Grade 12 Mathematics P1 Revision Guide 2026

HLAYISO STUDY NOTES · 2026 · GRADE 12 Mathematics Paper 1 Revision guide: every question type since 2018, and the first line to write. To the learner Many of you have told us: “I should know what to do, but I don’t.” That feel- ing 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. For every type of question you will learn how to recognise it, what to write first, and how a full solution looks. HOW TO USE IT Cover the working, try the example with a pencil, then read the reasons in grey. Start with any section. CONTENTS IN THE EXAM How to attack a hard paper 2 STRATEGY 1 Algebra: equations, inequalities and nature of roots 3 QUESTION 1 · 25 MARKS 2 Number patterns, sequences and series 7 QUESTIONS 2–3 · 25 MARKS 3 Functions and graphs 10 QUESTIONS 4–6 · 35 MARKS 4 Finance, growth and decay 14 QUESTION 7 · 15 MARKS 5 Differential calculus 18 QUESTIONS 8–10 · 35 MARKS 6 Probability and counting 22 LAST QUESTION · 15 MARKS If you see … write first … 25 QUICK REFERENCE

BEFORE YOU START · THREE HOURS, 150 MARKS How to attack a hard paper Six habits that turn ``I don't know'' into marks THE TIME RULE 1,2 minutes per mark. Twice the time and still stuck? Write the formula and the substitution, star the question, move on. Paper 1 counts 150 marks in three hours: 1,2 minutes per mark. A 4- mark question deserves about five minutes. Keep this number in your head; it stops one stubborn question from eating the time you need for ten easy ones. 1 The first five minutes: read the whole paper Mark each question with a tick (I can do it), a question mark (maybe) or a star (difficult). You do not have to start “from 1.1 nje”: number your answers cor- rectly and begin with the sure marks (derivatives by the rules, annuity formu- las, asymptotes, sequence formulas, probability rules). 2 Collect method marks, even when you are stuck SHOW THAT You may use the given result in the next part even if you could not prove it. Most memo marks are for method: the formula, the substitution, the standard form. Write and substitute even if you cannot see the end; a blank is always zero. A wrong answer in 6.1 used correctly in 6.2 still earns the marks for 6.2, so always attempt the follow-up parts. 3 When you are stuck, do this in order 1 Underline the key words: tangent, converge, outstanding balance, real roots, in- dependent, hence. Each points to a method. 2 Turn every fact into an equation: “passes through (2; 12)” is f (2) = 12; “touches the x-axis at 2” is f (2) = 0 and f ′(2) = 0. 3 Draw something: a sketch, a number line, a time line, a tree or a Venn dia- gram. Look at what the final answer must be. 4 “Show that”: work backwards from the given answer. A letter k: ask what you would do if k were a number, and do exactly that. 4 Why hard papers feel hard NAME THE DISGUISE A parameter (k, p, t) instead of a number; two topics joined; working backwards. Name it and the question becomes ordinary. A challenge paper hides a method you already know behind a disguise. Name the disguise (see the margin) and the question becomes ordinary. 5 The last ten minutes: check Reject impossible answers (n /∈N, a probability above 1, a negative length, a false root, 2x = −3). Round to two decimals, money to the cent, months in the sensible direction. Give the form asked for: y = . . ., (x; y), an interval. READING THE EXAMPLES Working left of the green rule, reasons in grey, answer highlighted. Standard: a normal exam question; Harder: the last marks; Challenge: Hlayiso challenge-paper level. Margin: First line, Spot it, Watch out. HLAYISO.COM 2

1 QUESTION 1 · ABOUT 25 MARKS Algebra Equations, inequalities and nature of roots IN THIS SECTION 1.1 Quadratics and hidden quadratics 1.2 Surd equations 1.3 Exponential equations 1.4 Inequalities 1.5 Simultaneous equations 1.6 Nature of roots 1.7 Exponents and surds 1.8 Proofs, max and min, word problems Almost every equation in Question 1 ends as a quadratic: your job is to find the substitution or the step that turns it into one. 1.1 Quadratic equations and hidden quadratics Make one side zero and factorise; “two decimal places” means the formula. The same block twice, once squared (x4 and x2, x and √x): let k = the block. An x in a denominator: write the restriction, then multiply every term by the LCD. Example 1.1 STANDARD FIRST LINE ax2 + bx + c = 0, or ``x̸ = 0: x4 + 18 = 11x2'', or ``Let k = . . .''. WATCH OUT (x + 3)(x −2) = 14 does not give x + 3 = 14: ``each factor = 0'' needs a zero on the other side. 3x2 = 48 gives x = ±4; never divide by x. Solve for x: (a) (x + 3)(x −2) = 14 (b) 2x2 = 5x + 4, correct to two decimal places. (a) x2 + x −20 = 0, so (x + 5)(x −4) = 0 multiply out first x = −5 or x = 4 (b) x = 5 ± √ (−5)2 −4(2)(−4) 2(2) = 5 ± √ 57 4 2x2 −5x −4 = 0 x ≈3,14 or x ≈−0,64 round at the end only Example 1.2 CHALLENGE SPOT IT (x2 −3x)2 −2(x2 −3x) −8 = 0: k = x2 −3x, k = 4 or −2, x ∈{−1; 1; 2; 4}. x −5√x + 6 = 0: k = √x, so x = 4 or 9. SPOT IT ``Hence'': a + 8 a = 6 gives a = 2 or 4; 2x + 23−x = 6 is the same with a = 2x: x = 1 or 2. Solve for x: (a) x2 + 18 x2 = 11 (b) x x −1 − 2 x + 1 = 4 x2 −1 (a) x4 −11x2 + 18 = 0 x̸ = 0; multiply by x2 (x2 −2)(x2 −9) = 0: x = ± √ 2 or x = ±3 a quadratic in x2 (b) x(x + 1) −2(x −1) = 4, x2 −x −2 = 0 x̸ = ±1; ×(x2 −1) (x −2)(x + 1) = 0: x = 2 x = −1 is not allowed 1.2 Surd equations Get the root alone, square, solve, and check every answer in the original equa- tion. Example 1.3 STANDARD SPOT IT Two surds, √ x + 5 = 1 + √x: square, the x terms cancel, 4 = 2√x, so x = 4 (check: 3 = 3). WATCH OUT Squaring creates false roots: the check carries a mark. A root is never negative, so √ 2x + 9 = x −3 needs x ⩾3. Solve for x: x −√ 2x + 9 = 3. √ 2x + 9 = x −3, 2x + 9 = x2 −6x + 9 isolate, then square x(x −8) = 0; x = 0: 0 − √ 9 = −3̸ = 3 reject x = 0 x = 8: 8 − √ 25 = 3, so x = 8 true HLAYISO.COM 3

