SL Analysis & Approaches Revision Notes

Unit 1 · Number & Algebra Unit 2 · Functions Unit 3 · Geometry & Trig Unit 4 · Stats & Probability Unit 5 · Calculus

Unit 1 · Number & Algebra

SL AA revision notes

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UNIT 1: NUMBER & ALGEBRA

1.1 Scientific Notation Scientific notation allows us to express very large or very small numbers compactly. It is foundational for estimation and standardizing values across different scales.

Scientific Notation Form $a \times 10^{k}$ Where $1 \leq a < 10$ and $k$ is an integer ( $k \in Z$ ).

Worked Example: Calculations with Scientific Notation Question: Given $x = 4.5 \times 10^{4}$ and $y = 3.0 \times 10^{5}$ , calculate the exact value of $x \times y$ . Give your answer in the form $a \times 10^{k}$ .
Solution:
Multiply the constants and add the exponents: $x \times y = (4.5 \times 10^{4}) \times (3.0 \times 10^{5}) = (4.5 \times 3.0) \times 10^{4 + 5} = 13.5 \times 10^{9}$ Since $13.5$ is not strictly between 1 and 10, adjust the decimal and the exponent: $= 1.35 \times 10^{10}$

1.2 & 1.3 Sequences and Series (Arithmetic & Geometric) A sequence is a list of numbers following a pattern. A series is the sum of those numbers.

Arithmetic Sequence: Changes by adding or subtracting a constant (common difference, $d$ ).
Geometric Sequence: Changes by multiplying by a constant (common ratio, $r$ ).

Sigma Notation ( $Σ$ ) is used to represent the sum of a sequence compactly: $\sum_{k = 1}^{n} u_{k}$ .

Sequences and Series Formulas Arithmetic:

$n^{th}$ term: $u_{n} = u_{1} + (n - 1) d$
Sum of $n$ terms: $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (u_{1} + u_{n})$

Geometric:

$n^{th}$ term: $u_{n} = u_{1} r^{n - 1}$
Sum of $n$ terms: $S_{n} = \frac{u_{1} (r^{n} - 1)}{r - 1} = \frac{u_{1} (1 - r^{n})}{1 - r}$ (where $r \neq 1$ )

Worked Example: Finding Geometric Terms and Sums Question: A geometric sequence has a 2nd term of 12 and a 5th term of 324. Find the common ratio $r$ and the sum of the first 8 terms.
Solution:
Write equations for the terms using the $n^{th}$ term formula: $u_{2} = u_{1} r = 12 and u_{5} = u_{1} r^{4} = 324$ Divide the larger equation by the smaller one to eliminate $u_{1}$ : $\frac{u_{1} r^{4}}{u_{1} r} = \frac{324}{12} ⟹ r^{3} = 27 ⟹ r = 3$ Substitute $r = 3$ back into the first equation to find $u_{1}$ : $u_{1} (3) = 12 ⟹ u_{1} = 4$ Calculate $S_{8}$ using the geometric series sum formula: $S_{8} = \frac{4 (3^{8} - 1)}{3 - 1} = \frac{4 (6561 - 1)}{2} = 2 (6560) = 13120$

Casio fx-CG50: Sigma Notation & Sequences

Sigma Sums: Run-Matrix $\to$ MATH (F4) $\to$ F6 $\to$ \Sigma( (F2). Enter the bounds and the formula to calculate exact sums.
Generating Sequences: MENU 8 (Recursion). Enter the explicit formula in $a_{n}$ or a recursive relationship. Press TABLE (F6) to view terms.

1.4 Financial Applications Compound interest models exponential growth in finance, where interest is earned on both the principal and previously accumulated interest. Depreciation works identically but with a negative interest rate.

Compound Interest Formula $F V = P V \times {(1 + \frac{r}{100 k})}^{k n}$ Where $F V$ = Future Value, $P V$ = Present Value, $r$ = nominal annual interest rate (%), $k$ = compounding periods per year, and $n$ = number of years.

Worked Example: Compound Interest Calculation Question: $$ 5000$ is invested at $4.2 %$ p.a., compounded monthly. Calculate its value after $5$ years.
Solution:
Identify variables: $P V = 5000$ , $r = 4.2$ , $k = 12$ , $n = 5$ . $F V = 5000 {(1 + \frac{4.2}{100 \times 12})}^{12 \times 5} = 5000 {(1 + \frac{4.2}{1200})}^{60} \approx $ 6166.52$

Casio fx-CG50: Financial App (TVM) Go to MENU C (Financial) $\to$ Compound Interest (F2).

n: Total compounding periods ( $k \times n$ )
I%: Annual interest rate ( $r$ )
PV: Present Value (enter as negative to represent money leaving your pocket)
P/Y and C/Y: Periods per year ( $k$ )

Press FV (F5) to instantly calculate the Future Value.

1.5 & 1.7 Exponents and Logarithms Logarithms are the inverse operations of exponents. They answer the question: "To what power must we raise the base to obtain a certain number?" When the base is Euler’s number ( $e$ ), we use the natural logarithm ( $\ln$ ).

Laws of Exponents and Logarithms

Exponents:

$a^{m} \times a^{n} = a^{m + n}$
$\frac{a^{m}}{a^{n}} = a^{m - n}$
$(a^{m})^{n} = a^{m n}$
$a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$

Logarithms:

$\log_{a} (x y) = \log_{a} x + \log_{a} y$
$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$
$\log_{a} (x^{m}) = m \log_{a} x$
Change of Base: $\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$
Definition: $a^{x} = b ⟺ x = \log_{a} b$

Worked Example: Solving Logarithmic Equations Question: Write the expression $3 \ln 2 - \ln 4$ in the form $\ln k$ , where $k \in Z$ . Hence, or otherwise, solve $3 \ln 2 - \ln 4 = - \ln x$ .
Solution:
Use the logarithm power law and subtraction law on the LHS: $3 \ln 2 - \ln 4 = \ln (2^{3}) - \ln 4 = \ln 8 - \ln 4 = \ln (\frac{8}{4}) = \ln 2$ Now, set it equal to the RHS ( $- \ln x$ ): $\ln 2 = - \ln x ⟹ \ln 2 = \ln (x^{- 1})$ Equating the arguments of the natural logarithm gives: $2 = x^{- 1} ⟹ 2 = \frac{1}{x} ⟹ x = \frac{1}{2}$

1.8 Infinite Convergent Geometric Series An infinite geometric series can only be summed if it is "convergent". This occurs when the terms get progressively smaller, specifically when the common ratio satisfies $| r | < 1$ .

