CAIE AS Level Mathematics PDF

ZNOTES.ORG UPDATED TO 2023-2025 SYLLABUS CAIE AS LEVEL MATHEMATICS SUMMARIZED NOTES ON THE THEORY SYLLABUS Prepared for Aqsa for personal use only. CAIE AS LEVEL MATHEMATICS Example 2 1. Quadratics...

ZNOTES.ORG UPDATED TO 2023-2025 SYLLABUS CAIE AS LEVEL MATHEMATICS SUMMARIZED NOTES ON THE THEORY SYLLABUS Prepared for Aqsa for personal use only. CAIE AS LEVEL MATHEMATICS Example 2 1. Quadratics Solve the quadratic equation 6x2 − 17x + 5 = 0 using factorisation. Answer 1.1. Quadratic Formula This can be factorised as The quadratic formula helps us find the solutions to a (2x − 5)(3x − 1) = 0 quadratic equation, ax2 + bx + c = 0. Logically, if we have ab = 0 then either a = 0 or b = 0. It helps us find the values of x for which the quadratic equation equals 0, which are also called the zeros of the 5 2x − 5 = 0 ⟹ x = quadratic equation for this reason. 2 ​ Quadratic formula (Given in MF19) 1 3x − 1 = 0 ⟹ x = 3 ​ −(b) ± (b)2 − 4(a)(c) x= ​ 2a 1.2. Completing the square ​ So each of the roots of the quadratic equation ax2 + bx + c = 0 are Completing the square allows us to write a quadratic, ax2 + bx + c, in the form p(x + q)2 + r. This is useful for a −b + b2 − 4ac variety of reasons, which we will see later on. x= ​ 2a ​ Given ax2 + bx + c, we can complete the square using And the formula: −b − b2 − 4ac b 2 b2 x= a (x + ) + c − ​ 2a ​ 2a 4a ​ ​ Factorisation Comparing coefficients in the expansion of (x + a)2 is Factorisation allows us to express a polynomial as the another way to complete the square. product of simpler terms. For example: Example Express 3x2 + 9x + 5 in the form of p(x + q)2 + r, where x2 − 2x − x + 2 = (x − 1)(x − 2) p, q and r are constants. Answer Factorisation of a quadratic can be done through two First we identify our coefficients, a = 3, b = 9, c = 5. Now methods: we can substitute in our values into the formula: splitting the middle term 9 2 92 Re-writing ax2 + bx + c = 0 as ax2 + dx + ex + 3 (x + ) +5− 2(3) 4(3) ​ ​ c = 0 [Where d × e = a × c and dx + ex = bx] Then factoring out the common factors in the Which simplifies to equation 2 A calculator 3 7 Using the “mode” button and then selecting the type 3 (x + ) − 2 4 ​ ​ of equation. This will give us the solutions to a quadratic equation, which can then be used to You can verify your answer by expanding it and factorise it. comparing it with your original quadratic. Example 1 1.3. Solving quadratic equations using Solve the quadratic equation 6x2 − 17x + 5 = 0 using the quadratic formula. completing the square form Answer It is possible to solve quadratic equations by completing the Here a = 6, b = −17, c = 5. Using the quadratic formula square. Given a quadratic equationax2 + bx + c = 0, it can −(−17) ± (−17)2 − 4(6)(5) be written as p(x − q)2 + r = 0. x= ​ 2(6) ​ By making x the subject of the formula, we get So the 2 values of x which satisfy the equation are −r x=± +q 5 1 ​ ​ q x = and x = 2 3 ​ ​ −r Note that if q ​ < 0, the quadratic has no real solutions. WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS Example By completing the square, find the solutions to the quadratic 7 r=− 4 ​ equation x2 − 5x + 6 = 0. Answer 3 7 The coordinates of the vertex are (− , − ) 2 4 ​ ​ First we complete the square 5 2 1 1.5. Quadratic inequalities x − 5x + 6 ⟹ (x − ) − 2 2 4 ​ ​ An inequality question will ask us to find the values of x for 5 2 1 which the inequality holds true. (x − ) − = 0 The most efficient way to solve these inequalities is to follow 2 4 ​ ​ these steps. Next, we make x the subject of the formula and solve for it Factorise the inequality. This will help us find the intersections of the 1 5 quadratic with the x-axis x=± + 4 2 Draw a simple sketch of the quadratic and label the x ​ ​ ​ coordinate of the intersections. x = 3 or x = 2 The quadratic curve will have a “smiley” or an U shape if the coefficient of x2 is positive. This point 1.4. Finding the coordinates of the is also known as a minimum point, or a minima. vertex The vertex of a quadratic equation is the highest or lowest point on the curve. We can find the coordinates of the vertex after completing the square Equation Coordinates of vertex y = p(x − q)2 + r (q, r) y = p(x + q)2 − r (−q, −r) Another way of finding coordinates of the vertex is through the general completing the square formula. Given a quadratic, y = ax2 + bx + c: The quadratic curve will have a “frown” or a n shape b 2 b2 y = a (x + ) + c − if the coefficient of x2 is negative. 