TABLE OF CONTENTS 2
CHAPTER 1
3
CHAPTER 2
3
CHAPTER 3
5
CHAPTER 4
5
CHAPTER 5
Algebra Logarithmic & Exponential Functions Trigonometry Differentiation Integration
8 Solving Equations Numerically CHAPTER 6
8 Vectors CHAPTER 7
13
CHAPTER 8
16
CHAPTER 9
Complex Numbers Differential Equations
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CIE A LEVEL- MATHEMATICS [9709] 1.4 Partial Fractions
1. ALGEBRA 1.1 The Modulus Function ο· No line with a modulus ever goes under the x-axis ο· 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 2 |π₯ | = |π₯|2 = π₯ 2 |π₯| = |π| β π₯ 2 = π2 βπ₯ 2 = |π₯|
1.2 Polynomials ο· To find unknowns in a given identity o Substitute suitable values of π₯ OR o Equalize given coefficients of like powers of π₯ ο· Factor theorem: If (π₯ β π‘) is a factor of the function π(π₯) then π(π‘) = 0 ο· 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
ππ₯ + π π΄ π΅ β‘ + (ππ₯ + π)(ππ₯ + π ) ππ₯ + π ππ₯ + π π ο· Multiply (ππ₯ + π), substitute π₯ = β π and find π΄ π
ο· Multiply (ππ₯ + π ), substitute π₯ = β π and find π΅ ππ₯ 2 + ππ₯ + π π΄ π΅ πΆ β‘ + + 2 (ππ₯ + π)(ππ₯ + π ) ππ₯ + π ππ₯ + π (ππ₯ + π )2 π ο· Multiply (ππ₯ + π), substitute π₯ = β π and find π΄ π
ο· Multiply (ππ₯ + π )2, substitute π₯ = β π and find πΆ ο· Substitute any constant e.g. π₯ = 0 and find π΅
ππ₯ 2 + ππ₯ + π π΄ π΅π₯ + πΆ β‘ + 2 2 (ππ₯ + π)(ππ₯ + π ) ππ₯ + π ππ₯ + π π ο· Multiply (ππ₯ + π), substitute π₯ = β and find π΄ π
ο· Take
π΄ ππ₯+π
to the other side, subtract and simplify.
ο· Linear eqn. left at top is equal to π΅π₯ + πΆ ο· 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}
Expanding (1 + π₯)π where |π₯| < 1 π π(π β 1) 2 π(π β 1)(π β 2) 3 1+ π₯+ π₯ + π₯ +β― 1 1Γ2 1Γ2Γ3 ο· 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. ο· 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. ο· 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
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: 5 β 5π₯ 2+ (π₯ + 1)(2π₯ β 3)
Now proceed with normal case for the fraction: π΄ π΅ 5 β 5π₯ + = π₯ + 1 2π₯ β 3 (π₯ + 1)(2π₯ β 3) Page 2 of 17
CIE A LEVEL- MATHEMATICS [9709] π΄(2π₯ β 3) + π΅(π₯ + 1) = 5 β 5π₯ When π₯ = β1 β5π΄ = 5 + 5 π΄ = β2 3 When π₯ = 2 5 15 π΅ =5β 2 2 π΅ = β1 Thus the partial fraction is: β2 β1 2+ + π₯ + 1 2π₯ β 3
3.3 Graphs
2. LOGARITHMIC & EXPONENTIAL FUNCTIONS π¦ = π π₯ β log π π¦ = π₯ log π 1 = 0 log π π π log π π + log π π
β‘ β‘
log π π β log π π
β‘
log π π
β‘
log π π
β‘
log π π = 1 π log π π log π ππ π log π π log π log π 1 log π π
2.1 Graphs of ln(x) and ex
3.4 Double Angle Identities sin 2π΄ β‘ 2 sin π΄ cos π΄ cos 2π΄ β‘ (cos π΄)2 β (sin π΄)2 β‘ 2(cos π΄)2 β 1 β‘ 1 β 2(sin π΄)2 2 tan π΄ tan 2π΄ β‘ 1 β (tan π΄)2
3.5 Addition Identities sin(π΄ Β± π΅) β‘ sin π΄ cos π΅ Β± cos π΄ sin π΅ cos(π΄ Β± π΅) β‘ cos π΄ cos π΅ β sin π΄ sin π΅ tan π΄ Β± tan π΅ tan(π΄ Β± π΅) β‘ 1 β tan π΄ tan π΅
3.6 Changing Forms π sin π₯ Β± π cos π₯ βΊ π
sin(π₯ Β± πΌ) π cos π₯ Β± π sin π₯ βΊ π
cos(π₯ β πΌ) Where π
= βπ2 + π 2 and
3. TRIGONOMETRY
π
cos πΌ = π, π
sin πΌ = π {S13-P33}
3.1 Ratios sin π
tan π = cos π cosec π =
1 sin π
with
1 2
0<πΌ< π Question 9:
1
sec π = cos π cot π =
3.2 Identities (cos π)2 + (sin π)2 β‘ 1 1 + (tan π)2 β‘ (sec π)2 (cot π)2 + 1 β‘ (cosec π)2
cos π sin π
Diagram shows curve, π¦ = sin2 2π₯ cos π₯, for 0 β€ π₯ β₯ π , and π is maximum point. Find the π₯ coordinate of 2 π.
