A Level H2 Mathematics Algebra Functions Quiz

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A-Level Maths H2 Quiz - Algebra Functions

Name: ____________________ Class: ____________________ Date: ____________________ Score: ________ / 65

Duration: 90 Minutes
Total Marks: 65
Instructions: Answer all questions. Show all necessary working. You may use an approved graphing calculator (GC).

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Section A: Basic Functions and Equations (Questions 1–7)

1. Given $f(x) = \frac{2x+3}{x-1}$, find the domain and range of $f$. [2]
   
   
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2. Let $g(x) = \sqrt{x-4}$ for $x \ge 4$. Find an expression for $g^{-1}(x)$ and state its domain. [3]
   
   
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3. Solve the inequality $\frac{x-3}{x+2} \le 0$. [2]
   
   
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4. Find the set of values of $x$ for which $|2x - 5| < 7$. [2]
   
   
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5. Given $h(x) = x^2 - 4x + 7$, find the range of $h$ for the domain $1 \le x \le 5$. [3]
   
   
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6. Solve the system of equations: $2x + 3y = 13$ $x^2 + y^2 = 13$ [4]
   
   
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7. Sketch the graph of $y = |2x - 3|$, clearly labelling the $x$-intercept and the vertex. [3]
   
   
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Section B: Composite and Inverse Functions (Questions 8–14)

8. Let $f(x) = \frac{1}{x+2}$ and $g(x) = x^2 - 1$. Determine if the composite function $fg$ exists for all $x \in \mathbb{R}$. Justify your answer. [4]
   
   
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9. Given $f(x) = 3x - 2$ and $g(x) = \frac{x+1}{x}$, find an expression for $fg(x)$ in its simplest form. [3]
   
   
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10. Let $f(x) = e^{2x}$ and $g(x) = \ln(x-1)$. Show that the composite function $fg$ exists and find its expression. [4]
    
    
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11. Given $f(x) = \frac{x}{x-2}$ for $x \neq 2$. Show that $f$ is a one-to-one function and find $f^{-1}(x)$. [4]
    
    
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12. Let $f(x) = \sqrt{x+3}$ and $g(x) = x^2 - 4$. Find the domain of $gf$ and the range of $gf$. [5]
    
    
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13. If $f(x) = \frac{2x}{x+1}$, find the value of $x$ such that $f(x) = f^{-1}(x)$. [4]
    
    
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14. Let $f(x) = \ln(x)$ and $g(x) = e^x + 1$. Determine the domain of $gf$. [3]
    
    
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Section C: Graphs, Parametrics, and Applications (Questions 15–20)

15. A curve $C$ is defined by the parametric equations $x = 2\cos t$ and $y = 3\sin t$ for $0 \le t \le 2\pi$. Find the Cartesian equation of $C$. [3]
    
    
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16. For the curve $C$ in Question 15, sketch the graph and state the coordinates of the points where $C$ meets the $x$-axis. [4]
    
    
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17. Given the function $f(x) = \frac{1}{x}$, describe the sequence of transformations that maps $y = f(x)$ to $y = \frac{3}{x-2} + 5$. [4]
    
    
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18. A curve $C$ is defined implicitly by $x^2 + 3xy + y^2 = 10$. Show that the gradient function $\frac{dy}{dx}$ is given by $\frac{dy}{dx} = \frac{-2x - 3y}{3x + 2y}$. [5]
    
    
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19. Let $f(x) = x^2 - 2x$. Sketch the graph of $y = f(|x|)$ and state the coordinates of its stationary points. [5]
    
    
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20. A population of bacteria $P$ grows at a rate proportional to the current population. Write down a differential equation relating $P$ and time $t$. If $P=100$ at $t=0$ and $P=400$ at $t=2$, find the expression for $P$ in terms of $t$. [6]
    
    
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A-Level Maths H2 Quiz - Algebra Functions (Answer Key)

1. Domain and Range of $f(x) = \frac{2x+3}{x-1}$

• Domain: $x \in \mathbb{R}, x \neq 1$
• Range: $y \in \mathbb{R}, y \neq 2$ (Horizontal asymptote $y=2$)
• Marks: 1 for domain, 1 for range.

