Sixth Term Examination Paper [STEP]

Mathematics 3 [9475]

2020

Mark Scheme
1. (i) Integrating by parts,

𝑢𝑢 = cos 𝑎𝑎 𝑥𝑥 𝑣𝑣 ′ = cos 𝑏𝑏𝑏𝑏

1
𝑢𝑢′ = −𝑎𝑎 cos 𝑎𝑎−1 𝑥𝑥 sin 𝑥𝑥 𝑣𝑣 = sin 𝑏𝑏𝑏𝑏 M1
𝑏𝑏

𝜋𝜋 𝜋𝜋
1 2 1
𝐼𝐼 (𝑎𝑎, 𝑏𝑏) = �cos 𝑎𝑎 𝑥𝑥 sin 𝑏𝑏𝑏𝑏� − ∫02 −𝑎𝑎 cos 𝑎𝑎−1 𝑥𝑥 sin 𝑥𝑥 sin 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑 M1 A1ft
𝑏𝑏 0 𝑏𝑏
𝜋𝜋 𝜋𝜋
2 2
1 𝑎𝑎
= 0 + � 𝑎𝑎 cos 𝑎𝑎−1 𝑥𝑥 sin 𝑥𝑥 sin 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑 = � cos 𝑎𝑎−1 𝑥𝑥 sin 𝑥𝑥 sin 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑
𝑏𝑏 𝑏𝑏
0 0
M1

cos(𝑏𝑏 − 1)𝑥𝑥 = cos 𝑏𝑏𝑏𝑏 cos 𝑥𝑥 + sin 𝑏𝑏𝑏𝑏 sin 𝑥𝑥 M1 A1

𝜋𝜋
𝑎𝑎
So 𝐼𝐼 (𝑎𝑎, 𝑏𝑏) =
𝑏𝑏
∫02 cos 𝑎𝑎−1 𝑥𝑥 (cos(𝑏𝑏 − 1)𝑥𝑥 − cos 𝑏𝑏𝑏𝑏 cos 𝑥𝑥 ) 𝑑𝑑𝑑𝑑
𝑎𝑎
= [𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 − 1) − 𝐼𝐼 (𝑎𝑎, 𝑏𝑏)] M1
𝑏𝑏

𝑎𝑎
Thus 𝐼𝐼 (𝑎𝑎, 𝑏𝑏) = 𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 − 1) as required. M1 A1* (9)
𝑎𝑎+𝑏𝑏

(ii) Suppose
2𝑘𝑘 𝑘𝑘! (2𝑚𝑚)! (𝑘𝑘 + 𝑚𝑚)!
𝐼𝐼 (𝑘𝑘, 𝑘𝑘 + 2𝑚𝑚 + 1) = (−1)𝑚𝑚
𝑚𝑚! (2𝑘𝑘 + 2𝑚𝑚 + 1)!
E1
Then by (i),
𝑘𝑘 + 1
𝐼𝐼 (𝑘𝑘 + 1, 𝑘𝑘 + 2𝑚𝑚 + 2) = 𝐼𝐼 (𝑘𝑘, 𝑘𝑘 + 2𝑚𝑚 + 1)
2𝑘𝑘 + 2𝑚𝑚 + 3
M1

𝑘𝑘 + 1 2𝑘𝑘 𝑘𝑘! (2𝑚𝑚)! (𝑘𝑘 + 𝑚𝑚)!

= (−1)𝑚𝑚
2𝑘𝑘 + 2𝑚𝑚 + 3 𝑚𝑚! (2𝑘𝑘 + 2𝑚𝑚 + 1)!
M1 A1

𝑘𝑘 + 1 2𝑘𝑘 𝑘𝑘! (2𝑚𝑚)! (𝑘𝑘 + 𝑚𝑚)! 2𝑘𝑘 + 2𝑚𝑚 + 2

= (−1)𝑚𝑚 ×
2𝑘𝑘 + 2𝑚𝑚 + 3 𝑚𝑚! (2𝑘𝑘 + 2𝑚𝑚 + 1)! 2𝑘𝑘 + 2𝑚𝑚 + 2
M1

2𝑘𝑘+1 (𝑘𝑘 + 1)! (2𝑚𝑚)! (𝑘𝑘 + 𝑚𝑚 + 1)!

= (−1)𝑚𝑚
𝑚𝑚! (2𝑘𝑘 + 2𝑚𝑚 + 3)!

2𝑘𝑘+1 (𝑘𝑘 + 1)! (2𝑚𝑚)! �(𝑘𝑘 + 1) + 𝑚𝑚�!

= (−1)𝑚𝑚
𝑚𝑚! (2(𝑘𝑘 + 1) + 2𝑚𝑚 + 1)!

which is the required result for 𝑘𝑘 + 1 A1

 𝜋𝜋
2
𝜋𝜋
1
𝐼𝐼 (0, 2𝑚𝑚 + 1) = � cos(2𝑚𝑚 + 1)𝑥𝑥 𝑑𝑑𝑑𝑑 = [sin(2𝑚𝑚 + 1)𝑥𝑥 ]02
2𝑚𝑚 + 1
0
M1
1 −1 1
= if m is even or = if m is odd , or alternatively (−1)𝑚𝑚 A1
2𝑚𝑚+1 2𝑚𝑚+1 2𝑚𝑚+1

If 𝑛𝑛 = 0 ,
2𝑛𝑛 𝑛𝑛! (2𝑚𝑚)! (𝑛𝑛 + 𝑚𝑚)! (2𝑚𝑚)! (𝑚𝑚)! 1
(−1)𝑚𝑚 = (−1)𝑚𝑚 = (−1)𝑚𝑚
𝑚𝑚! (2𝑛𝑛 + 2𝑚𝑚 + 1)! 𝑚𝑚! (2𝑚𝑚 + 1)! 2𝑚𝑚 + 1
M1 A1

so result is true for 𝑛𝑛 = 0 .

So by the principle of mathematical induction, the required result is true. E1 (11)

Alternative for (i)

cos 𝑎𝑎 𝑥𝑥 cos 𝑏𝑏𝑏𝑏 = cos 𝑎𝑎−1 𝑥𝑥 [cos 𝑥𝑥 cos 𝑏𝑏𝑏𝑏 ]
1
cos 𝑥𝑥 cos 𝑏𝑏𝑏𝑏 = [cos(𝑏𝑏 + 1)𝑥𝑥 + cos(𝑏𝑏 − 1)𝑥𝑥 ]
2
2𝐼𝐼 (𝑎𝑎, 𝑏𝑏) = 𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 + 1) + 𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 − 1)
Also
1
[cos(𝑏𝑏 − 1)𝑥𝑥 − cos(𝑏𝑏 + 1)𝑥𝑥 ]
sin 𝑥𝑥 sin 𝑏𝑏𝑏𝑏 =
2
so using integration by parts of main scheme,

𝑎𝑎
[𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 − 1) − 𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 + 1)]
2𝐼𝐼 (𝑎𝑎, 𝑏𝑏) =
𝑏𝑏
Eliminating 𝐼𝐼 (𝑎𝑎 − 1, 𝑏𝑏 + 1) between these results gives required result.
2. (i) sinh 𝑥𝑥 + sinh 𝑦𝑦 = 2𝑘𝑘
𝑑𝑑𝑑𝑑
Differentiating with respect to x, cosh 𝑥𝑥 + cosh 𝑦𝑦 =0 M1
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 0 ⇒ cosh 𝑥𝑥 = 0 which is not possible as cosh 𝑥𝑥 ≥ 1 ∀𝑥𝑥 , so there are no stationary points.
𝑑𝑑𝑑𝑑
E1 (2)

𝑑𝑑𝑑𝑑 2 𝑑𝑑 2 𝑦𝑦
Differentiating again with respect to x, sinh 𝑥𝑥 + sinh 𝑦𝑦 � � + cosh 𝑦𝑦 =0 M1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 2

𝑑𝑑𝑑𝑑 −cosh 𝑥𝑥 𝑑𝑑 2 𝑦𝑦 −cosh 𝑥𝑥 2

= and = 0 implies sinh 𝑥𝑥 + sinh 𝑦𝑦 � � =0
𝑑𝑑𝑑𝑑 cosh 𝑦𝑦 𝑑𝑑𝑥𝑥 2 cosh 𝑦𝑦

cosh2 𝑦𝑦 sinh 𝑥𝑥 + cosh2 𝑥𝑥 sinh 𝑦𝑦 = 0

(1 + sinh2 𝑦𝑦) sinh 𝑥𝑥 + (1 + sinh2 𝑥𝑥 ) sinh 𝑦𝑦 = 0

(sinh 𝑥𝑥 + sinh 𝑦𝑦)(1 + sinh 𝑥𝑥 sinh 𝑦𝑦) = 0

M1 A1
But sinh 𝑥𝑥 + sinh 𝑦𝑦 = 2𝑘𝑘 > 0 so

1 + sinh 𝑥𝑥 sinh 𝑦𝑦 = 0
as required. E1 (4)

𝑑𝑑 2 𝑦𝑦
At a point of inflection, = 0 , so sinh 𝑥𝑥 + sinh 𝑦𝑦 = 2𝑘𝑘 and sinh 𝑥𝑥 sinh 𝑦𝑦 = −1 and thus,
𝑑𝑑𝑥𝑥 2
sinh 𝑥𝑥 (and sinh 𝑦𝑦 as well) is a root of 𝜆𝜆2 − 2𝑘𝑘𝑘𝑘 − 1 = 0 M1
2𝑘𝑘 ± √4𝑘𝑘 2 + 4
𝜆𝜆 =
2
−1 −1 𝑘𝑘−√𝑘𝑘 2 +1 −�𝑘𝑘−√𝑘𝑘 2 +1�
sinh 𝑥𝑥 = 𝑘𝑘 + √𝑘𝑘 2 + 1 , sinh 𝑦𝑦 = = × = = 𝑘𝑘 − √𝑘𝑘 2 + 1
𝑘𝑘+√𝑘𝑘 2 +1 𝑘𝑘+√𝑘𝑘 2 +1 𝑘𝑘−√𝑘𝑘 2 +1 𝑘𝑘 2 −(𝑘𝑘 2 +1)
and vice versa.

So the points of inflection are

�sinh−1 �𝑘𝑘 + √𝑘𝑘 2 + 1� , sinh−1 �𝑘𝑘 − √𝑘𝑘 2 + 1�� and A1

−1 −1
�sinh �𝑘𝑘 − √𝑘𝑘 2 + 1� , sinh �𝑘𝑘 + √𝑘𝑘 2 + 1�� A1 (3)

(ii) 𝑥𝑥 + 𝑦𝑦 = 𝑎𝑎 ⇒ 𝑦𝑦 = 𝑎𝑎 − 𝑥𝑥 so as sinh 𝑥𝑥 + sinh 𝑦𝑦 = 2𝑘𝑘

𝑒𝑒 𝑥𝑥 − 𝑒𝑒 −𝑥𝑥 𝑒𝑒 𝑎𝑎−𝑥𝑥 − 𝑒𝑒 𝑥𝑥−𝑎𝑎

+ = 2𝑘𝑘
2 2
M1
Multiplying by 2𝑒𝑒 𝑥𝑥 ,

𝑒𝑒 2𝑥𝑥 − 1 + 𝑒𝑒 𝑎𝑎 − 𝑒𝑒 2𝑥𝑥 𝑒𝑒 −𝑎𝑎 = 4𝑘𝑘𝑒𝑒 𝑥𝑥

𝑒𝑒 2𝑥𝑥 (1 − 𝑒𝑒 −𝑎𝑎 ) − 4𝑘𝑘𝑒𝑒 𝑥𝑥 + (𝑒𝑒 𝑎𝑎 − 1) = 0

A1*
As 𝑒𝑒 𝑥𝑥 is real, ′𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 ≥ 0′ , so 16𝑘𝑘 2 − 4(1 − 𝑒𝑒 −𝑎𝑎 )(𝑒𝑒 𝑎𝑎 − 1) ≥ 0 M1

4𝑘𝑘 2 − 𝑒𝑒 𝑎𝑎 − 𝑒𝑒 −𝑎𝑎 + 2 ≥ 0
4𝑘𝑘 2 − 2 cosh 𝑎𝑎 + 2 ≥ 0

So cosh 𝑎𝑎 ≤ 2𝑘𝑘 2 + 1 A1

If 𝑎𝑎 = 0 , then 𝑥𝑥 = −𝑦𝑦 so sinh 𝑥𝑥 = - sinh 𝑦𝑦 and thus sinh 𝑥𝑥 + sinh 𝑦𝑦 = 2𝑘𝑘 = 0 but 𝑘𝑘 > 0 .