Page 3 of 17
CIE A LEVEL- MATHEMATICS [9709] Solution:
Use product rule to differentiate: π’ = sin2 2π₯ π£ = cos π₯ β² π’ = 4 sin 2π₯ cos 2π₯ π£ β² = β sin π₯ ππ¦ = π’β² π£ + π’π£ β² ππ₯ ππ¦ = (4 sin 2π₯ cos 2π₯)(cos π₯) + (sin2 2π₯)(β sin π₯) ππ₯ ππ¦ = 4 sin 2π₯ cos 2π₯ cos π₯ β sin2 2π₯ sin π₯ ππ₯ Use following identities: cos 2π₯ = 2 cos2 π₯ β 1 sin 2π₯ = 2 sin π₯ cos π₯ sin2 π₯ = 1 β cos 2 π₯ Equating to 0: ππ¦ =0 ππ₯ β΄ 4 sin 2π₯ cos 2π₯ cos π₯ β sin2 2π₯ sin π₯ = 0 4 sin 2π₯ cos 2π₯ cos π₯ = sin2 2π₯ sin π₯ Cancel sin 2π₯ on both sides 4 cos 2π₯ cos π₯ = sin 2π₯ sin π₯ Substitute identities 4(2 cos2 π₯ β 1) cos π₯ = (2 sin π₯ cos π₯) sin π₯ 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 3 cos2 π₯ = 5 cos π₯ = 0.7746 π₯ = cos β1 (0.7746) π₯ = 0.6847 β 0.685 {W13-P31}
Show that: cos 2π =
2 sin 2π β π 4π Solution:
First express area of sector ππ΅π΄πΆ 1 ππππ‘ππ π΄πππ = ππ 2 2 1 ππ΅π΄πΆ = (2π β 4π)π 2 = (π β 2π)π 2 2 Now express area of sector π΄π΅πΆ 1 π΄π΅πΆ = (2π)(πΏππππ‘β ππ π΅π΄)2 2 Express π΅π΄ using sine rule π sin(π β 2π) π΅π΄ = sin π Use double angle rules to simplify this expression π sin 2π π΅π΄ = sin π 2π sin π cos π = sin π = 2π cos π Substitute back into initial equation 1 π΄π΅πΆ = (2π)(2π cos π)2 2 π΄π΅πΆ = 4ππ 2 cos 2 π Now express area of kite π΄π΅ππΆ π΄π΅ππΆ = 2 Γ π΄πππ ππ ππππππππ 1 π΄π΅ππΆ = 2 Γ π 2 sin(π β 2π) 2 = π 2 sin(π β 2π) Finally, the expression of shaded region equated to half of circle 1 4π 2 π cos2 π + π 2 (π β 2π) β π 2 sin(π β 2π) = ππ 2 2 Cancel our π 2 on both sides for all terms 4π cos 2 π + π β 2π β (sin π cos 2π + sin 2π cos π) =
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.
Some things in the double angle cancel out 1 4π cos 2 π + π β 2π β sin 2π = π 2 Use identity here cos 2π + 1 1 4π ( ) + π β sin 2π β 2π = π 2 2 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 2 sin 2π β π cos 2π = 4π
Page 4 of 17
1 π 2
CIE A LEVEL- MATHEMATICS [9709] 4. DIFFERENTIATION
1 β cos(ππ₯ + π) π 1 cos(ππ₯ + π) sin(ππ₯ + π) π 1 sec 2 (ππ₯ + π) tan(ππ₯ + π) π (ππ₯ + π)π+1 (ππ₯ + π)π π(π + 1) ο· Use trigonometrical relationships to facilitate complex trigonometric integrals ο· Integrate by decomposing into partial fractions sin(ππ₯ + π)
4.1 Basic Derivatives π₯π
ππ₯ πβ1 ππ’ π’ π ππ₯ ππ’β ππ₯
ππ’ ln π’
π’ π cos ππ₯ β π sin ππ₯ π sec 2 ππ₯
sin ππ₯ cos ππ₯ tan ππ₯
4.2 Chain, Product and Quotient Rule
5.2 Integration by π-Substitution
ο· Chain Rule:
ππ₯ ππ’ ππ’ ο· Make π equal to something: when differentiated, multiply the substituted form directly ο· Make π equal to something: when differentiated, multiply the substituted form with its reciprocal ο· With definite integrals, change limits in terms of π’ β« π(π₯) ππ₯ = β« π(π₯)
ππ¦ ππ¦ ππ’ = Γ ππ₯ ππ’ ππ₯
ο· Product Rule:
π ππ£ ππ’ (π’π£) = π’ +π£ ππ₯ ππ₯ ππ₯
ο· Quotient Rule:
ππ’
ππ£
π£ βπ’ π π’ ( ) = ππ₯ 2 ππ₯ ππ₯ π£ π£
{W12-P33}
Question 7:
The diagram shows part of curve π¦ = sin3 2π₯ cos 3 2π₯. The shaded region shown is bounded by the curve and the π₯-axis and its exact area is denoted by π΄.
