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C2 notes
Algebraic and Functions
Factor Theorem


𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑥 − 1 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 2𝑥 3 − 3𝑥 2 − 𝑥 + 2
o 𝑅𝑒𝑎𝑎𝑟𝑎𝑛𝑔𝑒 𝑥 − 1 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑥, 𝑥 = 1
o 𝑆𝑢𝑏 𝑥 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 [2(1)3 − 3(1)2 − 1 + 2]
o 𝐼𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑙𝑠 𝑡𝑜 0 𝑡ℎ𝑒𝑛 𝑥 − 1 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟
o [2(1)3 − 3(1)2 − 1 + 2] = 0 ∴ 𝑥 − 1 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟

Long Division


𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 2𝑥 3 − 3𝑥 2 − 𝑥 + 2
o 𝑥 − 1 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 ∴ 𝑑𝑖𝑣𝑖𝑑𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑥 − 1
2𝑥 2 − 𝑥 − 2
o

o

o

𝑥 − 1⃒2𝑥 3 − 3𝑥 2 − 𝑥 + 2
- 2𝑥 3 − 2𝑥 2
𝑥2 − 𝑥
- 𝑥2 + 𝑥
−2𝑥 + 2
- −2𝑥 + 2
0
 Divide the highest term in the equation (2𝑥 3 ) by the highest term in the
factor (𝑥)
 Multiply the answer (2𝑥 2 ) the factor
 Subtract the answer from the 2 highest term in the equation ([2𝑥 3 −
3𝑥 2 ] − [2𝑥 3 − 2𝑥 2 ])
 The highest terms should cancel out
 Bring down the next term
 Repeat steps from the start until you are left with a remainder or 0
3
2𝑥 − 3𝑥 2 − 𝑥 + 2 dived by 𝑥 − 1 has 0 has a remainder so 𝑥 − 1 is one of the
solutions of the equation, the other 2 can be found by factorising the quadratic
equation got from long division (2𝑥 2 − 𝑥 − 2) )
Solutions are


1+√17 1−√17
, 4
4

−𝑏±√𝑏2 −4𝑎𝑐
2𝑎

If the equation has missing terms for example 3𝑥 4 − 2𝑥 2 + 4, simply
rewrite and put 0 for the co-efficient of the missing 𝑥 terms and use long
division to find solutions; so 3𝑥 4 − 2𝑥 2 + 4 = 3𝑥 4 − 0𝑥 3 + 2𝑥 2 + 0𝑥 + 4

Exponentials


𝑎𝑛𝑑 𝑥 − 1 – using the quadratic formula

Exponential functions : 𝑦 = 𝑎 𝑥

o



𝑥 axis is the asymptote

Logarithms


log 𝑎 𝐶
o
o
o
o

= 𝐵 ⟺ 𝑎𝐵= 𝐶
log 𝑎 𝑎 = 1 ⟺ 𝑎1 = 𝑎
log 𝑎 1 = 0 ⟺ 𝑎0 = 1
log 𝑎 𝑁 = log 𝑎 𝑀 ⇒ 𝑁 = 𝑀
If 𝑁 < 1 then log 𝑁 < 0
 Multiplication Rule : log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦
𝑥
 Division Rule : log 𝑎 = log 𝑎 𝑥 − log 𝑎 𝑦
𝑦



Power rule : log 𝑎 𝑥 𝑛 = 𝑛log 𝑎 𝑥



Changing log bases : log 𝑎 𝑏 =


Express the following in terms of log 𝑥 and log 𝑦
𝑥2

o



𝑥2 1

log √5𝑦2 = log(5𝑦2 )2
1
𝑥2
∴ log( 2 )
2
5𝑦
1
∴ (log 𝑥 2 − log 5𝑦 2 )
2
1
∴ [2 log 𝑥 − (log 5 + log 𝑦 2 )]
2
1
∴ [2 log 𝑥 − log 5 − 2 log 𝑦]
2

Simplify
log 81
log 27

o
∴

log 34
log 33

∴

4log 3
3log 3

∴


log 𝑐 𝑏
log 𝑐 𝑎

4
3

Solve
o

52𝑥−1 = 3 𝑥+1
∴ log 52𝑥−1 = log 3 𝑥+1
∴ (2𝑥 − 1) log 5 = (𝑥 + 1) log 3

∴ 2𝑥 log 5 − log 5 = 𝑥 log 3 + log 3
∴ 2𝑥 log 5 − 𝑥 log 3 = log 3 + log 5
∴ 𝑥(2 log 5 − log 3) = log 3 + log 5
log 3 + log 5
∴ 𝑥=
= 1
...
3 (1 𝑑
...
24 (2 𝑑
...