Sum to Infinity $S_{\infty} = \frac{u_{1}}{1 - r} provided | r | < 1$

Worked Example: Evaluating an Infinite Series Question: Find the sum to infinity of the series $18 + 6 + 2 + \frac{2}{3} + \dots$
Solution:
Identify $u_{1}$ and calculate $r$ : $u_{1} = 18$ , $r = \frac{6}{18} = \frac{1}{3}$ .
Because $| \frac{1}{3} | < 1$ , the sum converges. $S_{\infty} = \frac{18}{1 - \frac{1}{3}} = \frac{18}{\frac{2}{3}} = 18 \times \frac{3}{2} = 27$

1.9 The Binomial Theorem The Binomial Theorem is an efficient method to expand algebraic expressions of the form $(a + b)^{n}$ without undertaking manual, repetitive polynomial expansion.

Binomial Theorem & General Term $(a + b)^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + b^{n}$ General Term Formula: $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ Where the binomial coefficient $(\binom{n}{r}) =^{n} C_{r} = \frac{n!}{r! (n - r)!}$

Worked Example: Finding a Specific Binomial Term Question: Find the term containing $x^{3}$ in the expansion of $(2 x - 5)^{5}$ .
Solution:
Use the general term formula where $n = 5$ , $a = 2 x$ , $b = - 5$ . $T_{r + 1} = (\binom{5}{r}) (2 x)^{5 - r} (- 5)^{r}$ For the $x^{3}$ term, we need the power of $x$ to be 3. Thus, $5 - r = 3 ⟹ r = 2$ . $Term = (\binom{5}{2}) (2 x)^{3} (- 5)^{2} = 10 \times 8 x^{3} \times 25 = 2000 x^{3}$

Casio fx-CG50: Binomial Coefficients ( $n C r$ ) To find binomial coefficients quickly (e.g., $(\binom{5}{2})$ ), go to Run-Matrix. Press OPTN $\to$ PROB (F6) $\to$ nCr (F3). Type 5 C 2 and hit EXE. The calculator will display $10$ .

1.6 Simple Deductive Proof A deductive proof uses logical algebraic steps to prove a mathematical statement holds true universally (working clearly from LHS to RHS).

Even numbers: $2 n$ ( $n \in Z$ )
Odd numbers: $2 n + 1$ or $2 n - 1$
Consecutive integers: $n, n + 1, n + 2$

Worked Example: Algebraic Deductive Proof Question: Prove deductively that the sum of any three consecutive integers is always a multiple of 3.
Proof:
Let the first integer be $n$ , where $n \in Z$ .
The next two consecutive integers are $n + 1$ and $n + 2$ .
Find their sum algebraically: $Sum = n + (n + 1) + (n + 2) = 3 n + 3$ Factorise to show the multiple: $Sum = 3 (n + 1)$ Because $(n + 1)$ is an integer, the expression $3 (n + 1)$ is exactly divisible by 3.
Hence, the sum of any three consecutive integers is always a multiple of 3. $◼$

Practice Unit 1 questions → See SL AA plans

Unit 2 · Functions

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UNIT 2: FUNCTIONS

2.1 The Concept of a Function, Domain, and Range A function is a mathematical relationship assigning exactly one output ( $y$ ) to each valid input ( $x$ ).

Vertical Line Test: A graph represents a function if and only if no vertical line intersects the curve more than once.
Domain: The set of all possible input values ( $x$ ).
Range: The set of all possible resulting output values ( $y$ ).

Worked Example: Determining Domain and Range Question: Determine the domain and range of $f (x) = \sqrt{x - 3} + 2$ .
Solution:
For a real-valued square root, the expression inside the root must be non-negative: $x - 3 \geq 0 ⟹ x \geq 3$ Domain: ${x ∣ x \geq 3, x \in R}$ .
Since $\sqrt{x - 3} \geq 0$ , the minimum value is $0 + 2 = 2$ : $f (x) \geq 0 + 2 ⟹ f (x) \geq 2$ Range: ${y ∣ y \geq 2, y \in R}$ .

2.2 & 2.5 Inverse Functions An inverse function $f^{- 1} (x)$ reverses the effect of the original function $f (x)$ . It only exists for one-to-one functions (those passing a horizontal line test). Graphically, $y = f^{- 1} (x)$ is a direct reflection of $y = f (x)$ across the line $y = x$ .

Inverse Function Identity $(f \circ f^{- 1}) (x) = (f^{- 1} \circ f) (x) = x$ (The composite of a function and its inverse always yields the identity function, $x$ ).

Worked Example: Finding an Inverse Function Algebraically Question: Given $f (x) = \frac{3 x - 1}{x + 2}$ for $x \neq - 2$ , find $f^{- 1} (x)$ .
Solution:
1. Replace $f (x)$ with $y$ : $y = \frac{3 x - 1}{x + 2}$
2. Interchange $x$ and $y$ : $x = \frac{3 y - 1}{y + 2}$
3. Rearrange to solve for $y$ : $\begin{aligned} x (y + 2) & = 3 y - 1 \\ x y + 2 x & = 3 y - 1 \\ x y - 3 y & = - 2 x - 1 \\ y (x - 3) & = - (2 x + 1) \\ y & = \frac{- (2 x + 1)}{x - 3} = \frac{2 x + 1}{3 - x} \end{aligned}$ Thus, $f^{- 1} (x) = \frac{2 x + 1}{3 - x}$ for $x \neq 3$ .

