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

cheers @Evilhomer_93 for the link
if anyone can help fill in the rest cheers

Question 1
A) 63
B) 5 + √2 (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) √13
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) £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

I don’t agree with most of your questions

How many marks was question 3? I didn’t use b^2-4ac instead I equated the equations? Will I not get any marks?

What was the base length of the triangle?

cheers @Evilhomer_93 for the link
if anyone can help fill in the rest cheers

Question 1
A) 63
B) 5 + √2 (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) √13
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) £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

what does everyone think of the paper? easier than last year or harder?

How do you find t? Do you have to find the length ac and the length bc? And equal ?

I think for question 6b) the x coordinates were /-3 but I might be wrong

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!

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

Hmm idk they might just take one mark off if the rest of the working was right?