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TABLE OF CONTENTS 2
CHAPTER 1
Algebra
3 CHAPTER 2
Logarithmic & Exponential Functions
3 CHAPTER 3
Trigonometry
5 CHAPTER 4
Differentiation
5 CHAPTER 5
Integration
8 CHAPTER 6
Solving Equations Numerically
8 CHAPTER 7
Vectors
13 CHAPTER 8
Complex Numbers
16 CHAPTER 9
Differential Equations
Β© Copyright 2015 by Z Notes First edition Β© 2015, by Emir Demirhan, Saif Asmi & Zubair Junjunia. This document contain images and excerpts of text from educational resources available on the internet and printed books. If you are the owner of such media, text or visual, utilized in this document and do not accept its usage then I urge you to contact me and I would immediately replace said media. No part of this document may be copied or re-uploaded to another website without the express, written permission of the copyright owner. Under no conditions may this document be distributed under the name of false author(s) or sold for financial gain; the document is solely meant for educational purposes and it is to remain a property available to all at no cost. It is currently freely available from the website www.znotes.ml. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
CIE A LEVEL- MATHEMATICS [9709]
Page 2 of 17
1. ALGEBRA
1.1 The Modulus Function No line with a modulus ever goes under the x-axis
Any line that does go below the x-axis, when modulated
is reflected above it
|π x π| = |π|x |π|
|π
π| =
|π|
|π|
|π₯2| = |π₯|2 = π₯2 |π₯| = |π|
βπ₯2 = π2
βπ₯2 = |π₯|
1.2 Polynomials To find unknowns in a given identity
o Substitute suitable values of π₯
OR
o Equalize given coefficients of like powers of π₯
Factor theorem: If (π₯ β π‘) is a factor of the function
π(π₯) then π(π‘) = 0
Remainder theorem: If the function π(π₯) is divided by
(π₯ β π‘) then the remainder: π = π(π‘)
DIVIDEND = DIVISOR Γ QUOTIENT + REMAINDER
1.3 Binomial Series Expanding (1 + π₯)π where |π₯| < 1
1 +π
1π₯ +
π(π β 1)
1 Γ 2π₯2 +
π(π β 1)(π β 2)
1 Γ 2 Γ 3π₯3 + β―
Factor case: if constant is not 1, pull out a factor from
brackets to make it 1 & use general equation. Do not
forget the indices.
Substitution case: if bracket contains more than one π₯
term (e.g. (2 β π₯ + π₯2)) then make the last part π’,
expand and then substitute back in.
Finding the limit of π₯ in expansion:
E.g. (1 + ππ₯)π, limit can be found by substituting ππ₯
between the modulus sign in |π₯| < 1 and altering it to
have only π₯ in the modulus
1.4 Partial Fractions ππ₯ + π
(ππ₯ + π)(ππ₯ + π )β‘
π΄
ππ₯ + π+
π΅
ππ₯ + π
Multiply (ππ₯ + π), substitute π₯ = βπ
π and find π΄
Multiply (ππ₯ + π ), substitute π₯ = βπ
π and find π΅
ππ₯2 + ππ₯ + π
(ππ₯ + π)(ππ₯ + π )2β‘
π΄
ππ₯ + π+
π΅
ππ₯ + π +
πΆ
(ππ₯ + π )2
Multiply (ππ₯ + π), substitute π₯ = βπ
π and find π΄
Multiply (ππ₯ + π )2, substitute π₯ = βπ
π and find πΆ
Substitute any constant e.g. π₯ = 0 and find π΅
ππ₯2 + ππ₯ + π
(ππ₯ + π)(ππ₯2 + π )β‘
π΄
ππ₯ + π+
π΅π₯ + πΆ
ππ₯2 + π
Multiply (ππ₯ + π), substitute π₯ = βπ
π and find π΄
Take π΄
ππ₯+π to the other side, subtract and simplify.
Linear eqn. left at top is equal to π΅π₯ + πΆ
Improper fraction case: if numerator has π₯ to the
degree of power equivalent or greater than the
denominator then another constant is present. This can
be found by dividing denominator by numerator and
using remainder
{S12-P33} Question 8:
Express the following in partial fractions: 4π₯2 β 7π₯ β 1
(π₯ + 1)(2π₯ β 3)
Solution:
Expand the brackets 4π₯2 β 7π₯ β 1
2π₯2 β π₯ β 3
Greatest power of π₯ same in numerator and denominator, thus is an improper fraction case
Making into proper fraction:
2
2π₯2 β π₯ β 3 4π₯2 β 7π₯ β 1 4π₯2 β 2π₯ β 6 β5π₯ + 5
This is written as:
2 +5 β 5π₯
(π₯ + 1)(2π₯ β 3)
Now proceed with normal case for the fraction: π΄
π₯ + 1+
π΅
2π₯ β 3=
5 β 5π₯
(π₯ + 1)(2π₯ β 3)
CIE A LEVEL- MATHEMATICS [9709]
Page 3 of 17
π΄(2π₯ β 3) + π΅(π₯ + 1) = 5 β 5π₯ When π₯ = β1
β5π΄ = 5 + 5 π΄ = β2
When π₯ =3
2
5
2π΅ = 5 β
15
2
π΅ = β1 Thus the partial fraction is:
2 +β2
π₯ + 1+
β1
2π₯ β 3
2. LOGARITHMIC & EXPONENTIAL FUNCTIONS π¦ = ππ₯
β logπ π¦ = π₯
logπ 1 = 0 logπ π = 1 logπ ππ β‘ π logπ π
logπ π + logπ π β‘ logπ ππ
logπ π β logπ π β‘ logπ
π
π
logπ π β‘ log π
log π
logπ π β‘ 1
logπ π
2.1 Graphs of ln(x) and ex
3. TRIGONOMETRY
3.1 Ratios
tan π =sinπ
cosπ sec π =
1
cosπ
cosec π =1
sinπ cot π =
cosπ
sinπ
3.2 Identities (cos π)2 + (sinπ)2 β‘ 1
1 + (tanπ)2 β‘ (sec π)2
(cot π)2 + 1 β‘ (cosec π)2
3.3 Graphs
3.4 Double Angle Identities sin2π΄ β‘ 2 sinπ΄ cosπ΄
cos 2π΄ β‘ (cosπ΄)2 β (sinπ΄)2 β‘ 2(cosπ΄)2 β 1
β‘ 1 β 2(sinπ΄)2
tan 2π΄ β‘2 tanπ΄
1 β (tanπ΄)2
3.5 Addition Identities sin(π΄ Β± π΅) β‘ sinπ΄ cosπ΅ Β± cosπ΄ sinπ΅
cos(π΄ Β± π΅) β‘ cosπ΄ cosπ΅ β sinπ΄ sinπ΅
tan(π΄ Β± π΅) β‘tanπ΄ Β± tanπ΅
1 β tanπ΄ tanπ΅
3.6 Changing Forms π sin π₯ Β± π cos π₯ βΊ π sin(π₯ Β± πΌ)
π cos π₯ Β± π sinπ₯ βΊ π cos(π₯ β πΌ)
Where π = βπ2 + π2 and
π cosπΌ = π, π sinπΌ = π with 0 < πΌ <1
2π
{S13-P33} Question 9:
Diagram shows curve, π¦ = sin2 2π₯ cos π₯, for 0 β€ π₯ β₯π
2, and π is maximum point. Find the π₯ coordinate of
π.
