HSC-EXT2 · Proof

Proof

HSC HSC Mathematics Extension 2 · 2024 content registry · first HSC 2027

MEX-P1MEX-P2
Assessment evidence

What these examples cover

These examples show assessable skills, not an official HSC topic weighting or prediction. The 2024 NSW syllabuses have their first HSC examination in 2027; students sitting the 2026 HSC continue under the applicable 2017 syllabus.

Teacher check

Teacher verification note: check the mathematics, notation, permitted technology and syllabus fit for your cohort before classroom or assessment use.

3 deterministic examples

Proof questions with worked solutions

Each question is generated from an existing curriq QDSL archetype at a fixed seed. Marks and catalog content mappings travel with the generated question.

Worked example 1

3 marksext2-proof-1
An exponential inequality Prove that if x>0x>0, then e(3x)>1+3xe^(3x)>1+3x.
Part a · 3 marks

Give a calculus proof.

Worked solution

Answer: Apply derivatives to f(x)=e(3x)−1−3xf(x)=e^(3x)-1-3x.

  1. Auxiliary Function: Let f(x)=e(3x)−1−3xf(x)=e^(3x)-1-3x.
  2. Derivative Sign: f′(x)=3(e(3x)−1)>0f'(x)=3(e^(3x)-1)>0 for x>0x>0.
  3. Inequality: f(x)>f(0)=0f(x)>f(0)=0.

Catalog reference: ext2_morris_2025_q16a_exponential_inequality · seed 41001

Worked example 2

2 marksext2-proof-1
A chained absolute-value inequality Prove 3(abs(a−b)+abs(c−b)+abs(c−d))>=3(a−d)3(abs(a-b)+abs(c-b)+abs(c-d))>=3(a-d).
Part a · 2 marks

Give a direct proof.

Worked solution

Answer: Apply the triangle inequality to obtain abs(a−c)+abs(c−d)>=abs(a−d)abs(a-c)+abs(c-d)>=abs(a-d), then use abs(a−d)>=a−dabs(a-d)>=a-d.

  1. Triangle Inequality: abs(a−b)+abs(b−c)>=abs(a−c)abs(a-b)+abs(b-c)>=abs(a-c)
  2. Absolute Value Bound: abs(a−d)>=a−dabs(a-d)>=a-d

Catalog reference: ext2_nor25_chained_triangle_inequality · seed 41001

Worked example 3

3 marksext2-proof-1
A logarithmic inequality For a>1 and b>1, prove loga(ab(3))+logb(a(3)b)>=8log_a(a b^(3))+log_b(a^(3) b)>= 8.
Part a · 3 marks

Give a complete proof.

Worked solution

Answer: Set t=loga(b)>0t=log_a(b)>0. The left side is 2+3(t+1/t)2+3(t+1/t), and (t−1/t)2>=0(t-1/t)^2>=0 gives t+1/t>=2t+1/t>=2.

  1. Change Of Base: logb(a)=1/loga(b)log_b(a)=1/log_a(b).
  2. Nonnegative Square: t+1/t>=2t+1/t>=2.
  3. Substitute Bound: 8

Catalog reference: ext2_ind25_reciprocal_log_inequality · seed 41001

Content mapping and common errors

Proof content identifiers

MEX-P1

The Nature of Proof

Construct rigorous proofs using various techniques including contradiction, contrapositive and counterexample.

Common student mistakes
  • ·Assuming the result during a proof by contradiction instead of negating it
  • ·Confusing contrapositive (¬Q → ¬P) with converse (Q → P)
  • ·Not providing a valid counterexample that disproves both the statement and its converse
MEX-P2

Further Proof by Mathematical Induction

Prove results involving inequalities and series using strong and recursive induction.

Common student mistakes
  • ·Weak induction used where strong induction is required
  • ·Inequality direction errors in the inductive step

Source and methodology

Examples are deterministic curriq catalog generations from existing QDSL archetypes at the published fixed seed. The generator copies only student-facing prompts, mark allocations, topic mappings, display tables or options, checked answers and marking-path results into this resource.

Review the official NSW Curriculum source ↗

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