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A Level H2 Mathematics Practice Paper 4

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A Level H2 Mathematics From Real Exams Generated by Gemma 4 31B Updated 2026-09-16
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A-Level Maths H2 Quiz - Algebra Functions (Answer Key)

Section A: Functions and Composites

  1. fg(x)=f(x2−1)=2(x2−1)+3=2x2−2+3=2x2+1fg(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1 [2]

  2. f(x)=x+1x−2f(x) = \frac{x+1}{x-2}. f−1(x)f^{-1}(x): Let y=x+1x−2  ⟹  y(x−2)=x+1  ⟹  xy−2y=x+1  ⟹  x(y−1)=2y+1  ⟹  x=2y+1y−1y = \frac{x+1}{x-2} \implies y(x-2) = x+1 \implies xy - 2y = x+1 \implies x(y-1) = 2y+1 \implies x = \frac{2y+1}{y-1}. Therefore, f−1(x)=2x+1x−1f^{-1}(x) = \frac{2x+1}{x-1}. [3]

  3. f(x)=x2−4x+7f(x) = x^2 - 4x + 7. Completing the square: f(x)=(x−2)2+3f(x) = (x-2)^2 + 3. Domain: x∈Rx \in \mathbb{R}. Range: f(x)≥3f(x) \geq 3. [3]

  4. f(x)=x−3f(x) = \sqrt{x-3}. Domain: x−3≥0  ⟹  x≥3x-3 \geq 0 \implies x \geq 3. [2]

  5. f(x)=1x+2f(x) = \frac{1}{x+2}. f(f(x))=11x+2+2=11+2x+4x+2=x+22x+5f(f(x)) = \frac{1}{\frac{1}{x+2} + 2} = \frac{1}{\frac{1 + 2x + 4}{x+2}} = \frac{x+2}{2x+5}. [3]

  6. f(x)=3x−5f(x) = 3x - 5. f(x)=f−1(x)  ⟹  3x−5=x+53  ⟹  9x−15=x+5  ⟹  8x=20  ⟹  x=2.5f(x) = f^{-1}(x) \implies 3x - 5 = \frac{x+5}{3} \implies 9x - 15 = x + 5 \implies 8x = 20 \implies x = 2.5. [3]

  7. f(x)=x2+2x−3f(x) = x^2 + 2x - 3. f(x)=0  ⟹  (x+3)(x−1)=0  ⟹  x=−3,1f(x) = 0 \implies (x+3)(x-1) = 0 \implies x = -3, 1. [2]

  8. f(x)=e2xf(x) = e^{2x}. f−1(x)=12ln⁡xf^{-1}(x) = \frac{1}{2} \ln x. [2]

Section B: Modulus Functions and Inequalities

  1. ∣2x−5∣<3  ⟹  −3<2x−5<3  ⟹  2<2x<8  ⟹  1<x<4|2x - 5| < 3 \implies -3 < 2x - 5 < 3 \implies 2 < 2x < 8 \implies 1 < x < 4. [3]

  2. ∣x+2∣≥5  ⟹  x+2≥5|x + 2| \geq 5 \implies x + 2 \geq 5 or x+2≤−5  ⟹  x≥3x + 2 \leq -5 \implies x \geq 3 or x≤−7x \leq -7. [3]

  3. f(x)=∣x−3∣+∣x+1∣f(x) = |x - 3| + |x + 1|.

    • For x<−1x < -1: f(x)=−(x−3)−(x+1)=−2x+2f(x) = -(x-3) - (x+1) = -2x + 2.
    • For −1≤x<3-1 \leq x < 3: f(x)=−(x−3)+(x+1)=4f(x) = -(x-3) + (x+1) = 4.
    • For x≥3x \geq 3: f(x)=(x−3)+(x+1)=2x−2f(x) = (x-3) + (x+1) = 2x - 2. [5]

  4. ∣3x−2∣=∣x+4∣|3x - 2| = |x + 4|. 3x−2=x+4  ⟹  2x=6  ⟹  x=33x - 2 = x + 4 \implies 2x = 6 \implies x = 3. 3x−2=−(x+4)  ⟹  4x=−2  ⟹  x=−0.53x - 2 = -(x + 4) \implies 4x = -2 \implies x = -0.5. [4]

  5. f(x)=∣2x−1∣−3f(x) = |2x - 1| - 3. Vertex: 2x−1=0  ⟹  x=0.52x - 1 = 0 \implies x = 0.5. f(0.5)=−3f(0.5) = -3. x-intercepts: ∣2x−1∣=3  ⟹  2x−1=3|2x - 1| = 3 \implies 2x - 1 = 3 or 2x−1=−3  ⟹  x=22x - 1 = -3 \implies x = 2 or x=−1x = -1. [5]

  6. ∣x−1∣+∣x−4∣=3|x - 1| + |x - 4| = 3.

