Question 1:
1 | (a) | (i) | B1
B1
B1
[3] | 1.1b
1.1b
1.1b | Shape relative to axes, points on axes.
Shape relative to axes, points on axes.
Shape relative to axes, points on axes.
(ii) | Closed curve | B1
[1] | 1.2 | Bounded also allowed. | ‘Maximum’ B0
‘Continuous’ B0
(iii) | When a = 1 the curve is the points (x, y) which
satisfy
𝑥2+𝑦2+2𝑥𝑦 = 1
⇔ (𝑥+𝑦)2 = 1
⇔ 𝑥+𝑦 = 1 or 𝑥+𝑦 = −1
⇔ 𝑦 = 1−𝑥 or 𝑦 = −1−𝑥
Therefore the two straight lines are y = - x + 1
and y = -x – 1. | M1
A1
[2] | 1.1b
2.2a | Working must be shown for M1A1, otherwise
M0A0.
(b) | (i) | x 2 + y 2 + 2 a x y = 1
Setting x r c o s a n d y r s i n   = = gives
r 2 2 a r 2 s i n c o s 1   + = .
Therefore a polar form for the curve is
1
𝑟2 =
1+𝑎sin2𝜃
1
(So 𝑟 = (±)√ )
1+𝑎sin2𝜃
. | M1
M1
A1
[3] | 3.1a
1.1b
2.2a | Allow correct formula for r2. Allow formula
lwith +/- in front of the brackets. Allow
formula with 2sincosrather than s i n 2 for
A1.
Note entire curve is given for 0 2    
(taking values of θ for which 1 a s i n 2 0  + 
(ii) | When a = 2 the area is
 
1 2 1 2 1
r 2 d d    =  =
2 2 1 2 s i n ( 2 )  +
0 0
3 ( ) 3 ( )
ln 7 4 3 ln 7 4 3 − − = +
1 2 1 2 | M1
A1
[2] | 1.1a
1.1b | Need integration formula with correct limits.
Answer given is equivalent to
3 ( )
ln 2 + 3
6 . Note that exact form is
required. | √3 2−√3
ln( )
6 7−4√3
is also equivalent
to the answer.
(c) | Solving
y = m x
x 2 + y 2 + 2 a x y = 1
as a pair of simultaneous equations gives
solution pairs (x, y) of
 m 2 + + 2 a m 1 m m2 2 + 2 a m + 1 
,
m 2 + + 2 a m 1 m + 2 a m + 1
a n d
 m 2 + 2 a m + 1 m m2 2 + 2 a m + 1 
− , −
m 2 + 2 a m + 1 m + 2 a m + 1
These two points only exist if
m 2 + 2 a m + 1  0 , in which case they are
distinct.
m 2 + 2 a m + 1  0
 ( m + a 2 ) + 1 − a 2  0
 ( m + a 2 )  a 2 − 1
If −1a1 then this is true for all values of
m.
If a = 1 this is true only when m−a, so
when m−1
If a=−1this is true only when m−a, so
when m1 | M1
M1
M1
A1
B1
B1
B1 | 1.1a
2.5
3.1a
3.1a
1.1a
1.1a
1.1a | Sub mx as y to get
𝑥2+(𝑚𝑥)2+2𝑎𝑥(𝑚𝑥) = 1
or 𝑥2+𝑚2𝑥2+2𝑎𝑚𝑥2 = 1 (**) or better
Only x-coordinates required.
𝑥2(𝑚2+2𝑎𝑚+1) = 1 or better is needed for
this M1
Commenting on existence of the points (soi).
Allow use of discriminant
For finding conditions for existence of the
points (implies previous M1) | Putting
discriminant > 0
for (**) as a
quadratic in x is
another way to get
to 𝑚2+2𝑎𝑚+
1 > 0
If a < −1 or a > 1 then require that
m + a  a 2 − 1 o r m + a  − a 2 − 1
i.e m  − a + a 2 − 1 o r m  − a − a 2 − 1 | M1
A1
[9] | 3.2a
3.2a | For case descriptor and at least one inequality
(allow reasonable FT (e.g. sign errors or minor
coefficient errors) from earlier work.
For both inequalities.
Note that if this final case appears with
descriptor “a ≤ −1 or a ≥ 1” then the previous
two B1 marks are implied. | Must be strict
inequalities for
M1A1 but for
non-strict award
M1A0.