A Level H2 Mathematics Algebra Functions Quiz

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A-Level Maths H2 Quiz - Algebra Functions

Name: ________________________
Class: ________________________
Date: ________________________
Score: _______ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

• Answer all 20 questions.
• Show your working clearly where required.
• Use a graphing calculator where appropriate.
• Write your answers in the spaces provided.

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Section A: Functions, Domain and Range (Questions 1–5)

1. [2 marks] The function $f$ is defined by $f(x) = \sqrt{x - 3}$. State the domain and range of $f$.

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2. [2 marks] Explain why the function $g(x) = x^2 - 4x + 3$, with domain $x \in \mathbb{R}$, does not have an inverse function.

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3. [3 marks] The function $h$ is defined by $h(x) = 2x + 5$ for $x \in \mathbb{R}$. Find $h^{-1}(x)$ and state its domain.

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4. [2 marks] A function $p$ is given by $p(x) = \dfrac{1}{x + 2}$, $x \neq -2$. State the range of $p$.

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5. [3 marks] The function $q$ is defined as $q(x) = \ln(x - 1)$ for $x > 1$. Find $q^{-1}(x)$ and state the domain of $q^{-1}$.

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Section B: Composite and Inverse Functions (Questions 6–10)

6. [4 marks] The functions $f$ and $g$ are defined by $f(x) = x^2$, $x \geq 0$, and $g(x) = x + 1$, $x \in \mathbb{R}$. Show that the composite function $fg$ exists. Find an expression for $fg(x)$ and state its domain and range.

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7. [3 marks] Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$, find $gf(x)$ and state its range.

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8. [4 marks] The function $f: x \mapsto x^2 + 2x$, $x \geq -1$. (i) Explain why $f$ has an inverse. (ii) Find $f^{-1}(x)$ and state its domain.

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9. [3 marks] Functions $u$ and $v$ are defined by $u(x) = \dfrac{1}{x}$, $x > 0$, and $v(x) = x - 4$, $x > 4$. Determine whether the composite function $uv$ exists. Justify your answer.

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10. [4 marks] Let $f(x) = \dfrac{x}{x - 1}$, $x \neq 1$, and $g(x) = 2x + 3$, $x \in \mathbb{R}$. Find $fg(x)$ and $g f(x)$. State the domain of $fg$.

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Section C: Graphs, Transformations and Equations (Questions 11–15)

11. [2 marks] The graph of $y = f(x)$ has a vertical asymptote at $x = 2$ and horizontal asymptote $y = 0$. Write the equation of the graph of $y = f(x + 3)$.

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12. [3 marks] Given $f(x) = \dfrac{1}{x}$, describe the transformation from $y = f(x)$ to $y = -f(x) + 2$. State the new asymptotes.

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13. [3 marks] A curve $C$ has parametric equations $x = 2t$, $y = t^2 - 1$ for $t \in \mathbb{R}$. Find the Cartesian equation of $C$.

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14. [4 marks] The function $f(x) = \dfrac{2x + 1}{x - 3}$, $x \neq 3$. (i) Find the equations of the vertical and horizontal asymptotes. (ii) Sketch the graph of $y = f(x)$, indicating all intercepts.

Image pending generation: graph for Q14.

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15. [3 marks] The graph of $y = |f(x)|$ is obtained from $y = f(x)$ where $f(x) = x - 4$. Sketch the graph of $y = |x - 4|$ and state its range.

Image pending generation: graph for Q15.

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Section D: Inequalities and Modulus (Questions 16–20)

16. [2 marks] Solve the inequality $|x - 5| < 3$.

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17. [3 marks] Solve $\dfrac{x - 2}{x + 1} > 0$.

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18. [3 marks] Find the set of values of $x$ for which $|2x + 1| \geq 5$.

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19. [3 marks] Solve the inequality $x^2 - 3x - 4 < 0$.

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20. [4 marks] (a) [2 marks] State the relation $|x - a| > b \iff x < a - b$ or $x > a + b$. Use it to solve $|x - 2| > 4$.
(b) [2 marks] Hence write down the solution set.

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A-Level Maths H2 Quiz - Algebra Functions (Answer Key)

Total Marks: 50
Topic: Algebra & Functions (Syllabus 9758 Strand 1)

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Section A: Functions, Domain and Range

Q1. [2 marks]

• Domain: $x - 3 \geq 0 \Rightarrow x \geq 3$, so domain is $[3, \infty)$.
• Range: $\sqrt{x-3} \geq 0$, so range is $[0, \infty)$.
  Teaching note: Square root requires non-negative input; output is never negative.
  Marks: 1 for domain, 1 for range.

Q2. [2 marks]
$g(x) = x^2 - 4x + 3 = (x-2)^2 - 1$ is a parabola opening upwards with turning point at $(2,-1)$. It is not one-to-one on $\mathbb{R}$ (fails horizontal line test). Hence no inverse.
Teaching note: Only one-to-one functions have inverses. A quadratic over all reals is many-to-one.
Marks: 1 for identifying not one-to-one, 1 for correct reason (parabola / horizontal line test).