So cosh 𝑎𝑎 > 1 as required. E1 (5)

(iii) G6 (6)

Alternative
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 −cosh 𝑥𝑥 𝑑𝑑 2 𝑦𝑦 cosh 𝑦𝑦 sinh 𝑥𝑥−cosh 𝑥𝑥 sinh 𝑦𝑦 cosh2 𝑦𝑦 sinh 𝑥𝑥+cosh2 𝑥𝑥 sinh 𝑦𝑦
(i) = , = −� � = −�
𝑑𝑑𝑑𝑑
�
𝑑𝑑𝑑𝑑 cosh 𝑦𝑦 𝑑𝑑𝑥𝑥 2 cosh2 𝑦𝑦 cosh3 𝑦𝑦

then as before.
(ii) Substituting 𝑎𝑎 = 0 would imply 𝑒𝑒 𝑥𝑥 = 0 which is impossible.
3. (i)
𝑖𝑖𝑖𝑖
𝑘𝑘 − 𝑎𝑎 = (𝑏𝑏 − 𝑎𝑎)𝑒𝑒 − 3
M1 A1
Therefore,
1 𝑖𝑖𝑖𝑖
𝑔𝑔𝐴𝐴𝐴𝐴 = �𝑎𝑎 + 𝑏𝑏 + �𝑎𝑎 + (𝑏𝑏 − 𝑎𝑎)𝑒𝑒 − 3 ��
3
M1
𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖
2 − 𝑒𝑒 − 3 1 + 𝑒𝑒 − 3
= 𝑎𝑎 � � + 𝑏𝑏 � �
3 3
𝑖𝑖𝑖𝑖 √3 + 𝑖𝑖
𝜔𝜔 = 𝑒𝑒 6 =
2
and so
√3 − 𝑖𝑖
𝜔𝜔∗ =
2

1 − 𝑖𝑖√3
−
𝑖𝑖𝑖𝑖 2−� �
2 − 𝑒𝑒 3 2 3 + 𝑖𝑖√3 1 √3 + 𝑖𝑖 1
= = = = 𝜔𝜔
3 3 6 √3 2 √3
A1
and
1 − 𝑖𝑖√3
𝑖𝑖𝑖𝑖 1+� �
1 + 𝑒𝑒 − 3 2 3 − 𝑖𝑖√3 1 ∗
= = = 𝜔𝜔
3 3 6 √3
A1 (5)
1
Thus 𝑔𝑔𝐴𝐴𝐴𝐴 = (𝜔𝜔𝜔𝜔 + 𝜔𝜔∗ 𝑏𝑏) as required.
√3

1
(ii) 𝑔𝑔𝐴𝐴𝐴𝐴 = (𝜔𝜔𝜔𝜔 + 𝜔𝜔∗ 𝑏𝑏)
√3
1
𝑔𝑔𝐵𝐵𝐵𝐵 = (𝜔𝜔𝜔𝜔 + 𝜔𝜔∗ 𝑐𝑐 )
√3
1
𝑔𝑔𝐶𝐶𝐶𝐶 = (𝜔𝜔𝜔𝜔 + 𝜔𝜔∗ 𝑑𝑑 )
√3
1
𝑔𝑔𝐷𝐷𝐷𝐷 = (𝜔𝜔𝜔𝜔 + 𝜔𝜔∗ 𝑎𝑎)
√3
𝑄𝑄1 parallelogram ⇒ 𝑏𝑏 − 𝑎𝑎 = 𝑐𝑐 − 𝑑𝑑 ⇔ 𝑑𝑑 − 𝑎𝑎 = 𝑐𝑐 − 𝑏𝑏 B1

1 1
𝑔𝑔𝐵𝐵𝐵𝐵 − 𝑔𝑔𝐴𝐴𝐴𝐴 = �𝜔𝜔(𝑏𝑏 − 𝑎𝑎) + 𝜔𝜔∗ (𝑐𝑐 − 𝑏𝑏)� =
�𝜔𝜔(𝑐𝑐 − 𝑑𝑑 ) + 𝜔𝜔∗ (𝑑𝑑 − 𝑎𝑎)� = 𝑔𝑔𝐶𝐶𝐶𝐶 − 𝑔𝑔𝐷𝐷𝐷𝐷
√3 √3
⇒ 𝑄𝑄2 parallelogram. M1 A1
𝑄𝑄2 parallelogram ⇒ 𝑔𝑔𝐵𝐵𝐵𝐵 − 𝑔𝑔𝐴𝐴𝐴𝐴 = 𝑔𝑔𝐶𝐶𝐶𝐶 − 𝑔𝑔𝐷𝐷𝐷𝐷
1
{𝜔𝜔[(𝑏𝑏 − 𝑎𝑎) − (𝑐𝑐 − 𝑑𝑑 )] + 𝜔𝜔∗ [(𝑐𝑐 − 𝑏𝑏) − (𝑑𝑑 − 𝑎𝑎)]} = 0
√3
M1
1
(𝜔𝜔∗ − 𝜔𝜔)[(𝑎𝑎 − 𝑏𝑏) − (𝑑𝑑 − 𝑐𝑐 )] = 0
√3
A1
∗
As 𝜔𝜔 − 𝜔𝜔 ≠ 0 , (𝑎𝑎 − 𝑏𝑏) − (𝑑𝑑 − 𝑐𝑐 ) = 0 and so 𝑄𝑄1 is a parallelogram E1 (6)
(iii)
1
𝑔𝑔𝐵𝐵𝐵𝐵 − 𝑔𝑔𝐴𝐴𝐴𝐴 = �𝜔𝜔(𝑏𝑏 − 𝑎𝑎) + 𝜔𝜔∗ (𝑐𝑐 − 𝑏𝑏)�
√3
M1 A1
1
𝑔𝑔𝐶𝐶𝐶𝐶 − 𝑔𝑔𝐴𝐴𝐴𝐴 = �𝜔𝜔(𝑐𝑐 − 𝑎𝑎) + 𝜔𝜔∗ (𝑎𝑎 − 𝑏𝑏)� (1)
√3
1
𝜔𝜔2 (𝑔𝑔𝐵𝐵𝐵𝐵 − 𝑔𝑔𝐴𝐴𝐴𝐴 ) = �𝜔𝜔3 (𝑏𝑏 − 𝑎𝑎) + 𝜔𝜔(𝑐𝑐 − 𝑏𝑏)� (2)
√3
M1
1
= �𝑖𝑖 (𝑏𝑏 − 𝑎𝑎) + 𝜔𝜔(𝑐𝑐 − 𝑏𝑏)�
√3
A1
1 √3−𝑖𝑖 √3+𝑖𝑖
The coefficient of 𝑎𝑎 in (1) is 𝜔𝜔∗ − 𝜔𝜔 = − = −𝑖𝑖 B1
√3 2 2
1
The coefficient of 𝑏𝑏 in (1) is −𝜔𝜔∗ = (𝑖𝑖 − 𝜔𝜔) B1
√3
1
The coefficient of 𝑐𝑐 in (1) is 𝜔𝜔 B1
√3
𝜋𝜋
Thus 𝐺𝐺𝐴𝐴𝐴𝐴 𝐺𝐺𝐵𝐵𝐵𝐵 rotated through is 𝐺𝐺𝐴𝐴𝐴𝐴 𝐺𝐺𝐶𝐶𝐶𝐶 which means that 𝐺𝐺𝐴𝐴𝐴𝐴 𝐺𝐺𝐵𝐵𝐵𝐵 𝐺𝐺𝐶𝐶𝐶𝐶 is an equilateral
3
triangle. E2 (9)

(iii) Alternative
1
𝑥𝑥 = 𝑔𝑔𝐵𝐵𝐵𝐵 − 𝑔𝑔𝐴𝐴𝐴𝐴 = �𝜔𝜔(𝑏𝑏 − 𝑎𝑎) + 𝜔𝜔∗ (𝑐𝑐 − 𝑏𝑏)�
√3
M1 A1
1
𝑦𝑦 = 𝑔𝑔𝐶𝐶𝐶𝐶 − 𝑔𝑔𝐴𝐴𝐴𝐴 = �𝜔𝜔(𝑐𝑐 − 𝑎𝑎) + 𝜔𝜔∗ (𝑎𝑎 − 𝑏𝑏)�
√3
1 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 −𝑖𝑖𝑖𝑖 −𝑖𝑖𝑖𝑖
𝑥𝑥 = �𝑒𝑒 6 𝑏𝑏 − 𝑒𝑒 6 𝑎𝑎 + 𝑒𝑒 6 𝑐𝑐 − 𝑒𝑒 6 𝑏𝑏�
√3
M1
1 𝑖𝑖𝑖𝑖 −𝑖𝑖𝑖𝑖 𝑖𝑖7𝜋𝜋
= �𝑒𝑒 2 𝑏𝑏 + 𝑒𝑒 6 𝑐𝑐 + 𝑒𝑒 6 𝑎𝑎�
√3
A1
1 𝑖𝑖𝑖𝑖 𝑖𝑖3𝜋𝜋 𝑖𝑖5𝜋𝜋
𝑦𝑦 = �𝑒𝑒 6 𝑐𝑐 + 𝑒𝑒 2 𝑎𝑎 + 𝑒𝑒 6 𝑏𝑏�
√3
M1
𝑦𝑦 𝑖𝑖𝑖𝑖
= 𝑒𝑒 3
𝑥𝑥
M1 A1
𝑖𝑖𝑖𝑖
𝜋𝜋
𝑒𝑒 3 means y is x rotated through E1 and thus ABC is an equilateral triangle. E1 (9)
3

[or alternatively
𝑦𝑦 𝑧𝑧
� � = 1 and similarly � � = 1 M1 and thus all three sides are equal length E1 ]
𝑥𝑥 𝑦𝑦
4. 𝜋𝜋 has equation 𝑟𝑟. 𝑛𝑛 = 0 so 𝑛𝑛 is a vector perpendicular to this plane.