4.3 Parametric Equations ππ¦ ππ¦ ππ₯ = Γ· ππ₯ ππ‘ ππ‘ ο· 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 ο· These represent circles or lines with circular curves, on a Cartesian plane ο· Difficult to rearrange in form π¦ = β΄ differentiate as is ο· Differentiate π₯ terms as usual ο· For π¦ terms, differentiate the same as you would π₯ but ππ¦
multiply with ππ₯ ππ¦
ο· Then make ππ₯ the subject of formula for derivative
5. INTEGRATION 5.1 Basic Integrals ππ₯
π
π ππ₯+π 1 ππ₯ + π
π₯ π+1 π +π (π + 1) 1 ππ₯+π π π 1 ln|ππ₯ + π| π
Use the substitution π’ = sin 2π₯ 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 π 3π sin π₯ = 0 at π₯ = 0, π, 2π cos π₯ = 0 at π₯ = 2 , 4 Choose the two closest to 0 because the shaded area has gone through π¦ = 0 only twice π β΄ 0 and 2 Since it is sin 2π₯ and cos 2π₯, divide both by 2 π β΄ Limits are 0 and 4 Integrate by π’ substitution, let: ππ’ ππ₯ 1 π’ = sin 2π₯ = 2 cos 2π₯ = ππ₯ ππ’ 2 cos 2π₯ sin3 2π₯ cos3 2π₯ β‘ (sin 2π₯)3 (cos2 2π₯) cos 2π₯ β‘ (sin3 2π₯ Γ (1 β sin2 2π₯)) cos 2π₯
Page 5 of 17
CIE A LEVEL- MATHEMATICS [9709] {W13-P31}
1 β‘ (sin3 2π₯ β sin5 π₯) cos 2π₯ Γ 2 cos 2π₯ 1 3 5 β‘ (π’ β π’ ) 2 Now integrate: 1 1 π’4 π’6 β«(π’3 β π’5 ) = ( β ) 2 2 4 6 π The limits are π₯ = 0 and π₯ = 4 . In terms of π’,
Find the exact value of 4
β« 1
Substitute limits 1 14 16 1 04 06 1 ( β )β ( β )= 2 4 6 2 4 6 24 πβ² (π)
πβ² (π) ππ₯ = ln|π(π₯)| + π π(π) Question 10:
By splitting into partial fractions, show that: 2
β« 1
2π₯ 3 β 1 3 1 16 ππ₯ = + ln ( ) 2 π₯ (2π₯ β 1) 2 2 27 Solution:
Write as partial fractions 2
2
2π₯ 3 β 1 2 1 3 β« 2 ππ₯ β‘ β« 1 + + 2 + ππ₯ π₯ (2π₯ β 1) π₯ π₯ 2π₯ β 1 1
1
3 β‘ π₯ + 2 ln π₯ β π₯ β1 β ln|2π₯ β 1| 2 Substitute the limits 1 3 3 2 + 2 ln 2 β β ln 3 β 1 β 2 ln 1 + 1 + ln 1 2 2 2 3 1 1 1 3 1 16 + ln 16 + ln 3 β‘ + ln 2 2 2 3 2 2 27
5.4 Integrating By Parts ππ£ ππ’ ππ₯ = π’π£ β β« π£ ππ₯ ππ₯ ππ₯ For a definite integral: π π ππ£ ππ’ β« π’ ππ₯ = [π’π£]ππ β β« π£ ππ₯ ππ₯ ππ₯ π π What to make π: L A T E Logs Algebra Trig π β«π’
ππ₯
1
= π₯ 2 ln π₯
β‘ 2βπ₯ ln π₯ β 4βπ₯ Substitute limits = 4 ln 4 β 4
π(π)
{S10-P32}
βπ₯
βπ₯ Integrate by parts, let: 1 1 ππ’ 1 ππ£ π’ = ln π₯ = = π₯ β2 π£ = 2π₯ 2 ππ₯ π₯ ππ₯ 1 1 1 β1 2 2 β΄ ln π₯ 2π₯ β β« 2π₯ Γ π₯ β‘ 2βπ₯ ln π₯ β β« 2π₯ β2
π
β«
ln π₯
Solution:
Convert to index form: ln π₯
π’ = sin 2(0) = 0 and π’ = sin 2 ( 4 ) = 1
5.3 Integrating
Question 3:
5.5 Integrating Powers of Sine or Cosine To integrate sin π₯ or cos π₯ with a power: ο· If power is odd, pull out a sin π₯ or cos π₯ and use Pythagorean identities and double angle identities ο· If power is even, use the following identities 1 1 sin2 π₯ = β cos(2π₯) 2 2 1 1 2 cos π₯ = + cos(2π₯) 2 2
5.6 Integrating ππππ π ππππ π If π or π are odd and even, then: ο· Factor out one power from odd trig function ο· Use Pythagorean identities to transform remaining even trig function into the odd trig function ο· Let u equal to odd trig function and integrate If π and π are both even, then: ο· Replace all even powers using the double angle identities and integrate If π and π are both odd, then: ο· Choose one of the trig. functions & factor out one power ο· 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: ο· Let π’ equal to the trig function whose power doesnβt equal 1 then integrate ο· If both are 1, then let π’ equal either Page 6 of 17
CIE A LEVEL- MATHEMATICS [9709] {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 1 π 3