2.3 & 2.4 Key Features & Sign Diagrams Sign diagrams provide a rapid visual way to determine when a function is positive, negative, zero, or undefined without drawing a full to-scale graph.

Zeros (Roots): Solid vertical line markings.
Undefined Asymptotes: Dashed vertical line markings.

GDC Guide: Casio fx-CG50 - Analyzing Graphs

Graphing: Go to MENU 5 (Graph), enter functions, and press DRAW (F6). Use V-Window (SHIFT F3) to adjust the axes.
Key Features: Press G-Solv (SHIFT F5).
- ROOT (F1) for $x$ -intercepts.
- MAX (F2) or MIN (F3) for turning points.

2.7 Quadratic Functions and The Discriminant The discriminant ( $Δ = b^{2} - 4 a c$ ) determines the geometric nature of the roots of a parabola $y = a x^{2} + b x + c$ .

Worked Example: Using the Discriminant Question: Find the values of $k$ for which $k x^{2} + 4 x + (k - 3) = 0$ has two equal real roots.
Solution:
For two equal real roots, the discriminant must be exactly zero: $Δ = 0$ . $\begin{aligned} Δ = (4)^{2} - 4 (k) (k - 3) & = 0 \\ 16 - 4 k^{2} + 12 k & = 0 \\ - 4 (k^{2} - 3 k - 4) & = 0 \\ k^{2} - 3 k - 4 & = 0 \\ (k - 4) (k + 1) & = 0 ⟹ k = 4 or k = - 1 \end{aligned}$

2.6 Rational Functions and Asymptotes Rational functions $f (x) = \frac{a x + b}{c x + d}$ form hyperbolic curves broken by a Vertical Asymptote (where denominator $c x + d = 0$ ) and a Horizontal Asymptote (the ratio $y = \frac{a}{c}$ ).

2.9 Exponential and Logarithmic Functions Exponential functions ( $a^{x}, e^{x}$ ) and Logarithmic functions ( $\log_{a} x, \ln x$ ) are inverses of each other. The graph of an exponential function has a horizontal asymptote, while a logarithmic function has a vertical asymptote.

2.8 Transformations of Graphs Functions can be translated (shifted) or stretched. For example, $y = f (x - c) + d$ translates the graph right by $c$ and up by $d$ .

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Unit 3 · Geometry & Trig

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UNIT 3: GEOMETRY & TRIGONOMETRY

3.1 Coordinate Geometry & 3D Space The concepts of distance and midpoint in 2D coordinate geometry extend naturally into 3-dimensional space using $x, y,$ and $z$ axes. Working with volumes and surface areas of complex 3D shapes (cones, spheres, pyramids) is crucial for optimization problems.

3D Distance, Midpoint & Volumes Given two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ :

Distance: $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$
Midpoint: $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$

Volumes of Solids:

Right-Pyramid/Cone: $V = \frac{1}{3} \times base area \times height$
Sphere: $V = \frac{4}{3} π r^{3}$ , Surface Area $A = 4 π r^{2}$

Worked Example: 3D Coordinate Geometry Question: Two points $A (2, 1, 3)$ and $B (5, 2, 6)$ are located on an $x y z$ coordinate grid. Find the exact length of the line segment $A B$ and the coordinates of its midpoint.
Solution:

Use the 3D distance formula and expand: $\begin{aligned} A B & = \sqrt{(5 - 2)^{2} + (2 - 1)^{2} + (6 - 3)^{2}} \\ = \sqrt{(3)^{2} + (1)^{2} + (3)^{2}} \\ = \sqrt{9 + 1 + 9} = \sqrt{19} units \end{aligned}$ Use the midpoint formula: $M = (\frac{2 + 5}{2}, \frac{1 + 2}{2}, \frac{3 + 6}{2}) = (3.5, 1.5, 4.5)$

3.2 & 3.3 Non-Right Angled Trigonometry To solve triangles that do not possess a right angle, we rely on the Sine Rule and the Cosine Rule.

Use the Sine Rule when you have a "matching pair" (an angle and its known opposite side).
Use the Cosine Rule when you have 3 sides (SSS) or 2 sides and the included angle (SAS).

Sine Rule, Cosine Rule & Area of a Triangle

Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Cosine Rule (for sides): $c^{2} = a^{2} + b^{2} - 2 a b \cos C$
Cosine Rule (for angles): $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
Area of a Triangle: $Area = \frac{1}{2} a b \sin C$

Worked Example: Applying the Sine Rule Question: In quadrilateral $A B C D$ , it is given that $A B = 11 cm$ , $∠ B A D = 59^{\circ}$ , and $∠ A D B = 100^{\circ}$ . Find the length of the diagonal $D B$ .
Solution:

In $△ A B D$ , we know two angles and one side. We can find $D B$ using the Sine Rule because the side $A B$ is opposite $∠ A D B$ and the side $D B$ is opposite $∠ B A D$ . $\begin{aligned} \frac{D B}{\sin (∠ B A D)} & = \frac{A B}{\sin (∠ A D B)} \\ \frac{D B}{\sin 59^{\circ}} & = \frac{11}{\sin 100^{\circ}} \\ D B & = \frac{11 \times \sin 59^{\circ}}{\sin 100^{\circ}} \\ D B & \approx 9.57 cm (3 s.f.) \end{aligned}$

3.4 Radian Measure, Arcs and Sectors A radian is a standard unit of angular measure based on the radius of a circle. There are $2 π$ radians in a full circle ( $360^{\circ}$ ).

To convert from degrees to radians, multiply by $\frac{π}{180}$ .
To convert from radians to degrees, multiply by $\frac{180}{π}$ .