2a 4a ​ ​ This point is also known as a maximum point, or a −b maxima. x coordinate of vertex x= 2a ​ 2 b −4ac y coordinate of vertex y= 4a ​ The line x = −b2a acts as a line of symmetry for the ​ quadratic. It splits the quadratic curve into 2 equal and mirrored parts. Completing the square is also called the vertex form due to this. Example Find the coordinates of the vertex of the quadratic 3 2 (x + 32 ) − 74. ​ ​ Answer We observe that it has been written in the completing the square, or vertex, form. 2 3 7 3 (x + ) − = p(x − q)2 + r 2 4 ​ ​ 3 q=− 2 ​ WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS As we’re trying to find the values of x for which y < 0, we look at the blue region. Clearly any value that is greater than −4 and less than −2 is represented in the blue region. So the answer is −4 < x < −2. 1.6. Discriminant of a quadratic formula The discriminant helps us find the number of intersections of a quadratic equation with the x-axis. It can also be used to find the number of intersections of a quadratic curve with a line or another quadratic curve. Consider b2 − 4ac in quadratic equation Using your sketch, deduce the values of x for which the inequality holds true. −b ± b2 − 4ac x= ​ If a question asks you to find y ≤ 0 or y ≥ 0, then the 2a ​ points of intersections must be included in the range of If b2 − 4ac > 0 values. 2 This results in x = −b ± 2ab −4ac , meaning there are ​ ​ two distinct real roots that can be found. Example 2 If b − 4ac = 0 Find the values of x which satisfy the inequality x2 + 6x + As the square root is equal to 0, we get x = −b2a , 8 Two real distinct Two distinct points of 0 roots intersection 2 b − 4ac = Two equal real One point of intersection 0 roots b2 − 4ac < No real roots No points of intersection 0 Finding the number of intersection between The blue part of the curve represents the region where y < curves 0 and the red part of the curve represents the region where y > 0. WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS The theory of discriminants can be extended to find the This may require you to multiply the equations with an number of intersections between a quadratic and integer, or an unknown variable in more complicated another quadratic curve or linear line. problems. To find the number of intersections, we first equate both curves or line and then apply the discriminant Example formula. Solve the simultaneous equations For finding the number of intersections between a quadratic and another line or curve: 4a + 2b = 8 Discriminant Number of Nature of intersection a + 3b = 7 intersections with curve 2 b − 4ac > Two real distinct points Line meets curve at two Answer 0 of intersection points First we make an unknown variable have the same b2 − 4ac = One real point of Line is tangent to the coefficient. 0 intersection curve We can do this by multiplying the second equation by b2 − 4ac < No real points of Line does not meet the 4, resulting in 4a + 12b = 28, making the unknown 0 intersections curve variable a share the same coefficient in both equations. 4a + 2b = 8 4a + 12b = 28 Now we can subtract both equations to get rid of one unknown variable, resulting in 2b − 12b = 8 − 28. This Example will help us isolate one variable and find its value. Given a quadratic, kx2 + 4kx + 3k = 0, find the range of k values for which the equation has two distinct real roots. −10b = −20 Answer We need to satisfy the condition b2 − 4ac > 0 to have two b=2 distinct real roots. Substitute the value of b into any of the two equations. The coefficients of the quadratic equation can be labelled This is done to find the value of the other unknown a = k, b = 4k and c = 3k , leading to variable. (4k)2 − 4(k)(3k) > 0 4a + 2(2) = 8 2 2 16k − 12k > 0 a=1 4k 2 > 0 Substitution Thus we get the range of k as The second method, substitution, makes one unknown k>0 variable the subject of the equation. This is then substituted into the second equation to find its value. 