β« sin4 π ππ 1 π 6
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(sin 2π)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)
Solution: Part (i)
Change to index form: 1 = cos β1 π₯ cos π₯ Differentiate by chain rule: ππ¦ = β1(cos π₯)β2 Γ (β sin π₯) ππ₯ sin π₯ sin π₯ 1 β1(cos π₯)β2 Γ (β sin π₯) β‘ β‘ Γ 2 cos π₯ cos π₯ cos π₯ sin π₯ 1 Γ β‘ sec π₯ tan π₯ cos π₯ cos π₯ Part (ii)
Multiply numerator and denominator by sec π₯ + tan π₯ sec π₯ + tan π₯ sec π₯ + tan π₯ β‘ (sec π₯ β tan π₯) (sec π₯ + tan π₯) sec 2 π₯ β tan2 π₯ sec π₯ + tan π₯ sec π₯ + tan π₯ β‘ β‘ sec π₯ + tan π₯ 2 2 sec π₯ β tan π₯ 1 Part (iii)
Substitute identity from (part ii) 1 β‘ (sec π₯ + tan π₯)2 (sec π₯ β tan π₯)2 Open out brackets (sec π₯ + tan π₯)2 β‘ sec π₯ + 2 sec π₯ tan π₯ + tan2 π₯ β‘ sec 2 π₯ + 2 sec π₯ tan π₯ + sec 2 π₯ β 1 β‘ 2sec 2 π₯ + 2 sec π₯ tan π₯ β 1 β‘ 2 sec 2 π₯ β 1 + 2 sec π₯ tan π₯
Use identity from (part i):
2
1 π 3
1 β« cos 4π β 4 cos 2π + 3 8 1 π 6
Part (iv)
1
π 3 1 1 β‘ [ sin 4π β 2 sin π + 3π]1 8 4 π 6 Substitute limits 1 β‘ (2π β β3) 32
{W12-P32}
β«
β‘ β« 2 sec 2 π₯ β 1 + 2 sec π₯ tan π₯ ππ₯ β‘ 2 β« sec 2 π₯ β β« 1 + 2 β« sec 2 π₯ tan2 π₯
Question 5:
(i) By differentiating sec π₯ then
ππ¦ ππ₯
1 , cos π₯
show that if π¦ =
= sec π₯ tan π₯ 1
(ii) Show that sec π₯βtan π₯ β‘ sec π₯ + tan π₯ (iii) Deduce that: 1 β‘ 2 sec 2 π₯ β 1 + 2 sec π₯ tan π₯ (sec π₯ β tan π₯)2 (iv) Hence show that: 1 π 4
β«
0
1 1 ππ₯ = (8β2 β π) 2 (sec π₯ β tan π₯) 4
1 ππ₯ (sec π₯ β tan π₯)2
Using differential from part i: β‘ 2 tan π₯ β π₯ + 2 sec π₯ Substitute boundaries: 1 = (8β2 β π) 4
5.5 Trapezium Rule π΄πππ =
ππππ‘β ππ 1π π‘ ππ‘πππ Γ [1π π‘ βπππβπ‘ + πΏππ π‘ βπππβπ‘ 2 + 2(π π’π ππ β ππππππ)]
ππππ‘β ππ 1π π‘ ππ‘πππ =
Page 7 of 17
πβπ ππ. ππ πππ‘πππ£πππ
for
π
β«π ππ₯
CIE A LEVEL- MATHEMATICS [9709] 6. SOLVING EQUATIONS NUMERICALLY
7.2 Parallel, Skew or Intersects
6.1 Approximation
For the two lines: βββββ ππ΄ = πΜ + π πΜ
ο· To find root of a graph, find point where graph passes through π₯-axis β΄ look for a sign change ο· Carry out decimal search o Substitute values between where a sign change has occurred o Closer to zero, greater accuracy
6.2 Iteration ο· To solve equation π(π₯) = 0, you can rearrange π(π₯) into a form π₯ = β― ο· This function represents a sequence that starts at π₯β, moving to π₯π ο· Substitute a value for π₯β and put back into function getting π₯1 and so on. ο· As you increase π, value becomes more accurate ο· Sometimes iteration donβt work, these functions pare called divergent, and you must rearrange formula for π₯ in another way ο· For a successful iterative function, you need a convergent sequence
7. VECTORS
βββββ ππ΅ = Μπ + π‘πΜ
ο· Parallel: o For the lines to be parallel πΜ must equal πΜ or be in some ratio to it e.g. 1: 2 ο· Intersects: βββββ = ππ΅ βββββ o Make ππ΄ o If simultaneous works then intersects o If unknowns cancel then no intersection ο· Skew: o First check whether line parallel or not o If not, then make βββββ ππ΄ = βββββ ππ΅ 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 ο· Use dot product rule on the two direction vectors: π. π = cos π |π||π| ο· Note: π and π must be moving away from the point at which they intersect
7.1 Equation of a Line
7.4 Finding the Equation of a Line
