Year 11 HSC Maths - Mathematics Advanced sample questions
Year 11 HSC Maths - Mathematics Advanced sample questions
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Year 11 HSC Mathematics Advanced practice questions
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Skills covered
Practice functions, calculus, trigonometry, exponential and logarithmic relationships, probability, statistics, and modelling.
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Sample HSC Year 11 Mathematics Advanced questions
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HSC Year 11Mathematics AdvancedFunctions and transformationshardtransformation graph
1. The graph of y = f(x) has stationary point A(-1, 4). For g(x) = 2f(x - 3) - 5, which point is the corresponding stationary point on y = g(x)?
Choices
(2, 3)
(-4, 3)
(2, -13)
(-1, 3)
Explanation:
The input x - 3 must equal -1, so x = 2. The output becomes 2 x 4 - 5 = 3, so the corresponding stationary point is (2, 3).
HSC Year 11Mathematics AdvancedDifferential calculushardtangent graph
2. For f(x) = x3 - 3x2 - 9x + 2, the tangent T is drawn at P where x = 4. What is the gradient of T?
Choices
15
-15
5
29
Explanation:
Differentiate to get f'(x) = 3x2 - 6x - 9. At x = 4, f'(4) = 48 - 24 - 9 = 15, so the tangent gradient is 15.
HSC Year 11Mathematics AdvancedIntegral calculushardtangent and area graph
3. A particle has velocity v(t) = t(6 - t) for 0 ≤ t ≤ 6, measured in m/s. What is the displacement over this interval?
Choices
36 m
18 m
72 m
0 m
Explanation:
Displacement is the integral of velocity. The integral from 0 to 6 of t(6 - t) is [3t2 - t3/3] from 0 to 6 = 108 - 72 = 36 m.
HSC Year 11Mathematics AdvancedExponential modellinghardmodel graph
4. A population model is P(t) = 150e0.12t, where t is measured in years. When does the model first reach 300?
Choices
(25 ln 2)/3 years
3 ln 2 / 25 years
0.12 ln 2 years
150 ln 2 years
Explanation:
Solve 150e0.12t = 300, so e0.12t = 2. Taking logarithms gives 0.12t = ln 2, hence t = (25 ln 2)/3 years.
HSC Year 11Mathematics AdvancedContinuous probabilityharddensity graph
5. For the continuous random variable X with density f(x) = kx2, 0 ≤ x ≤ 3, what is P(X > 2)?
Choices
19/27
8/27
1/3
2/3
Explanation:
First 1 = the integral from 0 to 3 of kx2, so 1 = 9k and k = 1/9. Then P(X > 2) = the integral from 2 to 3 of x2/9, which is (27 - 8)/27 = 19/27.
HSC Year 11Mathematics AdvancedNormal distributionhardnormal curve
6. For X ~ N(μ, σ2), P(X < 58) = 0.1587 and P(X < 70) = 0.8413. Using Φ(-1) = 0.1587 and Φ(1) = 0.8413, what are μ and σ?
Choices
μ = 64, σ = 6
μ = 64, σ = 12
μ = 58, σ = 6
μ = 70, σ = 6
Explanation:
The probabilities place 58 one standard deviation below the mean and 70 one standard deviation above it. The mean is the midpoint (58 + 70)/2 = 64, and the standard deviation is 70 - 64 = 6.
For parents comparing HSC Year 11 Mathematics Advanced support
HSC Year 11 Mathematics Advanced practice should make functions, calculus, modelling, probability, and statistics feel exact but learnable. These examples preview graph-backed, single-best-answer questions before the no-login Mathematics Advanced demo.
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Practice Summary
Year 11 · HSC Maths - Mathematics Advanced · Test mode · 4 questions · 10 min
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Correct answersPending
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