Mathematics Survival Kit 1

Curriculum: FET MATHEMATICS GR12 REVISION BOOKLET 2022 SURVIVAL PACK PAGE FUNCTIONS 2 - 15 CALCULUS 16 – 19 FINANCIAL MATHEMATICS 20 - 24 TRIGONOMETRY 25 - 27 1

Curriculum: FET 1. MOTHER FUNCTION AND INFLUENCE OF PARAMETERS Influence of Mother function Influence of 𝒑 and 𝒒 𝒂>𝟎 𝒇(𝒙) = 𝒙² 𝒈(𝒙) = (𝒙 − 𝟐)² 𝒉(𝒙) = (𝒙 − 𝟐)𝟐 − 𝟒 Move 𝒂<𝟎 PARABOLA Moved to 2 units to the Moved 4 units down right (𝑝 < 0) (𝑞 < 0) 𝒂>𝟎 𝟏 𝟏 𝟏 𝒇(𝒙) = 𝒈(𝒙) = 𝒉(𝒙) = −𝟐 1st and 3rd 𝒙 𝒙−𝟐 𝒙 quadrant HYPERBOLA 𝒂<𝟎 2nd and 4th quadrant Moved to 2 units to the Moved 2 units down right (𝑝 < 0) (𝑞 < 0) 𝒃>𝟏 𝒇(𝒙) = 𝟐𝒙 𝒉(𝒙) = 𝟐𝒙+𝟐 𝒑(𝒙) = 𝟐𝒙 − 𝟒 increasing EXPONENTIAL GRAPH 𝟎<𝒃<𝟏 𝒂>𝟎 decreasing Above 𝒙 − 𝒂𝒙𝒊𝒔 𝒂<𝟎 Below 𝒙 − 𝒂𝒙𝒊𝒔 Moved to 2 units to the Moved 4 units down left (𝑝 > 0) (𝑞 < 0) 2

Curriculum: FET 2. SKETCH OF GRAPHS Parabola Hyperbola Exponential Graph Logarithmic graph Standard form 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 𝒂 𝒚 = 𝒂. 𝒃𝒙+𝒑 + 𝒒 𝒚 = 𝒍𝒐𝒈𝒂 𝒙 𝒚= +𝒒 𝐲 = 𝐚(𝐱 + 𝐩)𝟐 + 𝐪 𝒙+𝒑 𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) To sketch the Example 1 Example 2 Example 3 Example 4 graphs, you need: Sketch 𝑓(𝑥) = −2𝑥 2 − 8𝑥 + 10 −2 Sketch 𝑔(𝑥) = 𝑥+1 − 4 Sketch ℎ(𝑥) = 2𝑥 − 4 Sketch 𝑘(𝑥) = 𝑙𝑜𝑔2 𝑥 𝑦 −intercept (𝑥 = OR 0) 𝑓(𝑥) = −2(𝑥 + 2)2 + 18 𝑥 −intercept (𝑦 = 0) Turning point 𝑏 ∆ (− ; − ) 2𝑎 4𝑎 OR (−𝑝; 𝑞) Asymptote(s) • 𝑥 = −𝑝 • 𝑦=𝑞 Domain 𝑥∈𝑅 𝑥 ∈ 𝑅 ; 𝑥 ≠ −1 𝑥∈𝑅 𝑥 ∈ 𝑅 ;𝑥 > 0 Range 𝑦 ∈ 𝑅 ; 𝑦 ≤ 18 𝑦 ∈ 𝑅 ; 𝑦 ≠ −4 𝑦 ∈ 𝑅 ; 𝑦 > −4 𝑦∈𝑅 Symmetry axis 𝒃 𝒚 =𝒙+𝒄 𝒙=− 𝟐𝒂 Substitute (−𝑝; 𝑞) → (−1; −4) 𝑥 = −2 ∴ −4 = −1 + 𝑐 𝑐 = −3 𝑦=𝑥−3 OR 𝒚 = 𝒙 − +𝒄 Substitute (−𝑝; 𝑞) → (−1; −4) ∴ −4 = −(−1) + 𝑐 𝑐 = −5 𝑦 = −𝑥 − 5 3

Curriculum: FET 3. TO DETERMINE THE EQUATION OF A GRAPH Parabola Hyperbola Exponential Graph Logarithmic graph 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 𝒂 𝒚 = 𝒂. 𝒃𝒙+𝒑 + 𝒒 𝒚 = 𝒍𝒐𝒈𝒂 𝒙 𝒚= +𝒒 𝐲 = 𝐚(𝐱 + 𝐩)𝟐 + 𝐪 𝒙+𝒑 𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝒙 −intercepts given: Turning point given: For 𝒑 𝒂𝒏𝒅 𝒒: For 𝒒: For 𝒂: 𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝐲 = 𝐚(𝐱 + 𝐩)𝟐 + 𝐪 Start by substituting the asymptotes Start by substituting the Substitute a point given on the For 𝑎: into the equation. asymptote into the equation. graph into the equation. For 𝑎: Substitute a point given on the Substitute another point graph into the equation. For 𝒂: For any other: given on the graph into the Substitute a point given on the graph Substitute a point given on equation. into the equation. the graph into the equation. For this example, given 𝑦 = 𝑎𝑥 + 𝑞 𝑦 = 𝑎(𝑥 + 5)(𝑥 − 1) Subst. (0; 10) 𝑎 𝑦 = 𝑙𝑜𝑔𝑎 𝑥 10 = 𝑎(0 + 5)(0 − 1) 𝑦 = 𝑎(𝑥 − 1)2 − 6 𝑦= −4 Subst. (8; −3) 10 = −5𝑎 Subst. (2; −4): 𝑥+1 𝑎 = −2 −4 = 𝑎(2 − 1)2 − 6 −3 = 𝑙𝑜𝑔𝑎 8 subst. (0; −6) 𝑦 = 𝑎𝑥 − 4 𝑦 = −2(𝑥 + 5)(𝑥 − 1) 2 = 𝑎(1)² 𝑎 𝑦 = −2𝑥 2 − 8𝑥 + 10 𝑎=2 −6 = −4 Write in exponential form: 0+1 𝑠𝑢𝑏𝑠𝑡. (2; 0): 𝑦 = 2(𝑥 − 1)2 − 6 𝑎−3 = 8 0 = 𝑎2 − 4 −2 = 𝑎 𝑎2 = 4 1 OR 𝑎= 𝑎=2 2 𝑦 = 2𝑥 2 − 4𝑥 − 4 −2 𝑦= −4 𝑦 = 𝑙𝑜𝑔1 𝑥 𝑥+1 𝑦 = 2𝑥 − 4 2 4