CIE A LEVEL- MATHEMATICS [9709]
Page 4 of 17
Solution:
Use product rule to differentiate: π’ = sin2 2π₯ π£ = cos π₯
π’β² = 4 sin2π₯ cos2π₯ π£β² = βsinπ₯ ππ¦
ππ₯= π’β²π£ + π’π£β²
ππ¦
ππ₯= (4 sin2π₯ cos2π₯)(cos π₯) + (sin2 2π₯)(β sin π₯)
ππ¦
ππ₯= 4 sin 2π₯ cos 2π₯ cosπ₯ β sin2 2π₯ sin π₯
Use following identities: cos 2π₯ = 2 cos2 π₯ β 1 sin2π₯ = 2 sinπ₯ cos π₯ sin2 π₯ = 1 β cos2 π₯
Equating to 0: ππ¦
ππ₯= 0
β΄ 4 sin2π₯ cos2π₯ cos π₯ β sin2 2π₯ sinπ₯ = 0 4 sin2π₯ cos2π₯ cos π₯ = sin2 2π₯ sinπ₯
Cancel sin2π₯ on both sides 4 cos 2π₯ cosπ₯ = sin 2π₯ sin π₯
Substitute identities 4(2 cos2 π₯ β 1) cos π₯ = (2 sin π₯ cos π₯) sinπ₯
Cancel cos π₯ and constant 2 from both sides
4 cos2 π₯ β 2 = sin2 π₯ Use identity
4 cos2 π₯ β 2 = 1 β cos2 π₯ 5 cos2 π₯ = 3
cos2 π₯ =3
5
cosπ₯ = 0.7746 π₯ = cosβ1(0.7746) π₯ = 0.6847 β 0.685
{W13-P31} Question 6:
π΄ is a point on circumference of a circle center π, radius π. A circular arc, center π΄ meets circumference at π΅ & πΆ. Angle ππ΄π΅ is π radians. The area of the shaded region is equal to half the area of the circle.
Show that:
cos2π =2 sin2π β π
4π
Solution:
First express area of sector ππ΅π΄πΆ
ππππ‘ππ π΄πππ =1
2ππ2
ππ΅π΄πΆ =1
2(2π β 4π)π2 = (π β 2π)π2
Now express area of sector π΄π΅πΆ
π΄π΅πΆ =1
2(2π)(πΏππππ‘β ππ π΅π΄)2
Express π΅π΄ using sine rule
π΅π΄ =π sin(π β 2π)
sinπ
Use double angle rules to simplify this expression
π΅π΄ =π sin2π
sinπ
=2π sin π cos π
sin π
= 2π cos π Substitute back into initial equation
π΄π΅πΆ =1
2(2π)(2π cosπ)2
π΄π΅πΆ = 4ππ2 cos2 π Now express area of kite π΄π΅ππΆ
π΄π΅ππΆ = 2 Γ π΄πππ ππ ππππππππ
π΄π΅ππΆ = 2 Γ1
2π2 sin(π β 2π)
= π2 sin(π β 2π) Finally, the expression of shaded region equated to half of circle
4π2π cos2 π + π2(π β 2π) β π2 sin(π β 2π) =1
2ππ2
Cancel our π2 on both sides for all terms
4π cos2 π + π β 2π β (sin π cos 2π + sin 2π cos π) =1
2π
Some things in the double angle cancel out
4π cos2 π + π β 2π β sin 2π =1
2π
Use identity here
4π (cos2π + 1
2) + π β sin 2π β 2π =
1
2π
4π cos 2π + 4π + 2π β 2 sin 2π β 4π = π Clean up
4π cos 2π + 2π β 2 sin 2π = π 4π cos 2π = 2 sin 2π β π
cos 2π =2 sin 2π β π
4π
CIE A LEVEL- MATHEMATICS [9709]
Page 5 of 17
4. DIFFERENTIATION
4.1 Basic Derivatives π₯π ππ₯πβ1
ππ’ ππ’
ππ₯ππ’
ln π’ ππ’
ππ₯β
π’
sinππ₯ π cosππ₯
cos ππ₯ β π sin ππ₯
tan ππ₯ π sec2 ππ₯
4.2 Chain, Product and Quotient Rule Chain Rule:
ππ¦
ππ₯=
ππ¦
ππ’Γ
ππ’
ππ₯
Product Rule: π
ππ₯(π’π£) = π’
ππ£
ππ₯+ π£
ππ’
ππ₯
Quotient Rule:
π
ππ₯(π’
π£) =
π£ππ’
ππ₯β π’
ππ£
ππ₯
π£2
4.3 Parametric Equations ππ¦
ππ₯=
ππ¦
ππ‘Γ·
ππ₯
ππ‘
In a parametric equation π₯ and π¦ are given in terms of π‘
and you must use the above rule to find the derivative
4.4 Implicit Functions These represent circles or lines with circular curves, on a
Cartesian plane
Difficult to rearrange in form π¦ = β΄ differentiate as is
Differentiate π₯ terms as usual
For π¦ terms, differentiate the same as you would π₯ but
multiply with ππ¦
ππ₯
Then make ππ¦
ππ₯ the subject of formula for derivative
5. INTEGRATION
5.1 Basic Integrals
ππ₯π ππ₯π+1
(π + 1)+ π
πππ₯+π 1
ππππ₯+π
1
ππ₯ + π
1
πln|ππ₯ + π|
sin(ππ₯ + π) β1
πcos(ππ₯ + π)
cos(ππ₯ + π) 1
πsin(ππ₯ + π)
sec2(ππ₯ + π) 1
πtan(ππ₯ + π)
(ππ₯ + π)π (ππ₯ + π)π+1
π(π + 1)
Use trigonometrical relationships to facilitate complex
trigonometric integrals
Integrate by decomposing into partial fractions
5.2 Integration by π-Substitution
β«π(π₯) ππ₯ = β«π(π₯)ππ₯
ππ’ ππ’
Make π equal to something: when differentiated,
multiply the substituted form directly
Make π equal to something: when differentiated,
multiply the substituted form with its reciprocal
With definite integrals, change limits in terms of π’
{W12-P33} Question 7:
The diagram shows part of curve π¦ = sin3 2π₯ cos3 2π₯. The shaded region shown is bounded by the curve and the π₯-axis and its exact area is denoted by π΄.
