Grade 12 Science: Mathematics

In this unit, 'Relations, functions and inverse trigonometric functions', you learn to define and graph functions and their inverses. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Relations, functions and inverse trigonometric functions' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: define and graph functions and their inverses. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Find an inverse: f(x) = 2x - 3. Put y = 2x - 3, solve for x: x = (y + 3)/2. So the inverse is f^-1(x) = (x + 3)/2. Check: f(f^-1(5)) = f(4) = 5.
2. Step 2, Composition: f(x) = x + 1 and g(x) = x^2. Then (f o g)(2) = f(4) = 5 while (g o f)(2) = g(3) = 9. Composition is not commutative, so order matters.
3. Step 3, Principal values: sin^-1(1/2) = pi/6, cos^-1(-1/2) = 2 pi/3, tan^-1(1) = pi/4. The ranges are [-pi/2, pi/2] for sin^-1, [0, pi] for cos^-1.
4. Step 4, Pitfall: sin^-1(sin(5 pi/6)) is not 5 pi/6, since 5 pi/6 is outside the range. sin(5 pi/6) = 1/2, so the answer is sin^-1(1/2) = pi/6.
5. Step 5, Equivalence relation: R on integers, a R b if a - b is divisible by 3. Reflexive: a - a = 0. Symmetric: if 3 divides a - b it divides b - a. Transitive: sum of two multiples of 3 is a multiple of 3.
6. Step 6, Your turn: find the inverse of f(x) = 3x + 6 and tan^-1(root 3). Answer: x = (y - 6)/3, so f^-1(x) = (x - 6)/3; and tan^-1(root 3) = pi/3.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Define and graph functions and their inverses: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Matrices and determinants', you learn to operate on matrices and find inverses. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Matrices and determinants' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: operate on matrices and find inverses. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Determinant and inverse: A = [[2, 1], [5, 3]] has det = 2 x 3 - 1 x 5 = 1. The inverse is (1/1) x [[3, -1], [-5, 2]]. Swap the diagonal, change the sign of the other entries.
2. Step 2, Check: A x A^-1 = [[2 x 3 + 1 x (-5), 2 x (-1) + 1 x 2], [5 x 3 + 3 x (-5), 5 x (-1) + 3 x 2]] = [[1, 0], [0, 1]]. It gives the identity matrix, so the inverse is right.
3. Step 3, Solve a system: 2x + y = 7 and 5x + 3y = 18, written AX = B. Then X = A^-1 B = [[3, -1], [-5, 2]] x [7, 18] = [21 - 18, -35 + 36] = [3, 1].
4. Step 4, Check: x = 3, y = 1 gives 6 + 1 = 7 and 15 + 3 = 18. Both equations hold. A zero determinant would mean no unique solution.
5. Step 5, Triangular determinant: for [[1, 2, 3], [0, 1, 4], [0, 0, 2]] the determinant is the product of the diagonal, 1 x 1 x 2 = 2. Expanding along the first column gives the same, 1 x (1 x 2 - 4 x 0) = 2.
6. Step 6, Your turn: find the determinant and inverse of [[4, 3], [2, 2]]. Answer: det = 8 - 6 = 2; inverse = 1/2 x [[2, -3], [-2, 4]] = [[1, -3/2], [-1, 2]].

