1. No category

FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C1
Paper D
MARKING GUIDE
This guide is intended to be as helpful as possible to teachers by providing
concise solutions and indicating how marks could be awarded. There are
obviously alternative methods that would also gain full marks.
Method marks (M) are awarded for using a valid method.
Accuracy marks (A) can only be awarded when a correct method has been used.
(B) marks are independent of method marks.
Written by Shaun Armstrong
 Solomon Press
These sheets may be copied for use solely by the purchaser’s institute.
C1 Paper D – Marking Guide
1.
x = 4 ± 16 + 32
M1
2
= 4±4 3 = 2 ± 2 3
2
2.
3.
4.
quadratic, coeff of x2 = 1, minimum is (−2, 5)
∴ y = (x + 2)2 + 5
a = 4, b = 9
= x2 + 4x + 9,
(3)
M1 A1
M1 A1
(4)
x2 − 6x + 7 = 2x − 9
x2 − 8x + 16 = 0
(x − 4)2 = 0
x = 4, y = −1
M1
M1
A2
(ii)
the line is a tangent to the curve at the point (4, −1)
B1
(i)
= (6 +
(i)
4
1
16 ) 3
3
= (6 + 2) =
(ii)
3
x
x=
(i)
8 =2
A1
=4
M1
3
4
x =
(5)
B1 M1
1
3
5.
M1 A1
M1
9
16
A1
(6)
y
y = 2x − 1
B2
y = (x − 2)2
(0, 4)
B2
O
(2, 0)
(0, −1) ( 12 , 0)
x
(ii)
x2 − 4x + 4 > 2x − 1
x2 − 6x + 5 > 0
(x − 1)(x − 5) > 0
1
M1
M1
A1
5
x < 1 or x > 5
6.
(i)
3
y
2y +
=7
M1
2
(ii)
2y + 3 = 7y
2y2 − 7y + 3 = 0
M1
A1
(2y − 1)(y − 3) = 0
y = 12 , 3
M1
A1
1
x3 =
1
2
,3
1 3
x = ( 2 ) , 33 =
C1D MARKS page 2
(7)
1
8
, 27
M1 A1
 Solomon Press
(7)
7.
(i)
dy
=
dx
(ii)
d2 y
−3
−5
= − 14 x 2 − 3x 2
2
dx
(iii)
LHS = 4x2( − 14 x
1
2
x
− 12
+ 2x
− 32
− 32
1
2
= − x − 12x
=0
8.
(i)
(ii)
M1 A2
M1 A1
− 3x
− 12
− 52
1
2
+ 2x + 8x
+ 2x
− 32
1
2
− x + 4x
1
) − ( x 2 − 4x
− 12
− 12
)
M1 A1
A1
A1
f ′′(x) = 12 − 6x
f ′′(0) = 12, f ′′(x) > 0 ∴ (0, 2) minimum
f ′′(4) = −12, f ′′(x) < 0 ∴ (4, 34) maximum
M1
A1
A1
M1
y
x
(iv)
2 < k < 34
(i)
grad =
B2
B1
7−4
9−7
∴ y−4=
=
3
2
3
2
(x − 7)
M1
2y − 8 = 3x − 21
3x − 2y − 13 = 0
A1
y = 8x
B1
(iii)
at R,
3x − 2(8x) − 13 = 0
x = −1
∴ R (−1, −8)
M1
A1
OP =
(10)
M1 A1
(ii)
7 2 + 42 =
49 + 16 =
OR = (−1)2 + (−8) 2 =
∴ OP = OR
65
1 + 64 =
M1 A1
65
A1
(10)
(i)
8 4
= − − = −2
M1 A1
(ii)
= ( 2 +2 8 ,
M1 A1
(iii)
perp. grad =
8−2
4 −8
2
) = (5, −2)
−1
−2
=
1
2
M1
perp. bisector: y + 2 =
1
2
(x − 5)
M1 A1
centre where y = 0 ∴ x = 9 ⇒ (9, 0)
(iv)
(8)
M1 A1
O
10.
− 12
− 12
f ′(x) = 12x − 3x2
for SP, 12x − 3x2 = 0
3x(4 − x) = 0
x = 0, 4
∴ (0, 2), (4, 34)
(iii)
9.
) + 4x( 12 x
49 + 16 =
radius = dist. (2, 4) to (9, 0) =
2
2
∴ (x − 9) + (y − 0) = ( 65 )
x2 − 18x + 81 + y2 = 65
x2 + y2 − 18x + 16 = 0
M1 A1
65
2
B1
M1
 Solomon Press
A1
(12)
Total
(72)
C1D MARKS page 3
Performance Record – C1 Paper D
Question no.
1
2
3
4
5
6
7
8
9
10
Topic(s)
quad.
formula
compl.
square
simul.
eqn
indices
inequal.
quad. in
function
of x
diff.
SP
straight
lines
circle
Marks
3
4
5
6
7
7
8
10
10
12
Student
C1D MARKS page 4
 Solomon Press
Total
72