C1 Maths AS aqa 2016 (unofficial mark scheme new)

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)

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.

Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

Posted from TSR Mobile

Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

Posted from TSR Mobile

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

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

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

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 [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

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

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

See bold writing

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)