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Mathematics Grade 12

Mathematics Grade 12 Assignments 2021: Education Resource

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Mathematics Grade 12 Assignments 2021: Assessment is a continuous, planned process using various forms of tasks in order to identify, gather and interpret information about the performance of learners. It involves four steps: generating and collecting evidence of achievement, evaluating this evidence, recording the findings and using this information to understand and assist in the learner’s development in order to improve the process of learning and teaching. Assessment should be both informal (assessment for learning) and formal (assessment of learning). In both cases regular feedback should be provided to learners to enhance the learning experience.

Assignment Assessment types for Mathematics Grade 12


Although assessment guidelines are included in the annual teaching plan at the end of each term, the following general principles apply:

Tests and examinations are usually time-limited and assessed using a marking memorandum. Assignments are generally extended pieces of work in which time constraints have been relaxed and which may be completed at home. Assignments may be used to consolidate or deepen understanding of work done earlier. They may thus consist of a collection of past examination questions or innovative activities using any resource material. It is, however, advised that assignments be focused.

Projects are more extended tasks that may serve to deepen understanding of curricular mathematics topics. They may also involve extracurricular mathematical topics where the learner is expected to select appropriate mathematical content to solve context-based or real-life problems. The focus should be on the mathematical concepts and not on duplicated pictures and regurgitation of facts from reference material.

Investigations are set to develop the mathematical concepts or skills of systematic investigation into special cases with a view to observing general trends, making conjectures and proving them.

It is recommended that while the initial investigation can be done at home, the final write-up be done in the classroom, under supervision and without access to any notes. Investigations are marked using a rubric which can be specific to the task, or generic, listing the number of marks awarded to each skill as outlined below:

  • 40% for communicating individual ideas and discoveries, assuming the reader has not come across the task before. The appropriate use of diagrams and tables will enhance the assignment, investigation or project.
  • 35% for generalising, making conjectures and proving or disproving these conjectures;
  • 20% for the effective consideration of special cases; and
  • 5% for presentation, that is, neatness and visual impact.

Assignments topics and memo for 2021

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More Grade 12 Mathematics Past Papers for Previous Years

Hey, Grade 12 Learners. My Courses portal has everything you needed to ace your matric exams, tests, assessments, research tasks, and assignments. Feel free to explore all resources for grade 12 learners, such as Study Guides, Department of Basic Education Past Exam Papers with Memos, and Speech Topics.

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  • If you want to study at a University or College?
  • What career or course/qualification do you plan to pursue after school?
  • What subjects you will need to study for this career?
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    22 Aug 2021 at 3:15 pm

    I need the paper

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Mathematics Grade 12

Mathematics Grade 12 November 2020 Question Papers and Memos for study revision

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Mathematics Grade 12 November 2020 Question Papers and Memos pdf download: Paper 1 and Paper 2

List Mathematics Grade 12 November 2020 Question Papers and Memos:

Paper 1:

Paper 2:

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Mathematics Grade 11

Cubic Functions Grade 12 Questions and Answers pdf

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Cubic Functions Grade 12 Questions and Answers pdf: In mathematics, a cubic function is a function of the form f(x)=ax^3+bx^2+cx+d where the coefficients a, b, c, and d are real numbers, and the variable x takes real values, and a ≠ 0. In other words, it is both a polynomial function of degree three, and a real function.

Downloadable Cubic Functions Grade 12 Questions and Answers pdf

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Mathematics Grade 12

General Solution Trigonometric Grade 12 pdf

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General Solution Trigonometric Grade 12 pdf: This page contains a guide for Mathematics Grade 12 Students with the following Trigonometric notes: Trigonometric Functions, 3D Trigonometry, Identities Problems with Solutions, and Challenging Exam Practices.

Concepts and skills

  • Simplify trigonometric equations.
  • Find the reference angle.
  • Use reduction formulae to find other angles within each quadrant.
  • Find the general solution of a given trigonometric equation.

Prior knowledge:

  • Trigonometric ratios.
  • Trigonometric identities.
  • Solve problems in the Cartesian plane.
  • Co-functions.
  • Factorisation.
  • Revise the following trigonometric ratios for right-angled triangles, which you learnt in Grade 10.

Will start with the basic equation, sin x = 0. The principal solution for this case will be x = 0, π, 2π as these values satisfy the given equation lying in the interval [0, 2π]. But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.

Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc. Below here is the table defining the general solutions of the given trigonometric functions involved equations.

Solutions Trigonometric Equations

EquationsSolutions
sin x = 0 x = nπ
cos x = 0x = (nπ + π/2)
tan x = 0x = nπ
sin x = 1x = (2nπ + π/2) = (4n+1)π/2
cos x = 1x = 2nπ
sin x = sin θx = nπ + (-1)nθ, where θ ∈ [-π/2, π/2]
cos x = cos θx = 2nπ ± θ, where θ ∈ (0, π]
tan x = tan θx = nπ + θ, where θ ∈ (-π/2 , π/2]
sin2 x = sin2 θx = nπ ± θ
cos2 x = cos2 θx = nπ ± θ
tan2 x = tan2 θx = nπ ± θ

Watch: Video

How to solve general trigonometric equations formula and find solutions

The periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of 360360° and the tangent function has a period of 180180°.