1.7. Simultaneous equations Upon finding the value of one unknown variable, we can easily find the value of the other by substituting its value A simultaneous equation is a set of equations, with into any of the two equations. unknown variables, that satisfy a condition. These can be solved through the use of the methods of Example elimination and substitution. Solve the simultaneous equation Elimination 4a + 2b = 8 The method of elimination requires you to multiply one, or a + 3b = 7 both, equation such that one of the unknown variables shares the same coefficient in both equations. Answer Using the second equation, we make a the subject of the formula. a + 3b = 7 ⟹ a = 7 − 3b Substitute this expression into the first equation, 4a + 2b = 8. WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS 4(7 − 3b) + 2b = 8 16 31 (− , ) and (2, 3) 3 3 ​ ​ 2b − 12b = 8 − 28 Substitution b=2 Another way of solving simultaneous equations is through Substitute this value of b into any of the two equations. the use of a substitution. This allows us to write an equation This is done to find the value of the other unknown of one curve in terms of x or y. The new expression can variable. then be substituted into the second equation to get the a = 7 − 3(2) points of intersection. a=1 Example Find the coordinates of the intersections of the curve y 2 = 4x2 − 7 and line x + y = 5 using a substitution. 1.8. Simultaneous equations with Answer quadratics Make x the subject of the formula in the line equation to A simultaneous equation can include a quadratic curve get x = 5 − y. Now we can substitute this expression as and a linear line. x in the quadratic equation. Solving these can be thought of as finding the points of intersection of both equations. y 2 = 4(5 − y)2 − 7 Equating y 2 = 100 − 40y + 4y 2 − 7 One way of solving simultaneous equations is by equating 3y 2 − 40y + 93 = 0 both curves, or curve and line. This will result in a quadratic We can solve this quadratic equation in terms of y , to get the equation which can be solved to find the points of intersection. y coordinates of intersection. 3y 2 − 40y + 93 ⟹ (3y − 31)(y − 3) Example Find the coordinates of the intersections of the curve y 2 = 31 4x2 − 7 and line x + y = 5 by equating both equations. y= or y = 3 3 ​ Answer Substitute this value of y into the linear equation to find Make y the subject of the formula in the line equation, the x coordinates of intersection. and square both sides to get an equation in terms of y 2. 31 16 x+ =5 ⟹ x=− 2 2 3 3 ​ ​ y = 5 − x ⟹ y = (5 − x) x+3=5 ⟹ x=2 Now we can equate both equations, as their points of intersections must have the same y coordinate. The solutions to this simultaneous equation, or points of 2 2 intersection of both equations, is: (5 − x) = 4x − 7 16 31 x2 − 10x + 25 = 4x2 − 7 (− , ) and (2, 3) 3 3 ​ ​ 0 = 3x2 + 10x − 32 1.9. Substitutions We can solve this quadratic equation in terms of x, to get the x coordinates of intersection. Many questions may ask you to solve unique equations, which may not resemble a quadratic at first glance. These 3x2 + 10x − 32 ⟹ (3x + 16)(x − 2) can be solved by making a clever substitution, by the use of 16 a variable, and then factorising it to solve for the values of x x=− or x = 2. 3 ​ The table below shows an example of possible substitutions Substitute this value of x into the linear equation to find that one can use the y coordinates of intersection. 