C1 Maths AS aqa 2016 (unofficial mark scheme new)

Thanks

Unofficial Mark scheme for C1 AQA 2016

It would help if you could link answers to questions as I can't remember them

Questions:

1)
a)Asked to work out gradient of a tangent, m= $-{5\over3}$ [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) 75-32√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 were 4 parts to this question I can't remember part a and b
b)
c) Definite integral of $\int_{-2}^{1} 4-x^2-3x^3 = [ {{-3x^4}\over{4}} - {{x^3} \over {3}} + 4x + c]$ = 81 /4 [5]
d) Area of shaded region = ${84\over4} - {24\times2-{5\over4}\over2}$ = 45/4 [3]

8) a) can't 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-coordinate of p to get 45, 45>0 therefore minimum

Updated again.

Why was it not 6>k>-3/2

How do you work out ques 1. It was find the coordinate of B ... lies on diameter???

Posted from TSR Mobile

For 6d
I used the quadratic formula, I got -1+-squareroot|80

How did you get -1+-squareroot|5 ?
Sorry for poor quality post, I'm on my **** phone.

What was the question for k=4 and k=20?

Unofficial Mark scheme for C1 AQA 2016

It would help if you could link answers to questions as I can't remember them

Questions:

1)
a)Asked to work out gradient of a tangent, m= $-{5\over3}$ [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) 75-32√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 were 4 parts to this question I can't remember part a and b
b)
c) Definite integral of $\int_{-2}^{1} 4-x^2-3x^3 = [ {{-3x^4}\over{4}} - {{x^3} \over {3}} + 4x + c]$ = 81 /4 [5]
d) Area of shaded region = ${84\over4} - {24\times2-{5\over4}\over2}$ = 45/4 [3]

8) a) can't 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-coordinate of p to get 45, 45>0 therefore minimum

Updated again.

For 6d
I used the quadratic formula, I got -1+-squareroot|80

How did you get -1+-squareroot|5 ?
Sorry for poor quality post, I'm on my **** phone.

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

(x-5)^2-25+(y+3)^2-9=0
= (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34

for Q4 d) i swear the answer is (x-2)(x^2-3x+28) + 20

because 28+20= 48 ...

You know that C is the midpoint of the line AB.

To get from A to C, you go down by 4 and across by 7. So from C to B, you do the same. C(5,-3) so B(5+7,-3-4) = B(12,-7)

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.