Curriculum: FET 4. TRANSFORMATION 4.1 TRANSLATION • 𝒑 > 𝟎 horizontal shift left • 𝒑 < 𝟎 horizontal shift right • 𝒒 > 𝟎 vertical shift up • 𝒒 < 𝟎 vertical shift down EXAMPLE: Translate 3 units up. Translate 2 units to the right. Notation: 𝒇(𝒙) + 𝟑 Notation: 𝒇(𝒙 − 𝟐) PARABOLA 𝒇(𝒙) = 𝒙² 𝒉(𝒙) = 𝒙𝟐 + 𝟑 𝒈(𝒙) = (𝒙 − 𝟐)² 4.2 REFLECTION: • REFLECTION in the (𝒙 − 𝒂𝒙𝒊𝒔) OR in the line 𝒚 = 𝟎 Notation: 𝒇(𝒙) → −𝒇(𝒙) • REFLECTION in the (𝒚 − 𝒂𝒙𝒊𝒔) OR in the line 𝒙 = 𝟎 Notation: 𝒇(𝒙) → 𝒇(−𝒙) • REFLECTION in the line 𝒚 = 𝒙 Notation: 𝒇(𝒙) → 𝒇−𝟏 (𝒙) EXAMPLE: ℎ(𝑥) = 𝑥 2 + 3 𝑔(𝑥) = (𝑥 − 2)² 𝒇: 𝒚 = 𝒙² → 𝒌(𝒙) = −𝒙𝟐 − 𝟑 → 𝒑(𝒙) = (−𝒙 − 𝟐)² 𝒇−𝟏 : 𝒙 = 𝒚² PARABOLA ∴ 𝒚 = ±√𝒙 EXPONENTIAL 𝒇(𝒙) = 𝟐𝒙 + 𝟑 𝒇(𝒙) = 𝟐𝒙 − 𝟐 𝒇(𝒙) = 𝟐𝒙 FUNCTION → −𝒇(𝒙) = −(𝟐𝒙 ) − 𝟑 → 𝒇(−𝒙) = 𝟐−𝒙 − 𝟐 →𝒇 −𝟏 (𝒙) = 𝒍𝒐𝒈𝟐 𝒙 5

Curriculum: FET 5. GRAPHS AND INEQUALITIES: EXAMPLES PARABOLA HYPERBOLA EXPONENTIAL FUNCTION LOGARITMIC FUNCTION 6

Curriculum: FET 6. LOGARITHMIC 𝒇(𝒙) = 𝒂𝒙 and 𝒇−𝟏 (𝒙) = 𝒍𝒐𝒈𝒂 𝒙 GRAPH 𝒂>𝟏 𝟎<𝒂<𝟏 The inverse of the exponential function increasing decreasing 𝒇(𝒙) = 𝒂𝒙 (a > 0; a ≠1) is the logarithmic function 𝒇−𝟏 (𝒙) = 𝒍𝒐𝒈𝒂 𝒙. Note: Note: • The exponential graph f is an increasing graph around the • The exponential graph f is a decreasing graph around the y axis (𝒂 > 𝟏) y axis (𝟎 < 𝒂 < 𝟏) • The logarithmic graph 𝑓 is an increasing graph around −1 • The logarithmic graph f is a decreasing graph around the the x axis (𝒂 > 𝟏) x axis (𝟎 < 𝒂 < 𝟏) Domain and Range 𝑓 Domain: 𝑥 ∈ 𝑅 Range: 𝑦 > 0 −1 Range: 𝑦 ∈ 𝑅 Domain: 𝑥 > 0 𝑓 Inverse functions: Graph 𝑓 −1 is obtained by reflecting graph 𝑓 about the line 𝑦 = 𝑥. (the inverse of 𝑓) • If 𝒚 = 𝒂𝒙 then the inverse function is given by 𝒙 = 𝒂𝒚 which can also be written as 𝒚 = 𝒍𝒐𝒈𝒂 𝒙 • If 𝒚 = 𝒍𝒐𝒈𝒂 𝒙 then the inverse function is given by 𝒙 = 𝒍𝒐𝒈𝒂 𝒚 which can also be written as 𝒚 = 𝒂𝒙 Asymptotes: • The exponential graph f has an asymptote 𝒚 = 𝟎 ( 𝒙 − 𝒂𝒙𝒊𝒔) • The logarithmic graph 𝑓 −1 has an asymptote 𝒙 = 𝟎 ( 𝒚 − 𝒂𝒙𝒊𝒔) 7

