Grade 12 Mathematics P1 Cheat Sheet and Question Starter Notes 2026

Grade 12 Mathematics Paper 1 — Revision Notes Hlayiso Study Notes 2026 To the learner. Many of you have told us: “I should know what to do, but I don’t.” That feeling is normal, and it can be fixed. In a hard question the mathematics is usually something you already know; what is missing is the first line. These notes go through every type of question that has appeared in Paper 1 since 2018. For each type you will learn how to recognise it, what to write first, and how a full solution looks. Work through the examples with a pencil, cover the solution, and try them yourself before you read them. Start with any section — you do not have to begin at 1.1. Contents How to attack a hard paper 2 1 Algebra: equations, inequalities and nature of roots 4 2 Number patterns, sequences and series 20 3 Functions and graphs 28 4 Finance, growth and decay 37 5 Differential calculus 44 6 Probability and counting 53 If you see … write first … 59 www.hlayiso.com Hlayiso Study Notes 2026

Grade 12 Mathematics P1 — Revision Notes How to attack a hard paper How to attack a hard paper Paper 1 counts 150 marks and you have three hours: that is 1,2 minutes per mark. A 4-mark question deserves about five minutes. Keep this number in your head; it will stop one stubborn question from eating the time you need for ten easy ones. 1. The first five minutes: read the whole paper Page through every question before you write anything. Next to each question put a tick (I can do it), a question mark (maybe) or a star (difficult). You are not obliged to start at 1.1. If you begin “from 1.1 nje” and get stuck there, you lose both time and confidence. Number your answers correctly and you may answer the questions in any order. Start with the marks you are sure of: derivatives by the rules, annuity formulas, “write down the asymptotes”, arithmetic and geometric formulas, probability rules. 2. Collect method marks, even when you are stuck A memorandum gives most of its marks for method: the correct formula, the correct substitution, the standard form, and finally the answer. So write the formula and substitute into it, even if you cannot see the end. That alone is often one or two marks out of four. A blank page is always zero. 3. Consistent accuracy works for you If your answer to 6.1 is wrong but you use it correctly in 6.2, the marker gives you the marks for 6.2. In a “show that” question you may use the given result in the next part even if you could not prove it. So always attempt the follow-up questions. 4. When you are stuck, do this in order 1. Underline the key words: tangent, converge, outstanding balance, real roots, independent, hence. Each one points to a method. The checklist at the back of these notes lists them. 2. Turn every piece of information into an equation. “Passes through (2; 12)” means f (2) = 12. “Touches the x-axis at x = 2” means f (2) = 0 and f ′(2) = 0. 3. Draw something: a rough sketch, a number line, a time line for finance, a tree or Venn diagram. 4. Look at what the final answer must be: values of x, coordinates, a number of months? 5. In a “show that” question, work backwards from the given answer and ask what the line before it must be. 6. Ask yourself: “If the letter k were a number, what would I do?” Then do exactly that, carrying k along. 7. Still stuck after twice the time? Write the formula and substitution, star the question and move on. 5. Why hard papers feel hard A challenge paper hides a method you know behind three disguises: a parameter (k, p, t) instead of a number, two topics joined together (a sequence inside a logarithm, a tangent inside a parabola question), and working backwards (you are given the answer and must find the input). Name the disguise and the question becomes ordinary. www.hlayiso.com 2

Grade 12 Mathematics P1 — Revision Notes How to attack a hard paper 6. The last ten minutes: check – Did you reject impossible answers? (n /∈N, a probability bigger than 1, a negative length, a root that does not satisfy a surd equation, 2x = −3.) – Did you round to two decimal places unless told otherwise, money to the cent, and a “number of months” to a whole number in the sensible direction? – Is the answer in the form asked for: y = . . ., coordinates (x; y), an inequality or interval? How the examples are set out Each worked example shows the working on the left and the reason for each line on the right, in small italics. Read the reasons: they are what you say to yourself while you work. The final answer is framed. Examples are graded: the first ones in each subsection are standard examination questions, the later ones are at the level of the Hlayiso challenge paper. www.hlayiso.com 3

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots 1 Algebra: equations, inequalities and nature of roots Question 1 is worth about 25 marks. It usually opens with ordinary quadratics and builds up to equations in disguise. Almost every equation in this section ends as a quadratic: your job is to find the substitution or the step that turns it into one. 1.1 Quadratic equations: factorising and the formula Recognise it. – The equation has an x2 term and no higher power: make one side zero and factorise. – The question says “correct to two decimal places”: it will not factorise, so use the formula. – A product equals a number that is not zero, such as (x + 3)(x −2) = 14: multiply out first. Method. 1. Bring every term to one side so that the other side is 0 (standard form ax2 + bx + c = 0). First line: ax2 + bx + c = 0 2. Factorise, or, if it does not factorise, write the formula from the information sheet and substi- tute. 3. Set each factor equal to zero. Round only at the very end. Example 1.1 (standard) Solve for x: (a) x(2x −5) = 0 (b) 3x2 = 48 (c) (x + 3)(x −2) = 14 WORKING REASON (a) x = 0 or 2x −5 = 0 already factorised: each factor = 0 x = 0 or x = 5 2 (b) x2 = 16 divide by 3 x = ±4 two roots: never forget −4 (c) x2 + x −6 = 14 multiply out; the right side is not 0 x2 + x −20 = 0 standard form (x + 5)(x −4) = 0 factorise x = −5 or x = 4 Example 1.2 (standard) Solve for x, correct to two decimal places: 2x2 = 5x + 4. WORKING REASON 2x2 −5x −4 = 0 standard form: a = 2, b = −5, c = −4 x = −b ± √ b2 −4ac 2a = 5 ± √ (−5)2 −4(2)(−4) 2(2) write the formula, then substitute with brackets x = 5 ± √ 57 4 x ≈3,14 or x ≈−0,64 round at the end only Watch out: (x + 3)(x −2) = 14 does not mean x + 3 = 14 or x −2 = 14. The “each factor equals zero” rule only works when the other side is zero. www.hlayiso.com 4

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Watch out: Never divide both sides by x: you lose the root x = 0. Factorise instead. 1.2 Hidden quadratics and equations with fractions Recognise it. – The same block appears twice, once squared: x4 and x2, (x2 −3x)2 and (x2 −3x), x and √x, 1 x and 1 √x. Substitute k for the block. – There is an x in a denominator: write down the restriction, then multiply every term by the lowest common denominator. – A question says “hence”: the previous answer is the tool. Look for the same shape in the new equation. Method. 1. Write the restriction(s) first, for example x̸ = 0 or x̸ = ±1. 2. Multiply every term by the denominator, or let k = the repeated block. First line: “Let k = x2 −3x” or “x̸ = 0: x4 + 18 = 11x2” 3. Solve the quadratic in k, then go back to x. The value of k is not the answer. 4. Check every answer against the restrictions. Example 1.3 (standard — same type as a challenge question) Solve for x: x2 + 18 x2 = 11. WORKING REASON x4 + 18 = 11x2 x̸ = 0; multiply every term by x2 x4 −11x2 + 18 = 0 standard form: a quadratic in x2 (x2 −2)(x2 −9) = 0 factorise as if x2 were k x2 = 2 or x2 = 9 x = ± √ 2 or x = ±3 four answers Example 1.4 (standard) Solve for x: (x2 −3x)2 −2(x2 −3x) −8 = 0. WORKING REASON Let k = x2 −3x: k2 −2k −8 = 0 the block appears squared and on its own (k −4)(k + 2) = 0 x2 −3x = 4 or x2 −3x = −2 go back to x (x −4)(x + 1) = 0 or (x −1)(x −2) = 0 two more quadratics x ∈{−1 ; 1 ; 2 ; 4} www.hlayiso.com 5

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.5 (harder) Solve for x: (a) x −5√x + 6 = 0 (b) 6 x −5 √x + 1 = 0 WORKING REASON (a) Let k = √ x, so x = k2: k2 −5k + 6 = 0 x is the square of √x (k −2)(k −3) = 0, so √ x = 2 or √ x = 3 both are positive, so both are valid x = 4 or x = 9 square to get x (b) Let k = 1 √x, so 1 x = k2: 6k2 −5k + 1 = 0 x > 0; 1 x is the square of 1 √x (2k −1)(3k −1) = 0, so 1 √x = 1 2 or 1 3 √ x = 2 or √ x = 3 ⇒ x = 4 or x = 9 take reciprocals, then square Example 1.6 (standard — rejecting a root) Solve for x: x x −1 − 2 x + 1 = 4 x2 −1. WORKING REASON x̸ = 1 and x̸ = −1 restrictions: x2 −1 = (x −1)(x + 1) x(x + 1) −2(x −1) = 4 multiply every term by (x −1)(x + 1) x2 −x −2 = 0 (x −2)(x + 1) = 0 x = 2 or x = −1 x = −1 breaks the restriction: reject x = 2 Example 1.7 (challenge level — “hence”) (a) Solve for a: a + 8 a = 6. (b) Hence solve for x: 2x + 23−x = 6. WORKING REASON (a) a2 −6a + 8 = 0 a̸ = 0; multiply by a (a −2)(a −4) = 0, a = 2 or a = 4 (b) 2x + 23 2x = 6 23−x = 23 · 2−x This is (a) with a = 2x the “hence”: same shape 2x = 2 or 2x = 4 x = 1 or x = 2 www.hlayiso.com 6

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.8 (challenge level) Solve for x: (x2 + x) + 2 x2 + x = 3. Give irrational answers correct to two decimal places. WORKING REASON Let y = x2 + x: y + 2 y = 3 the block x2 + x appears twice y2 −3y + 2 = 0, so (y −1)(y −2) = 0 y̸ = 0 x2 + x = 1 or x2 + x = 2 back to x x2 + x −1 = 0: x = −1 ± √ 5 2 does not factorise: formula x2 + x −2 = 0: (x + 2)(x −1) = 0 x ≈0,62; x ≈−1,62; x = 1; x = −2 Watch out: After a k-substitution, always go back to x. “k = 4” scores nothing as a final answer. Watch out: Multiply every term by the denominator, including the terms that have no fraction. 1.3 Surd equations Recognise it. – The unknown is under a square root, for example x −√ 2x + 9 = 3. Isolate the root, square, and check. Method. 1. Get the square root alone on one side. First line: √ 2x + 9 = x −3 2. Square both sides. Remember (x −3)2 = x2 −6x + 9, not x2 + 9. 3. Solve the quadratic. 4. Check every answer in the original equation and write “reject” next to the false one. Example 1.9 (standard — same type as the prediction paper) Solve for x: x −√ 2x + 9 = 3. WORKING REASON √ 2x + 9 = x −3 isolate the surd 2x + 9 = x2 −6x + 9 square both sides x2 −8x = 0, so x(x −8) = 0 x = 0 or x = 8 Check x = 0: 0 − √ 9 = −3̸ = 3 reject Check x = 8: 8 − √ 25 = 3 true x = 8 www.hlayiso.com 7

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.10 (challenge level — two surds) Solve for x: √ x + 5 = 1 + √x. WORKING REASON x + 5 = 1 + 2√ x + x square both sides; (1 + √x)2 has a middle term 4 = 2√ x, so √ x = 2 the x terms cancel; one surd is left x = 4 square again Check: √ 9 = 3 and 1 + √ 4 = 3 true x = 4 Watch out: Squaring creates false roots. The check is part of the answer and usually carries a mark. Watch out: A square root is never negative: √ 2x + 9 = x −3 needs x −3 ⩾0. That is a quick way to reject roots. 1.4 Exponential equations Recognise it. – The unknown is in the exponent and both sides can be written with the same base: equate the exponents. – Several terms with the same base but different exponents, added or subtracted, such as 2x+1 + 2x−1: take out the common factor 2x. – Both 32x (or 9x) and 3x appear, or both 3x and 3−x: let k = 3x. – The unknown is the base, such as x 3 2 = 27: raise both sides to the reciprocal power. Method. 1. Write every number as a power of a prime: 8 = 23, 27 = 33, 1 9 = 3−2, √ 3 = 3 1 2 . 2. Same base on both sides: equate the exponents. First line: 23(x+1) = 22(2x) 3. Sum of terms: factorise. First line: 2x ( 2 + 1 2 ) = 20 4. A square of a power appears: substitute. First line: Let k = 3x: k2 −12k + 27 = 0 Example 1.11 (standard) Solve for x: (a) 8x+1 = 42x (b) 27x = 1 9 (c) x 3 2 = 27 (d) 5x = −25 WORKING REASON (a) 23x+3 = 24x 8 = 23 and 4 = 22 3x + 3 = 4x, x = 3 same base: equate exponents (b) 33x = 3−2, 3x = −2, x = −2 3 1 9 = 3−2 (c) x = 27 2 3 = ( 3√ 27 )2 , x = 9 raise both sides to the power 2 3 (d) no solution 5x > 0 for every real x www.hlayiso.com 8

