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Unit II: Algebra — Test 2

Reasoning, proof and modelling · 11 questions · 31 practice marks · 60 minutes.
MCQs are scored automatically; written answers use step-wise self/teacher review.

Section A
Question 1 (1 Mark)
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Full question paper and step-wise model answers

Practice marks are not official examination weightage. Written answers require self- or teacher assessment.

Question 1 · 1 mark

For P(x) = 4x³ − 2x + 9, the degree and coefficient of x² are:

  1. 3 and 0
  2. 2 and 4
  3. 3 and 2
  4. 9 and 0
Model answer and reasoning

(a) 3 and 0

  1. The largest exponent with non-zero coefficient is 3.
  2. There is no x² term, so its coefficient is 0.

Question 2 · 1 mark

The pair x + y = 3 and 2x + 2y = 8 has:

  1. One solution
  2. Infinitely many solutions
  3. No solution
  4. Exactly two solutions
Model answer and reasoning

(c) No solution

  1. The second equation reduces to x + y = 4.
  2. These are distinct parallel lines. No ordered pair can have both sums 3 and 4.

Question 3 · 1 mark

The expression (x² − 9)/(x − 3) simplifies to:

  1. x − 3 for all x
  2. x + 3 for x ≠ 3
  3. x + 3 for all x
  4. x² + 3
Model answer and reasoning

(b) x + 3 for x ≠ 3

  1. Factor x² − 9 = (x − 3)(x + 3).
  2. Cancel only when the original denominator is non-zero; retain x ≠ 3.

Question 4 · 1 mark

The minimum moves for four disks in the Tower of Hanoi puzzle are:

  1. 8
  2. 12
  3. 15
  4. 16
Model answer and reasoning

(c) 15

  1. To move n disks: move n − 1, move the largest disk, then move n − 1 again.
  2. Tₙ = 2Tₙ₋₁ + 1 with T₀ = 0; hence T₄ = 2⁴ − 1 = 15.

Question 5 · 4 marks

Factor x² + 5x + 6 by splitting its middle term. Describe how algebra tiles form a rectangle and derive a² − b² from (a + b)(a − b).

Model answer and reasoning

(x + 2)(x + 3); a² − b² = (a + b)(a − b).

  1. Numbers 2 and 3 have sum 5 and product 6, so write x² + 2x + 3x + 6.
  2. Group x(x + 2) + 3(x + 2) = (x + 2)(x + 3).
  3. For x ≥ 0, arrange one x² tile, five x tiles and six unit tiles as a rectangle of sides x + 2 and x + 3.
  4. Expand (a + b)(a − b): a² − ab + ab − b² = a² − b².

Question 6 · 3 marks

Simplify (a³ − b³)/(a − b), stating its restriction, and verify the resulting identity by multiplication.

Model answer and reasoning

a² + ab + b² for a ≠ b.

  1. Multiply (a − b)(a² + ab + b²).
  2. The mixed terms cancel, leaving a³ − b³.
  3. Divide by a − b only when a ≠ b; this restriction remains after simplification.

Question 7 · 4 marks

Graph y = x + 1 and y = −x + 5 using three points per line. Find their intersection. Explain what changes if x and y must be non-negative integers.

Model answer and reasoning

The lines intersect at (2,3); the restricted graphs consist only of allowed lattice points.

  1. For y = x + 1 use (0,1), (2,3), (4,5). For y = −x + 5 use (0,5), (2,3), (4,1).
  2. Plot and join each set for the real-variable graphs; they cross at (2,3).
  3. Verify 3 = 2 + 1 and 3 = −2 + 5.
  4. For integer counts retain only points with both coordinates non-negative integers. The common point is still allowed, but intermediate non-integer points are excluded.
Graph of y equals x plus one and y equals minus x plus five, meeting at two comma three(2,3)y = x + 1y = −x + 5xy0

Question 8 · 4 marks

A culture has 80 cells initially and doubles each hour. Give explicit and recursive rules, find its population after 5 hours, and contrast its plotted points with linear growth.

Model answer and reasoning

Nₙ = 80 × 2ⁿ, N₀ = 80, Nₙ₊₁ = 2Nₙ; N₅ = 2560.

  1. Each step multiplies by 2, so the values form a GP indexed from n = 0.
  2. The general rule is Nₙ = 80 × 2ⁿ and the recurrence is Nₙ₊₁ = 2Nₙ.
  3. After 5 hours the model gives 80 × 32 = 2560.
  4. Differences grow rather than staying constant. Its integer-time points are not collinear, unlike a linear model.

Question 9 · 4 marks

A line segment of length 1 is replaced at each Koch-curve step by four segments, each one third as long. After n steps, find the segment count, individual length and total length. Evaluate at n = 2.

Model answer and reasoning

4ⁿ segments of length 3⁻ⁿ; total length (4/3)ⁿ. At n = 2: 16 segments, each 1/9, total 16/9.

  1. Each stage multiplies the segment count by 4, starting with one segment.
  2. Each segment length is multiplied by 1/3, starting at length 1.
  3. Multiply count and length: 4ⁿ × 3⁻ⁿ = (4/3)ⁿ.
  4. Substitute n = 2. This illustrates GP scaling in a fractal construction.

Question 10 · 4 marks

Explain the recursive Tower of Hanoi algorithm for n disks and derive its minimum-move formula. Calculate it for 5 disks.

Model answer and reasoning

T₀ = 0; Tₙ = 2Tₙ₋₁ + 1 = 2ⁿ − 1; 31 moves for 5 disks.

  1. Move the top n − 1 disks to the spare peg using the same rules.
  2. Move the largest disk to the destination; then move the n − 1 disks onto it.
  3. This is necessary as well as sufficient: the largest disk can move only after the smaller disks are cleared, and they must be moved back. Thus Tₙ = 2Tₙ₋₁ + 1.
  4. Tₙ + 1 = 2(Tₙ₋₁ + 1); starting at T₀ + 1 = 1 gives Tₙ = 2ⁿ − 1. T₅ = 31.

Question 11 · 4 marks

Classify these pairs as uniquely solvable, consistent dependent, or inconsistent: (i) x + y = 4, x − y = 2; (ii) x + y = 4, 2x + 2y = 8; (iii) x + y = 4, x + y = 6. Describe the graphs and solve (i) by either algebraic method.

Model answer and reasoning

(i) Unique: (3,1). (ii) Infinitely many solutions. (iii) No solution.

  1. For (i), add the equations to get 2x = 6; x = 3 and y = 1. Two distinct non-parallel lines intersect once.
  2. For (ii), the second equation is twice the first. Their graphs coincide and every point on x + y = 4 is a solution.
  3. For (iii), the same left side cannot equal both 4 and 6. The graphs are distinct parallel lines.
  4. Pairs (i) and (ii) are consistent; pair (iii) is inconsistent.