AQA maths core 1, 13th May 2015

1a. -3/5 [2]
1b. 5x-3y+1=0 (I thought it was this form) [3]
1c. A(9,-4) [3?]
2. 7 + √15 [5]
3a. y-6=-10(x+1) [4]
3bi. 108/5 [5]
3bii. 162/5 [3]
4a. (x+1)^2 + (y-3)^2 = 50 [2?]
4bi. C(-1,3) [1]
4bii. 5√2 [2]
4c. k=-8, 2 [2]
4d. minimum distance =7 [2]
5a. p = 3/2, q = -¼ [3]
5bi. (-3/2, -¼) [2]
5bii. x=-3/2 [1]
5c. y = x^2 - x + 4 [3?]
6ai. h = 24/r -r/2 [3]
6aii. show V = 24πr -π/2 r^3 [3]
6bi. 24π -3π/2 r^2 [2]
6bii. =-12π therefore max (r=+4) [4]
7a. draw x^2(x-2) [3]
7bi. R = 36 [2]
7bii. R= 0 therefore root [2]
7biii.(x-2)(x^2-5x+10) [2]
7biv. x=2 [3]
8a. Show that x^2 + 3(k-2)x -13-k=0 [1]
8b. 9k^2 -32k -16 < 0 [3]
8c. -4/9 < k <4 [4]

Edit: added marks, not sure about a few of them.

(edited 8 years ago)

Reply 161

Hexaneandheels

4

For the graph, which I now know is probably the only error I have made, I drew the line beginning in bottom left quadrant, up through the origin and continued up and then down and just touched the x-axis at 3 and then back up again. Will I get any mark for this?

Original post by Heffalump .

What do you mean by 3 roots? it depends on where they were, it was 3 marks, but if you plotted on of the roots correctly (the root 3) then you would have only lost two I think...

Reply 162

SayeedahR

Original post by SaadMuhammad

I got 108/5 or something similar not 31.

yeah thats similar to what i got

Reply 163

Moe21

6

Original post by Heffalump .

What do you mean by 3 roots? it depends on where they were, it was 3 marks, but if you plotted on of the roots correctly (the root 3) then you would have only lost two I think...

-3, 0 and 3, instead of a double root at 0 i plotted one at -3

Reply 164

MHenton11

1

For question 5bii (line of sim question) like an idiot I wrote y = -3/2 instead of x =. Reckon I would still be awarded the marks?

Reply 165

Heffalump .

8

Original post by Hexaneandheels

For the graph, which I now know is probably the only error I have made, I drew the line beginning in bottom left quadrant, up through the origin and continued up and then down and just touched the x-axis at 3 and then back up again. Will I get any mark for this?

You may get one mark, as there's probably a mark for the right numbers (0 and 3) and then the other two will be based on touching the origin and going through the 3.. so you might still get a mark for that

Reply 166

username1933511

2

Qn 8.
B i) 9k^2 - 32k -16
= (9k+4)(k-4)
I) Inequality = -4/9<k<4

Posted from TSR Mobile

Reply 167

student0042

7

Original post by MHenton11

For question 5bii (line of sim question) like an idiot I wrote y = -3/2 instead of x =. Reckon I would still be awarded the marks?

It's only 1 mark, so probably not.

Reply 168

Hexaneandheels

4

Thank you! But I recall checking at the end, that graph was just worth 2 marks not three. Unless I had a total blank moment when reading that!

Original post by Heffalump .

You may get one mark, as there's probably a mark for the right numbers (0 and 3) and then the other two will be based on touching the origin and going through the 3.. so you might still get a mark for that

Reply 169

Heffalump .

8

Original post by MHenton11

For question 5bii (line of sim question) like an idiot I wrote y = -3/2 instead of x =. Reckon I would still be awarded the marks?

You won't be, it was only one mark, and I remember a past paper question that specifically stated it must be x=... not y=...

Reply 170

tanyapotter

20

if i did all the integration correctly but ended up with 163/5 for the first bit and subsequently 54-163/5 = 107/5 rather than 108/5, how many marks would i lose? absolutely gutted, this brings my whole mark down to 68 or 67/75 depending on whether or not i get penalised twice.

Reply 171

B82

1

I drew the graph touching the x axis at 3 and crossing the x at 0, why is it the other way around? And how many marks would I get for this

Also can someone explain the question about finding the values for k thanks

Reply 172

tadit

2

For the circle equation question, how were the two possible values for k found? I subbed in the coordinates of P(4,k) into the circle equation (x+1)^2+(y-3)^2=50 and got a quadratic like x^2-6x-24=0 which I had to use the quadratic equation to get something like 3 plus or minus root 33... it seems some people got 8 or -2 or something

(edited 8 years ago)

Reply 173

Heffalump .

8

Original post by Hexaneandheels

Thank you! But I recall checking at the end, that graph was just worth 2 marks not three. Unless I had a total blank moment when reading that!

Oh you're probably right then, like I said you might get one mark for it, but it's only a two mark loss if not, so it's not too bad :smile:

)

Reply 174

Hexaneandheels

4

The worst part about Core one is that, regardless of how gifted we are mathematically, just one little error CAN cost us the 100 UMS, as the 100 UMS is usually 71+. Core 2 is a little more forgiving.

Reply 175

Programming_Nerd

1

Original post by Moe21

-3, 0 and 3, instead of a double root at 0 i plotted one at -3

The equation was

(X^2)(X - 3)

So the only roots would be 0 and +3

Where X = 2 is also a stationary point and y = -4

Reply 176

Heffalump .

8

Original post by tanyapotter

if i did all the integration correctly but ended up with 163/5 for the first bit and subsequently 54-163/5 = 107/5 rather than 108/5, how many marks would i lose? absolutely gutted, this brings my whole mark down to 68 or 67/75 depending on whether or not i get penalised twice.