7. REVISION EXERCISES SECTION A – GR 12 INVERSE FUNCTIONS (Routine questions) • Note that some of the questions in this section is taken from MIND THE GAP. QUESTION 1 1.1 • A function is a relationship between x and y, where for every x-value there is only one y-value. • One way to decide whether or not a graph represents a function is to use the vertical line test. Say if the following graphs are functions or not. GRAPH A GRAPH B GRAPH C GRAPH D 1.2 Complete the following sentences: (i) A function must be reflected along the line __________to form the inverse of the function. (ii) The notation for the inverse of a function 𝑓 is written as _____ __

GR 12 – REVISION QUESTION 2 Given: 𝑓(𝑥) = 2𝑥 + 6. 2.1 Determine 𝑓 −1 (𝑥). 2.2 Sketch the graphs of 𝑓, 𝑓 −1 and 𝑦 = 𝑥 on the same set of axes. Indicate clearly the intercepts with the axes of the graph of each of 𝑓 and 𝑓 −1 . 2.3 Determine the domain and range for 𝑓 𝑎𝑛𝑑 𝑓 −1 QUESTION 3 3.1 Sketch 𝑓(𝑥) = 2𝑥 2 Indicate clearly the intercepts with the axes, as well as another point on the graph of 𝑓. 3.2 Determine the inverse of 𝑓(𝑥). 3.3 Sketch the graphs of 𝑓 −1 and 𝑦 = 𝑥 on the same set of axes. Indicate clearly the intercepts with the axes, as well as the coordinates of another point on the graph of 𝑓 −1 . 3.4 Determine the domain and range for 𝑓 𝑎𝑛𝑑 𝑓 −1 QUESTION 4 Given 𝑓(𝑥) = −3𝑥 2 4.1 Determine the inverse of 𝑓(𝑥) in the form 𝑦 =……….. 4.2 Determine the domain and range for 𝑓 𝑎𝑛𝑑 𝑓 −1 QUESTION 5 Given: 𝑓(𝑥) = 2𝑥 5.1 Determine 𝑓 −1 (𝑥) in the form 𝑦 =……….. 5.2 Sketch the graphs of 𝑓, 𝑓 −1 and 𝑦 = 𝑥 on the same set of axes. Indicate clearly the intercepts with the axes, as well as the coordinates of another point on the graph of each of 𝑓 and 𝑓 −1 . 5.3 Determine the domain and range for 𝑓 𝑎𝑛𝑑 𝑓 −1 9

GR 12 – REVISION SECTION B − GR 12 INVERSE FUNCTIONS (Complex questions) • Note that the questions in this section is taken from past exam papers. (The year from which the question is, is indicated in brackets next to the question.) QUESTION 1 (March 2016 – Question 6) 1 Given: 𝑓(𝑥) = 4 𝑥 2 , 𝑥 ≤ 0 1.1 Determine the equation of 𝑓 −1 in the form 𝑓 −1 (𝑥) = ………… 1.2 On the same axes, sketch the graphs of 𝑓 , 𝑓 −1 , indicate clearly the intercepts with the axes, as well as the coordinates of another point on the graph of each of 𝑓 and 𝑓 −1 . 1.3 Is 𝑓 −1 a function? Give a reason for your answer. QUESTION 2 (Nov 2015 – Question 5) Given: ℎ(𝑥) = 2𝑥 − 3 for −2 ≤ 𝑥 ≤ 4. The 𝑥 −intercept of ℎ is Q. 2.1 Determine the coordinates of Q. 2.2 Write down the domain of ℎ−1. 2.3 Sketch the graphs of ℎ−1 , clearly indicating the 𝑦 − intercept and the coordinates of the end points. 10

GR 12 – REVISION 2.4 For which value(s) of 𝑥 will ℎ(𝑥) = ℎ−1 (𝑥) ? QUESTION 3 (DBE 2015 – Question 5) 1 𝑥 Given: 𝑓(𝑥) = (4) 3.1 Determine the value of 𝑓(−2). 3.2 Write down the equation of 𝑓 −1 (𝑥) in the form 𝑦 =……... 3.3 Andrew has no idea how to draw the graph of 𝑓 −1 . Explain to Andrew how he may use the graph of 𝑓 to draw the graph of 𝑓 −1 . 3.4 Hence or otherwise, sketch the graph of 𝑓 −1 . Clearly indicate ALL intercepts with the axes. 3.5 Write down the domain of 𝑓 −1. 3.6 For which value(s) of 𝑥 will 𝑓 −1 (𝑥) ≥ −2 ? 3.7 1 Given 𝑞 = 𝑙𝑜𝑔1 2 4 3.7.1 Determine the value of 𝑞. 3.7.2 Hence or otherwise, determine the coordinates of the point of intersection of 𝑓 and 𝑓 −1 . QUESTION 4 (March 2015-Question 5) The graph of 𝑓(𝑥) = 𝑎 𝑥 , 𝑎 > 1 is shown below. T (2 ; 9) lies on 𝑓. 4.1 Calculate the value of 𝑎. 4.2 Determine the equation of 𝑔(𝑥) if 𝑔(𝑥) = 𝑓(−𝑥) 4.3 Determine the value(s) of 𝑥 for which 𝑓 −1 (𝑥) ≥ 2 11