Q lies on 𝜋𝜋 if 𝑥𝑥 − (𝑥𝑥. 𝑛𝑛)𝑛𝑛 satisfies 𝑟𝑟. 𝑛𝑛 = 0

(𝑥𝑥 − (𝑥𝑥. 𝑛𝑛)𝑛𝑛). 𝑛𝑛 = 𝑥𝑥. 𝑛𝑛 − (𝑥𝑥. 𝑛𝑛)𝑛𝑛. 𝑛𝑛 = 𝑥𝑥. 𝑛𝑛 − 𝑥𝑥. 𝑛𝑛 = 0 so Q lies on 𝜋𝜋 as required. M1 A1

𝑃𝑃𝑃𝑃 = (𝑥𝑥 − (𝑥𝑥. 𝑛𝑛)𝑛𝑛) − 𝑥𝑥 = −(𝑥𝑥. 𝑛𝑛)𝑛𝑛 which is parallel to 𝑛𝑛 and so is perpendicular to 𝜋𝜋 . E1 (3)

𝑎𝑎
(i) The image of a point with position vector 𝑥𝑥 under T is 𝑥𝑥 − 2 𝑥𝑥. 𝑛𝑛 𝑛𝑛 , so as 𝑛𝑛 = �𝑏𝑏� and
( )
𝑐𝑐
M1
1 1 1 𝑎𝑎 𝑎𝑎 1 𝑎𝑎 1 − 2𝑎𝑎2
𝑖𝑖 = �0� , the image of 𝑖𝑖 under T is �0� − 2 �0� . �𝑏𝑏 � �𝑏𝑏 � = �0� − 2𝑎𝑎 �𝑏𝑏 � = � −2𝑎𝑎𝑎𝑎 � A1
0 0 0 𝑐𝑐 𝑐𝑐 0 𝑐𝑐 −2𝑎𝑎𝑎𝑎

But 𝑎𝑎2 + 𝑏𝑏 2 + 𝑐𝑐 2 = 1 so 1 − 2𝑎𝑎2 = 𝑎𝑎2 + 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 𝑎𝑎2 dM1

𝑏𝑏 2 + 𝑐𝑐 2 − 𝑎𝑎2
Thus, the image of 𝑖𝑖 under T is � −2𝑎𝑎𝑎𝑎 � as required.
−2𝑎𝑎𝑎𝑎
−2𝑎𝑎𝑎𝑎 −2𝑎𝑎𝑎𝑎
Similarly, the images of 𝑗𝑗 and 𝑘𝑘 are �𝑐𝑐 2 + 𝑎𝑎2 − 𝑏𝑏 2 � and � −2𝑏𝑏𝑏𝑏 � respectively.
−2𝑏𝑏𝑏𝑏 𝑎𝑎2 + 𝑏𝑏 2 − 𝑐𝑐 2
B1 B1

𝑏𝑏 2 + 𝑐𝑐 2 − 𝑎𝑎2 −2𝑎𝑎𝑎𝑎 −2𝑎𝑎𝑎𝑎

Thus 𝑀𝑀 = � −2𝑎𝑎𝑎𝑎 𝑐𝑐 + 𝑎𝑎2 − 𝑏𝑏 2
2
−2𝑏𝑏𝑏𝑏 � B1 (6)
2 2 2
−2𝑎𝑎𝑎𝑎 −2𝑏𝑏𝑏𝑏 𝑎𝑎 + 𝑏𝑏 − 𝑐𝑐

(ii) 1 − 2𝑎𝑎2 = 0.∙ 64 ⇒ 𝑎𝑎 = ±0 ∙ 3√2 and thus as −2𝑎𝑎𝑎𝑎 = 0 ∙ 48 and −2𝑎𝑎𝑎𝑎 = 0 ∙ 6 ,

M1 A1
𝑏𝑏 = ∓0 ∙ 4√2 and 𝑐𝑐 = ∓0 ∙ 5√2 and the plane is 3𝑥𝑥 − 4𝑦𝑦 − 5𝑧𝑧 = 0 (or −3𝑥𝑥 + 4𝑦𝑦 + 5𝑧𝑧 = 0)
A1ft A1ft (4)

(iii) Suppose the position vector of the point Q on the given line such that 𝑃𝑃𝑃𝑃 is perpendicular to
𝑎𝑎 𝑎𝑎
that line is 𝑦𝑦 , then 𝑦𝑦 = 𝜆𝜆 �𝑏𝑏 � for some 𝜆𝜆 and (𝑦𝑦 − 𝑥𝑥 ). �𝑏𝑏 � = 0 E1
𝑐𝑐 𝑐𝑐
𝑎𝑎 𝑎𝑎 𝑎𝑎
So, 𝑦𝑦. �𝑏𝑏 � − 𝑥𝑥. �𝑏𝑏 � = 0 , i.e. 𝜆𝜆 = 𝑥𝑥. �𝑏𝑏 � M1
𝑐𝑐 𝑐𝑐 𝑐𝑐

𝑎𝑎 𝑎𝑎
So, the image of P under the rotation, is 𝑥𝑥 + 2(𝑦𝑦 − 𝑥𝑥 ) = 2𝑦𝑦 − 𝑥𝑥 = 2𝑥𝑥. �𝑏𝑏 � �𝑏𝑏� − 𝑥𝑥 M1
𝑐𝑐 𝑐𝑐
2𝑎𝑎2 − 1 𝑎𝑎2 − 𝑏𝑏 2 − 𝑐𝑐 2
The image of 𝑖𝑖 under the rotation is thus � 2𝑎𝑎𝑎𝑎 � = � 2𝑎𝑎𝑎𝑎 � , A1 and of 𝑗𝑗 and 𝑘𝑘
2𝑎𝑎𝑎𝑎 2𝑎𝑎𝑎𝑎
are
 2𝑎𝑎𝑎𝑎 2𝑎𝑎𝑎𝑎
�𝑏𝑏 2 − 𝑐𝑐 2 − 𝑎𝑎2 � and � 2𝑏𝑏𝑏𝑏 � respectively.
2𝑏𝑏𝑏𝑏 𝑐𝑐 2 − 𝑎𝑎2 − 𝑏𝑏 2

𝑎𝑎2 − 𝑏𝑏 2 − 𝑐𝑐 2 2𝑎𝑎𝑎𝑎 2𝑎𝑎𝑎𝑎

Thus 𝑁𝑁 = � 2𝑎𝑎𝑎𝑎 𝑏𝑏 2 − 𝑐𝑐 2 − 𝑎𝑎2 2𝑏𝑏𝑏𝑏 � , which, incidentally = −𝑀𝑀 . A1 (5)
2𝑎𝑎𝑎𝑎 2𝑏𝑏𝑏𝑏 𝑐𝑐 2 − 𝑎𝑎2 − 𝑏𝑏 2

(iv) 𝑁𝑁𝑁𝑁 = −𝑀𝑀𝑀𝑀 = −𝐼𝐼 as 𝑀𝑀 is self-inverse. B1

Thus the single transformation is an enlargement, scale factor -1 , with centre of enlargement the
origin. B1 (2)

alternative for (iii)

𝑥𝑥 = 𝑢𝑢 + 𝑣𝑣 where 𝑢𝑢 ∈ Π and 𝑣𝑣 ⊥ Π M1

𝑀𝑀𝑀𝑀 = −𝑣𝑣 𝑀𝑀𝑀𝑀 = 𝑢𝑢 𝑀𝑀𝑀𝑀 = 𝑢𝑢 − 𝑣𝑣 M1

𝑁𝑁𝑁𝑁 = 𝑣𝑣 − 𝑢𝑢 = −𝑀𝑀𝑀𝑀 A1

𝑁𝑁 = −𝑀𝑀 A1 (4)

alternative for (ii) the matrix represents a reflection, an invariant point under the reflection lies on
the plane of reflection. E1
Therefore,
0.64 0.48 0.6 𝑥𝑥 𝑥𝑥
�0.48 0.36 −0.8� �𝑦𝑦� = �𝑦𝑦�
0.6 −0.8 0 𝑧𝑧 𝑧𝑧
B1
Taking the simplest component equation
0.6𝑥𝑥 − 0.8𝑦𝑦 = 𝑧𝑧
(although the other two give equivalent equations). B1

This simplifies to
3𝑥𝑥 − 4𝑦𝑦 − 5𝑧𝑧 = 0 B1
5. (𝑥𝑥 − 𝑦𝑦)(𝑥𝑥 𝑛𝑛−1 + 𝑥𝑥 𝑛𝑛−2 𝑦𝑦 + ⋯ + 𝑦𝑦 𝑛𝑛−1 ) = 𝑥𝑥 𝑛𝑛 − 𝑥𝑥 𝑛𝑛−1 𝑦𝑦 + 𝑥𝑥 𝑛𝑛−1 𝑦𝑦 − 𝑥𝑥 𝑛𝑛−2 𝑦𝑦 + ⋯ + 𝑥𝑥𝑦𝑦 𝑛𝑛−1 − 𝑦𝑦 𝑛𝑛
M1
𝑛𝑛 𝑛𝑛
= 𝑥𝑥 − 𝑦𝑦
as each even numbered term cancels with its subsequent term. E1 (2)

(i) If
1 𝐴𝐴 𝑓𝑓(𝑥𝑥 )
𝐹𝐹 (𝑥𝑥 ) = = + 𝑛𝑛
𝑥𝑥 𝑛𝑛 (𝑥𝑥− 𝑘𝑘) 𝑥𝑥 − 𝑘𝑘 𝑥𝑥
then multiplying by 𝑥𝑥 𝑛𝑛 (𝑥𝑥 − 𝑘𝑘)

1 = 𝐴𝐴𝑥𝑥 𝑛𝑛 + (𝑥𝑥 − 𝑘𝑘)𝑓𝑓(𝑥𝑥 )

M1
1
𝑥𝑥 = 𝑘𝑘 ⇒ 𝐴𝐴 =
𝑘𝑘 𝑛𝑛
M1 A1

so
𝑥𝑥 𝑛𝑛
1= + (𝑥𝑥 − 𝑘𝑘)𝑓𝑓(𝑥𝑥 )
𝑘𝑘 𝑛𝑛
and
1 𝑥𝑥 𝑛𝑛
𝑓𝑓 (𝑥𝑥 ) = �1 − � � �
𝑥𝑥 − 𝑘𝑘 𝑘𝑘
as required. A1*

Thus
1 1 𝑥𝑥 𝑛𝑛
𝑛𝑛 �1 − � � �
𝐹𝐹 (𝑥𝑥 ) = 𝑘𝑘 + 𝑥𝑥 − 𝑘𝑘 𝑛𝑛 𝑘𝑘
𝑥𝑥 − 𝑘𝑘 𝑥𝑥
M1
1 𝑥𝑥 𝑛𝑛 − 𝑘𝑘 𝑛𝑛
= −
𝑘𝑘 𝑛𝑛 (𝑥𝑥 − 𝑘𝑘) 𝑘𝑘 𝑛𝑛 𝑥𝑥 𝑛𝑛 (𝑥𝑥 − 𝑘𝑘)
and so, by the result of the stem,
𝑛𝑛
1 1
𝐹𝐹 (𝑥𝑥 ) = 𝑛𝑛 − 𝑛𝑛 𝑛𝑛 � 𝑥𝑥 𝑛𝑛−𝑟𝑟 𝑘𝑘 𝑟𝑟−1
𝑘𝑘 (𝑥𝑥 − 𝑘𝑘) 𝑘𝑘 𝑥𝑥
𝑟𝑟=1

M1
𝑛𝑛
1 1 1
= 𝑛𝑛 − � 𝑛𝑛−𝑟𝑟 𝑟𝑟
(
𝑘𝑘 𝑥𝑥 − 𝑘𝑘 ) 𝑘𝑘 𝑘𝑘 𝑥𝑥
𝑟𝑟=1

A1* (7)
(ii)
𝑛𝑛
𝑛𝑛
1 𝑥𝑥 𝑛𝑛 1 𝑥𝑥 𝑛𝑛−𝑟𝑟
𝑥𝑥 𝐹𝐹 (𝑥𝑥 ) = = 𝑛𝑛 − � 𝑛𝑛−𝑟𝑟
𝑥𝑥 − 𝑘𝑘 𝑘𝑘 (𝑥𝑥 − 𝑘𝑘) 𝑘𝑘 𝑘𝑘
𝑟𝑟=1

