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IGCSE Further Maths/C1 Inequalities

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Presentation on theme: "IGCSE Further Maths/C1 Inequalities"β€” Presentation transcript:

1 IGCSE Further Maths/C1 Inequalities
Dr J Frost Objectives: Be able to solve both linear and quadratic inequalities. Be able to manipulate inequalities (including squared terms). Last modified: 20th March 2016

2 RECAP :: Linear Inequalities
Bromonology: Remember that β€˜linear’ just means that if we plotted the equation/inequality we’d end up with a straight line/region bounded by a straight line. Jan 2013 Paper 2 Solve 5π‘‘βˆ’3>𝑑+17 ? 𝑑>5 June 2012 Paper 1 Work out the greatest integer value of π‘₯ that satisfies the inequality 3π‘₯+10<1 ? π‘₯<βˆ’3 Thus greatest integer is -4. Solve 5<3π‘₯βˆ’1≀17 πŸ”<πŸ‘π’™β‰€πŸπŸ– 𝟐<π’™β‰€πŸ” ?

3 𝟎 πŸ— Manipulating Inequalities βˆ’3≀π‘₯≀2 πŸŽβ‰€ 𝒙 𝟐 β‰€πŸ— ? ? ?
What is the smallest value of π‘₯ 2 ? 𝟎 What is the largest value of π‘₯ 2 ? πŸ— Hence determine an inequality for π‘₯ 2 . πŸŽβ‰€ 𝒙 𝟐 β‰€πŸ— ? ? ?

4 Test Your Understanding
? ? βˆ’1≀π‘₯≀ β†’ ≀ π‘₯ 2 ≀16 βˆ’3≀π‘₯<βˆ’ β†’ < π‘₯ 2 ≀9 βˆ’10≀π‘₯< β†’ ≀ π‘₯ 2 ≀100 ? ? ? ? June 2012 Paper 1 1β‰€π‘šβ‰€5 and βˆ’9≀𝑛≀2 (a) Work out an inequality for π‘š+𝑛. βˆ’πŸ–β‰€π’Ž+π’β‰€πŸ• (b) Work out an inequality for π‘š+𝑛 2 πŸŽβ‰€ π’Ž+𝒏 𝟐 β‰€πŸ”πŸ’ ? ? ? ?

5 Further Example ? ? ? What’s the least π‘Žβˆ’π‘ can be? πŸβˆ’πŸ=βˆ’πŸ
Given that 1β‰€π‘Žβ‰€4 and βˆ’3≀𝑏≀2, work out an inequality for π‘Žβˆ’π‘. What’s the least π‘Žβˆ’π‘ can be? πŸβˆ’πŸ=βˆ’πŸ What’s the greatest π‘Žβˆ’π‘ can be? πŸ’βˆ’βˆ’πŸ‘=πŸ• Thus inequality for π‘Žβˆ’π‘: βˆ’πŸβ‰€π’‚βˆ’π’ƒβ‰€πŸ• ? ? ?

6 Exercise 1 1 Solve the following: 5+3π‘₯β‰₯ 𝒙β‰₯𝟐 6𝑦+1≀4𝑦 π’šβ‰€πŸ’ π‘βˆ’3≀5𝑏 𝒃β‰₯βˆ’πŸ‘ 2π‘₯βˆ’3 3 < 𝒙<𝟏𝟐 5βˆ’3π‘₯ 4 ≀ 𝒙β‰₯βˆ’πŸ“ 2βˆ’4π‘₯ 3 β‰₯ π’™β‰€βˆ’πŸ’ 4≀5π‘₯βˆ’6≀ πŸβ‰€π’™β‰€πŸ’ 5<7βˆ’2π‘₯< βˆ’πŸ‘<𝒙<𝟏 5>9βˆ’4π‘₯> 𝟏<𝒙<𝟐 Given that 0≀𝑝≀3 and 2β‰€π‘žβ‰€5, work out the inequality for π‘βˆ’π‘ž. βˆ’πŸ“β‰€π’‘βˆ’π’’<𝟏 Given that 1β‰€π‘Žβ‰€6 and βˆ’3≀𝑏≀3 work out inequalities for: βˆ’πŸβ‰€π’‚+π’ƒβ‰€πŸ— βˆ’πŸβ‰€π’‚βˆ’π’ƒβ‰€πŸ— 4 Given 0<π‘₯<1 and 𝑦>0, decide whether the following statements are ALWAYS TRUE< SOMETIMES TRUE, or NEVER TRUE. 1 π‘₯ > Always true π‘₯+𝑦< Never true π‘₯𝑦> Sometimes true. π‘₯ 2 > Never true. π‘₯βˆ’π‘¦<0 Sometimes true. [June 2013 Paper 2] 𝑀 is an integer such that 6≀3𝑀<18. π‘₯ is an integer such that βˆ’4≀π‘₯≀3. What is the highest possible value of π‘₯ What is the lowest possible value of π‘€βˆ’π‘₯. -1 Given that βˆ’1<𝑛<2 state a value of 𝑛 for which: 𝑛 2 >1 e.g. 1.5 1 𝑛 >1 e.g. 0.5 1βˆ’π‘›>1 e.g. -0.5 a ? b ? ? c ? a b ? d ? c ? ? d ? e e ? ? f 5 ? g ? h ? ? i ? 6 2 a ? b ? ? c ? 3 ? ? ? ?

