Question 6:
--- 6(a) ---
6(a) | Selects a method leading to any
calculation pertaining to one of the
following methods seen (not
necessarily correct); gradients of
sides, lengths of sides or
intersection or lengths of diagonals | AO3.1a | M1 | Grad BC = -5/2 = Grad DA
Grad AB = 2/5 = Grad DC
Both pairs of opposite sides have
equal gradient so parallel, so
ABCD is a parallelogram
Grad BC × grad AB = -1
ABC = 90o therefore all angles in
ABCD are 90 o so ABCD is a
rectangle
Finds gradients of all 4 sides or
lengths of all 4 sides or midpoints
of both diagonals correctly | AO1.1b | A1
Proves one angle is 90° by using
gradients or Pythagoras | AO1.1a | M1
Completes proof that ABCD is a
rectangle. There must be a clear
statement that there are 2 pairs of
parallel sides and all angles are 90◦ | AO2.1 | R1
Note, there are various ways of proving that ABCD is a rectangle
(1 – 5 below score M1 A1 M1 before final required statement for relevant R1 stating how their
method used proves a rectangle)
1. As in the typical solution shown: show that both pairs of opposite sides are parallel, show that
one angle is 90°.
2. Show that each pair of opposite sides is equal in length, show that one angle is 90°.
3. Show that one pair of opposite sides is parallel and equal in length, show that one angle is
90°.
4. Show that the diagonals bisect (the midpoint of one is also the midpoint of the other) and are
equal in length.
5. Show that each pair of opposite sides are parallel and length of the two diagonals are the
same
NB May be expressed using vectors
NB Diagonals AC and BD = √377
(b) | Finds correct lengths of two
adjacent sides (accept to at least
1dp accuracy) | AO1.1a | M1 | AB ( = DC ) = √261 = 3√29
BC ( = DA ) = √116 = 2√29
Area = 174
Obtains correct area (AWRT) | AO1.1b | A1
Total | 6
Q 7 | Marking Instructions | AO | Marks | Typical solution
(a) | Takes out a factor of 2, or obtains
a = 2 by equating coefficients.
Note 2(x2 – ‘anything’)
or 2(x – ‘anything’)2 scores M1 | AO1.1a | M1 | 2(𝑥2− 5𝑥 ) + k
2
5 2 25
2(𝑥− ) +𝑘−
4 8
2
5
Expresses as (𝑥− ) or obtains
4
5
b = by equating coefficients
4 | AO1.1a | M1
Obtains correct expression, If
using equating coefficients must
be put back in given form required | AO1.1b | A1
(b) | Selects a method using completed
square form (recognition that
vertex occurs when x = ‘their’ b),
discriminant (any use of b2 – 4ac
seen) or calculus (finds y
coordinate of stationary point). | AO3.1a | M1 | Typical solution 1
When x= 5 (x - 5 )2 = 0
4 4
25
(𝑘− )> 3
8
49
𝑘 >
8
Forms an appropriate correct
inequality for their chosen method.
(first time inequality sign seen in
each typical solution scores A1) | AO1.1b | A1
49
Obtains 𝑘 > (ACF)(OE)
8 | AO1.1b | A1
Typical solution 2
2x2 – 5x + k = 3
2x2 – 5x + k – 3 = 0
52 – 4 × 2 × (k – 3) < 0
49
𝑘 >
8
Typical solution 3
𝑑𝑦
= 4x – 5
𝑑𝑥
𝑑𝑦
At stationary point = 0
𝑑𝑥
4x – 5 = 0
5
x =
4
5 25
when x = y = k -
4 8
25
k - > 3
8
49
𝑘 >
8
Total | 6
Q 8 | Marking Instructions | AO | Marks | Typical Solution
(a) | Produces a combined diagram
showing circles intersecting at
origin and (0, 10) or two separate
diagrams. Allow reasonable ‘hand
drawn’ circles which illustrate
symmetry. Circles must cut the x
axis again. Do not accept circles
that go off the page. | AO2.2a | B1
(b) | Deduces that y coordinate of centre
is 5. (PI by any use of (y – 5) in any
circle equation or marked on
diagram or seen as a y coordinate
or used in Pythagoras) | AO2.2a | B1 | 62 = 52 + a2
a = +√11 or –√11
( x – √11)2 + (y – 5)2 = 62 = 36
( x + √11)2 + (y – 5)2 = 62 = 36
Or
x2 + 2√11x + y2 – 10y = 0
x2 – 2√11x + y2 – 10y = 0
Forms correct equation for x
coordinate of centres using
Pythagoras (PI) | AO1.1a | M1
Obtains two correct circle
equations (either form)
Condone 3.3 or better provided
𝑎 = √11 seen earlier | AO1.1b | A1
Total | 4
Q | Marking Instructions | AO | Marks | Typical Solution