MATHEMATICS P1 · REVISION GUIDE 1 · ALGEBRA 1.3 Exponential equations Write every number as a prime power and equate exponents. Added terms (2x+1 + 2x−1): take out 2x. Both 9x and 3x, or 3x and 3−x: let k = 3x. Example 1.4 STANDARD WATCH OUT 2x+1 + 2x̸ = 22x+1: add exponents only when you multiply. 3 · 3x = 3x+1, not 9x. 5x = −25 has no solution: 5x > 0. Solve for x: (a) 8x+1 = 42x (b) x 3 2 = 27 (c) 2x+1 + 2x−1 = 20 (d) 3x+2 −3x = 72 (a) 23x+3 = 24x, 3x + 3 = 4x, x = 3 8 = 23, 4 = 22 (b) x = 27 2 3 = ( 3√ 27 )2 , x = 9 both sides to the 2 3 (c) 2x ( 2 + 1 2 ) = 20, 2x = 8, x = 3 common factor 2x (d) 3x(9 −1) = 72, 3x = 9, x = 2 common factor 3x Example 1.5 CHALLENGE SPOT IT 9x −4 · 3x+1 + 27 = 0 is (3x)2 −12 · 3x + 27 = 0: with k = 3x, x = 1 or x = 2. SPOT IT x, y natural, (√ 2 )x + ( 3√ 3 )y = 25: 2 x 2 + 3 y 3 must be two whole numbers; only 16 + 9 works: x = 8, y = 6, xy = 48. Solve for x: (a) 52x −5x −20 = 0 (b) 2x+1 + 22−x = 9 (a) k = 5x: k2 −k −20 = 0, (k −5)(k + 4) = 0 52x = (5x)2 5x = 5, x = 1 5x = −4 is impossible (b) 2 · 2x + 4 2x = 9; k = 2x: 2k2 −9k + 4 = 0 22−x = 22 2x ; ×k (2k −1)(k −4) = 0, 2x = 1 2 or 2x = 4 x = −1 or x = 2 1.4 Quadratic inequalities and ``when is it real?'' SPOT IT x −4 3 (x + 4)(x −3) < 0: between > 0: outside Example 1.6 STANDARD TO HARDER FIRST LINE One side zero, factorise, critical values, sketch. ``Real'': under a root ⩾0, denominator̸ = 0. A fraction against 0signs of top and bottom. WATCH OUT x2 ⩽9 gives −3 ⩽x ⩽3, not just x ⩽3. A value that makes a denominator zero is always excluded. (a) (x −2)(x + 3) > 6 (b) 3x −x2 ⩾−10 (c) For which x is √ 8 + 2x −x2 x + 1 real? (d) x + 3 x −1 ⩽0 (a) x2 + x −12 > 0, (x + 4)(x −3) > 0 multiply out x < −4 or x > 3 outside the roots (b) x2 −3x −10 ⩽0, (x −5)(x + 2) ⩽0 ×(−1): the sign turns −2 ⩽x ⩽5 between the roots (c) 8 + 2x −x2 ⩾0, so (x −4)(x + 2) ⩽0 under the root ⩾0 −2 ⩽x ⩽4, x̸ = −1 denominator̸ = 0 (d) −3 < x < 1 gives + −< 0 critical values −3, 1 −3 ⩽x < 1 x = 1 never allowed HLAYISO.COM 4

MATHEMATICS P1 · REVISION GUIDE 1 · ALGEBRA 1.5 Simultaneous equations, including unknowns in exponents Linear and quadratic: make x or y the subject of the linear one and substitute. Unknowns in exponents: same base, so each equation becomes linear. A log: use the definition. Every term of degree 2: factorise. Example 1.7 CHALLENGE FIRST LINE 3x · 33y = 32, so x + 3y = 2. SPOT IT y = 2x −3, x2 + xy = 6: 3x2 −3x −6 = 0: (2; 1) or (−1; −5). 3x+y = 27, 2x−y = 1 2 : x + y = 3, x −y = −1: x = 1, y = 2. WATCH OUT Give matched pairs. No calculator logs: match the bases. Solve for x and y: (a) 3x · 27y = 9 and x2 + 3y2 = 4 (b) log3(x −y) = 2 and x2 −5xy + 4y2 = 0 (a) x + 3y = 2; (2 −3y)2 + 3y2 = 4 substitute x = 2 −3y 12y2 −12y = 0, so y = 0 or y = 1 the 4s cancel (2 ; 0) or (−1 ; 1) x = 2 −3y (b) x −y = 32 = 9; (x −y)(x −4y) = 0 definition of log x = 4y, 3y = 9: y = 3 and x = 12 x −y = 9̸ = 0 1.6 Nature of roots and the discriminant THE TABLE ∆< 0: non-real ∆= 0: real, equal, rational ∆> 0, a square: real, unequal, rational ∆> 0, not a square: irrational ∆⩾0: real “Nature of the roots” or “values of k”: write ∆= b2 −4ac in terms of k and ap- ply the condition. “For all k”: ∆is a perfect square, or a square plus a positive number. Roots given as 3± √ k−2 4 : the part under the root is ∆. Example 1.8 STANDARD TO HARDER SPOT IT 2x2 −3x + 5 = 0: ∆= −31, non-real. 4x2 −12x + 9 = 0: ∆= 0, real, equal, rational. WATCH OUT If b = −(k + 1), then b2 = (k + 1)2: keep the brackets. Say ``non-real'', not ``no roots''. (a) For which k does x2 −(k + 1)x + 4 = 0 have equal roots? Find the root in each case. (b) For which k does x2 + kx + k + 3 = 0 have non-real roots? (c) Show that x2 −(p + 1)x + p −1 = 0 has real, unequal roots for all real p. (d) One root of ax2 + bx + c = 0 is 1, the other is −4−√ 16+20a 2a . Find a, b, c. (a) (k + 1)2 −4(1)(4) = 0: k = 3 or k = −5 equal roots: ∆= 0 Root −b 2a = k + 1 2 : x = 2 or x = −2 (b) k2 −4(k + 3) < 0, (k −6)(k + 2) < 0 non-real: ∆< 0 −2 < k < 6 between (c) ∆= (p + 1)2 −4(p −1) = p2 −2p + 5 brackets around b = (p −1)2 + 4 > 0: real and unequal for every p a square +4 (d) −b = −4, b = 4; −4ac = 20a, c = −5 compare with the formula a + 4 −5 = 0: a = 1, b = 4, c = −5 x = 1 is a root HLAYISO.COM 5

MATHEMATICS P1 · REVISION GUIDE 1 · ALGEBRA Example 1.9 CHALLENGE SPOT IT Integer roots from a fraction: split it, 2k k+1 = 2 − 2 k+1 , so k + 1 must divide 2. (k + 1)x2 −(3k + 1)x + 2k = 0, k̸ = −1. (a) Prove that the roots are rational for every rational k. (b) Find all integers k for which both roots are integers. (a) ∆= (3k + 1)2 −4(k + 1)(2k) b = −(3k + 1), c = 2k = k2 −2k + 1 = (k −1)2 a perfect square ∆is a perfect square and a,b,c are rational, so the roots are rational (b) x = (3k + 1) ± (k −1) 2(k + 1) : x = 2k k + 1 or x = 1 √ ∆= k −1 2k k + 1 = 2 − 2 k + 1, so k + 1 ∈{−2; −1; 1; 2} then 2 k + 1 is an integer k ∈{−3 ; −2 ; 0 ; 1} k = 0: roots 0; 1 1.7 Exponents and surds without a calculator Example 1.10 STANDARD AND CHALLENGE SPOT IT √ 48 + √ 75 − √ 27 = 4 √ 3 + 5 √ 3 −3 √ 3 = 6 √ 3. ( √ 12 + √ 3)2 = (3 √ 3)2 = 27. (a) Simplify 32025 −32023 32024 + 32022 . (b) If x = √ 5 + √ 3, y = √ 5 − √ 3, find x y + y x. (a) = 32022(33 −3) 32022(32 + 1) = 24 10 = 12 5 smallest power out (b) 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 = 8 common denominator 1.8 Proofs, maximum and minimum, word problems Example 1.11 HARDER AND CHALLENGE SPOT IT (2 + 1)(22 + 1)(24 + 1)(28 + 1): multiply by (2 −1); it collapses to 216 −1 = 65 535. Largest k with 2k | 20!: 10 + 5 + 2 + 1 = 18. WATCH OUT End a proof with a sentence. (a) Prove that 3n+2 −3n+1 + 4 · 3n is divisible by 10 for every natural n. (b) Simplify ( 1 + 1 3 ) ( 1 + 1 4 ) · · · ( 1 + 1 n ) and find all n ⩾3 for which it is an inte- ger. (a) = 3n ( 32 −3 + 4 ) = 10 · 3n take out 3n a multiple of 10, so divisible by 10 (b) 4 3 · 5 4 · 6 5 · · · n + 1 n = n + 1 3 it telescopes n = 5; 8; 11; . . . (n = 3m −1, m ⩾2) n + 1 a multiple of 3 Example 1.12 STANDARD TO CHALLENGE FIRST LINE x2 −6x = (x −3)2 −9. The maximum value is the y-value 14, not x = 3. WORD PROBLEMS Perimeter 46 m, diagonal 17 m: ℓ2 + (23 −ℓ)2 = 172: 15 m by 8 m. 30 v − 30 v+5 = 1: v = 10 km/h. (a) Find the maximum value of 5 + 6x −x2. (b) Find the maximum value of 12 x2 −4x + 7. (c) Prove x2 −4xy + 5y2 ⩾0; when is it 0? (a) = − [ (x −3)2 −9 ] + 5 = −(x −3)2 + 14 take out −1 maximum 14 at x = 3 −(x −3)2 ⩽0 (b) x2 −4x + 7 = (x −2)2 + 3 ⩾3: max 4 smallest denominator (c) = (x −2y)2 + y2 ⩾0: 0 only if x = y = 0 5y2 = 4y2 + y2 HLAYISO.COM 6