Q3. [3 marks]
$y = 2x + 5 \Rightarrow x = \dfrac{y - 5}{2}$, so $h^{-1}(x) = \dfrac{x - 5}{2}$. Domain: $x \in \mathbb{R}$.
Teaching note: Linear function is one-to-one; inverse found by swapping x,y and rearranging.
Marks: 1 inverse expr, 1 domain, 1 working.

Q4. [2 marks]
$p(x) = \dfrac{1}{x+2}$, $x \neq -2$. As $x \to \pm\infty$, $p \to 0$; never zero. Range: $y \neq 0$ (all real except 0).
Marks: 2 for correct range.

Q5. [3 marks]
$y = \ln(x-1) \Rightarrow e^y = x - 1 \Rightarrow x = e^y + 1$. So $q^{-1}(x) = e^x + 1$. Domain of $q^{-1}$ is range of $q$: $x \in \mathbb{R}$.
Marks: 1 inverse, 1 domain, 1 working.

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Section B: Composite and Inverse Functions

Q6. [4 marks]
$g(x)=x+1$ range is $\mathbb{R}$; domain of $f$ is $x \geq 0$. Range of $g$ not subset of domain of $f$ for all x, but $fg(x)=f(g(x))$ requires $g(x) \geq 0 \Rightarrow x \geq -1$. So $fg$ exists for domain $x \geq -1$.
$fg(x) = (x+1)^2$. Domain: $x \geq -1$. Range: $[0,\infty)$.
Marks: 1 existence, 1 expr, 1 domain, 1 range.

Q7. [3 marks]
$gf(x) = g(f(x)) = (3x-2)^2 + 1 = 9x^2 -12x +5$. Since square $\geq 0$, min at x=2/3 gives 1. Range: $[1,\infty)$.
Marks: 2 expr, 1 range.

Q8. [4 marks]
(i) $f(x)=x^2+2x=(x+1)^2-1$, domain $x\geq -1$ is strictly increasing, hence one-to-one → inverse exists.
(ii) $y=(x+1)^2-1 \Rightarrow (x+1)^2 = y+1 \Rightarrow x = -1 + \sqrt{y+1}$ (since $x\geq -1$). So $f^{-1}(x) = -1 + \sqrt{x+1}$. Domain: $x \geq -1$.
Marks: 2 (i), 2 (ii).

Q9. [3 marks]
$v(x)=x-4$, for $x>4$ range is $(0,\infty)$. $u$ domain $x>0$. Range of $v \subseteq$ domain of $u$, so $uv$ exists.
Marks: 1 conclusion, 2 justification.

Q10. [4 marks]
$fg(x) = f(g(x)) = \dfrac{2x+3}{2x+3-1} = \dfrac{2x+3}{2x+2}$, $x \neq -1$.
$g f(x) = g(f(x)) = 2\left(\dfrac{x}{x-1}\right)+3 = \dfrac{2x+3x-3}{x-1} = \dfrac{5x-3}{x-1}$, $x\neq1$.
Domain of $fg$: $x \neq -1$.
Marks: 1 each expression, 1 domain, 1 overall.

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Section C: Graphs, Transformations and Equations

Q11. [2 marks]
Replace $x$ by $x+3$: vertical asymptote becomes $x+3=2 \Rightarrow x=-1$; horizontal unchanged $y=0$. Equation: $y = f(x+3)$ with asymptote $x=-1$.
Marks: 1 asymptote, 1 eq.

Q12. [3 marks]
$-f(x)$ reflects in x-axis; $+2$ translates up 2. New asymptotes: $y=2$ (horizontal), $x=0$ (vertical).
Marks: 1 transform, 2 asymptotes.

Q13. [3 marks]
$t = x/2 \Rightarrow y = (x/2)^2 - 1 = x^2/4 - 1$. Cartesian: $y = \frac{x^2}{4} - 1$.
Marks: 3 for correct elimination.

Q14. [4 marks]
(i) Vertical: $x=3$; horizontal: $y=2$ (ratio of coefficients).
(ii) Intercepts: x: $2x+1=0 \Rightarrow x=-0.5$; y: $x=0 \Rightarrow y=-1/3$. Sketch shows two branches.
Marks: 2 asymptotes, 2 sketch/intercepts.

Q15. [3 marks]
Graph V-shape vertex (4,0). Range: $[0,\infty)$.
Marks: 2 sketch, 1 range.

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Section D: Inequalities and Modulus

Q16. [2 marks]
$|x-5|<3 \iff 2 < x < 8$.
Marks: 2.

Q17. [3 marks]
Critical points: $x=2$, $x=-1$. Sign chart: positive for $x<-1$ or $x>2$. Solution: $x<-1$ or $x>2$.
Marks: 1 pts, 2 intervals.

Q18. [3 marks]
$2x+1 \geq 5 \Rightarrow x \geq 2$; or $2x+1 \leq -5 \Rightarrow x \leq -3$. Set: $x \leq -3$ or $x \geq 2$.
Marks: 3.

Q19. [3 marks]
$(x-4)(x+1)<0 \Rightarrow -1 < x < 4$.
Marks: 3.

Q20. [4 marks]
(a) $|x-2|>4 \Rightarrow x< -2$ or $x>6$.
(b) Solution set: $\{x: x<-2 \text{ or } x>6\}$.
Marks: 2 (a), 2 (b).