    • x<1x < 1: −(x−1)−(x−4)=3  ⟹  −2x+5=3  ⟹  x=1-(x-1) - (x-4) = 3 \implies -2x + 5 = 3 \implies x = 1 (not in range).
    • 1≤x<41 \leq x < 4: −(x−1)+(x−4)=3  ⟹  −3=3-(x-1) + (x-4) = 3 \implies -3 = 3 (no solution). Wait, check: (x−1)(x-1) is positive, (x−4)(x-4) is negative. Correct: (x−1)−(x−4)=3  ⟹  3=3(x-1) - (x-4) = 3 \implies 3 = 3. This is true for all x∈[1,4]x \in [1, 4].
    • x≥4x \geq 4: (x−1)+(x−4)=3  ⟹  2x−5=3  ⟹  x=4(x-1) + (x-4) = 3 \implies 2x - 5 = 3 \implies x = 4. Solution: 1≤x≤41 \leq x \leq 4. [5]

  7. ∣x2−4∣<3|x^2 - 4| < 3. −3<x2−4<3  ⟹  1<x2<7-3 < x^2 - 4 < 3 \implies 1 < x^2 < 7. 1<x<71 < x < \sqrt{7} or −7<x<−1-\sqrt{7} < x < -1. [6]

Section C: Advanced Algebra and Polynomials

  1. f(x)=x3−6x2+11x−6f(x) = x^3 - 6x^2 + 11x - 6. Possible roots: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6. f(1)=1−6+11−6=0f(1) = 1 - 6 + 11 - 6 = 0. (x−1)(x2−5x+6)=(x−1)(x−2)(x−3)(x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3). Roots: x=1,2,3x = 1, 2, 3. [5]

  2. f(x)=2x3+ax2+bx+4f(x) = 2x^3 + ax^2 + bx + 4. f(1)=0  ⟹  2+a+b+4=0  ⟹  a+b=−6f(1) = 0 \implies 2 + a + b + 4 = 0 \implies a + b = -6. f(−2)=0  ⟹  −16+4a−2b+4=0  ⟹  4a−2b=12  ⟹  2a−b=6f(-2) = 0 \implies -16 + 4a - 2b + 4 = 0 \implies 4a - 2b = 12 \implies 2a - b = 6. Adding: 3a=0  ⟹  a=03a = 0 \implies a = 0. b=−6b = -6. [6]

  3. P(x)=x4−5x2+4P(x) = x^4 - 5x^2 + 4. (x2−1)(x2−4)=(x−1)(x+1)(x−2)(x+2)(x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2). [4]

  4. f(x)=x3−3x+2f(x) = x^3 - 3x + 2. f′(x)=3x2−3f'(x) = 3x^2 - 3. 3x2−3=0  ⟹  x=±13x^2 - 3 = 0 \implies x = \pm 1. f(1)=1−3+2=0f(1) = 1 - 3 + 2 = 0 (Local Min). f(−1)=−1+3+2=4f(-1) = -1 + 3 + 2 = 4 (Local Max). [6]

  5. f(x)=x3+px+qf(x) = x^3 + px + q. f′(x)=3x2+pf'(x) = 3x^2 + p. For 3 distinct real roots, f′(x)=0f'(x) = 0 must have 2 distinct roots   ⟹  p<0\implies p < 0. Local max f(−−p/3)f(-\sqrt{-p/3}) and local min f(−p/3)f(\sqrt{-p/3}) must have opposite signs. f(−p/3)⋅f(−−p/3)<0f(\sqrt{-p/3}) \cdot f(-\sqrt{-p/3}) < 0. [8]