ο· The column vector form: 1 1 π = ( 3 ) + π‘ ( 1) β2 3 ο· The linear vector form: π = π’ + 3π£ β 2π€ + π‘(π’ + π£ + 3π€) ο· The parametric form: π₯ = 1 + π‘, π¦ = 2 + π‘, π§ = β2 + 3π‘ ο· The cartesian form; rearrange parametric π₯β1 π¦β3 π§+2 = = 1 1 3
ο· 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 ο· 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 ο· AKA: shortest distance from the point to the line ο· Find vector for the point, π΅, on the line 1 1 Vector equation of the line: π«Μ = ( 3 ) + π‘ (1) β2 3 Page 8 of 17
CIE A LEVEL- MATHEMATICS [9709] 1+π‘ β΄ βββββ ππ΅ = ( 3 + π‘ ) 3π‘ β 2 ο· π΄ is the point given 2 βββββ ππ΄ = (3) 4 1+π‘β2 π‘β1 βββββ = ( 3 + π‘ β 3 ) = ( π‘ ) β΄ π΄π΅ 3π‘ β 2 β 4 3π‘ β 6 ο· Use Dot product of π΄π΅ and the direction vector βββββ π΄π΅. π = cos 90 π‘β1 1 ( π‘ ) . (1) = 0 3π‘ β 6 3 1(π‘ β 1) + 1(π‘) + 3(3π‘ β 6) = 0 11π‘ β 19 = 0 19 π‘= 11 ο· Substitute π‘ into equation to get foot ο· Use Pythagorasβ Theorem to find distance {S08-P3}
Substitute 1 into 2: 2 + 2π‘ = 5 + π‘ β΄ π‘ = 3 and then π = β6 Equation 3: 3=2βπ‘ Substitute the value of π‘ 3 = 2 β 3 so 3 = β1 This is incorrect therefore lines donβt intersect Part (ii)
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΄ 2 1 1 = (1) β (2) = (β1) 3 3 0 1 1 1 β2 ππ = (2) + π (β1) and π = (5) + π‘ ( 1 ) 3 0 2 β1 Equating the two lines 1+π 1 β 2π‘ (2 β π ) = ( 5 + π‘ ) 3 2βπ‘ Equation 1: 1 + π = 1 β 2π‘ so π = β2π‘ Equation 2: 2 β π = 5 + π‘
Angle ππ΄π΅ is formed by the intersection of the lines π΄π and π΄π΅ 1 β 2π‘ π =(5+π‘ ) 2βπ‘ π΄π = ππ β ππ΄ 1 β 2π‘ 1 β2π‘ π΄π = ( 5 + π‘ ) β (2) = ( 3 + π‘ ) 2βπ‘ 3 β1 β π‘ 1 π΄π΅ = (β1) 0 Now use the dot product rule to form an eqn. |π΄π. π΄π΅| β3π‘ β 3 ; = cos 60 |π΄π||π΄π΅| β6π‘ 2 + 8π‘ + 10 Γ β2 1 β3π‘ β 3 = β6π‘ 2 + 8π‘ + 10 Γ β2 2 36π‘ 2 + 72π‘ + 36 = 12π‘ 2 + 16π‘ + 20 24π‘ 2 + 56π‘ + 16 = 0 1 π‘ = β 3 or π‘ = β2 {W11-P31}
Question:
With respect to the origin π, the position vectors of βββββ = π’ + 2π£ + 2π€ two points π΄ and π΅ are given by ππ΄ and βββββ ππ΅ = 3π’ + 4π£. The point π lies on the line through π΄ βββββ and π΅, and βββββ π΄π = ππ΄π΅ βββββ = (1 + 2π)π’ + (2 + 2π)π£ + (2 β (i) ππ 2π) π€ (ii) By equating expressions for cos π΄ππ and cos π΅ππ in terms of π, find the value of π for which ππ bisects the angle π΄ππ΅. Solution: Part (i)
βββββ βββββ = π(ππ΅ β ππ΄) π΄π = ππ΄π΅ 3 1 2 = π (4) β (2) = ( 2 ) 0 2 β2 2π β΄ π΄π = ( 2π ) β2π Page 9 of 17
CIE A LEVEL- MATHEMATICS [9709] o πΜ is what we want to find o πΜ is the cross product of 2 vectors parallel to the plane
1 2π 2π ππ = ππ΄ + ( 2π ) = (2) + ( 2π ) 2 β2π β2π
Part (ii)
Interpreting the question gives the information that π΄ππ is equal to π΅ππ β΄ cos π΄ππ is equal to cos π΅ππ. Now you can equate the two dot product equations ππ΄. ππ 9 + 2π cos π΄ππ = = |ππ΄||ππ| 3β9 + 4π + 12π2 ππ΅. ππ 11 + 14π cos π΅ππ = = |ππ΅||ππ| 5β9 + 4π + 12π2 9 + 2π 11 + 14π = 3β9 + 4π + 12π2 5β9 + 4π + 12π2 Cancel out the denominator to give you 9 + 2π 11 + 14π = 3 5 45 + 10π = 33 + 42π 3 12 = 32π and β΄ π = 8
7.6 Equation of a Plane
Μ and π΄πΆ Μ then πΜ = ππ΄ o If we use π΄π΅ β4 o β΄ πΜ = π΄π΅ Γ π΄πΆ = (β5) β1 o Substitute point π΄ to get πΜ. πΜ β4 o β΄ πΜ . (β5) = β13 β1 ο· Given a point and a line on the plane: 2 β2 o π΄(1,2,3) and πΜ = (1) + π ( 1 ) 0 1 o Make 2 points on the line o Substitute different values for π o Repeat 3 point process ο· 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 ο· 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 ο· If a line is parallel to plane, the dot product of the direction vector and normal of the plane is zero