2 QUESTIONS 2–3 · ABOUT 25 MARKS Patterns Number patterns, sequences and series IN THIS SECTION 2.1 Quadratic patterns 2.2 Arithmetic series and sigma 2.3 Geometric series 2.4 Sum to infinity 2.5 Combined sequences and proofs First decide which kind of pattern you have. The test takes ten seconds and tells you which formula to use. TEST KIND GENERAL TERM first differences constant arithmetic Tn = a + (n −1)d second differences constant quadratic Tn = an2 + bn + c, 2a = second diff. ratios T2 T1 = T3 T2 constant geometric Tn = arn−1 2.1 Quadratic patterns Example 2.1 CHALLENGE FIRST LINE 2a = second difference, 3a + b = T2 −T1, a + b + c = T1. Unknown terms: Tn = an2 + bn + c from the known terms. Here a, b, c are not the first term and differences. SPOT IT 4; 9; 18; 31: 2a = 4, 3a + b = 5, a + b + c = 4, so Tn = 2n2 −n + 3. Given the quadratic pattern 2 ; p ; 16 ; q ; 54 ; . . . (a) Show that p = 6 and q = 32 and find Tn. (b) Prove that 777 is not a term. (a) a + b + c = 2, 9a + 3b + c = 16, 25a + 5b + c = 54 T1, T3, T5 8a + 2b = 14, 16a + 2b = 38: a = 3, b = −5, c = 4 subtract pairs p = T2 = 6, q = T4 = 32; Tn = 3n2 −5n + 4 (b) Tn = n(3n −5) + 4; n or 3n −5 is even one factor is even Tn is always even and 777 is odd: 777 is not a term (or n ≈16,91 /∈N) Example 2.2 STANDARD TO HARDER FIGURES 1, 2, 3 FIRST LINE Tn+1 −Tn = 4n −1; largest term at n = −b 2a . (a) 5 ; 8 ; 15 ; 26 ; . . .: between which terms is the first difference 199? (b) Tn = −n2 + 12n −8: find the largest term and the first negative term. (c) How many tiles in figure n (see the margin), and which figure uses 239? (a) 3; 7; 11; . . .: 4n −1 = 199, n = 50 first differences between T50 and T51 (b) n = −12 −2 = 6: T6 = 28 a < 0: a maximum n2 −12n + 8 > 0: n > 6 + √ 28 ≈11,29 negative terms T12 = −8 T11 = 3 > 0 (c) 5; 11; 19: a = 1, b = 3, c = 1 second difference 2 n2 + 3n + 1 = 239, (n + 17)(n −14) = 0 Tn = n2 + 3n + 1 figure 14 n = −17 /∈N HLAYISO.COM 7

MATHEMATICS P1 · REVISION GUIDE 2 · PATTERNS 2.2 Arithmetic series and sigma notation Example 2.3 STANDARD FIRST LINE Tn = a + (n −1)d; Sn = n 2 [2a + (n −1)d] or n 2 (a + ℓ). Sigma: first term, last term, top − bottom +1 terms. SPOT IT Three terms 2x + 1; 3x + 2; 5x −1: T2 −T1 = T3 −T2 gives x = 4. 5 + 9 + 13 + . . . = 945: (2n + 45)(n −21) = 0, n = 21. 3 + 8 + 13 + . . . + 248: (a) How many terms? (b) Find the sum. (c) Find the sum of the even terms. (a) 3 + (n −1)5 = 248, n = 50 Tn = 248 (b) S50 = 50 2 (3 + 248) = 6 275 Sn = n 2 (a + ℓ) (c) 8; 18; . . . ; 248: 8 + (m −1)10 = 248, m = 25 every second term S25 = 25 2 (8 + 248) = 3 200 Example 2.4 HARDER AND CHALLENGE SPOT IT Saving R200, then R50 more each month: n2 + 7n −400 > 0, n > 16,80: the 17th month. WATCH OUT From n = 11 to n = 40 there are 30 terms, not 29. (a) 40 ∑ n=11 (4n −3) (b) Write 2 + 5 + 8 + . . . + 59 in sigma notation. (c) Solve m ∑ k=1 (2k + 3) = 320. (d) Find k if k ∑ r=1 log3 ( 9r 3 ) = 144. (a) 41 + . . . + 157: 30 2 (41 + 157) = 2 970 40 −11 + 1 = 30 terms (b) 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 (d) log3 32r 3 = 2r −1: 1 + 3 + 5 + . . . simplify Tr first Sk = k2 = 144, k = 12 k = −12 not allowed 2.3 Geometric sequences and series Example 2.5 STANDARD FIRST LINE Two terms given: divide to remove a. n in an exponent: isolate the power, then logs. WATCH OUT When 0 < r < 1, log r < 0: dividing by it turns ``<'' into ``>''. (a) Which term of 3; 6; 12; . . . is 1 536? (b) T3 = 18 and T6 = 486: find a and r. (c) How many terms of 2 + 6 + 18 + . . . exceed 10 000? (a) 3 · 2n−1 = 1 536, 2n−1 = 29, n = 10 (b) ar5 ar2 = 27, r = 3, 9a = 18: a = 2, r = 3 divide to remove a (c) 2(3n −1) 3 −1 > 10 000, 3n > 10 001 Sn, a = 2, r = 3 n > log 10 001 log 3 ≈8,38: 9 terms round up Example 2.6 HARDER A ball is dropped from 8 m and each bounce rises to 75% of the previous height. (a) How high is the 4th bounce? (b) After which bounce is the rise first below 1 m? (c) Sipho says the ball travels more than 50 m. Is he right? (a) 8(0,75)4 ≈2,53 m bounce n: 8(0,75)n (b) 8(0,75)n < 1, n > log 0,125 log 0,75 ≈7,23 log 0,75 < 0: turn after the 8th bounce (c) 8 + 2 · 6 1 −0,75 = 56 m: yes, 56 > 50 every rise is fallen again HLAYISO.COM 8

MATHEMATICS P1 · REVISION GUIDE 2 · PATTERNS 2.4 Sum to infinity and convergence Example 2.7 CHALLENGE FIRST LINE ``Converges'': −1 < r < 1, S∞= a 1−r . ``Differs from S∞by less than'': S∞−Sn = arn 1−r . SPOT IT 54 + 18 + 6 + . . . = 54 1−1 3 = 81. 0,˙2˙7 = 0,27 1−0,01 = 3 11 . 15 + 15(3x −2) + 15(3x −2)2 + . . . (a) For which x does it converge? (b) S∞= 24. Find the smallest n for which Sn differs from S∞by less than 0,001. (a) −1 < 3x −2 < 1: 1 3 < x < 1 r in the middle (b) 15 1 −r = 24, r = 3 8; S∞−Sn = 24rn Sn = 24(1 −rn) 24(0,375)n < 0,001, n > 10,28 logs; the sign turns 11 terms 2.5 Combined sequences, Sn formulas and proofs Example 2.8 CHALLENGE SPOT IT Arithmetic terms that are geometric: equal ratios. Sn given: Tn = Sn −Sn−1. Alternating terms: two sequences. The 1st, 2nd and 5th 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 arithmetic terms is 192. Find the 6th geometric term. (a) a + d a = a + 4d a + d , (a + d)2 = a(a + 4d) equal ratios d2 = 2ad, d̸ = 0, so d = 2a (b) 8 2 [2a + 7(2a)] = 64a = 192, a = 3 d = 6 3 ; 9 ; 27: T6 = 3(3)5 = 729 r = 3 Example 2.9 HARDER (a) Sn = 3n2 + 2n. Find Tn. (b) 1; 3; 4; 9; 7; 27; 10; 81; . . . continues in the same way. Find the sum of the first 30 terms. (a) 3n2 + 2n − [ 3(n −1)2 + 2(n −1) ] = 6n −1 Tn = Sn −Sn−1 (b) 15 2 [2 + 14(3)] = 330 odd places: d = 3 3(315 −1) 3 −1 = 21 523 359; total 21 523 689 even places: r = 3 Example 2.10 STANDARD Prove (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. (a) Sn = a + (a + d) + . . . + [a + (n −1)d] forwards Sn = [a + (n −1)d] + . . . + (a + d) + a backwards 2Sn = n [2a + (n −1)d]: Sn = n 2 [2a + (n −1)d] add: equal pairs (b) rSn = ar + ar2 + . . . + arn multiply by r rSn −Sn = arn −a: Sn = a(rn−1) r−1 subtract; r̸ = 1 HLAYISO.COM 9