o 𝑥 2 + 𝑦 2 + 10𝑥 + 2𝑦 + 13 = 0
∴ 𝑥 2 + 10𝑥 + 𝑦 2 + 2𝑦 + 13 = 0
∴ (𝑥 + 5)2 − 25 + (𝑦 + 1)2 − 1 + 13 = 0
∴ (𝑥 + 5)2 + (𝑦 + 1)2 − 13 = 0
∴ 𝑐𝑒𝑛𝑡𝑟𝑒 𝑖𝑠 − 5, −1
∴ 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑟𝑒 𝑡𝑜 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
2 − −1
3
=
−3 − −5 2
∴ 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑡𝑜
𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
2
−
3
2
∴ 𝑦 − 2 = − (𝑥 − −3)
3
2
∴ 𝑦−2=− 𝑥−2
3
2
∴ 𝑦=− 𝑥
3

Equation of a Circle Given Three Points on the Circumference


𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 (−1,5), (3,5) 𝑎𝑛𝑑 (3, −1)

o

𝐷𝑟𝑎𝑤 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑛𝑑 𝑚𝑎𝑟𝑘 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝐷𝑟𝑎𝑤 𝑡ℎ𝑒 𝑐ℎ𝑜𝑟𝑑𝑠 𝑓𝑟𝑜𝑚 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴(−1,5) 𝑡𝑜 𝐵(3,5)𝑎𝑛𝑑 𝐵 𝑡𝑜 𝐶(3, −1)
𝐷𝑟𝑎𝑤 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑜𝑟𝑑𝑠 𝐴𝐵 𝑎𝑛𝑑 𝐵𝐶
𝑊ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 2 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑜𝑟𝑑𝑠 𝑚𝑒𝑒𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒
𝑇ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑖𝑠 ℎ𝑒𝑟𝑒𝑏𝑦 𝑡ℎ𝑒 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐵 𝑎𝑛𝑑 𝐵𝐶
−1 + 3
5 + −1
∴ 𝑐𝑒𝑛𝑡𝑟𝑒 = (
) 𝑎𝑛𝑑 (
)
2
2
∴ 𝑐𝑒𝑛𝑡𝑟𝑒 = (1,2)

𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑡𝑜 𝑎𝑛𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠
∴ 𝑈𝑠𝑖𝑛𝑔 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠, 𝑟 2 = (3 − 1)2 + (5 − 2)2
∴ 𝑟 2 = 13
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 𝑖𝑠 𝑟 2 = (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 "𝑤ℎ𝑒𝑟𝑒 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒"
∴ 13 = (𝑥 − 1)2 + (𝑦 − 2)2

Circle Theorems



Sequences and Series
Geometric Sequences


𝑛 𝑡ℎ = 𝑎𝑟 𝑛−1



𝑆𝑛 =

𝑎(𝑟 𝑛 −1)
𝑟−1

o
o
o


=

𝑎(1−𝑟 𝑛 )
1− 𝑟

𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
𝑟 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜
𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠

𝐶𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 ⇒
o

2 𝑛𝑑 𝑡𝑒𝑟𝑚
1 𝑠𝑡 𝑡𝑒𝑟𝑚

=

3 𝑟𝑑 𝑡𝑒𝑟𝑚
2 𝑛𝑑 𝑡𝑒𝑟𝑚

𝐼𝑓 2𝑥 + 5, 6𝑥 − 10 𝑎𝑛𝑑 8𝑥 + 20 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑡𝑒𝑟𝑚𝑠 𝑖𝑛 𝑎 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐
𝑠𝑒𝑟𝑖𝑒𝑠
...
5% 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟
...
𝐴𝑓𝑡𝑒𝑟 ℎ𝑜𝑤 𝑚𝑎𝑛𝑦 𝑦𝑒𝑎𝑟𝑠 ℎ𝑎𝑠 ℎ𝑖𝑠 𝑎𝑐𝑐𝑜𝑢𝑛𝑡
𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 £20000?
𝑎(𝑟 𝑛 − 1)
𝑎(1 − 𝑟 𝑛 )
𝑆𝑛 =
=
𝑟−1
1− 𝑟
(500 × 1
...
035 𝑛 − 1)
∴ 20000 ≥
1
...
035)(1
...
035
∴ 700 ≥ 517
...
035 𝑛 − 1)
∴ 700 ≥ 517
...
035 𝑛 ) − 517
...
5 ≥ 517
...
035 𝑛 )
∴ 2
...
035 𝑛
log 𝑎 𝐶 = 𝐵 ⟺ 𝑎 𝐵 = 𝐶
∴ log1
...
35265 = 𝑛
∴ 𝑛 = 24
...
5100 = 7
...
5101 = 3
...