2. Inverse of $g(x) = \sqrt{x-4}$

• $y = \sqrt{x-4} \implies y^2 = x-4 \implies x = y^2 + 4$
• $g^{-1}(x) = x^2 + 4$
• Domain of $g^{-1}$ is range of $g$: $x \ge 0$.
• Marks: 2 for expression, 1 for domain.

3. Inequality $\frac{x-3}{x+2} \le 0$

• Critical points: $x=3, x=-2$.
• Testing intervals: $(-2, 3]$.
• Solution: $-2 < x \le 3$.
• Marks: 1 for critical points, 1 for correct interval.

4. Modulus $|2x - 5| < 7$

• $-7 < 2x - 5 < 7$
• $-2 < 2x < 12$
• $-1 < x < 6$.
• Marks: 1 for inequality setup, 1 for final range.

5. Range of $h(x) = x^2 - 4x + 7$ for $1 \le x \le 5$

• Vertex: $x = -(-4)/2 = 2$. $h(2) = 4 - 8 + 7 = 3$.
• Endpoints: $h(1) = 1 - 4 + 7 = 4$; $h(5) = 25 - 20 + 7 = 12$.
• Range: $[3, 12]$.
• Marks: 1 for vertex, 1 for endpoints, 1 for range.

6. System $2x + 3y = 13$ and $x^2 + y^2 = 13$

• $x = \frac{13-3y}{2} \implies (\frac{13-3y}{2})^2 + y^2 = 13$
• $\frac{169 - 78y + 9y^2}{4} + y^2 = 13 \implies 169 - 78y + 13y^2 = 52$
• $13y^2 - 78y + 117 = 0 \implies y^2 - 6y + 9 = 0 \implies (y-3)^2 = 0 \implies y=3$.
• $x = \frac{13-9}{2} = 2$.
• Solution: $(2, 3)$.
• Marks: 2 for substitution/quadratic, 2 for final coordinates.

7. Sketch $y = |2x - 3|$

• V-shape with vertex at $(1.5, 0)$.
• $y$-intercept at $(0, 3)$.
• Marks: 1 for vertex, 1 for $y$-intercept, 1 for correct shape.

8. Existence of $fg$ for $f(x) = \frac{1}{x+2}, g(x) = x^2 - 1$

• Range of $g$: $x^2 - 1 \ge -1$.
• Domain of $f$: $x \neq -2$.
• Since the range of $g$ includes values other than $-2$ (specifically, $g(x)$ can never be $-2$ because $x^2-1 = -2 \implies x^2 = -1$, impossible for real $x$), the range of $g$ is a subset of the domain of $f$.
• Yes, $fg$ exists for all $x \in \mathbb{R}$.
• Marks: 2 for range of $g$, 2 for comparison with domain of $f$.

9. $fg(x)$ for $f(x) = 3x - 2, g(x) = \frac{x+1}{x}$

• $fg(x) = 3(\frac{x+1}{x}) - 2 = \frac{3x+3-2x}{x} = \frac{x+3}{x}$.
• Marks: 2 for substitution, 1 for simplification.

10. $fg$ for $f(x) = e^{2x}, g(x) = \ln(x-1)$

• Domain of $g$: $x > 1$. Range of $g$: $\mathbb{R}$.
• Domain of $f$: $\mathbb{R}$.
• Range of $g \subseteq$ Domain of $f$, so $fg$ exists.
• $fg(x) = e^{2\ln(x-1)} = e^{\ln(x-1)^2} = (x-1)^2$.
• Marks: 2 for existence, 2 for expression.

11. $f(x) = \frac{x}{x-2}$

• One-to-one: $f(a)=f(b) \implies \frac{a}{a-2} = \frac{b}{b-2} \implies ab - 2a = ab - 2b \implies a=b$.
• Inverse: $y = \frac{x}{x-2} \implies xy - 2y = x \implies x(y-1) = 2y \implies x = \frac{2y}{y-1}$.
• $f^{-1}(x) = \frac{2x}{x-1}$.
• Marks: 2 for 1-to-1 proof, 2 for inverse.