3 QUESTIONS 4–6 · ABOUT 35 MARKS Functions Functions and graphs IN THIS SECTION 3.1 The hyperbola 3.2 Parabola, tangents and lines 3.3 Exponential and log graphs 3.4 Inverses and transformations 3.5 Two graphs together Most of the marks come from a few moves that work for every graph. Learn the left column as a reflex. YOU ARE ASKED FOR WRITE FIRST y-intercept / x-intercepts put x = 0 / put y = 0 and solve a point lies on the graph / intersections f (2) = 12 / f (x) = g(x) average gradient from a to b / inverse f (b)−f (a) b−a / swap x and y 3.1 The hyperbola Split x+3 x−1 into a x+p + q. Asymptotes x = −p, y = q; axes of symmetry y = ±(x + p) + q. “The line does not cut the graph”: equate, standard form, ∆< 0. Example 3.1 STANDARD SKETCH −3 −3 f y = x x y FIRST LINE x + 3 x −1 = (x −1) + 4 x −1 f (x) = x + 3 x −1. (a) Show f (x) = 4 x −1 + 1. (b) Give the asymptotes, domain and range. (c) Find the intercepts and axes of symmetry; sketch f. (a) (x −1) + 4 x −1 = x −1 x −1 + 4 x −1 = 1 + 4 x −1 split the numerator (b) x = 1 and y = 1 ; domain x̸ = 1; range y̸ = 1 (c) (0; −3) and (−3; 0) original form y = x and y = −x + 2 through (1; 1), m = ±1 Example 3.2 CHALLENGE For which values of m will the line y = mx NOT intersect the graph of f (x) = x + 3 x −1? mx2 −mx = x + 3, mx2 −(m + 1)x −3 = 0 equate; x̸ = 1 ∆= (m + 1)2 + 12m < 0, m2 + 14m + 1 < 0 no real roots −7 −4 √ 3 < m < −7 + 4 √ 3 roots −7 ± 4 √ 3 HLAYISO.COM 10

MATHEMATICS P1 · REVISION GUIDE 3 · FUNCTIONS Example 3.3 HARDER SPOT IT x · h(x) ⩽0: zeros of each factor (0, 4), the asymptote (2), then signs interval by interval. The asymptotes of h(x) = a x+p + q meet at (2; −1) and h passes through (3; 1). (a) Find a, p, q. (b) Give the axis of symmetry with negative gradient. (c) For which x is x · h(x) ⩽0? (a) h(x) = a x −2 −1; 1 = a 1 −1, a = 2 p = −2, q = −1 (b) y = −(x −2) −1: y = −x + 1 gradient −1 (c) h > 0 only for 2 < x < 4 x-intercept 4 0 ⩽x < 2 or x ⩾4 x = 2 excluded 3.2 The parabola, tangents and lines Roots given: y = a(x −x1)(x −x2), axis halfway. “y = 8x + k is a tangent”: f ′(x) = 8. “Maximum length MN” (vertical): top −bottom, then maximise. g(x) = f (x + t): every x moves t left. Example 3.4 CHALLENGE SPOT IT Both roots positive: the line lies between the tangent and the line through f's y-intercept. f (x) = a(x + 3)(x −1) has its turning point on y = 2x −6. (a) Show a = 2. (b) y = 8x + k is a tangent to f: find k. (c) For which k does y = 8x + k cut f twice, both with positive x? (d) g(x) = f (x + t): for which t are both x-intercepts of g negative? (e) Find the average gradient of f from x = 0 to x = 2. (a) x = −3 + 1 2 = −1, y = 2(−1) −6 = −8 on the line a(2)(−2) = −8, so a = 2 substitute (−1; −8) (b) f ′(x) = 4x + 4 = 8, x = 1, f (1) = 0 gradient 8 0 = 8(1) + k, k = −8 (c) 2x2 −4x −(6 + k) = 0: two points if k > −8 ∆= 64 + 8k > 0 −(6 + k) 2 > 0, k < −6: −8 < k < −6 product of the roots (d) 1 −t < 0: t > 1 roots −3 −t, 1 −t (e) f (2) −f (0) 2 −0 = 10 −(−6) 2 = 8 Example 3.5 HARDER GRAPH A C D M N g f x y WATCH OUT A maximum at y = 9 gives range y ⩽9, not y ⩾9. f (x) = −x2 + 4x + 5 and the line g meet at A(−1; 0) and C(4; 5); MN is verti- cal, M on f, N on g. (a) Find the turning point D and the range of f. (b) Find g. (c) For which x is f ⩾g and f increasing? (d) Find the maximum length of MN. (e) Find the tangent to f parallel to g. (a) x = −4 −2 = 2, f (2) = 9: D(2; 9); y ⩽9 x = −b 2a (b) m = 5 −0 4 + 1 = 1: g(x) = x + 1 through A (c) −1 ⩽x ⩽4 and x < 2: −1 ⩽x ⩽2 left of D (d) MN = −x2 + 3x + 4; −2x + 3 = 0, x = 3 2 top minus bottom MNmax = −9 4 + 9 2 + 4 = 25 4 units (e) −2x + 4 = 1, x = 3 2, f (3 2 ) = 35 4 gradient 1 35 4 = 3 2 + c: y = x + 29 4 y = x + c HLAYISO.COM 11

MATHEMATICS P1 · REVISION GUIDE 3 · FUNCTIONS 3.3 Exponential and logarithmic graphs y = bx+p + q has asymptote y = q. Inverse (reflection in y = x): swap x and y, and y = logb x ⇐⇒ x = by. Domain of f −1 = range of f. A log inequality: use the definition, then add the domain. Example 3.6 CHALLENGE FIRST LINE Swap: x = 3y+1 −3, so 3y+1 = x + 3. WATCH OUT logb x exists only for x > 0: a log inequality answer needs this lower limit. h(x) = 3x+p + q passes through the origin and (1; 6). (a) Show p = 1, q = −3. (b) Find h−1. (c) Give the domain of h−1. (d) Solve h−1(x) ⩽1. (a) 3p + q = 0 and 3p+1 + q = 6 both points 3 · 3p −3p = 6, 3p = 3: p = 1, q = −3 subtract (b) y + 1 = log3(x + 3): y = log3(x + 3) −1 definition of log (c) x > −3 range of h (d) log3(x + 3) ⩽2, x + 3 ⩽9: −3 < x ⩽6 include the domain Example 3.7 HARDER SKETCH 1 1 f g y = x x y f (x) = bx passes through ( −2; 1 16 ) . (a) Show b = 4. (b) g is the reflection of f in y = x: give g. (c) h is f moved 1 right and 2 down: give h and its x-intercept. (d) For which x is g(x) · h(x) ⩾0? (a) b−2 = 1 16, b2 = 16, b = 4; (b) g(x) = log4 x b > 0 (c) h(x) = 4x−1 −2 ; 22x−2 = 21, x = 3 2 right 1, down 2 (d) g < 0 on 0 < x < 1; h < 0 for x < 3 2 from the sketch 0 < x ⩽1 or x ⩾3 2 same sign; x > 0 Example 3.8 STANDARD Determine the point of intersection of f (x) = 2x and g(x) = ( 1 2 )x−1 . 2x = 2−(x−1) = 21−x both with base 2 x = 1 −x, x = 1 2; y = 2 1 2 = √ 2: ( 1 2; √ 2 ) equate exponents 3.4 Inverses, transformations, domain and range NEW FUNCTION WHAT HAPPENS TO THE GRAPH (x; y) BECOMES f (x −a) / f (x) + b shift a right / b up (x + a; y) / (x; y + b) −f (x) / f (−x) reflect in the x-axis / in the y-axis (x; −y) / (−x; y) −f (−x) / f −1(x) both axes (180◦about O) / in y = x (−x; −y) / (y; x) HLAYISO.COM 12

