CAPE 2017 Paper 2 SOLUTIONS
QUESTION 1
(a) (i) ~𝑝 ∨ 𝑞
(ii) 𝑝 → ~𝑞
(b) (i) ∗ is commutative since 2 ∗ 1 = 1 ∗ 2
The table is symmetric about the diagonal hence * is commutative or
showing that a*b =b*a for all a,b.
(ii) 1 ∗ 3 = 1
2∗3 =2
3*a=a*3 for all a in {1, 2,3,4} therefore 3 is the identity element of *.
3∗3 =3
4∗3 =4
(c) (i) 𝑓(𝑥) = 𝑎𝑥 + 9𝑥 − 11𝑥 + 𝑏
(ii) 𝑓(𝑥) = 2𝑥 + 9𝑥 − 11𝑥 − 30
𝑓(2) = 0
𝑎(2) + 9(2) − 11(2) + 𝑏 = 0
8𝑎 + 𝑏 = −14
𝑓(−2) = 12
𝑎(−2) + 9(−2) − 11(−2) + 𝑏 = 12
𝑓(𝑥) = (𝑥 − 2)(2𝑥 + 13𝑥 + 15)
𝑓(𝑥) = (𝑥 − 2)(2𝑥 + 3)(𝑥 + 5)
−8𝑎 + 𝑏 = −46
3
𝑥 = −5, − , 2
2
2𝑏 = −60
𝑏 = −30
𝑎=2
(d) Let 𝑃 : ∑
8𝑟 = 4𝑛(𝑛 + 1)
∀𝑛∈ℕ
𝑃 : 8(1) = 4(1)(1 + 1)
8=8
Therefore 𝑃 is true.
40
Assume 𝑃 is true for 𝑛 = 𝑘
𝑃 :
8𝑟 = 4𝑘(𝑘 + 1)
𝑃
:
8𝑟 = 4(𝑘 + 1)(𝑘 + 2)
𝑃
= 𝑃 + (𝑘 + 1)term
= 4𝑘(𝑘 + 1) + 8(𝑘 + 1)
= (𝑘 + 1)(4𝑘 + 8)
= 4(𝑘 + 1)(𝑘 + 2)
Therefore 𝑃
is true ∀ 𝑃 is true.
Hence by mathematical induction 8 + 11 + 24 + 32 + ⋯ + 8𝑛 = 4𝑛(𝑛 + 1) for all 𝑛 ∈ ℕ.
QUESTION 2
2.
(𝑎 + 𝑏) = 𝑎 + 2𝑎𝑏 + 𝑏
(a) (i)
ln
𝑎+𝑏
1
= (ln 𝑎 + ln 𝑏)
4
2
𝑎+𝑏
1
ln
= (ln 𝑎𝑏)
4
2
ln
𝑎+𝑏
= ln(𝑎𝑏)
4
𝑎+𝑏
= √𝑎𝑏
4
Therefore
𝑎 + 𝑏 = 14𝑎𝑏
𝑎 + 𝑏 − 14𝑎𝑏 = 0
𝑎 + 𝑏 + 2𝑎𝑏 − 16𝑎𝑏 = 0
(𝑎 + 𝑏) = 16𝑎𝑏
(𝑎 + 𝑏)
= 𝑎𝑏
16
𝑎 + 𝑏 = 4√𝑎𝑏
(𝑎 + 𝑏) = 16𝑎𝑏
𝑎 + 2𝑎𝑏 + 𝑏 = 16𝑎𝑏
𝑎+𝑏
4
ln
𝑎 + 𝑏 = 14𝑎𝑏
𝑎+𝑏
4
2 ln
Alternately
ln
= 𝑎𝑏
= ln 𝑎𝑏
𝑎+𝑏
= ln 𝑎 + ln 𝑏
4
𝑎+𝑏
1
= (ln 𝑎 + ln 𝑏)
4
2
41
(ii) 2
+ 3(2 ) = 4
2 =
1
+ 3(2 ) = 4
2
ln 2 = ln
1
+ 3𝑦 − 4 = 0
𝑦
1
3
1
3 = − ln 3
𝑥=
ln 2
ln 2
ln
3𝑦 − 4𝑦 + 1 = 0
(3𝑦 − 1)(𝑦 − 1) = 0
𝑦=
1
3
2 =1
𝑥=0
1
,1
3
(b)
(c) 𝛼 + 𝛽 + 𝛾 = − = 0
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 =
𝛼𝛽𝛾 = −
𝑐
=3
𝑎
𝑑
= −2
𝑎