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.12 (standard — common factor) Solve for x: (a) 2x+1 + 2x−1 = 20 (b) 3x+2 −3x = 72 WORKING REASON (a) 2x · 2 + 2x · 2−1 = 20 split the exponents: 2x+1 = 2x · 2 2x ( 2 + 1 2 ) = 20 common factor 2x 2x = 8, x = 3 20 ÷ 5 2 = 8 (b) 3x(9 −1) = 72, 3x = 9, x = 2 common factor 3x Example 1.13 (harder — k-substitution) Solve for x: (a) 9x −4 · 3x+1 + 27 = 0 (b) 52x −5x −20 = 0 WORKING REASON (a) (3x)2 −12 · 3x + 27 = 0 9x = (3x)2 and 4 · 3x+1 = 12 · 3x Let k = 3x: k2 −12k + 27 = 0 (k −3)(k −9) = 0 3x = 3 or 3x = 9, x = 1 or x = 2 back to x (b) Let k = 5x: k2 −k −20 = 0, (k −5)(k + 4) = 0 5x = 5 or 5x = −4 5x = −4 has no solution x = 1 Example 1.14 (challenge level — bx and b−x together) Solve for x: 2x+1 + 22−x = 9. WORKING REASON 2 · 2x + 4 2x = 9 22−x = 22 2x Let k = 2x: 2k + 4 k = 9 2k2 −9k + 4 = 0 multiply by k (k > 0) (2k −1)(k −4) = 0 2x = 1 2 or 2x = 4 x = −1 or x = 2 Watch out: 2x+1 + 2x is not 22x+1. You may only add exponents when powers are multiplied. Watch out: 3 · 3x = 3x+1, not 9x. 1.5 Simultaneous equations, including unknowns in the exponents Recognise it. – One linear equation and one quadratic: make x or y the subject of the linear one and substitute. – Both x and y are in exponents, such as 3x · 27y = 9 or 3x+y = 27: write each side with the same base. Each exponential equation becomes a linear equation in x and y. – A logarithm such as log3(x −y) = 2: use the definition to turn it into x −y = 32. – A quadratic in which every term has degree 2, such as x2 −5xy + 4y2 = 0: try to factorise it into two linear factors. www.hlayiso.com 9

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Method. 1. Turn every exponential or logarithmic equation into a linear one. First line: 3x · 33y = 32 so x + 3y = 2 2. Make one letter the subject of the linear equation. First line: x = 2 −3y 3. Substitute into the other equation, simplify to standard form and solve. 4. Find the other letter for each value, and write the answers as matched pairs. Example 1.15 (standard) Solve for x and y: y = 2x −3 and x2 + xy = 6. WORKING REASON x2 + x(2x −3) = 6 y is already the subject: substitute 3x2 −3x −6 = 0, so x2 −x −2 = 0 divide by 3 (x −2)(x + 1) = 0 x = 2, y = 1 or x = −1, y = −5 use the linear equation for y (2 ; 1) or (−1 ; −5) matched pairs Example 1.16 (standard — both unknowns in the exponents) Solve for x and y: 3x+y = 27 and 2x−y = 1 2. WORKING REASON 3x+y = 33, so x + y = 3 same base: equate exponents 2x−y = 2−1, so x −y = −1 2x = 2 add the two linear equations x = 1 and y = 2 Example 1.17 (challenge level — same type as the challenge paper) Solve for x and y: 3x · 27y = 9 and x2 + 3y2 = 4. WORKING REASON 3x · 33y = 32 27 = 33, 9 = 32 x + 3y = 2, so x = 2 −3y multiply powers: add exponents (2 −3y)2 + 3y2 = 4 substitute into the quadratic 4 −12y + 9y2 + 3y2 = 4 12y2 −12y = 0, so 12y(y −1) = 0 y = 0, x = 2 or y = 1, x = −1 x = 2 −3y (2 ; 0) or (−1 ; 1) www.hlayiso.com 10

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.18 (challenge level — a logarithm and a factorising quadratic) Solve for x and y: log3(x −y) = 2 and x2 −5xy + 4y2 = 0. WORKING REASON x −y = 32 = 9 definition: logb A = c ⇐⇒A = bc (x −y)(x −4y) = 0 factorise the second equation x −y = 0 is impossible, so x = 4y x −y = 9̸ = 0 4y −y = 9, y = 3, x = 12 substitute into x −y = 9 x = 12 and y = 3 Example 1.19 (challenge level — natural-number exponents) x and y are natural numbers such that (√ 2 )x + ( 3√ 3 )y = 25. Determine xy. WORKING REASON 2 x 2 + 3 y 3 = 25 write roots as fractional exponents Both terms must be whole numbers an irrational root cannot cancel to give 25 2 x 2 ∈{2; 4; 8; 16}, 3 y 3 ∈{3; 9} the values less than 25 16 + 9 = 25 is the only sum that works try the pairs x 2 = 4, x = 8 and y 3 = 2, y = 6 xy = 48 Watch out: Give answers as pairs. Writing x = 2 or x = −1 and then y = 0 or y = 1 separately loses the matching. Watch out: When both unknowns are in exponents, do not use logs on a calculator. Match the bases; you get a linear system. 1.6 Quadratic inequalities and “for which values is it real?” Recognise it. – There is an inequality sign and an x2 term: make one side zero, find the critical values, use a sketch or number line. – “For which values of x is the expression real (defined)?”: the expression under a square root must be ⩾0, and a denominator must not be 0. If the root is in the denominator, the expression under it must be > 0. – A fraction compared with 0, such as x+3 x−1 ⩽0: use the signs of numerator and denominator. www.hlayiso.com 11

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Method. 1. Bring everything to one side. If the x2 coefficient is negative, multiply by −1 and turn the sign around. First line: x2 + x −12 > 0 2. Factorise and write the critical values. 3. Sketch the parabola (roots on the x-axis) or draw a number line with signs. For a > 0: “< 0” is between the roots; “> 0” is outside the roots. x −4 3 (x + 4)(x −3) < 0: between > 0: outside Example 1.20 (standard) Solve for x: (a) (x −2)(x + 3) > 6 (b) 3x −x2 ⩾−10 WORKING REASON (a) x2 + x −6 −6 > 0, x2 + x −12 > 0 multiply out; one side must be 0 (x + 4)(x −3) > 0 critical values −4 and 3 x < −4 or x > 3 “> 0”: outside the roots (b) −x2 + 3x + 10 ⩾0 everything to the left x2 −3x −10 ⩽0 multiply by −1: the sign turns (x −5)(x + 2) ⩽0, −2 ⩽x ⩽5 “⩽0”: between the roots Example 1.21 (harder — same type as a challenge question) For which values of x is √ 8 + 2x −x2 x + 1 a real number? WORKING REASON 8 + 2x −x2 ⩾0 under the root: not negative x2 −2x −8 ⩽0, (x −4)(x + 2) ⩽0 multiply by −1, turn the sign −2 ⩽x ⩽4 between the roots and x + 1̸ = 0 denominator not zero −2 ⩽x ⩽4, x̸ = −1 www.hlayiso.com 12

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.22 (harder — a fraction and zero) Solve for x: x + 3 x −1 ⩽0. WORKING REASON Critical values: x = −3 (top = 0), x = 1 (bottom = 0) Signs: x < −3: − −> 0; −3 < x < 1: + −< 0; x > 1: + + > 0 test each interval −3 ⩽x < 1 x = −3 allowed; x = 1 never allowed Watch out: x2 ⩽9 gives −3 ⩽x ⩽3, not just x ⩽3. Always sketch. Watch out: An endpoint that makes a denominator zero is always excluded, even with ⩽or ⩾. 1.7 Nature of roots and the discriminant Recognise it. – “Determine the nature of the roots” or “without solving”: calculate ∆= b2 −4ac. – “For which values of k are the roots equal / real / non-real?”: write ∆in terms of k and apply the condition. – “Show that the roots are real (or rational) for all values of k”: show ∆is a perfect square, or a square plus a positive number. – “The roots are x = 3± √ k−2 4 ”: the part under the root is the discriminant. Discriminant Nature of the roots ∆< 0 non-real ∆= 0 real, equal and rational ∆> 0 and a perfect square real, unequal and rational ∆> 0 and not a perfect square real, unequal and irrational ∆⩾0 real Method. 1. Write the equation in standard form and read off a, b and c with their signs. Keep brackets around expressions. First line: ∆= b2 −4ac = [−(k + 1) ]2 −4(1)(4) 2. Simplify ∆and apply the condition (= 0, ⩾0, < 0, …). This usually gives an equation or a quadratic inequality in k. www.hlayiso.com 13

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.23 (standard) Without solving, determine the nature of the roots of (a) 2x2 −3x + 5 = 0 (b) 4x2 −12x + 9 = 0. WORKING REASON (a) ∆= (−3)2 −4(2)(5) = −31 ∆< 0 non-real roots (b) ∆= (−12)2 −4(4)(9) = 0 ∆= 0 real, equal, rational roots Example 1.24 (standard — equal roots) For which values of k does x2 −(k + 1)x + 4 = 0 have equal roots? Determine the root in each case. WORKING REASON ∆= (k + 1)2 −4(1)(4) = 0 equal roots: ∆= 0 (k + 1)2 = 16, k + 1 = ±4 k = 3 or k = −5 Equal root: x = −b 2a = k + 1 2 when ∆= 0 the formula gives −b 2a k = 3: x = 2; k = −5: x = −2 Example 1.25 (harder — a quadratic inequality in k) For which values of k does x2 + kx + k + 3 = 0 have non-real roots? WORKING REASON ∆= k2 −4(1)(k + 3) < 0 non-real: ∆< 0 k2 −4k −12 < 0 (k −6)(k + 2) < 0 critical values −2 and 6 −2 < k < 6 “< 0”: between Example 1.26 (harder — for all values) Show that x2 −(p + 1)x + p −1 = 0 has real, unequal roots for all real values of p. WORKING REASON ∆= [−(p + 1) ]2 −4(1)(p −1) brackets around b = p2 + 2p + 1 −4p + 4 = p2 −2p + 5 = (p −1)2 + 4 complete the square (p −1)2 ⩾0, so ∆⩾4 > 0 for all p a square is never negative ∆> 0 for every p, so the roots are real and unequal www.hlayiso.com 14

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots Example 1.27 (challenge level — same type as the challenge paper) Given (k + 1)x2 −(3k + 1)x + 2k = 0, k̸ = −1. (a) Prove that the roots are rational for every rational value of k. (b) Determine all integer values of k for which both roots are integers. WORKING REASON (a) ∆= (3k + 1)2 −4(k + 1)(2k) a = k + 1, b = −(3k + 1), c = 2k = 9k2 + 6k + 1 −8k2 −8k = k2 −2k + 1 = (k −1)2 a perfect square for every k ∆is a perfect square and a,b,c are rational, so the roots are rational (b) x = (3k + 1) ± (k −1) 2(k + 1) use √ ∆= k −1 in the formula x = 4k 2(k + 1) = 2k k + 1 or x = 2k + 2 2(k + 1) = 1 one root is always 1 2k k + 1 = 2(k + 1) −2 k + 1 = 2 − 2 k + 1 split the fraction k + 1 must divide 2: k + 1 ∈{−2; −1; 1; 2} only then is 2 k+1 an integer k ∈{−3 ; −2 ; 0 ; 1} check: k = 0 gives x2 −x = 0, roots 0; 1 Example 1.28 (harder — reading the formula backwards) One root of ax2 + bx + c = 0 is x = 1 and the other is x = −4 −√ 16 + 20a 2a . Determine a, b, c and the other root. WORKING REASON x = −b − √ b2 −4ac 2a compare with the formula −b = −4, so b = 4 b2 −4ac = 16 + 20a, so −4ac = 20a, c = −5 a̸ = 0 a(1)2 + 4(1) −5 = 0, so a = 1 x = 1 is a root: substitute a = 1, b = 4, c = −5; other root x = −5 −4− √ 36 2 = −5 Watch out: If b = −(k + 1), then b2 = (k + 1)2. Losing the brackets is the most common error in this topic. Watch out: “Non-real” is the correct word. Do not write “no roots” or “imaginary”. 1.8 Exponents and surds without a calculator Recognise it. – Huge powers such as 32025: factorise out the smallest power. Everything cancels to a small fraction. – x = √ 5 + √ 3 and y = √ 5 − √ 3 (conjugates): find xy and x2 + y2 first; xy is always rational. – “Prove that … is divisible by” or “is even for all n”: factorise out the smallest power of the base. Method. 1. Take out the smallest power as a common factor in numerator and denominator. First line: 32022(33 −3) 32022(32 + 1) 2. With conjugates use (√a + √ b)(√a − √ b) = a −b and (√a ± √ b)2 = a + b ± 2 √ ab. www.hlayiso.com 15

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots 3. Simplify surds with square factors: √ 48 = √ 16 · 3 = 4 √ 3. Example 1.29 (standard — same type as the prediction paper) Without using a calculator, determine the value of 32025 −32023 32024 + 32022 . WORKING REASON = 32022(33 −31) 32022(32 + 1) smallest power 32022 out of every term = 27 −3 9 + 1 = 24 10 12 5 Example 1.30 (standard) Simplify without a calculator: (a) √ 48 + √ 75 − √ 27 (b) (√ 12 + √ 3 )2 WORKING REASON (a) 4 √ 3 + 5 √ 3 −3 √ 3 = 6 √ 3 48 = 16 · 3, 75 = 25 · 3, 27 = 9 · 3 (b) ( 2 √ 3 + √ 3 )2 = ( 3 √ 3 )2 = 27 simplify inside the bracket first Example 1.31 (challenge level — same type as the challenge paper) Given x = √ 5 + √ 3 and y = √ 5 − √ 3. Without a calculator, determine x y + y x. WORKING REASON xy = ( √ 5)2 −( √ 3)2 = 2 difference of squares x2 + y2 = (8 + 2 √ 15) + (8 −2 √ 15) = 16 the surds cancel x y + y x = x2 + y2 xy = 16 2 common denominator xy 8 Example 1.32 (challenge level) A rectangular box has faces with areas 2 √ 3 cm2, 6 √ 2 cm2 and 2 √ 6 cm2. Without a calculator, determine its volume. WORKING REASON Let the edges be a, b, c: ab = 2 √ 3, bc = 6 √ 2, ac = 2 √ 6 name what you do not know (abc)2 = 2 √ 3 · 6 √ 2 · 2 √ 6 = 24 √ 36 = 144 multiply the three areas: each edge appears twice V = abc = 12 cm3 Watch out: √ a + b̸ = √a + √ b and (√a + √ b)2̸ = a + b: the middle term 2 √ ab is always there. www.hlayiso.com 16

