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C1 Maths AS aqa 2016 (unofficial mark scheme new)

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Reply 200
Oh thank god, i found the area of triangle to be 9 too. And the area of the shaded region was 29-9=20, is that right?

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Original post by Jin99
Oh thank god, i found the area of triangle to be 9 too. And the area of the shaded region was 29-9=20, is that right?

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I didn't get the area of triangle, but area under curve was 20.25, so area shaded region was 20.25-9=11.25 from what I've seen on here.
Reply 202
For question 3 c) Translation of (1/2 , 41/4), I got -1/2 , could someone explain why I'm wrong please, thanks :biggrin:
Original post by Igzzy__
For question 3 c) Translation of (1/2 , 41/4), I got -1/2 , could someone explain why I'm wrong please, thanks :biggrin:

These are the two lines I think...
y=(x-7/2)^2 - 41/4 ---> y=(x-4)^2
So the vertex of the first is (7/2, -41/4)
So to move from that to the vertex of the other curve (4,0)
We need to move from 3.5 -> 4 therefore a translation of 0.5 in the x-direction
and upwards of 41/4 from -41/4.
Reply 204
Original post by Pentaquark
These are the two lines I think...
y=(x-7/2)^2 - 41/4 ---> y=(x-4)^2
So the vertex of the first is (7/2, -41/4)
So to move from that to the vertex of the other curve (4,0)
We need to move from 3.5 -> 4 therefore a translation of 0.5 in the x-direction
and upwards of 41/4 from -41/4.


yh so if you move from 3.5 -> 4 won't it be -0.5 because you're moving to the right in the x-direction and because its the x value you have to give it in negative
For the complete the square questions i used decimals, would i still get full marks?
So here is a mark scheme which has the equations any everything. Anyone can edit it - so if you wanna colaberate on it just jump in and add the marks you know. We're missing 7. a) and b)

https://docs.google.com/document/d/1iNmYSXMwLW0RID8A4Yk__VQINLjQZuPljHsUdbsfoj4/edit?usp=sharing
Original post by timtjtim
Okay, I've added a little bit.

1) a)Asked to work out gradient of a parallel line, m= - 3/2
b) Asked to find co-ordinates of B, B( - 3,4)
c) Asked to find K, K= - 30

2) a) simplify (3√5)^2 = 45
b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5 = 75 - 325

3) a) y=(x-7/2)^2 - 41/4
b) Translation of (1/2 , 41/4)

4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48. p(-3) = -3^3 - 5 * 3-^2 - 8 * -3 + 48 = 0. As p(-3) = 0, (x+3) is a factor of p(x).

b) Three linear factors goes to (x+3)(x-4)(x-4)
c) find the remainder when x^3 - 5x^2 -8x + 48 was divided by (x+2) R=20
d) factorise p(x) using your answer to part c) as R = (x-2)(x^2-3x+14) + 20

5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2)
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7)
c) asked to find the equation of the tangent at A 7x-4y+18=0
d) asked to find the length of CT = 9

6) a)

7) a) I think there were 4 parts to this question I can't remember part a and b
b)
c) Definite integral (from 1 to -2) = 81/4 (20.25)
d) Area of shaded region (integral - triangle area) = 45/4 (11.25)

8) a) can't remember anything except the answers were k>6 and k<-3/2 (NOT 6 > k > -3/2 )

Answers that need a question to be assigned to
- Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
- k=4 and k=20
- x = -1±√5
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum


Why did you change 1a? and how many marks was each question worth?
Original post by Rit101
Nope pretty sure the top students in my Sixth Form got root65 as the radius


I got the radius as 7? Then the distance between C and T as root65?
Original post by Igzzy__
yh so if you move from 3.5 -> 4 won't it be -0.5 because you're moving to the right in the x-direction and because its the x value you have to give it in negative

If you plot the vertex of equation 1 and the vertex of equation 2.
It is simply a translation of 1/2 across and 41/4 up.
I can't see why it would be -0.5 since that means it will move to the left in the negative direction?

e.g. moving y=x^2 across by +4 units will give the equation y=(x-4)^2
This is a translation of [4] across. Not -4.
Its describing a movement of one graph to another.