Arc Length and Sector Area (Radians) When an angle $θ$ is measured strictly in radians:

Arc Length: $l = r θ$
Area of a Sector: $A = \frac{1}{2} r^{2} θ$

Worked Example: Radians and Sectors Question: A circle with centre $O$ has radius $4 cm$ . The points $P, Q,$ and $R$ lie on the circumference such that the arc length $P Q R$ is $10 cm$ . Find the angle $P \hat{O} R$ in radians, and calculate the exact area of the shaded sector.
Solution:

Use the arc length formula to find the angle $θ$ : $\begin{aligned} l & = r θ \\ 10 & = 4 θ \\ θ & = 2.5 radians \end{aligned}$ Now substitute $θ$ into the area formula: $\begin{aligned} A & = \frac{1}{2} r^{2} θ \\ A & = \frac{1}{2} (4)^{2} (2.5) \\ A & = \frac{1}{2} (16) (2.5) = 20 {cm}^{2} \end{aligned}$

3.5 & 3.6 The Unit Circle and Trigonometric Identities The Unit Circle is defined on the Cartesian plane with a radius of exactly 1. For any point $(x, y)$ on the unit circle at an angle $θ$ from the positive $x$ -axis, the coordinates define the trigonometric ratios: $x = \cos θ$ and $y = \sin θ$ .

Key Trigonometric Identities

Pythagorean Identity: $\cos^{2} θ + \sin^{2} θ = 1$
Tangent Definition: $\tan θ = \frac{\sin θ}{\cos θ}$
Double Angle (Sine): $\sin 2 θ = 2 \sin θ \cos θ$
Double Angle (Cosine): $\begin{aligned} \cos 2 θ & = \cos^{2} θ - \sin^{2} θ \\ = 1 - 2 \sin^{2} θ \\ = 2 \cos^{2} θ - 1 \end{aligned}$

Worked Example: Using Double Angle Identities Question: Given that $\sin θ = \frac{3}{5}$ and $θ$ is acute, find the exact value of $\sin 2 θ$ .
Solution:

First, find $\cos θ$ using the Pythagorean Identity: $\begin{aligned} \cos^{2} θ + {(\frac{3}{5})}^{2} & = 1 \\ \cos^{2} θ & = 1 - \frac{9}{25} = \frac{16}{25} \end{aligned}$ Since $θ$ is acute (1st quadrant), $\cos θ$ is positive: $\cos θ = \frac{4}{5}$ .
Substitute into the double angle formula: $\begin{aligned} \sin 2 θ & = 2 \sin θ \cos θ \\ = 2 (\frac{3}{5}) (\frac{4}{5}) = \frac{24}{25} \end{aligned}$

3.7 & 3.8 Trigonometric Functions and Graphs Trigonometric functions model periodic phenomena (e.g., tides, springs, pendulums). The general transformation form is: $f (x) = A \sin (B (x - C)) + D$

Amplitude ( $| A |$ ): Half the distance between the maximum and minimum values.
Period ( $\frac{2 π}{B}$ or $\frac{360^{\circ}}{B}$ ): The length of one complete wave cycle.
Phase Shift ( $C$ ): The horizontal translation.
Principal Axis ( $y = D$ ): The vertical translation or center line.

GDC Guide: Casio fx-CG50 - Radian/Degree Mode & Graphing

Check Your Mode: Press SHIFT $\to$ MENU (SET UP). Scroll down to Angle and select Rad (F2) or Deg (F1). Always check this before an exam!
Viewing Windows: In MENU 5 (Graph), ensure your V-Window (SHIFT F3) matches your angle mode.
- If in Radians: Set Xmin to $0$ and Xmax to $2 π$ .
- If in Degrees: Set Xmin to $0$ and Xmax to $360$ .
Finding Features: Press G-Solv (SHIFT F5). Use MAX/MIN to find the crests and troughs of the wave (useful for finding Amplitude $A$ and Principal Axis $D$ ).

Worked Example: Identifying Features of Trig Graphs Question: The depth of water in a harbour is modelled by $D (t) = 3 \cos (\frac{π}{6} t) + 5$ , where $t$ is the time in hours after midnight. Find the maximum depth, minimum depth, and the time it takes for one complete tidal cycle (the period).
Solution:

Identify the constants: $A = 3$ , $B = \frac{π}{6}$ , $D = 5$ . $\begin{aligned} Max Depth & = D + A = 5 + 3 = 8 m \\ Min Depth & = D - A = 5 - 3 = 2 m \end{aligned}$ The period is calculated using $B$ : $\begin{aligned} Period & = \frac{2 π}{B} \\ = \frac{2 π}{(\frac{π}{6})} = 12 hours \end{aligned}$

3.8 Solving Trigonometric Equations Trigonometric equations often have infinite solutions due to the periodic nature of the waves. Exams will specify a bounded interval (e.g., $0 \leq x \leq 2 π$ ) for which you must find all valid solutions. Use technology (GDC) to efficiently find intersections unless an analytical/exact answer is specifically requested.

GDC Guide: Casio fx-CG50 - Solving Trig Equations To solve a complex equation like $2 \sin (2 x) = \cos (x)$ for $0 \leq x \leq 2 π$ :

Verify your calculator is in Radian mode (SHIFT -> MENU).
Go to MENU 5 (Graph).
Enter the left side of the equation into Y1: ‘2sin(2x)‘
Enter the right side of the equation into Y2: ‘cos(x)‘
Press SHIFT -> F3 (V-Window) and set Xmin = 0 and Xmax = 2\pi. Set Ymin = -3 and Ymax = 3 to ensure you can see the intersections clearly.
Press DRAW (F6).
Press G-Solv (SHIFT F5) $\to$ ISCT (F5).
The calculator will jump to the first intersection point. Note the $x$ -value.
Press the right directional arrow on the D-pad to jump to the next valid intersection within your domain. Repeat this until no more intersections are found.

Worked Example: Solving Trig Equations Graphically Question: Solve the equation $e^{\sin x} = x^{2}$ for $0 \leq x \leq π$ . Give your answers to 3 significant figures.
Solution:
Because this mixes an exponential trigonometric function with a polynomial, it cannot be solved analytically. We must use a GDC intersection method. $\begin{aligned} Let Y_{1} & = e^{\sin x} \\ Let Y_{2} & = x^{2} \end{aligned}$ Set the GDC to Radian mode. Set the V-Window domain to $X m i n = 0$ and $X m a x = π$ . Draw both graphs and use the ISCT tool.