𝑥 − (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)𝑥 + (𝛼𝛽)(𝛼𝛾) + (𝛼𝛽)(𝛽𝛾) + (𝛼𝛾)(𝛽𝛾) 𝑥 − (𝛼𝛽)(𝛼𝛾)(𝛽𝛾) = 0
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = 3
(𝛼𝛽)(𝛼𝛾) + (𝛼𝛽)(𝛽𝛾) + (𝛼𝛾)(𝛽𝛾)
= 𝛼 𝛽𝛾 + 𝛽 𝛼𝛾 + 𝛾 𝛼𝛽
= (𝛼𝛽𝛾)(𝛼 + 𝛽 + 𝛾)
= (−2)(0)
42
=0
(𝛼𝛽)(𝛼𝛾)(𝛽𝛾)
= (𝛼𝛽𝛾)
= (−2)
=4
𝑥 − 3𝑥 − 4 = 0
QUESTION 3
3.
(ii) sin 𝐴 =
(a) (i) tan(𝐴 + 𝐵) =
tan 𝐴 =
sin(𝐴 + 𝐵)
cos(𝐴 + 𝐵)
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
=
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
sin 𝐴 cos 𝐵 sin 𝐵 cos 𝐴
=
+
cos 𝐴 cos 𝐵 cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 sin 𝐴 sin 𝐵
÷
−
cos 𝐴 cos 𝐵 cos 𝐴 cos 𝐵
tan 𝐴 + tan 𝐵
=
1 − tan 𝐴 tan 𝐵
3
4
cos 𝐵 = −
1
2
tan 𝐵 = −√3
tan(𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
3
+ −√3
= 4
3
1−
−√3
4
ALTERNATELY
=
3
3√3
− √3 ÷ 1 +
4
4
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
=
3 − 4√3
4 + 3√3
×
4
4
=
sin 𝐴 sin 𝐵
sin 𝐴
+
÷ 1−
cos 𝐴 cos 𝐵
cos 𝐴
sin 𝐵
cos 𝐵
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
=
÷
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵
=
=
=
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
sin(𝐴 + 𝐵)
=
cos(𝐴 + 𝐵)
4 + 3√3
3 − 4√3 4 − 3√3
4 + 3√3 4 − 3√3
=
12 − 9√3 − 16√3 + 36
16 − 27
=
48 − 25√3
−11
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵
×
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
=
3 − 4√3
=−
48 25√3
+
11
11
= tan(𝐴 + 𝐵)
43
(b) sin 𝜃 − 2 cos 𝜃 + 3 cos 𝜃 + 5 = 0
(1 − cos 𝜃) − 2 cos 𝜃 + 3 cos 𝜃 + 5 = 0
−3 cos 𝜃 + 3 cos 𝜃 + 6 = 0
cos 𝜃 − cos 𝜃 − 2 = 0
(cos 𝜃 − 2)(cos 𝜃 + 1) = 0
cos 𝜃 = −1
𝜃=𝜋
𝜃 = 3𝜋
By using the graph of cos 𝜃.