12. $f(x) = \sqrt{x+3}, g(x) = x^2 - 4$

• $gf(x) = (\sqrt{x+3})^2 - 4 = x+3-4 = x-1$.
• Domain of $gf$: Domain of $f \cap \{x : f(x) \in \text{Dom } g\}$.
• $x+3 \ge 0 \implies x \ge -3$.
• Range of $gf$: Since $x \ge -3$, $gf(x) \ge -3-1 \implies gf(x) \ge -4$.
• Marks: 2 for expression, 1 for domain, 2 for range.

13. $f(x) = \frac{2x}{x+1}$

• $f^{-1}(x) = \frac{x}{2-x}$.
• $f(x) = f^{-1}(x) \implies \frac{2x}{x+1} = \frac{x}{2-x}$
• $2x(2-x) = x(x+1) \implies 4x - 2x^2 = x^2 + x \implies 3x^2 - 3x = 0 \implies 3x(x-1) = 0$.
• $x=0$ or $x=1$.
• Marks: 2 for inverse, 2 for solving.

14. Domain of $gf$ for $f(x) = \ln(x), g(x) = e^x + 1$

• $gf(x) = e^{\ln x} + 1 = x + 1$.
• Domain is restricted by the inner function $f(x)$.
• Domain: $x > 0$.
• Marks: 3 for identifying inner function restriction.

15. Cartesian equation of $x = 2\cos t, y = 3\sin t$

• $\cos t = x/2, \sin t = y/3$.
• $\cos^2 t + \sin^2 t = 1 \implies (\frac{x}{2})^2 + (\frac{y}{3})^2 = 1 \implies \frac{x^2}{4} + \frac{y^2}{9} = 1$.
• Marks: 2 for identity, 1 for final form.

16. Sketch of $\frac{x^2}{4} + \frac{y^2}{9} = 1$

• Ellipse centered at $(0,0)$.
• $x$-intercepts: set $y=0 \implies x^2=4 \implies x = \pm 2$. Points: $(2,0), (-2,0)$.
• Marks: 2 for sketch, 2 for intercepts.

17. Transformations of $f(x) = 1/x$ to $y = \frac{3}{x-2} + 5$

• 1. Translation by vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ (Right 2).
• 2. Vertical stretch by scale factor 3.
• 3. Translation by vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ (Up 5).
• Marks: 1 mark per correct transformation.

18. Implicit differentiation of $x^2 + 3xy + y^2 = 10$

• $\frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) = 0$
• $2x + 3(x\frac{dy}{dx} + y) + 2y\frac{dy}{dx} = 0$
• $\frac{dy}{dx}(3x + 2y) = -2x - 3y \implies \frac{dy}{dx} = \frac{-2x - 3y}{3x + 2y}$.
• Marks: 2 for product rule, 3 for rearrangement.

19. Sketch $y = f(|x|)$ for $f(x) = x^2 - 2x$

• $f(|x|) = |x|^2 - 2|x| = x^2 - 2|x|$.
• Graph is symmetric about $y$-axis. For $x>0$, it is $x^2-2x$ (vertex $(1, -1)$).
• For $x<0$, it is $x^2+2x$ (vertex $(-1, -1)$).
• Stationary points: $(1, -1), (-1, -1)$ and a cusp/point at $(0, 0)$.
• Marks: 2 for symmetry, 2 for sketch, 1 for points.

20. Population Growth

• DE: $\frac{dP}{dt} = kP$.
• Solution: $P = Ae^{kt}$.
• $t=0, P=100 \implies A=100$.
• $t=2, P=400 \implies 400 = 100e^{2k} \implies 4 = e^{2k} \implies 2k = \ln 4 \implies k = \ln 2$.
• $P = 100e^{(\ln 2)t} = 100(2^t)$.
• Marks: 1 for DE, 2 for general solution, 3 for constants.