The graphs intersect at exactly two points in this interval. Thus, the solutions are: $x = 1.15 and x = 1.70 (3 s.f.)$

Practice Unit 3 questions → See SL AA plans

Unit 4 · Stats & Probability

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UNIT 4: STATISTICS & PROBABILITY

4.1 - 4.3 Descriptive Statistics & Outliers Data is classified as discrete (counted, exact values) or continuous (measured, given in intervals). We analyse data using measures of central tendency (mean, median, mode) and dispersion (variance, standard deviation, interquartile range). Box and whisker diagrams visually display the five-number summary and flag mathematical outliers.

Outlier Boundaries & Spread An outlier is defined mathematically as any data value that lies $1.5 \times IQR$ below the lower quartile ( $Q_{1}$ ) or above the upper quartile ( $Q_{3}$ ).

Interquartile Range (IQR): $Q_{3} - Q_{1}$
Lower Outlier Boundary: $Q_{1} - 1.5 \times IQR$
Upper Outlier Boundary: $Q_{3} + 1.5 \times IQR$

GDC Guide: Casio fx-CG50 - 1-Variable Statistics

Go to MENU 2 (Stat). Enter your data into List 1 (and frequencies into List 2 if grouped).
Press CALC (F2) $\to$ SET (F6). Ensure 1Var XList is List 1 and 1Var Freq is 1 (or List 2 if using frequencies).
Press 1-VAR (F1). Scroll to find $\bar{x}$ (mean), $σ x$ (population standard deviation), $M e d$ (median), and the quartiles. Square $σ x$ manually to find the variance.

Worked Example: Calculating Outliers Question: A dataset of student test scores has a lower quartile ( $Q_{1}$ ) of $42$ and an upper quartile ( $Q_{3}$ ) of $66$ . Determine the outlier boundaries and state whether a score of $15$ is an outlier.
Solution:

Casio fx-CG50 Method:
To find quartiles quickly from raw data, go to MENU 2 (Stat), enter data in List 1, press CALC $\to$ 1-VAR and scroll down to $Q_{1}$ and $Q_{3}$ .
Calculate the Interquartile Range (IQR): $\begin{aligned} IQR & = Q_{3} - Q_{1} = 66 - 42 = 24 \end{aligned}$ Calculate the boundaries: $\begin{aligned} Lower & = 42 - 1.5 (24) = 42 - 36 = 6 \\ Upper & = 66 + 1.5 (24) = 66 + 36 = 102 \end{aligned}$ Since $15 \geq 6$ , it lies within the boundary. Therefore, $15$ is not an outlier.

4.4 & 4.10 Bivariate Data and Linear Regression When exploring the relationship between two variables, we plot a scatter diagram.

Pearson’s correlation coefficient ( $r$ ): Measures the strength and direction of the linear relationship ( $- 1 \leq r \leq 1$ ). It is only meaningful for linear relationships.
Regression Line ( $y = a x + b$ ): The line of best fit. It always passes through the mean point $(\bar{x}, \bar{y})$ . It can be used for interpolation (reliable predictions within the data range) but is unreliable for extrapolation (predicting outside the data range).

GDC Guide: Casio fx-CG50 - Linear Regression & Pearson’s $r$

Enter your $x$ -values into List 1 and $y$ -values into List 2 via MENU 2 (Stat).
Press CALC (F2) $\to$ REG (F3) $\to$ X (F1) $\to$ ax+b (F1).
The screen will display the gradient ( $a$ ), the $y$ -intercept ( $b$ ), and Pearson’s correlation coefficient ( $r$ ).

Worked Example: Linear Regression & Predictions Question: The regression line of $y$ on $x$ for a set of bivariate data is $y = 1.2 x + 3.4$ . The mean of the $x$ -values is $\bar{x} = 5$ . Find the mean of the $y$ -values, $\bar{y}$ . Predict the value of $y$ when $x = 10$ , and state whether this prediction is reliable given the original $x$ -values range from $2$ to $8$ .
Solution:

Casio fx-CG50 Method:
If given raw data, use MENU 2 $\to$ CALC $\to$ REG $\to$ X to find the equation. Here it is given.
The regression line always passes exactly through the mean point $(\bar{x}, \bar{y})$ . Substitute $\bar{x}$ : $\begin{aligned} \bar{y} & = 1.2 (\bar{x}) + 3.4 \\ \bar{y} & = 1.2 (5) + 3.4 = 9.4 \end{aligned}$ Substitute $x = 10$ into the equation to predict $y$ : $y = 1.2 (10) + 3.4 = 15.4$ This prediction is unreliable because $x = 10$ lies outside the original data range ( $2 \leq x \leq 8$ ), making it an extrapolation.

4.5 & 4.6 Probability Rules and Events Probability calculates the likelihood of events. You must confidently navigate combined events using Venn diagrams, tree diagrams, and algebraic formulas.

Probability Formulas

Combined Events: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Mutually Exclusive Events: Cannot happen at the same time. $P (A \cap B) = 0$
Conditional Probability: The probability of $A$ given that $B$ has already occurred: $P (A | B) = \frac{P (A \cap B)}{P (B)} ⟹ P (A \cap B) = P (B) P (A | B)$
Independent Events: The outcome of one does not affect the other. $P (A \cap B) = P (A) P (B)$

Worked Example: Conditional Probability (2-Set Venn) Question: Events $A$ and $B$ are such that $P (A) = 0.4$ , $P (A | B) = 0.25$ , and $P (A \cup B) = 0.55$ . Find $P (B)$ .
Solution:

Use the conditional probability formula to link the intersection to $P (B)$ : $\begin{aligned} P (A | B) & = \frac{P (A \cap B)}{P (B)} \\ 0.25 & = \frac{P (A \cap B)}{P (B)} \\ P (A \cap B) & = 0.25 P (B) \end{aligned}$ Now substitute this into the combined events formula: $\begin{aligned} P (A \cup B) & = P (A) + P (B) - P (A \cap B) \\ 0.55 & = 0.4 + P (B) - 0.25 P (B) \\ 0.15 & = 0.75 P (B) \\ P (B) & = \frac{0.15}{0.75} = 0.2 \end{aligned}$

4.6 Extended Probability & 3-Set Venn Diagrams When dealing with three intersecting events in a sample space $U$ , drawing a 3-sectioned Venn diagram is often the most efficient way to solve the problem without relying purely on complex formulas. Always start by filling in the center intersection (all three events) and work your way outwards.