Use the substitution π’ = sin2π₯ in a suitable integral to find the value of π΄
Solution:
To find the limit, you are trying to the find the points at which π¦ = 0
sin π₯ = 0 at π₯ = 0, π, 2π cos π₯ = 0 at π₯ =π
2,3π
4
Choose the two closest to 0 because the shaded area has gone through π¦ = 0 only twice
β΄ 0 and π
2
Since it is sin 2π₯ and cos 2π₯, divide both by 2
β΄ Limits are 0 and π
4
Integrate by π’ substitution, let:
π’ = sin2π₯ ππ’
ππ₯= 2 cos 2π₯
ππ₯
ππ’=
1
2 cos 2π₯
sin3 2π₯ cos3 2π₯ β‘ (sin2π₯)3(cos2 2π₯) cos2π₯
β‘ (sin3 2π₯ Γ (1 β sin2 2π₯)) cos 2π₯
CIE A LEVEL- MATHEMATICS [9709]
Page 6 of 17
β‘ (sin3 2π₯ β sin5 π₯) cos 2π₯ Γ1
2 cos 2π₯
β‘1
2(π’3 β π’5)
Now integrate: 1
2β«(π’3 β π’5) =
1
2(π’4
4β
π’6
6)
The limits are π₯ = 0 and π₯ =π
4. In terms of π’,
π’ = sin 2(0) = 0 and π’ = sin 2 (π
4) = 1
Substitute limits 1
2(14
4β
16
6) β
1
2(04
4β
06
6) =
1
24
5.3 Integrating πβ²(π)
π(π)
β«πβ²(π)
π(π)ππ₯ = ln|π(π₯)| + π
{S10-P32} Question 10:
By splitting into partial fractions, show that:
β«2π₯3 β 1
π₯2(2π₯ β 1)ππ₯
2
1
=3
2+
1
2ln (
16
27)
Solution:
Write as partial fractions
β«2π₯3 β 1
π₯2(2π₯ β 1)ππ₯
2
1
β‘ β«1 +2
π₯+
1
π₯2+
3
2π₯ β 1ππ₯
2
1
β‘ π₯ + 2 ln π₯ β π₯β1 β3
2ln|2π₯ β 1|
Substitute the limits
2 + 2 ln 2 β1
2β
3
2ln 3 β 1 β 2 ln 1 + 1 +
3
2ln 1
3
2+
1
2ln 16 +
1
2ln
1
33β‘
3
2+
1
2ln
16
27
5.4 Integrating By Parts
β«π’ππ£
ππ₯ ππ₯ = π’π£ β β«π£
ππ’
ππ₯ππ₯
For a definite integral:
β« π’ππ£
ππ₯ ππ₯
π
π
= [π’π£]ππ β β« π£
ππ’
ππ₯ππ₯
π
π
What to make π:
L A T E Logs Algebra Trig π
{W13-P31} Question 3:
Find the exact value of
β«ln π₯
βπ₯
4
1
ππ₯
Solution:
Convert to index form: ln π₯
βπ₯= π₯
1
2 ln π₯
Integrate by parts, let:
π’ = ln π₯ ππ’
ππ₯=
1
π₯
ππ£
ππ₯= π₯β
1
2 π£ = 2π₯1
2
β΄ ln π₯ 2π₯1
2 β β«2π₯1
2 Γ π₯β1 β‘ 2βπ₯ ln π₯ β β«2π₯β1
2
β‘ 2βπ₯ ln π₯ β 4βπ₯ Substitute limits
= 4 ln 4 β 4
5.5 Integrating Powers of Sine or Cosine To integrate sin π₯ or cos π₯ with a power:
If power is odd, pull out a sin π₯ or cos π₯ and use
Pythagorean identities and double angle identities
If power is even, use the following identities
sin2 π₯ = 1
2β
1
2cos(2π₯)
cos2 π₯ = 1
2+
1
2cos(2π₯)
5.6 Integrating ππππ π ππππ π If π or π are odd and even, then:
Factor out one power from odd trig function
Use Pythagorean identities to transform remaining even
trig function into the odd trig function
Let u equal to odd trig function and integrate
If π and π are both even, then:
Replace all even powers using the double angle
identities and integrate
If π and π are both odd, then:
Choose one of the trig. functions & factor out one power
Use Pythagorean identity to transform remaining even
power of chosen trig function into other trig. function
If either π or π or both = 1, then:
Let π’ equal to the trig function whose power doesnβt
equal 1 then integrate
If both are 1, then let π’ equal either
CIE A LEVEL- MATHEMATICS [9709]
Page 7 of 17
{W09-P31} Question 5:
(i) Prove the identity
cos 4π β 4 cos 2π + 3 β‘ 8 sin4 π
(ii) Using this result find, in simplified form, the
exact value of
β« sin4 π ππ
1
3π
1
6π
Solution: Part (i)
Use double angle identities cos 4π β 4 cos 2π + 3
β‘ 1 β 2 sin2 2π β 4(1 β 2 sin2 π)+ 3
Open everything and clean
β‘ 1 β 2 sin2 2π β 4 + 8 sin2 π + 3 β‘ 1 β 2(sin2π)2 β 4 + 8 sin2 π + 3
β‘ 1 β 2(2 sin π cos π)2 β 4 + 8 sin2 π + 3 β‘ 1 β 2(4 sin2 π cos2 π) β 4 + 8 sin2 π + 3
β‘ 1 β 2(4 sin2 π (1 β sin2 π)) β 4 + 8 sin2 π + 3 β‘ 1 β 8 sin2 π + 8 sin4 π β 4 + 8 sin2 π + 3
β‘ 8 sin4 π Part (ii)
Use identity from (part i):
1
8β« cos 4π β 4 cos 2π + 3
1
3π
1
6π
β‘1
8[1
4sin4π β 2 sin π + 3π]
1
6π
1
3π
Substitute limits
β‘1
32(2π β β3)
{W12-P32} Question 5:
(i) By differentiating 1
cosπ₯, show that if π¦ =
sec π₯ then ππ¦
ππ₯= sec π₯ tan π₯
(ii) Show that 1
secπ₯βtanπ₯β‘ sec π₯ + tan π₯
(iii) Deduce that:
1
(secπ₯ β tan π₯)2β‘ 2 sec2 π₯ β 1 + 2 secπ₯ tan π₯
(iv) Hence show that:
β«1
(secπ₯ β tan π₯)2
1
4π
0
ππ₯ =1
4(8β2 β π)
Solution: Part (i)
Change to index form: 1
cos π₯= cosβ1 π₯
Differentiate by chain rule: ππ¦
