AQA Core 1 June 2010 Question paper & Mark Scheme

oh you thank you very much! +rep from me

i thought u said you had c4 june 2010 ,Pim not doing c1

1 bii to find equation AD we know AD is perpendicular to AB hence we use the gradient AB to be the reference gradient. i.e say if gradient AB is -1/2,then gradient of AD will be 2.
For part c once you got the equation. Use substitution to get the co-ordinate.
Question 5: r = square root 61 or r^2 = 61.
I had difficulty with part 6 (i) too for the you need to consider all the sides:
so you get:3x(y) +4x(y)+5x(y)+3x (4x) = 144
sort them out and you should end up with the given equation.

One way of solving part 7 ii is by finding out the determinant.If determinant < 0 then we have no real roots.

Okay, so I just did this paper and I can explain those two questions.

For part 1bii) you know that BAD is a right angle. Therefore the gradient of AD is 3/2 as m1*m2=-1 and m1 is -2/3. The line AD passes through both the points A and D so you can calculate the line equation with either of them. As we are given D it is easier to use that point.
You should get y-7=3/2(x-3),
which simplifies to give 2x-3y+15=0
It should then be easy to use for the next question as you can substitute x=5y-6 into this line equation...

For 7ai) the 2 in front of the brackets means that it's factorised. Therefore you divide what you would normally put in by 2 so by completing the square normally without 2x^2 we would put (x-10)^2 as you've done but becasue there is a coefficient of 2 we would put 2(x-5)^2. You got -47 as 10^2 =100. But as the correct answer is 2(x-5)^2 we would do 5^2 instead - but we also have to multiply that by the 2 in front of the brackets. THis gives 2(x-5)^2-50+53 so if simplifies to give 2(x-5)^2+3.

Hope that helped, just ask if you are stuck at all!!

1c - You have to do simultaneous equations -

(BC) 5y - x = 6
(AB) 3y + 2x = 14

From that you can work out x and y

So sorry about the delay of copying this up.
Here is the mark scheme that i copied down from class:

1a) y=-2/3x +14/3
gradient is -2/3
b) y=-2/3x+9
ii) you'll know if you got this one right or not if you got to the right answer.. if you didn't quote me and i'll do it later

c) y=2, x=4 --> (4,2)

2a) 14-6root5
b)-11+5root5


3ai) likewise you'll know if yuo got this one right.. you must say x+3 is a factor

ii) long division correct --> (x+3)(x+5)(x-1)

b)p(2)=35
ci)p(-1) = -16
p(0) = -15
therefore p(-1) is less, meaning the lowest point on the graph is nearer to when y=-1, to y = 0

cii) draw correctly, going through -5, -3, and 1 on the x axis, and -15 at the bottom
IMPORTANT: because for part ci we found that the lowest point (the vertex) is around where y=-1, the lowest point on yourgraph CAN'T be where it crosses the y axis. the vertex must be slightly to the left of the y axis. quote me if you need help.

4a) 1/5x^5 - 4x^2 + 9x (+x)
42/5
4aii) y=9 integrate 9
9x
sub in 2
=18
18-42/5 = 48/5

b) dy/dx = 4x^3 -8
where x = 1
= -4
ii) where x=1, y=2
so equation is y=-4x+6

5a) (x+5)^2 + (y-6)^2 = 5^2
bi) 25=25
bii) gradient = -4/3
y=-4/3x-2/3
iii) m = -7/2,4
PM^2 =2^2 + (3/2)^2
pm = 5/2
PO = 2root2

sp p is closer to M


6ai)you'll know if you got this one right or not
ii)
v=6x^2y
rearrange 12=x^2 + y -> y=12/x -x
v = 6x^2(12/x-x)
v = 72x - 6x^3

bi) dv/dx = 72 - 18x^2.. when x=2 it equals 0
c) d^2v/dx^2 = -36x
sub in when x=2 = -72 <0 so a maximum

7) 2(x-5)^2 +3
no real roots - when try to solve can't squareroot a negative number

bi) you'll know if you got this right, ask me if you want me to go through it.
ii) (7k+1)(k-1) <0 (PLEASE NOTE THERE SHOULD BE A LESS THAN OR EQUALS SIGN, WHERE I HAVE JUST PUT A LESS THAN because i dont know how to do it otherwise!)
iii) -1/7 < k < 1

again, there should be a less than or equals sign, not just less than. !!

Hope this helped everyone, sorry again about the delay, and if you're stuck quote me and i'll try and help

what do you mean about 3cii? The y intercept is not -15?

Yes it is?
Y intercept is where x = 0
so x^3 + 7x^2 + 7x -15

so, when x is 0, y=-15

?

Or have i made a really silly mistake? what do you think it is?

Yes it is?
Y intercept is where x = 0
so x^3 + 7x^2 + 7x -15

so, when x is 0, y=-15

?

Or have i made a really silly mistake? what do you think it is?

does anyone know the link of where i can get the c1 aqa maths paper june 2010 online ?