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots 1.9 Proofs, products and other algebra These appear as the last question of Question 1 and look unfamiliar. Each one still uses a single idea. Example 1.33 (harder — divisibility) Prove that 3n+2 −3n+1 + 4 · 3n is divisible by 10 for every natural number n. WORKING REASON = 3n ( 32 −3 + 4 ) take out the smallest power 3n = 3n(10) = 10 · 3n 10 · 3n is a multiple of 10, so the expression is divisible by 10 Example 1.34 (challenge level — telescoping product) Simplify ( 1 + 1 3 ) ( 1 + 1 4 ) ( 1 + 1 5 ) · · · ( 1 + 1 n ) and determine all values of n (n ⩾3) for which the product is an integer. WORKING REASON = 4 3 · 5 4 · 6 5 · · · n + 1 n write each bracket as one fraction = n + 1 3 each numerator cancels the next denominator n + 1 must be a multiple of 3 n = 5; 8; 11; . . . (n = 3m −1, m ⩾2) Example 1.35 (challenge level — repeated difference of squares) Calculate (2 + 1)(22 + 1)(24 + 1)(28 + 1) without a calculator. WORKING REASON Multiply by (2 −1) = 1: (2 −1)(2 + 1) = 22 −1 multiplying by 1 changes nothing (22 −1)(22 + 1) = 24 −1, (24 −1)(24 + 1) = 28 −1 each step is a difference of squares (28 −1)(28 + 1) = 216 −1 65 535 Example 1.36 (challenge level — counting prime factors) Determine the largest value of k such that 2k is a factor of 20! = 1 × 2 × 3 × · · · × 20. WORKING REASON Multiples of 2 up to 20: 10 each gives at least one factor 2 Multiples of 4: 5; of 8: 2; of 16: 1 each gives one extra factor 2 10 + 5 + 2 + 1 k = 18 Watch out: For a proof, end with a sentence that answers the question. “= 10 · 3n” alone does not say why it is divisible by 10. www.hlayiso.com 17

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots 1.10 Maximum and minimum values: completing the square Recognise it. – “Maximum or minimum value of the expression”, “range of y”, “show it can never be nega- tive”: complete the square. – The expression is a fraction with a quadratic in the denominator, such as 12 x2−4x+7: the fraction is largest when the denominator is smallest. Method. 1. Make the x2 coefficient 1 (take out a common factor if needed). 2. Add and subtract ( b 2 )2 . First line: x2 −6x + . . . = (x −3)2 −9 + . . . 3. A square is never negative: (x −3)2 ⩾0, with equality when x = 3. Example 1.37 (standard) Determine the maximum value of 5 + 6x −x2 and the value of x at which it occurs. WORKING REASON = −(x2 −6x) + 5 take out −1 = − [(x −3)2 −9 ] + 5 = −(x −3)2 + 14 complete the square −(x −3)2 ⩽0 a negative square is at most 0 maximum 14 when x = 3 Example 1.38 (harder) Determine the maximum value of 12 x2 −4x + 7. WORKING REASON x2 −4x + 7 = (x −2)2 + 3 ⩾3 complete the square in the denominator The fraction is largest when the denominator is smallest 12 is positive maximum 12 3 = 4 (at x = 2) Example 1.39 (challenge level) Prove that x2 −4xy + 5y2 ⩾0 for all real x and y. When is it equal to zero? WORKING REASON = x2 −4xy + 4y2 + y2 split 5y2 = 4y2 + y2 = (x −2y)2 + y2 a perfect square plus a square Both squares are ⩾0, so the sum is ⩾0 = 0 only when y = 0 and x = 0 both squares must be 0 Watch out: The maximum value is the y-value (14), not the x-value (3). Read which one is asked. www.hlayiso.com 18

Grade 12 Mathematics P1 — Revision Notes 1 Algebra: equations, inequalities and nature of roots 1.11 Word problems Method. 1. Name the unknowns with letters and write down what each letter means. 2. Turn each sentence into an equation. Perimeter, area, Pythagoras and “time = distance speed ” are the usual ones. 3. Solve, and reject answers that make no sense in the context (negative lengths or speeds). Example 1.40 (standard — same type as the prediction paper) A rectangle has a perimeter of 46 m and a diagonal of 17 m. Calculate its length and breadth. WORKING REASON 2ℓ+ 2b = 46, so b = 23 −ℓ perimeter ℓ2 + b2 = 172 Pythagoras on the diagonal ℓ2 + (23 −ℓ)2 = 289 substitute 2ℓ2 −46ℓ+ 240 = 0, ℓ2 −23ℓ+ 120 = 0 (ℓ−15)(ℓ−8) = 0 15 m by 8 m both roots give the same rectangle Example 1.41 (harder) A learner cycles 30 km. If she cycled 5 km/h faster, the trip would take one hour less. Determine her speed. WORKING REASON Let her speed be v km/h: 30 v − 30 v + 5 = 1 time = distance speed 30(v + 5) −30v = v(v + 5) multiply by v(v + 5) v2 + 5v −150 = 0, (v + 15)(v −10) = 0 v = 10 km/h v = −15: a speed cannot be negative Example 1.42 (standard — “show that”) A closed box has a square base of side x cm and height y cm. Its total surface area is 200 cm2. Show that x2 + 2xy −100 = 0. WORKING REASON Area = 2x2 + 4xy top and bottom: 2x2; four sides: 4xy 2x2 + 4xy = 200 x2 + 2xy −100 = 0 divide by 2: the required result www.hlayiso.com 19

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series 2 Number patterns, sequences and series Questions 2 and 3 are worth about 25 marks together. Before you do anything, decide which kind of pattern you have: Test Kind General term first differences constant arithmetic Tn = a + (n −1)d second differences constant quadratic Tn = an2 + bn + c, with 2a = second difference ratios T2 T1 = T3 T2 constant geometric Tn = arn−1 2.1 Quadratic patterns: the general term Recognise it. – The second differences are constant: the pattern is quadratic. – Some terms are unknown (p, q): use the condition “second differences are equal”, or set up Tn = an2 + bn + c from the known terms. – “Prove that 777 is not a term”: solve Tn = 777 and show n is not a natural number, or use a parity (odd/even) argument. Method. 1. Write the first and second differences under the pattern. 2. Use the three equations: 2a = second difference, 3a + b = T2 −T1, a + b + c = T1. First line: 2a = 4, so a = 2 Example 2.1 (standard) Given the quadratic pattern 4 ; 9 ; 18 ; 31 ; . . . Determine the general term and T20. WORKING REASON First differences: 5 ; 9 ; 13; second difference: 4 write them under the terms 2a = 4, a = 2 3a + b = 5, b = −1 3a + b is the first first-difference a + b + c = 4, c = 3 a + b + c = T1 Tn = 2n2 −n + 3; T20 = 783 check: T3 = 18 −3 + 3 = 18 www.hlayiso.com 20

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.2 (challenge level — same type as the challenge paper) Given the quadratic pattern 2 ; p ; 16 ; q ; 54 ; . . . (a) Show that p = 6 and q = 32. (b) Determine Tn. (c) Prove that 777 is not a term of the pattern. WORKING REASON (a) Tn = an2 + bn + c: a + b + c = 2; 9a + 3b + c = 16; 25a + 5b + c = 54 use the three known terms T1, T3, T5 8a + 2b = 14 and 16a + 2b = 38 subtract pairs to remove c 8a = 24: a = 3, b = −5, c = 4 p = T2 = 12 −10 + 4 = 6; q = T4 = 48 −20 + 4 = 32 the required values (b) Tn = 3n2 −5n + 4 from (a) (c) Tn = n(3n −5) + 4 If n is even, n(3n −5) is even; if n is odd, 3n −5 is even one factor is always even So Tn is always even; 777 is odd 777 is not a term (or: 3n2 −5n −773 = 0 gives n ≈16,91, not a natural number) Watch out: a, b, c in Tn = an2 + bn + c are not the first term and differences. Use the three equations. 2.2 First differences, largest terms, and patterns in context Recognise it. – “Between which two terms is the first difference 199?”: the first difference is itself an arithmetic sequence; find its general term. – “Which term is the largest?” or “from which term are the terms negative?”: a quadratic Tn with a < 0. Use the turning point or an inequality. – A diagram of tiles, dots or matches: count the first three figures, then treat the counts as a pattern. Method. 1. First difference between Tn and Tn+1: Tn+1 −Tn, or write the first differences as an arithmetic sequence. First line: Tn+1 −Tn = 4n −1 2. Largest term: n = −b 2a, then round to the nearest natural number and check both neighbours. Example 2.3 (standard — same type as the prediction paper) Given the quadratic pattern 5 ; 8 ; 15 ; 26 ; . . . with Tn = 2n2 −3n + 6. Between which two consecutive terms is the first difference equal to 199? WORKING REASON First differences: 3 ; 7 ; 11 ; . . . arithmetic: a = 3, d = 4 nth first difference: 3 + (n −1)4 = 4n −1 this is Tn+1 −Tn 4n −1 = 199, n = 50 between T50 and T51 www.hlayiso.com 21

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.4 (harder) The pattern 3 ; 12 ; 19 ; 24 ; . . . has Tn = −n2 + 12n −8. Determine the largest term, and the first term that is negative. WORKING REASON Turning point: n = −b 2a = −12 −2 = 6 a < 0: a maximum T6 = −36 + 72 −8 = 28, so the largest term is 28 −n2 + 12n −8 < 0, so n2 −12n + 8 > 0 negative terms n > 6 + √ 28 ≈11,29 critical values 6 ± √ 28 T12 = −8 is the first negative term check: T11 = 3 > 0 Example 2.5 (harder — pattern from a diagram) Tiles are arranged as shown. Determine a formula for the number of tiles in figure n, and which figure uses 239 tiles. Figure 1 Figure 2 Figure 3 WORKING REASON Tiles: 5 ; 11 ; 19; first differences 6 ; 8; second 2 count the figures a = 1, 3a + b = 6 so b = 3, c = 5 −4 = 1 Tn = n2 + 3n + 1 (each figure is an (n + 1) × (n + 2) rectangle minus one tile) n2 + 3n + 1 = 239, n2 + 3n −238 = 0 (n + 17)(n −14) = 0 n = −17 is not natural figure 14 2.3 Arithmetic sequences and series Recognise it. – “How many terms?”: use Tn = a + (n −1)d with the last term. – “Sum of the series”, or “how many terms give a sum of 945?”: use Sn = n 2 [2a + (n −1)d], or Sn = n 2(a + ℓ) when you know the last term ℓ. – “Sum of the even terms” (or every second term): those terms form a new arithmetic sequence. – Three terms in terms of x that are arithmetic: T2 −T1 = T3 −T2. www.hlayiso.com 22

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.6 (standard — same type as the prediction paper) Given the arithmetic series 3 + 8 + 13 + . . . + 248. (a) How many terms are there? (b) Calculate the sum. (c) Calculate the sum of the terms that are even numbers. WORKING REASON (a) 3 + (n −1)5 = 248, n = 50 Tn = a + (n −1)d with Tn = 248 (b) S50 = 50 2 (3 + 248) = 6 275 Sn = n 2 (a + ℓ) (c) Even terms: 8 ; 18 ; 28 ; . . . ; 248 every second term: a = 8, d = 10 8 + (m −1)10 = 248, m = 25 S25 = 25 2 (8 + 248) = 3 200 Example 2.7 (harder) (a) How many terms of 5 + 9 + 13 + . . . must be added to give 945? (b) 2x + 1 ; 3x + 2 ; 5x −1 are the first three terms of an arithmetic sequence. Determine x. WORKING REASON (a) n 2 [2(5) + (n −1)4] = 945 Sn formula with a = 5, d = 4 n(2n + 3) = 945, 2n2 + 3n −945 = 0 (2n + 45)(n −21) = 0, n = 21 n must be natural (b) (3x + 2) −(2x + 1) = (5x −1) −(3x + 2) common difference x + 1 = 2x −3, x = 4 terms 9 ; 14 ; 19 Example 2.8 (harder — context) Thandi saves R200 in January. Each month she saves R50 more than the month before. In which month will her total savings first exceed R10 000? WORKING REASON n 2 [2(200) + (n −1)50] > 10 000 total savings is Sn; a = 200, d = 50 n(350 + 50n) > 20 000, n2 + 7n −400 > 0 n > −7 + √ 1 649 2 ≈16,80 positive critical value the 17th month S16 = 9 200; S17 = 10 200 2.4 Sigma notation Recognise it. – ∑40 n=11(4n −3): write out the first term, the last term and the number of terms (top −bottom +1). – “Write in sigma notation”: find Tn and the value of n for the last term. – The unknown is the top number: write Sn in terms of it and solve. – A log or exponent inside the sigma: simplify the general term first. It usually becomes arith- metic. www.hlayiso.com 23