ππ₯= β1(cos π₯)β2 Γ (βsin π₯)
β1(cos π₯)β2 Γ (βsin π₯) β‘sin π₯
cos2 π₯β‘
sinπ₯
cosπ₯Γ
1
cos π₯
sin π₯
cos π₯Γ
1
cos π₯β‘ sec π₯ tan π₯
Part (ii)
Multiply numerator and denominator by sec π₯ + tan π₯ sec π₯ + tan π₯
(sec π₯ β tan π₯) (sec π₯ + tan π₯)β‘
sec π₯ + tan π₯
sec2 π₯ β tan2 π₯
sec π₯ + tan π₯
sec2 π₯ β tan2 π₯β‘
sec π₯ + tan π₯
1β‘ sec π₯ + tan π₯
Part (iii)
Substitute identity from (part ii) 1
(secπ₯ β tan π₯)2β‘ (sec π₯ + tan π₯)2
Open out brackets (sec π₯ + tan π₯)2
β‘ sec2 π₯ + 2 secπ₯ tan π₯ + tan2 π₯ β‘ sec2 π₯ + 2 sec π₯ tan π₯ + sec2 π₯ β 1
β‘ 2sec2 π₯ + 2 sec π₯ tan π₯ β 1 β‘ 2 sec2 π₯ β 1 + 2 sec π₯ tan π₯
Part (iv)
β«1
(sec π₯ β tan π₯)2ππ₯
β‘ β«2sec2 π₯ β 1 + 2 sec π₯ tan π₯ ππ₯
β‘ 2β«sec2 π₯ β β«1 + 2β«sec2 π₯ tan2 π₯
Using differential from part i: β‘ 2 tan π₯ β π₯ + 2 sec π₯
Substitute boundaries:
=1
4(8β2 β π)
5.5 Trapezium Rule
π΄πππ =ππππ‘β ππ 1π π‘ ππ‘πππ
2Γ [1π π‘ βπππβπ‘ + πΏππ π‘ βπππβπ‘
+ 2(π π’π ππ β ππππππ)]
ππππ‘β ππ 1π π‘ ππ‘πππ =πβπ
ππ. ππ πππ‘πππ£πππ for β« ππ₯
π
π
CIE A LEVEL- MATHEMATICS [9709]
Page 8 of 17
6. SOLVING EQUATIONS NUMERICALLY
6.1 Approximation To find root of a graph, find point where graph passes
through π₯-axis β΄ look for a sign change
Carry out decimal search
o Substitute values between where a sign change has
occurred
o Closer to zero, greater accuracy
6.2 Iteration To solve equation π(π₯) = 0, you can rearrange π(π₯)
into a form π₯ = β―
This function represents a sequence that starts at π₯β,
moving to π₯π
Substitute a value for π₯β and put back into function
getting π₯1 and so on.
As you increase π, value becomes more accurate
Sometimes iteration donβt work, these functions pare
called divergent, and you must rearrange formula for
π₯ in another way
For a successful iterative function, you need a
convergent sequence
7. VECTORS
7.1 Equation of a Line
The column vector form:
π = (13
β2) + π‘ (
113)
The linear vector form:
π = π’ + 3π£ β 2π€ + π‘(π’ + π£ + 3π€)
The parametric form:
π₯ = 1 + π‘, π¦ = 2 + π‘, π§ = β2 + 3π‘
The cartesian form; rearrange parametric π₯ β 1
1=
π¦ β 3
1=
π§ + 2
3
7.2 Parallel, Skew or Intersects For the two lines:
ππ΄ββββ β = οΏ½ΜοΏ½ + π οΏ½ΜοΏ½ ππ΅ββ ββ β = οΏ½ΜοΏ½ + π‘οΏ½ΜοΏ½
Parallel:
o For the lines to be parallel οΏ½ΜοΏ½ must equal οΏ½ΜοΏ½ or be in
some ratio to it e.g. 1: 2
Intersects:
o Make ππ΄ββββ β = ππ΅ββ ββ β
o If simultaneous works then intersects
o If unknowns cancel then no intersection
Skew:
o First check whether line parallel or not
o If not, then make ππ΄ββββ β = ππ΅ββ ββ β
o Carry out simultaneous
o When a pair does not produce same answers as
another, then lines are skew
7.3 Angle between Two Lines Use dot product rule on the two direction vectors:
π. π
|π||π|= cos π
Note: π and π must be moving away from the point at
which they intersect
7.4 Finding the Equation of a Line Given 2 points:
o Find the direction vector using
e.g. π΄π΅ = ππ΅ β ππ΄
o Place either of the points as a given vector
To check if a point lies on a line, check if constant of the
direction vector is the same for π₯, π¦ and π§ components
7.5 β₯ Distance from a Line to a Point
AKA: shortest distance from the point to the line
Find vector for the point, π΅, on the line
Vector equation of the line: οΏ½ΜοΏ½ = (13
β2) + π‘ (
113)
CIE A LEVEL- MATHEMATICS [9709]
Page 9 of 17
β΄ ππ΅ββ ββ β = (1 + π‘3 + π‘3π‘ β 2
)
π΄ is the point given
ππ΄ββββ β = (234)
β΄ π΄π΅ββββ β = (1 + π‘ β 23 + π‘ β 33π‘ β 2 β 4
) = (π‘ β 1
π‘3π‘ β 6
)
Use Dot product of π΄π΅ and the direction vector
π΄π΅ββββ β. π = cos90
(π‘ β 1
π‘3π‘ β 6
) . (113) = 0
1(π‘ β 1) + 1(π‘) + 3(3π‘ β 6) = 0
11π‘ β 19 = 0
π‘ =19
11
Substitute π‘ into equation to get foot
Use Pythagorasβ Theorem to find distance
{S08-P3} Question:
The points π΄ and π΅ have position vectors, relative to the origin π, given by
ππ΄ = π’ + 2π£ + 3π€ and ππ΅ = 2π’ + π£ + 3π€ The line π has vector equation
π« = (1 β 2π‘)π’ + (5 + π‘)π£ + (2 β π‘)π€ (i) Show that π does not intersect the line
passing through π΄ and π΅.