Differentiating with respect to 𝑥𝑥 ,

𝑛𝑛
−1 𝑛𝑛𝑛𝑛 𝑛𝑛−1 𝑥𝑥 𝑛𝑛 1 (𝑛𝑛 − 𝑟𝑟)𝑥𝑥 𝑛𝑛−𝑟𝑟−1
= − − �
(𝑥𝑥 − 𝑘𝑘)2 𝑘𝑘 𝑛𝑛 (𝑥𝑥 − 𝑘𝑘) 𝑘𝑘 𝑛𝑛 (𝑥𝑥 − 𝑘𝑘)2 𝑘𝑘 𝑘𝑘 𝑛𝑛−𝑟𝑟
𝑟𝑟=1

M1 A1
−1
Multiplying by
𝑥𝑥 𝑛𝑛
𝑛𝑛
1 −𝑛𝑛 1 𝑛𝑛 − 𝑟𝑟
= 𝑛𝑛 + 𝑛𝑛 + � 𝑛𝑛+1−𝑟𝑟 𝑟𝑟+1
𝑥𝑥 (𝑥𝑥 − 𝑘𝑘)
𝑛𝑛 2 𝑥𝑥𝑥𝑥 (𝑥𝑥 − 𝑘𝑘) 𝑘𝑘 (𝑥𝑥 − 𝑘𝑘) 2 𝑘𝑘 𝑥𝑥
𝑟𝑟=1

A1* (3)

(iii)
𝑁𝑁 𝑁𝑁 3
1 −3 1 3 − 𝑟𝑟
� 3 𝑑𝑑𝑑𝑑 = � + + � 𝑟𝑟+1 𝑑𝑑𝑑𝑑
𝑥𝑥 (𝑥𝑥 − 1) 2 𝑥𝑥 (𝑥𝑥 − 1) (𝑥𝑥 − 1) 2 𝑥𝑥
2 2 𝑟𝑟=1

M1

𝑁𝑁 3
3 3 1 3 − 𝑟𝑟
=� − + + � 𝑟𝑟+1 𝑑𝑑𝑑𝑑
𝑥𝑥 (𝑥𝑥 − 1) (𝑥𝑥 − 1) 2 𝑥𝑥
2 𝑟𝑟=1

dM1
3 𝑁𝑁
1 3 − 𝑟𝑟
= �3ln 𝑥𝑥 − 3 ln(𝑥𝑥 − 1) − −� �
𝑥𝑥 − 1 𝑟𝑟𝑟𝑟 𝑟𝑟
𝑟𝑟=1 2

3 𝑁𝑁
𝑥𝑥 1 3 − 𝑟𝑟
= �3 ln � �− −� �
𝑥𝑥 − 1 𝑥𝑥 − 1 𝑟𝑟𝑟𝑟 𝑟𝑟
𝑟𝑟=1 2

M1 A1
3 3
𝑁𝑁 1 3 − 𝑟𝑟 3 − 𝑟𝑟
= 3 ln � �− −� 𝑟𝑟
− 3 ln(2) + 1 + �
𝑁𝑁 − 1 𝑁𝑁 − 1 𝑟𝑟𝑟𝑟 𝑟𝑟2𝑟𝑟
𝑟𝑟=1 𝑟𝑟=1

A1
𝑁𝑁−1 𝑁𝑁−1 1 1
As 𝑁𝑁 → ∞ , � � → 1 , so 3 ln � � → 0 , and →0 , →0
𝑁𝑁 𝑁𝑁 𝑁𝑁−1 𝑁𝑁𝑟𝑟

E1

So the limit of the integral is,

 2 1 17
−3 ln 2 + 1 + + = −3 ln 2 +
2 8 8
M1 A1 (8)

Alternatives for stem using sum of GP or proof by induction M1 A1

6. (i)

𝑦𝑦 = cos 𝑥𝑥 + √cos 2𝑥𝑥

𝜋𝜋 1
𝑥𝑥 = 0 , 𝑦𝑦 = 2 There is symmetry in 𝑥𝑥 = 0 G1 𝑥𝑥 = ± , 𝑦𝑦 = G1
4 √2

𝑑𝑑𝑑𝑑 sin 2𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

= − sin 𝑥𝑥 − M1 so 𝑥𝑥 = 0 , = 0 𝑥𝑥 > 0 , < 0 and vice versa
𝑑𝑑𝑑𝑑 √cos 2𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝜋𝜋 𝑑𝑑𝑑𝑑
as 𝑥𝑥 → , → −∞ A1
4 𝑑𝑑𝑑𝑑

G1 (5)

(ii)

G1 G1 (2)
 𝜋𝜋 1
(iii) 𝜃𝜃 = ± , 𝑟𝑟 =
4 √2

𝑟𝑟 2 − 2𝑟𝑟 cos 𝜃𝜃 + sin2 𝜃𝜃 = 0

(𝑟𝑟 − cos 𝜃𝜃)2 = cos 2 𝜃𝜃 − sin2 𝜃𝜃 = cos 2𝜃𝜃

Therefore, 𝑟𝑟 − cos 𝜃𝜃 = ± √cos 2𝜃𝜃 , i.e. 𝑟𝑟 = cos 𝜃𝜃 ± √cos 2𝜃𝜃 M1

From (i), 𝑟𝑟 is only small on the branch, 𝑟𝑟 = cos 𝜃𝜃 − √cos 2𝜃𝜃 . For 𝜃𝜃 = 0 , 𝑟𝑟 = 0

Otherwise, cos 𝜃𝜃 − √cos 2𝜃𝜃 = 0 , cos 2𝜃𝜃 = cos 2 𝜃𝜃 , 2cos 2 𝜃𝜃 − 1 = cos 2 𝜃𝜃 , cos 𝜃𝜃 = ±1 so for
𝜋𝜋 𝜋𝜋
− ≤ 𝜃𝜃 ≤ , 𝜃𝜃 = 0 is the only value for which 𝑟𝑟 = 0
4 4

So 𝑟𝑟 small implies 𝑟𝑟 = cos 𝜃𝜃 − √cos 2𝜃𝜃 and 𝜃𝜃 is small E1

1
𝜃𝜃2 (2𝜃𝜃)2 2 𝜃𝜃2 𝜃𝜃2
Thus 𝑟𝑟 ≈ 1 − − �1 − � ≈1− − 1 + 𝜃𝜃 2 = as required. B1*
2 2 2 2

G1 G1 G1 (6)

Area required is
𝜋𝜋 𝜋𝜋
1 4 2 1 4 2
� �cos 𝜃𝜃 + √cos 2𝜃𝜃� 𝑑𝑑𝑑𝑑 − � �cos 𝜃𝜃 − √cos 2𝜃𝜃� 𝑑𝑑𝑑𝑑
2 0 2 0

M1
𝜋𝜋
4
= 2 � cos 𝜃𝜃 √cos 2𝜃𝜃 𝑑𝑑𝑑𝑑
0

A1
𝜋𝜋
4
= 2 � cos 𝜃𝜃 �1 − 2 sin2 𝜃𝜃 𝑑𝑑𝑑𝑑
0

𝑑𝑑𝑑𝑑
Let √2 sin 𝜃𝜃 = sin 𝑢𝑢 , M1 then √2 cos 𝜃𝜃 = cos 𝑢𝑢 ,
𝑑𝑑𝑑𝑑
So the integral becomes
𝜋𝜋 𝜋𝜋 𝜋𝜋
2 cos 2 𝑢𝑢 2 cos 2𝑢𝑢 +1 sin 2𝑢𝑢 𝑢𝑢 2 𝜋𝜋
2� 𝑑𝑑𝑑𝑑 = √2 � 𝑑𝑑𝑑𝑑 = √2 � + � =
0 √2 0 2 4 2 0 2√2

M1 A1 M1 A1* (7)

(iii) alternative
𝑟𝑟 ≪ 1 ⇒ −2𝑟𝑟 cos 𝜃𝜃 + sin2 𝜃𝜃 ≈ 0
sin2 𝜃𝜃 1 𝜃𝜃 2
𝑟𝑟 ≈ = sin 𝜃𝜃 tan 𝜃𝜃 ≈
2 cos 𝜃𝜃 2 2
 𝑑𝑑𝑑𝑑
7. (i) 𝑢𝑢 = + 𝑔𝑔(𝑥𝑥 )𝑦𝑦
𝑑𝑑𝑑𝑑

Thus
𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑
= 2 + 𝑔𝑔(𝑥𝑥 ) + 𝑔𝑔′ (𝑥𝑥 )𝑦𝑦
𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑
M1
𝑑𝑑𝑑𝑑
As + 𝑓𝑓(𝑥𝑥 )𝑢𝑢 = ℎ(𝑥𝑥 )
𝑑𝑑𝑑𝑑

𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
2
+ 𝑔𝑔(𝑥𝑥 ) + 𝑔𝑔′ (𝑥𝑥 )𝑦𝑦 + 𝑓𝑓(𝑥𝑥 ) � + 𝑔𝑔(𝑥𝑥 )𝑦𝑦� = ℎ(𝑥𝑥 )
𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
M1

that is

𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑
+ �𝑔𝑔(𝑥𝑥 ) + 𝑓𝑓(𝑥𝑥 )� + �𝑔𝑔′ (𝑥𝑥 ) + 𝑓𝑓(𝑥𝑥 )𝑔𝑔(𝑥𝑥 )�𝑦𝑦 = ℎ(𝑥𝑥 )
𝑑𝑑𝑥𝑥 2 𝑑𝑑𝑑𝑑
as required. A1* (3)

(ii)
4
𝑔𝑔(𝑥𝑥 ) + 𝑓𝑓(𝑥𝑥 ) = 1 +
𝑥𝑥
4
and so 𝑓𝑓(𝑥𝑥 ) = 1 + − 𝑔𝑔(𝑥𝑥 )
𝑥𝑥

2 2
𝑔𝑔′ (𝑥𝑥 ) + 𝑓𝑓(𝑥𝑥 )𝑔𝑔(𝑥𝑥 ) = +
𝑥𝑥 𝑥𝑥 2
M1

so

4 2 2
𝑔𝑔′ (𝑥𝑥 ) + �1 + − 𝑔𝑔(𝑥𝑥 )� 𝑔𝑔(𝑥𝑥 ) = + 2
𝑥𝑥 𝑥𝑥 𝑥𝑥

as requested. A1 (2)

If 𝑔𝑔(𝑥𝑥 ) = 𝑘𝑘𝑥𝑥 𝑛𝑛 , 𝑔𝑔′ (𝑥𝑥 ) = 𝑘𝑘𝑘𝑘𝑥𝑥 𝑛𝑛−1

4 2 2
𝑘𝑘𝑘𝑘𝑥𝑥 𝑛𝑛−1 + �1 + − 𝑘𝑘𝑥𝑥 𝑛𝑛 � 𝑘𝑘𝑥𝑥 𝑛𝑛 = + 2
𝑥𝑥 𝑥𝑥 𝑥𝑥
M1 A1

−𝑘𝑘 2 𝑥𝑥 2𝑛𝑛+2 + 𝑘𝑘𝑥𝑥 𝑛𝑛+2 + 𝑘𝑘(𝑛𝑛 + 4)𝑥𝑥 𝑛𝑛+1 − 2𝑥𝑥 − 2 = 0

Considering the 𝑥𝑥 2𝑛𝑛+2 term,

either it is eliminated by the 𝑥𝑥 𝑛𝑛+2 term, in which case, 2𝑛𝑛 + 2 = 𝑛𝑛 + 2 and −𝑘𝑘 2 + 𝑘𝑘 = 0

which would imply 𝑛𝑛 = 0 and 𝑘𝑘 = 0 or 𝑘𝑘 = 1

𝑘𝑘 = 0 is not possible (−2𝑥𝑥 − 2 = 0) ; 𝑛𝑛 = 0 , 𝑘𝑘 = 1 would give 4𝑥𝑥 − 2𝑥𝑥 − 2 = 0 so not possible

