January 2005
6663 Core Mathematics C1 Mark Scheme Scheme
Question
Marks
number
1.
(a) 4
(or ±4) --
Bl
1
(b) 16 2
-
—j- and any attempt to find 162
Ml
162 — (or exact equivalent, e.g. 0.015625) 64
(or ± —) 64
Al
(3) 3
2.
(i)(a)15x2+7
Ml Al Al
(3)
(i) (b) 30*
Blft
(1)
3
3
(ii) x + 2x2+x~l + C
Al:x+C, A1:+2jc2, A1:+x~'
Ml Al Al Al(4) 8
3.
Attempt to use discriminant b2 - Aac (Need not be equated to zero)
Should have no jc's
Ml
(Could be within the quadratic formula)
144_4xJtx£=0 or Jl44-4xkxk = 0
Al
Attempt to solve for k
Ml
(Could be an inequality)
k=6
Al
(4)
4 4.
x2+2(2-x) = 12
or
(2-y)2+2y = l2
x2-2x-8 = 0
or
y2-2y-S = 0
(Eqn. in x or j; only)
(Correct 3 term version)
Ml Al
(Allow, e.g. x2-2x = 8) (x-4)(x + 2) = 0
x = 4,
x=...
x = -2
y = -2, y = 4
or
(y-4)(y + 2) = 0
or
y = 4, y = -2
or
x = -2, x = 4
y= ...
Ml
Al
(M: attempt one, A: both)
Ml Alft
(6) 6
Marks
Scheme
Question
number _3
(a)
5
_i
i
(1)
(ft only if terms in (a) are in arithmetic progression) Blft
(b)
2
(c)
Sum=-U{2(-3) + (n-l)(2)} or in{(-3) + (2«-5)} = -n\2n - 8} = n(n - 4)
(a)
Ml Alft
(Not just n2 - An )
2 l
6.
(2)
B1B1
Bl: One correct
Al
(*)
(3) 6
▲
(3t2)
->/^\*
Reflection in x-axis, cutting x-axis twice.
Bl
2 and 4 labelled (or (2, 0) and (4, 0) seen)
Bl
* Image ofP (3, 2)
1
\
(3)!
Bl
(b) ,
\
f~ i\ /2
7.
(a)
Stretch parallel to x-axis
Ml
1 and 2 labelled (or (1, 0) and (2, 0) seen)
Al
Image olr (1/2,
Al
/;
5"* 5 x{-5 l) (-Sx-'-l) X
X
Ay
dx" X'
X ^
X
Ml
)
|
2
Ml A1.A1
dv ^ Whenx= 1,-^ = 3
f*\ v ;
dx:
(b)
(c)
(3)
Al cso
(5)
Bl
AtP,j=8
Equation of tangent: y - 8 = 3(x -1)
(7 = 3x + 5)
Where y = 0, x = --
(or exact equiv.)
(= it)
(or equiv.)
Ml Alft
(3)
Ml Al
(2) 10
Marks
Scheme
Question number 8.
B1B1
(a)
(b)
Grad. of line ADC: m = - -,
Grad. of perp. line =
7
m
\
B1.M1
=7
5)
7
Ml Alft
Equation of /: y-2 = -(x-S)
jx-Sy-46 = 0 (c)
(a)
Ml
81 — 7
Al
4 11— 7
(or exact equiv.)
Evaluate gradient at x = 1 to get 4,
(3x-l)2=9x2-6x + Integrate:
6x2 3
Grad. of normal =
i
(
2
r
=—
(May be seen elsewhere)
(2)
B1,M1
Ml Al
(4)
Bl
Ml Alft
+ x (+C)
Substitute (1, 4) to find c= ...,
(c)
(5)
9
Equation of normal: j;-4 = --(x-l)
(b)
Al
(Allow rearrangements, e.g. 5y = Ix - 46 )
Substitute y = 7 into equation of / and find x = ... or
(2)
c= 3
Ml , Alcso
Gradient of given line is -2
Bl
Gradient of (tangent to) C is > 0 (allow >0), so can never equal -2.
Bl
(5)
(2) 11
Marks
Scheme
Question number 10.
= (x-3)2, +9
(a) (b)
(c)
x2-6x + 18 = 41
"U"-shaped parabola
Ml
Vertex in correct qua'drant
Alft
P: (0, 18) (or 18on.y-axis)
Bl
fi: (3, 9)
Blft
(x-3)2+9 = 41
Ml
Attempt to solve 3 term quadratic x= ...
Ml
x =
or
B1,M1 Al
6ÂąJ36-(4x-23) 2
Vl28 = V64 x V2
3 + 4V2
.
(or equiv.)
(or surd manipulation V2a =
(3)
Nu
1.
(4)
2.
Al
Ml Al
Qu
(5) 12