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.9 (standard) (a) Calculate 40 ∑ n=11 (4n −3). (b) Write 2 + 5 + 8 + . . . + 59 in sigma notation. (c) Solve for m: m ∑ k=1 (2k + 3) = 320. WORKING REASON (a) 41 + 45 + . . . + 157; 40 −11 + 1 = 30 terms substitute n = 11 and n = 40 S30 = 30 2 (41 + 157) = 2 970 (b) Tn = 3n −1 = 59, n = 20: 20 ∑ n=1 (3n −1) (c) m 2 [2(5) + (m −1)2] = 320 a = 5, d = 2 m2 + 4m −320 = 0, (m + 20)(m −16) = 0, m = 16 Example 2.10 (challenge level — same type as the challenge paper) Determine k if k ∑ r=1 log3 (9r 3 ) = 144. WORKING REASON log3 (32r 3 ) = log3 32r−1 = 2r −1 simplify the general term first Series: 1 + 3 + 5 + . . . a = 1, d = 2 k 2 [2 + (k −1)2] = k2 = 144 k = 12 k = −12 is not allowed Watch out: The number of terms from n = 11 to n = 40 is 30, not 29. 2.5 Geometric sequences and series Recognise it. – Each term is a fixed multiple of the one before (“90% of the previous swing”, “doubles”): geometric. – “Which term is …” or “how many terms must be added to exceed …”: the unknown n is in an exponent, so use logs. – Two terms are given, such as T3 and T6: divide the equations to remove a. Method. 1. Write Tn = arn−1 or Sn = a(rn−1) r−1 with the values substituted. 2. Isolate the power, then n = log(...) log r . If you divide by a negative log (because 0 < r < 1), the inequality turns. www.hlayiso.com 24

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.11 (standard) (a) Which term of 3 ; 6 ; 12 ; . . . is 1 536? (b) In a geometric sequence T3 = 18 and T6 = 486. Determine a and r. (c) How many terms of 2 + 6 + 18 + . . . must be added for the sum to exceed 10 000? WORKING REASON (a) 3 · 2n−1 = 1 536, 2n−1 = 512 = 29, n = 10 (b) ar5 ar2 = 486 18 , r3 = 27, r = 3 divide to remove a a(3)2 = 18, a = 2, r = 3 (c) 2(3n −1) 3 −1 > 10 000, 3n > 10 001 Sn with a = 2, r = 3 n > log 10 001 log 3 ≈8,38, 9 terms round up Example 2.12 (harder — same type as the prediction paper) A ball is dropped from a height of 8 m. After each bounce it rises to 75% of the previous height. (a) How high does it rise after the 4th bounce? (b) After which bounce does it first rise less than 1 m? (c) Sipho says the ball will travel more than 50 m before it stops. Is he correct? WORKING REASON (a) 8(0,75)4 ≈2,53 m after bounce n: 8(0,75)n (b) 8(0,75)n < 1, (0,75)n < 0,125 n > log 0,125 log 0,75 ≈7,23 log 0,75 < 0: the sign turns after the 8th bounce (c) Down 8 m, then up and down: 2(6 + 4,5 + . . .) each rise is also fallen again Total = 8 + 2 · 6 1 −0,75 = 8 + 48 = 56 m S∞with a = 6, r = 0,75 Yes: 56 m > 50 m Watch out: When 0 < r < 1, log r is negative. Dividing by it turns “<” into “>”. 2.6 Sum to infinity and convergence Recognise it. – “Converges”, “sum to infinity”, “if the process continues forever”: geometric with −1 < r < 1 and S∞= a 1−r. – “For which values of x does the series converge?”: r is an expression in x; solve −1 < r < 1. – “Smallest n so that the sum differs from S∞by less than …”: S∞−Sn = arn 1−r, then logs. www.hlayiso.com 25

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.13 (standard) (a) Calculate 54 + 18 + 6 + . . . (to infinity). (b) Write 0,˙2˙7 as a fraction using a series. WORKING REASON (a) r = 18 54 = 1 3, S∞= 54 1 −1 3 = 81 −1 < r < 1, so it converges (b) 0,27 + 0,0027 + . . .: a = 0,27, r = 0,01 S∞= 0,27 0,99 = 3 11 Example 2.14 (challenge level — same type as the challenge paper) Given 15 + 15(3x −2) + 15(3x −2)2 + . . . (a) For which values of x does the series converge? (b) The sum to infinity is 24. Determine the smallest number of terms that must be added so that the sum differs from the sum to infinity by less than 0,001. WORKING REASON (a) r = 3x −2: −1 < 3x −2 < 1 convergence condition 1 < 3x < 3, 1 3 < x < 1 (b) 15 1 −r = 24, 1 −r = 5 8, r = 3 8 S∞= a 1−r S∞−Sn = 24 −15 (1 −rn) 1 −r = 24rn 15 1−r = 24, so Sn = 24(1 −rn) 24(0,375)n < 0,001 n > log ( 0,001 24 ) log 0,375 ≈10,28 sign turns: log 0,375 < 0 11 terms x itself (x = 19 24) is not needed Watch out: Write the convergence condition with r in the middle: −1 < r < 1. Then solve both sides at once. 2.7 Combined sequences, Sn formulas and proofs Recognise it. – Terms of an arithmetic sequence form a geometric sequence: write them as a, a + d, … and set the ratios equal. – Sn is given as a formula in n: Tn = Sn −Sn−1 (and T1 = S1). – The terms alternate between two patterns (odd and even positions): split into two sequences. – “Prove that Sn = . . .”: learn the two standard proofs below; they are often 4 free marks. www.hlayiso.com 26

Grade 12 Mathematics P1 — Revision Notes 2 Number patterns, sequences and series Example 2.15 (challenge level — same type as the challenge paper) The first, second and fifth terms of an arithmetic sequence (d̸ = 0) are the first three terms of a geometric sequence. (a) Show that d = 2a. (b) The sum of the first 8 terms of the arithmetic sequence is 192. Deter- mine the 6th term of the geometric sequence. WORKING REASON (a) T1 = a, T2 = a + d, T5 = a + 4d a + d a = a + 4d a + d , (a + d)2 = a(a + 4d) geometric: equal ratios a2 + 2ad + d2 = a2 + 4ad, d2 = 2ad d(d −2a) = 0, d̸ = 0, so d = 2a (b) 8 2 [2a + 7(2a)] = 64a = 192, a = 3, d = 6 Geometric: 3 ; 9 ; 27 ; . . ., r = 3 T1 = 3, T2 = 9, T5 = 27 T6 = 3(3)5 = 729 Example 2.16 (harder) (a) For a series, Sn = 3n2 + 2n. Determine Tn. (b) The sequence 1 ; 3 ; 4 ; 9 ; 7 ; 27 ; 10 ; 81 ; . . . continues in the same way. Calculate the sum of the first 30 terms. WORKING REASON (a) Tn = Sn −Sn−1 = 3n2 + 2n − [ 3(n −1)2 + 2(n −1) ] = 6n −1 check: T1 = 5 = S1 (b) Odd positions: 1 ; 4 ; 7 ; . . . (d = 3); even positions: 3 ; 9 ; 27 ; . . . (r = 3) two sequences, 15 terms each S15 = 15 2 [2 + 14(3)] = 330 arithmetic part S15 = 3(315 −1) 3 −1 = 21 523 359 geometric part 21 523 689 add Example 2.17 (standard — the two proofs) Prove that (a) Sn = n 2 [2a + (n −1)d] for an arithmetic series, and (b) Sn = a(rn−1) r−1 for a geometric series (r̸ = 1). WORKING REASON (a) Sn = a + (a + d) + . . . + [a + (n −1)d] write the series forwards Sn = [a + (n −1)d] + [a + (n −2)d] + . . . + a and backwards 2Sn = [2a + (n −1)d] + . . . (n times) add: every pair has the same sum Sn = n 2 [2a + (n −1)d] (b) Sn = a + ar + ar2 + . . . + arn−1 rSn = ar + ar2 + . . . + arn−1 + arn multiply by r rSn −Sn = arn −a subtract: the middle terms cancel Sn = a(rn −1) r −1 divide by r −1̸ = 0 www.hlayiso.com 27

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs 3 Functions and graphs Questions 4 to 6 carry about 35 marks. Most of the marks come from a small set of moves that work for every graph. You are asked for Write first y-intercept put x = 0 x-intercept(s) put y = 0 and solve a point lies on the graph substitute the point: f (2) = 12 points of intersection of f and g f (x) = g(x) average gradient from x = a to x = b f (b) −f (a) b −a inverse of f swap x and y, then make y the subject 3.1 The hyperbola y = a x+p + q Recognise it. – f (x) = 3x+1 x−1 (a fraction with x at the top and bottom): rewrite it as a x+p + q by splitting the numerator. – Asymptotes: x = −p and y = q. Domain x̸ = −p; range y̸ = q. – Axes of symmetry pass through the point where the asymptotes meet, with gradients 1 and −1: y = (x + p) + q and y = −(x + p) + q. – “For which values of m (or k) does the line not cut the graph?”: equate, get a quadratic, use ∆< 0. Method. 1. To split the fraction, write the numerator in terms of the denominator. First line: x + 3 x −1 = (x −1) + 4 x −1 2. To find a, p, q from a graph: read p and q from the asymptotes, then substitute a point to find a. 3. To sketch: asymptotes (dotted), intercepts, and one more point on each branch. www.hlayiso.com 28

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs Example 3.1 (standard — same type as the challenge paper) Given f (x) = x + 3 x −1. (a) Show that f (x) = 4 x −1 + 1. (b) Write down the asymptotes, domain and range. (c) Determine the intercepts and the axes of symmetry, and sketch f. WORKING REASON (a) x + 3 x −1 = (x −1) + 4 x −1 = x −1 x −1 + 4 x −1 = 1 + 4 x −1 write x + 3 as (x −1) + 4 (b) x = 1 and y = 1 denominator 0; the constant q Domain x ∈R, x̸ = 1; range y ∈R, y̸ = 1 (c) f (0) = −3: (0; −3); x + 3 = 0: (−3; 0) use the original form for intercepts y = (x −1) + 1 = x and y = −(x −1) + 1 = −x + 2 through (1; 1), gradients ±1 −3 −3 f x y Example 3.2 (challenge level — same type as the challenge paper) For which values of m will the line y = mx NOT intersect the graph of f (x) = x + 3 x −1? WORKING REASON mx = x + 3 x −1, mx2 −mx = x + 3 equate; x̸ = 1 mx2 −(m + 1)x −3 = 0 standard form No intersection: ∆= (m + 1)2 + 12m < 0 no real roots m2 + 14m + 1 < 0; critical values m = −7 ± 4 √ 3 formula: −14± √ 192 2 −7 −4 √ 3 < m < −7 + 4 √ 3 (−13,93 < m < −0,07) between the roots Check m = 0: y = 0 cuts f at x = −3 so m = 0 is correctly excluded www.hlayiso.com 29

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs Example 3.3 (harder — same type as the prediction paper) The asymptotes of h(x) = a x+p + q meet at (2; −1) and the graph passes through (3; 1). (a) De- termine a, p and q. (b) Write down the axis of symmetry with a negative gradient. (c) For which values of x is x · h(x) ⩽0? WORKING REASON (a) p = −2, q = −1: h(x) = a x −2 −1 vertical asymptote x = 2 1 = a 3 −2 −1, a = 2 substitute (3; 1) (b) y = −(x −2) −1, y = −x + 1 through (2; −1), gradient −1 (c) x-intercept: 2 x −2 = 1, x = 4 critical values: 0, 2, 4 h > 0 for 2 < x < 4; h < 0 for x < 2 or x > 4 read from the sketch x · h(x) ⩽0: one factor positive, one negative (or zero) 0 < x < 2: (+)(−); x > 4: (+)(−) 0 ⩽x < 2 or x ⩾4 x = 2 excluded; x = 0, x = 4 give 0 Example 3.4 (harder — describing a transformation) f (x) = 4 x−1 + 1 and g(x) = 4 x+1 −1. Describe two different ways in which f can be transformed to g. WORKING REASON f (x + 2) −2 = 4 x + 1 + 1 −2 = g(x) try a translation first translate 2 units left and 2 units down −f (−x) = − ( 4 −x −1 + 1 ) = 4 x + 1 −1 = g(x) −f (−x): reflect in both axes reflect in the y-axis, then in the x-axis (a rotation of 180◦about O) Watch out: For intercepts use the original form x+3 x−1 if it is easier; for asymptotes use a x+p + q. Watch out: Reflecting a hyperbola in its horizontal asymptote y = q gives the same graph as reflecting it in its vertical asymptote x = −p. 3.2 The parabola, tangents and lines Recognise it. – x-intercepts are given: use y = a(x −x1)(x −x2). The turning point is given: use y = a(x − p)2 + q. Then substitute one more point for a. – The axis of symmetry lies halfway between the x-intercepts: x = x1+x2 2 . – “The line y = 8x + k is a tangent”: f ′(x) = 8 gives the point of contact (or equate and use ∆= 0). – “Maximum length of MN” with MN vertical: length = top graph −bottom graph, then max- imise. – g(x) = f (x + t): the graph moves t units left; every x-value decreases by t. www.hlayiso.com 30

