
4 days ago
#17
What did u get for area question
0
4 days ago
#18
Hmm idk they might just take one mark off if the rest of the working was right?
(Original post by Virolite)
bro i did the whole thing with (5+t) instead of (5-t) how many marks do you reckon i dropped?

it was 3 marks for t and 2 for area
0
4 days ago
#19
what does everyone think grade boundaries are going to be for an A? 67/68? also i got every answer correct however my working out was very very very messy do you think ill drop any marks because of that?
0
4 days ago
#20
they will take off about 3 marks sorry mate
(Original post by Mackenziekav)
Hmm idk they might just take one mark off if the rest of the working was right?
0
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