# MATHSCLINIC SMARTPREP GR12 ENG 2023 V2.1 > Grade 12 question paper from the Mathematics, 75 pages. Read online or get the free PDF download, memo linked when available. - Page: https://www.hlayiso.com/library/grade-12-mathematics-question-paper/0270e359-f94f-4f32-a9ff-0221924b6d14 - Updated: 2026-09-27 - Subject: Grade 12 - Grade: Multiple grades - Document type: Question paper - Exam period: Mathematics - Curriculum: CAPS - Pages: 75 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [MATHSCLINIC SMARTPREP GR12 ENG 2023 V2.1](https://www.hlayiso.com/library/grade-12-mathematics-question-paper/0270e359-f94f-4f32-a9ff-0221924b6d14) ## About this document Grade 12 · Mathematics. Question paper, 75 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 © Lesson videos for every section of these SmartPrep theory summaries available on www.scienceclinic.co.za ### Page 2 VIDEO LESSONS! We have free video lessons for every topic in this book! These videos can only be accessed via Science Clinic's website - follow these steps to access this amazing resource! DOWNLOAD A BOOK Visit our website to download a SmartPrep book. We'll send you an email with download links. VIDEOS WEBPAGE Go to our Videos webpage. Log in with the email address that you used to download a book. SELECT YOUR TOPIC From your videos page, select a topic. You'll be redirected to a playlist with video lessons. ABOUT OUR SMARTPREP BOOKS Our SmartPrep books provide crystal-clear summaries of the essential theory for Grade 8-12 Core Mathematics and Grade 10-12 Physical Science. Our books are available in English and Afrikaans. 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You may distribute this material, provided that you adhere to the conditions in our copyright notice and license: ¢ If you copy and distribute this book electronically or in paper form, you must keep this copyright notice intact. ¢ If you use questions or syllabus summaries from this book, you must acknowledge the author. ¢ This Smart Prep Book is meant for the benefit of the community and you may not use or distribute it for commercial purposes. ¢ You may not broadcast, publicly perform, adapt, transform, remix, add on to this book and distribute it on your own terms. ° By exercising any of the rights to this Smart Prep Book, you accept and agree to the terms and conditions of the license provided on this link: CONTENT ACKNOWLEDGEMENT Thank you to all the amazing people involved in the production, translation and moderation of this book: S Bouwer, E Britz, G Kyle, D Kotze, Q Meades, S Sapsford, S Stevens, G Swanepoel, GM van Onselen, L Vosloo QD ww sciencectinie.coza © exciencectinic EB ew acodook.comisciencectiniesa ### Page 3 TABLE OF CONTENTS CORE THEORY SUMMARIES Paper 1 Nature of Roots 2 Quadratic Equations 5 Exponents and Surds 6 Sequences and Series 10 Finance, Growth and Decay 15 Functions, Graphs and Inverses 20 Differential Calculus 29 Probability 34 Paper 2 Euclidean Geometry 39 Trigonometry 49 Analytical Geometry 57 Statistics 62 www ### Page 4 Mathematics/P 1 DBE/2021 SC/NSC INFORMATION SHEET ee bt vb? = Aae 2a A= P(+ni) A= P(l-ni) A=P(l-i)’ A= PQ+i)" T, =a+(n—ld 8, =F fpa+(n—Nd] T, =ar"™" s, -a"=1) sr#l S.= “s-ler0 h | TC ne? X+xX, Yt d=\G2—x) 402-9 ar ie = —y- _ 2 7W y=mx+e Y-y, =Mx-X) n= m=tand % —% (x-al +(y-by =r? mn AABC: 4_=—° c sind sinB sinC a =b? +c’ —2be.cos A area AABC = Sabin Cc sin(a+ ) =sin a.cos $+cosa.sinB cos(a+ f)=cosa.cos S-sinasinfB cos’ a-sin’ a cos2@ =41-2sin’ a 2cos’a-1 Copyright reserved sin(a- f)= sin a@.cos f-cosa.sinB cos (a —f) =cosa@.cos #+sina.sinZ sin 2a = 2sina.cosa Yo, -» = i=l 2 oj = n P(A or B)= P(A) + P(B) — P(A and B) X@-*)0-7 Sexy ### Page 5 Mathematics 9 DBE/2021 Examination Guidelines 4. ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY In order to have some kind of uniformity, the use of the following shortened versions of the theorem statements is encouraged. 