GR 12 – REVISION 4.4 Is the inverse of 𝑓 a function? Explain your answer. SECTION C− GR 11 FUNCTIONS (from past exam papers) QUESTION 1 (GR11 DBE NOV 2017) −3 Given 𝑓(𝑥) = +1 and 𝑔(𝑥) = 2−𝑥 − 4 𝑥+2 1.1 Determine 𝑓(−3). 1.2 Determine 𝑥 if 𝑔(𝑥) = 4. 1.3 Write down the asymptotes of 𝑓. 1.4 Write down the range of 𝑔. 1.5 Determine the coordinates of the 𝑥 −and 𝑦 −intercepts of 𝑓. 1.6 Determine the equation of the axis of symmetry of 𝑓 which has a negative gradient. Leave your answer in the form 𝑦 = 𝑚𝑥 + 𝑐. 1.7 Sketch the graphs of 𝑓 and g on the same axes. Clearly show ALL the intercepts with the axes and the asymptotes. QUESTION 2 (GR11 DBE NOV 2015) −9 The sketch below represents the graphs of 𝑓(𝑥) = 𝑥−1 − 2 A is the point of intersection of the asymptotes of 𝑓. 2.1 Write down the coordinates of A. 2.2 Give the domain of 𝑓. 2.3 Give the equation of the horizontal asymptote for 𝑓(𝑥) − 1. 12

GR 12 – REVISION 2.4 Determine the coordinates of the 𝑥 − and 𝑦 −intercepts of 𝑓. 2.5 Write down the equation of the axis of symmetry of 𝑓 that has a negative gradient. 2.6 Determine the values of 𝑥 if 𝑓(𝑥) ≥ 0 QUESTION 3 (GR 12 DBE 2019 Question 5) Sketched below are the graphs of 𝑘(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 and ℎ(𝑥) = −2𝑥 + 4. Graph 𝑘 has a turning point at (−1 ; 18). S is the 𝑥 −intercept of ℎ and 𝑘. Graphs ℎ and 𝑘 also intersect at T 3.1 Calculate the coordinates of S. 3.2 Determine the equation of 𝑘 in the form 𝑦 = 𝑎(𝑥 + 𝑝)2 + 𝑞 3.3 If 𝑘(𝑥) = −2𝑥 2 − 4𝑥 + 16 , determine the coordinates of T. 3.4 Determine the value(s) of 𝑥 for which 𝑘(𝑥) < ℎ(𝑥) 3.5 It is further given that 𝑘 is the graph of 𝑔′(𝑥) (a) For which values of 𝑥 will the graph of 𝑔 be concave up? (b) Sketch the graph of 𝑔, showing clearly the 𝑥 −values of the turning points and the point of inflection. 3.6 Determine the values of 𝑘 such that 𝑓(𝑥) = 𝑘 has equal roots. 3.7 If the graph 𝑓 is shifted TWO units to the right and ONE unit upwards to form ℎ , determine the equation ℎ in the form 𝑦 = 𝑎(𝑥 + 𝑝)2 + 𝑞. 13

GR 12 – REVISION QUESTION 4 (NSC March 2011) A parabola f intersects the x-axis at B and C and the y-axis at E. The axis of symmetry of the x 7 parabola has equation x = 3 . The line through E and C has equation g ( x) = − . 2 2 y x B O 3 C g E f 4.1 Show that the coordinates of C are (7 ; 0). 4.2 Calculate the x-coordinate of B. 4.3 Determine the equation of f in the form y = a( x − p ) 2 + q. 4.4 Write down the equation of the graph of h, the reflection of f in the x-axis. 4.5 Write down the maximum value of t(x) if t(x) = 1 – f(x). 4.6 Determine the values of 𝑥 if 𝑓(𝑥) ≥ 0 4.7 Determine the values of 𝑥 if 𝑓 ′ (𝑥) < 0 4.8 Determine the values of 𝑥 if 𝑓(𝑥) < 𝑔(𝑥) 14

GR 12 – REVISION QUESTION 5 (GR12 NOV 2019 Question 4) 𝑎 Below are the graphs of 𝑓(𝑥) = 𝑥 2 + 𝑏𝑥 − 3 and 𝑔(𝑥) = 𝑥+𝑝 • 𝑓 has a turning point at C and passes through the 𝑥 − axis at (1; 0) • 𝐷 is the 𝑦 −intercept of both 𝑓 and 𝑔. The graphs 𝑓 and 𝑔 also intersect each other at E and J. • The vertical asymptote of 𝑔 passes through the 𝑥 −intercept of 𝑓. 5.1 Write down the values of 𝑝. 5.2 Show that 𝑎 = 3 and 𝑏 = 2. 5.3 Calculate the coordinates of C. 5.4 Write down the range of 𝑓. 5.5 Determine the equation of the line through C that makes an angle of 45° with the positive 𝑥 −axis. Write your answer in the form 𝑦 =…….. 5.6 Is the straight line, determined in QUESTION 5.5, a tangent to 𝑓? Explain your answer. 5.7 The function ℎ(𝑥) = 𝑓(𝑚 − 𝑥) + 𝑞 has only one 𝑥 −iintercept at 𝑥 = 0. Determine the values of 𝑚 and 𝑞. 15