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs Example 3.5 (standard — same type as the challenge paper) The graph of f (x) = a(x + 3)(x −1) has its turning point on the line y = 2x −6. Show that a = 2. WORKING REASON Axis of symmetry: x = −3 + 1 2 = −1 halfway between the roots y = 2(−1) −6 = −8: turning point (−1; −8) it lies on the line a(2)(−2) = −8, −4a = −8, a = 2 substitute into f Example 3.6 (challenge level — same type as the challenge paper) f (x) = 2(x + 3)(x −1) = 2x2 + 4x −6. (a) The line y = 8x + k is a tangent to f. Determine k. (b) For which values of k does y = 8x + k cut f at two distinct points, both with positive x- coordinates? (c) g(x) = f (x + t). For which values of t are both x-intercepts of g negative? (d) Calculate the average gradient of f between x = 0 and x = 2. WORKING REASON (a) f ′(x) = 4x + 4 = 8, x = 1; f (1) = 0 gradient of tangent = 8 0 = 8(1) + k, k = −8 point (1; 0) on the line (b) 2x2 + 4x −6 = 8x + k, 2x2 −4x −(6 + k) = 0 equate Two points: k > −8 line above the tangent (∆= 64 + 8k > 0) When k = −6 the roots are x = 0 and x = 2 the line passes through f’s y-intercept (0; −6) Both positive: product −(6 + k) 2 > 0, so k < −6 below that position −8 < k < −6 (c) Intercepts of g: x = −3 −t and x = 1 −t every x moves t to the left 1 −t < 0, t > 1 the larger one must be negative (d) f (2) −f (0) 2 −0 = 10 −(−6) 2 = 8 Example 3.7 (harder — same type as the prediction paper) f (x) = −x2 + 4x + 5 and the line g meet at A(−1; 0) and C(4; 5). MN is a vertical line segment with M on f and N on g, between A and C. (a) Find the turning point D and the range of f. (b) Find the equation of g. (c) For which x is f (x) ⩾g(x) and f increasing? (d) Calculate the maximum length of MN. (e) Determine the equation of the tangent to f parallel to g. A C D M N g f x y www.hlayiso.com 31

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs WORKING REASON (a) x = −b 2a = −4 −2 = 2, f (2) = 9: D(2; 9); range y ⩽9 (b) m = 5 −0 4 −(−1) = 1, g(x) = x + 1 through A(−1; 0) (c) f ⩾g for −1 ⩽x ⩽4; f increasing for x < 2 between the intersections; left of D −1 ⩽x ⩽2 both conditions (d) MN = (−x2 + 4x + 5) −(x + 1) = −x2 + 3x + 4 top minus bottom d dx(MN) = −2x + 3 = 0, x = 3 2 maximise MNmax = −9 4 + 9 2 + 4 = 25 4 units (e) f ′(x) = −2x + 4 = 1, x = 3 2; f (3 2 ) = 35 4 parallel to g: gradient 1 y = x + c: 35 4 = 3 2 + c, y = x + 29 4 Watch out: The range of a parabola with a maximum at y = 9 is y ⩽9, not y ⩾9. Look at which way it opens. 3.3 Exponential and logarithmic graphs, and inverses Recognise it. – y = bx+p + q: horizontal asymptote y = q; it is increasing when b > 1 and decreasing when 0 < b < 1. – “Reflection in the line y = x” or “inverse”: swap x and y. The inverse of y = bx is y = logb x. – Domain of the inverse = range of f; and a point (a; b) on f gives (b; a) on f −1. – Inequalities such as h−1(x) ⩽1: rewrite with the definition of the log, then remember the domain. Method. 1. Definition of a logarithm: y = logb x ⇐⇒x = by (b > 0, b̸ = 1, x > 0). First line: x = 3y+1 −3 (swap), then 3y+1 = x + 3 2. Laws you may need: logb(AB) = logb A + logb B, logb A B = logb A −logb B, logb An = n logb A, logb b = 1, logb 1 = 0. www.hlayiso.com 32

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs Example 3.8 (harder — same type as the challenge paper) The graph of h(x) = 3x+p + q passes through the origin and through (1; 6). (a) Show that p = 1 and q = −3. (b) Determine h−1 in the form y = . . . (c) Write down the domain of h−1. (d) For which x is h−1(x) ⩽1? WORKING REASON (a) 3p + q = 0 …(1); 3p+1 + q = 6 …(2) substitute both points (2)−(1): 3 · 3p −3p = 6, 3p = 3, p = 1 3p+1 = 3 · 3p q = −31 = −3 from (1) (b) x = 3y+1 −3, 3y+1 = x + 3 swap x and y y + 1 = log3(x + 3), y = log3(x + 3) −1 definition of log (c) x > −3 range of h is y > −3 (d) log3(x + 3) ⩽2, x + 3 ⩽9, x ⩽6 h−1 is increasing −3 < x ⩽6 include the domain Example 3.9 (harder — same type as the prediction paper) f (x) = bx passes through (−2; 1 16 ) . (a) Show that b = 4. (b) g is the reflection of f in y = x; write down g. (c) h is f translated 1 unit right and 2 units down. Find h and its x-intercept. (d) For which x is g(x) · h(x) ⩾0? WORKING REASON (a) b−2 = 1 16, b2 = 16, b = 4 b > 0 (b) g(x) = log4 x inverse of 4x (c) h(x) = 4x−1 −2 right: x →x −1; down: −2 4x−1 = 2, 22x−2 = 21, x = 3 2 (d) g < 0 for 0 < x < 1, g > 0 for x > 1; h < 0 for x < 3 2, h > 0 for x > 3 2 signs from the sketch below Same sign (or zero): 0 < x ⩽1 or x ⩾3 2 g exists only for x > 0 1 1 f g y = x x y www.hlayiso.com 33

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs Example 3.10 (standard) Determine the point of intersection of f (x) = 2x and g(x) = ( 1 2 )x−1. WORKING REASON 2x = 2−(x−1) = 21−x write both with base 2 x = 1 −x, x = 1 2; y = 2 1 2 = √ 2 equate exponents (1 2; √ 2 ) Watch out: logb x only exists for x > 0. Every inequality answer with a log needs this lower limit. 3.4 Inverses of lines and parabolas Recognise it. – Inverse of a straight line: swap and make y the subject. A line and its inverse meet on y = x. – Inverse of y = ax2: y = ± √x a. It is only a function if the domain of f is restricted to x ⩾0 or x ⩽0. – “Is f −1 a function?”: one x-value may give only one y-value (vertical line test on f −1, horizontal line test on f). Example 3.11 (standard) g(x) = 3x −6. (a) Determine g−1. (b) Find the point where g and g−1 intersect. (c) Solve g−1(x) · g(x) = 0. WORKING REASON (a) x = 3y −6, g−1(x) = x 3 + 2 swap, then make y the subject (b) On y = x: 3x −6 = x, x = 3, (3; 3) they meet on the mirror line (c) g−1(x) = 0: x = −6; g(x) = 0: x = 2 a product is 0 if a factor is 0 x = −6 or x = 2 Example 3.12 (harder) f (x) = 2x2, x ⩽0. (a) Determine f −1. (b) Explain why the inverse of y = 2x2, x ∈R, is not a function. (c) p(x) = 2√x, x ⩾0: determine p−1. WORKING REASON (a) x = 2y2, y2 = x 2, y = ± √ x 2 swap and solve f −1(x) = − √ x 2, x ⩾0 y ⩽0 because the domain of f was x ⩽0 (b) for one x (e.g. x = 2) there are two y-values, ±1: one-to-many (c) x = 2√y, p−1(x) = x2 4 , x ⩾0 the restriction stays www.hlayiso.com 34

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs 3.5 Transformations, reflections, domain and range New function What happens to the graph Point (x; y) becomes f (x −a) shift a units right (x + a; y) f (x) + b shift b units up (x; y + b) −f (x) reflect in the x-axis (x; −y) f (−x) reflect in the y-axis (−x; y) −f (−x) reflect in both axes (180◦about O) (−x; −y) f −1(x) reflect in the line y = x (y; x) Example 3.13 (standard) P(1; −3) lies on f (x) = x2 −4x. Give the image of P on each of: (a) g(x) = f (x −2) + 1 (b) h(x) = −f (x) (c) k(x) = f (−x) (d) f −1. Also give the range of g. WORKING REASON (a) (3; −2) 2 right, 1 up (b) (1; 3) (c) (−1; −3) (d) (−3; 1) use the table Turning point of f: (2; −4), so of g: (4; −3) f has range y ⩾−4 Range of g: y ⩾−3 Watch out: f (x −2) moves the graph to the right. The sign inside the bracket works backwards. 3.6 Two graphs together: inequalities from graphs Recognise it. – “For which x is f (x) ⩾g(x)?”: find the intersections; read where f is above g. – “f (x) · g(x) < 0” or “x · f (x) ⩾0”: find where each factor is zero, then compare signs interval by interval. – Watch for asymptotes: a hyperbola changes sign at its vertical asymptote without touching the axis. Example 3.14 (challenge level — parabola and hyperbola) The graphs of f (x) = x2 −7 and g(x) = −6 x are shown. They intersect at x = −3, x = 1 and x = 2. For which x is f (x) ⩾g(x)? f g x y www.hlayiso.com 35

Grade 12 Mathematics P1 — Revision Notes 3 Functions and graphs WORKING REASON x2 −7 = −6 x ⇐⇒x3 −7x + 6 = 0 ⇐⇒ (x −1)(x −2)(x + 3) = 0 check the given intersections Left of −3: f above g; between −3 and 0: g above g has an asymptote at x = 0 0 < x < 1: f above (g →−∞); 1 < x < 2: g above; x > 2: f above read interval by interval x ⩽−3 or 0 < x ⩽1 or x ⩾2 x = 0 is never included Watch out: When you read an inequality from a graph, include the endpoints only for ⩽or ⩾, and never at an asymptote. www.hlayiso.com 36

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay 4 Finance, growth and decay Question 7 is worth about 15 marks. Every finance question is solved with the formulas on the information sheet. The skill is choosing the right one, and the habit that makes this easy is drawing a time line first. Situation Formula Clue words one amount grows / shrinks A = P(1 ± ni), A = P(1 ± i)n invest, depreciate equal payments, value at the end F = x [(1 + i)n −1] i save, sinking fund, annuity equal payments, value now P = x [1 −(1 + i)−n] i loan, bond, instalment, balance Always write i and n for the compounding period: 9% p.a. compounded monthly gives i = 0,09 12 and n in months. Keep at least four decimals (or the whole calculator value) until the end. 4.1 Simple and compound growth, changing rates, effective rates Recognise it. – One sum of money, no regular payments: A = P(1 + ni) (simple) or A = P(1 + i)n (com- pound). – “How long?” or “after how many months?”: n is the unknown in an exponent, so use logs. – The rate changes part of the way through: work in stages; the end value of one stage is P of the next. – “Effective” rate: the rate that, compounded once a year, gives the same growth: 1 + ieff = ( 1 + inom m )m . Example 4.1 (standard) R12 000 is invested for 4 years at 7,5% p.a. (a) What is it worth with simple interest? (b) With interest compounded monthly? (c) At 9% p.a. compounded monthly, how many months does it take for an investment to double? (d) R5 000 grows to R8 000 in 6 years, compounded annually. Find the rate. WORKING REASON (a) A = 12 000(1 + 0,075 × 4) = R15 600 simple: A = P(1 + ni) (b) A = 12 000 ( 1 + 0,075 12 )48 = R16 183,19 n = 4 × 12 (c) 2P = P(1,0075)n, n = log 2 log 1,0075 = 92,77 P cancels: any amount 93 months the first whole month after it doubles (d) 8 000 = 5 000(1 + i)6, 1 + i = 6√ 1,6 isolate the bracket i = 0,0815, 8,15% p.a. www.hlayiso.com 37

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.2 (standard) (a) Convert 10% p.a. compounded monthly to an effective annual rate. (b) Find the nominal rate, compounded quarterly, that is equivalent to an effective rate of 9% p.a. WORKING REASON (a) 1 + ieff = ( 1 + 0,1 12 )12 , ieff = 10,47% (b) 1,09 = ( 1 + i 4 )4 , i 4 = 4√ 1,09 −1 i = 8,71% p.a. compounded quarterly Example 4.3 (challenge level — same type as the challenge paper) R30 000 is invested. For the first 2 years it earns 7% p.a. compounded monthly; after that 10% p.a. compounded quarterly. After how many months, from the start, is the investment first worth more than R50 000? WORKING REASON After 2 years: 30 000 ( 1 + 0,07 12 )24 = R34 494,18 stage 1: n = 24 months 50 000 = 34 494,18(1,025)q stage 2: i = 0,10 4 , q quarters q = log ( 50 000 34 494,18 ) log 1,025 = 15,03 16 quarters are needed interest is only added at the end of a quarter 24 + 16 × 3 = 72 months after 15 quarters it is only R49 957,86 Example 4.4 (standard — comparing options) Lerato has R20 000 to invest for 5 years. Option A: 9% p.a. simple interest. Option B: 8,4% p.a. compounded monthly. Which is better? WORKING REASON A: 20 000(1 + 0,09 × 5) = R29 000 B: 20 000 ( 1 + 0,084 12 )60 = R30 394,73 Option B, by R1 394,73 always answer the question in words 4.2 Depreciation Recognise it. – “Straight-line” (or “simple decay”): the same rand amount is lost every year, A = P(1 −ni). – “Reducing balance”: a percentage of the current value is lost every year, A = P(1 −i)n. www.hlayiso.com 38