4.1 ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY (ENGLISH) THEOREM STATEMENT ACCEPTABLE REASON(S) LINES The adjacent angles on a straight line are supplementary. ‘s on a str line If the adjacent angles are supplementary, the outer arms of these angles adj ‘s supp form a straight line. The adjacent angles in a revolution add up to 360q. ‘s round a pt OR ‘s in a rev Vertically opposite angles are equal. vert opp ‘s = If AB || CD, then the alternate angles are equal. alt ‘s; AB || CD If AB || CD, then the corresponding angles are equal. corresp ‘s; AB || CD If AB || CD, then the co-interior angles are supplementary. co-int ‘s; AB || CD If the alternate angles between two lines are equal, then the lines are alt ‘s = parallel. If the corresponding angles between two lines are equal, then the lines corresp ‘s = are parallel. If the co-interior angles between two lines are supplementary, then the coint ‘s supp lines are parallel. TRIANGLES The interior angles of a triangle are supplementary. ‘ sum in ' OR sum of ‘s in ∆ OR Int ‘s ' The exterior angle of a triangle is equal to the sum of the interior ext ‘ of ' opposite angles. The angles opposite the equal sides in an isosceles triangle are equal. ‘s opp equal sides The sides opposite the equal angles in an isosceles triangle are equal. sides opp equal ‘s In a right-angled triangle, the square of the hypotenuse is equal to the Pythagoras OR sum of the squares of the other two sides. Theorem of Pythagoras If the square of the longest side in a triangle is equal to the sum of the Converse Pythagoras squares of the other two sides then the triangle is right-angled. OR Converse Theorem of Pythagoras If three sides of one triangle are respectively equal to three sides of SSS another triangle, the triangles are congruent. If two sides and an included angle of one triangle are respectively SAS OR S‘S equal to two sides and an included angle of another triangle, the triangles are congruent. If two angles and one side of one triangle are respectively equal to two AAS OR ‘‘S angles and the corresponding side in another triangle, the triangles are congruent. If in two right-angled triangles, the hypotenuse and one side of one RHS OR 90°HS triangle are respectively equal to the hypotenuse and one side of the other, the triangles are congruent Copyright reserved Please turn over ### Page 6 Mathematics 10 DBE/2021 Examination Guidelines THEOREM STATEMENT ACCEPTABLE REASON(S) The line segment joining the midpoints of two sides of a triangle is Midpt Theorem parallel to the third side and equal to half the length of the third side The line drawn from the midpoint of one side of a triangle, parallel to line through midpt || to 2nd side another side, bisects the third side. A line drawn parallel to one side of a triangle divides the other two line || one side of ' sides proportionally. OR prop theorem; name || lines If a line divides two sides of a triangle in the same proportion, then the line divides two sides of ∆ in prop line is parallel to the third side. If two triangles are equiangular, then the corresponding sides are in ||| 's OR equiangular ∆s proportion (and consequently the triangles are similar). If the corresponding sides of two triangles are proportional, then the Sides of ∆ in prop triangles are equiangular (and consequently the triangles are similar). If triangles (or parallelograms) are on the same base (or on bases of same base; same height OR equal length) and between the same parallel lines, then the triangles (or equal bases; equal height parallelograms) have equal areas. CIRCLES The tangent to a circle is perpendicular to the radius/diameter of the tan A radius circle at the point of contact. tan A diameter If a line is drawn perpendicular to a radius/diameter at the point where line A radius OR the radius/diameter meets the circle, then the line is a tangent to the converse tan A radius OR circle. converse tan A diameter The line drawn from the centre of a circle to the midpoint of a chord is line from centre to midpt of chord perpendicular to the chord. The line drawn from the centre of a circle perpendicular to a chord line from centre A to chord bisects the chord. The perpendicular bisector of a chord passes through the centre of the perp bisector of chord circle; The angle subtended by an arc at the centre of a circle is double the ‘ at centre = 2 ב at circumference size of the angle subtended by the same arc at the circle (on the same side of the chord as the centre) The angle subtended by the diameter at the circumference of the circle ‘s in semi-circle OR is 90q. diameter subtends right angle OR 1 ‘ in ~ 2 If the angle subtended by a chord at the circumference of the circle is chord subtends 90q OR 90q, then the chord is a diameter. converse ‘s in semi-circle Angles subtended by a chord of the circle, on the same side of the ‘s in the same seg chord, are equal If a line segment joining two points subtends equal angles at two points line subtends equal ‘s OR on the same side of the line segment, then the four points are concyclic. converse ‘s in the same seg Equal chords subtend equal angles at the circumference of the circle. equal chords; equal ‘s Equal chords subtend equal angles at the centre of the circle. equal chords; equal ‘s Equal chords in equal circles subtend equal angles at the circumference equal circles; equal chords; equal ‘s of the circles. Copyright reserved Please turn over ### Page 7 Mathematics 11 DBE/2021 Examination Guidelines THEOREM STATEMENT ACCEPTABLE REASON(S) Equal chords in equal circles subtend equal angles at the centre of the equal circles; equal chords; equal ‘s circles. The opposite angles of a cyclic quadrilateral are supplementary opp ‘s of cyclic quad If the opposite angles of a quadrilateral are supplementary then the opp ‘s quad supp OR quadrilateral is cyclic. converse opp ‘s of cyclic quad The exterior angle of a cyclic quadrilateral is equal to the interior ext ‘ of cyclic quad opposite angle. If the exterior angle of a quadrilateral is equal to the interior opposite ext ‘ = int opp ‘ OR angle of the quadrilateral, then the quadrilateral is cyclic. converse ext ‘ of cyclic quad Two tangents drawn to a circle from the same point outside the circle Tans from common pt OR are equal in length Tans from same pt The angle between the tangent to a circle and the chord drawn from the tan chord theorem point of contact is equal to the angle in the alternate segment. If a line is drawn through the end-point of a chord, making with the converse tan chord theorem OR chord an angle equal to an angle in the alternate segment, then the line ‘ between line and chord is a tangent to the circle. QUADRILATERALS The interior angles of a quadrilateral add up to 360q. sum of ‘s in quad The opposite sides of a parallelogram are parallel. opp sides of ||m If the opposite sides of a quadrilateral are parallel, then the opp sides of quad are || quadrilateral is a parallelogram. The opposite sides of a parallelogram are equal in length. opp sides of ||m If the opposite sides of a quadrilateral are equal , then the quadrilateral opp sides of quad are = is a parallelogram. OR converse opp sides of a parm The opposite angles of a parallelogram are equal. opp ‘s of ||m If the opposite angles of a quadrilateral are equal then the quadrilateral opp ‘s of quad are = OR is a parallelogram. converse opp angles of a parm The diagonals of a parallelogram bisect each other. diag of ||m If the diagonals of a quadrilateral bisect each other, then the diags of quad bisect each other quadrilateral is a parallelogram. OR converse diags of a parm If one pair of opposite sides of a quadrilateral are equal and parallel, pair of opp sides = and || then the quadrilateral is a parallelogram. The diagonals of a parallelogram bisect its area. diag bisect area of ||m The diagonals of a rhombus bisect at right angles. diags of rhombus The diagonals of a rhombus bisect the interior angles. diags of rhombus All four sides of a rhombus are equal in length. sides of rhombus All four sides of a square are equal in length. sides of square The diagonals of a rectangle are equal in length. diags of rect The diagonals of a kite intersect at right-angles. diags of kite A diagonal of a kite bisects the other diagonal. diag of kite A diagonal of a kite bisects the opposite angles diag of kite Copyright reserved Please turn over ### Page 8 Paper 1 (fw). scienceclinic.co.za © 2022@)- ### Page 9 Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © NUMBERS NATURE: Refers to the type of numbers that the roots are. Non-real (R’) (Sometimes referred to as imaginary) ROOTS: The x-intercepts/solutions/zeros of a quadratic equation. Real (R) Two real roots One real root No (0) real roots Rational (Q) Irrational (Q’) y y y Integers and Fractions EXAMPLE Integers (Z) Positive and negative whole numbers π 2 Whole numbers (N0) x 0+N All decimals that cannot x x be written as a fraction. Natural numbers (N) 1, 2, 3... OR OR OR y y y x x x QUADRATIC FORMULA THE DISCRIMINANT −b ± b 2 − 4ac x= −b ± b 2 − 4ac 2a x= 2a a = the coefficient of x2 DETERMINING THE NATURE OF THE ROOTS b = the coefficient of x Indicated by Δ . c = the constant term The DISCRIMINANT is used to determine the nature of the roots. 2 ∴ Δ = b − 4ac Used to factorise quadratic equations. Δ EXAMPLE EXAMPLE Δ<0 Δ≥0 3x 2 + 2x − 4 = 0 NO Real Real 3x 2 +2x − 4 = 0 a = 3 roots roots b = 2 a = 3 c = −4 b = 2 c = −4 Δ>0 Δ=0 2 unequal 2 equal, −b ± b 2 − 4ac Δ = b 2 − 4a c roots real roots x = 2a Δ = (2)2 − 4(3)(−4) −(2) ± (2) 2 − 4(3)(− 4) Then solve for Δ x = Δ = perfect square Δ ≠ perfect square 2(3) irrational roots rational roots Then solve for x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 2 ### Page 10 Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © DISCRIMINANT b2 - 4 a c EXAMPLES NATURE OF ROOTS NUMBER OF REAL ROOTS (∆ = b2 - 4ac) a>0 a<0 y y x2 + x + 1 = 0 Δ = b2 – 4ac x = (1)2 – 4(1)(1) 0 =1–4 Non real =–3 a b c Δ<0 x y y x2 – 6x + 9 = 0 Δ = b2 – 4ac x = (–6)2 – 4(1)(9) Real (Δ = +) 1 (2 of the same) = 36 – 36 Rational (Δ = perfect =0 square) a b c Δ=0 Equal (Δ = 0) x x2 – 5x – 6 = 0 Δ = b2 – 4ac = (–5)2 – 4(1)(–6) Real (Δ = +) 2 y y = 25 + 24 Rational (Δ = perfect = 49 square) a b c Δ > 0 (perfect square) Unequal (Δ ≠ 0) x x 2x2 + 3x – 7 = 0 Δ = b2 – 4ac = (3)2 – 4(2)(–7) Real (Δ = +) 2 = 9 + 56 Irrational (Δ ≠ = 65 perfect square) a b c Δ > 0 (not perfect square) Unequal (Δ ≠ 0) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 3 ### Page 11 Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © DETERMINING THE NATURE FOR WHICH VALUES OF k PROVE THE NATURE OF THE ROOTS OF ROOTS WITHOUT WILL THE EQUATION HAVE The nature of the roots will be supplied and the discriminant can be used to prove the nature, with either one, or no, SOLVING THE EQUATION EQUAL ROOTS? unknown value. The roots of an equation can be deter- The discriminant (Δ) can be used to calculate Steps to prove the nature of roots (NO unknown): Steps to prove the nature of roots (ONE unknown): mined by calculating the value of the the unknown value of k. (e.g. Ask yourself, for discriminant (Δ). which values of k will the discriminant be 0?) 1. Put the equation in its standard form 1. Put the equation in its standard form 2. Substitute the correct values in and calculate the dis- 2. Substitute the correct values in and calculate the dis- Steps to determine the roots using Steps to determine the values of k using criminant criminant the discriminant: the discriminant: 3. Determine the roots and confirm whether they are as 3. Determine the roots and confirm whether they are as 1. Put the equation in its standard form 1. Put the equation in its standard form supplied supplied 2. Substitute the correct values in and 2. Substitute the correct values in and calculate EXAMPLE EXAMPLE calculate the discriminant the discriminant Prove the equation has two, unequal, irrational roots: For the equation x(6x – 7m) = 5m2, prove that the roots 3. Determine the nature of the roots of 3. Equate the discriminant to 0 and solve for k x2 = 2x + 9 are real, rational and unequal if m > 0 the equation (quadratic equation) EXAMPLE 1. Standard form 1. Standard form EXAMPLE x2 – 2x – 9 = 0 6x2 – 7mx – 5m2 = 0 Determine the nature of the roots of For which values of k the equation will have x2 = 2x +1 without solving the equa- a b c a b c equal roots? tion REMEMBER: Δ = 0 for equal roots 2. Calculate the discriminant 2. Calculate the discriminant 1. Standard form Δ = b2 – 4ac Δ = b2 – 4ac x2 = 2x +1 Δ = (–2)2 – 4(1)(–9) Δ = (–7m)2 – 4(6)(– 5m2) x2 – 2x – 1 = 0 1. Standard form Δ = 4 + 