ο· Scalar product form: β4 π«Μ. (β5) = β13 β1 The vector after π«Μ is the normal to the plane ο· 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 ο· Cartesian form: 4π₯ + 5π¦ + π§ = 13
7.7 Cross Product Rule ππ β ππ π π (π) Γ (π ) = ( ππ β ππ ) ππ β ππ π π
7.8 Finding the Equation of a Plane ο· Given 3 points on a plane: o π΄(1,2, β1), π΅(2,1,0), πΆ(β1,3,2) o Use this equation: πΜ . πΜ = πΜ. πΜ
7.10 Finding the Point of Intersection between Line and Plane ο· Form Cartesian equation for line ο· Form Cartesian equation for plane ο· 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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CIE A LEVEL- MATHEMATICS [9709] Solution: Part (i)
7.12 β₯ Distance from a Point to a Plane
3 π΄π΅ = ππ΅ β ππ΄ = ( 1 ) β1 The equation of the line π΄π΅ = ππ΄ + π΄π΅ 2 3 2 + 3π = (β3) + π ( 1 ) = ( π β 3 ) 2 β1 2βπ Substitute values into plane equation π₯ + π¦ = 5 βΉ 2 + 3π + π β 3 = 5 3 4π β 1 = 5 βΉ π = 2 Substitute lambda back into general line equation 6.5 2 + (3 Γ 1.5) ( 1.5 β 3 ) = (β1.5) 0.5 2 β 1.5
π·=
βπ2 + π 2 + π 2 ο· Point πΉ is the foot of the perpendicular {S12-P32}
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. 3 1 ( 1 ) . (π) = 0 β1 π 3+πβπ =0 so π = 3 + π Using the fact that the plane π and π intersect at an angle of 60Β° 1 1 (1) . (π) 1 0 π = cos 60 = 2 β2 Γ β1 + π 2 + π 2 2 2 2 + 2π = β2 + 2π + 2π 4π 2 + 8π + 4 = 2π 2 + 2π 2 + 2 Substitute the first equation into π 2π 2 + 8π + 2 β 18 β 12π β 2π 2 = 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 NonParallel Planes Μπ Γ π§ Μπ ο· The direction vector of this line is π§ Μπ is the normal of the first plane ο·π§ Μπ is the normal of the second plane ο·π§
|ππ₯1 + ππ¦1 + ππ§1 + π|
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 2 1 (1) . ( 2 ) = 0 2 β2 Part (ii)
Any point on π would have the value 1 + 2π (1+π ) 2π 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 4
β8+4π
π·π = |3| and π·π = | 3 | Equate them as they equal the same distance 4 β8 + 4π | |=| | βΉ |4| = |β8 + 4π| 3 3 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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CIE A LEVEL- MATHEMATICS [9709] Substitute lambda values back into vector general line π equation to get the two points π1 and π2 7 3 π1 = (4) and π2 = (2) 5 1 Use Pythagorasβs Theorem to find the distance β42 + 22 + 42 = β36 = 6
(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 1
from π΄ to ππ· is 3 β65 Part (i)
Solution:
First find two vectors on the plane e.g. π΄π΅ and π΄πΆ β2 1 π΄π΅ = ππ΅ β ππ΄ = ( 4 ) and π΄πΆ = ππΆ β ππ΄ = (1) β1 2 Find the common perpendicular of the two β2 1 9 ( 4 ) Γ (2) = ( 3 ) β1 2 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
7.13 Angle between Two Planes Μ. Μπ π§ π π§ |π§ Μ||π§ π Μ| π Μβs here represent the normals of each plane ο· The π§ ο· Ignore any negative signs cos π =
7.14 Angle between a Line and a Plane
Part (ii)
(i) πΆπ· = 2π·π΅ ππ· β ππΆ = 2ππ΅ β 2ππ· 3 1 1 ππ· = (2ππ΅ + ππΆ) = (6) = (2) 3 6 2
ο· First find β
:
Part (iii)
Μ. πΜ π§ cos β
= |π§ Μ||πΜ| ο· π = 90 β β
ο· 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 2 0 3 βββββ = (β1), ππ΅ βββββ = (3) and ππΆ βββββ = (0). ππ΄ 2 1 4 The point π· lies on π΅πΆ, between π΅ and πΆ, and is such that πΆπ· = 2π·π΅.