(ii) The point π lies on π and is such that angle
ππ΄π΅ is equal to 60Β°. Given that the
position vector of π is (1 β 2π‘)π’ +
(5 + π‘)π£ + (2 β π‘)π€, show that 3π‘2 +
7π‘ + 2 = 0. Hence find the only possible
position vector of π Solution:
Part (i)
Firstly, we must find the equation of line π΄π΅ π΄π΅ = ππ΅ β ππ΄
= (213) β (
123) = (
1β10
)
ππ = (123) + π (
1β10
) and π = (152) + π‘ (
β21
β1)
Equating the two lines
(1 + π 2 β π
3) = (
1 β 2π‘5 + π‘2 β π‘
)
Equation 1: 1 + π = 1 β 2π‘ so π = β2π‘ Equation 2: 2 β π = 5 + π‘
Substitute 1 into 2: 2 + 2π‘ = 5 + π‘
β΄ π‘ = 3 and then π = β6 Equation 3:
3 = 2 β π‘ Substitute the value of π‘ 3 = 2 β 3 so 3 = β1
This is incorrect therefore lines donβt intersect Part (ii)
Angle ππ΄π΅ is formed by the intersection of the lines π΄π and π΄π΅
π = (1 β 2π‘5 + π‘2 β π‘
)
π΄π = ππ β ππ΄
π΄π = (1 β 2π‘5 + π‘2 β π‘
) β (123) = (
β2π‘3 + π‘
β1 β π‘)
π΄π΅ = (1
β10
)
Now use the dot product rule to form an eqn. |π΄π. π΄π΅|
|π΄π||π΄π΅|;
β3π‘ β 3
β6π‘2 + 8π‘ + 10 Γ β2= cos 60
β3π‘ β 3 =1
2β6π‘2 + 8π‘ + 10 Γ β2
36π‘2 + 72π‘ + 36 = 12π‘2 + 16π‘ + 20 24π‘2 + 56π‘ + 16 = 0
π‘ = β1
3 or π‘ = β2
{W11-P31} Question:
With respect to the origin π, the position vectors of
two points π΄ and π΅ are given by ππ΄ββββ β = π’ + 2π£ + 2π€ and
ππ΅ββ ββ β = 3π’ + 4π£. The point π lies on the line through π΄
and π΅, and π΄πββββ β = ππ΄π΅ββββ β
(i) ππββββ β = (1 + 2π)π’ + (2 + 2π)π£ + (2 β
2π) π€
(ii) By equating expressions for cosπ΄ππ and
cosπ΅ππ in terms of π, find the value of π
for which ππ bisects the angle π΄ππ΅. Solution:
Part (i)
π΄πββββ β = ππ΄π΅ββββ β = π(ππ΅ β ππ΄)
= π(340) β (
122) = (
22
β2)
β΄ π΄π = (2π2π
β2π)
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ππ = ππ΄ + (2π2π
β2π) = (
122) + (
2π2π
β2π)
Part (ii)
Interpreting the question gives the information that π΄ππ is equal to π΅ππ β΄ cosπ΄ππ is equal to cosπ΅ππ. Now you can equate the two dot product equations
cosπ΄ππ =ππ΄.ππ
|ππ΄||ππ|=
9 + 2π
3β9 + 4π + 12π2
cosπ΅ππ =ππ΅.ππ
|ππ΅||ππ|=
11 + 14π
5β9 + 4π + 12π2
9 + 2π
3β9 + 4π + 12π2=
11 + 14π
5β9 + 4π + 12π2
Cancel out the denominator to give you 9 + 2π
3=
11 + 14π
5
45 + 10π = 33 + 42π
12 = 32π and β΄ π =3
8
7.6 Equation of a Plane
Scalar product form:
οΏ½ΜοΏ½. (β4β5β1
) = β13
The vector after οΏ½ΜοΏ½ is the normal to the plane
The components of the normal vector of the plane are
the coefficients of π₯, π¦ and π§ in the Cartesian form. You
must substitute a point to find π
Cartesian form:
4π₯ + 5π¦ + π§ = 13
7.7 Cross Product Rule
(πππ) Γ (
πππ) = (
ππ β ππππ β ππππ β ππ
)
7.8 Finding the Equation of a Plane Given 3 points on a plane:
o π΄(1,2, β1), π΅(2,1,0), πΆ(β1,3,2)
o Use this equation: οΏ½ΜοΏ½. οΏ½ΜοΏ½ = οΏ½ΜοΏ½. οΏ½ΜοΏ½
o οΏ½ΜοΏ½ is what we want to find
o οΏ½ΜοΏ½ is the cross product of 2 vectors parallel to the plane
o If we use π΄οΏ½ΜοΏ½ and π΄οΏ½ΜοΏ½ then οΏ½ΜοΏ½ = ππ΄
o β΄ οΏ½ΜοΏ½ = π΄π΅ Γ π΄πΆ = (β4β5β1
)
o Substitute point π΄ to get οΏ½ΜοΏ½. οΏ½ΜοΏ½
o β΄ οΏ½ΜοΏ½. (β4β5β1
) = β13
Given a point and a line on the plane:
o π΄(1,2,3) and οΏ½ΜοΏ½ = (210) + π (
β211
)
o Make 2 points on the line
o Substitute different values for π
o Repeat 3 point process
Given 2 lines on a plane:
o Find a point on one line
o Find 2 points on the other line
o Repeat 3 point process
7.9 A Line and a Plane If a line lies on a plane then any two points on the line
(π‘ = 0 and π‘ = 1)should satisfy the plane equation β
substitute and see if equation works
If a line is parallel to plane, the dot product of the
direction vector and normal of the plane is zero
7.10 Finding the Point of Intersection
between Line and Plane Form Cartesian equation for line
Form Cartesian equation for plane
Solve for π₯, π¦ and π§
{S13-P32} Question:
The points π΄ and π΅ have position vectors 2π’ β 3π£ + 2π€ and 5π’ β 2π£ + π€ respectively. The plane π has equation π₯ + π¦ = 5
(i) Find position vector of the point of
intersection of the line through π΄ and π΅ and
the plane π.