16 31 − +y =5 ⟹ y = 3 3 ​ ​ 2+y =5 ⟹ y =3 The solutions to this simultaneous equation, or points of intersection of both equations, is: WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS Equation Substitution ax4 + bx2 + c u = x2 and u2 = x4 Types of functions and relations ax6 + bx3 + c u = x3 and u2 = x6 There are many types of relations a function can have: ax + b x + c u = x and u2 = x Graphical Function Definition ​ ​ 2 1 1 2 ax 3 + bx 3 + c ​ ​ u = x 3 and u2 = x 3 ​ ​ representation ax−2 + bx−1 + c u = x−1 and u2 = x−2 One x-value input Example 1 gives one y -value Solve the equation x4 − 5x2 + 4 = 0 output. As an example, One to Answer all linear functions of One Let u = x , such that u = x4 and x = ± u 2 2 ​ the type f (x) = Using this substitution, we get u2 − 5u + 4 = 0. This mx + c are One to quadratic equation can be factorised easily. One. Graph of y = x, also (u − 4)(u − 1) = 0 written as f (x) = x. u = 4 ⟹ x2 = 4 ⟹ x = ±2 u = 1 ⟹ x2 = 1 ⟹ x = ±1 Two x-value inputs Example 2 Many to give the same y -value Solve the equation 6x + x − 1 = 0 One output. Answer Let u = x, such that u2 = x and x = u2 ​ Using this substitution, we get 6u2 + u − 1 = 0. This Graph of y = x2 , also quadratic can be factorised easily. written as f (x) = x2. (3u − 1)(2u + 1) = 0 Another type of relation, that is not a function is: 1 1 1 Graphical Relation Definition u= ⟹ x= ⟹ x= representation 3 3 9 ​ ​ ​ ​ One x-value input −1 −1 1 gives two y -value u= ⟹ x= ⟹ x= 2 2 4 outputs. This is not ​ ​ ​ ​ a function as it 2. Functions One to many gives us more than [Not a function] one y -value output with one x-value A function assigns a collection of x values to only one y input, contradicting value. the definition of a function. Graph of y 2 = x. A function can be denoted by f (x), g(x), h(x) etc. Inputting a x value into the function f outputs a y value [also written as f (x)] Vertical Line Test to check for Existence of a Function To check for the existence of a function, we make use of the vertical line test. For example f (x) = x2 + 4x + 4 When x = 2 f (2) = (2)2 + 4(2) + 4 = 16 WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS If a vertical line, x = a, cuts the the function at one point only then the function exists Note that a is a constant which is included within the interval of x-values the function is defined for. Curve passes the vertical line test As the curve y = x passes the horizontal line test, the function has a One to One relation. Curve fails the horizontal line test As the curve y = x2 passes the vertical line test, it is a function. Curve fails the vertical line test As the curve y = x2 fails the horizontal line test, the function does not have a One to One relation. It instead has a Many to One relation in this case. 2.1. Domain and Range As the curve y 2 = x fails the vertical line test, it is not a Domain function. The domain of a function is the set of x-values, or Horizontal Line Test to check for One to One inputs, that the function f is defined for. Relation It consists of the smallest and biggest x-value the function accepts as an input, or is defined for. To check if a function has a One to One relation, we make Example: x ∈ R implies the function the function is use of the horizontal line test. defined for all real values of x. 4 < x < 10 implies the function is defined between If a horizontal line, y = a, cuts the function at one point x = 4 and x = 10. only then the function has a One to One relation. Else it will have a Many to One relation. Range Note that a is a constant which is included within the interval of y -values the function is defined for. Curve passes the horizontal line test WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS Coordinates of The range of a function is the set of corresponding y - Equation Nature of vertex Range vertex values, or outputs, that the function f is defined for. y = p(x − Maximum point ( r ≥ y or It consists of the lowest and the highest y -value the (q, r) q)2 + r a < 0 or p < 0) r>y function outputs for a given domain. y = p(x − Minimum point ( r ≤ y or Example: y ∈ R implies the function is defined for all (q, r) q)2 + r a > 0 or p > 0) r or ≤) y = −2 and y = 5, where both values are inclusive. depends on if the domain includes the x coordinate of the vertex, q , using ≤ and ≥. Example Example 1 Given a quadratic, f (x) = 2x2 − 2x + 3 with a domain of a ≤ x, find the smallest value of a for which the function has a One to One relation. Answer First we complete the square for f (x) 1 2 5 2 (x − ) + 2 2 ​ ​ As a > 0 and p > 0, the vertex is a minimum point with coordinates ( 12 , 52 ). ​ ​ So the smallest value of a, for which the function has a One For this curve, y = f (x): to One relation, is 12 and the