7 Quadratic Inequalities
Solve π‘₯ 2 βˆ’4π‘₯βˆ’5<0 π‘₯ 2 βˆ’4π‘₯βˆ’5<0 π‘₯+1 π‘₯βˆ’5 <0 ! Step 1: Get 0 on one side. Step 2 ? ! Step 2: Factorise. ! Step 3: Sketch 𝑦=𝐿𝐻𝑆. Step 3 ? 𝑦 𝑦= π‘₯+1 π‘₯βˆ’5 ! Step 4: Identify parts of line where 𝑦 value (i.e. LHS of inequality) satisfies inequality. π‘₯ -1 5 Since 𝑦= π‘₯+1 π‘₯βˆ’5 <0, we’re interested in the parts of the line where 𝑦<0. Therefore: βˆ’πŸ<𝒙<πŸ“ Step 4? ?

8 Quadratic Inequalities
Solve π‘₯ 2 βˆ’4π‘₯βˆ’5>0 Now suppose we changed < for >… π‘₯ 2 βˆ’4π‘₯βˆ’5>0 π‘₯+1 π‘₯βˆ’5 >0 ! Step 1: Get 0 on one side. ! Step 2: Factorise. ! Step 3: Sketch 𝑦=𝐿𝐻𝑆. 𝑦 𝑦= π‘₯+1 π‘₯βˆ’5 ! Step 4: Identify parts of line where 𝑦 value (i.e. LHS of inequality) satisfies inequality. π‘₯ -1 5 Since 𝑦= π‘₯+1 π‘₯βˆ’5 >0, we’re interested in the parts of the line where 𝑦>0. Therefore: 𝒙<βˆ’πŸ or 𝒙>πŸ“ Step 4?

9 Test Your Understanding
Solve π‘₯ 2 +π‘₯βˆ’6≀0 Solve 2π‘₯+ π‘₯ 2 >3 ? ? π‘₯+3 π‘₯βˆ’2 ≀0 π‘₯ 2 +2π‘₯βˆ’3>0 π‘₯+3 π‘₯βˆ’1 >0 𝑦 𝑦 π‘₯ -3 2 π‘₯ -3 1 βˆ’πŸ‘β‰€π’™β‰€πŸ (note that ≀ has to be consistent with original question) 𝒙<βˆ’πŸ‘ or 𝒙>𝟏

10 Exercise 2 ? ? ? ? ? ? ? ? ? ? ? 1 Solve the following inequalities:
(ii) π‘Ž 2 +3π‘Žβˆ’4≀0 βˆ’πŸ’β‰€π’‚β‰€πŸ (iii) 2 𝑦 2 +π‘¦βˆ’3<0 βˆ’ πŸ‘ 𝟐 <π’š<𝟏 (iv) 4βˆ’ 𝑦 2 β‰₯0 βˆ’πŸβ‰€π’šβ‰€πŸ (v) π‘₯ 2 βˆ’4π‘₯+4>0 𝒙<𝟐 𝒐𝒓 𝒙>𝟐 (vi) 𝑝 2 βˆ’3π‘β‰€βˆ’2 πŸβ‰€π’‘β‰€πŸ (vii) π‘Ž+2 π‘Žβˆ’1 >4 𝒂<βˆ’πŸ‘ 𝒐𝒓 𝒂>𝟐 (viii) 8βˆ’2π‘Žβ‰₯ π‘Ž 2 βˆ’πŸ’β‰€π’‚β‰€πŸ (ix) 3 𝑦 2 +2π‘¦βˆ’1>0 π’š<βˆ’πŸ 𝒐𝒓 π’š> 𝟏 πŸ‘ (x) 𝑦 2 β‰₯4𝑦+5 π’šβ‰€βˆ’πŸ 𝒐𝒓 π’šβ‰₯πŸ“ The area of the square is less than the area of the rectangle. Work out an inequality for π‘₯. ? ? ? ? ? ? ? ? ? ? 2 π‘₯+1 ? 2π‘₯βˆ’1 𝒙+𝟏 𝒙+𝟏 < πŸπ’™βˆ’πŸ π’™βˆ’πŸ 𝒙 𝟐 +πŸπ’™+𝟏<𝟐 𝒙 𝟐 βˆ’πŸ‘π’™+𝟏 𝒙 𝟐 βˆ’πŸ“π’™>𝟎 𝒙 π’™βˆ’πŸ“ >𝟎 𝒙<𝟎 𝒐𝒓 𝒙>πŸ“ (but clearly 𝒙 can’t be less than 0) π‘₯+1 π‘₯βˆ’1

11 C1 Discriminants We now (hopefully!) have the sufficient skills to tackle more questions concerning discriminants: Edexcel C1 Jan 2013 a ? π‘˜+3 π‘₯ 2 +6π‘₯+ π‘˜βˆ’5 =0 π‘Ž=π‘˜ 𝑏= 𝑐=π‘˜βˆ’5 Discriminant: 36βˆ’4 π‘˜+3 π‘˜βˆ’5 >0 36βˆ’4 π‘˜ 2 βˆ’2π‘˜βˆ’15 >0 36βˆ’4 π‘˜ 2 +8π‘˜+60>0 4 π‘˜ 2 βˆ’8π‘˜βˆ’96<0 π‘˜ 2 βˆ’2π‘˜βˆ’24<0 π‘˜+4 π‘˜βˆ’6 <0 After sketching: βˆ’4<π‘˜<6 ? Reminder: No solutions: 𝑏 2 βˆ’4π‘Žπ‘<0 Equal solutions: 𝑏 2 βˆ’4π‘Žπ‘=0 Distinct solutions: 𝑏 2 βˆ’4π‘Žπ‘>0 ? b ? ? ?

12 Test Your Understanding
Edexcel C1 Jan 2011 ? ?

13 Combining Inequalities
Edexcel C1 June 2009 π‘₯>2 βˆ’ 3 2 <π‘₯<4 c) It may help to draw number lines for both and combine. Otherwise use common sense! 𝟐<𝒙<πŸ’ ?

14 Exercise 3 Edexcel C1 Jan 2009 Q7 ? ? Edexcel C1 Jan Q10


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