*UNOFFICIAL MARK SCHEME* Edexcel Maths C1 2017

1) +C
2)a=1/16
3) a2 = 2k
a3 = k+1/2
k=17/6
4) d=6
10316 total
5) x=3 and x=-1
6) (x-4)^2+3
Minimum point at (4,3) intercept at (0,19)
PQ : 4sqrt(17) units
7)
8) 5x-4y-49=0
Area = 369/10
9) line theough (0,C)
Asymptote at y= 5
equate and disciminant greater than 0
c<5-2sqrt(3) and c> 5+sqrt(3)
10) k = 75
c = 5/2
just differentiate
x = -5/3

(some questions might be missing; reply with edits)

Hey man,
For queston 1 I remember getting x^6/3 + 1/8x^2 -5x + C
Also, do you know how many marks Q2 and Q3 were worth?

The easiest C1 exam in the last 3 years?
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Nah K was 17/2. a3 2k^2 + 0.5 or whatever the thread starter put. k/2k is 1/2 not k

S was (-5/2,0) and T was (49/5,0). Y intercept was (0,6).

So area = 1/2*6*(49/5--5/2) = 369/10

1) +C
2)a=1/16
3) a2 = 2k
a3 = k+1/2
k=17/6
4) d=6
10316 total
5) x=3 and x=-1
6) (x-4)^2+3
Minimum point at (4,3) intercept at (0,19)
PQ : 4sqrt(17) units
7)
8) 5x-4y-49=0
Area = 369/10
9) line theough (0,C)
Asymptote at y= 5
equate and disciminant greater than 0
c<5-2sqrt(3) and c> 5+sqrt(3)
10) k = 75
c = 5/2
just differentiate
x = -5/3

(some questions might be missing; reply with edits)

For Q9,
the quadratic (5-c)^2 - 12 is a U shape, which means that for > 0, you are looking at the leftmost and rightmost sides (not the inner bits).
Also, bear in mind that c is POSITIVE (this means that there is a second bound, c >= 0).
This gives us final bounds as
0 <= c < 5-2sqrt(3), and
c < 5+2sqrt(3)


(could be wrong here, but since in the question it mentions that c is indeed a positive constant, it should be bound by c >= 0)

Is it? Sorry if it is but I just didn't see where the 6 is coming from? Where did you get 6?

The terms given and found were:

a = the value I think you said it was 140?

difference was 11? as they wanted 11 more bikes sold per week.

the n values were 12 and 40 (52-12 for the latter half)

plugging those into n/2(2a+(n-1)d) gave that calculation mentioned? 10406?

Where did I and in that case, many others go wrong?

Ur wet smh

I do road fact

Question 4b is wrong

140 bikes on the first week, increases output of bikes by 6 a week

a=140
d=6
n= 52

s(n) = n/2(2a + (n-1)d)

s(52) = 26*(280 + 306)

s(52) = 15236

D = 6
As week 12 is 206
So 206 = 140 + (n-1)d
206=140 + 11d
66 = 11d
D = 6
By the looks of it the rest of your working is fine, so you'll probably get method marks on part b)