GR 12 – REVISION REVISION GR 12 – CALCULUS CALCULUS – SUMMARY Question Type Summary of procedure Example question 1. Find the limits of simple Clear the unknown in 4ℎ−ℎ 2 𝑎 2 −36 lim ℎ ; lim 𝑎−6 functions denominator and substitute the ℎ→0 𝑎→6 value being approached in the expression. 2. Find the average gradient Need the coordinates of two If 𝑓(𝑥) = 2𝑥 3 , find the between two points points, find gradient using gradient between 𝑥 = 1 and formula 𝑥=2 3. Find the gradient at a point Find the derivative of the If 𝑓(𝑥) = 2𝑥 3 , find the expression, substitute 𝑥 value gradient where 𝑥 = 2 of point 4. Find the derivative from first Have 𝑓(𝑥), find 𝑓(𝑥 + ℎ) and Find 𝑓′(𝑥) if 𝑓(𝑥) = 2𝑥 2 from principles use derivative formula first principles 5. Find derivatives using the Simplify the expression and 𝑑𝑦 Find 𝑑𝑥 if rules then differentiate. Be careful of 2 layout of the expression. 𝑦 = − √𝑥 = 2(𝑥 + 1) 𝑥 6. Use different notations for Be careful of layout of the 𝑑𝑦 𝑑𝐴 𝑑𝑠 𝑓 ′ (𝑥), 𝑑𝑥 , 𝐷𝑥 , 𝑑𝑥 , 𝑑𝑡 differentiation expression. 7. Find turning points, Make derivative equal to zero, horizontal points of substitute back into original 𝑦 = 𝑥 2 − 3𝑥 − 4 inflection of curves equation. 8. Finding 𝑥 intercepts of cubic Make 𝑦 = 0, solve equation. Determine where curves Guess 1𝑠𝑡 factor, factorize 𝑦 = 𝑥 3 − 5𝑥 2 − 8𝑥 + 12 cuts further , may need quadratic the 𝑥-axis equation formula. 9. Sketch cubic graphs 𝑦 cut, 𝑥 cuts using factor Sketch the curve of theorem, turning points using 𝑦 = 𝑥 3 − 5𝑥 2 − 8𝑥 + 12 derivative = 0, sketch 10. Find points of inflection Let 2𝑛𝑑 derivative = 0 and Find the point of inflection substitute 𝑥 value back into the between the 2 stationary points original equation of 𝑦 = 𝑥 3 − 5𝑥 2 − 8𝑥 + 12 11. Find the equation of a 𝑦 = 𝑚𝑥 + 𝑐 , Substitute Find the equation of a tangent tangent to a graph 𝑥 value into the derivative to to the curve 𝑦 = 35𝑥 2 − 4𝑥 + get 𝑚. Substitute 𝑥 value into 1 at the point where 𝑥 = 1 original equation to get a point on the curve. Use point to find 𝑐 12. Maximize and minimize Find expression to represent lengths, areas, volumes etc. situation described, make the derivative = 0, substitute 𝑥 value back into original equation. 16

GR 12 – REVISION NOTE: • Questions in this revision were compiled from past exam National papers for GR 12. SECTION A QUESTION 1. Determine 𝑓′(𝑥) from first principles if 1.1 𝑓(𝑥) = 3𝑥 2 − 5 (5) 1.2 𝑓(𝑥) = −𝑥 2 + 4 (5) 1.3 4 (5) 𝑓(𝑥) = 𝑥 1.4 𝑓(𝑥) = 𝑥 3 (5) QUESTION 2. Determine : 2.1 𝑑𝑦 4 (3) if 𝑦 = 2𝑥 5 + 𝑥 3 𝑑𝑥 2.2 𝑑𝑦 2 (4) if 𝑦 = (√𝑥 − 𝑥 2 ) 𝑑𝑥 2.3 𝑑𝑦 1 2 (3) if 𝑦 = (𝑥 2 − ) 𝑑𝑥 𝑥2 2.4 3 2 (3) the derivative of 𝑓(𝑥) = (𝑥 − ) 𝑥 2.5 1 2 (4) the derivative of 𝑓(𝑥) = ( 3 + 4𝑥) 𝑥 2.6 𝑥 3 −1 (3) 𝐷𝑥 ( 𝑥−1 ) 2.7 3 1 (3) 𝐷𝑥 [ √𝑥 2 − 2 𝑥] QUESTION 3. 3.1 𝑑𝑦 (3) If 𝑦 = (𝑥 6 − 1)2 , prove that = 12𝑥 5 √𝑦 , if 𝑥 > 1. 𝑑𝑥 3.2 Given: 𝑓(𝑥) = 2𝑥 3 − 2𝑥 2 + 4𝑥 − 1. Determine the interval on which 𝑓 is concave (4) up. 3.3 𝑔(𝑥) = −8𝑥 + 20 is a tangent to 𝑓(𝑥) = 𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 18 at 𝑥 = 1. (5) Calculate the values of 𝑎 and 𝑏. 17

GR 12 – REVISION QUESTION 4 Given 𝑓(𝑥) = 2𝑥 3 − 23𝑥 2 + 80𝑥 − 84 4.1 Prove that (𝑥 − 2) is a factor of 𝑓. (2) 4.2 Hence, or otherwise, factorise 𝑓(𝑥) fully. (2) 4.3 Determine the 𝑥 − coordinates of the turning points of 𝑓. (4) 4.4 Sketch the graph of 𝑓, clearly labelling ALL intercepts with the axes and turning points. (3) 4.5 Determine the coordinates of the 𝑦 − intercept of the tangent to 𝑓 that has a slope of 40 (6) and touches 𝑓 at a point where the 𝑥 − coordinate is an integer. QUESTION 5 The sketch below are the graphs of 𝑓(𝑥) = (𝑥 −)2 (𝑥 − 𝑘) and 𝑔(𝑥) = 𝑚𝑥 + 12. • A and D are the 𝑥 − intercepts of 𝑓. • B is the common 𝑦 − intercepts of 𝑓 and 𝑔 . • C and D are turning points of 𝑓. • The straight line 𝑔 passes through A. 5.1 Write down the 𝑦 −coordinate of B. (1) 5.2 Calculate the 𝑥 − coordinates of A. (3) 5.3 If 𝑘 = −3, calculate the coordinates of C. (6) 5.4 For which values of 𝑥 will 𝑓 be concave down. (3) 18