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Operate on matrices and find inverses: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Continuity, differentiability and applications of derivatives', you learn to differentiate and find maxima and minima. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Continuity, differentiability and applications of derivatives' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: differentiate and find maxima and minima. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Continuity: f(x) = (x^2 - 9)/(x - 3) for x not equal to 3, and f(3) = 6. The limit as x approaches 3 is x + 3 = 6, which equals f(3), so f is continuous at 3.
2. Step 2, Chain rule: y = (3x + 1)^5 gives dy/dx = 5(3x + 1)^4 x 3 = 15(3x + 1)^4. Also d/dx of e^(2x) = 2e^(2x). Differentiate the outer function, then multiply by the inner derivative.
3. Step 3, Maxima and minima: f(x) = x^3 - 3x + 2. f'(x) = 3x^2 - 3 = 0 gives x = 1 or -1. f''(x) = 6x. At x = 1, f'' = 6 > 0, a minimum with f(1) = 0. At x = -1, f'' = -6 < 0, a maximum with f(-1) = 4.
4. Step 4, Check values: f(1) = 1 - 3 + 2 = 0 and f(-1) = -1 + 3 + 2 = 4. Also f(0) = 2 lies between them, as expected for the curve between the turning points.
5. Step 5, Optimisation: a rectangle has perimeter 40, so sides x and 20 - x. Area A = x(20 - x), A' = 20 - 2x = 0, so x = 10 and A = 100. It is a square, and A'' = -2 < 0.
6. Step 6, Your turn: find the minimum of x^2 - 6x + 5. Answer: f' = 2x - 6 = 0 gives x = 3, f'' = 2 > 0, and f(3) = 9 - 18 + 5 = -4.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Differentiate and find maxima and minima: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Integrals and applications of integrals', you learn to integrate and find areas under curves. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Integrals and applications of integrals' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: integrate and find areas under curves. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Basic integral: the integral of 3x^2 + 2x + 1 dx = x^3 + x^2 + x + C. And the integral of sin 2x dx = -cos 2x / 2 + C. Differentiate to check: d/dx of -cos 2x / 2 = sin 2x.
2. Step 2, Substitution: the integral of 2x(x^2 + 1)^3 dx. Let u = x^2 + 1, then du = 2x dx. The integral becomes u^3 du = u^4/4 + C = (x^2 + 1)^4 / 4 + C.
3. Step 3, Definite integral: from 0 to 1 of x^2 dx = [x^3/3] = 1/3. From 0 to 2 of 3x^2 + 1 dx = [x^3 + x] = 8 + 2 = 10.
4. Step 4, Area under y = x^2 from 0 to 3: the integral is [x^3/3] from 0 to 3 = 27/3 = 9 square units. Rough check: the curve is below the line y = 3x, whose area is 13.5.
5. Step 5, Area between curves: y = x and y = x^2 meet at x = 0 and x = 1, with the line above. Area = integral of (x - x^2) = [x^2/2 - x^3/3] = 1/2 - 1/3 = 1/6.
6. Step 6, Your turn: integrate cos 3x and evaluate the integral of 2x from 0 to 3. Answer: sin 3x / 3 + C; and [x^2] from 0 to 3 = 9.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Integrate and find areas under curves: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Differential equations', you learn to solve first-order differential equations. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Differential equations' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: solve first-order differential equations. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Order and degree: (d^2y/dx^2)^3 + dy/dx = 0 has highest derivative d^2y/dx^2, so order 2, and that derivative is raised to power 3, so degree 3.
2. Step 2, Direct integration: dy/dx = 2x gives y = x^2 + C. With y(0) = 3 we get C = 3, so y = x^2 + 3. Check: derivative 2x, and y(0) = 3.
3. Step 3, Separable: dy/dx = x/y gives y dy = x dx. Integrate: y^2/2 = x^2/2 + c, so y^2 = x^2 + k. At (0, 2), k = 4, so y^2 = x^2 + 4.
4. Step 4, Exponential: dy/dx + 2y = 0 gives dy/y = -2 dx, so ln|y| = -2x + c and y = Ce^(-2x). With y(0) = 5, C = 5. Check: y' = -10e^(-2x) and 2y = 10e^(-2x), which sum to 0.
5. Step 5, Growth model: dy/dx = ky has solution y = Ae^(kx). If a population of 1,000 grows with k = 0.1, then after 10 time units y = 1,000 x e^1, about 2,718.
6. Step 6, Your turn: solve dy/dx = 3x^2 with y(1) = 2, and give the general solution of dy/dx = 4y. Answer: y = x^3 + C, then 1 + C = 2 so y = x^3 + 1; and y = Ae^(4x).

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Solve first-order differential equations: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Vectors and three-dimensional geometry', you learn to use dot and cross products and equations of lines and planes. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Vectors and three-dimensional geometry' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: use dot and cross products and equations of lines and planes. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Dot product: a = (1, 2, 3) and b = (2, -1, 4). a.b = 1 x 2 + 2 x (-1) + 3 x 4 = 2 - 2 + 12 = 12. And |a| = root of 1 + 4 + 9 = root 14.
2. Step 2, Cross product: a x b = (2x4 - 3x(-1), 3x2 - 1x4, 1x(-1) - 2x2) = (11, 2, -5). Check: a . (11, 2, -5) = 11 + 4 - 15 = 0, and b . (11, 2, -5) = 22 - 2 - 20 = 0.
3. Step 3, Angle: |b| = root of 4 + 1 + 16 = root 21. So cos theta = 12 / root 294, about 12 / 17.15 = 0.70, and theta is about 45.6 degrees.
4. Step 4, Line: through (1, 2, 3) with direction (2, 1, -1): (x - 1)/2 = (y - 2)/1 = (z - 3)/(-1). Setting each ratio to t gives x = 1 + 2t, y = 2 + t, z = 3 - t.
5. Step 5, Plane: through (1, 0, 0) with normal (2, 3, 1): 2(x - 1) + 3y + z = 0, so 2x + 3y + z = 2. Its distance from the origin is 2/root 14, about 0.53.
6. Step 6, Your turn: find the dot and cross product of (1, 0, 2) and (3, 1, -1). Answer: dot = 3 + 0 - 2 = 1; cross = (0 - 2, 6 + 1, 1 - 0) = (-2, 7, 1), and (1, 0, 2) . (-2, 7, 1) = 0.