(c) (i) 𝑓(𝜃) = 6 cos 𝜃 + 8 sin 𝜃
𝑟 = √6 + 8 = 10
6
𝛼 = tan
= 36.87°
8
𝑓(𝜃) = 10 sin(𝜃 + 36.87°)
(ii) 10 sin(𝜃 + 36.87°) = 2
sin(𝜃 + 36.87°) = 0.2
𝑅𝐴 = sin (0.2) = 11.54°
𝐼: 𝜃 + 36.87° = 11.54°
𝜃 = −25.33°
𝐼𝐼: 𝜃 + 36.87° = 180° − 11.54°
= 168.46°
𝜃 = 131.59°
General solution:
𝜃 = −25.33 + 360°𝑛, 𝑛 ∈ ℤ
𝜃 = 131.69° + 360°𝑛, 𝑛 ∈ ℤ
QUESTION 4
(a) (i) 𝑥 + 𝑦 − 4𝑥 + 2𝑦 − 2 = 0
(ii) 𝑥 + 𝑦 − 4𝑥 + 2𝑦 + 2 = 0
(𝑥 − 2) + (𝑦 + 1) = 2 + 4 + 1
𝑥 + 3𝑦 = 3
(𝑥 − 2) + (𝑦 + 1) = 7
(3 − 3𝑦) + 𝑦 − 4(3 − 3𝑦) + 2𝑦 − 2 = 0
Centre (2, −1)
+ −1 − (−2)
𝑥 = 3 − 3𝑦
9𝑦 − 18𝑦 + 9 + 𝑦 − 12 + 12𝑦 + 2𝑦 − 2 = 0
Radius of 𝐶 =
2 − (−1)
→
10𝑦 − 4𝑦 − 5 = 0
= √10
𝐶 : (𝑥 − 2) + (𝑦 + 1) = 10
𝑏 − 4𝑎𝑐
(−4) − 4(10)(−5) > 0
Since discriminant is positive the
quadratic has 2 distinct roots therefore
the 𝐿 cannot be a tangent to 𝐶 .
(b) (i) 𝑃𝑄⃗ = −2𝑖 − 3𝑗 + 2𝑘
(ii) 𝑟. 𝑛 = 𝑎. 𝑛
𝑥
−2
1
−2
𝑦 . −3 = −2 . −3
𝑧
2
4
2
−2𝑥 − 3𝑦 + 2𝑧 = −2 + 6 + 8
−2𝑥 − 3𝑦 + 2𝑧 = 12
44
(c) 𝐿 = −𝑖 + 𝑗 − 2𝑘 + 𝛼(−2𝑖 + 𝑗 − 3𝑘)
Solving (1) and (2) simultaneously
𝐿 = −2𝑖 + 𝑗 − 4𝑘 + 𝛽(𝑖 − 𝑗 + 𝑘)
2𝛼 + 𝛽 = 1
−𝑖 + 𝑗 − 2𝑘 − 2𝛼𝑖 + 𝛼𝑗 − 3𝛼𝑘
= −2𝑖 + 𝑗 − 4𝑘 + 𝛽𝑖 − 𝛽𝑗
+ 𝛽𝑘
𝛼+𝛽 =0
(−1 − 2𝛼)𝑖 = (−2 + 𝛽)𝑖
𝛽 = −1
−1 − 2𝛼 = −2 + 𝛽
Checking in (3)
2𝛼 + 𝛽 = 1
(1)
𝛼=1
3(1) + (−1) = 2
Therefore 𝐿 and 𝐿 intersect.
(ii) 𝐿 = −𝑖 + 𝑗 − 2𝑘 − 2𝑖 + 𝑗 − 3𝑘 since 𝛼 =
1
(1 + 𝛼)𝑗 = (1 − 𝛽)𝑗
1+𝛼 =1−𝛽
𝐿 = −3𝑖 + 2𝑗 − 5𝑘
𝛼+𝛽 =0
(2)
When 𝛽 = −1
𝐿 = −2𝑖 + 𝑗 − 4𝑘 − 𝑖 + 𝑗 − 𝑘
(−2 − 3𝛼)𝑘 = (−4 + 𝛽)𝑘
𝐿 = −3𝑖 + 2𝑗 − 5𝑘
−2 − 3𝛼 = −4 + 𝛽
Point of intersection is −3𝑖 + 2𝑗 − 5𝑘
3𝛼 + 𝛽 = 2
(3)
QUESTION 5
(a) lim
→
lim
→
=𝑘