Worked Example: Three-Sectioned Venn Diagram Question: In a team of 30 judo players, 13 have won a match by throwing ( $T$ ), 12 have won by hold-down ( $H$ ), and 13 have won by points decision ( $P$ ). 2 have won matches by all three methods. 5 have won matches by throwing and hold-down. 4 have won matches by hold-down and points decision. 3 have won matches by throwing and points decision. Find the probability that a randomly selected player has won a match by exactly one method.
Solution:

Start from the center ( $T \cap H \cap P = 2$ ) and subtract this from the two-event intersections:

$T$ and $H$ only: $5 - 2 = 3$
$H$ and $P$ only: $4 - 2 = 2$
$T$ and $P$ only: $3 - 2 = 1$

Now find the players who only won by one method:

Only $T$ : $13 - (3 + 1 + 2) = 7$
Only $H$ : $12 - (3 + 2 + 2) = 5$
Only $P$ : $13 - (1 + 2 + 2) = 8$

The number of players who won by exactly one method is $7 + 5 + 8 = 20$ . $P (exactly one) = \frac{20}{30} = \frac{2}{3}$

4.7 Discrete Random Variables A discrete random variable ( $X$ ) has specific, countable outcomes, each with an associated probability. The sum of all probabilities in the sample space must equal exactly $1$ .

Expected Value $E (X)$ The expected value is the theoretical long-run average (mean) of the distribution. $E (X) = \sum x P (X = x)$ Note: In the context of gambling, if $E (X) = 0$ , the game is considered mathematically "fair".

Worked Example: Expected Value and Unknown Constants Question: A discrete random variable $X$ has a probability distribution given by $P (X = n) = \frac{k}{n}$ , where $n \in {1, 2, 3, 4}$ and $k$ is a positive constant. Find $k$ , and calculate $E (X)$ .
Solution:

The sum of all probabilities must equal $1$ . $\begin{aligned} \frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} & = 1 \\ k (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) & = 1 \\ k (\frac{12 + 6 + 4 + 3}{12}) & = 1 \\ k (\frac{25}{12}) & = 1 ⟹ k = \frac{12}{25} = 0.48 \end{aligned}$ Calculate Expected Value $E (X)$ : $\begin{aligned} E (X) & = \sum x P (X = x) \\ = 1 (\frac{k}{1}) + 2 (\frac{k}{2}) + 3 (\frac{k}{3}) + 4 (\frac{k}{4}) \\ = k + k + k + k = 4 k \\ E (X) & = 4 (0.48) = 1.92 \end{aligned}$

4.8 The Binomial Distribution The binomial distribution models situations with a fixed number of independent trials ( $n$ ), where each trial has only two possible outcomes (Success or Failure) and a constant probability of success ( $p$ ). We write this as $X \sim B (n, p)$ .

Binomial Mean and Variance If $X \sim B (n, p)$ , then:

Expected Value (Mean): $E (X) = n p$
Variance: $Var (X) = n p (1 - p)$

GDC Guide: Casio fx-CG50 - Binomial Probabilities Go to MENU 2 (Stat) $\to$ DIST (F5) $\to$ BINOMIAL (F5).

Bpd (F1): Use for exact probabilities, e.g., $P (X = 4)$ . Set Data to Variable, enter $x$ (successes), $N u m t r i a l$ ( $n$ ), and $p$ .
Bcd (F2): Use for cumulative probabilities, e.g., $P (X \leq 4)$ . Enter Lower bound, Upper bound, $N u m t r i a l$ , and $p$ . Note: For "at least 3" ( $P (X \geq 3)$ ), set Lower = 3 and Upper = $n$ .

Worked Example: Binomial Probabilities Question: An archer has a $90 %$ chance of hitting a target with each independent arrow. If $5$ arrows are fired, find the probability that she hits the target exactly $4$ times, and the probability she hits it at least $3$ times.
Solution:

Let $X$ be the number of hits. The distribution is binomial: $X \sim B (n = 5, p = 0.9)$ .
Casio fx-CG50 Method - Exact Probability:
To find $P (X = 4)$ , go to MENU 2 $\to$ DIST $\to$ BINOMIAL $\to$ Bpd. Set Data: Variable, x: 4, Numtrial: 5, p: 0.9. $P (X = 4) = 0.328 (3 s.f.)$ Casio fx-CG50 Method - Cumulative Probability:
To find "at least 3" ( $P (X \geq 3)$ ), use Bcd. Set Data: Variable, Lower: 3, Upper: 5, Numtrial: 5, p: 0.9. $P (X \geq 3) = 0.991 (3 s.f.)$

4.9 & 4.12 The Normal Distribution The normal distribution models continuous, naturally occurring data (e.g., heights, weights) in a symmetrical, bell-shaped curve defined by the population mean ( $μ$ ) and standard deviation ( $σ$ ). We write this as $X \sim N (μ, σ^{2})$ .

Standardization ( $Z$ -values) To compare different distributions or find unknown means/standard deviations, we transform the variable $X$ into the standardized normal variable $Z \sim N (0, 1)$ . The $Z$ -value represents the number of standard deviations from the mean. $Z = \frac{x - μ}{σ}$

GDC Guide: Casio fx-CG50 - Normal Distribution Go to MENU 2 (Stat) $\to$ DIST (F5) $\to$ NORM (F1).

