Specialist Mathematics Exam 1: Mon 9 Nov, 9:00am — 30 days away

ATARMAxxing · Specialist Mathematics

VCE Specialist Mathematics Practice Questions

64 exam-style questions · full worked solutions

The 64 practice questions inside the VCE Specialist Mathematics Mastery Pack, grouped by area of study. Every question comes with a full worked solution.

  1. AOS19 questions · 20 marks
    • Multiple choice × 5
    • Prove × 4
  2. AOS28 questions · 19 marks
    • Multiple choice × 5
    • Analyse × 1
    • Sketch × 1
    • Express × 1
  3. AOS38 questions · 18 marks
    • Multiple choice × 5
    • Solve × 2
    • Find × 1
  4. AOS57 questions · 17 marks
    • Multiple choice × 4
    • Prove × 1
    • Find × 2
  5. AOS4a8 questions · 19 marks
    • Multiple choice × 5
    • Integrate by parts × 1
    • Use partial fractions × 1
    • Arc length of a parametric curve × 1
  6. AOS4b7 questions · 20 marks
    • Multiple choice × 4
    • Form and solve a differential equation × 1
    • Separation of variables × 1
    • Logistic model application × 1
  7. AOS4c6 questions · 20 marks
    • Multiple choice × 3
    • Rectilinear motion with calculus × 1
    • Acceleration as a function of position × 1
    • Vector calculus: projectile motion × 1
  8. AOS6a5 questions · 12 marks
    • Multiple choice × 3
    • Linear combination of random variables × 1
    • Difference of two independent normals × 1
  9. AOS6b6 questions · 19 marks
    • Multiple choice × 3
    • Construct a confidence interval × 1
    • Two-tailed hypothesis test × 1
    • One-tailed test with sample standard deviation × 1
Sample question
Prove by mathematical induction that for all integers n ≥ 1, 1·1! + 2·2! + 3·3! + … + n·n! = (n+1)! − 1. (4 marks)
Show the worked answer

Answer: Worked solution

Let P(n) be the statement 1·1! + 2·2! + … + n·n! = (n+1)! − 1. Base step (n = 1): LHS = 1·1! = 1; RHS = 2! − 1 = 2 − 1 = 1. So LHS = RHS and P(1) is true. Inductive step: assume P(k) holds for some integer k ≥ 1, i.e. 1·1! + … + k·k! = (k+1)! − 1. Consider the sum to k+1 terms: 1·1! + … + k·k! + (k+1)(k+1)! = [(k+1)! − 1] + (k+1)(k+1)! (by the assumption) = (k+1)!·[1 + (k+1)] − 1 = (k+1)!·(k+2) − 1 = (k+2)! − 1 = ((k+1)+1)! − 1. This is exactly P(k+1). Hence P(k) ⇒ P(k+1). Since P(1) is true and the implication holds, by the principle of mathematical induction P(n) is true for all integers n ≥ 1. ∎
Included in the VCE Specialist Mathematics Mastery Pack

20 full-length practice exams with worked solutions, 20 revision notes, 64 practice questions and 200 flashcards.

Unlock Specialist Mathematics — $20

Preview a sample note and question free on the VCE Specialist Mathematics hub →

Specialist Mathematics · 64 practice questions