Secondary 3/4 A-Math

A-Math is taken alongside E-Math from Secondary 3, for students continuing with Mathematics beyond the general syllabus. It moves quickly into calculus, and into more formal algebra, coordinate geometry and trigonometry, plus a first taste of writing a formal geometric proof.

Algebra, extended

Quadratic functions, including finding maximum and minimum values by completing the square.
Surds, and solving equations that involve them.
Polynomials, the remainder and factor theorems, and partial fractions.
The Binomial Theorem, and exponential and logarithmic functions.

Equations and coordinate geometry

Conditions on a quadratic equation for real, equal, or no real roots.
Simultaneous equations where one equation is linear and the other quadratic, solved by substitution.
Quadratic inequalities.
Conditions for two lines to be parallel or perpendicular, the midpoint of a line segment, the area of a rectilinear figure, and the equation of a circle.

Trigonometric identities and equations

A-Math trigonometry moves past the basic ratios into identities that need to be proved, not just applied, and equations that can have several correct answers within a given range. The common trap is stopping after finding one solution instead of checking the full range the question asks for.

Qn 1

Solve 2cos²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

cos²θ = 1/2, so cos θ = ±1/√2

Within 0° to 360°, cos θ = ±1/√2 at four points

θ = 45°, 135°, 225°, 315° ✓

Differentiation: a genuinely new idea

Differentiation is the one topic on this syllabus with no direct lead-in from earlier levels — it's a new way of thinking about a curve's gradient, not an extension of algebra students already know. Once the mechanics click, it opens up gradients, tangents and rates of change.

Qn 2

Differentiate y = x³ − 4x² + 5, and find the gradient of the curve at x = 2.

dy/dx = 3x² − 8x

At x = 2: 3(2)² − 8(2) = 12 − 16

Gradient at x = 2 is −4 ✓

Integration: differentiation in reverse

Integration undoes what differentiation does, and the mechanics — reversing the power rule — are usually the easy part. The real hurdle is the "+ c": undoing a derivative gives a whole family of curves, all with the same gradient function, and it takes one extra piece of information, like a point the curve passes through, to pin down exactly one of them.

Qn 3

Given that dy/dx = 6x² − 4, and the curve passes through (1, 3), find y in terms of x.

y = 2x³ − 4x + c

At (1, 3): 3 = 2(1) − 4(1) + c, so c = 5

y = 2x³ − 4x + 5 ✓

The Linear Law: turning a curve into a straight line

Some relationships aren't straight lines on an ordinary graph, but become one after the right substitution. The Linear Law takes an equation like y = kxⁿ, applies logarithms to both sides, and turns it into the equation of a straight line. The technique itself is short; the usual mistake is misreading which transformed quantity is the gradient and which is the intercept.

Qn 4

Variables x and y are related by y = kxⁿ. When lg y is plotted against lg x, a straight line is obtained with gradient 3 and vertical intercept 0.7. Find n and k.

lg y = lg k + n(lg x)

Gradient = n, so n = 3. Intercept = lg k = 0.7, so k = 10^0.7

n = 3, k ≈ 5.01 ✓

Proofs in plane geometry

This is the one part of A-Math that isn't really about calculating an answer — it's about writing a logically complete argument, using congruence, similarity or the properties of parallel lines and circles as stated reasons for each step, the way a formal proof is built. Marks here are lost less often for wrong mathematics than for skipped steps or reasons left unstated.

Qn 5

In triangle ABC, AB = AC, and D is the midpoint of BC. Prove that AD is perpendicular to BC.

In △ABD and △ACD: AB = AC (given), BD = DC (D is midpoint), AD = AD (common)

So △ABD ≅ △ACD (SSS), hence ∠ADB = ∠ADC. Since ∠ADB + ∠ADC = 180°, each angle = 90°

AD ⊥ BC (proven) ✓

Also see E-Math

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