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AQA C1 Unofficial Markscheme

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Reply 40

queenleelee

Original post by smitherito

Core 1 was much easier than previous papers

I agree it was a lot easier than the 2015 paper.

Reply 41

TheGreatPumpkin

Original post by rndmtngr99

And I got root 55 for the circle question?!?

Yeah!!! I'm not alone! :biggrin:

Reply 42

AyyS

Quick question, you know question 1. Did it give us the equation of AB or did we have to create it with the c=7. Or was the y=mx+7 there to confuse us??

Reply 43

musslih

I thought the equation for the tangent at P is y=-40x-56..... since Y-24=-40(x-2), how did you guys get gradient as -32...?

Reply 44

timtjtim

Original post by BlackthornRose

Here are my answers to the questions I can remember:

Surd question: 75-32 square root 5
Circle question: equation tangent I got 7x-4y+8=0
Equation circle (x-5)^2 + (y+3)^2 = 65 (I think I forgot to add on the additional numbers from completing the square and so got this wrong, please let me know)
Coordinates B I got (12, -7)
Length CT I got 9.
Integration question: equation of tangent I got y=-32x-40
x coordinate of Q I got -5/4
Area under curve
I think I correctly got the area under the curve which was 20.25 but after that I just messed up, how many marks do you think I lost for the question? I didn't recognise the whole area of triangle and minus that from the area under the curve.
Completing the square I got (x-7/2)^2 - 41/4
There was a transformation question too and I got a translation of 0.5 and -41/4, but looking back I'm not sure if I got the signs right, I think the y value was supposed to be positive in the translation? I'm sure I got that wrong.
Something about values for k I got k=4 and k=20.
Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
Another one about finding roots of an equation and I got x= -1+/- square root of 5.
Remainder theorem I got 20.
The linear factors I got (x+3)(x-4)^2 (can't remember if the first factor was plus or minus actually, just used the one they gave)
That's all I can remember at the moment; if anyone else can shed some light on the ones I'm not sure about and tell me roughly how many marks they think I lost?

Also, what do you suppose the grade boundaries are this year?

No, that's the right circle equation. The area under the curve was 20.25, subtract 9 (area of triangle) to get 11.25. The translation seems right. (x+3)(x-4)^2 was right, but I expect the mark scheme will require (x+3)(x-4)(x-4) - as it wanted 3 linear factors.Marks lost is hard to say tbh.

Reply 45

TeeEm

Original post by TheOtherSide.

You're back! :eek3:

...guest appearance

Reply 46

timtjtim

Original post by musslih

I thought the equation for the tangent at P is y=-40x-56..... since Y-24=-40(x-2), how did you guys get gradient as -32...?

It should be been -32 - I can't remember the dy/dx tho. Then y-24 = -32(x + 2) as it was x--2.

Reply 47

BlackthornRose

Original post by timtjtim

Thanks, I just edited my post as well as i just realised it was in fact correct. Yeah I i got the area totally wrong. I wrote down both? Like, i wrote (x+3)(x-4)(x-4) or (x+3)(x-4)^2 just to make sure.

What do you suppose the grade boundaries will be like?

Reply 48

username2237201

Original post by BlackthornRose

Here are my answers to the questions I can remember:

Surd question: 75-32 square root 5
Circle question: equation tangent I got 7x-4y+8=0
Equation circle (x-5)^2 + (y+3)^2 = 65 (I think I forgot to add on the additional numbers from completing the square and so got this wrong, please let me know)
Coordinates B I got (12, -7)
Length CT I got 9.
Integration question: equation of tangent I got y=-32x-40
x coordinate of Q I got -5/4
Area under curve
I think I correctly got the area under the curve which was 20.25 but after that I just messed up, how many marks do you think I lost for the question? I didn't recognize the whole area of triangle and minus that from area under curve thing.
Completing the square I got (x-7/2)^2 - 41/4
There was a transformation question too and I got a translation of 0.5 and -41/4, but looking back I'm not sure if I got the signs right, I think the y value was supposed to be positive in the translation? I'm sure I got that wrong.
Something about values for k I got k=4 and k=20.
Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
Another one about finding roots of an equation and I got x= -1+/- square root of 5.
Remainder theorem I got 20.
The linear factors I got (x+3)(x-4)^2 (can't remember if the first factor was plus or minus actually, just used the one they gave)
That's all I can remember at the moment, if anyone else can shed some light on the ones I'm not sure about and tell me roughly how many marks they think I lost?

Also, what do you suppose the grade boundaries are this year?

Edit: I actually just realised that I'm pretty sure my equation for the circle is right, as the equation is (x-a)^2 + (y-b)^2=r^2 and i worked out the radius using coordinates for centre and point A then just plugged it into equation.

Also, there was another question that asked for the value of k at a specific point like (3+5k, 4-3k) to the best of my memory, and I got k=-30.

Anyone else?

what about the question where you had to sketch the -2x"2 graph or something and state the y-intercept, any1 know how many marks that was? 2 or 3?

(edited 7 years ago)

Reply 49

BearackObama

How did people get root 65 for the radius of the circle when the centre co-ordinates were (5,-3). On expanding this came out as root 34 surely?

Reply 50

BlackthornRose

Original post by AyyS

Quick question, you know question 1. Did it give us the equation of AB or did we have to create it with the c=7. Or was the y=mx+7 there to confuse us??

They gave the equation without the gradient and I'm sure we just had to find the gradient using the 2 coordinates they gave us and the gradient formula (m=y2-y1/x2-x1). I can't remember it properly but i think i got something like 4/7? Really can't remember.