GR 12 – REVISION QUESTION 6 Given : ℎ(𝑥) = −𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 and 𝑔(𝑥) = −12𝑥. P and Q(2; 10) are the turning points of ℎ. The graph of ℎ passes through the origin. 6.1 3 (5) Show that 𝑎 = 2 and 𝑏 = 6. 6.2 Calculate the average gradient of ℎ between P and Q, if it is given that 𝑥 = −1 at P? (4) 6.3 1 (3) Show that the concavity of ℎ changes at 𝑥 = 2 . 6.4 Explain the significance of the change in QUESTION 6.3 with respect to ℎ. (1) 6.5 Determine the value of 𝑥, given 𝑥 < 0, at which the tangent to ℎ is parallel to 𝑔. (4) QUESTION 7 Given : 𝑓(𝑥) = (𝑥 + 2)(𝑥 2 − 6𝑥 + 9) = 𝑥 3 − 4𝑥 2 − 3𝑥 + 18 7.1 Calculate the coordinates of the turning points of the graph of 𝑓 (6) 7.2 Sketch the graph of , clearly indicating the intercepts with the axes and the turning points. (4) 7.3 For which value(s) of 𝑥 will 𝑥. 𝑓′(𝑥) < 0 (3) QUESTION 8 Given : 𝑓(𝑥) = 3𝑥 3 8.1 Solve 𝑓(𝑥) = 𝑓′(𝑥) (3) 8.2 The graphs 𝑓, 𝑓 ′ and 𝑓′′ , all passes through the point (0 ; 0 ) 8.2.1 For which of the graphs will ( 0;0) be a stationary point. (1) 8.2.2 Explain the difference , if any , in the stationary points referred to in QUESTION 8.2.1 (2) 8.3 Determine the vertical distance between the graphs of 𝑓′ and 𝑓′′ at 𝑥 = 1 (3) 8.4 For which value(s) of 𝑥 is 𝑓(𝑥) − 𝑓 ′ (𝑥) < 0 (4) 19

GR 12 – REVISION Financial Mathematics FINANCE – SUMMARY FORMULES Simple Interest 𝐴 = 𝑃(1 + 𝑖𝑛) Compound Interest 𝐴 = 𝑃(1 + 𝑖)𝑛 Straight line depreciation 𝐴 = 𝑃(1 − 𝑖𝑛) Reducing Balance depreciation 𝐴 = 𝑃(1 − 𝑖)𝑛 𝑖 𝑚 Effective and Nominal interest 1 + 𝑖𝑒𝑓𝑓 = (1 + 𝑛𝑜𝑚 ) 𝑚 Sum of Geometric series 𝑎(𝑟 𝑛 −1) 𝑎(1−𝑟 𝑛 ) 𝑆𝑛 = 𝑜𝑟 𝑆𝑛 = ,𝑟 ≠ 1 𝑟−1 1−𝑟 𝐴 Final amount 𝑃 Original amount 𝑖 Is the interest rate per period 𝑛 Is the number of periods 𝑖𝑒𝑓𝑓 Is the effective annual interest rate 𝑖𝑛𝑜𝑚 Nominal interest is the quoted annual interest rate 𝑚 Is the number of times interest is compounded in one year. Future value annuity 𝑥[(1+𝑖)𝑛 −1] 𝐹= 𝑖 Present value annuity 𝑥[1−(1+𝑖)−𝑛 ] 𝑃= 𝑖 𝑛 Is the number of payments NOTE: • Questions in this revision were compiled from past exam National papers for GR 12. 20