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Use dot and cross products and equations of lines and planes: at least 70 percent on an exam-style task, with the method shown.

In this unit, 'Linear programming and probability', you learn to solve optimisation and conditional probability problems. You work with exam-style questions and written reasoning. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Linear programming and probability' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: solve optimisation and conditional probability problems. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Set up: maximise Z = 4x + 3y subject to 2x + y <= 10, x + 2y <= 8, x, y >= 0. The corner points are (0, 0), (5, 0), (0, 4), and the meeting point of the two lines.
2. Step 2, Intersection: from 2x + y = 10, y = 10 - 2x. Substitute into x + 2y = 8: x + 20 - 4x = 8, so x = 4 and y = 2. Check: 8 + 2 = 10 and 4 + 4 = 8.
3. Step 3, Evaluate Z: (0, 0) gives 0, (5, 0) gives 20, (0, 4) gives 12, (4, 2) gives 16 + 6 = 22. The maximum is 22 at (4, 2). Your answer must come from a corner of the region.
4. Step 4, Conditional probability: P(A) = 0.6, P(B) = 0.5, P(A and B) = 0.3. Then P(A given B) = 0.3/0.5 = 0.6 = P(A), and 0.6 x 0.5 = 0.3, so A and B are independent.
5. Step 5, Total probability: bag 1 has 3 red and 2 black, bag 2 has 1 red and 4 black, and a bag is chosen at random. P(red) = 1/2 x 3/5 + 1/2 x 1/5 = 4/10 = 2/5. By Bayes, P(bag 1 given red) = (3/10)/(4/10) = 3/4.
6. Step 6, Your turn: maximise Z = 2x + 3y with x + y <= 5, x <= 3. Answer: corners (0,0), (3,0), (3,2), (0,5) give 0, 6, 12, 15, so the maximum is 15 at (0, 5).

Activity. About 40 to 50 minutes of class work, then one short practice task for home. Keep evidence: a timed paper and an error log.

Standard to meet. Solve optimisation and conditional probability problems: at least 70 percent on an exam-style task, with the method shown.

On the slide: L1 Basic · Count and name shapes

Notes: Count objects up to 20 and name a circle, square and triangle. Notes: give ten buttons and two trays; learners count aloud, move three across and say how many are left (7), then hunt the room for one shape of each kind.

On the slide: L2 Awareness · Add, subtract and measure

Notes: Add and subtract within 100 and measure a length with a ruler. Notes: work 46 + 27 = 73 on a number line, jumping 20 then 7; learners measure a pencil in centimetres and explain how they checked their answer a second way.

On the slide: L3 Foundation · Fractions and decimals

Notes: Compare fractions and decimals and use the four operations with whole numbers. Notes: shade 3/4 and 6/8 on equal strips to show they match, then add 0.5 + 0.25 = 0.75 and write it as 75 per cent of one whole.

On the slide: L4 Practitioner · Equations and percentages

Notes: Solve one-step and two-step equations and percentage problems, showing each step. Notes: practice case is 3x + 5 = 20, so x = 5, checked by substituting back; then a 15 per cent discount on 400 leaves 340, because 400 x 0.85 = 340.

On the slide: L5 Specialist · Geometry, data and graphs

Notes: Use the right-angled triangle rule, the mean and a straight-line graph to solve a problem. Notes: legs of 6 and 8 give a longest side of 10; the data 4, 6, 6, 8, 11 have mean 7; plot y = 2x + 1 and read y = 7 at x = 3.

This course: Grade 12 Science Mathematics is at this level. Its units include Relations, functions and inverse trigonometric functions; Matrices and determinants; Continuity, differentiability and applications of derivatives; Integrals and applications of integrals and more.

On the slide: L6 Advanced · Proof, calculus and algebra

Notes: Prove results and use calculus and matrices to model change. Notes: differentiate f(x) = x^3 - 3x to get 3x^2 - 3, so turning points sit at x = 1 and x = -1; the second derivative 6x shows a minimum then a maximum. Learners prove the square root of 2 is irrational.

On the slide: L7 Super expert · Research-level modelling

Notes: Formulate and defend a proof or statistical model, stating assumptions and limits. Notes: for 25 values with mean 50 and standard deviation 10, the standard error is 2 and a 95 per cent interval is about 45.9 to 54.1 (t = 2.064). Learners check residuals and write a referee-style critique.