(E1 can be awarded here if neither of next two B marks are obtained)
Or it is eliminated by the 𝑥𝑥 𝑛𝑛+1 term, in which case, 2𝑛𝑛 + 2 = 𝑛𝑛 + 1 which implies 𝑛𝑛 = −1 and
thus −𝑘𝑘 2 + 𝑘𝑘 (𝑛𝑛 + 4) − 2 = 0 and considering the other two terms 𝑘𝑘 − 2 = 0

𝑘𝑘 = 2 and 𝑛𝑛 = −1 satisfy −𝑘𝑘 2 + 𝑘𝑘(𝑛𝑛 + 4) − 2 = 0 so these are possible values. B1 B1

2 4 2
So 𝑔𝑔(𝑥𝑥 ) = and as 𝑓𝑓(𝑥𝑥 ) = 1 + − 𝑔𝑔(𝑥𝑥 ) , 𝑓𝑓 (𝑥𝑥 ) = 1 + ℎ(𝑥𝑥 ) = 4𝑥𝑥 + 12 M1
𝑥𝑥 𝑥𝑥 𝑥𝑥

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2
+ 𝑓𝑓(𝑥𝑥 )𝑢𝑢 = ℎ(𝑥𝑥 ) is thus + �1 + � 𝑢𝑢 = 4𝑥𝑥 + 12 M1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥

The integrating factor is

2
� 𝑑𝑑𝑑𝑑
𝑒𝑒 ∫�1+𝑥𝑥 = 𝑒𝑒 𝑥𝑥+2 ln 𝑥𝑥 = 𝑥𝑥 2 𝑒𝑒 𝑥𝑥
M1

Thus
𝑑𝑑𝑑𝑑
𝑥𝑥 2 𝑒𝑒 𝑥𝑥 + (𝑥𝑥 2 + 2𝑥𝑥 )𝑒𝑒 𝑥𝑥 𝑢𝑢 = (4𝑥𝑥 + 12)𝑥𝑥 2 𝑒𝑒 𝑥𝑥 = (4𝑥𝑥 3 + 12𝑥𝑥 2 )𝑒𝑒 𝑥𝑥
𝑑𝑑𝑑𝑑
Integrating with respect to x

𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑢𝑢 = �(4𝑥𝑥 3 + 12𝑥𝑥 2 )𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 = 4𝑥𝑥 3 𝑒𝑒 𝑥𝑥 + 𝑐𝑐

M1 A1
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑
As 𝑢𝑢 = + 𝑔𝑔(𝑥𝑥 )𝑦𝑦 , and 𝑔𝑔(𝑥𝑥 ) = , when 𝑥𝑥 = 1 , 𝑦𝑦 = 5 , = −3 , we have 𝑢𝑢 = −3 + 2 × 5
𝑑𝑑𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑

That is 𝑢𝑢 = 7 , so 7𝑒𝑒 = 4𝑒𝑒 + 𝑐𝑐 , which means 𝑐𝑐 = 3𝑒𝑒 M1 A1

𝑒𝑒 −𝑥𝑥
So 𝑢𝑢 = 4𝑥𝑥 + 3𝑒𝑒
𝑥𝑥 2

𝑑𝑑𝑑𝑑 2 𝑒𝑒 −𝑥𝑥
+ 𝑦𝑦 = 4𝑥𝑥 + 3𝑒𝑒 2
𝑑𝑑𝑑𝑑 𝑥𝑥 𝑥𝑥
This has integrating factor
2
𝑒𝑒 ∫𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒 2 ln 𝑥𝑥 = 𝑥𝑥 2

M1

So
𝑑𝑑𝑑𝑑
𝑥𝑥 2 + 2𝑥𝑥𝑥𝑥 = 4𝑥𝑥 3 + 3𝑒𝑒 𝑒𝑒 −𝑥𝑥
𝑑𝑑𝑑𝑑
Integrating with respect to x

𝑥𝑥 2 𝑦𝑦 = � 4𝑥𝑥 3 + 3𝑒𝑒 𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑥𝑥 4 − 3𝑒𝑒𝑒𝑒 −𝑥𝑥 + 𝑐𝑐′

M1 A1

when 𝑥𝑥 = 1 , 𝑦𝑦 = 5 so 5 = 1 − 3 + 𝑐𝑐′ which means 𝑐𝑐 ′ = 7 B1 (15)

Therefore,
7 3𝑒𝑒 −𝑥𝑥+1
𝑦𝑦 = 𝑥𝑥 2 + 2 −
𝑥𝑥 𝑥𝑥 2
8. (i) All terms of the sequence are positive integers because they are all either equal to a previous
term or the sum of two previous terms which are positive integers. E1

Thus, for 𝑘𝑘 ≥ 1 , as 𝑢𝑢2𝑘𝑘 = 𝑢𝑢𝑘𝑘 and 𝑢𝑢2𝑘𝑘+1 = 𝑢𝑢𝑘𝑘 + 𝑢𝑢𝑘𝑘+1 , 𝑢𝑢2𝑘𝑘+1 − 𝑢𝑢2𝑘𝑘 = 𝑢𝑢𝑘𝑘+1 ≥ 1 E1

Also, 𝑢𝑢2𝑘𝑘+1 − 𝑢𝑢2𝑘𝑘+2 = 𝑢𝑢𝑘𝑘 + 𝑢𝑢𝑘𝑘+1 − 𝑢𝑢𝑘𝑘+1 = 𝑢𝑢𝑘𝑘 ≥ 1 . Thus, the required result is proved for terms
from the third onwards. (The only terms not included in this proof are the first two, which are in
case both equal to 1). E1 (3)

(ii) Suppose that 𝑢𝑢2𝑘𝑘 = 𝑐𝑐 , and that 𝑢𝑢2𝑘𝑘+1 = 𝑑𝑑 , for 𝑘𝑘 ≥ 1 , where d and c share a common factor
greater than one, then 𝑢𝑢𝑘𝑘 = 𝑐𝑐 , as 𝑢𝑢2𝑘𝑘 = 𝑢𝑢𝑘𝑘 , and 𝑢𝑢𝑘𝑘+1 = 𝑑𝑑 − 𝑐𝑐 ≥ 1 as 𝑢𝑢2𝑘𝑘+1 = 𝑢𝑢𝑘𝑘 + 𝑢𝑢𝑘𝑘+1 and
using (i). Then as d and c share a common factor greater than one, d-c and c share a common factor
greater than one. So, two earlier terms in the sequence do share the same common factor. E1

Likewise, suppose that 𝑢𝑢2𝑘𝑘+2 = 𝑐𝑐 , and that 𝑢𝑢2𝑘𝑘+1 = 𝑑𝑑 , for 𝑘𝑘 ≥ 1 , where d and c share a
common factor greater than one, then 𝑢𝑢𝑘𝑘+1 = 𝑐𝑐 and 𝑢𝑢𝑘𝑘 = 𝑑𝑑 − 𝑐𝑐 giving the same result. E1

This is true for pairs of consecutive terms from the second term (and third) onwards. Repeating this
argument, E1 we find that it would imply that the first two terms would share a common factor
greater than one, which is a contradiction. E1 Hence any two consecutive terms are co-prime. (4)

(iii) For 𝑘𝑘 ≥ 1 , and 𝑚𝑚 ≥ 1 suppose that 𝑢𝑢2𝑘𝑘 = 𝑐𝑐 and 𝑢𝑢2𝑘𝑘+1 = 𝑑𝑑 , and that 𝑢𝑢2𝑘𝑘+𝑚𝑚 = 𝑐𝑐 and
𝑢𝑢2𝑘𝑘+𝑚𝑚+1 = 𝑑𝑑 , then as 𝑑𝑑 > 𝑐𝑐 , 2𝑘𝑘 + 𝑚𝑚 is even, so 𝑚𝑚 is even, say 2𝑛𝑛 . Thus, 𝑢𝑢𝑘𝑘 = 𝑐𝑐 and 𝑢𝑢𝑘𝑘+1 =
𝑑𝑑 − 𝑐𝑐 , and 𝑢𝑢𝑘𝑘+𝑛𝑛 = 𝑐𝑐 and 𝑢𝑢𝑘𝑘+𝑛𝑛+1 = 𝑑𝑑 − 𝑐𝑐 . That is, an earlier pair of terms would appear
consecutively. E1

Likewise, if 𝑢𝑢2𝑘𝑘+2 = 𝑐𝑐 and 𝑢𝑢2𝑘𝑘+1 = 𝑑𝑑 , and that 𝑢𝑢2𝑘𝑘+𝑚𝑚+2 = 𝑐𝑐 and 𝑢𝑢2𝑘𝑘+𝑚𝑚+1 = 𝑑𝑑, the same
argument applies. E1

So the argument can be repeated down to the first two terms, which are of course equal, E1 and it
would imply a later pair are likewise which contradicts (i). E1 (4)

(iv) If (𝑎𝑎, 𝑏𝑏) does not occur, where 𝑎𝑎 and 𝑏𝑏 are coprime and 𝑎𝑎 > 𝑏𝑏 , then there does not exist 𝑘𝑘
such that 𝑢𝑢2𝑘𝑘+1 = 𝑎𝑎 and 𝑢𝑢2𝑘𝑘+2 = 𝑏𝑏 . E1 Therefore there cannot exist a k such that 𝑢𝑢𝑘𝑘+1 = 𝑏𝑏
and 𝑢𝑢𝑘𝑘 = 𝑎𝑎 − 𝑏𝑏 , the sum of which is 𝑎𝑎 , which is smaller than 𝑎𝑎 + 𝑏𝑏 . E1

If (𝑎𝑎, 𝑏𝑏) does not occur, where 𝑎𝑎 and 𝑏𝑏 are coprime and 𝑎𝑎 < 𝑏𝑏 , then there does not exist 𝑘𝑘 such
that 𝑢𝑢2𝑘𝑘 = 𝑎𝑎 and 𝑢𝑢2𝑘𝑘+1 = 𝑏𝑏 . E1 Therefore there cannot exist a k such that 𝑢𝑢𝑘𝑘 = 𝑎𝑎 and 𝑢𝑢𝑘𝑘+1 =
𝑏𝑏 − 𝑎𝑎 , the sum of which is b, which is smaller than 𝑎𝑎 + 𝑏𝑏 . E1 (4)

(v) Suppose that there exists an ordered pair of coprime integers (a,b) which does not occur
consecutively in the sequence. Then by part (iv) E1 the pair (a-b, b) [if a>b] or (a, b-a) [if b>a] (which
has a smaller sum) does not occur. Repeating this means that a coprime pair with sum <3 does not
occur. The only coprime pair of integers with sum <3 is (1, 1) which are the first two
terms. Contradiction and so every ordered pair of coprime integers occurs in the sequence E1
and by (iii) only occurs once. E1 Therefore, there exists an 𝑛𝑛 , and that 𝑛𝑛 is unique such that
𝑢𝑢𝑛𝑛
𝑞𝑞 = , for any positive rational 𝑞𝑞 E1 (which is expressed in lowest form). So the inverse of f
𝑢𝑢𝑛𝑛+1
exists. E1 (5)
9. (i)

G1 G1

Resolving vertically, (1) M1

𝑅𝑅 cos 𝛼𝛼 + 𝑆𝑆 cos 𝛽𝛽 = 𝑊𝑊
Resolving horizontally, (2) M1

𝑅𝑅 sin 𝛼𝛼 = 𝑆𝑆 sin 𝛽𝛽
Taking moments about Q, (3) M1 A1

𝑊𝑊𝑊𝑊 cos 𝜃𝜃 = 2𝑅𝑅𝑅𝑅 cos(𝛼𝛼 − 𝜃𝜃)