(ii) A second plane π has an equation of the form
π₯ + ππ¦ + ππ§ = π. The plane π contains the
line π΄π΅, and the acute angle between the
planes π and π is 60Β°. Find the equation of π.
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Solution: Part (i)
π΄π΅ = ππ΅ β ππ΄ = (31
β1)
The equation of the line π΄π΅ = ππ΄ + π΄π΅
= (2
β32
) + π (31
β1) = (
2 + 3ππ β 32 β π
)
Substitute values into plane equation π₯ + π¦ = 5 βΉ 2 + 3π + π β 3 = 5
4π β 1 = 5 βΉ π =3
2
Substitute lambda back into general line equation
(2 + (3 Γ 1.5)
1.5 β 32 β 1.5
) = (6.5
β1.50.5
)
Part (ii)
Using the fact that line π΄π΅ lies on the plane, the direction vector of π΄π΅ is perpendicular to the plane. Remember there is no coefficient for π₯ which means that it is equal to 1.
(31
β1) . (
1ππ) = 0
3 + π β π = 0 so π = 3 + π Using the fact that the plane π and π intersect at an angle of 60Β°
(110) . (
1ππ)
β2 Γ β1 + π2 + π2= cos 60 =
1
2
2 + 2π = β2 + 2π2 + 2π2 4π2 + 8π + 4 = 2π2 + 2π2 + 2
Substitute the first equation into π 2π2 + 8π + 2 β 18 β 12π β 2π2 = 0
β4π β 16 = 0 π = β4 and π = β1 We have found the normal to the plane, now we must find π
π₯ β 4π¦ β π§ = 0 Substitute the point π΄ into the equation because the point lies on it
(2) β 4(β3) β 2 = π π = 12 π₯ β 4π¦ β π§ = 12
7.11 Finding Line of Intersection of Two Non-
Parallel Planes The direction vector of this line is π§οΏ½ΜοΏ½ Γ π§οΏ½ΜοΏ½
π§οΏ½ΜοΏ½ is the normal of the first plane
π§οΏ½ΜοΏ½ is the normal of the second plane
7.12 β₯ Distance from a Point to a Plane
π· =|ππ₯1 + ππ¦1 + ππ§1 + π|
βπ2 + π2 + π2
Point πΉ is the foot of the perpendicular
{S12-P32} Question:
Two planes, π and π, have equations π₯ + 2π¦ β 2π§ =1 and 2π₯ β 2π¦ + π§ = 7 respectively. The line π has equation π« = π’ + π£ β π€ + π(2π’ + π£ + 2π€)
(i) Show that π is parallel to π
(ii) A point π lies on π such that its perpendicular
distances from π and π are equal. Find the
position vectors of the two possible positions
for π and calculate the distance between
them. Solution:
Part (i)
If π is parallel to π, then the direction vector of π would be perpendicular to the normal of π β΄ their dot product is equal to zero
(212) . (
12
β2) = 0
Part (ii)
Any point on π would have the value
(1 + 2π1 + π2π β 1
)
Using the distance formula of a point to a plane, find the perpendicular distance of the general point on π from the plane π and π
π·π = |4
3| and π·π = |
β8+4π
3|
Equate them as they equal the same distance
|4
3| = |
β8 + 4π
3| βΉ |4| = |β8 + 4π|
Remove modulus sign by taking into consideration the positive and negative
4 = β8 + 4π and β4 = β8 + 4π π = 3 and π = 1
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Substitute lambda values back into vector general line π equation to get the two points π1 and π2
π1 = (745) and π2 = (
321)
Use Pythagorasβs Theorem to find the distance
β42 + 22 + 42 = β36 = 6
7.13 Angle between Two Planes
cos π =π§οΏ½ΜοΏ½. π§οΏ½ΜοΏ½
|π§οΏ½ΜοΏ½||π§οΏ½ΜοΏ½|
The οΏ½ΜοΏ½βs here represent the normals of each plane
Ignore any negative signs
7.14 Angle between a Line and a Plane
First find β :
cosβ =οΏ½ΜοΏ½. οΏ½ΜοΏ½
|οΏ½ΜοΏ½||οΏ½ΜοΏ½|
π = 90 β β
π is the angle between the line and the plane
{W13-P32} Question:
The diagram shows three points π΄, π΅ and πΆ whose position vectors with respect to the origin π are given by
ππ΄ββββ β = (2
β12
), ππ΅ββ ββ β = (031) and ππΆββββ β = (
304).
The point π· lies on π΅πΆ, between π΅ and πΆ, and is such that πΆπ· = 2π·π΅.
(i) Find the equation of the plane π΄π΅πΆ, giving
your answer in the form ππ₯ + ππ¦ + ππ§ = π
(ii) Find the position vector of π·.
