Bac Général Djibouti - independent subject foundations study materials
Bac Général Djibouti - independent subject foundations study resources

Bac Général Djibouti - independent subject foundations: Upper-secondary functions, calculus, geometry and probability

Bac Général Djibouti - independent subject foundations · Work through your exam, one topic at a time.

All guide sections

Upper-secondary functions, calculus, geometry and probability

Algebra and Functions

What this domain tests

Algebra and function practice asks you to move fluently between symbolic, graphical, and verbal representations of relationships. Expect manipulation of polynomial, rational, radical, exponential, and logarithmic expressions; solving equations and inequalities; and reasoning about domain, range, inverse functions, and transformations. Questions often present a function in one form and require an answer in another, so translation skill matters as much as raw computation.

Core rules and concepts

  • Domain restrictions come first. Denominators cannot be zero; even-index radicals require nonnegative radicands; logarithm arguments must be positive. State restrictions before simplifying rational expressions, because cancelled factors still exclude original zeros.
  • Function composition is not commutative. f(g(x)) and g(f(x)) generally differ; compute inside-out.
  • Inverses undo. f⁻¹ exists only if f is one-to-one on its stated domain; swap x and y, then solve. Domains and ranges swap too.
  • Transformations: af(b(x−h))+k gives vertical stretch by |a|, horizontal compression by 1/|b|, shift right h, up k. Horizontal changes behave opposite to intuition.
  • Exponential vs. logarithmic: y = bˣ and y = log_b x are inverses; log_b(MN) = log_b M + log_b N, log_b(M/N) = log_b M − log_b N, log_b(Mᵖ) = p log_b M.
  • Rational functions: vertical asymptotes at non-cancelled denominator zeros, holes at cancelled ones, and end behavior from degree comparison.

Worked example

Solve log₂(x) + log₂(x − 2) = 3.

Combine: log₂[x(x − 2)] = 3, so x(x − 2) = 2³ = 8. Then x² − 2x − 8 = 0, giving (x − 4)(x + 2) = 0, so x = 4 or x = −2. Now check the domain: both log arguments must be positive, so x > 0 and x − 2 > 0, meaning x > 2. Therefore x = −2 is extraneous and the only solution is x = 4. Verification: log₂4 + log₂2 = 2 + 1 = 3. ✓

Traps that cost marks

  • Reporting extraneous roots from squaring or from log equations without domain checking.
  • Cancelling a factor in a rational expression and then claiming the cancelled value is in the domain.
  • Applying a horizontal shift in the wrong direction, or stretching horizontally when the coefficient is inside the argument.
  • Forgetting that log_b x is undefined for x ≤ 0 and that b must be positive and not 1.
  • Losing the ± when solving x² = k.

Calculus

Calculus practice develops the ability to move fluently between a function, its rate of change, and its accumulated total. Expect differentiation of composite, implicit and parametric forms, evaluation of definite and improper integrals, and questions that require you to justify a limit before differentiating.

Differentiation

  • Chain rule: d/dx f(g(x)) = f'(g(x))·g'(x). The inner derivative is the most commonly dropped factor.
  • Implicit differentiation: differentiate both sides with respect to x, treating y as y(x), then collect the dy/dx terms.
  • Logarithmic differentiation pays off for products of powers such as x^x or (sin x)^x.
  • L'Hôpital applies only to 0/0 or ∞/∞; verify the indeterminate form first.

Integration

  • Substitution reverses the chain rule; for a definite integral, change the limits at the same time.
  • Integration by parts: ∫u dv = uv − ∫v du. Choose u so that du simplifies.
  • The Fundamental Theorem converts a definite integral into an antiderivative difference; improper integrals need an explicit limit.
  • Partial fractions for rational functions, after comparing the degrees of numerator and denominator.

Worked example

Evaluate ∫₀¹ x·e^{2x} dx. Take u = x, dv = e^{2x} dx, so du = dx and v = e^{2x}/2. ∫ x·e^{2x} dx = x·e^{2x}/2 − ∫ (e^{2x}/2) dx = x·e^{2x}/2 − e^{2x}/4. At x = 1: e²/2 − e²/4 = e²/4. At x = 0: 0 − 1/4. Subtract: e²/4 − (−1/4) = (e² + 1)/4 ≈ 2.097.