36 Δ = 49m2 + 120m2 x2 + 2kx = –4x – 9k Δ = 40 Δ = 169m2 a b c x2 + 2kx + 4x + 9k = 0 3. Determine the roots 3. Determine the roots 2. Calculate the discriminant a b c The Roots are: The Roots are: Δ = b2 – 4ac Real (Δ > 0) Real (Δ > 0) Δ = (–2)2 – 4(1)(–1) 2. Calculate the discriminant Unequal (Δ ≠ 0) Unequal (Δ ≠ 0) Δ=4+4 Δ = b2 – 4ac Irrational (Δ ≠ perfect square) Rational (Δ = perfect square) Δ=8 Δ = (2k + 4)2 – 4(1)(9k) Δ = 4k2 +16k +16 – 36k 3. Determine the nature of the Δ = 4k2 – 20k + 16 roots The Roots are: 3. Equate to zero (0) and solve for k Real (Δ > 0) 0 = 4k2 – 20k + 16 (÷ 4) Unequal (Δ ≠ 0) 0 = k2 – 5k + 4 Irrational (Δ ≠ perfect square) 0 = (k – 1)(k – 4) Therefore k = 1 or k = 4 k needs to either be 1 or 4 to ensure that the discriminant of the equation is 0 (the discriminant must be 0 in order for equal roots) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 4 ### Page 12 Grade 12 Maths Essentials Quadratic Equations SCIENCE CLINIC 2022 © Quadratic Equations are equations of the second degree (i.e. the highest exponent of the variable is 2). The degree of the equation determines the maximum number of real roots/solutions/x-intercepts/zeros. The standard form of a quadratic equation is: a x 2 + bx + c = 0 wh ere a ≠ 0 SOLVING QUADRATIC EQUATIONS DIFFERENCE OF TWO COMPLETE THE FRACTIONS AND FACTORING QUADRATIC FORMULA ONE ROOT TWO ROOTS SQUARES SQUARE RESTRICTIONS 1. Find the LCD and list restrictions 1. Put in standard Substitute into the quadratic 1. Substitute the known 2. Solve for x 1. Write in standard form 1. Substitute the roots into form formula: 2 root −b ± b − 4a c 2. Move C across the equation 3. Check your answers against your 2. Apply the zero x = 2. Solve for the variable 2a 3. Divide both sides by A restrictions factor law 3. Substitute the value 2. Use “FOIL” for the 4. Add (½ x b)2 to both sides REMEMBER: 3. List possible where a = coefficient of x2, b = of the variable and quadratic equation 5. Factorise and solve 0 x solutions coefficient of x, c = constant term solve for the root = 0 BUT = u nd e f i ne d x 0 x Eg. x2 = –2x + 63 Eg. –3x2 = –12 + 7x Either method may be used Eg. x2 + 2x = 1 Eg. x2 + mx – 15 = 0, Eg. –9 and 7 are the Eg. x−2 = x −1 3 − 2 −2 x x2 + 2x – 63 = 0 –3x2 – 7x + 12 = 0 (½ . 2)2 where 5 is a root. roots of a quadratic x a = –3; b = –7; c = 12 Eg. x2 = 25 (1) (5)2 + (5)m – 15 = 0 equation x−2 = x −1 3 + x −2 2 Find factors of 63 so x2 – 25 = 0 x2 = ± 25 x 2 + 2x + 1 = 1 + 1 25 + 5m – 15 = 0 x = –9 or x=7 LCD: (x − 2)(x − 3) −(−7) ± (−7) 2 − 4(−3)(12) (x – 5)(x + 5) = 0 5m = –10 x+9=0 x–7=0 that F1 x F2 = –63 x = (x + 1)2 = ± 2 Restrictions: x − 2 ≠ 0; x − 3 ≠ 0 2(−3) x = 5 or x = –5 x = ±5 m = –2 and F1 + F2 = 2 x(x − 2)(x − 3) 1(x − 2)(x − 3) 2(x − 2)(x − 3) 7± 49 + 144 x + 1= ± 2 (x + 9)(x – 7) = 0 = + (x − 2) (x − 3) (x − 2) x = Therefore x = ±5 x2 – 2x – 15 = 0 x2 – 7x + 9x – 63 = 0 (x + 9)(x – 7) = 0 −6 x + 1= − 2 or x + 1= 2 (x – 5)(x + 3) = 0 7± 193 x2 + 2x – 63 = 0 x(x – 3) = 1(x – 2) + 2(x – 3) x + 9 = 0 or x – 7 = 0 x = −1− 2 x = −1+ 2 x=5 or x = –3 x = x2 – 3x = x – 2 + 2x – 6 x = –9 x=7 −6 (given) 7+ 193 7− 193 x2 – 3x – x – 2x + 2 + 6 = 0 x = or x = x2 – 6x + 8 = 0 −6 −6 (x – 4)(x – 2) = 0 x–4=0 or x–2=0 Answer in surd form or can be x=4 x=2 calculated/rounded off to 2 decimals Check restrictions: x ≠ 2, x ≠ 3 x = –3,48 OR x = 1,15 Thus, x = 4 is the only solution. Remember: General Steps for solving Quadratic Equations Cannot Sum * LHS = Left Hand Side * RHS = Right Hand Side factorise * Fractions and their restrictions 2 perfect squares Finding factors: Binomial Difference * Factor 1 + Factor 2 = b * Factor 1 x Factor 2 = c 2. Factor out Highest Zero factor law: 1. Standard Form Common Factor If A x B = 0 then A, B or both = 0 Trinomial Factoring EXAMPLE Factorise Quadratic Formula x.y=0 x = 0 or y = 0 or (x + 1)(x – 3) = 0 Complete the x + 1 = 0 or x – 3 = 0 square Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 5 Text truncated here. Open the document page to read the rest. ## Collection - [Past papers & memos](https://www.hlayiso.com/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.