Grade 12 — Term 1
5 study notes · 95 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 5 lessons (NIE Units 3, 4, 5, 6, 7). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Quadratic functions
What you must know
- Vertex coordinates: Completing the square f(x) = a(x + b/(2a))^2 + (4ac - b^2)/(4a) gives the vertex at (-b/(2a), -(b^2 - 4ac)/(4a)); line of symmetry is x = -b/(2a).
- Common root condition: For two quadratics a1*x^2 + b1*x + c1 = 0 and a2*x^2 + b2*x + c2 = 0 to have a common root α, eliminate x^2 and constant terms using cross-multiplication: (c1*a2 - c2*a1)^2 = (a1*b2 - a2*b1)(b1*c2 - b2*c1).
- Sign of roots: Both roots positive if Δ ≥ 0, α + β > 0, and αβ > 0; both roots negative if Δ ≥ 0, α + β < 0, and αβ > 0; opposite signs if αβ < 0 (no need to check Δ as ac < 0 ensures Δ > 0).
Try one question
If α and β are the distinct real roots of the quadratic equation x² - 5x + 3 = 0, what is the exact value of α² + β²?
- A. 19
- B. 22
- C. 25
- D. 31
Show the answer
Answer: A. 19. From Vieta's formulas for ax² + bx + c = 0: Sum of roots α + β = -b/a = 5, Product of roots αβ = c/a = 3. Using the algebraic identity α² + β² = (α + β)² - 2αβ = (5)² - 2(3) = 25 - 6 = 19.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Polynomial functions
What you must know
- Repeated factor condition: (x - a)^2 is a factor of f(x) if and only if f(a) = 0 AND f'(a) = 0.
- Quadratic divisor remainder: When f(x) is divided by a quadratic divisor (x - a)(x - b), the remainder is linear: r(x) = Ax + B; find A and B by substituting x = a and x = b into f(x) = (x - a)(x - b)q(x) + (Ax + B).
- Rational Root Theorem: If a polynomial with integer coefficients has a rational root p/q (in lowest terms), then p must divide the constant term a0, and q must divide the leading coefficient an.
Try one question
According to the Remainder Theorem, what is the remainder when the polynomial P(x) = 2x³ - 3x² + 4x - 5 is divided by (x - 2)?
- A. -3
- B. 9
- C. 5
- D. 7
Show the answer
Answer: D. 7. By the Remainder Theorem, the remainder when a polynomial P(x) is divided by (x - a) is simply P(a). Here, a = 2. Evaluating: P(2) = 2(2)³ - 3(2)² + 4(2) - 5 = 2(8) - 3(4) + 8 - 5 = 16 - 12 + 8 - 5 = 7.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Partial fractions
What you must know
- Cover-up rule (Heaviside method): For distinct linear factors (x - a), the numerator coefficient A is obtained instantly by covering up (x - a) in the original fraction and substituting x = a into the remaining expression.
- Degrees check before decomposition: Always verify that deg(P) < deg(Q); attempting partial fractions directly on an improper fraction produces erroneous results lacking the polynomial term.
- Coefficient comparison vs substitution: Use substitution of root values x = a, b to find constants quickly, and compare highest power (or constant) coefficients to solve for remaining parameters.
Try one question
What is the partial fraction decomposition of the rational function f(x) = (2x + 3) / [(x + 1)(x + 2)]?
- A. 1 / (x + 1) + 1 / (x + 2)
- B. 1 / (x + 1) - 1 / (x + 2)
- C. 2 / (x + 1) + 3 / (x + 2)
- D. 3 / (x + 1) - 1 / (x + 2)
Show the answer
Answer: A. 1 / (x + 1) + 1 / (x + 2). Let (2x + 3) / [(x + 1)(x + 2)] = A / (x + 1) + B / (x + 2). Then 2x + 3 = A(x + 2) + B(x + 1). Setting x = -1: 2(-1) + 3 = A(1) => A = 1. Setting x = -2: 2(-2) + 3 = B(-1) => -1 = -B => B = 1. Thus, f(x) = 1 / (x + 1) + 1 / (x + 2).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Indices and logarithms
What you must know
- Logarithmic domain constraints: For log_a(f(x)) to be defined, strictly enforce f(x) > 0, a > 0, and a ≠ 1; always verify potential solutions against original domain to eliminate extraneous roots.
- Inequality base reversal: If a > 1, then log_a(x) < log_a(y) <=> x < y; but if 0 < a < 1, the inequality direction REVERSES: log_a(x) < log_a(y) <=> x > y.
- Natural log and Euler's e: ln(x) = log_e(x) where e ≈ 2.71828; e^(ln x) = x and ln(e^x) = x for all valid x.
Try one question
What is the real solution to the logarithmic equation log₂(x) + log₂(x - 2) = 3?
- A. x = 4 and x = -2
- B. x = 4
- C. x = 8
- D. x = -2
Show the answer
Answer: B. x = 4. Using log_b(u) + log_b(v) = log_b(uv): log₂[x(x - 2)] = 3 => x(x - 2) = 2³ = 8 => x² - 2x - 8 = 0 => (x - 4)(x + 2) = 0. Since the argument of a logarithm must be strictly positive (x > 0 and x - 2 > 0 => x > 2), the only valid real root is x = 4 (x = -2 is rejected).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Inequalities
What you must know
- Strict exclusion of denominator zeros: The solutions of P(x)/Q(x) ≥ 0 can include roots where P(x) = 0, but MUST STRICTLY EXCLUDE points where Q(x) = 0 to prevent division by zero.
- Triangle Inequality: For all real numbers x and y, |x + y| ≤ |x| + |y|; equality holds if and only if xy ≥ 0 (x and y have the same sign or at least one is zero).
- AM-GM Inequality: For non-negative real numbers a and b, (a + b)/2 ≥ √(ab); equality holds if and only if a = b; useful for finding minimum/maximum values.
Try one question
What is the complete set of real values of x satisfying the rational inequality (x - 1) / (x + 2) ≤ 0?
- A. x ≤ -2 or x > 1
- B. -2 < x ≤ 1
- C. -2 ≤ x ≤ 1
- D. x < -2 or x ≥ 1
Show the answer
Answer: B. -2 < x ≤ 1. The critical points are x = 1 (numerator zero) and x = -2 (denominator zero, where expression is undefined). Testing sign intervals: for x < -2, (-)/(-) = +; for -2 < x < 1, (-)/(+) = -; for x > 1, (+)/(+) = +. The inequality is non-positive for x in (-2, 1]. Note x = -2 is excluded because denominator cannot be zero.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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