Dividing (3) by 𝑙𝑙 gives (4)
𝑊𝑊 cos 𝜃𝜃 = 2𝑅𝑅 cos(𝛼𝛼 − 𝜃𝜃)
Multiplying (1) by cos 𝜃𝜃 sin 𝛽𝛽 gives

𝑅𝑅 cos 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 + 𝑆𝑆 cos 𝛽𝛽 cos 𝜃𝜃 sin 𝛽𝛽 = 𝑊𝑊 cos 𝜃𝜃 sin 𝛽𝛽

Using (2) and (4) to substitute for 𝑆𝑆 sin 𝛽𝛽 and 𝑊𝑊 cos 𝜃𝜃 respectively, M1

𝑅𝑅 cos 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 + 𝑅𝑅 sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜃𝜃 = 2𝑅𝑅 cos(𝛼𝛼 − 𝜃𝜃) sin 𝛽𝛽

Thus
cos 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 + sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜃𝜃 = 2 cos 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 + 2 sin 𝛼𝛼 sin 𝜃𝜃 sin 𝛽𝛽

and so

sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜃𝜃 − cos 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 = 2 sin 𝛼𝛼 sin 𝜃𝜃 sin 𝛽𝛽

Dividing by sin 𝛼𝛼 cos 𝜃𝜃 sin 𝛽𝛽 gives

cot 𝛽𝛽 − cot 𝛼𝛼 = 2 tan 𝜃𝜃

as required. M1 A1* (9)
(ii)

G1

Resolving vertically, (5) M1

𝑅𝑅 cos 𝛼𝛼 + 𝑆𝑆 cos 𝛽𝛽 = 𝑊𝑊 + 𝜇𝜇𝜇𝜇 sin 𝛽𝛽

Resolving horizontally, (6) M1 A1 M1

𝑅𝑅 sin 𝛼𝛼 = 𝑆𝑆 sin 𝛽𝛽 + 𝜇𝜇𝜇𝜇 cos 𝛽𝛽

Taking moments about Q, and dividing by by 𝑙𝑙 gives as before (4)

𝑊𝑊 cos 𝜙𝜙 = 2𝑅𝑅 cos(𝛼𝛼 − 𝜙𝜙)

Multiplying (5) by (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙 gives

𝑅𝑅 cos 𝛼𝛼 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙 + 𝑆𝑆 cos 𝛽𝛽 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙

= 𝑊𝑊 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙 + 𝜇𝜇𝜇𝜇 sin 𝛽𝛽 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙
Using (6) and (4) to substitute for 𝑆𝑆(sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) and 𝑊𝑊 cos 𝜙𝜙 respectively, M1 A1

𝑅𝑅 cos 𝛼𝛼 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙 + 𝑅𝑅 sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜙𝜙

= 2𝑅𝑅 cos(𝛼𝛼 − 𝜙𝜙)(sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) + 𝜇𝜇 𝑅𝑅 sin 𝛼𝛼 sin 𝛽𝛽 cos 𝜙𝜙
Thus
cos 𝛼𝛼 (sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) cos 𝜙𝜙 + sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜙𝜙
= 2(cos 𝛼𝛼 cos 𝜙𝜙 + sin 𝛼𝛼 sin 𝜙𝜙)(sin 𝛽𝛽 + 𝜇𝜇 cos 𝛽𝛽 ) + 𝜇𝜇 sin 𝛼𝛼 sin 𝛽𝛽 cos 𝜙𝜙

So, dividing by sin 𝛼𝛼 cos 𝛽𝛽 cos 𝜙𝜙 , M1 A1

cot 𝛼𝛼 (tan 𝛽𝛽 + 𝜇𝜇) + 1 = 2 cot 𝛼𝛼 (tan 𝛽𝛽 + 𝜇𝜇) + 2 tan 𝜙𝜙 (tan 𝛽𝛽 + 𝜇𝜇) + 𝜇𝜇 tan 𝛽𝛽
1 = cot 𝛼𝛼 (tan 𝛽𝛽 + 𝜇𝜇) + 2 tan 𝜙𝜙 (tan 𝛽𝛽 + 𝜇𝜇) + 𝜇𝜇 tan 𝛽𝛽

From (i), cot 𝛼𝛼 = cot 𝛽𝛽 − 2 tan 𝜃𝜃

so M1
1 = (cot 𝛽𝛽 − 2 tan 𝜃𝜃 )(tan 𝛽𝛽 + 𝜇𝜇) + 2 tan 𝜙𝜙 (tan 𝛽𝛽 + 𝜇𝜇) + 𝜇𝜇 tan 𝛽𝛽
1 − 𝜇𝜇 tan 𝛽𝛽 − 1 − 𝜇𝜇 cot 𝛽𝛽 = 2(tan 𝜙𝜙 − tan 𝜃𝜃)(tan 𝛽𝛽 + 𝜇𝜇)
Hence,
 2(tan 𝜃𝜃 − tan 𝜙𝜙)(tan 𝛽𝛽 + 𝜇𝜇) = 𝜇𝜇 tan 𝛽𝛽 + 𝜇𝜇 cot 𝛽𝛽 = 𝜇𝜇(tan 𝛽𝛽 + cot 𝛽𝛽 )

sin 𝛽𝛽 cos 𝛽𝛽 sin2 𝛽𝛽 + cos 2 𝛽𝛽 1

tan 𝛽𝛽 + cot 𝛽𝛽 = + = =
cos 𝛽𝛽 sin 𝛽𝛽 sin 𝛽𝛽 cos 𝛽𝛽 sin 𝛽𝛽 cos 𝛽𝛽
and so,
𝜇𝜇
tan 𝜃𝜃 − tan 𝜙𝜙 =
2 sin 𝛽𝛽 cos 𝛽𝛽(tan 𝛽𝛽 + 𝜇𝜇)
That is
𝜇𝜇
tan 𝜃𝜃 − tan 𝜙𝜙 =
(𝜇𝜇 + tan 𝛽𝛽 ) sin 2𝛽𝛽

as required. A1* (11)

Alternative method using concurrency principle

10. If the extension in the equilibrium position is 𝑑𝑑 , then
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
𝑚𝑚𝑚𝑚 =
𝑎𝑎
𝑎𝑎
Thus, 𝑑𝑑 =
𝑘𝑘

If the extension when the particle is released is 𝑑𝑑 + 𝑥𝑥 , then the equation of motion is
𝑘𝑘𝑘𝑘𝑘𝑘(𝑑𝑑 + 𝑥𝑥 ) 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
𝑚𝑚𝑥𝑥̈ = 𝑚𝑚𝑚𝑚 − = 𝑚𝑚𝑚𝑚 − − =−
𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎
M1 A1
𝑘𝑘𝑘𝑘𝑘𝑘
𝑥𝑥̈ = −
𝑎𝑎
A1
2𝜋𝜋 𝑘𝑘𝑘𝑘
This is simple harmonic motion with period where Ω2 = , i.e. 𝑘𝑘𝑘𝑘 = 𝑎𝑎Ω2 as required. E1 (4)
Ω 𝑎𝑎

Let 𝑦𝑦 be the displacement of the platform below the centre point of its oscillation,

then, 𝑦𝑦 = 𝑏𝑏 − 𝑥𝑥 and 𝑦𝑦̈ = −𝜔𝜔2 𝑦𝑦 = −𝜔𝜔2 (𝑏𝑏 − 𝑥𝑥 ) (x newly defined as in question) M1 A1

Thus, the equation of motion of the particle becomes

𝑘𝑘𝑘𝑘𝑘𝑘(ℎ − 𝑎𝑎 − 𝑥𝑥 )
𝑚𝑚 𝑦𝑦̈ = 𝑚𝑚𝑚𝑚 − 𝑅𝑅 −
𝑎𝑎
M1A1

So,
𝑚𝑚𝑚𝑚Ω2 (ℎ − 𝑎𝑎 − 𝑥𝑥 )
−𝑚𝑚𝜔𝜔2 (𝑏𝑏 − 𝑥𝑥 ) = 𝑚𝑚𝑚𝑚 − 𝑅𝑅 −
𝑎𝑎
M1 A1

That is,
𝑅𝑅 = 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 + 𝑥𝑥 − ℎ) + 𝑚𝑚𝜔𝜔2 (𝑏𝑏 − 𝑥𝑥 )

as required. A1* (7)

To remain in contact, 𝑅𝑅 ≥ 0 for 0 ≤ 𝑥𝑥 ≤ 2𝑏𝑏

𝑅𝑅 = 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 + 𝑚𝑚(Ω2 − 𝜔𝜔2 )𝑥𝑥

M1

so if 𝜔𝜔 < Ω , the minimum value of 𝑅𝑅 is 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 , (when 𝑥𝑥 = 0) M1

thus 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 ≥ 0

Rearranging,
𝑔𝑔 + 𝜔𝜔2 𝑏𝑏 𝑎𝑎 𝜔𝜔2 𝑏𝑏 1 𝜔𝜔2 𝑏𝑏
ℎ≤ + 𝑎𝑎 = + + 𝑎𝑎 = 𝑎𝑎 �1 + � +
Ω2 𝑘𝑘 Ω2 𝑘𝑘 Ω2
as required. A1* (3)

If 𝜔𝜔 > Ω , minimum value of 𝑅𝑅 is 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 + 2𝑚𝑚𝑚𝑚(Ω2 − 𝜔𝜔2 ) (at 𝑥𝑥 = 2𝑏𝑏) M1
Thus,
𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 + 2𝑚𝑚𝑚𝑚(Ω2 − 𝜔𝜔2 ) ≥ 0
and so,

𝑔𝑔 + 𝜔𝜔2 𝑏𝑏 2𝑏𝑏(Ω2 − 𝜔𝜔2 ) 1 𝜔𝜔2 𝑏𝑏 2𝜔𝜔2 𝑏𝑏 1 𝜔𝜔2 𝑏𝑏

ℎ≤ + 𝑎𝑎 + = 𝑎𝑎 �1 + � + + 2𝑏𝑏 − = 𝑎𝑎 �1 + � − + 2𝑏𝑏
Ω2 Ω2 𝑘𝑘 Ω2 Ω2 𝑘𝑘 Ω2
A1 (2)
1 𝜔𝜔2 𝑏𝑏 1
Thus, if 𝜔𝜔 < Ω , ℎ ≤ 𝑎𝑎 �1 + � + < 𝑎𝑎 �1 + � + 𝑏𝑏 ; M1
𝑘𝑘 Ω2 𝑘𝑘

1 𝜔𝜔2 𝑏𝑏 1
if 𝜔𝜔 > Ω , ℎ ≤ 𝑎𝑎 �1 + � − + 2𝑏𝑏 < 𝑎𝑎 �1 + � + 𝑏𝑏 M1
𝑘𝑘 Ω2 𝑘𝑘

If 𝜔𝜔 = Ω , then 𝑅𝑅 = 𝑚𝑚𝑚𝑚 + 𝑚𝑚Ω2 (𝑎𝑎 − ℎ) + 𝑚𝑚𝜔𝜔2 𝑏𝑏 ≥ 0 , M1

so,
1
ℎ ≤ 𝑎𝑎 �1 + � + 𝑏𝑏
𝑘𝑘
A1* (4)

