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# C1 Maths AS aqa 2016 (unofficial mark scheme new) watch

1. (Original post by BlackthornRose)
Do you happen to remember the simultaneous equations? I'd hate to lose marks on the first question and it's the one I remember least as I probably gave it the least thought.
um i think it was the first equation 5x +3y+3=0
and um think the second equatuion was 3x + or - 2y or sumn +17 i think
2. I got question 5a right but I didn't cross out my working before it which was wrong. Does that mean I get no marks?
3. (Original post by GabbytheGreek_48)
um i think it was the first equation 5x +3y+3=0
and um think the second equatuion was 3x + or - 2y or sumn +17 i think
i got y=-76/19

y=-4
so x=3

(3,-4)
4. (Original post by money-for-all)
i got y=-76/19

y=-4
so x=3

(3,-4)
yeah i think that was what i got i was just saying the equations that you had to use in that question
5. Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

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6. (Original post by BlackthornRose)
Do you happen to remember the simultaneous equations? I'd hate to lose marks on the first question and it's the one I remember least as I probably gave it the least thought.
I'll try to work backwards from the answer I got but I'm not sure.
7. (Original post by Jin99)
Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

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Yes, you do.
8. (Original post by Jin99)
Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

Posted from TSR Mobile
Yes, I ended up getting 135/12=45/4
9. For question 7 I think...
(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
10. I'm pretty sure the answer for 1b was (-3,4).
11. (Original post by Applebananaaaa)
(x-5)^2-25+(y+3)^2-9=0
= (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34
No, you have to understand that you only minus the 'extra numbers' when you go from the 'multiplied out' version of the equation of the circle and are asked to complete the square to get the 'factorised/completed square' version. Here they already gave you the coordinates for centre, you simply have to work out the radius and put it all into the general equation of a circle which is (x-a)^2+(y-b)^2 = r^2 where (a,b) are the coordinates of the centre.

It is definitely 65.

(Original post by GabbytheGreek_48)
um i think it was the first equation 5x +3y+3=0
and um think the second equatuion was 3x + or - 2y or sumn +17 i think
Oh yes! Thank you very much, i didn't in fact get it right. The two simultaneous equations were 5x+3y+3=0 and 3x-2y+17=0, and I solved them to get 19x=-57 hence x=-3, then sub that into either on of equations to find y which equals to 4, hence coordinates B are (-3,4) (for anyone who wanted to know).

Once again thank you
12. (Original post by Pentaquark)
For question 7 I think...
(a) You had to find the derivative of the equation 4-x^2-3x^3 using point (-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

Then you had to find that integral.

(b) (ii)
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
Indeed.
13. In question 8 they made you calculate (d^2y) / (dx^2) in part a) i) for 2 marks i think. They then asked you in part a) ii) to verify that p (x = -3/2) was a minimum point, for which there were 4 marks available. You had to first prove dy/dx = 0 at P and then calculate (d^2y) / (dx^2) which > 0 --> minimum point at P.

In part b) you had to show that when y was increasing, dy/dx > 0 with an equation of something like 2x^2 + 4x + 8 or whatever

In part c) you had to find the values of k or something for which the final answer was k < -3/2 and k > 6
Hope this helps, good luck w/ mark scheme and hope everyone did ok
14. i got 801/36 for the integral would that be okay
15. (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

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 was75 - 32√5 [5]

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
2b. This was definitely only 4 mark, not 5
16. (Original post by Rager6amer)
In question 8 they made you calculate (d^2y) / (dx^2) in part a) i) for 2 marks i think. They then asked you in part a) ii) to verify that p (x = -3/2) was a minimum point, for which there were 4 marks available. You had to first prove dy/dx = 0 at P and then calculate (d^2y) / (dx^2) which > 0 --> minimum point at P.

In part b) you had to show that when y was decreasing, dy/dx < 0 with an equation of something like 2x^2 + 4x + 8 or whatever

In part c) you had to find the values of k or something for which the final answer was k < -3/2 and k > 6
Hope this helps, good luck w/ mark scheme and hope everyone did ok
See bold writing
17. (Original post by BlackthornRose)
No, you have to understand that you only minus the 'extra numbers' when you go from the 'multiplied out' version of the equation of the circle and are asked to complete the square to get the 'factorised/completed square' version. Here they already gave you the coordinates for centre, you simply have to work out the radius and put it all into the general equation of a circle which is (x-a)^2+(y-b)^2 = r^2 where (a,b) are the coordinates of the centre.

It is definitely 65.

Oh yes! Thank you very much, i didn't in fact get it right. The two simultaneous equations were 5x+3y+3=0 and 3x-2y+17=0, and I solved them to get 19x=-57 hence x=-3, then sub that into either on of equations to find y which equals to 4, hence coordinates B are (-3,4) (for anyone who wanted to know).

Once again thank you
yeah thats ok i did that question wrong the first time but tried to find for x instead of y first and got it
18. (Original post by Sniperdon227)
See bold writing
oh yeah you're right lol, my brain has temporarily become numb to maths/numbers in general atm :P
19. https://www.youtube.com/watch?v=JGdwNwi4G80 reaction to that paper
20. Did the question where you had to complete the square for 8-4x-2x^2 have two parts where you had to complete the square then find the roots or was it one question?

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