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.5 (standard — same type as the prediction paper) (a) A machine costs R250 000 and depreciates at 12% p.a. on the straight-line method. Find its value after 5 years. (b) A taxi bought for R520 000 has a book value of R230 000 after 6 years (reducing balance). Find the rate of depreciation. (c) At 15% p.a. reducing balance, after how many full years is an asset first worth less than half its cost? WORKING REASON (a) A = 250 000(1 −0,12 × 5) = R100 000 (b) 230 000 = 520 000(1 −i)6 1 −i = 6 √ 230 000 520 000 = 0,8729, i = 12,71% p.a. (c) 0,5P = P(0,85)n, n = log 0,5 log 0,85 = 4,27 after 5 years Watch out: For depreciation the bracket is (1 −i). Writing (1 + i) is the most common way to lose all the marks. 4.3 Future value annuities and sinking funds Recognise it. – Regular deposits and the question asks for the amount at the end: future value F. – “Sinking fund”: money saved now to replace an asset later. Target = future cost −trade-in value. – Deposits start now (immediately), not in one month: there is one extra payment; use n + 1 payments, or multiply by (1 + i) if the value is measured one period after the last payment. – Money stays invested after the deposits stop: grow F further with (1 + i)n. Method. 1. Draw a time line: mark the first and last payment and the point where the value is required. 2. Count payments: the first payment one month from now and the last at the end of 6 years gives n = 72. Example 4.6 (standard) Sipho deposits R1 500 at the end of every month for 10 years into an account paying 8% p.a. com- pounded monthly. (a) How much is in the account after 10 years? (b) He then stops depositing but leaves the money for 2 more years. What is it worth then? (c) Suppose instead that the first deposit was made immediately and the last one month before the end of the 10 years. What would the account be worth after 10 years? WORKING REASON (a) F = 1 500 [( 1 + 0,08 12 )120 −1 ] 0,08 12 n = 120, i = 0,08 12 R274 419,05 (b) 274 419,05 ( 1 + 0,08 12 )24 = R321 862,80 one amount growing: A = P(1 + i)n (c) 274 419,05 ( 1 + 0,08 12 ) = R276 248,51 every deposit earns one extra month: ×(1 + i) www.hlayiso.com 39

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.7 (challenge level — same type as the prediction paper) A farmer buys a tractor for R400 000. He will replace it in 6 years. The price of tractors rises at 7% p.a. compounded annually. The old tractor depreciates at 18% p.a. on the reducing-balance method and will be traded in. He makes monthly payments into a sinking fund at 10,2% p.a. com- pounded monthly, the first one month from now and the last at the end of the 6 years. Calculate the monthly payment. WORKING REASON Cost in 6 years: 400 000(1,07)6 = R600 292,14 inflation: compound growth Trade-in: 400 000(0,82)6 = R121 602,67 reducing balance Sinking fund target: R478 689,47 difference 478 689,47 = x [(1 + 0,0085)72 −1 ] 0,0085 F formula, i = 0,102 12 , n = 72 x = R4 847,62 Watch out: The future-value formula gives the value on the day of the last payment. If the value is wanted later, grow it further. 4.4 Loans: the instalment and a delayed first payment Recognise it. – Loan, bond, “pay off”, “instalment”: present value P (the amount borrowed today). – “First instalment 3 months after the loan is granted”: interest runs for 2 months with no pay- ment. Grow the loan to one month before the first payment, then use the P formula. Example 4.8 (standard — same type as the prediction paper) A home loan of R1 200 000 is repaid over 20 years by equal monthly instalments starting one month after it is granted. Interest is 10,5% p.a. compounded monthly. Calculate (a) the instalment and (b) the total interest paid over the 20 years. WORKING REASON (a) 1 200 000 = x [ 1 − ( 1 + 0,105 12 )−240] 0,105 12 P formula, n = 240 x = R11 980,56 (b) 240 × 11 980,56 −1 200 000 = R1 675 334,40 total paid minus amount borrowed www.hlayiso.com 40

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.9 (challenge level — same type as the challenge paper) A loan of R250 000 is taken at 15% p.a. compounded monthly. It is repaid with 48 equal monthly instalments, the first one 4 months after the loan is granted. Show that the instalment is R7 221,88. 0 1 2 3 4 51 · · · R250 000 x x P here WORKING REASON i = 0,15 12 = 0,0125 Balance after 3 months: 250 000(1,0125)3 = R259 492,68 one month before the first payment 259 492,68 = x [ 1 −(1,0125)−48] 0,0125 now an ordinary loan of 48 payments x = R7 221,88 Watch out: If the first payment is in month 4, grow the loan for 3 months, not 4. The P formula always places the first payment one period after P. 4.5 Outstanding balance, extra payments, final payment Recognise it. – “Outstanding balance after k payments”: the present value of the payments that are left, P = x[1−(1+i)−(n−k)] i . – An extra lump sum is paid: subtract it from the balance, then find the new number of payments with logs. – The instalment is increased: use the balance as the new P with the new x, and find n. – “How many full payments” and “final payment”: n is not whole; the integer part is the number of full payments. – Missed payments or a rate change: work out the balance at that moment, then start a new calculation from there. Method. 1. Find the balance at the moment something changes. First line: Balance = x [ 1 −(1 + i)−(remaining)] i 2. Adjust it (subtract the lump sum, or note the new rate or instalment). 3. Solve for n: (1 + i)−n = 1 −Pi x , then n = −log(...) log(1+i) . www.hlayiso.com 41

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.10 (challenge level — same type as the challenge paper) For the loan in Example 4.9 (instalment R7 221,88, 48 payments, i = 0,0125): immediately after the 20th instalment an additional R40 000 is paid into the account. The same instalment continues. How many more full instalments will be paid? WORKING REASON Remaining payments: 48 −20 = 28 Balance = 7 221,88 [ 1 −(1,0125)−28] 0,0125 = R169 732,36 PV of what is left New balance: 169 732,36 −40 000 = R129 732,36 subtract the lump sum 129 732,36 = 7 221,88 [1 −(1,0125)−n] 0,0125 same x, unknown n (1,0125)−n = 1 −129 732,36 × 0,0125 7 221,88 = 0,77545 . . . isolate the power n = −log 0,77545 . . . log 1,0125 = 20,47 20 more full instalments and a smaller final payment Example 4.11 (challenge level — same type as the prediction paper) For the home loan in Example 4.8 (R11 980,56 per month, 240 payments): immediately after the 60th payment the owner increases the instalment to R14 000. (a) How many months sooner is the loan paid off? (b) Calculate the final payment. (c) Instead, suppose the interest rate rose to 12% p.a. after the 60th payment. What would the new instalment be? WORKING REASON (a) Balance = 11 980,56 [ 1 − ( 1 + 0,105 12 )−180] 0,105 12 = R1 083 822,30 180 payments left 1 083 822,30 = 14 000 [ 1 − ( 1 + 0,105 12 )−n] 0,105 12 , n = 129,86 new x; solve with logs 129 full payments + 1 final payment = 130 180 −130 = 50 months sooner (b) Balance after 129 payments: 1 083 822,30(1 + i)129 −14 000 [(1 + i)129 −1 ] i grow the debt, subtract the payments’ value Final payment = that balance ×(1 + i) = R12 007,62 one more month of interest (c) 1 083 822,30 = x [ 1 −(1,01)−180] 0,01 new i = 0,12 12 , same 180 months x = R13 007,69 www.hlayiso.com 42

Grade 12 Mathematics P1 — Revision Notes 4 Finance, growth and decay Example 4.12 (harder — missed payments) For the home loan in Example 4.8: immediately after the 60th payment the owner misses the next three payments. From month 64 he pays a new, larger instalment so that the loan is still paid off at the end of the 20 years. Calculate the new instalment. WORKING REASON Balance after 60 payments: R1 083 822,30 as in Example 4.11 Three months with no payment: 1 083 822,30 ( 1 + 0,105 12 )3 = R1 112 522,30 the debt only grows Payments left: 240 −63 = 177 months 64 to 240 1 112 522,30 = x [ 1 − ( 1 + 0,105 12 )−177] 0,105 12 P one month before the first new payment x = R12 384,16 Example 4.13 (standard — the “other” questions) (a) A car costs R300 000 and the buyer pays a 15% deposit. What is the loan amount? (b) For the loan in Example 4.9, how much interest was paid up to and including the 20th instalment? WORKING REASON (a) 300 000 × 0,85 = R255 000 loan = price −deposit (b) Paid: 20 × 7 221,88 = R144 437,60 Debt reduced from R250 000 to R169 732,36: by R80 267,64 principal repaid Interest = 144 437,60 −80 267,64 = R64 169,96 payments minus reduction in debt Watch out: “How many payments” is rounded down for full payments; “how many months to reach” a savings target is rounded up. Watch out: Use the unrounded balance (calculator memory) in the next step. Rounding in the middle can change the cents in the final answer. www.hlayiso.com 43

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus 5 Differential calculus Questions 8 to 10 carry about 35 marks. The first question (first principles and the rules) is the safest set of marks in the paper; learn it until it is automatic. 5.1 Derivative from first principles Recognise it. – “From first principles”: you must use the limit definition. Using the rules scores nothing. Method. 1. Find f (x + h) on its own first, with brackets. 2. Write f (x + h) −f (x) and simplify until every term has a factor h. With a fraction such as 2 x, use a common denominator. 3. Write the definition with the limit, divide by h, then let h →0. First line: f ′(x) = lim h→0 f (x + h) −f (x) h Example 5.1 (standard — same type as the prediction paper) Determine f ′(x) from first principles if f (x) = −2x2 + 3x. WORKING REASON f (x + h) = −2(x + h)2 + 3(x + h) = −2x2 −4xh −2h2 + 3x + 3h expand with brackets f (x + h) −f (x) = −4xh −2h2 + 3h every term now has h f ′(x) = lim h→0 −4xh −2h2 + 3h h = lim h→0(−4x −2h + 3) definition, then divide by h f ′(x) = −4x + 3 let h →0 Example 5.2 (challenge level — same type as the challenge paper) Determine f ′(x) from first principles if f (x) = 3x2 + 2 x. WORKING REASON f (x + h) = 3x2 + 6xh + 3h2 + 2 x + h f (x + h) −f (x) = 6xh + 3h2 + 2 x + h −2 x 2 x + h −2 x = 2x −2(x + h) x(x + h) = −2h x(x + h) common denominator f ′(x) = lim h→0 6xh + 3h2 − 2h x(x+h) h = lim h→0 ( 6x + 3h − 2 x(x + h) ) divide every term by h f ′(x) = 6x −2 x2 Watch out: Write “limh→0” on every line until h has gone, and drop it on the last line. Wrong notation costs a mark. www.hlayiso.com 44

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus 5.2 The rules, and simplifying first Recognise it. – Each term is a power of x: use d dx (axn) = anxn−1. – There are brackets, roots, or a division: rewrite first as a sum of powers of x. Never differentiate a quotient or product directly. – A fraction like x3+27 x+3 : factorise and cancel. – “ dy da”: differentiate with respect to a; x is treated as a constant. Method. 1. Rewrite: √x = x 1 2 , 1 x2 = x−2, x2+4x √x = x 3 2 + 4x 1 2 . First line: y = x 3 2 + 4x 1 2 2. Differentiate term by term. Use the notation of the question: dy dx, f ′(x) or Dx[. . .]. Example 5.3 (standard — same type as the prediction paper) Determine: (a) dy dx if y = 4x3 −2 x2 + 5√x (b) f ′(x) if f (x) = (x + 2)(3x −1) (c) dy dx if y = x3+27 x+3 WORKING REASON (a) y = 4x3 −2x−2 + 5x 1 2 rewrite as powers dy dx = 12x2 + 4x−3 + 5 2x−1 2 (b) f (x) = 3x2 + 5x −2, f ′(x) = 6x + 5 multiply out first (c) y = (x + 3)(x2 −3x + 9) x + 3 = x2 −3x + 9 sum of cubes; cancel (x̸ = −3) dy dx = 2x −3 Example 5.4 (challenge level — same type as the challenge paper) Determine: (a) dy dx if y = (x+2)2−4 √x (b) Dx [( 2√x + 1 x )2] (c) dy dx if xy = 3x2 −5 (d) dy da if y = ax3 + a WORKING REASON (a) (x + 2)2 −4 = x2 + 4x simplify the numerator y = x2 + 4x x 1 2 = x 3 2 + 4x 1 2 divide each term dy dx = 3 2x 1 2 + 2x−1 2 (b) ( 2√ x + 1 x )2 = 4x + 4x−1 2 + x−2 middle term: 2 · 2x 1 2 · x−1 4 −2x−3 2 −2x−3 (c) y = 3x −5 x = 3x −5x−1, dy dx = 3 + 5 x2 make y the subject first (d) dy da = x3 + 1 a is the variable here Watch out: 1 3x2 = 1 3x−2, not 3x−2. The 3 stays in the denominator. www.hlayiso.com 45

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus 5.3 Tangents Recognise it. – “Equation of the tangent at x = 2”: you need a point and a gradient; the gradient is f ′(2). – “Tangent parallel to y = 9x + 2” or “with gradient 9”: solve f ′(x) = 9. – “Angle of inclination” of the tangent: tan θ = m. – “The tangent at P passes through a point not on the curve”: let P= (p; f (p)), write the tangent in terms of p, and substitute the point. – “The line is a tangent to both graphs”: equate the line with each graph; each gives ∆= 0 (a perfect square). Method. 1. Point: y-value from f. Gradient: from f ′. First line: m = f ′(2) = . . . and f (2) = . . . 2. Substitute into y −y1 = m(x −x1). Example 5.5 (standard) (a) Determine the equation of the tangent to y = x3 −2x2 + 1 at x = 2, and its angle of inclination. (b) Determine the points on y = x3 −3x where the tangent is parallel to y = 9x + 2. (c) Explain why no tangent to p(x) = 2x3 + x has a negative gradient. WORKING REASON (a) y(2) = 8 −8 + 1 = 1; y′ = 3x2 −4x, m = y′(2) = 4 point (2; 1) y −1 = 4(x −2), y = 4x −7 tan θ = 4, θ = 75,96◦ (b) 3x2 −3 = 9, x2 = 4, x = ±2 parallel: same gradient 9 (2; 2) and (−2; −2) y = x3 −3x (c) p′(x) = 6x2 + 1 ⩾1 > 0 for all x x2 ⩾0 Example 5.6 (challenge level — same type as the challenge paper) The tangent to y = x3 −3x2 at the point P(p ; p3 −3p2) passes through (0; 1), which is not on the curve. Determine the possible coordinates of P. WORKING REASON m = 3p2 −6p gradient at P Tangent: y −(p3 −3p2) = (3p2 −6p)(x −p) point P, gradient m (0; 1): 1 −p3 + 3p2 = −3p3 + 6p2 substitute x = 0, y = 1 2p3 −3p2 + 1 = 0 (p −1)2(2p + 1) = 0 p = 1 is a root (factor theorem) P(1; −2) or P ( −1 2; −7 8 ) check: tangents y = −3x + 1 and y = 15 4 x + 1 www.hlayiso.com 46