Reply 51

BearackObama

Original post by BlackthornRose

Here are my answers to the questions I can remember:

Surd question: 75-32 square root 5
Circle question: equation tangent I got 7x-4y+8=0
Equation circle (x-5)^2 + (y+3)^2 = 65 (I think I forgot to add on the additional numbers from completing the square and so got this wrong, please let me know)
Coordinates B I got (12, -7)
Length CT I got 9.
Integration question: equation of tangent I got y=-32x-40
x coordinate of Q I got -5/4
Area under curve
I think I correctly got the area under the curve which was 20.25 but after that I just messed up, how many marks do you think I lost for the question? I didn't recognize the whole area of triangle and minus that from area under curve thing.
Completing the square I got (x-7/2)^2 - 41/4
There was a transformation question too and I got a translation of 0.5 and -41/4, but looking back I'm not sure if I got the signs right, I think the y value was supposed to be positive in the translation? I'm sure I got that wrong.
Something about values for k I got k=4 and k=20.
Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
Another one about finding roots of an equation and I got x= -1+/- square root of 5.
Remainder theorem I got 20.
The linear factors I got (x+3)(x-4)^2 (can't remember if the first factor was plus or minus actually, just used the one they gave)
That's all I can remember at the moment, if anyone else can shed some light on the ones I'm not sure about and tell me roughly how many marks they think I lost?

Also, what do you suppose the grade boundaries are this year?

Edit: I actually just realised that I'm pretty sure my equation for the circle is right, as the equation is (x-a)^2 + (y-b)^2=r^2 and i worked out the radius using coordinates for centre and point A then just plugged it into equation.

Also, there was another question that asked for the value of k at a specific point like (3+5k, 4-3k) to the best of my memory, and I got k=-30.

Anyone else?

I agree with the K=-30 but still unsure how you came to root 65 for the radius?

Reply 52

vyyzx

Was the values for k ----> (-3/2<k<6 ) or (-3/2>k or k>6)

*Brackets show the 2 different answers*

Reply 53

timtjtim

Original post by BearackObama

How did people get root 65 for the radius of the circle when the centre coordinates were (5,-3). On expanding this came out as root 34 surely?

To find the circle given C, you don't complete the square. The centre was 5,-3. You can say those are (a,b).

Now plug a,b into (x-a)^2 + (y-b)^2 = r^2

We get (x-5)^2 + (y+3)^2 = r^2.

Now for r - we know that A lies on this circle and is at coordinates A(-2,1). That gives a triangle, with sides 4 and 7. 7^2 + 4^2 = 65, so r = √65. But we want r^2, and √65^2 = 65 so k = 65.

My image may help - see attached.

circle.png23.0KB

Reply 54

Sayless

Original post by BearackObama

How did people get root 65 for the radius of the circle when the centre co-ordinates were (5,-3). On expanding this came out as root 34 surely?

A(-2,1) C(5,-3)
(x-5)^2 + (y+3)^2 = k
sub A into it
(-2-5)^2 + (1+3)^2 = k
(-7)^2 + (4)^2 = k
49 + 16 = k
k = 65

Reply 55

username2237201

Original post by BlackthornRose

Here are my answers to the questions I can remember:

Surd question: 75-32 square root 5
Circle question: equation tangent I got 7x-4y+8=0
Equation circle (x-5)^2 + (y+3)^2 = 65 (I think I forgot to add on the additional numbers from completing the square and so got this wrong, please let me know)
Coordinates B I got (12, -7)
Length CT I got 9.
Integration question: equation of tangent I got y=-32x-40
x coordinate of Q I got -5/4
Area under curve
I think I correctly got the area under the curve which was 20.25 but after that I just messed up, how many marks do you think I lost for the question? I didn't recognize the whole area of triangle and minus that from area under curve thing.
Completing the square I got (x-7/2)^2 - 41/4
There was a transformation question too and I got a translation of 0.5 and -41/4, but looking back I'm not sure if I got the signs right, I think the y value was supposed to be positive in the translation? I'm sure I got that wrong.
Something about values for k I got k=4 and k=20.
Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
Another one about finding roots of an equation and I got x= -1+/- square root of 5.
Remainder theorem I got 20.
The linear factors I got (x+3)(x-4)^2 (can't remember if the first factor was plus or minus actually, just used the one they gave)
That's all I can remember at the moment, if anyone else can shed some light on the ones I'm not sure about and tell me roughly how many marks they think I lost?

Also, what do you suppose the grade boundaries are this year?

Edit: I actually just realised that I'm pretty sure my equation for the circle is right, as the equation is (x-a)^2 + (y-b)^2=r^2 and i worked out the radius using coordinates for centre and point A then just plugged it into equation.

Also, there was another question that asked for the value of k at a specific point like (3+5k, 4-3k) to the best of my memory, and I got k=-30.

Anyone else?

Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
what question was this ?

Reply 56

tma39

Original post by Agent R5

what about the question where you had to sketch the -2x"2 graph or something and state the y-intercept, any1 know how many marks that was? 2 or 3?

it was 2 marks

Reply 57

timtjtim

Original post by vyyzx

Was the values for k ----> (-3/2<k<6 ) or (-3/2>k or k>6)

*Brackets show the 2 different answers*

The latter. See image.

ineq.png29.1KB

Reply 58

username2592579

to work out radius of circle u needed to find the "difference" in x and y directions and use Pythagoras to find the actual length of it
if you do M1 you would be very familiar with this.
think of it like this, how do you get from co ordinates (10,10) to (5,6)
well you go -5 in x direction and -4 in y direction
now think of x and y as being horizontal and vertical respectively.
what is the hypotenuse of that triangle u have? ( hypotenuse = radius of the circle)