𝑎
∴ 𝑆∞ =
1− 𝑟

Binomial Expansion
Binomial Expansion Method[


𝒏

𝑪𝒓 =
o



𝒏

𝑪 𝒓 𝒐𝒓 ( 𝒓𝒏)]

𝑛!

( 𝑛−𝑟)! 𝑟!
𝟓

𝟓

𝑪 𝟐 = ( 𝟐)

5!

(5−2)! 2!

=

5×4×3×2×1
(3×2×1)2×1

=

5×4
2×1

= 10

Expanding (𝑎 + 𝑏) 𝑛 when 𝑛 is a positive integer
o (𝑎 + 𝑏) 𝑛 ≡ 𝒏 𝑪0 𝑎 𝑛 𝑏 0 + 𝒏 𝑪1 𝑎 𝑛−1 𝑏1 + 𝒏 𝑪2 𝑎 𝑛−2 𝑏 2 + 𝒏 𝑪3 𝑎 𝑛−3 𝑏3 + ⋯ +
𝒏
𝑪 𝑛 𝑎0 𝑏 𝑛
 Expand (3 + 2𝑥)4
 (3 + 2𝑥)4 = 4 𝐶0 (3)4 (2𝑥)0 + 4 𝐶1 (3)4−1 (2𝑥)1 + 4 𝐶2 (3)4−2 (2𝑥)2 +
4 (3)4−3 (2𝑥)3
𝐶3
+ 4 𝐶4 (3)0 (2𝑥)4
∴ (3 + 2𝑥)4 = 1 (3)4 (2𝑥)0 + 4(3)3 (2𝑥)1 + 6(3)2 (2𝑥)2 + 4(3)1 (2𝑥)3 +
1(3)0 (2𝑥)4
∴ (3 + 2𝑥)4 = 1 (81)(1) + 4(27)(2𝑥) + 6(9)(4𝑥 2 ) + 4(3)(8𝑥 3 ) +
1(1)(16𝑥 4 )
∴ (3 + 2𝑥)4 = 1 (81) + 4(54𝑥) + 6(36𝑥 2 ) + 4(24𝑥 3 ) + 1(16𝑥 4 )
∴ (3 + 2𝑥)4 = 81 + 216𝑥 + 216𝑥 2 + 96𝑥 3 + 16𝑥 4
 You can use Pascal’s Triangle to replace 𝒏 𝑪 𝒓 , use the (𝑛 + 1)𝑡ℎ row







Expand (2 − 3𝑥)5 up to 𝑥 3
(2 − 3𝑥)5 = 5 𝐶0 (2)5 (−3𝑥)0 + 5 𝐶1 (2)5−1 (−3𝑥)1 + 5 𝐶2 (2)5−2 (−3𝑥)2
+ 5 𝐶3 (2)5−3 (−3𝑥)3
∴ (2 − 3𝑥)5 = 1(2)5 (−3𝑥)0 + 5(2)4 (−3𝑥)1 + 10(2)3 (−3𝑥)2
+ 10(2)2 (−3𝑥)3
∴ (2 − 3𝑥)5 = 1(32)(1) + 5(16)(−3𝑥) + 10(8)(9𝑥 2 ) + 10(4)(−27𝑥 3 )
∴ (2 − 3𝑥)5 = 1(32) + 5(−48𝑥) + 10(72𝑥 2 ) + 10(−108𝑥 3 )
∴ (2 − 3𝑥)5 = 32 − 240𝑥 + 720𝑥 2 − 1080𝑥 3
Find the term in 𝑥 5 in the expansion (5 − 2𝑥)8
𝒏
𝑪1 𝑎 𝑛−1 𝑏1
∴ 8 𝐶5 (5)8−5 (−2𝑥)5
∴ 56(5)3 (−2𝑥)5
∴ 56(125)(−32𝑥 5 )
∴ 56(−4000𝑥 5 )
∴ −224000𝑥 5

Binomial Expansion Formula


(1 + 𝑎) 𝑛 ≡ 1 + 𝑛𝑎 +

𝑛(𝑛−1)
2!

o

(1 + 𝑥)4

o

(1 + 𝑥)4 = 1 + 4𝑥 +

o
o
o

𝑎2 +

𝑛(𝑛−1)(𝑛−2)
3!

4(4−1)
2!