Traps that cost marks

  • Omitting +C on an indefinite integral.
  • Forgetting to change the limits after substitution, then substituting back into x-limits.
  • Splitting an improper integral across a singularity without taking one-sided limits.
  • Sign slips when subtracting the lower limit; write the bracket out explicitly.
  • Losing a factor when the constant multiplies the whole term rather than just the outer function.

Geometry and Trigonometry

What this domain tests

This geometry and trigonometry chapter covers triangle congruence and similarity, circle properties, coordinate geometry, area and volume, and the trigonometric ratios extended to the unit circle. You should be able to prove relationships, apply the sine and cosine laws to non-right triangles, work in radians, and connect trigonometric functions to their graphs and identities.

Core rules and concepts

  • Similarity: corresponding angles equal and sides proportional; areas scale as the square of the linear ratio, volumes as the cube.
  • Circle facts: an inscribed angle is half its intercepted arc; a tangent is perpendicular to the radius at the point of contact; the perpendicular from the centre bisects a chord.
  • Right-triangle ratios: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
  • Sine law: a/sin A = b/sin B = c/sin C, with the ambiguous case (two possible triangles) when given two sides and a non-included angle.
  • Cosine law: c² = a² + b² − 2ab cos C, used for SAS and SSS.
  • Radians: π radians = 180°; arc length s = rθ and sector area A = ½r²θ for θ in radians.
  • Identities: sin²θ + cos²θ = 1; tan θ = sin θ/cos θ.

Worked example

A triangle has sides a = 7, b = 9, and included angle C = 60°. Find side c.

Because we know two sides and the included angle (SAS), use the cosine law: c² = 7² + 9² − 2(7)(9)cos 60° c² = 49 + 81 − 126(0.5) c² = 130 − 63 = 67 c = √67 ≈ 8.19

Reasoning note: the minus sign in the cosine law is essential; using the sine law here would be invalid because no side–opposite-angle pair is known.

Traps that cost marks

  • Using degree mode when the problem is in radians (or vice versa) in calculator work.
  • Ignoring the ambiguous case of the sine law and reporting only one triangle.
  • Confusing the linear scale factor with the area or volume factor in similarity problems.
  • Mixing up inscribed angle and central angle relationships (central is double, not half).
  • Applying arc length or sector formulas with θ in degrees instead of radians.

Probability and Statistics

Probability and statistics practice develops conditional reasoning, standard distributions, and correct interpretation of inference - the last of which is where most marks are lost.

Probability toolkit

  • P(A∪B) = P(A) + P(B) − P(A∩B); independence means P(A∩B) = P(A)P(B).
  • Conditional probability P(A|B) = P(A∩B)/P(B); Bayes reverses the conditioning: P(A|B) = P(B|A)P(A)/P(B).
  • Law of total probability: sum over a partition of the sample space.
  • E[aX + b] = aE[X] + b; Var(aX + b) = a²Var(X); Var(X) = E[X²] − (E[X])².

Distributions the paper uses

  • Binomial(n, p): mean np, variance np(1 − p).
  • Poisson(λ): mean and variance both λ, used for counts.
  • Normal: standardise with z = (x − μ)/σ before using tables.
  • Sampling: X̄ has mean μ and standard error σ/√n; for independent identically distributed observations with finite variance, the CLT supports a normal approximation to the standardized sample mean as n grows; the required sample size depends on population shape.

Inference

  • A confidence interval describes the procedure, not a probability attached to a fixed parameter.
  • A p-value is P(data at least as extreme | H₀), not P(H₀ true).
  • Test statistic for a mean with known σ: z = (x̄ − μ₀)/(σ/√n).
  • Unbiasedness, efficiency and consistency are estimator properties; the sample variance divides by n − 1 to be unbiased.

Worked example

A disease affects 1% of a population. A test has sensitivity 0.95 and specificity 0.90. Given a positive test, find P(disease). P(+) = 0.95(0.01) + 0.10(0.99) = 0.0095 + 0.099 = 0.1085. P(D|+) = 0.0095 / 0.1085 ≈ 0.0876, about 8.8%. The low prior dominates - the base-rate effect.

Traps that cost marks

  • Confusing P(A|B) with P(B|A).
  • Using σ instead of σ/√n, or dropping the √n entirely.
  • Treating 'fail to reject H₀' as proof that H₀ is true.
  • Adding variances of dependent random variables.
  • For discrete data, using the wrong tail or omitting a continuity correction.