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus Example 5.7 (challenge level — a common tangent) The line y = 2x −1 is a tangent to f (x) = x2 and also to g(x) = −x2 + bx −5, touching g at a point with a positive x-coordinate. Determine b. WORKING REASON x2 −(2x −1) = (x −1)2: it touches f at x = 1 a perfect square means a tangent −x2 + bx −5 = 2x −1, x2 −(b −2)x + 4 = 0 equate with g ∆= (b −2)2 −16 = 0, b −2 = ±4 tangent: ∆= 0 Contact point x = b −2 2 > 0, so b −2 = 4 b = 6 (contact at x = 2) 5.4 Cubic graphs Recognise it. – “Show that x = 2 is a root” or “hence factorise”: factor theorem, then divide (or compare coefficients). – “Touches the x-axis at x = −1”: a double root, so f (−1) = 0 and f ′(−1) = 0; or write f (x) = a(x + 1)2(x −c). – “Turning points”: f ′(x) = 0. “Point of inflection”: f ′′(x) = 0 (it lies halfway between the turning points). – The graph of f ′ is given instead of f: the x-intercepts of f ′ are the turning points of f; where f ′ > 0, f increases; the turning point of f ′ is the point of inflection of f. – “For which k does f (x) = k have three / two positive / one root(s)?”: draw horizontal lines y = k across your sketch. Method. 1. Intercepts: f (0), then factorise. Turning points: f ′(x) = 0. Concavity: sign of f ′′. First line: f ′(x) = 3x2 −6x −9 = 0 2. Sketch: shape from the sign of a (a > 0 ends bottom-left to top-right), then all intercepts and turning points with coordinates. Example 5.8 (standard — factor theorem) (a) Show that x = 2 is a root of f (x) = x3 −4x2 + x + 6 = 0 and hence factorise f completely. (b) The graph of g(x) = ax3 + bx2 + cx + d cuts the x-axis at −2, 1 and 3, and the y-axis at 12. Determine g. WORKING REASON (a) f (2) = 8 −16 + 2 + 6 = 0, so (x −2) is a factor factor theorem f (x) = (x −2)(x2 −2x −3) divide, or compare: x2 + bx −3 with −3 · (−2) = 6 f (x) = (x −2)(x −3)(x + 1) (b) g(x) = a(x + 2)(x −1)(x −3); g(0) = 6a = 12, a = 2 roots give the factors g(x) = 2x3 −4x2 −10x + 12 www.hlayiso.com 47

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus Example 5.9 (challenge level — same type as the challenge paper) f (x) = x3 + px2 + qx −5 touches the x-axis at x = −1. (a) Show that p = −3 and q = −9. (b) Find the turning points and the point of inflection, and sketch f. (c) For which values of k does f (x) = k have two distinct positive roots? (d) For which x is g(x) = f (x + 2) decreasing? WORKING REASON (a) f ′(x) = 3x2 + 2px + q; f ′(−1) = 0: 3 −2p + q = 0 touching: f ′(−1) = 0 f (−1) = 0: −1 + p −q −5 = 0, so p −q = 6 and f (−1) = 0 q = 2p −3, p −(2p −3) = 6, p = −3, q = −9 solve simultaneously (b) f ′(x) = 3x2 −6x −9 = 3(x −3)(x + 1) = 0 f (−1) = 0; f (3) = −32: (−1; 0) and (3; −32) substitute into f f ′′(x) = 6x −6 = 0, x = 1, f (1) = −16: (1; −16) halfway between −1 and 3 f (x) = (x + 1)2(x −5): intercepts −1, 5 and (0; −5) for the sketch −1 5 −5 (3; −32) (1; −16) y = k f x y WORKING REASON (c) A line y = k with −32 < k < 0 cuts f three times: once left of −1 move the dotted line up and down The middle root is positive only while the line is below (0; −5) at k = −5 one root is x = 0 −32 < k < −5 at k = −32 the roots are equal (d) f decreases for −1 < x < 3; g is f moved 2 left f (x + 2): 2 units left −3 < x < 1 Example 5.10 (challenge level — same type as the prediction paper) The graph of f ′(x) for f (x) = ax3 + bx2 + cx + d is a parabola that cuts the x-axis at −1 and 3 and the y-axis at 9. Also f (0) = 5. (a) Determine f. (b) Give the turning points of f and say which is a maximum. (c) Find the point of inflection, the tangent there and its angle of inclination. (d) For which k does f (x) = k have three distinct real roots? (e) Where is f concave up? WORKING REASON (a) f ′(x) = m(x + 1)(x −3); f ′(0) = −3m = 9, m = −3 x-intercepts of f ′ f ′(x) = −3x2 + 6x + 9 = 3ax2 + 2bx + c compare coefficients a = −1, b = 3, c = 9, d = 5: f (x) = −x3 + 3x2 + 9x + 5 d = f (0) (b) f (−1) = 0, f (3) = 32 f ′: −to + at −1; + to −at 3 www.hlayiso.com 48

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus minimum (−1; 0), maximum (3; 32) (c) x = −1 + 3 2 = 1, f (1) = 16; m = f ′(1) = 12 inflection: turning point of f ′ y −16 = 12(x −1), y = 12x + 4 ; tan θ = 12, θ = 85,24◦ (d) 0 < k < 32 between the turning point y-values (e) f ′′(x) = −6x + 6 > 0, x < 1 concave up: f ′′ > 0 (f ′ increasing) Watch out: A sketch must show the shape and all intercepts and turning points with coordinates. Each one carries a mark. Watch out: If the graph of f ′ is given, do not read the turning points of f ′ as turning points of f. 5.5 Increasing, decreasing and inequalities with f and f ′ Recognise it. – “Increasing”: f ′(x) > 0. “Decreasing”: f ′(x) < 0. “Concave down”: f ′′(x) < 0. – “x · f ′(x) < 0” or “f (x) · f ′(x) > 0”: the two factors must have opposite (or the same) signs. Find where each is zero, then test the intervals. Example 5.11 (harder) f (x) = x3 −6x2 + 9x. (a) For which x is f increasing? (b) Concave down? (c) For which x is x · f ′(x) < 0? WORKING REASON (a) f ′(x) = 3x2 −12x + 9 = 3(x −1)(x −3) > 0 x < 1 or x > 3 outside the roots (b) f ′′(x) = 6x −12 < 0, x < 2 (c) Zeros: x = 0 (from x), x = 1 and x = 3 (from f ′) critical values x < 0: (−)(+); 0 < x < 1: (+)(+); 1 < x < 3: (+)(−); x > 3: (+)(+) signs of x and f ′(x) x < 0 or 1 < x < 3 opposite signs Example 5.12 (challenge level — same type as the challenge paper) f (x) = 2(x + 3)(x −1). For which values of x is f (x) · f ′(x) > 0? WORKING REASON f (x) > 0 for x < −3 or x > 1; f (x) < 0 for −3 < x < 1 a parabola opening upwards f ′(x) = 4x + 4: f ′ > 0 for x > −1; f ′ < 0 for x < −1 Both positive: x > 1; both negative: −3 < x < −1 same sign gives a positive product −3 < x < −1 or x > 1 www.hlayiso.com 49

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus 5.6 Values of a parameter Recognise it. – “Tangent with the smallest (or largest) gradient”: the gradient function f ′ is a parabola; its turning point is at f ′′(x) = 0. – “f (x) = k has one real root”: the line y = k lies above the maximum or below the minimum. Example 5.13 (harder) (a) Determine the equation of the tangent to f (x) = x3 −3x2 + 5x that has the smallest gradient. (b) For f (x) = x3 −3x2 −9x −5 (Example 5.9), for which k does f (x) = k have only one real root? WORKING REASON (a) f ′(x) = 3x2 −6x + 5; f ′′(x) = 6x −6 = 0, x = 1 minimum of the gradient function m = f ′(1) = 2, f (1) = 3: y = 2x + 1 point (1; 3) (b) Turning points (−1; 0) and (3; −32) k > 0 or k < −32 above the maximum or below the minimum 5.7 Optimisation Recognise it. – “Maximum”, “minimum”, “largest”, “least cost” in a story: write the quantity as a function of ONE variable, differentiate, set equal to zero. – Two variables appear (radius and height): use the other piece of information (volume, perime- ter, similar triangles) to remove one. Method. 1. Write the given constraint and make the unwanted variable the subject. First line: πr2h = 250π, so h = 250 r2 2. Substitute into the quantity to be optimised (“show that” usually ends here). 3. Differentiate, set the derivative equal to 0, solve, and calculate the required value. Example 5.14 (challenge level — same type as the challenge paper) An open cylindrical container (no lid) must hold 250π cm3. The base costs R0,06 per cm2 and the curved side R0,03 per cm2. (a) Show that the cost is C(r) = 0,06πr2 + 15π r . (b) Find the radius for the minimum cost, and that cost. WORKING REASON (a) πr2h = 250π, h = 250 r2 volume of a cylinder C = 0,06(πr2) + 0,03(2πrh) one base only; side = 2πrh = 0,06πr2 + 0,06πr · 250 r2 = 0,06πr2 + 15π r substitute h (b) C′(r) = 0,12πr −15π r2 = 0 15π r = 15πr−1 0,12r3 = 15, r3 = 125, r = 5 cm C(5) = 1,5π + 3π = 4,5π = R14,14 www.hlayiso.com 50

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus Example 5.15 (challenge level — same type as the prediction paper) A cylinder of radius r and height h stands inside a cone of base radius 4 cm and height 10 cm, with its top edge touching the cone. (a) Show that V = 10πr2 −5 2πr3. (b) Calculate the maximum volume. r h 4 10 WORKING REASON 10 −h r = 10 4 , h = 10 −5 2r similar triangles above the cylinder V = πr2h = πr2 ( 10 −5 2r ) = 10πr2 −5 2πr3 V′(r) = 20πr −15 2 πr2 = 0, r ( 20 −15 2 r ) = 0, r = 8 3 r̸ = 0 V = π · 64 9 · 10 3 = 640π 27 ≈74,47 cm3 h = 10 3 Example 5.16 (standard) Equal squares of side x cm are cut from the corners of a 30 cm by 30 cm sheet, and the sides are folded up to make an open box. Find x for the maximum volume. WORKING REASON V = x(30 −2x)2 base (30 −2x)2, height x V′ = (30 −2x)2 −4x(30 −2x) = (30 −2x)(30 −6x) = 0 expanding first also works: V = 4x3 −120x2 + 900x x = 15 (no box) or x = 5 ; V = 5(20)2 = 2 000 cm3 Watch out: Read what is asked: the radius, the minimum cost, or both. Give units. 5.8 Rates of change and working backwards Recognise it. – “Rate of change”, “velocity”, “how fast”: the derivative. “Acceleration”: the second derivative. – “Initially”: substitute t = 0. “When is it empty / on the ground”: set the quantity equal to 0. – The derivative is given and you need the function: think backwards (which function has this derivative?), then use the extra information for the constant. www.hlayiso.com 51

Grade 12 Mathematics P1 — Revision Notes 5 Differential calculus Example 5.17 (standard) A ball is thrown upwards; its height after t seconds is h(t) = 30t −5t2 metres. (a) Find its velocity after 2 s. (b) Find its maximum height. (c) With what velocity does it hit the ground? WORKING REASON (a) h′(t) = 30 −10t, h′(2) = 10 m/s velocity = h′(t) (b) 30 −10t = 0, t = 3; h(3) = 45 m maximum: velocity 0 (c) 30t −5t2 = 0, t = 6; h′(6) = −30 m/s negative: moving downwards Example 5.18 (harder — working backwards) (a) g is a straight line with g(1) = 5 and g′(3) = 2. Find g. (b) f (x) = 3x2 + bx + c, with f ′(1) = 9 and f (1) = 0. Find b and c. (c) p′(x) = 4x −8 and the minimum value of p is −3. Determine p(x). WORKING REASON (a) gradient 2: g(x) = 2x + c, 5 = 2 + c, g(x) = 2x + 3 the derivative of a line is its gradient (b) f ′(x) = 6x + b, 6 + b = 9, b = 3; 3 + 3 + c = 0, b = 3, c = −6 (c) p(x) = 2x2 −8x + c d dx (2x2 −8x) = 4x −8 Minimum at p′(x) = 0: x = 2; 8 −16 + c = −3, p(x) = 2x2 −8x + 5 www.hlayiso.com 52