Ncd (F2): Used to find the probability (area) between bounds. Enter Lower, Upper, $σ$ , and $μ$ . If the boundary is infinity, use a massive number like $10^{99}$ or $- 10^{99}$ .
InvN (F3): Used when you already know the area (probability) and need to find the boundary $x$ -value. Set Data to Variable and Tail to Left, Right, or Central based on where the shaded region is.

Worked Example: Normal Probability Calculations Question: Apples from a grower’s crop are normally distributed with a mean weight of $173 g$ and a standard deviation of $34 g$ . Apples weighing less than $130 g$ are considered too small to sell. Find the proportion of apples that are too small to sell.
Solution:

Let $X$ be the weight of an apple.
The distribution is continuous: $X \sim N (173, 34^{2})$ .
We want to find the probability that $X < 130$ . $P (X < 130)$ Casio fx-CG50 Method:
Go to MENU 2 $\to$ DIST $\to$ NORM $\to$ Ncd.
Set Lower: $- 10^{99}$ (representing negative infinity), Upper: $130$ , $σ : 34$ , $μ : 173$ . $\begin{aligned} P (X < 130) & \approx 0.103 (3 s.f.) \\ = 10.3 % \end{aligned}$

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Unit 5 · Calculus

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UNIT 5: CALCULUS

5.1 & 5.6 Limits and Differentiation Rules Differentiation calculates the instantaneous rate of change (gradient) of a function at a specific point. For polynomial terms, we use the power rule. For composite, multiplied, or divided functions, we must apply the Chain, Product, and Quotient rules.

Rules of Differentiation

Power Rule: If $f (x) = a x^{n}$ , then $f^{'} (x) = a n x^{n - 1}$ .
Chain Rule: $y = g (u)$ and $u = h (x) ⟹ \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ .
Product Rule: $y = u v ⟹ y^{'} = u v^{'} + v u^{'}$ .
Quotient Rule: $y = \frac{u}{v} ⟹ y^{'} = \frac{v u^{'} - u v^{'}}{v^{2}}$ .
Standard Derivatives: $\frac{d}{d x} (\sin x) = \cos x$ , $\frac{d}{d x} (\cos x) = - \sin x$ , $\frac{d}{d x} (e^{x}) = e^{x}$ , $\frac{d}{d x} (\ln x) = \frac{1}{x}$ .

GDC Guide: Casio fx-CG50 - Evaluating Gradients at a Point You can evaluate exact derivatives at a specific $x$ -value without doing algebra:

Go to MENU 1 (Run-Matrix).
Press OPTN $\to$ CALC (F4) $\to$ d/dx (F2).
Type the function and the specific $x$ -value to instantly find the gradient.

Worked Example: Applying Differentiation Rules Question: Given $f (x) = 3 x^{4} - 5 x^{2} + 2 x$ , find the gradient of the curve at $x = 2$ .
Solution:

Using the power rule: multiply by the power and subtract one from the power. $\begin{aligned} f^{'} (x) & = 3 (4) x^{3} - 5 (2) x^{1} + 2 (1) x^{0} \\ f^{'} (x) & = 12 x^{3} - 10 x + 2 \end{aligned}$ Substitute $x = 2$ into the derivative function: $\begin{aligned} f^{'} (2) & = 12 (2)^{3} - 10 (2) + 2 \\ = 12 (8) - 20 + 2 \\ = 96 - 20 + 2 = 78 \end{aligned}$

Casio fx-CG50 Check:
In Run-Matrix, press OPTN $\to$ CALC $\to$ d/dx.
Enter: d/dx(3x^4 - 5x^2 + 2x, 2)
The GDC will return exactly 78.

5.4 Tangents and Normals A tangent is a straight line that touches a curve at a single point, possessing the exact same gradient as the curve at that point ( $m = f^{'} (x_{1})$ ). A normal is a straight line that intersects the curve at that point but is completely perpendicular to the tangent ( $m_{T} \times m_{N} = - 1$ ).

Equations of Lines To find the equation of a tangent or normal line, you need a point $(x_{1}, y_{1})$ and a gradient $m$ . Substitute these into the point-gradient formula: $y - y_{1} = m (x - x_{1})$ Remember: Gradient of Normal = $- \frac{1}{m_{T}}$ .

GDC Guide: Casio fx-CG50 - Finding Tangent/Normal Equations

Go to MENU 5 (Graph), enter your function, and press DRAW (F6).
Press Sketch (F4). Choose Tang (F2) for the tangent or Norm (F3) for the normal.
Type the $x$ -coordinate using the keypad and press EXE. The line will be drawn and the exact equation ( $y = m x + c$ ) will appear on screen.

Worked Example: Finding a Normal Equation Question: Find the equation of the normal to the curve $f (x) = e^{2 x}$ at the point where $x = 0$ .
Solution:

1. Find the $y$ -coordinate: $f (0) = e^{2 (0)} = e^{0} = 1 ⟹ Point is (0, 1)$ 2. Find the gradient of the curve ( $m_{T}$ ): $\begin{aligned} f^{'} (x) & = 2 e^{2 x} (using chain rule) \\ f^{'} (0) & = 2 e^{0} = 2 ⟹ m_{T} = 2 \end{aligned}$ 3. Find the gradient of the normal ( $m_{N}$ ): $m_{N} = - \frac{1}{2}$ 4. Substitute into the line equation: $\begin{aligned} y - 1 & = - \frac{1}{2} (x - 0) \\ y & = - \frac{1}{2} x + 1 \end{aligned}$

GDC Check:
Use Sketch \rightarrow Norm at $x = 0$ . GDC reads $Y = - 0.5 X + 1$ .

5.7 & 5.8 Curve Sketching, Stationary Points & Optimization Stationary points (turning points) occur when $f^{'} (x) = 0$ . We use the second derivative $f^{″} (x)$ to classify these points:

Local Maximum: $f^{″} (x) < 0$ (Concave down).
Local Minimum: $f^{″} (x) > 0$ (Concave up).
Point of Inflexion: Occurs when $f^{″} (x) = 0$ AND changes sign.