The new equation of (x-7/2)-41/4 moving to (x-4)^2
will be (x-3.5-0.5)-41/4 + 41/2 => (x-4)^2
If it moves +0.5 then inside the bracket it will be -0.5
Original post by timtjtim
So here is a mark scheme which has the equations any everything. Anyone can edit it - so if you wanna colaberate on it just jump in and add the marks you know. We're missing 7. a) and b)

https://docs.google.com/document/d/1iNmYSXMwLW0RID8A4Yk__VQINLjQZuPljHsUdbsfoj4/edit?usp=sharing


Due to spam, we are locking the document. It is Viewable only (commenting off).

https://docs.google.com/document/d/10VHknriHhsTcJxHsn2jf2TUFBX9Yhy_dEB3r6SToYcc/edit?usp=sharing

If you wish to "chat" with each other and me, go to https://www.yourworldoftext.com/aqac12016ms
(edited 7 years ago)
sorry could someone post the most updated mark scheme so i can update it? im kind of busy to search around :P
thanks
Original post by beanigger
Unoffical Mark scheme for C1 AQA 2016

It would help if you could link answers to questions as i cant remember them :smile:

Questions:

1) a)Asked to work out gradient of a tangent, m= - 5/3 [2]
b)Asked to find co-ordinates of B B( - 3,4) [3]
c)Asked to find K K= - 30 [2]

2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was 75 - 32√5 [4]

3) a) y=(x-7/2)^2 - 41/4
b) min value = -41/4
c) Translation of (1/2 , 41/4)

4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48, p(-3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x-4)(x-4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x-2)(x^2-3x-14) + 20 [3]

5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7) [2]
c) asked to find equation of tangent at A 7x-4y+18=0 [5]
d) asked to find length of CT = 9 [2]

6) a) y=-32x-40
b)Q(-5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = -1±√5

7) a) i think there was 4 parts to this question i cant remember part a and b
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]


8) a) cant remember anything except the answers were k>6 and k<-3/2

Answers that need a question to be assigned to
- k=4 and k=20
- d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum


what was the working for the last question on 1
http://prntscr.com/b5o7n1

Do you need to put "+c" for the integration? I was taught you don't have to if you find the integral between x and y. :'(
Calculating coordinate Q was in Q7 I am sure because you had to use that to find the area of the triangle.
I dont understand why it would be in 6 (a) & 6 (b) ?

(a) You had to find the derivative of the equation 4-x^2-3x^3 using point P(-2,24)
dy/dx= (-2x-9x^2) so subbing in -2 gives -32.
y-24=-32(x+2)
y=-32x-64+24
y=-32x-40

(b) then you'd use that to find coordinate Q by subbing in y=0
into y=-32x-40
0=-32x-40 to get
x=-5/4

(c)? Then you had to find that integral.

(d)?
To find the area of the triangle you use Coordinate Q -5/4
You had to subtract -5/4 from the limit (-2)
So -5/4 - - 2 = 3/4
Base of the triangle is then 3/4

So the area of the triangle is 0.5 * 0.75 * 24 = 9
(edited 7 years ago)
Original post by Pentaquark
Calculating coordinate Q was in Q7 I am sure because you had to use that to find the area of the triangle.
I dont understand why it would be in 6 (a) & 6 (b) ?

(a) You had to find the derivative of the equation 4-x^2-3x^3 using point P(-2,24)
dy/dx= (-2x-9x^2) so subbing in -2 gives -32.
y-24=-32(x+2)
y=-32x-64+24
y=-32x-40

(b) then you'd use that to find coordinate Q by subbing in y=0
into y=-32x-40
0=-32x-40 to get
x=-5/4

(c)? Then you had to find that integral.

(d)?
To find the area of the triangle you use Coordinate Q -5/4
You had to subtract -5/4 from the limit (-2)
So -5/4 - - 2 = 3/4
Base of the triangle is then 3/4

So the area of the triangle is 0.5 * 0.75 * 24 = 9


the part c and d you are talking about was definitely question 7 but not part a and b
i remember that because there was so much god dam space for question 7 and to work out x=-5/4 i ran out of space to write that and had to get extra paper so there is no way that was in question 7
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It definitely said 3 linear factors. But one was a repeated root. So I guess as long as you put (x-4)^2 I guess you'll get the mark
Original post by Jaspal3131
what was the working for the last question on 1


im not sure :tongue:
Anyone got a hard copy?
Original post by beanigger
the part c and d you are talking about was definitely question 7 but not part a and b
i remember that because there was so much god dam space for question 7 and to work out x=-5/4 i ran out of space to write that and had to get extra paper so there is no way that was in question 7

Ahh ok fair enough :smile:
I wish a teacher would put up model solutions
I'm so impatient

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