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting 6 Probability and counting The last question carries about 15 marks. Write every rule you use; the rule itself often earns the first mark. Words in the question Rule “or” P(A or B) = P(A) + P(B) −P(A and B) mutually exclusive P(A and B) = 0 independent P(A and B) = P(A) × P(B) complementary, “not”, “at least one” P(not A) = 1 −P(A) arrange n different objects n! n objects with p alike and q alike n! p! q! 6.1 The rules: addition, independence, unknown probabilities Recognise it. – “Are A and B independent?”: calculate P(A) × P(B) and compare it with P(A and B). Write the conclusion. – The probabilities contain x: substitute into the addition rule; independence gives P(A and B) = x(x + 0,1); solve the quadratic. – “Exactly one”: P(A or B) −P(A and B). Example 6.1 (standard) P(A) = 0,3, P(B) = 0,5 and P(A or B) = 0,65. Are A and B (a) mutually exclusive? (b) indepen- dent? WORKING REASON 0,65 = 0,3 + 0,5 −P(A and B), P(A and B) = 0,15 addition rule (a) No: P(A and B)̸ = 0 (b) P(A) × P(B) = 0,3 × 0,5 = 0,15 = P(A and B) Yes, they are independent Example 6.2 (challenge level — same type as the challenge paper) A and B are independent. P(A) = x, P(B) = x + 0,1 and P(A or B) = 0,58. (a) Find x. (b) Find the probability that exactly one of A and B occurs. WORKING REASON (a) P(A and B) = x(x + 0,1) independent 0,58 = x + (x + 0,1) −x(x + 0,1) addition rule x2 −1,9x + 0,48 = 0 standard form x = 1,9 ± √3,61 −1,92 2 = 1,9 ± 1,3 2 x = 0,3 x = 1,6 > 1: reject (b) P(A and B) = 0,3 × 0,4 = 0,12 0,58 −0,12 = 0,46 or 0,3(0,6) + 0,7(0,4) www.hlayiso.com 53

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting 6.2 Venn diagrams Method. 1. Fill the diagram from the middle outwards: all three, then the “two only” regions, then the “one only” regions, then “none”. 2. Use letters for unknown regions and make an equation from the total. 3. Independence of two events in a Venn diagram: n(A ∩B) = n(A)×n(B) n(S) . Example 6.3 (challenge level — same type as the prediction paper) 100 learners were asked about three clubs: Maths (M), Science (S) and Debating (D). 50 belong to M, 40 to S and 30 to D; 5 belong to all three; 6 belong to S and D but not M; 10 belong to none. The events “belongs to M” and “belongs to D” are independent. (a) Show that 15 learners belong to both M and D. (b) Draw a Venn diagram and find how many belong to M and S only. (c) Find the probability that a learner belongs to exactly one club. WORKING REASON (a) P(M and D) = 50 100 × 30 100 = 0,15, so n = 15 independent (b) M and D only: 15 −5 = 10; D only: 30 −10 −5 −6 = 9 middle outwards Let M and S only = x: M only = 35 −x, S only = 29 −x (35 −x) + (29 −x) + 9 + x + 10 + 6 + 5 + 10 = 100 total 104 −x = 100, x = 4 (c) one club only: 31 + 25 + 9 = 65, P = 0,65 M S D 31 25 4 5 10 6 9 10 6.3 Tree diagrams and two-way tables Recognise it. – One event happens after another (two draws, two days): tree diagram. Multiply along the branches, add between branches. – “Without replacement”: the numbers on the second set of branches change. – Data in rows and columns (gender against result): two-way table; test independence with the totals. www.hlayiso.com 54

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting Example 6.4 (standard) A bag contains 5 red and 3 blue balls. Two balls are taken out one after the other without replace- ment. Find the probability that (a) they are the same colour, (b) at least one is blue. R B R 5 8 · 4 7 = 20 56 B R B 3 8 · 2 7 = 6 56 5 8 3 8 4 7 3 7 5 7 2 7 WORKING REASON (a) 5 8 · 4 7 + 3 8 · 2 7 = 26 56 = 13 28 RR or BB (b) 1 −P(RR) = 1 −20 56 = 9 14 “at least one”: use the complement Example 6.5 (harder — an unknown number) A bag contains 4 blue balls and n red balls. Two balls are taken without replacement. The proba- bility that both are blue is 2 15. Find n. WORKING REASON 4 n + 4 · 3 n + 3 = 2 15 total n + 4, then one fewer 180 = 2(n + 4)(n + 3), n2 + 7n −78 = 0 (n + 13)(n −6) = 0, n = 6 Example 6.6 (standard — two-way table) The results of 200 learners are shown. Is passing independent of gender? Pass Fail Total Male 48 32 80 Female 72 48 120 Total 120 80 200 WORKING REASON P(male) × P(pass) = 80 200 × 120 200 = 0,24 use the totals P(male and pass) = 48 200 = 0,24 the cell Equal, so passing is independent of gender www.hlayiso.com 55

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting 6.4 Counting: arrangements, letters and codes Recognise it. – Arrange n different things in a row: n!. Fill positions one at a time and multiply (fundamental counting principle). – Letters of a word with repeated letters: divide by the factorial of each repeat. – Codes and numbers: draw boxes for the positions. “Repetition allowed” keeps the number of choices; “not allowed” reduces it by one each time. – A condition on a digit (even, odd, greater than 4 000): fill the restricted positions first. If two restrictions overlap, split into cases. Example 6.7 (standard) (a) In how many ways can the letters of LIMPOPO be arranged? (b) How many 4-digit PIN codes (digits 0 to 9) are there with repetition allowed, and without? WORKING REASON (a) 7 letters, P twice and O twice: 7! 2! 2! = 1 260 divide out the repeats (b) 10 × 10 × 10 × 10 = 10 000 ; 10 × 9 × 8 × 7 = 5 040 Example 6.8 (harder — same type as the prediction paper) Four-digit numbers are formed from 1; 2; 3; 4; 5; 6 without repetition. (a) How many numbers are possible? (b) How many are odd AND greater than 4 000? (c) Find the probability that a number contains 1 and 2 next to each other. WORKING REASON (a) 6 × 5 × 4 × 3 = 360 (b) First digit 5: last digit 1 or 3: 1 × 4 × 3 × 2 = 24 5 is both large and odd: a separate case First digit 4 or 6: last digit 1, 3 or 5: 2 × 4 × 3 × 3 = 72 first and last first, then the middle 24 + 72 = 96 (c) Block “12” or “21”: 3 positions × 2 orders treat the pair as one unit Other two places: 4 × 3 = 12; total 3 × 2 × 12 = 72 P = 72 360 = 1 5 6.5 Arrangements with restrictions Recognise it. – “Must sit together”: glue them into one unit, arrange the units, then arrange inside the unit. – “No two … next to each other”: place the others first, then put the restricted ones into the gaps (including both ends). – “Do NOT sit next to each other” (two people): 1 −P(together). – “At each end” or “starts with”: fill those positions first. www.hlayiso.com 56

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting Example 6.9 (challenge level — same type as the challenge paper) Three teachers (one is Mr Dube) and five learners (one is Sipho) sit at random in a row of 8 chairs. (a) In how many ways can they sit if the teachers sit together? (b) Find the probability that no two teachers sit next to each other. (c) Find the probability that Mr Dube and Sipho do not sit next to each other. (d) How many arrangements have a learner at each end? WORKING REASON (a) 6 units: 6!; teachers inside the unit: 3! glue the teachers 6! × 3! = 4 320 (b) Learners: 5! = 120; this makes 6 gaps _ L _ L _ L _ L _ L _ Teachers into 3 different gaps: 6 × 5 × 4 = 120 one teacher per gap P = 120 × 120 8! = 14 400 40 320 = 5 14 n(S) = 8! (c) Together: 2 × 7!; P = 1 −2 × 7! 8! = 1 −1 4 = 3 4 complement (d) Ends: 5 × 4; middle: 6!; 14 400 fill the ends first Example 6.10 (harder) (a) 4 boys and 4 girls sit in a row, alternating boy and girl. How many arrangements? (b) The letters of PRETORIA are arranged. How many arrangements are there in total, how many start with P, and how many have all four vowels together? WORKING REASON (a) BGBGBGBG or GBGBGBGB: 2 × 4! × 4! = 1 152 two patterns (b) 8 letters, R twice: 8! 2! = 20 160 Start with P: 7! 2! = 2 520 the other 7 letters, R still repeated Vowels E, O, I, A as one unit: units P, R, T, R, [vowels] 5 units, R twice 5! 2! × 4! = 60 × 24 = 1 440 arrange inside the unit Watch out: In a probability question, the total n(S) uses the same counting method with no restriction (here 8!). 6.6 Other probability questions Example 6.11 (challenge level — a geometric series) Thabo and Lindiwe take turns to shoot at a target, Thabo first. Thabo hits with probability 0,4 and Lindiwe with probability 0,5. The first to hit wins. Find the probability that Thabo wins. WORKING REASON Thabo wins on shot 1, 3, 5, …: 0,4 + (0,6)(0,5)(0,4) + (0,6 · 0,5)2(0,4) + . . . both miss once per round Geometric: a = 0,4, r = 0,3 S∞= 0,4 1 −0,3 = 4 7 www.hlayiso.com 57

Grade 12 Mathematics P1 — Revision Notes 6 Probability and counting Example 6.12 (harder — money and probability) A game costs R5 to play. Two dice are thrown and the player wins if the total is 10 or more. 300 people play per hour. The owner wants to keep at least 60% of the income. What is the largest prize per winner? WORKING REASON Winning totals: (4;6), (5;5), (6;4), (5;6), (6;5), (6;6): P = 6 36 = 1 6 list the outcomes Income: 300 × 5 = R1 500; prizes at most 40%: R600 Expected winners: 1 6 × 300 = 50 R12 per winner 600 ÷ 50 www.hlayiso.com 58

Grade 12 Mathematics P1 — Revision Notes If you see … write first … If you see … write first … Read this list the night before the examination. Cover the right-hand column and say the first line aloud. If you see … write first … Algebra a quadratic that is not = 0 standard form: ax2 + bx + c = 0 “correct to two decimal places” x = −b± √ b2−4ac 2a with brackets x4 and x2; x and √x; a repeated block “Let k = . . .” x in a denominator the restriction, then multiply by the LCD a surd equation isolate the root, square, check 2x+1 + 2x−1 2x(2 + 1 2) 9x and 3x; 2x and 2−x “Let k = 3x” (and reject k ⩽0) x and y in exponents same base on each side ⇒linear equations log3(x −y) = 2 x −y = 32 linear + quadratic simultaneous make x or y the subject of the linear one inequality with x2 one side 0, factorise, critical values, sketch “for which x is √ A B real” A ⩾0 and B̸ = 0 nature of roots / values of k ∆= b2 −4ac = . . . “rational for all k” ∆= (. . .)2 “real for all p” ∆= (. . .)2+ positive large powers, no calculator take out the smallest power x = √a + √ b, y = √a − √ b xy = a −b and x2 + y2 maximum / minimum of an expression complete the square Sequences and series second differences constant 2a = . . ., 3a + b = . . ., a + b + c = . . . unknown terms p, q in a quadratic pattern equal second differences, or Tn = an2 + bn + c “is not a term” solve Tn = . . . and show n /∈N (or odd/even) “first difference is 199” Tn+1 −Tn = . . . “how many terms” Tn = a + (n −1)d or Tn = arn−1 a sum is given Sn formula = the sum ∑notation first term, last term, number of terms log inside a ∑ simplify the general term first n in an exponent n = log(...) log r (watch the sign) “converges” −1 < r < 1 “sum to infinity” S∞= a 1−r S∞−Sn < . . . arn 1−r < . . . Sn given as a formula Tn = Sn −Sn−1 arithmetic terms that form a geometric sequence Tb Ta = Tc Tb Functions 3x+1 x−1 split: 3(x−1)+4 x−1 = 3 + 4 x−1 axes of symmetry of a hyperbola y = ±(x + p) + q line does not cut / touches a graph equate, standard form, ∆< 0 / ∆= 0 parabola with given roots y = a(x −x1)(x −x2) turning point of a parabola x = −b 2a (or f ′(x) = 0) “tangent with gradient m” f ′(x) = m “maximum length of MN” top −bottom, then maximise g(x) = f (x + t) every x-value moves t to the left www.hlayiso.com 59

Grade 12 Mathematics P1 — Revision Notes If you see … write first … If you see … write first … “passes through (2; 12)” f (2) = 12 inverse swap x and y; y = logb x ⇐⇒x = by domain of f −1 range of f f (x) ⩾g(x) intersections, then read from the graph x · f (x) ⩽0, f (x) · g(x) > 0 zeros of each factor, signs interval by interval average gradient f (b)−f (a) b−a Finance any finance question a time line; i and n per period depreciation (1 −i), never (1 + i) “effective rate” 1 + ieff = ( 1 + i m )m rate changes during the term stage by stage saving towards a target, sinking fund F = x[(1+i)n−1] i ; target = cost −trade-in loan, instalment P = x[1−(1+i)−n] i first payment after k months grow the loan for k −1 months first outstanding balance PV of the payments that are left extra lump sum / bigger instalment new balance, then solve for n with logs Calculus “first principles” f (x + h) first, then limh→0 f (x+h)−f (x) h brackets, roots or division rewrite as powers of x, then differentiate tangent at a point m = f ′(x1), y −y1 = m(x −x1) tangent through a point not on the curve P(p; f (p)), tangent in terms of p, substitute “touches the x-axis at a” f (a) = 0 and f ′(a) = 0 turning points f ′(x) = 0 point of inflection, concavity f ′′(x) = 0; sign of f ′′ graph of f ′ given roots of f ′ = turning points of f f (x) = k has n roots horizontal lines across the sketch optimisation constraint ⇒one variable; derivative = 0 rate, velocity the derivative; “initially” t = 0 Probability and counting “or” P(A or B) = P(A) + P(B) −P(A and B) “independent” P(A and B) = P(A) × P(B) “at least one” 1 −P(none) three overlapping groups Venn diagram from the middle outwards one event after another tree diagram; without replacement changes the numbers repeated letters n! p! q! “together” glue into one unit, × arrangements inside “no two next to each other” place the others, use the gaps digit conditions (even, > 4 000) restricted positions first; split into cases You have done the work. Read each question slowly, write the first line, and trust your method. Good luck. www.hlayiso.com 60