(𝑥 − 1)(𝑥 + 𝑥 + 𝑥 + 𝑥 + 1)
=𝑘
𝑥−1
lim 𝑥 + 𝑥 + 𝑥 + 𝑥 + 1 = 𝑘
→
1 +1 +1 +1+1=𝑘
5=𝑘
45
(b) (i) 𝑥 = 5𝑡 + 3, 𝑦 = 𝑡 − 𝑡 + 2
When 𝑡 = 0
𝑑𝑥
=5
𝑑𝑡
𝑥 = 5(0) + 3 = 3
𝑦=0 −0 +2=2
𝑑𝑦
= 3𝑡 − 2𝑡
𝑑𝑡
(3, 2)
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
When 𝑡 =
𝑑𝑦 3𝑡 − 2𝑡
=
𝑑𝑥
5
(ii)
𝑥=5
=0
𝑦=
3𝑡 − 2𝑡 = 0
2
3
(c) (i) (a) 𝑦 = √2 + 2𝑥 = (2 + 2𝑥 )
+2=
50
27
+ 2𝑥 − (2 + 2𝑥 ) (4𝑥)
− 4𝑥 (2 + 2𝑥 )
𝑑 𝑦 4
−
𝑑𝑥
𝑦
𝑑𝑦
𝑦
− 2𝑥
𝑑𝑥
=0
2
3
𝑑 𝑦
= 2(2 + 2𝑥)
𝑑𝑥
𝑑𝑦
= 2𝑥(2 + 2𝑥 )
𝑑𝑥
= 2𝑥 − 2𝑥
−
= 2(2 + 2𝑥 )
(b)
𝑑𝑦 1
= (2 + 2𝑥 ) (4𝑥)
𝑑𝑥 2
= (2 + 2𝑥 ) 2𝑥(2 + 2𝑥 )
2
3
19 50
,
3 27
𝑡(3𝑡 − 2) = 0
𝑡 = 0,
2
19
+3=
3
3
= 2(2 + 2𝑥 )
− 4𝑥 (2 + 2𝑥 )
− 4 (2 + 2𝑥 )
= 2(2 + 2𝑥 )
− 4𝑥 (2 + 2𝑥 )
− 4(2 + 2𝑥 )
− 2𝑥
= 2(2 + 2𝑥 ) [2 + 2𝑥 − 2𝑥 − 2]
= 2(2 + 2𝑥 ) (0)
=0
(ii)
when 𝑥 = 0
= 2 2 + 2(0)
− 4(0) (2 + 2(0) )
= √2
46
QUESTION 6
Equation of 𝑄𝑅
(a) (i)
𝑚=
3−2
= −1
3−4
𝑦 = 𝑚𝑥 + 𝑐
(3, 3) 𝑚 = −1
3 = −1(3) + 𝑐
𝑐=6
𝑦 = −𝑥 + 6
Equation of 𝑃𝑅
(ii) Equation of 𝑃𝑄
𝑚=
3−1 2
=
3−0 3
𝑚=
2−1 1
=
4−0 4
𝑦=
2
𝑥+1
3
𝑦=
1
𝑥+1
4
(iii) ∫
=
𝑥 + 1 𝑑𝑥 + ∫ −𝑥 + 6 𝑑𝑥 − ∫
2 𝑥
3 2
+𝑥
3
𝑥
4
1 𝑥
+ − + 6𝑥 −
0
2
3
4 2
(3)
0
+3 −
+0
3
3
=
= (6 − 0) + 16 −
=
𝑥 + 1 𝑑𝑥
+
−
+𝑥
4
0
4
3
+ 6(4) − − + 6(3)
2
2
−
4
0
+4 −
+0
8
8
27
− (6 − 0)
2
5
2
47
(c) ∫ [3𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = 5
and ∫ [5𝑓(𝑥) − 2𝑔(𝑥)] 𝑑𝑥 = 1
(1) × 2: 6𝑚 + 2𝑛 = 10
(2)
5𝑚 − 2𝑛 = 1
11𝑚 = 11
[3𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = 5
𝑚=1
𝑛=2
3
𝑓(𝑥) 𝑑𝑥 +
𝑔(𝑥) 𝑑𝑥 = 5
𝑓(𝑥) 𝑑𝑥 = 1
5
𝑓(𝑥) 𝑑𝑥 − 2
𝑔(𝑥) 𝑑𝑥 = 1
𝑔(𝑥) 𝑑𝑥 = 2
Let ∫ 𝑓(𝑥) 𝑑𝑥 = 𝑚 and ∫ 𝑔(𝑥) 𝑑𝑥 = 𝑛
3𝑚 + 𝑛 = 5
(1)
5𝑚 − 2𝑛 = 1
(2)
48