Method for finding the solution:

  1. Simplify the equation using algebraic methods and trigonometric identities.
  2. Determine the reference angle (use a positive value).
  3. Use the CAST diagram to determine where the function is positive or negative (depending on the given equation/information).
  4. Restricted values: find the angles that lie within a specified interval by adding/subtracting multiples of the appropriate period.
  5. General solution: find the angles in the interval [0°;360°][0°;360°] that satisfy the equation and add multiples of the period to each answer.
  6. Check answers using a calculator.

Questions and Answers

My Courses has a large Questions and Answers repository for the most popular High School and Tertiary Schools subjects. This comes in handy when doing your revision or preparing for exams, tests, research tasks, and assignments.

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Identities Problems with Solutions for exam Practice

1. (1 – sin A)/(1 + sin A) = (sec A – tan A)2

Solution:

L.H.S = (1 – sin A)/(1 + sin A)

= (1 – sin A)2/(1 – sin A) (1 + sin A),[Multiply both numerator and denominator by (1 – sin A)

= (1 – sin A)2/(1 – sin2 A)

= (1 – sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 – sin2 θ]

= {(1 – sin A)/cos A}2

= (1/cos A – sin A/cos A)2

= (sec A – tan A)2 = R.H.S. Proved.

2. Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ – cot θ.

Solution:

L.H.S.= √{(sec θ – 1)/(sec θ + 1)}

= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]; [multiplying numerator and denominator by (sec θ – l) under radical sign]

= √{(sec θ – 1)2/(sec2 θ – 1)}

=√{(sec θ -1)2/tan2 θ}; [since, sec2 θ = 1 + tan2 θ ⇒ sec2 θ – 1 = tan2 θ]

= (sec θ – 1)/tan θ

= (sec θ/tan θ) – (1/tan θ)

= {(1/cos θ)/(sin θ/cos θ)} – cot θ

= {(1/cos θ) × (cos θ/sin θ)} – cot θ

= (1/sin θ) – cot θ

= cosec θ – cot θ = R.H.S. Proved.

3. tan4 θ + tan2 θ = sec4 θ – sec2 θ

Solution:

L.H.S = tan4 θ + tan2 θ

= tan2 θ (tan2 θ + 1)

= (sec2 θ – 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]

= (sec2 θ – 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]

= sec4 θ – sec2 θ = R.H.S. Proved.

More problems on trigonometric identities are shown where one side of the identity ends up with the other side.

4. . cos θ/(1 – tan θ) + sin θ/(1 – cot θ) = sin θ + cos θ

Solution:

L.H.S = cos θ/(1 – tan θ) + sin θ/(1 – cot θ)

= cos θ/{1 – (sin θ/cos θ)} + sin θ/{1 – (cos θ/sin θ)}

= cos θ/{(cos θ – sin θ)/cos θ} + sin θ/{(sin θ – cos θ/sin θ)}

= cos2 θ/(cos θ – sin θ) + sin2 θ/(cos θ – sin θ)

= (cos2 θ – sin2 θ)/(cos θ – sin θ)

= [(cos θ + sin θ)(cos θ – sin θ)]/(cos θ – sin θ)

= (cos θ + sin θ) = R.H.S. Proved.

5. Show that, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A)

Solution:

We have,

1/(csc A – cot A) + 1/(csc A + cot A)

= (csc A + cot A + csc A – cot A)/(csc2 A – cot2 A)

= (2 csc A)/1; [since, csc2 A = 1 + cot2 A ⇒ csc2A – cot2 A = 1]

= 2/sin A; [since, csc A = 1/sin A]

Therefore,

1/(csc A – cot A) + 1/(csc A + cot A) = 2/sin A

⇒ 1/(csc A – cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A

Therefore, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A) Proved.

6. (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ)/cos θ

Solution:

L.H.S = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)

= [(tan θ + sec θ) – (sec2 θ – tan2 θ)]/(tan θ – sec θ + 1), [Since, sec2 θ – tan2 θ = 1]

= {(tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)}/(tan θ – sec θ + 1)

= {(tan θ + sec θ) (1 – sec θ + tan θ)}/(tan θ – sec θ + 1)

= {(tan θ + sec θ) (tan θ – sec θ + 1)}/(tan θ – sec θ + 1)

= tan θ + sec θ

= (sin θ/cos θ) + (1/cos θ)

= (sin θ + 1)/cos θ

= (1 + sin θ)/cos θ = R.H.S. Proved.

Sources:

https://www.math-only-math.com/problems-on-trigonometric-identities.html

https://intl.siyavula.com/read/maths/grade-12/trigonometry/04-trigonometry-04

https://byjus.com/maths/trigonometric-equations/

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