MATHEMATICS P1 · REVISION GUIDE 3 · FUNCTIONS Example 3.9 STANDARD AND HARDER SPOT IT Inverse of y = 2x2, x ∈R: one x gives two y-values (x = 2: y = ±1), so it is not a function. Restrict to x ⩽0 or x ⩾0. (a) g(x) = 3x −6: find g−1, where g meets g−1, and solve g−1(x) · g(x) = 0. (b) Find f −1 for f (x) = 2x2, x ⩽0, and p−1 for p(x) = 2√x, x ⩾0. (a) x = 3y −6: g−1(x) = x 3 + 2 swap; y the subject 3x −6 = x: (3; 3) ; x = −6 or x = 2 on y = x; a factor is 0 (b) x = 2y2, y ⩽0: f −1(x) = − √ x 2, x ⩾0 domain of f: x ⩽0 x = 2√y: p−1(x) = x2 4 , x ⩾0 the restriction stays Example 3.10 HARDER WATCH OUT f (x −2) moves the graph right. A hyperbola reflected in y = q equals its reflection in x = −p. (a) P(1; −3) lies on f (x) = x2 −4x. Give its image on f (x −2) + 1, −f (x), f (−x), f −1, and the range of g(x) = f (x −2) + 1. (b) Describe two ways to transform f (x) = 4 x−1 + 1 to g(x) = 4 x+1 −1. (a) (3; −2) ; (1; 3) ; (−1; −3) ; (−3; 1) use the table range of g: y ⩾−3 turning point (4; −3) (b) f (x + 2) −2 = 4 x + 1 + 1 −2 = g(x) a translation 2 units left and 2 down or −f (−x) = g(x): reflect in both axes 180◦about O 3.5 Two graphs together: inequalities from graphs Example 3.11 CHALLENGE GRAPH f g x y WATCH OUT Endpoints only for ⩽or ⩾, never at an asymptote. f (x) = x2 −7 and g(x) = −6 x meet at x = −3, 1 and 2 (margin). For which x is f (x) ⩾g(x)? x3 −7x + 6 = (x −1)(x −2)(x + 3) = 0 check the intersections f above g: x < −3, 0 < x < 1, x > 2 asymptote at x = 0 x ⩽−3 or 0 < x ⩽1 or x ⩾2 x = 0 never included HLAYISO.COM 13

4 QUESTION 7 · ABOUT 15 MARKS Finance Finance, growth and decay IN THIS SECTION 4.1 Growth and effective rates 4.2 Depreciation 4.3 Future value, sinking funds 4.4 Loans and balances ALWAYS i and n per period: 9% p.a. monthly gives i = 0,09 12 , n in months. Keep the whole calculator value. Every finance question uses the formulas on the information sheet. The skill is choosing the right one: draw a time line first. SITUATION FORMULA CLUE WORDS one amount grows or 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 equal payments, value now P = x[1−(1+i)−n] i loan, bond, balance 4.1 Simple and compound growth, changing rates, effective rates “How long?”: n is in an exponent, so use logs. A rate change: work in stages. Effective rate: 1 + ieff = ( 1 + inom m )m . Example 4.1 STANDARD R12 000 for 4 years at 7,5% p.a.: (a) simple; (b) compounded monthly. (c) At 9% p.a. monthly, when does money double? (d) R5 000 grows to R8 000 in 6 years (annually): find the rate. (e) 10% p.a. monthly as an effective rate. (f) The quarterly nominal rate equal to 9% p.a. effective. (a) 12 000(1 + 0,075 × 4) = R15 600 A = P(1 + ni) (b) 12 000 ( 1 + 0,075 12 )48 = R16 183,19 n = 4 × 12 (c) n = log 2 log 1,0075 = 92,77: 93 months 2P = P(1,0075)n (d) 1 + i = 6√ 1,6: 8,15% p.a. isolate the bracket (e) ( 1 + 0,1 12 )12 −1 = 10,47% (f) 1,09 = ( 1 + i 4 )4 : 8,71% p.a. i = 4 ( 4√ 1,09 −1 ) Example 4.2 STANDARD Lerato has R20 000 to invest for 5 years. Option A: 9% p.a. simple interest. Op- tion B: 8,4% p.a. compounded monthly. Which is better? A: 20 000(1 + 0,09 × 5) = R29 000 simple B: 20 000 ( 1 + 0,084 12 )60 = R30 394,73 n = 60 months Option B, by R1 394,73 answer in words HLAYISO.COM 14

MATHEMATICS P1 · REVISION GUIDE 4 · FINANCE Example 4.3 CHALLENGE R30 000 earns 7% p.a. compounded monthly for 2 years, then 10% p.a. com- pounded quarterly. After how many months is it first worth more than R50 000? 30 000 ( 1 + 0,07 12 )24 = R34 494,18 stage 124 months 50 000 = 34 494,18(1,025)q, q = 15,03 stage 2quarters 16 quarters: 24 + 16 × 3 = 72 months 15 quarters: R49 957,86 4.2 Depreciation Example 4.4 STANDARD WATCH OUT Depreciation uses (1 −i). Writing (1 + i) loses all the marks. (a) R250 000, 12% p.a. straight-line: the value after 5 years. (b) R520 000 falls to R230 000 in 6 years, reducing balance: the rate. (c) At 15% p.a. reducing bal- ance, after how many full years is an asset first worth less than half its cost? (a) 250 000(1 −0,12 × 5) = R100 000 A = P(1 −ni) (b) 1 −i = 6 √ 230 000 520 000 = 0,8729, i = 12,71% A = P(1 −i)n (c) 0,5 = (0,85)n, n = log 0,5 log 0,85 = 4,27 after 5 years 4.3 Future value annuities and sinking funds Deposits, value at the end: F. Sinking fund target = future cost −trade-in. De- posits starting now: one more payment, or ×(1 + i). Example 4.5 STANDARD FIRST LINE Time line: first payment, last payment, and when the value is wanted. Months 1 to 120n = 120. WATCH OUT F is the value on the day of the last payment. Wanted later? Grow it further. R1 500 at the end of every month for 10 years at 8% p.a. monthly. (a) The value after 10 years. (b) Left 2 more years without deposits. (c) Instead the first de- posit is now and the last one month before the end: the value after 10 years. (a) F = 1 500 [(1 + i)120 −1 ] i = R274 419,05 i = 0,08 12 (b) 274 419,05(1 + i)24 = R321 862,80 one amount grows (c) 274 419,05(1 + i) = R276 248,51 one month more each Example 4.6 CHALLENGE A tractor costs R400 000 and is replaced in 6 years. Prices rise 7% p.a.; the old one depreciates 18% p.a. (reducing balance) and is traded in. Monthly pay- ments into a sinking fund at 10,2% p.a. monthly start in one month and end after 6 years. Find the payment. 400 000(1,07)6 = R600 292,14 new price 400 000(0,82)6 = R121 602,67 trade-in 478 689,47 = x [(1,0085)72 −1 ] 0,0085 target; i = 0,102 12 x = R4 847,62 HLAYISO.COM 15

MATHEMATICS P1 · REVISION GUIDE 4 · FINANCE 4.4 Loans, outstanding balance, extra and final payments Loan: P. First payment after 4 months: grow the loan 3 months first. Balance = PV of the payments left. Lump sum, new instalment, missed payments, new rate: find the balance then, adjust it, start again. n not whole: its integer part is the number of full payments. Example 4.7 STANDARD AND CHALLENGE TIME LINE FOR (C) 0 1 2 3 4 51 · · · R250 000 P x x WATCH OUT First payment in month 4grow for 3 months, not 4. The P formula puts the first payment one period after P. (a) R1 200 000 over 20 years, monthly from one month on, at 10,5% p.a. monthly: find the instalment and (b) the total interest. (c) R250 000 at 15% p.a. monthly, 48 instalments, the first 4 months after the loan: show the instal- ment is R7 221,88. (a) 1 200 000 = x [ 1 −(1 + i)−240] i , i = 0,105 12 n = 240 x = R11 980,56 (b) 240 × 11 980,56 −1 200 000 = R1 675 334,40 paid minus borrowed (c) 250 000(1,0125)3 = R259 492,68 3 months, i = 0,0125 259 492,68 = x [ 1 −(1,0125)−48] 0,0125 : x = R7 221,88 an ordinary loan Example 4.8 CHALLENGE FIRST LINE Balance = x[1−(1+i)−left] i ; then (1 + i)−n = 1 −Pi x . SPOT IT Loan after a 15% deposit on R300 000300 000 × 0,85 = R255 000. For the loan in Example 4.7(c): after the 20th instalment an extra R40 000 is paid in and the instalment stays the same. (a) How many more full instal- ments? (b) How much interest was paid up to the 20th instalment? (a) 7 221,88 [ 1 −(1,0125)−28] 0,0125 = R169 732,36 28 payments left 169 732,36 −40 000 = R129 732,36 minus the lump sum (1,0125)−n = 1 −129 732,36 × 0,0125 7 221,88 same x, new n n = 20,47: 20 more full instalments then a smaller one (b) 20 × 7 221,88 = R144 437,60 paid debt fell R80 267,64 144 437,60 −80 267,64 = R64 169,96 paid minus reduction HLAYISO.COM 16

