This document provides the final mark scheme for Edexcel's Core Mathematics C1 exam from January 2012. It lists the questions, schemes for awarding marks, and total marks for each question. The six mark questions cover topics like algebra, inequalities, coordinate geometry, and calculus. The longer questions involve multi-step problems applying these concepts, including sketching curves, finding equations of tangents and normals, and solving word problems involving formulas.

1. EDEXCEL CORE MATHEMATICS C1 (6663) – JANUARY 2012 FINAL MARK SCHEME
Question Scheme Marks
1
1. (a) 4 x 3 + 3x
−
2 M1A1A1 (3)
3
x5
(b) + 4x 2 + C M1A1A1 (3)
5
6 marks
2. (a) √32 =4 √2 or √18 =3 √2 B1
( 32 + 18 = ) 7√ 2 B1 (2)
3− √ 2 −3 + 2
(b) ×
3− √ 2
or × seen M1
−3 + 2
 32 + 18 3 − 2  a 2 3 − 2
× =
( ) →
3a 2 − 2 a
(or better) M1

 3+ 2 3− 2  [ 9 − 2] [ 9 − 2]
= 3 2, −2 A1, A1 (4)
6 marks
3. (a) 5x > 20 M1
x>4 A1 (2)
(b) x2 − x −
4 12 =0
( x +2 ) ( x −6 ) [ = 0] M1
x = 6, −2
A1
x <−2
,x>6 M1, A1ft (4)
6 marks
4. (a) ( x2 = ) a+5 B1 (1)
(b) ( x3 ) =a "( a +5) "+5 M1
= a + a+
52
5 (*) A1 cso (2)
(c) 41 = a + a+
2
5 5 ⇒ + a−
a 5 2
36( =0) or 36 = 2 + a
a 5 M1
(a + 9)( a – 4) = 0 M1
a = 4 or −9
A1 (3)
6 marks
1

2. EDEXCEL CORE MATHEMATICS C1 (6663) – JANUARY 2012
y 8 FINAL MARK SCHEME
6
Question 4
Scheme Marks
1
5. (a) x (5 − x ) =
2
2
(5 x + 4) (o.e.) M1
x
−3 −2 −1
2 x 2 − x +4( =0)
5
1 2 3 4
(o.e.) e.g.
5 6
x 2 −2.5 x +2 ( =0 ) A1
−2
b 2 −4ac =( − ) −4 ×2 ×4 M1
2
5
−4
=25 −32 <0
, so no roots or no intersections or no solutions A1 (4)
(b)
Curve: ∩
shape and passing through (0, 0) B1
∩
shape and passing through (5, 0) B1
Line : +ve gradient and no intersections
with C. If no C drawn score B0 B1
Line passing through (0, 2) and
( 0.8, 0) marked on axes
−
B1 (4)
8 marks
6. (a) ( m =) 2
3 (or exact equivalent) B1 (1)
(b) B: (0, 4) [award when first seen – may be in (c)] B1
−1 3
Gradient: m
=−
2
M1
3x  3x 
y−4 = − or equiv. e.g. y =− + 4, 3x + 2 y −8 = 0  A1 (3)
2  2 
(c) A: ( −6, 0 ) [award when first seen – may be in (b)] B1
3x 8
C: 2
=4 ⇒ x=
3
[award when first seen – may be in (b)] B1ft
1
Area: Using ( xC − x A ) y B M1
2
1 8  52  1
=  + 6 4 =  = 17  A1 cso (4)
2 3  3  3
8 marks
2

3. EDEXCEL CORE MATHEMATICS C1 (6663) – JANUARY 2012 FINAL MARK SCHEME
Question Scheme Marks
3x 3
3x 2
 3 3 2 
7. [ f ( x) =] − + 5 x [ +c ] or x − x + 5 x (+c )  M1A1
3 2  2 
10 = 8 – 6 + 10 + c M1
6 y
c= −2
A1
34 5
f(1) = 1− + 5 "− 2"
22
= 2
(o.e.) A1ft (5)
x
−3 −2 −1 1 2 5 marks
−2
dy
8. (a) [ y = x 3 +2x 2
−4 ] so dx
= 3x 2 + 4 x M1A1 (2)
−6
(b)
B1
Shape
Touching x-axis at origin B1
Through and not touching or stopping at −2 B1
on x –axis. Ignore extra intersections. (3)
4 y
2
dy
(c) At x = −2:
x
−1 1 2
dx
= 3(−42) 2 + 4( −2) = 4
3 5 M1
−2
dy
At x = 0: =0 (Both values correct) A1 (2)
−4
−6
dx
−8
(d)
Horizontal translation (touches x-axis still)
M1
k −2
and k marked on positive x-axis
B1
k (2 −k )
2
(o.e) marked on negative y-axis B1
(3)
10 marks
3

4. EDEXCEL CORE MATHEMATICS C1 (6663) – JANUARY 2012 FINAL MARK SCHEME
Question Scheme Marks
10 10
9. (a) S10 = [ 2 P +9 ×2T ] or ( P +[ P +18T ]) M1
2 2
e.g. 5[2 P + T ]
18
= (£) (10P + 90T) or (£) 10P + 90T (*) A1cso (2)
10
(b) Scheme 2: S10 = [ 2( P +1800) +9T ] ={10 P +18000 +45T } M1A1
2
10P + 90T = 10P + 18000 + 45T M1
90T = 18000 + 45T
T = 400 (only) A1 (4)
(c) Scheme 2, Year 10 salary: [ a +(n − d = ( P +
1) ] 1800) +9T B1ft
P + 1800 + “3600” = 29850 M1
P = (£) 24450 A1 (3)
9 marks
1 
10. (a)  , 0
2 
B1 (1)
(b) dy
dx
= x −2 M1A1
−2
1 dy  1 
At x=
2
, = 
dx  2 
=4 (= m) A1
1  1
Gradient of normal =− = −  M1
m  4
1 1
Equation of normal: y −0 = − x −  M1
4 2
2x + 8y – 1 = 0 (*) A1cso (6)
1 1 1
(c) 2−
x
=− x+
4 8
M1
[= 2x 2 + x − =
15 8 0 ] or [ 8 y 2 − y =0
17 ]
( 2 x − )( x + ) =
1 8 0
leading to x = … M1
1 
x =   or − 8 A1
2
17
y= 8
(or exact equivalent) A1ft (4)
11 marks
4