Alternative

stem measuring y below A

𝑘𝑘𝑘𝑘𝑘𝑘(𝑦𝑦 − 𝑎𝑎)
𝑚𝑚𝑦𝑦̈ = 𝑚𝑚𝑚𝑚 −
𝑎𝑎
𝑘𝑘𝑘𝑘 1
𝑦𝑦̈ = − 𝑦𝑦 + 𝑘𝑘𝑘𝑘 �1 + �
𝑎𝑎 𝑘𝑘

and for platform introduced

𝑘𝑘𝑘𝑘𝑘𝑘(𝑦𝑦 − 𝑎𝑎)
𝑚𝑚 𝑦𝑦̈ = 𝑚𝑚𝑚𝑚 − 𝑅𝑅 −
𝑎𝑎
11. (i)
𝑃𝑃(𝑌𝑌 ≤ 𝑦𝑦) = 𝑃𝑃(𝑓𝑓 (𝑋𝑋) ≤ 𝑦𝑦)

= 𝑃𝑃�𝑋𝑋 ≥ 𝑓𝑓 −1 (𝑦𝑦)�

E1

as f is a strictly decreasing function

= 𝑃𝑃�𝑋𝑋 ≥ 𝑓𝑓(𝑦𝑦)�

E1
𝑏𝑏 − 𝑓𝑓 (𝑦𝑦)
=
𝑏𝑏 − 𝑎𝑎
because X is uniformly distributed on [a,b]. M1 A1* (4)

Thus, the pdf of Y is

𝑑𝑑 𝑏𝑏 − 𝑓𝑓(𝑦𝑦) −𝑓𝑓′(𝑦𝑦)
� �=
𝑑𝑑𝑑𝑑 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎

𝑦𝑦 ∈ [𝑎𝑎, 𝑏𝑏] M1
𝑏𝑏 𝑏𝑏𝑏𝑏
2)
−𝑓𝑓′(𝑦𝑦)
2
−𝑓𝑓(𝑦𝑦) −𝑓𝑓(𝑦𝑦)
𝐸𝐸 (𝑌𝑌 = � 𝑦𝑦 𝑑𝑑𝑑𝑑 = �𝑦𝑦 2 � − � 2𝑦𝑦 𝑑𝑑𝑑𝑑
𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑏𝑏 − 𝑎𝑎

by integration by parts M1 M1
𝑏𝑏 𝑏𝑏 𝑏𝑏
−𝑎𝑎𝑏𝑏 2 + 𝑎𝑎2 𝑏𝑏 𝑓𝑓 (𝑥𝑥 ) 𝑎𝑎𝑎𝑎(𝑎𝑎 − 𝑏𝑏) 𝑓𝑓(𝑥𝑥 ) 𝑓𝑓 (𝑥𝑥 )
= + � 2𝑥𝑥 𝑑𝑑𝑑𝑑 = + � 2𝑥𝑥 𝑑𝑑𝑑𝑑 = −𝑎𝑎𝑎𝑎 + � 2𝑥𝑥 𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑏𝑏 − 𝑎𝑎

as required. M1 A1* (5)

(ii) Considering Z as a function of X, it satisfies the three conditions for the function f in part (i), as
trivially by the definition of c the first is satisfied, considering the graph or the derivative the second
is, and by symmetry, the third is. E1
1 1 1
+ =
𝑍𝑍 𝑋𝑋 𝑐𝑐
so
1 1 1 𝑋𝑋 − 𝑐𝑐
= − =
𝑍𝑍 𝑐𝑐 𝑋𝑋 𝑐𝑐𝑐𝑐
and therefore
𝑐𝑐𝑐𝑐
𝑍𝑍 =
𝑋𝑋 − 𝑐𝑐
Therefore,
𝑏𝑏 𝑏𝑏 𝑏𝑏
𝑐𝑐𝑐𝑐 1 𝑐𝑐 𝑥𝑥 − 𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑐𝑐
𝐸𝐸 (𝑍𝑍) = � 𝑑𝑑𝑑𝑑 = � + 𝑑𝑑𝑑𝑑 = � 1+ 𝑑𝑑𝑑𝑑
𝑎𝑎 𝑥𝑥 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑥𝑥 − 𝑐𝑐 𝑥𝑥 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑥𝑥 − 𝑐𝑐
 𝑐𝑐
= [𝑥𝑥 + c ln(𝑥𝑥 − 𝑐𝑐 )]𝑏𝑏𝑎𝑎
𝑏𝑏 − 𝑎𝑎
M1

𝑐𝑐 2 𝑏𝑏 − 𝑐𝑐
= 𝑐𝑐 + ln � �
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐
A1

From (i),
𝑏𝑏
𝑐𝑐𝑐𝑐 2𝑥𝑥
𝐸𝐸 (𝑍𝑍 2 ) = −𝑎𝑎𝑎𝑎 + � 𝑑𝑑𝑑𝑑
𝑎𝑎 𝑥𝑥 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎

M1
𝑏𝑏 𝑏𝑏 2
2𝑐𝑐 𝑥𝑥 2 2𝑐𝑐 𝑥𝑥 − 𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥 − 𝑐𝑐 2 𝑐𝑐 2
= −𝑎𝑎𝑎𝑎 + � 𝑑𝑑𝑑𝑑 = −𝑎𝑎𝑎𝑎 + � + + 𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑥𝑥 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑥𝑥 − 𝑐𝑐 𝑥𝑥 − 𝑐𝑐 𝑥𝑥 − 𝑐𝑐

M1
𝑏𝑏
2𝑐𝑐 𝑥𝑥 2
= −𝑎𝑎𝑎𝑎 + � + 𝑐𝑐𝑐𝑐 + 𝑐𝑐 2 ln(𝑥𝑥 − 𝑐𝑐 )�
𝑏𝑏 − 𝑎𝑎 2 𝑎𝑎

2𝑐𝑐 3 𝑏𝑏 − 𝑐𝑐
= −𝑎𝑎𝑎𝑎 + 𝑐𝑐 (𝑎𝑎 + 𝑏𝑏) + 2𝑐𝑐 2 + ln � �
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐
A1

2𝑐𝑐 3 𝑏𝑏 − 𝑐𝑐 2𝑐𝑐 3 𝑏𝑏 − 𝑐𝑐
= −𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 2𝑐𝑐 2 + ln � � = 2𝑐𝑐 2 + ln � �
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐
E1

Thus,
2
2𝑐𝑐 3
2
𝑏𝑏 − 𝑐𝑐 𝑐𝑐 2 𝑏𝑏 − 𝑐𝑐
𝑉𝑉𝑉𝑉𝑉𝑉(𝑍𝑍) = 2𝑐𝑐 + ln � � − �𝑐𝑐 + ln � ��
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐

M1
2
2
𝑐𝑐 2 𝑏𝑏 − 𝑐𝑐
= 𝑐𝑐 − � ln � ��
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐

A1ft

As 𝑉𝑉𝑉𝑉𝑉𝑉(𝑍𝑍) > 0 , (Z is not a constant)

2
2
𝑐𝑐 2 𝑏𝑏 − 𝑐𝑐
𝑐𝑐 − � ln � �� > 0
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐

M1
𝑐𝑐 2 𝑏𝑏−𝑐𝑐
and so as 𝑐𝑐 and ln � � are both positive,
𝑏𝑏−𝑎𝑎 𝑎𝑎−𝑐𝑐
 𝑐𝑐 2 𝑏𝑏 − 𝑐𝑐
𝑐𝑐 > ln � �
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 − 𝑐𝑐
and similarly,
𝑏𝑏 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎
ln � � <
𝑎𝑎 − 𝑐𝑐 𝑐𝑐
as required. A1* (11)
12. (i) 𝑃𝑃(𝑋𝑋 = 𝑥𝑥 ) = 𝑞𝑞 𝑥𝑥−1 𝑝𝑝 and 𝑃𝑃(𝑌𝑌 = 𝑦𝑦) = 𝑞𝑞 𝑦𝑦−1 𝑝𝑝 for 𝑥𝑥, 𝑦𝑦 ≥ 1
𝑠𝑠−1 𝑠𝑠−1

𝑃𝑃(𝑆𝑆 = 𝑠𝑠) = 𝑃𝑃(𝑋𝑋 + 𝑌𝑌 = 𝑠𝑠) = � 𝑃𝑃(𝑋𝑋 = 𝑥𝑥, 𝑌𝑌 = 𝑠𝑠 − 𝑥𝑥 ) = � 𝑞𝑞 𝑥𝑥−1 𝑝𝑝𝑞𝑞 𝑠𝑠−𝑥𝑥−1 𝑝𝑝

𝑥𝑥=1 𝑥𝑥=1
𝑠𝑠−1

= � 𝑞𝑞 𝑠𝑠−2 𝑝𝑝2 = (𝑠𝑠 − 1)𝑝𝑝2 𝑞𝑞 𝑠𝑠−2

𝑥𝑥=1

for 𝑠𝑠 ≥ 2 . M1 A1 (2)

𝑃𝑃(𝑇𝑇 = 𝑡𝑡) = 𝑃𝑃(𝑋𝑋 = 𝑡𝑡, 𝑌𝑌 ≤ 𝑡𝑡) + 𝑃𝑃(𝑌𝑌 = 𝑡𝑡, 𝑋𝑋 ≤ 𝑡𝑡) − 𝑃𝑃(𝑋𝑋 = 𝑡𝑡, 𝑌𝑌 = 𝑡𝑡)

M1
𝑡𝑡
𝑡𝑡−1
= 2𝑞𝑞 𝑝𝑝 � 𝑞𝑞 𝑦𝑦−1 𝑝𝑝 − 𝑞𝑞 𝑡𝑡−1 𝑝𝑝𝑞𝑞 𝑡𝑡−1 𝑝𝑝
𝑦𝑦=1

M1

1 − 𝑞𝑞 𝑡𝑡
= 2𝑞𝑞 𝑡𝑡−1 𝑝𝑝2 − 𝑞𝑞 2𝑡𝑡−2 𝑝𝑝2 = 2𝑞𝑞 𝑡𝑡−1 𝑝𝑝(1 − 𝑞𝑞 𝑡𝑡 ) − 𝑞𝑞 2𝑡𝑡−2 𝑝𝑝2
1 − 𝑞𝑞
A1

= 𝑞𝑞 𝑡𝑡−1 𝑝𝑝 (2 − 2𝑞𝑞 𝑡𝑡 − 𝑞𝑞 𝑡𝑡−1 𝑝𝑝) = 𝑝𝑝𝑞𝑞 𝑡𝑡−1 (2 − 2𝑞𝑞 𝑡𝑡 − (1 − 𝑞𝑞 )𝑞𝑞 𝑡𝑡−1 ) = 𝑝𝑝𝑞𝑞 𝑡𝑡−1 (2 − 𝑞𝑞 𝑡𝑡−1 − 𝑞𝑞 𝑡𝑡 )
for 𝑡𝑡 ≥ 1 as required. A1* (4)

(ii)
∞ ∞

𝑃𝑃(𝑈𝑈 = 𝑢𝑢) = � 𝑃𝑃(𝑋𝑋 = 𝑥𝑥, 𝑌𝑌 = 𝑥𝑥 + 𝑢𝑢) + � 𝑃𝑃(𝑌𝑌 = 𝑥𝑥, 𝑋𝑋 = 𝑥𝑥 + 𝑢𝑢)

𝑥𝑥=1 𝑥𝑥=1

for 𝑢𝑢 ≥ 1
∞ ∞
𝑥𝑥−1 𝑥𝑥+𝑢𝑢−1
1
= 2 � 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 = 2𝑝𝑝 𝑞𝑞 � 𝑞𝑞 2𝑥𝑥−2 = 2𝑝𝑝2 𝑞𝑞 𝑢𝑢
2 𝑢𝑢
1 − 𝑞𝑞 2
𝑥𝑥=1 𝑥𝑥=1

