C1 Maths AS aqa 2016 (unofficial mark scheme new)

My brain stopped working on question 7 😠😢😠😢 because i made a mistake somewhere so i spent about 20 minutes on it(mainly staring blankly...and crying inside 😥) so I'm pretty sure it was as follows:

a) answer was y=-32x - 40

b)find x coordinate of axis intersect (of Q i think?)

c) the 5 marker integration question: 81/4

d) shaded area (I believe you used the x coordinate you found of on part b to find the area of the triangle) .: area= 45/4

The point is, i somehow got y=-32x - 26 for part a so i got part b and c wrong too. Anybody know how many marks was it for QUESTION 7 PART A, B & D?!?

Thank you 😄

Mark Scheme with some workings etc.

Lol!!! According to ur answers I prob failed this exam. I don't know how u guys managed to get x=-1+_ root5. Because I got root-5, I know that is wrong.

well you cant root a minus number so your obviously wrong so you made some amateur mistake there my g.

Love you bruv, i love people who make mark schemes, i imagine them as snails working away slowly on their computer, for hours on end.

You are mark scheme champion

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 was 75 - 32√5 [4]

3) a) y=(x-7/2)^2 - 41/4
b) min value = -41/4 [1]
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)
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]


8) a) d^2y/dx^2= - 2x - 9x^2 [2]
b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c) show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)
d) 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

No, unfortunately you are expected to give things in the simplest form.

Does anyone remember how much of Q1 would be screwed up by giving the wrong gradient for 1a? I think every other answer I gave on the paper was correct, but I'm genuinely panicking about this one. I don't remember even doing Q1, which says it all really...

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 was 75 - 32√5 [4]

3) a) y=(x-7/2)^2 - 41/4
b) min value = -41/4 [1]
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)
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]


8) a) d^2y/dx^2= - 2x - 9x^2 [2]
b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c) show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)
d) 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

a) was like 5 marks i think, (cos you had to find the gradient and the equation no?) and d) was only 3. But not to worry even if you made a mistake you should get some method marks e.g. for finding gradient etc.. and might get some follow through marks in part d) . Im sorry to hear about your issue with question 7, i think the grade boundaries will be a bit lower than usual and also a lot of people lost many marks on this question, including myself. Good luck though

As I said I know I got it wrong, but don't know how to get the answer.

I know the number of marks for each question. I also know some questions. Let me know what question you want.

Marks:
q1) 6 mark total
q2) 5 mark total
q3) 6 mark total
q4) 10 mark total
q5) 13 mark total
q6) 8 mark total
q7)14 mark total
q8) 12 mark total