Finding a perpendicular from π΄ to ππ·; find the equation of the line ππ΄ 1 ππ· = π (2) 2 A point π lies on ππ· and is perpendicular to π΄. First we must find the vector π΄π πβ2 π΄π = ππ β ππ΄ = (2π + 1) 2π β 2 Dot product of the point π΄π and the direction vector of ππ· is equal to zero as it is perpendicular 1 πβ2 (2π + 1) . (2) = 0 2 2π β 2 4 9π = 4 β΄ π = 9 Substitute back into general equation of ππ· to find π 4 8 8 π=( , , ) 9 9 9 To find the shortest distance, use Pythagoras theorem to find the distance from point π΄ to π
Page 12 of 17
β(
14 2 17 2 10 2 65 1 ) + (β ) + ( ) = β = β65 9 9 9 9 3
CIE A LEVEL- MATHEMATICS [9709] 7.15 Angles
8.3 Square Roots
ο· When using dot product rule to fine an angle, Question asks for acute angle
Question asks for obtuse angle
Example:
Find square roots of: 4 + 3π
Question asks for both angles
Solution:
We can say that: Use +ve value of dot product
Use -ve value of dot product
Use +ve and ve value of dot product
8. COMPLEX NUMBERS 8.1 The Basics π 2 = β1 ο· General form for all complex numbers: π + ππ ο· From this we say: π
π(π + ππ) = π & πΌπ(π + ππ) = π ο· Conjugates: o The complex number π§ and its conjugate π§ β π§ = π + ππ & π§ β = π β ππ ο· Arithmetic: o Addition and Subtraction: add and subtract real and imaginary parts with each other o Multiplication: carry out algebraic expansion, if π 2 present convert to β1 o Division: rationalize denominator by multiplying conjugate pair o Equivalence: equate coefficients
β4 + 3π = π + ππ Square both sides π2 β π 2 + 2πππ = 4 + 3π Equate real and imaginary parts π2 β π 2 = 4 2ππ = 3 Solve simultaneous equation: 3β2 2 3β2 β2 β΄ β4 + 3π = + π 2 2 π=
π=
β2 2
ππ
β
3β2 β2 β π 2 2
8.4 Argand Diagram For the complex number π§ = π + ππ ο· Its magnitude is defined as the following: |π§| = βπ2 + π 2 ο· Its argument is defined as the following: π arg π§ = tanβ1 π ο· Simply plot imaginary (π¦-axis) against real (π₯-axis):
8.2 Quadratic ο· Use the quadratic formula: o π 2 β 4ππ is a negative value o Pull out a negative and replace with π 2 o Simplify to general form ο· Use sum of 2 squares: consider the example
Arguments: Always: βπ < π < π
Example:
Solve: π§ 2 + 4π§ + 13 = 0 Solution:
Convert to completed square form: (π§ + 2)2 + 9 = 0 2 Utilize π as β1 to make it difference of 2 squares: (π§ + 2)2 β 9π 2 = 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
ο· The position of π§ β is a reflection in the π₯-axis of π§
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CIE A LEVEL- MATHEMATICS [9709] 8.5 Locus
{W11-P31}
|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.
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.
Question 10:
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)
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.
The triangle magnified |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£ 2 3 πΌ = 0.730 sin πΌ =
π=πΌ+ {W11-P31}
i.
π π = 0.730 + = 2.30 2 2 Question 10:
On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities 1 |π§ β 2 + 2π| β€ 2, arg π§ β€ β π and π
π π§ β₯ 1, 4 ii. Calculate the greatest possible value of π
π π§ for points lying in the shaded region.