MATHEMATICS P1 · REVISION GUIDE 4 · FINANCE Example 4.9 CHALLENGE WATCH OUT Full payments: round down; months to reach a target: round up. For the loan in Example 4.7(a): after the 60th payment the instalment rises to R14 000. (a) How many months sooner is it paid off? (b) Find the final pay- ment. (c) If instead the rate rose to 12% p.a. after the 60th payment, find the new instalment. (a) 11 980,56 [ 1 −(1 + i)−180] i = R1 083 822,30 180 left, i = 0,105 12 1 083 822,30 = 14 000 [1 −(1 + i)−n] i , n = 129,86 new x 180 −130 = 50 months sooner 129 full + 1 final (b) 1 083 822,30(1 + i)129 −14 000 [(1 + i)129 −1 ] i balance after 129 that ×(1 + i) = R12 007,62 one more month (c) 1 083 822,30 = x [ 1 −(1,01)−180] 0,01 i = 0,12 12 x = R13 007,69 Example 4.10 HARDER For the loan in Example 4.7(a): after the 60th payment the owner misses the next three payments. From month 64 a new instalment is paid so that the loan still ends after 20 years. Find the new instalment. Balance after 60 payments: R1 083 822,30 as in Example 4.9 1 083 822,30(1 + i)3 = R1 112 522,30 3 months unpaid 1 112 522,30 = x [ 1 −(1 + i)−177] i 240 −63 = 177 left x = R12 384,16 i = 0,105 12 HLAYISO.COM 17

5 QUESTIONS 8–10 · ABOUT 35 MARKS Calculus Differential calculus IN THIS SECTION 5.1 First principles and rules 5.2 Tangents 5.3 Cubic graphs 5.4 Increasing and inequalities 5.5 Optimisation 5.6 Rates and working backwards First principles and the rules are the safest marks in the paper: learn them until they are automatic. 5.1 First principles and the rules First principles: f (x + h) first, simplify f (x + h) −f (x) until every term has h, then the limit. The rules: rewrite brackets, roots and divisions as powers of x first. Example 5.1 STANDARD AND CHALLENGE FIRST LINE f ′(x) = lim h→0 f (x + h) −f (x) h WATCH OUT Write ``lim'' on every line until h has gone. Using the rules here scores nothing. From first principles: (a) f (x) = −2x2 + 3x (b) f (x) = 3x2 + 2 x (a) f (x + h) −f (x) = −4xh −2h2 + 3h expand f (x + h) first f ′(x) = lim h→0(−4x −2h + 3) = −4x + 3 divide by h (b) 2 x + h −2 x = 2x −2(x + h) x(x + h) = −2h x(x + h) common denominator f (x + h) −f (x) = 6xh + 3h2 − 2h x(x + h) every term has h f ′(x) = lim h→0 ( 6x + 3h − 2 x(x + h) ) divide by h f ′(x) = 6x −2 x2 let h →0 Example 5.2 STANDARD AND CHALLENGE SPOT IT f (x) = (x + 2)(3x −1): multiply out first, f ′(x) = 6x + 5. WATCH OUT 1 3x2 = 1 3 x−2, not 3x−2. (a) dy dx: y = 4x3 −2 x2 + 5√x (b) dy dx: y = x3+27 x+3 (c) dy dx: y = (x+2)2−4 √x (d) Dx [( 2√x + 1 x )2] (e) dy dx: xy = 3x2 −5 (f) dy da: y = ax3 + a (a) y = 4x3 −2x−2 + 5x 1 2 powers first dy dx = 12x2 + 4x−3 + 5 2x−1 2 (b) y = (x + 3)(x2 −3x + 9) x + 3 = x2 −3x + 9 sum of cubes dy dx = 2x −3 x̸ = −3 (c) y = x2 + 4x x 1 2 = x 3 2 + 4x 1 2 divide each term dy dx = 3 2x 1 2 + 2x−1 2 (d) 4x + 4x−1 2 + x−2: 4 −2x−3 2 −2x−3 middle: 2 · 2x 1 2 · x−1 (e) y = 3x −5x−1: 3 + 5 x2 (f) x3 + 1 y the subject; a varies HLAYISO.COM 18

MATHEMATICS P1 · REVISION GUIDE 5 · CALCULUS 5.2 Tangents Point from f, gradient from f ′, then y −y1 = m(x −x1). Parallel to y = 9x + 2: f ′(x) = 9. Smallest gradient: f ′′(x) = 0. Through a point not on the curve: P(p; f (p)). Tangent to two graphs: ∆= 0 for each. Example 5.3 STANDARD AND HARDER FIRST LINE m = f ′(2) and f (2); angle of inclination: tan θ = m. (a) Tangent to y = x3 −2x2 + 1 at x = 2, and its inclination. (b) Points on y = x3 −3x where the tangent is parallel to y = 9x + 2. (c) Why has no tangent to p(x) = 2x3 + x a negative gradient? (d) Tangent to f (x) = x3 −3x2 + 5x with the smallest gradient. (a) y(2) = 1; y′ = 3x2 −4x, m = y′(2) = 4 point (2; 1) y = 4x −7 ; tan θ = 4, θ = 75,96◦ (b) 3x2 −3 = 9, x = ±2: (2; 2) and (−2; −2) gradient 9 (c) p′(x) = 6x2 + 1 ⩾1 > 0 for all x x2 ⩾0 (d) f ′′(x) = 6x −6 = 0, x = 1; m = 2, f (1) = 3 minimum of f ′ y = 2x + 1 Example 5.4 CHALLENGE (a) The tangent to y = x3 −3x2 at P(p; p3 −3p2) passes through (0; 1), not on the curve. Find P. (b) y = 2x −1 is a tangent to f (x) = x2 and to g(x) = −x2 + bx −5, touching g at a positive x. Find b. (a) y −(p3 −3p2) = (3p2 −6p)(x −p) tangent in terms of p (0; 1): 2p3 −3p2 + 1 = 0, (p −1)2(2p + 1) = 0 factor theorem P(1; −2) or P ( −1 2; −7 8 ) (b) x2 −(2x −1) = (x −1)2: touches f a perfect square x2 −(b −2)x + 4 = 0; (b −2)2 −16 = 0 equate with g; ∆= 0 contact x = b −2 2 > 0: b = 6 5.3 Cubic graphs Root given: factor theorem. “Touches the x-axis at −1”: f (−1) = 0 and f ′(−1) = 0. Turning points f ′ = 0; inflection f ′′ = 0 (halfway). Graph of f ′: its roots are the turning points of f. “f (x) = k has n roots”: slide y = k across the sketch. Example 5.5 STANDARD (a) Show that x = 2 is a root of x3 −4x2 + x + 6 = 0 and factorise. (b) g(x) = ax3 + bx2 + cx + d cuts the x-axis at −2, 1, 3 and the y-axis at 12. Find g. (a) f (2) = 8 −16 + 2 + 6 = 0: (x −2) is a factor factor theorem (x −2)(x2 −2x −3) = (x −2)(x −3)(x + 1) divide (b) g(x) = a(x + 2)(x −1)(x −3), 6a = 12 a = 2 g(x) = 2x3 −4x2 −10x + 12 HLAYISO.COM 19

