Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Determine the equation of an inverse function (exponential and logarithmic).
• Apply Euclidean Geometry theorems to solve problems.
• Sketch graphs of logarithmic functions, identifying key features.
• Solve geometric problems involving riders, providing reasons.
• Apply transformations to functions and their inverses.

Lesson summary

This week's focus is on consolidating core mathematical concepts and applying them to examination-style questions. Success in the final Grade 12 Mathematics examination requires not only understanding the theory but also the ability to apply that knowledge quickly and accurately under pressure. This revision period is crucial for honing these skills. We will be concentrating on areas frequently encountered in past papers and those known to cause students difficulty. This week’s specific focus will be on Functions (specifically inverse and logarithmic functions) and Euclidean Geometry.

Lesson notes

Functions: Inverse Functions and Logarithms Inverse Functions: The inverse of a function, denoted as f⁻¹(x), "undoes" what the original function f(x) does. To find the inverse, we typically swap x and y in the function's equation and then solve for y.

Key Concepts: A function must be one-to-one (pass the horizontal line test) to have an inverse function. If it's not, we restrict the domain. The domain of f(x) becomes the range of f⁻¹(x), and the range of f(x) becomes the domain of f⁻¹(x). The graphs of f(x) and f⁻¹(x) are reflections of each other over the line y = x.

Example 1: Finding the Inverse of an Exponential Function Find the inverse of f(x) = 2 x +

1. Replace f(x) with y: y = 2 x + 1 Swap x and y: x = 2 y + 1 Solve for y: x - 1 = 2 y log 2 (x - 1) = y (Taking the logarithm base 2 of both sides) Replace y with f⁻¹(x): f⁻¹(x) = log 2 (x - 1)

Explanation: We swapped x and y to reflect the function across the line y=x. We then used the definition of logarithms to isolate 'y', thus finding the inverse function. The base of the exponential function becomes the base of the logarithm.

Logarithmic Functions: Logarithmic functions are the inverses of exponential functions. The logarithmic function log b (x) asks the question: "To what power must we raise b to get x?".

Key Concepts: log b (x) is defined only for x > 0 (the argument must be positive). b > 0 and b ≠ 1 (the base must be positive and not equal to 1). log b (1) = 0 log b (b) = 1 log b (x n ) = n log b (x) log b (xy) = log b (x) + log b (y) log b (x/y) = log b (x) - log b (y)

Example 2: Solving Logarithmic Equations Solve for x: log 3 (x + 2) + log 3 (x - 4) = 2 Combine logarithms using the product rule: log 3 ((x + 2)(x - 4)) = 2 Rewrite in exponential form: (x + 2)(x - 4) = 3 2 Expand and simplify: x 2 - 2x - 8 = 9 Rearrange into a quadratic equation: x 2 - 2x - 17 = 0 Solve for x using the quadratic formula: x = (2 ± √((-2) 2 - 4(1)(-17))) / (2(1)) x = (2 ± √(4 + 68)) / 2 x = (2 ± √72) / 2 x = (2 ± 6√2) / 2 x = 1 ± 3√2 Check for extraneous solutions: We must ensure that x + 2 > 0 and x - 4 > 0. x = 1 + 3√2 ≈ 5.24 (Valid, since 5.24 + 2 > 0 and 5.24 - 4 > 0) x = 1 - 3√2 ≈ -3.24 (Invalid, since -3.24 + 2 x-1 -

2. Solution: Replace f(x) with y: y = 3 x-1 - 2 Swap x and y: x = 3 y-1 - 2 Solve for y: x + 2 = 3 y-1 log 3 (x + 2) = y - 1 y = log 3 (x + 2) + 1 Replace y with f⁻¹(x): f⁻¹(x) = log 3 (x + 2) + 1

Commentary: This question tests the ability to manipulate exponential equations and use the definition of logarithms to find the inverse. Remember to isolate the exponential term before taking the logarithm.

Question 2: Solve for x: 2log(x + 1) - log(x - 2) = log(7) (Assume base 10)

Solution: Use the power rule: log(x + 1) 2 - log(x - 2) = log(7)

Use the quotient rule: log((x + 1) 2 / (x - 2)) = log(7)

Equate the arguments: (x + 1) 2 / (x - 2) = 7 Expand and simplify: (x 2 + 2x + 1) = 7(x - 2)

Rearrange into a quadratic equation: x 2 + 2x + 1 = 7x - 14 x 2 - 5x + 15 = 0 Calculate the discriminant: Δ = b 2 -4ac = (-5) 2 - 4(1)(15) = 25 - 60 = -35 Since the discriminant is negative, there are no real solutions.

Commentary: This question tests the properties of logarithms and solving equations. The key here is to combine the logarithmic terms into a single logarithm before equating the arguments. We must also remember to consider the domain of logarithms. In this specific problem, the discriminant is negative, therefore no real roots exist and thus there is no solution.

Question 3: In the diagram, O is the centre of the circle. A, B, and C are points on the circumference of the circle. If ∠AOC = 130°, calculate the size of ∠AB

C. Solution: ∠ABC = ½ ∠AOC (∠ at centre = 2 ∠ at circumference) ∠ABC = ½ (130°) ∠ABC = 65°

Commentary: A straightforward application of the "angle at centre" theorem. The key is recognising the relationship between the central angle and the angle at the circumference subtended by the same arc. Independent Practice (Questions Only) Determine the inverse of the function g(x) = (x - 3) 2 + 1, where x ≥

3. State the domain and range of g⁻¹(x).

Solve for x: log 2 (x) + log 2 (x - 2) =

3. Sketch the graph of y = log 3 (x + 1) - 2, showing all intercepts and asymptotes. In the diagram, PQ is a tangent to the circle at Q. R and S are points on the circumference of the circle. If ∠RQS = 50°, calculate the size of ∠RSQ if RQ is a diameter. In cyclic quadrilateral ABCD, ∠A = 75°. Determine the size of ∠

C. If f(x) = 2 x and g(x) = x + 3, find f(g(x)) and g(f(x)).

Solve for x: 3 2x+1 = 27 x-1 Prove that the opposite angles of a cyclic quadrilateral are supplementary. (Full rider)