M1
1 2𝑝𝑝 𝑞𝑞 𝑢𝑢
= 2𝑝𝑝2 𝑞𝑞 𝑢𝑢 =
𝑝𝑝(1 + 𝑞𝑞 ) (1 + 𝑞𝑞 )
A1

and
∞ ∞ ∞
𝑥𝑥−1 𝑥𝑥−1 2 2𝑥𝑥−2
𝑝𝑝2 𝑝𝑝
𝑃𝑃(𝑈𝑈 = 0) = � 𝑃𝑃(𝑋𝑋 = 𝑥𝑥, 𝑌𝑌 = 𝑥𝑥 ) = � 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 = 𝑝𝑝 � 𝑞𝑞 = =
1 − 𝑞𝑞 2 1 + 𝑞𝑞
𝑥𝑥=1 𝑥𝑥=1 𝑥𝑥=1

B1 (3)
 𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = 𝑃𝑃(𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 ≥ 𝑤𝑤) + 𝑃𝑃(𝑌𝑌 = 𝑤𝑤, 𝑋𝑋 ≥ 𝑤𝑤) − 𝑃𝑃(𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 = 𝑤𝑤)

for 𝑤𝑤 ≥ 1
∞ ∞
𝑤𝑤−1 𝑤𝑤+𝑦𝑦−1 𝑤𝑤−1 𝑤𝑤−1 2 2𝑤𝑤−2
= 2 � 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 − 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 = 2𝑝𝑝 𝑞𝑞 � 𝑞𝑞 𝑦𝑦 − 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2
𝑦𝑦=0 𝑦𝑦−0

M1
2 2 1 + 𝑞𝑞
= 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 − 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 � − 1� = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2
1 − 𝑞𝑞 1 − 𝑞𝑞 1 − 𝑞𝑞
= 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 )

A1 (2)

(iii) 𝑆𝑆 = 2 ⇒ 𝑋𝑋 = 1 , 𝑌𝑌 = 1 and 𝑇𝑇 = 3 ⇒ 𝑋𝑋 = 3 or 𝑌𝑌 = 3 or both

Thus,

𝑃𝑃(𝑆𝑆 = 2 , 𝑇𝑇 = 3) = 0 E1

However,

𝑃𝑃(𝑆𝑆 = 2) = 𝑝𝑝2 ≠ 0 and 𝑃𝑃(𝑇𝑇 = 3) = 𝑝𝑝𝑞𝑞 2 (2 − 𝑞𝑞 2 − 𝑞𝑞 3 ) = 𝑝𝑝𝑞𝑞 2 (1 − 𝑞𝑞 2 + 1 − 𝑞𝑞 3 )

= 𝑝𝑝𝑞𝑞 2 �𝑝𝑝(1 + 𝑞𝑞 ) + 𝑝𝑝(1 + 𝑞𝑞 + 𝑞𝑞 2 )�

= 𝑝𝑝2 𝑞𝑞 2 (2 + 2𝑞𝑞 + 𝑞𝑞 2 ) ≠ 0

and so, 𝑃𝑃(𝑆𝑆 = 2 , 𝑇𝑇 = 3) ≠ 𝑃𝑃(𝑆𝑆 = 2) × 𝑃𝑃(𝑇𝑇 = 3) E1 (2)

(iv)

𝑈𝑈 = 𝑢𝑢 and 𝑊𝑊 = 𝑤𝑤 ⇒ 𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 = 𝑤𝑤 + 𝑢𝑢 or 𝑌𝑌 = 𝑤𝑤 , 𝑋𝑋 = 𝑤𝑤 + 𝑢𝑢 for 𝑢𝑢 > 0 M1

So 𝑃𝑃(𝑈𝑈 = 𝑢𝑢 , 𝑊𝑊 = 𝑤𝑤) = 2𝑞𝑞 𝑤𝑤−1 𝑝𝑝𝑞𝑞 𝑤𝑤+𝑢𝑢−1 𝑝𝑝 = 2𝑝𝑝2 𝑞𝑞 2𝑤𝑤+𝑢𝑢−2 M1

2𝑝𝑝 𝑞𝑞 𝑢𝑢
𝑃𝑃(𝑈𝑈 = 𝑢𝑢) =
(1 + 𝑞𝑞 )
and
𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 )

so
2𝑝𝑝 𝑞𝑞 𝑢𝑢
𝑃𝑃(𝑈𝑈 = 𝑢𝑢) × 𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = × 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 ) = 2𝑝𝑝2 𝑞𝑞 2𝑤𝑤+𝑢𝑢−2 = 𝑃𝑃(𝑈𝑈 = 𝑢𝑢 , 𝑊𝑊 = 𝑤𝑤)
(1 + 𝑞𝑞 )

A1

In the case 𝑢𝑢 = 0 ,

𝑈𝑈 = 0 and 𝑊𝑊 = 𝑤𝑤 ⇒ 𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 = 𝑤𝑤

so 𝑃𝑃(𝑈𝑈 = 0 , 𝑊𝑊 = 𝑤𝑤) = 𝑞𝑞 𝑤𝑤−1 𝑝𝑝 𝑞𝑞 𝑤𝑤−1 𝑝𝑝 = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 M1

𝑝𝑝
whereas 𝑃𝑃(𝑈𝑈 = 0) = and 𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 )
1+𝑞𝑞
so
𝑝𝑝
𝑃𝑃(𝑈𝑈 = 0) × 𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = × 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 ) = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 = 𝑃𝑃 (𝑈𝑈 = 0 , 𝑊𝑊 = 𝑤𝑤)
1 + 𝑞𝑞
Thus, U and W are independent. A1 (5)

Pairing S and U – consider 𝑆𝑆 = 2 , 𝑈𝑈 = 3

𝑆𝑆 = 2 ⇒ 𝑋𝑋 = 1 , 𝑌𝑌 = 1 which would imply 𝑈𝑈 = 0

2𝑝𝑝 𝑞𝑞 3
Thus 𝑃𝑃(𝑆𝑆 = 2) = 𝑝𝑝2 ≠ 0 and 𝑃𝑃(𝑈𝑈 = 3) = (1+𝑞𝑞) ≠ 0 whereas 𝑃𝑃(𝑆𝑆 = 2 , 𝑈𝑈 = 3) = 0 so

𝑃𝑃(𝑆𝑆 = 2 , 𝑈𝑈 = 3) ≠ 𝑃𝑃(𝑆𝑆 = 2) × 𝑃𝑃(𝑈𝑈 = 3) and S and U are not independent.

Pairing S and W - consider 𝑆𝑆 = 2 , 𝑊𝑊 = 3

𝑆𝑆 = 2 ⇒ 𝑋𝑋 = 1 , 𝑌𝑌 = 1 which would imply 𝑊𝑊 = 1

so 𝑃𝑃(𝑆𝑆 = 2 , 𝑊𝑊 = 3) = 0 whereas 𝑃𝑃(𝑆𝑆 = 2) ≠ 0 and 𝑃𝑃(𝑊𝑊 = 3) ≠ 0 so S and W are not

independent.

Pairing T and U - consider 𝑇𝑇 = 1 , 𝑈𝑈 = 1

𝑇𝑇 = 1 ⇒ 𝑋𝑋 = 1 , 𝑌𝑌 = 1 which would imply 𝑈𝑈 = 0

so 𝑃𝑃(𝑇𝑇 = 1 , 𝑈𝑈 = 1) = 0 whereas 𝑃𝑃(𝑇𝑇 = 1) ≠ 0 and 𝑃𝑃(𝑈𝑈 = 1) ≠ 0 so T and U are not

independent.

Pairing T and W - consider 𝑇𝑇 = 1 , 𝑊𝑊 = 2

𝑇𝑇 = 1 ⇒ 𝑋𝑋 = 1 , 𝑌𝑌 = 1 which would imply 𝑊𝑊 = 1

so 𝑃𝑃(𝑇𝑇 = 1 , 𝑊𝑊 = 2) = 0 whereas 𝑃𝑃(𝑇𝑇 = 1) ≠ 0 and 𝑃𝑃(𝑊𝑊 = 2) ≠ 0 so T and W are not

independent. B2 (2)

Alternative (i)

𝑃𝑃(𝑇𝑇 = 𝑡𝑡) = 𝑃𝑃(𝑋𝑋 = 𝑡𝑡, 𝑌𝑌 < 𝑡𝑡) + 𝑃𝑃(𝑌𝑌 = 𝑡𝑡, 𝑋𝑋 < 𝑡𝑡) + 𝑃𝑃(𝑋𝑋 = 𝑡𝑡, 𝑌𝑌 = 𝑡𝑡)

[or 𝑃𝑃(𝑇𝑇 = 𝑡𝑡) = 𝑃𝑃(𝑋𝑋 = 𝑡𝑡, 𝑌𝑌 < 𝑡𝑡) + 𝑃𝑃(𝑌𝑌 = 𝑡𝑡, 𝑋𝑋 ≤ 𝑡𝑡) ]

M1
𝑡𝑡−1
𝑡𝑡−1
= 2𝑞𝑞 𝑝𝑝 � 𝑞𝑞 𝑦𝑦−1 𝑝𝑝 + 𝑞𝑞 𝑡𝑡−1 𝑝𝑝𝑞𝑞 𝑡𝑡−1 𝑝𝑝
𝑦𝑦=1

M1

1 − 𝑞𝑞 𝑡𝑡−1
= 2𝑞𝑞 𝑡𝑡−1 𝑝𝑝2 + 𝑞𝑞 2𝑡𝑡−2 𝑝𝑝2 = 2𝑞𝑞 𝑡𝑡−1 𝑝𝑝(1 − 𝑞𝑞 𝑡𝑡−1 ) + 𝑞𝑞 2𝑡𝑡−2 𝑝𝑝2
1 − 𝑞𝑞
A1
 = 𝑞𝑞 𝑡𝑡−1 𝑝𝑝 (2 − 2𝑞𝑞 𝑡𝑡−1 + 𝑞𝑞 𝑡𝑡−1 𝑝𝑝) = 𝑝𝑝𝑞𝑞 𝑡𝑡−1 (2 − 2𝑞𝑞 𝑡𝑡−1 + (1 − 𝑞𝑞 )𝑞𝑞 𝑡𝑡−1 ) = 𝑝𝑝𝑞𝑞 𝑡𝑡−1 (2 − 𝑞𝑞 𝑡𝑡−1 − 𝑞𝑞 𝑡𝑡 )
for 𝑡𝑡 ≥ 1 as required. A1* (4)

(ii)

𝑃𝑃(𝑊𝑊 = 𝑤𝑤) = 𝑃𝑃(𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 > 𝑤𝑤) + 𝑃𝑃(𝑌𝑌 = 𝑤𝑤, 𝑋𝑋 > 𝑤𝑤) + 𝑃𝑃 (𝑋𝑋 = 𝑤𝑤, 𝑌𝑌 = 𝑤𝑤)

for 𝑤𝑤 ≥ 1
∞ ∞
𝑤𝑤−1 𝑤𝑤+𝑦𝑦−1 𝑤𝑤−1 𝑤𝑤−1 2 2𝑤𝑤−2
= 2 � 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 + 𝑞𝑞 𝑝𝑝 𝑞𝑞 𝑝𝑝 = 2𝑝𝑝 𝑞𝑞 � 𝑞𝑞 𝑦𝑦 + 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2
𝑦𝑦=1 𝑦𝑦−1

M1
2 2𝑞𝑞 1 + 𝑞𝑞
= 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−1 + 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2 � + 1� = 𝑝𝑝2 𝑞𝑞 2𝑤𝑤−2
1 − 𝑞𝑞 1 − 𝑞𝑞 1 − 𝑞𝑞
= 𝑝𝑝𝑞𝑞 2𝑤𝑤−2 (1 + 𝑞𝑞 )

A1 (2)

Multiple alternatives for counter-examples for (iv)

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