𝑎3 + ⋯

4(4−1)(4−2)

4(4−1)(4−2)(4−3)

𝑥2 +
𝑥3 +
𝑥4
3!
4!
4(3) 2
4(3)(2) 3
4(3)(2)(1) 4
∴ (1 + 𝑥)4 = 1 + 4𝑥 +
𝑥 +
𝑥 +
𝑥
2×1
3×2×1
4×3×2×1
12 2 24 3 24 4
∴ (1 + 𝑥)4 = 1 + 4𝑥 +
𝑥 +
𝑥 +
𝑥
2
6
24
4
2
3
4
∴ (1 + 𝑥) = 1 + 4𝑥 + 6𝑥 + 4𝑥 + 𝑥
(2 − 3𝑥)3
3
3
(2 − 3𝑥)3 = [2(1 − 2 𝑥)]3 = 23 (1 − 2 𝑥)3
3

3

3(3−1)

3

3(3−1)(3−2)

3

23 (1 − 2 𝑥)3 = 23 [1 + 3(− 2)𝑥 + 2! (− 2)𝑥 2 +
(− 2)𝑥 3 ]
3!
3
3
3(2)
3
3(2)(1)
3
∴ 23 (1 − 𝑥)3 = 23 [1 + 3(− )𝑥 +
(− )𝑥 2 +
(− )𝑥 3 ]
2
2
2×1
2
3×2×1 2
3 3
3
6 3 2 6
3
∴ 23 (1 − 𝑥) = 23 [1 + 3(− )𝑥 + (− 𝑥) + (− 𝑥)3 ]
2
2
2 2
6
2
3 3
3
6 9 2
6
27 3
3
3
∴ 2 (1 − 𝑥) = 2 [1 + 3(− )𝑥 + ( 𝑥 ) + (−
𝑥 )]
2
2
2 4
6
8
3
9
27 2 27 3
∴ 23 (1 − 𝑥)3 = 23 [1 − 𝑥 +
𝑥 −
𝑥 ]
2
2
4
8
3
9
27 2 27 3
∴ 23 (1 − 𝑥)3 = 8[1 − 𝑥 +
𝑥 −
𝑥 ]
2
2
4
8
∴ (2 − 3𝑥)3 = 8 − 36𝑥 + 54𝑥 2 − 27𝑥 3

Trigonometry
Trigonometry Ratios

Trigonometry Graphs

Transformation of Trigonometry Graphs





𝑦 = 𝑘𝑓(𝑥) → Stretch parallel to the 𝑦 − 𝑎𝑥𝑖𝑠
𝑦 = 𝑓(𝑘𝑥) → Stretch parallel to the 𝑥 − 𝑎𝑥𝑖𝑠
𝑦 = −𝑓(𝑥) → Reflection in the 𝑥 − 𝑎𝑥𝑖𝑠
𝑦 = 𝑓(−𝑥) → Reflection in the 𝑦 − 𝑎𝑥𝑖𝑠




𝑦 = 𝑓(𝑥 + 𝑎) → Translation parallel 𝑥 − 𝑎𝑥𝑖𝑠 by −𝑎 units
𝑦 = 𝑓(𝑥) + 𝑎 → Translation parallel 𝑦 − 𝑎𝑥𝑖𝑠 by 𝑎 units

Applications of Trigonometry functions
Area of a Triangle



1

𝐴𝑟𝑒𝑎 = 2 𝑎𝑏 sin 𝐶

Find Length of a side or size of an angle
Sine Rule



sin 𝐴
𝑎

=

sin 𝐵
𝑏

=

sin 𝐶
𝑐

Cosine Rule



𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos 𝐴 𝑜𝑟 cos 𝐴 =

𝑏2 +𝑐 2−𝑎2
2𝑏𝑐

Radians





A radian (1 𝑐 ) is the angle subtended at the centre of the circle of the length r and arc r, r is
the radius

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 𝑖𝑛 𝑜𝑛𝑒 𝑡𝑢𝑟𝑛 (360°) =
𝐷𝑒𝑔𝑟𝑒𝑒𝑠 𝑝𝑒𝑟 𝑟𝑎𝑑𝑖𝑎𝑛 𝑖𝑠

360°
2𝜋
𝑐

360° ≡ 2𝜋



o

=

2𝜋𝑟
𝑟

= 2𝜋 ≈ 6
...
6𝑐𝑚2



o

𝐴𝑟𝑒𝑎 =

𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = (

142
×
360

2𝜋6) + (6 + 6) = 14
...
9𝑐𝑚

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑛𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 (𝑠ℎ𝑎𝑑𝑒𝑑 𝑏𝑒𝑙𝑜𝑤), 𝑔𝑖𝑣𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑎𝑛𝑠𝑤𝑒𝑟 𝑡𝑜 1 𝑑𝑝




𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 =


o

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 − 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

70
× 𝜋5
...
2𝑐𝑚2
360
1
= 2 (5
...
3) sin 1
...
2𝑐𝑚2

 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 17
...
2 = 4
...






𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑎𝑗𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 − 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑖𝑛𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋3
...
36𝑐𝑚2
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑖𝑛𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 − 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒



𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 =



𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒




𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑖𝑛𝑜𝑟 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 8
...
72𝑐𝑚2 = 1
...
36𝑐𝑚2 − 1
...
4𝑐𝑚2

1
...
82 = 8
...
8 × 3
...
2 𝑐
2
2

= 6
...

o 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝑦 = 𝑥 3 + 3𝑥 2 −
9𝑥 + 4
𝑑𝑦
∴
= 3𝑥 2 + 6𝑥 − 9
𝑑𝑥
𝑑𝑦
𝑎𝑡 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠
=0
𝑑𝑥
∴ 3𝑥 2 + 6𝑥 − 9 = 0
∴ 𝑥 = −3 𝑜𝑟 𝑥 = 1
𝑠𝑢𝑏 𝑥 𝑏𝑎𝑐𝑘 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
∴ 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒 (−3,23) 𝑎𝑛𝑑 (1, −9)
To know the nature of a stationary point (whether it is a minimum, maximum or a turning
point) you have to find

𝑑𝑦
𝑑𝑥

= 0 to get 𝑥 value of the stationary point and sub in ± 1𝑥 values

from the stationary point into the
𝑥 values
𝑑𝑦
𝑑𝑥

= 0 𝑜𝑟 𝑓 ′ (𝑥) = 0 𝑜𝑟 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 0; for

2
-ve
+ve
+ve
-ve

𝑑𝑦
𝑑𝑥

equation to determine the nature of the point
...
𝐼𝑓 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑥
𝑖𝑠 𝑡𝑤𝑖𝑐𝑒 𝑡ℎ𝑒 𝑤𝑖𝑑𝑡ℎ
...
when 𝑡 → ∞,

1
=0
𝑡3

Area under a curve



When find the area bounded by a curve and the 𝑥-axis, use the integration formula to get
the area using the limits provided
...
Then subtract the
area below the line from the area below the curve
o 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑦 = 𝑥 2 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑦 = 𝑥
𝑥2 = 𝑥
∴ 𝑥2 − 𝑥 = 0
∴ 𝑥(𝑥 − 1) = 0
∴ 𝑥 = 0 𝑜𝑟 𝑥 = 1
∴ 𝑙𝑖𝑚𝑖𝑡𝑠 𝑎𝑟𝑒 0 𝑎𝑛𝑑 1
1

𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑙𝑖𝑛𝑒 ∫ 𝑥 𝑑𝑥
𝑥2
∴
2
12
02
1
∴[ ]−[ ]=
2
2
2

0

1

𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 ∫ 𝑥 2
0

𝑥3
3
13
03
1
∴[ ]−[ ]=
3
3
3
1 1 1
∴ 𝐴𝑟𝑒𝑎 = − = 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
2 3 6
∴

Trapezium Rule



The formula for the area of a trapezium is: 𝐴 =

𝑎+𝑏
ℎ,
2

when trying to find the area under a

curve using the trapezium rule; 𝑛 is ℎ, 𝑦1 is 𝑎 and 𝑦2 is 𝑏



𝑏

𝑛
2

𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑓𝑜𝑟 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑟𝑢𝑙𝑒: ∫𝑎 𝑓(𝑥) 𝑑𝑥 = [(𝑦1 + 𝑦 𝑛 ) + 2(𝑦2 + 𝑦3 + ⋯ + 𝑦 𝑛−1 )]

o

7

𝑈𝑠𝑒 𝑡ℎ𝑒 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝑟𝑢𝑙𝑒 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 ∫ √𝑥 − 1 𝑑𝑥 𝑡𝑎𝑘𝑖𝑛𝑔 𝑠𝑡𝑟𝑖𝑝𝑠 𝑜𝑓
3
𝑤𝑖𝑑𝑡ℎ 1 𝑢𝑛𝑖𝑡
𝑥

𝑦 = √𝑥 − 1

3
4
5
6
7

𝑦1 = √2
𝑦2 = √3
𝑦3 = 2
𝑦4 = √5
𝑦5 = √6

7
1
∴ ∫ √𝑥 − 1 𝑑𝑥 = [(√2 + √6) + 2(√3 + 2 + √5)]
2
3
∴ 𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑑 𝐴𝑟𝑒𝑎 𝑖𝑠 7

---