Finding turning points allows us to solve Optimization (maximum/minimum) problems in real-world contexts like volume, area, and profit.

GDC Guide: Casio fx-CG50 - Solving $f^{'} (x) = 0$ for Turning Points If $f^{'} (x)$ is a quadratic, you can instantly find its roots (which are the $x$ -values of the turning points) using the built-in Polynomial solver:

Go to MENU A (Equation) $\to$ Polynomial (F2) $\to$ Degree 2 (F1).
Enter your $a$ , $b$ , and $c$ coefficients and press EXE to solve for $x$ .

Worked Example: Optimization and Turning Points Question: A company’s daily profit is modeled by $P (x) = - 0.1 x^{3} + 12 x^{2} - 60 x + 100$ , where $x$ is the number of items sold. Find the number of items they should sell to maximize their profit, and prove it is a maximum.
Solution:

1. Find $P^{'} (x)$ and set it to zero: $\begin{aligned} P^{'} (x) & = - 0.3 x^{2} + 24 x - 60 \\ 0 & = - 0.3 x^{2} + 24 x - 60 \end{aligned}$ Casio fx-CG50 Method:
Use Equation mode $\to$ Polynomial. Enter $a = - 0.3, b = 24, c = - 60$ .
This gives $x = 77.4$ or $x = 2.58$ . Since they must sell whole items, let’s test $x \approx 77$ .
2. Use the second derivative to test for a maximum: $\begin{aligned} P^{″} (x) & = - 0.6 x + 24 \\ P^{″} (77.4) & = - 0.6 (77.4) + 24 \\ = - 22.44 \end{aligned}$ Since $P^{″} (x) < 0$ , the curve is concave down, confirming this is a maximum.

5.5, 5.10 & 5.11 Integration and Area Integration is the reverse process of differentiation (anti-differentiation).

Indefinite Integrals ( $\int f (x) d x$ ): Result in a family of curves, requiring a constant of integration ( $+ C$ ). If given a coordinate, you can solve for $C$ .
Definite Integrals ( $\int_{a}^{b} f (x) d x$ ): Result in a numerical value. Geometrically, this calculates the exact area enclosed between the curve $y = f (x)$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ .

Key Integration Rules

Reverse Power Rule: $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$
Special Functions: $\int \frac{1}{x} d x = \ln | x | + C$ , $\int e^{x} d x = e^{x} + C$
Trig Functions: $\int \sin x d x = - \cos x + C$ , $\int \cos x d x = \sin x + C$
Reverse Chain Rule (Linear): $\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$

GDC Guide: Casio fx-CG50 - Definite Integrals and Area

Run-Matrix Mode: Press MATH (F4) $\to$ \int dx (F6 \rightarrow F1). Fill in the function and the lower/upper bounds to instantly calculate the exact area/value.
Graph Mode: Draw the curve. Press G-Solv (SHIFT F5) $\to$ \int dx (F6 \rightarrow F3). Enter the lower bound, press EXE, enter the upper bound, press EXE. It will visibly shade the area!

Worked Example: Indefinite Integration & Finding $C$ Question: The gradient function of a curve is given by $f^{'} (x) = 6 x^{2} - 4 x + 3$ . Given that the curve passes through the point $(1, 5)$ , find the full equation of the curve $f (x)$ .
Solution:

Integrate $f^{'} (x)$ to find the general equation for $f (x)$ : $\begin{aligned} f (x) & = \int (6 x^{2} - 4 x + 3) d x \\ = \frac{6 x^{3}}{3} - \frac{4 x^{2}}{2} + 3 x + C \\ = 2 x^{3} - 2 x^{2} + 3 x + C \end{aligned}$ Substitute the known point $(x = 1, y = 5)$ to solve for $C$ : $\begin{aligned} 5 & = 2 (1)^{3} - 2 (1)^{2} + 3 (1) + C \\ 5 & = 2 - 2 + 3 + C \\ 5 & = 3 + C ⟹ C = 2 \end{aligned}$ The full equation is $f (x) = 2 x^{3} - 2 x^{2} + 3 x + 2$ .

5.11 Kinematics (Motion in a Straight Line) Calculus connects displacement ( $s$ ), velocity ( $v$ ), and acceleration ( $a$ ).

Differentiating: $s (t) \overset{\frac{d}{d t}}{\to} v (t) \overset{\frac{d}{d t}}{\to} a (t)$
Integrating: $a (t) \overset{\int d t}{\to} v (t) \overset{\int d t}{\to} s (t)$
Total Distance Travelled: Found by integrating the absolute value (speed) of velocity: $Distance = \int_{t_{1}}^{t_{2}} | v (t) | d t$ .

Worked Example: Kinematics via Differentiation & Roots Question: The height of a projectile is modeled by $h (t) = - 4.9 t^{2} + 15 t + 2$ , where $t$ is the time in seconds. Find the initial velocity, and determine the exact time the object reaches its maximum height.
Solution:
1. Find the velocity function by differentiating displacement $h (t)$ : $\begin{aligned} v (t) & = h^{'} (t) \\ v (t) & = 2 (- 4.9) t + 15 \\ v (t) & = - 9.8 t + 15 \end{aligned}$ 2. "Initial" means at time $t = 0$ : $v (0) = - 9.8 (0) + 15 = 15 m/s$ 3. Maximum height occurs when the object stops moving upwards and turns around (stationary point), which means velocity equals zero: $\begin{aligned} v (t) & = 0 \\ - 9.8 t + 15 & = 0 \\ 9.8 t & = 15 \\ t & = \frac{15}{9.8} \approx 1.53 seconds \end{aligned}$ Casio fx-CG50 Method Check:
To find the maximum height time graphically: Graph $Y 1 = - 4.9 X^{2} + 15 X + 2$ , press G-Solv $\to$ MAX. The $x$ -value will be exactly $1.5306$ .

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SL Analysis & Approaches Revision Notes

Unit 1 · Number & Algebra

Unit 2 · Functions

Unit 3 · Geometry & Trig

Unit 4 · Stats & Probability

Unit 5 · Calculus