MATHEMATICS P1 · REVISION GUIDE 5 · CALCULUS Example 5.6 CHALLENGE SKETCH −1 5 −5 (3; −32) (1; −16) y = k f x y SPOT IT Only one real root: y = k above the maximum or below the minimum, k > 0 or k < −32. f (x) = x3 + px2 + qx −5 touches the x-axis at x = −1. (a) Show p = −3, q = −9. (b) Find the turning points and the inflection, and sketch f. (c) For which k has f (x) = k two distinct positive roots? (d) Where is g(x) = f (x + 2) decreasing? (a) f ′(−1) = 0: 3 −2p + q = 0 touching f (−1) = 0: p −q = 6; so p = −3, q = −9 solve together (b) f ′(x) = 3x2 −6x −9 = 3(x −3)(x + 1) = 0 (−1; 0) and (3; −32) substitute into f f ′′(x) = 6x −6 = 0: (1; −16) halfway f (x) = (x + 1)2(x −5): intercepts −1, 5, (0; −5) for the sketch (c) three roots for −32 < k < 0 slide y = k middle root positive below −5: −32 < k < −5 k = −5: root x = 0 (d) f decreases on −1 < x < 3: −3 < x < 1 g is f moved 2 left Example 5.7 CHALLENGE GRAPH OF f ′ −1 3 9 f ′ x y WATCH OUT The turning point of f ′ is the inflection of f, not a turning point of f. The graph of f ′ for f (x) = ax3 + bx2 + cx + d cuts the x-axis at −1 and 3 and the y-axis at 9; f (0) = 5. (a) Find f. (b) Find the turning points. (c) Find the inflection, the tangent there and its inclination. (d) For which k has f (x) = k three real roots? (e) Where is f concave up? (a) f ′(x) = m(x + 1)(x −3), −3m = 9, m = −3 f ′(0) = 9 −3x2 + 6x + 9 = 3ax2 + 2bx + c compare f (x) = −x3 + 3x2 + 9x + 5 d = f (0) (b) minimum (−1; 0), maximum (3; 32) f ′: −to + at −1 (c) x = 1, f (1) = 16, m = f ′(1) = 12 halfway y = 12x + 4 ; tan θ = 12, θ = 85,24◦ (d) 0 < k < 32 between turning points (e) f ′′(x) = −6x + 6 > 0: x < 1 concave up: f ′′ > 0 5.4 Increasing, decreasing and inequalities with f and f ′ Example 5.8 HARDER AND CHALLENGE SPOT IT Increasing: f ′ > 0. Concave down: f ′′ < 0. A product: zeros of each factor, then signs. (a) f (x) = x3 −6x2 + 9x: where is f increasing, concave down, and where is x · f ′(x) < 0? (b) f (x) = 2(x + 3)(x −1): where is f (x) · f ′(x) > 0? (a) f ′(x) = 3(x −1)(x −3) > 0: x < 1 or x > 3 outside the roots f ′′(x) = 6x −12 < 0: x < 2 concave down x · f ′(x) < 0: x < 0 or 1 < x < 3 zeros 0, 1, 3 (b) f < 0 on −3 < x < 1; f ′ < 0 for x < −1 f ′(x) = 4x + 4 −3 < x < −1 or x > 1 same signs HLAYISO.COM 20

MATHEMATICS P1 · REVISION GUIDE 5 · CALCULUS 5.5 Optimisation Write the quantity in ONE variable (use the constraint to remove the other), differentiate, set equal to zero. Example 5.9 CHALLENGE FIRST LINE πr2h = 250π, so h = 250 r2 . SPOT IT Box from a 30 cm sheet, corners x: V = x(30 −2x)2, V′ = (30 −2x)(30 −6x) = 0, x = 5, V = 2 000 cm3. An open cylinder holds 250π cm3. The base costs R0,06 per cm2, the side R0,03 per cm2. (a) Show C(r) = 0,06πr2 + 15π r . (b) Find the minimum cost. (a) C = 0,06(πr2) + 0,03(2πrh) one base; side 2πrh = 0,06πr2 + 0,06πr · 250 r2 = 0,06πr2 + 15π r h = 250 r2 (b) C′(r) = 0,12πr −15π r2 = 0, r3 = 125 r = 5: C(5) = 4,5π = R14,14 Example 5.10 CHALLENGE CROSS-SECTION r h 4 10 A cylinder stands inside a cone (base radius 4 cm, height 10 cm), its top edge touching the cone. (a) Show V = 10πr2 −5 2πr3. (b) Find the maximum vol- ume. (a) 10 −h r = 10 4 , h = 10 −5 2r similar triangles V = πr2 ( 10 −5 2r ) = 10πr2 −5 2πr3 (b) V′(r) = 20πr −15 2 πr2 = 0, r = 8 3 r̸ = 0 V = 640π 27 ≈74,47 cm3 h = 10 3 5.6 Rates of change and working backwards Example 5.11 STANDARD AND HARDER SPOT IT Velocity: h′(t). Initially: t = 0. Derivative given: which function has it? Then use the extra fact. (a) h(t) = 30t −5t2: velocity at 2 s, maximum height, velocity on landing. (b) Line g: g(1) = 5, g′(3) = 2. (c) f (x) = 3x2 + bx + c, f ′(1) = 9, f (1) = 0. (d) p′(x) = 4x −8 and the minimum of p is −3. (a) h′(t) = 30 −10t, h′(2) = 10 m/s velocity t = 3: h(3) = 45 m ; t = 6: h′(6) = −30 m/s h′ = 0; h = 0 (b) g(x) = 2x + c, 5 = 2 + c: g(x) = 2x + 3 gradient 2 (c) 6 + b = 9, 3 + 3 + c = 0: b = 3, c = −6 f ′(x) = 6x + b (d) p(x) = 2x2 −8x + c; p(2) = −3 minimum: p′(x) = 0 p(x) = 2x2 −8x + 5 HLAYISO.COM 21

6 LAST QUESTION · ABOUT 15 MARKS Probability Probability and counting IN THIS SECTION 6.1 The rules and tables 6.2 Venn and tree diagrams 6.3 Counting 6.4 Restrictions and other questions Write every rule you use: the rule itself often earns the first mark. In counting, draw boxes for the positions before you multiply. WORDS IN THE QUESTION RULE “or” P(A or B) = P(A) + P(B) −P(A and B) mutually exclusive / independent P(A and B) = 0 / P(A) × P(B) “not”, “at least one” P(not A) = 1 −P(A) n different objects / p and q alike n! / n! p! q! 6.1 The rules, unknown probabilities and two-way tables Example 6.1 STANDARD AND CHALLENGE TABLE FOR (C) Pass Fail Total Male 48 32 80 Female 72 48 120 Total 120 80 200 WATCH OUT A probability above 1 is rejected: say so. (a) P(A) = 0,3, P(B) = 0,5, P(A or B) = 0,65: mutually exclusive? indepen- dent? (b) A, B independent, P(A) = x, P(B) = x + 0,1, P(A or B) = 0,58: find x and P(exactly one). (c) Is passing independent of gender (table in the margin)? (a) 0,65 = 0,3 + 0,5 −P(A and B) addition rule P(A and B) = 0,15̸ = 0: not mutually exclusive 0,3 × 0,5 = 0,15: independent P(A) × P(B) (b) 0,58 = x + (x + 0,1) −x(x + 0,1) independent x2 −1,9x + 0,48 = 0, x = 1,9 ± 1,3 2 : x = 0,3 reject x = 1,6 > 1 0,58 −0,3 × 0,4 = 0,46 or minus P(A and B) (c) 80 200 × 120 200 = 0,24 = 48 200: independent totals vs the cell HLAYISO.COM 22

