AQA AS Further Maths Paper 1 2025

cameronholt1

2

How did everyone find AS AQA Further Maths Paper 1 2025??

Reply 1

Encryptox

3

im not doing as fm but apparently the paper was horrid

Reply 2

al2025

horrible and terrible gna cancel aqa

Reply 3

nisha101

7

Original post by cameronholt1

How did everyone find AS AQA Further Maths Paper 1 2025??

I don’t think the actual content itself was really hard the whole paper just felt so different to every past one!! Like normally there is 12 questions, this one had 3 parts to 16 questions which was crazy
Hopefully the grade boundaries make it a bit better im glad we all thought similar tho :smile:

Reply 4

nisha101

7

Original post by nisha101

I don’t think the actual content itself was really hard the whole paper just felt so different to every past one!! Like normally there is 12 questions, this one had 3 parts to 16 questions which was crazy
Hopefully the grade boundaries make it a bit better im glad we all thought similar tho :smile:

How did everyone else find it??

Reply 5

Yousafshinvari

2

Original post by nisha101

How did everyone else find it??

I think it was tough. Definitely different to past years. What did you get for the last question on complex loci?

Reply 6

Yusuf777

3

Original post by nisha101

How did everyone else find it??

*****es paper of my life counted my marks its between 48-52 and I Nedd a B
I left 10 marks blank due to time
1hr 30 for that **** AQA need to f off and sort their timing of this ****

Reply 7

RDODERS

1

paper started off quite nice. but to many weird matrix's questions. and the final questions on polar coordinates and loci in the complex plane i just bugged. I'm still thinking on how you would solve them, really weird anyone know what the circle center one was? sort of ran out of time aswell. going to have to cluch up on the discrete and statistics...

Reply 8

RDODERS

1

Original post by Yousafshinvari

I think it was tough. Definitely different to past years. What did you get for the last question on complex loci?

i dident get it but i think that the circle center was (3,0)
so you knew the circle center lied on the +ve real axes
so you could set the that distance from the origin to the circle center equal to x say
then this means for the 2+iroot 15 thing you have a right angle triangle with the radius of the circle, your root15 and 2+x
you would then use Pythagoras to find the radius
and do the same for the 7+iroot14, so you get values radius, root14 and 7-x
you then let the radius's equal each other and rearage for x
this came out as 3 which i think was the answer

should be more the 2 marks in my opinion

Reply 9

Sloth_sloth

1

what do you think the grade boundary will be like?

Reply 10

Yousafshinvari

2

Original post by RDODERS

i dident get it but i think that the circle center was (3,0)
so you knew the circle center lied on the +ve real axes
so you could set the that distance from the origin to the circle center equal to x say
then this means for the 2+iroot 15 thing you have a right angle triangle with the radius of the circle, your root15 and 2+x
you would then use Pythagoras to find the radius
and do the same for the 7+iroot14, so you get values radius, root14 and 7-x
you then let the radius's equal each other and rearage for x
this came out as 3 which i think was the answer
should be more the 2 marks in my opinion

I left that question blank unfortunately good thing it hasn’t more than 2 marks.

Reply 11

sm1929387

1

Original post by Yousafshinvari

I think it was tough. Definitely different to past years. What did you get for the last question on complex loci?

I agree the past papers were easy this year different style of questions i reckon i got between 40-45 lmao i did so bad

Reply 12

Yusuf777

3

Original post by sm1929387

I agree the past papers were easy this year different style of questions i reckon i got between 40-45 lmao i did so bad

I counted mine after my teacher had the paper I got between 49-54 hopefully a B

Reply 13

mqb2766

21

Original post by RDODERS

i dident get it but i think that the circle center was (3,0)
so you knew the circle center lied on the +ve real axes
so you could set the that distance from the origin to the circle center equal to x say
then this means for the 2+iroot 15 thing you have a right angle triangle with the radius of the circle, your root15 and 2+x
you would then use Pythagoras to find the radius
and do the same for the 7+iroot14, so you get values radius, root14 and 7-x
you then let the radius's equal each other and rearage for x
this came out as 3 which i think was the answer
should be more the 2 marks in my opinion

If the question was those two points lying on a circle, you could do it using pythagoras or you could do the perpendicular bisector of the chord which intersects with the real axis. The chord bisector is (4.5, (sqrt(15)+sqrt(14))/2) and the gradient is -(sqrt(15)-sqrt(14))/5 so the grad of the perp is 5/(sqrt(15)-sqrt(14)) so the x correction is 1/10 so 4.4 for the center.

With pythagoras for the first point it should be (x-2)? Even doing it that way, it shouldnt be more than a few lines?

(edited 1 month ago)

Reply 14

RDODERS

1

Original post by mqb2766

If the question was those two points lying on a circle, you could do it using pythagoras or you could do the perpendicular bisector of the chord which intersects with the real axis. The chord bisector is (4.5, (sqrt(15)+sqrt(14))/2) and the gradient is -(sqrt(15)-sqrt(14))/5 so the grad of the perp is 5/(sqrt(15)-sqrt(14)) so the x correction is 1/10 so 4.4 for the center.
With pythagoras for the first point it should be (x-2)? Even doing it that way, it shouldnt be more than a few lines?

i thought first one would be x+2 as the center lied on the +ve real axies and the corordiate was -2+sqrt5
so therefore the lengh would have to be x+2
but idk may be wrong.

and wouldent for you midpoint the first value be 2.5 as you dont minus the 2 values?

Reply 15

mqb2766

21

Original post by RDODERS

i thought first one would be x+2 as the center lied on the +ve real axies and the corordiate was -2+sqrt5
so therefore the lengh would have to be x+2
but idk may be wrong.
and wouldent for you midpoint the first value be 2.5 as you dont minus the 2 values?

What was the values of the two complex numbers that lay on the circumference? If the first one was -2+isqrt(5) and the other 7+isqrt(14), then 3 sounds good. The previous post had 2+isqrt(15) for the first one.

Going down the gradient route, so the equidistant locus between two points is the perpendicular bisector,
Dx = 9
Dy = sqrt(14) - sqrt(5)
and the differences are 9 (dots for Dy) so there is a +1/2 correction to the mean (7-2)/2.

(edited 1 month ago)

Reply 16

Rpro

1

Does anyone have the paper or unofficial ms to AQA As level maths paper 1?

AQA AS Further Maths Paper 1 2025

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