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CIE A LEVEL- MATHEMATICS [9709] Solution: Part (i)
Polar Form to General Form: Example:
Argand diagram:
π
Convert from polar to general, π§ = 4π 4 π π π
=4 arg π§ = 4 π π β΄ π§ = 4 (cos + π sin ) 4 4 β2 β2 π§ = 4( + π) 2 2 π§ = 2β2 + (2β2)π General Form to Polar Form:
Solution:
Example:
Convert from general to polar, π§ = 2β2 + (2β2)π Solution:
π§ = 2β2 + (2β2)π 2
2
π
= β(2β2) + (2β2) = 4
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:
π = tanβ1
2β2
=
π 4
2β2 π π π β΄ 4 (cos + π sin ) = 4π 4 π 4 4
8.7 Multiplication and Division in Polar Form ο· To find product of two complex numbers in polar form: o Multiply their magnitudes o Add their arguments π§1 π§2 = |π§1 ||π§2 |(arg π§1 + arg π§2 ) Example:
Find π§1 π§2 in polar form given, π π π π π§1 = 2 (cos + π sin ) π§2 = 4 (cos + π sin ) 4 4 8 8
Solution:
Now using circle and Pythagoras theorems we can find the value of π₯: 1 π₯ = 2 Γ cos π 4 π₯ = β2 β΄ greatest value of π
π π§ = 2 + β2
8.6 Polar Form ο· For a complex number π§ with magnitude π
and argument π: π§ = π
(cos π + π sin π) = π
π ππ β΄ cos π + π sin π = π ππ
π π π π π§1 π§2 = (2 Γ 4) (cos ( + ) + π sin ( + )) 4 8 4 8 3π 3π π§1 π§2 = 8 (cos + π sin ) 8 8
ο· To find quotient of two complex numbers in polar form: o Divide their magnitudes o Subtract their arguments |π§1 | π§1 (arg π§1 β arg π§2 ) = |π§2 | π§2
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CIE A LEVEL- MATHEMATICS [9709] Example:
When π‘ = 10:
π§
Find π§1 in polar form given, 2 π π π§1 = 2 (cos + π sin ) 4 4
1
π π π§2 = 4 (cos + π sin ) 8 8
Solution:
π§1 2 π π π π = ( ) (cos ( β ) + π sin ( β )) π§2 4 4 8 4 8 π§1 1 π π = (cos + π sin ) π§2 2 8 8
10π + 3 = (2(27) β 5)2 10π = β49 β 3 π = 0.4 Now substitute 20 as π‘ and then find π΄: 1
0.4(20) + 3 = (2π΄ β 5)2 1
11 = (2π΄ β 5)2 121 = 2π΄ β 5 π΄ = 63π2
8.8 De Moivreβs Theorem π§ π = π
π (cos ππ + π sin ππ) = π
π π πππ
{S13-P31}
9. DIFFERENTIAL EQUATIONS ο· 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 ο· Separate variables, bring ππ₯ and ππ‘ on opposite sides ο· Integrate both sides to form an equation ο· Add arbitrary constant ο· 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)
Part (ii)
ππ΄ β β2π΄ β 5 = πβ2π΄ β 5 ππ‘
Proceed to form an equation in π΄ and π‘: ππ΄ = πβ2π΄ β 5 ππ‘ Separate variables 1 ππ΄ = πππ‘ β2π΄ β 5 Integrate both side 1
ππ‘ + π = (2π΄ β 5)2 When π‘ = 0: π΄=7
β΄
π=3 1
ππ‘ + 3 = (2π΄ β 5)2
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: 1 π = (80 β 80π βππ‘ ) π ii. π = 500 when π‘ = 15, show: 4 β 4π β15π π= 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 = 80 β ππ ππ‘ Proceed to solve the differential equation: ππ‘ 1 = ππ 80 β ππ 1 ππ‘ = ππ 80 β ππ 1 β«(1)ππ‘ = β« ππ 80 β ππ 1 π‘ + π = β ln|80 β ππ| π Use the given information; when π‘ = 0, π = 0: 1 β΄ π = β ln(80) π Substitute back into equation: 1 1 π‘ β ln(80) = β ln|80 β ππ| π π 1 1 π‘ = ln(80) β ln|80 β ππ| π π
Page 16 of 17
CIE A LEVEL- MATHEMATICS [9709] Solution:
1 80 ln ( ) π 80 β ππ 80 ππ‘ = ln ( ) 80 β ππ 80 π ππ‘ = 80 β ππ 80 80 β ππ = ππ‘ π ππ = 80 β 80π βππ‘ 1 π = (80 β 80π βππ‘ ) π π‘=
Part (ii)
Part (i)
First represent info they give us as an equation: 1 π = ππ 2 β 3 π = tan 60 Γ β = ββ3 1 2 β΄ π = π(ββ3) β = πβ3 3 ππ = 3πβ2 πβ 1 ππ β βββ = βπβ2 ππ‘ Find the rate of change of β: πβ ππ ππ = Γ· ππ‘ 1ππ‘ πβ 3 πβ βπβ2 π = = β β β2 2 ππ‘ 3πβ 3π
You did the mishwaar iterations and found: β΄ π = 0.14 (2π. π. ) Part (iii)
Simply substitute into the equationβs π‘: 1 π= (80 β 80π β0.14(20) ) = 537 ππ3 0.14 The volume of liquid in the tank after a long time approaches the max volume: 1 (80) = 571 ππ3 π= 0.14 {W13-P31}
Part (ii)
1
ππ‘ =
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: 3 πβ = βπ΄ββ2 ππ‘ 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».
3 πβ βπ΄ββ2 1 β« π΄ππ‘ = β« 3 πβ βββ2 2 5 π΄π‘ + π = β β2 5 Use given information to find unknowns; when π‘ = 0: 5 2 2 5 βπ΄(0) + π = (π»)2 β΄ π = π» 2 5 5 When π‘ = 60: βπ΄(60) + π = 0 π = 60π΄ 1 5 π΄= π»2 150 Thus the initial equation becomes: 1 5 2 5 2 5 β π» 2 π‘ + π» 2 = β2 150 5 5 5 π‘ 2 2 5 π» 2 (β + ) = β2 150 5 5 5 π‘ 2 2β2 β + = 150 5 5π» 52 5
π‘ 2 2β2 = β 150 5 5π» 52 5
5 5 2 2β2 β 2 2 π‘ = 150 ( β 5 ) = 60 β 60β π» 5 5π» 2 5
β 2 π‘ = 60 (1 β ( ) ) π»
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