MATHEMATICS P1 · REVISION GUIDE 6 · PROBABILITY 6.2 Venn diagrams and tree diagrams Example 6.2 CHALLENGE FIRST LINE Venn: fill from the middle outwards; a letter for the unknown region; an equation from the total. VENN DIAGRAM M S D 31 25 4 5 10 6 9 10 100 learners: 50 in Maths (M), 40 in Science (S), 30 in Debating (D); 5 in all three; 6 in S and D only; 10 in none. M and D are independent. (a) Show that 15 are in M and D. (b) Find M and S only. (c) Find P(exactly one club). (a) 50 100 × 30 100 = 0,15: n = 15 independent (b) M and D only: 10; D only: 30 −10 −5 −6 = 9 middle outwards (35 −x) + (29 −x) + 9 + x + 10 + 6 + 5 + 10 = 100 total; x = M and S only x = 4 (c) 31 + 25 + 9 = 65: P = 0,65 Example 6.3 STANDARD AND HARDER TREE DIAGRAM R B R 20 56 B R B 6 56 5 8 3 8 4 7 3 7 5 7 2 7 SPOT IT Tree: multiply along the branches, add between them. Without replacement the second-stage numbers change. (a) 5 red and 3 blue balls; two are drawn without replacement. Find the prob- ability of the same colour, and of at least one blue. (b) 4 blue and n red balls; the probability of two blue (without replacement) is 2 15. Find n. (a) 5 8 · 4 7 + 3 8 · 2 7 = 26 56 = 13 28 RR or BB 1 −P(RR) = 1 −20 56 = 9 14 ``at least one'' (b) 4 n + 4 · 3 n + 3 = 2 15, n2 + 7n −78 = 0 one fewer each time (n + 13)(n −6) = 0: n = 6 6.3 Counting: letters, codes and digits Example 6.4 HARDER FIRST LINE Draw a box per position. Restricted positions first; overlapping conditions: split into cases. SPOT IT LIMPOPO 7! 2! 2! = 1 260. 4-digit PINs: 104 = 10 000; without repetition 10 · 9 · 8 · 7 = 5 040. 4-digit numbers from 1; 2; 3; 4; 5; 6, no repetition. (a) How many? (b) How many are odd and greater than 4 000? (c) P(1 and 2 next to each other). (a) 6 × 5 × 4 × 3 = 360 (b) first digit 5: 1 × 4 × 3 × 2 = 24 last digit 1 or 3 first digit 4 or 6: 2 × 4 × 3 × 3 = 72; 96 last digit 1, 3 or 5 (c) 3 × 2 × (4 × 3) = 72, P = 72 360 = 1 5 ``12'' as one unit HLAYISO.COM 23

MATHEMATICS P1 · REVISION GUIDE 6 · PROBABILITY 6.4 Arrangements with restrictions, and other questions Example 6.5 CHALLENGE FIRST LINE Together: glue into one unit, then arrange inside. No two together: place the others, use the gaps. Apart: 1 −P(together). WATCH OUT n(S) is the count with no restriction (here 8!). 3 teachers (one is Mr Dube) and 5 learners (one is Sipho) sit in a row of 8. (a) How many ways with the teachers together? (b) The probability that no two teachers are adjacent? (c) The probability that Mr Dube and Sipho are not ad- jacent? (d) How many ways with a learner at each end? (a) 6! × 3! = 4 320 6 units; inside 3! (b) P = 5! × 6 × 5 × 4 8! = 14 400 40 320 = 5 14 learners, then 6 gaps (c) 1 −2 × 7! 8! = 1 −1 4 = 3 4 complement (d) 5 × 4 × 6! = 14 400 ends first Example 6.6 HARDER AND CHALLENGE SPOT IT 4 boys and 4 girls alternating: 2 × 4! × 4! = 1 152. (a) PRETORIA: all arrangements, those starting with P, those with the four vowels together. (b) Thabo (hits 0,4) and Lindiwe (0,5) shoot in turn, Thabo first; the first hit wins. Find the probability that Thabo wins. (c) A game costs R5; two dice are thrown and a total of 10 or more wins. 300 people play per hour and the owner keeps at least 60% of the income. What is the largest prize per winner? (a) 8! 2! = 20 160 ; 7! 2! = 2 520 R twice 5! 2! × 4! = 1 440 vowels as one unit (b) 0,4 + 0,3(0,4) + 0,32(0,4) + . . . = 0,4 1 −0,3 = 4 7 r = 0,6 × 0,5 (c) (4;6), (5;5), (6;4), (5;6), (6;5), (6;6): P = 6 36 = 1 6 list the outcomes prizes at most R600; 1 6 × 300 = 50 winners 40% of R1 500 R12 per winner 600 ÷ 50 HLAYISO.COM 24

QUICK REFERENCE If you see … write first … The night before: cover the right-hand column and say the first line aloud IF YOU SEE … WRITE FIRST … 1 ALGEBRA a quadratic that is not = 0 / “two decimal places” standard form ax2 + bx + c = 0 / the formula, with brackets x4 and x2; x and √x; a repeated block “Let k = . . .”; then go back to x x in a denominator / a surd equation restriction, multiply every term by the LCD / isolate, square, check 2x+1 + 2x−1 / 9x and 3x; 2x and 2−x 2x(2 + 1 2) / “Let k = 3x” (reject k ⩽0) x and y in exponents / log3(x −y) = 2 same base ⇒linear equations / x −y = 32 linear + quadratic simultaneous make x or y the subject of the linear one inequality with x2 / “when is √ A B real” one side 0, factorise, critical values, sketch / A ⩾0, B̸ = 0 nature of roots, values of k ∆= b2 −4ac = . . . “rational for all k” / “real for all p” ∆= (. . .)2 / ∆= (. . .)2+ positive large powers; “divisible by” / √a ± √ b take out the smallest power / xy = a −b and x2 + y2 maximum / minimum of an expression complete the square 2 SEQUENCES AND SERIES second differences constant 2a = . . ., 3a + b = . . ., a + b + c = . . . unknown terms p, q / “is not a term” Tn = an2 + bn + c from the known terms / n /∈N, or odd/even “first difference is 199” / “largest term” Tn+1 −Tn = . . . / n = −b 2a “how many terms” / a sum is given Tn = a + (n −1)d or arn−1 / Sn formula = the sum ∑notation / a log inside the ∑ first term, last term, number of terms / simplify Tr first n in an exponent n = log(...) log r (watch the sign when r < 1) “converges” / “sum to infinity” / S∞−Sn −1 < r < 1 / S∞= a 1−r / arn 1−r Sn as a formula / arithmetic terms that are geometric Tn = Sn −Sn−1 / Tb Ta = Tc Tb 3 FUNCTIONS 3x+1 x−1 / axes of symmetry of a hyperbola 3(x−1)+4 x−1 = 3 + 4 x−1 / y = ±(x + p) + q line does not cut / touches a graph equate, standard form, ∆< 0 / ∆= 0 parabola with given roots / its turning point y = a(x −x1)(x −x2) / x = −b 2a or f ′(x) = 0 “tangent with gradient m” / “maximum length MN” f ′(x) = m / top −bottom, then maximise g(x) = f (x + t) / “passes through (2; 12)” every x moves t left / f (2) = 12 inverse / domain of f −1 swap x and y; y = logb x ⇐⇒x = by / range of f f (x) ⩾g(x) / x · f (x) ⩽0, f (x) · g(x) > 0 intersections, read the graph / zeros of each factor, signs HLAYISO.COM 25

MATHEMATICS P1 · REVISION GUIDE QUICK REFERENCE IF YOU SEE … WRITE FIRST … average gradient f (b)−f (a) b−a 4 FINANCE any finance question / depreciation a time line; i and n per period / (1 −i), never (1 + i) “effective rate” / the rate changes 1 + ieff = ( 1 + i m )m / stage by stage saving towards a target, sinking fund F = x[(1+i)n−1] i ; target = cost −trade-in loan, instalment / first payment after k months P = x[1−(1+i)−n] i / grow the loan k −1 months first outstanding balance PV of the payments that are left lump sum, bigger instalment, missed payments new balance, then solve for n (logs) or the new x 5 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 / through a point not on the curve m = f ′(x1), y −y1 = m(x −x1) / P(p; f (p)), tangent in p “touches the x-axis at a” f (a) = 0 and f ′(a) = 0 turning points / inflection, concavity f ′(x) = 0 / f ′′(x) = 0; sign of f ′′ graph of f ′ given / f (x) = k has n roots roots of f ′ = turning points of f / slide y = k across the sketch optimisation / rate, velocity constraint ⇒one variable, derivative = 0 / the derivative; “initially” t = 0 6 PROBABILITY AND COUNTING “or” / “independent” P(A) + P(B) −P(A and B) / P(A and B) = P(A) × P(B) “at least one” / three overlapping groups 1 −P(none) / Venn diagram from the middle outwards one event after another tree diagram; without replacement changes the numbers repeated letters / digit conditions n! p! q! / restricted positions first; split into cases “together” / “no two next to each other” glue into one unit, × arrangements inside / place the others, use the gaps You have done the work. Read each question slowly, write the first line, and trust your method. Good luck. HLAYISO.COM 26