Year 12 HSC Maths - Mathematics Extension 2 sample questions
Year 12 HSC Maths - Mathematics Extension 2 sample questions
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Year 12 HSC Mathematics Extension 2 practice questions
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Skills covered
Practice complex numbers, proof, vectors, mechanics, integration techniques, and further calculus.
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Sample HSC Year 12 Mathematics Extension 2 questions
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HSC Year 12Mathematics Extension 2Complex numbershardArgand diagram
1. On the Argand diagram, z = 2 + 3i and w = (1 + i)z. Which Cartesian form gives w?
HSC Year 12Mathematics Extension 2Further integrationhardtangent and area graph
2. What is the exact value of the integral from 0 to 1 of x e2x dx?
Choices
(e2 + 1)/4
(e2 - 1)/4
e2/2
(2e2 + 1)/4
Explanation:
Using integration by parts, the antiderivative is e2x(x/2 - 1/4). Evaluating from 0 to 1 gives e2/4 - (-1/4) = (e2 + 1)/4.
HSC Year 12Mathematics Extension 2Vectorshardvector diagram
3. Points A(1, 2, 0) and B(4, -1, 2) define a line. Which vector equation represents the line AB?
Choices
r = (1, 2, 0) + λ(3, -3, 2)
r = (1, 2, 0) + λ(5, 1, 2)
r = (4, -1, 2) + λ(3, 3, -2)
r = (3, -3, 2) + λ(1, 2, 0)
Explanation:
The direction vector is AB = (4 - 1, -1 - 2, 2 - 0) = (3, -3, 2). A valid line equation is r = (1, 2, 0) + λ(3, -3, 2).
HSC Year 12Mathematics Extension 2Mechanicshardgraph
4. A particle has velocity v(t) = kt(4 - t). If v(1) = 6, what is the acceleration at t = 3?
Choices
-4 m/s2
4 m/s2
6 m/s2
-2 m/s2
Explanation:
From v(1) = 3k = 6, k = 2. Thus v(t) = 8t - 2t2 and a(t) = v'(t) = 8 - 4t. At t = 3, a(3) = -4 m/s2.
HSC Year 12Mathematics Extension 2Proof and counterexamplehardtransformation graph
5. Which counterexample disproves the claim: if f'(0) = 0, then f has a local maximum at x = 0?
Choices
f(x) = x3
f(x) = -x2
f(x) = 1 - x2
f(x) = 4
Explanation:
For f(x) = x3, f'(x) = 3x2, so f'(0) = 0. But x = 0 is a stationary point of inflection, not a local maximum, so the claim is false.
HSC Year 12Mathematics Extension 2Complex rootshardroot diagram
6. The diagram marks point P, the root of z4 = 16 with argument 3π/4. Which exact Cartesian form is P?
Choices
-√2 + √2i
√2 + √2i
-√2 - √2i
2 + √2i
Explanation:
Each root has modulus 161/4 = 2. At argument 3π/4, P = 2(cos 3π/4 + i sin 3π/4) = -√2 + √2i.
For parents comparing HSC Year 12 Mathematics Extension 2 support
HSC Year 12 Mathematics Extension 2 practice should feel rigorous but readable across complex numbers, vectors, further integration, mechanics, and proof. These examples preview that style before the no-login Extension 2 demo.
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Practice Summary
Year 12 · HSC Maths - Mathematics Extension 2 · Test mode · 5 questions · 10 min
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HSC Mathematics Extension 2 practice questions FAQ
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