Question 1:
1 | (a) | (i) | B1
B1
B1
[3] | 1.1
1.1
1.1 | Shape including near asymptotes; position of curves
in quadrants relative to axes; points on axes clearly
indicated, asymptote (no need sketch the asymptotes
themselves).
Shape including near asymptotes; position of curves
in quadrants relative to axes; points on axes clearly
indicated, asymptote (no need to sketch the
asymptotes themselves).
Shape including near asymptotes; position relative to
axes, quadrants; passes through origin, asymptote (no
need to sketch the asymptotes themselves). Position
of maximum.
(a) | (ii) | The curves all have a vertical asymptote, the
line x = −1. | B1
[1] | 1.2 | Any reasonable common feature acceptable.
Other acceptable responses include unbounded and
pass through the origin.
(a) | (iii) | The curve when a = 0 has stationary points
whereas there are none in the case of a = −2 and
a = −1. | B1
[1] | 1.2 | Any reasonable distinguishing feature acceptable. For
example, the curve crossing the 𝑥-axis for a = −2 and
a = −1 but is tangent to 𝑥-axis when a = 0.
(b) | (i) | x2 (a+1)x2 +ax
y =ax+ =
1+x 1+x
(1+x)((a+1)x−1)+1 1
= =(a+1)x−1+
1+x 1+x
1
As x →   , → 0 .
x + 1
Therefore, the oblique asymptote is
y = ( a + 1 ) x − 1 . | M1
M1
B1
[3] | 1.1a
1.1
1.1 | This can be implied by the correct answer. To be
implied the answer must include a statement that the
final equation is the oblique asymptote.
If “oblique asymptote” statement is missing, must see
at least one correct limit taken e.g. 𝑥 → ∞.
(b) | (ii) | The line between the two points is
a b 2 + b 2 − a b2 2
y = x − .
b 2 − 1 b − 1
The point where this crosses the y-axis is
 b2 
0,− 
b2−1

This is independent of a. | M1
M1
A1
[3] | 1.1a
1.1
2.2a | Line can be obtained using CAS. Can formulate with
𝑥 = 0 provided working shows reasoning is due to
finding the point on 𝑦-axis.
Accept consideration of the midpoint of the line
between the two points.
Accept statement of the 𝑦-coordinate of point.
(b) | (iii) | x 2 x ( ( a + 1 ) x + a )
y = a x + = (*)
1 + x 1 + x
If a  0 and x  0 then ( a + 1 ) x + a  0 .
If a  − 1 and x  0 then ( a + 1 ) x + a  0 .
If − 1  a  0 then
− a
( a + 1 ) x + a  0 when 0  x 
1 + a
− a
and ( a + 1 ) x + a  0 when x 
1 + a
Therefore by considering (*), y  0 for all 𝑥 ≥
0 whenever a  0 and y  0 for all 𝑥 ≥
0 whenever a  − 1 and neither of these
statements is true for −1 < 𝑎 < 0.
Therefore
• y  0 for all x  0 precisely when a  0 .
• y  0 for all x  0 precisely when a  − 1 | M1
M1
M1
A1
B1
[5] | 1.1a
2.1
2.1
2.2a | Must see 𝑥 factored out in preparation for sign
argument. Can be implied by subsequent sign
argument working.
Alternative method
x 2 x ( ( a + 1 ) x + a )
y = a x + =
1 + x 1 + x | M1 | M1 | 1.1a | 1.1a
−𝑎
Roots are 𝑥 = 0 and 𝑥 = .
𝑎+1
If 𝑎 ≥ 0, there are no positive roots. | M1 | 2.1 | Condone use of weak inequalities.
−𝑎
If 𝑎 ≤ −1, < 0 so there are no positive
𝑎+1
roots.
−𝑎
> 0 if and only if −1 < 𝑎 < 0.
𝑎+1 | −𝑎
> 0 if and only if −1 < 𝑎 < 0.
𝑎+1 | M1 | 2.1 | Condone lack of precise reasoning with
biconditional/if and only if statement.
Condone use of weak inequalities
𝑑𝑦 2𝑥(𝑥+1)−𝑥2
= 𝑎+
𝑑𝑥 (𝑥+1)2 | A1 | 2.2a | 𝑥(𝑥+2)
= 𝑎+ .
(𝑥+1)2 | 𝑥(𝑥+2)
= 𝑎+ .
(𝑥+1)2
𝑑𝑦
At 𝑥 = 0, = 𝑎.
𝑑𝑥
𝑑𝑦 𝑑𝑦
If 𝑎 > 0, = 𝑎 > 0. If 𝑎 ≤ −1, = 𝑎 < 0.
𝑑𝑥 𝑑𝑥
𝑑𝑦
At 𝑥 = 0, 𝑦 = 0 and if 𝑎 > 0, = 𝑎 > 0, so
𝑑𝑥 | B1 | 2.2a
y  0 for all x  0 precisely when a  0
𝑑𝑦
At 𝑥 = 0, 𝑦 = 0 and if 𝑎 ≤ −1, = 𝑎 < 0, so
𝑑𝑥
y  0 for all x  0 precisely when a  − 1
[5]
2.1
(c) | By plotting the curve it can be seen that the
cusp is at the origin. The existence of a cusp at
the origin is fully justified as follows.
1 3
 x2  4 d y x ( x + 2 )2  x + 1  4
If y=  then = .
1+x d x 4 ( x + 1 ) 2 x
dy d y
By CAS, lim =− and li→ m =  .
x→0− dx x 0 + d x
Hence the gradient tends to being infinitely
steep (negative on the left and positive on the
right) as x tends to zero from above and below.
When x = 0, y = 0 and so (0,0) is on the curve.
The curve has a cusp at (0,0). | M1
M1
M1
M1
B1
[5] | 1.1a
1.1
2.1
1.1a
2.2a | Must clearly consider the consequence of limits being
±∞ from above and below in geometric terms. For
example, the tangent(s) is vertical.