GR 12 – REVISION Section A: QUESTION 1 1.1 Anisha and Lindiwe each received R12 000 to invest for a period of 5 years. They invested the money at the same time according to the following options: • Anisha: 8,5% p.a. simple interest. At the end of the 5 years she will receive an additional bonus pay-out of exactly 7,5% of the original amount invested. • Lindiwe: 8,5% p.a. compounded quarterly. Who will have the larger final amount after 5 years? Justify your answer with appropriate calculations. (6) 1.2 A company bought office furniture that cost R120 000. After how many years will the furniture depreciate to a value of R41 611,57 according to the reducing-balance method, if the rate of depreciation is 12,4% p.a.? (4) 1.3 Tebogo opened a savings account with a single deposit of R5 000 at the beginning of June 2015. He then made 24 monthly deposits of R800 at the end of every month, starting at the end of June 2015. The account earned interest at 15% p.a. compounded monthly. Calculate the amount that should be in his savings account immediately after he makes the (5) last deposit. [15] QUESTION 2 2.1 Exactly five years ago Mpume bought a new car for R145 000. The current book value of this car is R72 500. If the car depreciates by a fixed annual rate according to the reducing-balance method, calculate the rate of depreciation. (3) 2.2 Samuel took out a home loan for R500 000 at an interest rate of 12% per annum, compounded monthly. He plans to repay this loan over 20 years and his first payment is made one month after the loan is granted. 2.2.1 Calculate the value of Samuel's monthly instalment. (4) 2.2.2 Melissa took out a loan for the same amount and at the same interest rate as Samuel. Melissa decided to pay R6 000 at the end of every month. Calculate how many months it took for Melissa to settle the loan. (4) 2.2.3 Who pays more interest, Samuel or Melissa? Justify your answer. (2) [13] QUESTION 3 On 1 June 2016 a bank granted Thabiso a loan of R250 000 at an interest rate of 15% p.a. compounded monthly, to buy a car. Thabiso agreed to repay the loan in monthly instalments commencing on 1 July 2016 and ending 4 years later on 1 June 2020. However, Thabiso was unable to make the first two instalments and only commenced with the monthly instalments on 1 September 2016. 3.1 Calculate the amount Thabiso owed the bank on 1 August 2016, a month before he paid his first monthly instalment. (2) 3.2 Having paid the first monthly instalment on 1 September 2016, Thabiso will still pay his last monthly instalment on 1 June 2020. Calculate his monthly instalment. (4) 3.3 If Thabiso paid R9 000 as his monthly instalment starting on 1 September 2016, how many months sooner will he repay the loan? (5) 3.4 If Thabiso paid R9 000 as a monthly instalment starting on 1 September 2016, calculate the final instalment to repay the loan. (4) [15] 21

GR 12 – REVISION QUESTION 4 4.1 How long would the price of an asset take to reduce by a third of its original value if it depreciates on a reducing balance at a rate of 4,7% p.a.? (4) 4.2 Lebogo bought a tractor for Rx on 1 April 2016. • She will trade in this tractor when she replaces it with a similar one in 5 years' time on 1 April 2021. • The tractor depreciates by 20% p.a. according to the reducing-balance method. • The price of a similar tractor increases by 18% annually. • Lebogo calculated that if she deposited R8 000 per month into a sinking fund, which paid interest at 10% p.a. compounded monthly, she would have enough money to cover the replacement cost of the tractor. She made the first deposit in this fund on 30 April 2016 and will continue to do so at the end of every month until 31 March 2021. 4.2.1 Determine, in terms of x, what the book value of the current tractor will be on 1 April 2021 (that is, 5 years after it was bought). Give your answer correct to FIVE decimal places. (2) 4.2.2 Determine, in terms of x, what the price of a similar new tractor will be on 1 April 2021. Give your answer correct to FIVE decimal places. (2) 4.2.3 Calculate the amount accumulated in the sinking fund on 1 April 2021. (4) 4.2.4 Calculate the value of x, the price of the current tractor. Round off your answer to the nearest thousand (4) [16] QUESTION 5 The graph of 𝑓 shows the book value of a vehicle 𝑥 years after the time Joe bought it. The graph of 𝑔 shows the cost price of a similar new vehicle 𝑥 years later. Time (in years) 5.1 How much did Joe pay for the vehicle? (1) 5.2 Use the reducing-balance method to calculate the percentage annual rate of depreciation of the vehicle that Joe bought. (4) 5.3 If the average rate of the price increase of the vehicle is 8,1% p.a., calculate the value of a. (3) 5.4 A vehicle that costs R450 000 now, is to be replaced at the end of 4 years. The old vehicle will be used as a trade-in. A sinking fund is created to cover the replacement cost of this vehicle. Payments will be made at the end of each month. The first payment will be made at the end of (5) the 13th month and the last payment will be made at the end of the 48th month. The sinking fund earns interest at a rate of 6,2% p.a., compounded monthly. Calculate the monthly payment to the fund. [13] 22

GR 12 – REVISION QUESTION 6. Siphokazi bought a house. She paid a deposit of R102 000, which is equivalent to 12% of the selling price of the house. She obtained a loan from the bank to pay the balance of the selling price. The bank charges her interest of 9% per annum, compounded monthly. 6.1 Determine the selling price of the house. (1) 6.2 The period of the loan is 20 years and she starts repaying the loan one month after it was granted. Calculate her monthly instalment. (4) 6.3 How much interest will she pay over the period of 20 years? Round your answer correct to the nearest rand. (2) 6.4 Calculate the balance of her loan immediately after her 85th instalment. (3) 6.5 She experienced financial difficulties after the 85th instalment and did not pay any instalments for 4 months (that is months 86 to 89). Calculate how much Siphokazi owes on her bond at the end of the 89th month. (2) 6.6 She decides to increase her payments to R8 500 per month from the end of the 90th month. How many months will it take to repay her bond after the new payment of R8 500 per month? (4) [16] 23