(iii) Show that the length of the perpendicular
from π΄ to ππ· is 1
3β65
Solution: Part (i)
First find two vectors on the plane e.g. π΄π΅ and π΄πΆ
π΄π΅ = ππ΅ β ππ΄ = (β24
β1) and π΄πΆ = ππΆ β ππ΄ = (
112)
Find the common perpendicular of the two
(β24
β1) Γ (
122) = (
93
β6)
We have now found the normal to the plane and now must find π
9π₯ + 3π¦ β 6π§ = π Substitute a point that lies on the plane e.g. π΄
9(2) + 3(β1) β 6(2) = π π = 3 9π₯ + 3π¦ β 6π§ = 3
Part (ii)
(i) πΆπ· = 2π·π΅ ππ· β ππΆ = 2ππ΅ β 2ππ·
ππ· =1
3(2ππ΅ + ππΆ) = (
366) = (
122)
Part (iii)
Finding a perpendicular from π΄ to ππ·; find the equation of the line ππ΄
ππ· = π(122)
A point π lies on ππ· and is perpendicular to π΄. First we must find the vector π΄π
π΄π = ππ β ππ΄ = (π β 22π + 12π β 2
)
Dot product of the point π΄π and the direction vector of ππ· is equal to zero as it is perpendicular
(π β 22π + 12π β 2
) . (122) = 0
9π = 4 β΄ π =4
9
Substitute back into general equation of ππ· to find π
π = (4
9,8
9,8
9)
To find the shortest distance, use Pythagoras theorem to find the distance from point π΄ to π
β(14
9)2
+ (β17
9)2
+ (10
9)2
= β65
9=
1
3β65
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7.15 Angles
When using dot product rule to fine an angle,
8. COMPLEX NUMBERS
8.1 The Basics π2 = β1
General form for all complex numbers:
π + ππ
From this we say:
π π(π + ππ) = π & πΌπ(π + ππ) = π
Conjugates:
o The complex number π§ and its conjugate π§β
π§ = π + ππ & π§β = π β ππ
Arithmetic:
o Addition and Subtraction: add and subtract real and
imaginary parts with each other
o Multiplication: carry out algebraic expansion, if π2
present convert to β1
o Division: rationalize denominator by multiplying
conjugate pair
o Equivalence: equate coefficients
8.2 Quadratic Use the quadratic formula:
o π2 β 4ππ is a negative value
o Pull out a negative and replace with π2
o Simplify to general form
Use sum of 2 squares: consider the example
Example:
Solve: π§2 + 4π§ + 13 = 0 Solution:
Convert to completed square form:
(π§ + 2)2 + 9 = 0 Utilize π2 as β1 to make it difference of 2 squares:
(π§ + 2)2 β 9π2 = 0 Proceed with general difference of 2 squares method:
(π§ + 2 + 3π)(π§ + 2 β 3π) = 0 π§ = β2 + 3π πππ π§ = β2 β 3π
8.3 Square Roots
Example:
Find square roots of: 4 + 3π
Solution:
We can say that:
β4 + 3π = π + ππ Square both sides
π2 β π2 + 2πππ = 4 + 3π Equate real and imaginary parts
π2 β π2 = 4 2ππ = 3 Solve simultaneous equation:
π =3β2
2 π =
β2
2
β΄ β4 + 3π =3β2
2+
β2
2π ππ β
3β2
2β
β2
2π
8.4 Argand Diagram For the complex number π§ = π + ππ
Its magnitude is defined as the following:
|π§| = βπ2 + π2
Its argument is defined as the following:
arg π§ = tanβ1π
π
Simply plot imaginary (π¦-axis) against real (π₯-axis):
Arguments:
Always: βπ < π < π
The position of π§β is a reflection in the π₯-axis of π§
Question asks for acute angle
Use +ve value of dot product
Question asks for obtuse angle
Use -ve value of dot product
Question asks for both angles
Use +ve and -ve value of dot
product
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8.5 Locus |π β π| = π
The locus of a point π§ such that |π§ β π€| = π, is a circle
with its centre at π€ and with radius π.
ππ«π (π β π) = π½
The locus of a point π§ such that arg(π§ β π€) = π is a ray
from π€, making an angle π with the positive real axis.
|π§ β π€| = |π§ β π£|
The locus of a point π§ such that |π§ β π€| = |π§ β π£| is the
perpendicular bisector of the line joining π€ and π£
{W11-P31} Question 10:
On a sketch of an Argand diagram, shade the region whose points represent the complex numbers π§ which satisfy the inequality|π§ β 3π| β€ 2. Find the greatest value of arg π§ for points in this region.
Solution:
The part shaded in blue is the answer. To find the greatest value of arg π§ within this region we must use the tangent at point on the circle which has the greatest value of π from the horizontal (red line)
The triangle magnified
sinπΌ =2
3
πΌ = 0.730
π = πΌ +π
2= 0.730 +
π
2= 2.30
{W11-P31} Question 10:
i. On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities
|π§ β 2 + 2π| β€ 2, arg π§ β€ β1
4π and π π π§ β₯ 1,
ii. Calculate the greatest possible value of π π π§ for points lying in the shaded region.
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Solution: Part (i)
Argand diagram:
Part (ii)
The greatest value for the real part of π§ would be the one which is furthest right on the π π axis but within the limits of the shaded area. Graphically:
Now using circle and Pythagoras theorems we can find the value of π₯:
π₯ = 2 Γ cos1
4π
π₯ = β2
β΄ greatest value of π π π§ = 2 + β2
8.6 Polar Form For a complex number π§ with magnitude π and
argument π:
π§ = π (cos π + π sin π) = π πππ
β΄ cos π + π sinπ = πππ
Polar Form to General Form:
Example:
Convert from polar to general, π§ = 4ππ
4π
Solution:
π = 4 arg π§ =π
4
β΄ π§ = 4 (cosπ
4+ π sin
π
4)
π§ = 4(β2
2+
β2
2π)
π§ = 2β2 + (2β2)π
General Form to Polar Form:
Example:
Convert from general to polar, π§ = 2β2 + (2β2)π Solution:
π§ = 2β2 + (2β2)π
π = β(2β2)2+ (2β2)
2= 4
π = tanβ12β2
2β2=
π
4
β΄ 4 (cosπ
4+ π sin
π
4) = 4π
π
4π
8.7 Multiplication and Division in Polar Form To find product of two complex numbers in polar form:
o Multiply their magnitudes
o Add their arguments
π§1π§2 = |π§1||π§2|(arg π§1 + arg π§2)
Example:
Find π§1π§2 in polar form given,
π§1 = 2(cosπ
4+ π sin
π
4) π§2 = 4(cos
π
8+ π sin
π
8)
Solution:
π§1π§2 = (2 Γ 4) (cos (π
4+
π
8) + π sin (
π
4+
π
8))
π§1π§2 = 8(cos3π
8+ π sin
3π
8)
To find quotient of two complex numbers in polar form:
o Divide their magnitudes
o Subtract their arguments
π§1
π§2 =
|π§1|
|π§2|(arg π§1 β arg π§2)
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Example:
Find π§1
π§2 in polar form given,
π§1 = 2(cosπ
4+ π sin
π
4) π§2 = 4(cos
π
8+ π sin
π
8)
Solution: π§1
π§2= (
2
4) (cos (
π
4β
π
8) + π sin (
π
4β
π
8))
π§1
π§2=
1
2(cos
π
8+ π sin
π
8)
8.8 De Moivreβs Theorem
π§π = π π(cos ππ + π sin ππ) = π πππππ
9. DIFFERENTIAL EQUATIONS Form a differential equation using the information given
o If something is proportional, add constant of
proportionality π
o If rate is decreasing, add a negative sign
Separate variables, bring ππ₯ and ππ‘ on opposite sides
Integrate both sides to form an equation
Add arbitrary constant
Use conditions given to find π and/or π
{W10-P33} Question 9:
A biologist is investigating the spread of a weed in a particular region. At time π‘ weeks, the area covered by the weed is π΄π2. The biologist claims that rate of
increase of π΄ is proportional to β2π΄ β 5. i. Write down a differential equation given info
ii. At start of investigation, area covered by weed was 7π2
. 10 weeks later, area covered = 27π2
Find the area covered 20 weeks after the start of the investigation.