domain is ​ The range is y1 ≤ f (x) ≤ y2 ​ ​ 1 Also written as y1 ≤ y ≤ y2 ≤x 2 ​ ​ ​ The domain is x1 ≤ x ≤ x3 The graph is shown below ​ ​ 2.2. Domain and Range of a Quadratic Domain of a Quadratic The domain of a quadratic is x ∈ R, unless it is restricted. The line x = − 2a b , or x = q (where q is the x coordinate ​ of the vertex), is the line of symmetry We can find a restricted domain such that the function has a One to One relation. Range of a Quadratic The range of a quadratic depends on its minimum or maximum y value. Example 2 Find the range of f (x) = −2x2 + x + 4, where x ∈ R. Consider a quadratic equation written in the form Answer ax2 + bx + c = 0 First we complete the square for f (x) We can complete the square for this equation to get the y - 1 2 33 f (x) = −2 (x − ) + coordinate of the vertex, or the minimum or maximum y 4 8 ​ ​ value. WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS As a < 0 and p < 0, the vertex is a maximum point. We can g(x) now find the range as f (x) = a + for x ∈ R ​ h(x) 33 ≥y Where a is a non-zero constant, and g(x) and h(x) are 8 ​ functions. The graph is shown below Then the range is f (x) < a or f (x) > a depending on if the function is approaching the value a from above or below the line y = a. This can be checked by inputting a large x value and seeing if the output is greater than or less than a. Note that the inequality can change if the domain changes. Example Find the range of 2 2 1 f (x) = + for x > 3 3(3x − 1) 3 ​ ​ ​ 2.3. Different types of Functions Answer Inputting in a large x value, such as 999, gives us Square root function 2 2 f (999) = + Consider a function f (x) = ax + b 3 3(3(999) − 1) ​ ​ ​ The domain must be chosen such that ax + b > 0 always. In this case the domain is x > −b Which is greater than 23 , so the function is approaching this a. ​ value from above. This gives us the range of the function as ​ The range of such a function is always y > 0 For a more general way of solving, consider a function 2 y> f (x) = g(x) [where g(x) is any function such as a 3 ​ ​ linear function, quadratic, or even a higher degree polynomial] 2.4. Inverse Function The domain must be chosen such that g(x) ≥ 0 always An inverse function reverses another function. The range of function is always a positive, non-zero, Commonly denoted by f −1 (x). interval. It is also important to note that f −1 (x) does not 1 imply f (x). Rational function ​ Inputting x into the function f outputs f (x) while (x) inputting f (x) into f −1 outputs x [also written as A rational function is a function written in the form fg(x) , f −1 (x)] ​ where g(x) = 0. ​ To find the domain Consider a function f and its inverse function f −1. These Find the values of x for which g(x) = 0. These values two functions can be represented as of x cannot be inputted into the function and must be excluded in the domain. To find the range Input a large x value, such as 999 and −999. This will help you find the y value which the rational function is approaching, but never reaches. This limits the range of the rational function. If the function is written in the form And WWW.ZNOTES.ORG Copyright © 2024 ZNotes Education & Foundation. All Rights Reserved. This document is authorised for personal use only by Aqsa at undefined on 07/12/24. CAIE AS LEVEL MATHEMATICS We can find the equation for an inverse function by “swapping” all the x variables to y , and y variables to x, and then making y the subject of the formula again. Example Given f (x) = 2x + 3, find the inverse of this function. Answer This gives us an important result: We start by swapping the x and y values f −1 ( f (x) ) = f ( f −1 (x) ) = x y = 2x + 3 ⟹ x = 2y + 3 An inverse function, f −1 , can only exist if the function f Now we can make y the subject of the formula has a One to One relation. To check if an inverse function exists, we can use the x−3 y= horizontal line test. 2 ​ So we get the inverse function as Domain and Range of an Inverse Function x−3 Using the previous result, f −1 ( f (x) ) = f ( f −1 (x) ) = x, f −1 (x) = 2 ​ we can find the domain and range of an inverse function: Function Domain Range Example 1 f (x) a