Western Cape Government FOR YOU Education TRIGONOMETRY TRIG SUMMARY (Grade 12) » ANGLES IN STANDARD POSITIONS * Positive /* te ay 4 : {anticlockwise from 0° to 360") a Yo. ¢ ‘ & w w y * ny vhoat ee & wv m + Nogative /* (clockwise from 0° to ~360") GR 12— REVISION S ANSWER ESERIES Your Key to Exam Success nso | Mt! ,. possible r : >» THE RATIOS »+ sin® tan® & their + Definitions: (€] Cl * Signs: ult ' 1 sin 0 is positive in 1 & cos is positive in 1 & IV v ™ tan @ is positive in 1 & Mt 1 0 ot * Critical values: ° o}4 ie i 0 4 0 Ne + Minimum & Maximum values of sin@ & cos@ The valves of sin 8 8 cos @ range rom ~1 to 4 | Note that tan @ has] Se — no minimum or minimum maximum maximum values. value is 4 values | the range of values \of tan Bis from = t0 «, All values are proper fractions or 0 or £4 Graphs: y=sine y=cos@ y=tan@ Y > IDENTITIES con tan@ = 3in® 1-cos?@ | cos @ & cos? = 1~sin?0 | > SPECIAL /5 yex Asp 1 v2 ‘ 2 a, (and cos 60° = 1] lay , 1 & thoirtamiies: | QD tamity 4 180°+0 wv 4 360°- @ »GENERALFORMS Y fin.$) Yoyo y vir = that SAME ratioo! @ -@ » CO-RATIOS (sine and cosine) ny + 90°=@ (an acute angle) {| * 90°+8 (an obtuse angie) [bay ree sin(90°-@) = cos@ / |*__ sin(90"+8) = cos@ ») c0s(90°- 6) = sin@ ios c0s(90° +6) = -sin@ x ‘The ratio CHANGES to the CO-ratio, SOLUTION OF A$ In Right-angled 4°, we use v + Regular rig. ratios the Theorem of Pythagoras. + Area= bh In Non-Right-angled 4°, wo use: + Sine Rule: + Cosine Rule + Area Rule: sinc 7 [c= at+b?-2abcos¢] [AREA = tab sinc But also: Area of 4 = bh sin. sinB . a b > COMPOUND / > DOUBLE /§ {+ sin2x=2sin.x cos x * sin(A £ B) = sin A 00s B cos Asin B . cos 2c = coe x= sts - 8 =2sin® x * cos(A # B) = cos A cos B sin A sin B eae 24

GR 12 – REVISION Section A: NOTE: Questions in this revision were compiled from past exam National papers for GR 12. SECTION A (Routine questions) QUESTION 1 (GR12 NSC NOV 2019) 1.1 1.1.1 Simplify the following expression to ONE trigonometric term: sin𝑥 + sin(180° + 𝑥) cos (90° − 𝑥) cos𝑥. tan𝑥 1.1.2 Without using a calculator, determine the value of: sin2 35° − cos 2 35° 4sin10°cos10° (GR12 NSC NOV 2017) 1.2 sin(A − 360°) . cos (90° + A) Given: cos(90° − A) . tan (−A) Simplify the expression to a single trigonometric ratio. (GR12 NSC NOV 2017) 1.3 Prove: 1.3.1 Sin(A+B)−sin(A−B) = 2 cosA.sinB 1.3.2 Without using a calculator, that sin17° − sin 43° = sin 17° (GR12 NSC March 2018) 1.4 Simplify: sin (180°−𝑥). cos (−𝑥) + cos (90°+𝑥). cos (𝑥 −180°) to a single trigonometric ratio. 1.5 Determine the value of sin3𝑥. cosy+cos3𝑥. sin𝑦 if 3𝑥 + 𝑦 = 270° QUESTION 2 (GR12 NSC NOV 2018)-Q5 2.1 In the diagram 𝑃(𝑘; 1) is a point in the 2nd quadrant and is √5 units from the origin. R is a point on the positive 𝑥 − axis and obtuse RÔ P= 𝜃 2.1.1 Calculate the value of 𝑘. 2.1.2 Without using a calculator, calculate the value of: a) tan 𝜃 b) Cos (180° + 𝜃) c) 𝑎+𝑏 sin (𝜃 + 60°) in the form 20 √ 25

GR 12 – REVISION (GR12 NSC NOV 2017)-Q5 2.2 In the diagram, P(3; 𝑡) is a point in the Cartesian plane. OP= √34 and HO ̂ P = 𝛽 is a reflex angle. Without using a calculator, determine the value of: 2.2.1. 𝑡 2.2.2. tan 𝛽 2.2.3. cos 2𝛽 (GR12 NSC March 2018)-Q5 2.3 If cos2𝜃 = − 5 , where 2𝜃𝜖[180°;270°], calculate without using a calculator, the values in 6 simplest form of 2.3.1 sin2𝜃 2.3.2 sin2 𝜃 QUESTION 3 (GR 12 NSC November 2019) 3.1 In the diagram, the graphs of 𝑓(𝑥) = sin𝑥 − 1 and 𝑔(𝑥) = cos2𝑥 are drawn for the interval 𝑥 ∈ [−90° ; 360°]. Graphs 𝑓 and 𝑔 intersect at A. B (360° ; −1) is a point on 𝑓. 3.1.1 Write down the range of 𝑓. 3.1.2 Write down the values of 𝑥 in the interval 𝑥𝜖[−90°; 360°] for which graph 𝑓 is decreasing. (GR 12 NSC November 2017) 3.2 In the diagram, the graph of 𝑓(𝑥) = cos2𝑥 is drawn for the interval 𝑥 ∈ [−270° ; 90°]. 3.2.1 Draw the graph of 𝑔(𝑥) = 2sin𝑥 − 1 for the interval 𝑥 ∈ [−270° ; 90°] on the grid given. Show ALL intercepts with the axes, as well as the turning points. 3.2.2 Let A be a point of intersection of the graphs 𝑓 and 𝑔. Show that the 𝑥 −coordinate of A −1+√5 satisfies the equation sin 𝑥 = 2 . 3.2.3 Hence, calculate the coordinates of the points of intersection of graphs of 𝑓 and 𝑔 for the interval 𝑥 ∈ [−270° ; 90°]. 26