Solution: Part (i)
ππ΄
ππ‘β β2π΄ β 5 = πβ2π΄ β 5
Part (ii)
Proceed to form an equation in π΄ and π‘: ππ΄
ππ‘= πβ2π΄ β 5
Separate variables 1
β2π΄ β 5ππ΄ = πππ‘
Integrate both side
ππ‘ + π = (2π΄ β 5)1
2 When π‘ = 0:
π΄ = 7 β΄ π = 3
ππ‘ + 3 = (2π΄ β 5)1
2
When π‘ = 10:
10π + 3 = (2(27) β 5)1
2
10π = β49 β 3 π = 0.4
Now substitute 20 as π‘ and then find π΄:
0.4(20) + 3 = (2π΄ β 5)1
2
11 = (2π΄ β 5)1
2 121 = 2π΄ β 5
π΄ = 63π2
{S13-P31} Question 10:
Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, π‘ minutes later, the volume of liquid in the tank is π ππ3. The liquid is flowing into the tank at a constant rate of 80 ππ3 per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to ππ ππ3 per minute where π is a positive constant.
i. Write down a differential equation describing this situation and solve it to show that:
π =1
π(80 β 80πβππ‘)
ii. π = 500 when π‘ = 15, show:
π =4 β 4πβ15π
25
Find π using iterations, initially π = 0.1 iii. Work out volume of liquid at π‘ = 20 and state
what happens to volume after a long time. Solution:
Part (i)
Represent the given information as a derivative: ππ
ππ‘= 80 β ππ
Proceed to solve the differential equation: ππ‘
ππ=
1
80 β ππ
ππ‘ =1
80 β ππππ
β«(1)ππ‘ = β«1
80 β ππππ
π‘ + π = β1
πln|80 β ππ|
Use the given information; when π‘ = 0, π = 0:
β΄ π = β1
πln(80)
Substitute back into equation:
π‘ β1
πln(80) = β
1
πln|80 β ππ|
π‘ =1
πln(80) β
1
πln|80 β ππ|
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π‘ =1
πln (
80
80 β ππ)
ππ‘ = ln (80
80 β ππ)
πππ‘ =80
80 β ππ
80 β ππ =80
πππ‘
ππ = 80 β 80πβππ‘
π =1
π(80 β 80πβππ‘)
Part (ii)
You did the mishwaar iterations and found: β΄ π = 0.14 (2π. π. )
Part (iii)
Simply substitute into the equationβs π‘:
π =1
0.14(80 β 80πβ0.14(20)) = 537 ππ3
The volume of liquid in the tank after a long time approaches the max volume:
π =1
0.14(80) = 571 ππ3
{W13-P31} Question 10:
A tank containing water is in the form of a cone with vertex πΆ. The axis is vertical and the semi-vertical angle is 60Β°, as shown in the diagram. At time π‘ = 0, the tank is full and the depth of water is π». At this instant, a tap at πΆ is opened and water begins to flow out. The volume of water in the tank decreases at a
rate proportional to ββ, where β is the depth of water
at time t. The tank becomes empty when π‘ = 60. i. Show that β and π‘ satisfy a differential
equation of the form: πβ
ππ‘= βπ΄ββ
3
2
Where π΄ is a positive constant. ii. Solve differential equation given in part i and
obtain an expression for π‘ in terms of β and π».
Solution: Part (i)
First represent info they give us as an equation:
π =1
3ππ2β
π = tan60 Γ β = ββ3
β΄ π =1
3π(ββ3)
2β = πβ3
ππ
πβ= 3πβ2
ππ
ππ‘β βββ = βπβ
1
2
Find the rate of change of β: πβ
ππ‘=
ππ
ππ‘Γ·
ππ
πβ
πβ
ππ‘=
βπβ1
2
3πβ2= β
π
3πββ
3
2
Part (ii)
ππ‘ =1
βπ΄ββ3
2
πβ
β«π΄ππ‘ = β«1
βββ3
2
πβ
π΄π‘ + π = β2
5β
5
2
Use given information to find unknowns; when π‘ = 0:
βπ΄(0) + π =2
5(π»)
5
2 β΄ π =2
5π»
5
2
When π‘ = 60: βπ΄(60) + π = 0
π = 60π΄
π΄ =1
150π»
5
2
Thus the initial equation becomes:
β1
150π»
5
2π‘ +2
5π»
5
2 =2
5β
5
2
π»5
2 (βπ‘
150+
2
5) =
2
5β
5
2
βπ‘
150+
2
5=
2β5
2
5π»5
2
π‘
150=
2
5β
2β5
2
5π»5
2
π‘ = 150(2
5β
2β5
2
5π»5
2
) = 60 β 60β5
2π»β5
2
π‘ = 60(1 β (β
π»)
5
2
)