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.
• You may use an approved Graphing Calculator (GC) without CAS.
• Show all necessary working.
• Give your answers in exact form unless otherwise stated.

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

1. Let $f(x) = \sqrt{4 - x^2}$. State the domain and range of $f$. [2]
   
   
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2. Given $g(x) = \frac{3}{x-2}$, find the value of $k$ such that $g(x)$ is undefined at $x=k$. [1]
   
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3. Let $h(x) = 2x^2 - 5x + 3$. Find the range of $h$ for the domain $x \in [0, 3]$. [3]
   
   
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4. Determine if the function $f(x) = x^3 - x$ is a one-to-one function for the domain $x \in \mathbb{R}$. Justify your answer. [3]
   
   
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5. Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{2x+1}{x-3}$, $x \neq 3$. [3]
   
   
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Section B: Composite and Inverse Functions (Questions 6–12)

6. Given $f(x) = \ln(x)$ for $x > 0$ and $g(x) = e^{2x}$, find $fg(x)$ and state its range. [3]
   
   
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7. Let $f(x) = x^2 + 1$ for $x \ge 0$ and $g(x) = \sqrt{x-1}$ for $x \ge 1$. Show that the composite function $fg$ exists. [3]
   
   
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8. Using the functions in Question 7, find an expression for $fg(x)$ and simplify. [2]
   
   
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9. Given $h(x) = 3x - 4$ and $k(x) = \frac{1}{x}$, find the domain of the composite function $hk$. [2]
   
   
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10. Let $f(x) = \frac{1}{x+1}$ for $x > -1$. Find the domain of $f^{-1}$. [2]
    
    
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11. Show that for $f(x) = \frac{x+2}{x-1}$, $ff(x) = x$ for all $x$ in the domain. [4]
    
    
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12. If $f(x) = 2x + 3$ and $g(x) = x^2 - 1$, solve for $x$ such that $fg(x) = gf(x)$. [4]
    
    
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Section C: Graphs and Transformations (Questions 13–20)

13. The graph of $y = f(x)$ is translated by the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. Write the equation of the new graph in terms of $f(x)$. [2]
    
    
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14. Given $y = 2f(x+1) - 3$, describe the sequence of transformations that maps $y = f(x)$ onto this graph. [3]
    
    
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15. Let $f(x) = e^x$. Sketch the graph of $y = |f(x)|$ and $y = f(|x|)$ on the same axes. [4]
    
    
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16. Find the coordinates of the turning point of $y = 3(x-2)^2 + 5$ and state its nature. [2]
    
    
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17. A function is defined by $y = \frac{1}{x^2 - 4}$. State the equations of all vertical and horizontal asymptotes. [3]
    
    
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18. Given the parametric equations $x = 2t$ and $y = t^2 - 1$, find the Cartesian equation of the curve. [3]
    
    
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19. For the curve in Question 18, find the coordinates of the point where the curve meets the $x$-axis. [3]
    
    
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20. Let $f(x) = \frac{x}{x+1}$. Sketch the graph of $y = \frac{1}{f(x)}$ for $x > 0$. [4]
    
    
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Answer Key - A-Level Maths H2 Quiz (Algebra Functions)

1. Domain: $[-2, 2]$. Range: $[0, 2]$. (2 marks)
2. $k = 2$. (1 mark)
3. Vertex at $x = -(-5)/(2 \cdot 2) = 1.25$. $h(1.25) = 2(1.25)^2 - 5(1.25) + 3 = -0.125$. Endpoints: $h(0)=3, h(3)=6$. Range: $[-0.125, 6]$. (3 marks)
4. Not one-to-one. $f(x) = x(x-1)(x+1)$. $f(0)=0, f(1)=0, f(-1)=0$. Since multiple $x$ values map to the same $y$, it is not one-to-one. (3 marks)
5. $y = \frac{2x+1}{x-3} \implies yx - 3y = 2x + 1 \implies x(y-2) = 3y+1 \implies x = \frac{3y+1}{y-2}$. $f^{-1}(x) = \frac{3x+1}{x-2}$. (3 marks)
6. $fg(x) = \ln(e^{2x}) = 2x$. Since $g(x) > 0$ for all $x$, and $2x$ is linear, Range: $\mathbb{R}$. (3 marks)
7. Range of $g(x) = \sqrt{x-1}$ is $[0, \infty)$. Domain of $f(x) = x^2+1$ is $[0, \infty)$. Since Range($g$) $\subseteq$ Domain($f$), $fg$ exists. (3 marks)
8. $fg(x) = (\sqrt{x-1})^2 + 1 = x - 1 + 1 = x$. (2 marks)
9. $k(x) = 1/x$ requires $x \neq 0$. Domain: $\{x \in \mathbb{R} : x \neq 0\}$. (2 marks)
10. Range of $f(x) = \frac{1}{x+1}$ for $x > -1$ is $(0, \infty)$. Domain of $f^{-1}$ is $(0, \infty)$. (2 marks)
11. $ff(x) = \frac{\frac{x+2}{x-1} + 2}{\frac{x+2}{x-1} - 1} = \frac{x+2 + 2x-2}{x+2 - (x-1)} = \frac{3x}{3} = x$. (4 marks)
12. $fg(x) = 2(x^2-1)+3 = 2x^2+1$. $gf(x) = (2x+3)^2 - 1 = 4x^2 + 12x + 8$. $2x^2+1 = 4x^2 + 12x + 8 \implies 2x^2 + 12x + 7 = 0$. $x = \frac{-12 \pm \sqrt{144 - 56}}{4} = \frac{-12 \pm \sqrt{88}}{4} = -3 \pm \frac{\sqrt{22}}{2}$. (4 marks)
13. $y = f(x+2) + 3$. (2 marks)
14. 1. Translation by vector $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$. 2. Stretch parallel to $y$-axis scale factor 2. 3. Translation by vector $\begin{pmatrix} 0 \\ -3 \end{pmatrix}$. (3 marks)
15. $|f(x)|$ is same as $e^x$ (since $e^x > 0$). $f(|x|)$ is symmetric about $y$-axis (mirror image of $e^x$ for $x<0$). (4 marks)
16. $(2, 5)$, Minimum. (2 marks)
17. Vertical: $x=2, x=-2$. Horizontal: $y=0$. (3 marks)
18. $t = x/2 \implies y = (x/2)^2 - 1 \implies y = \frac{x^2}{4} - 1$. (3 marks)
19. $0 = \frac{x^2}{4} - 1 \implies x^2 = 4 \implies x = \pm 2$. Points: $(2, 0)$ and $(-2, 0)$. (3 marks)
20. $y = \frac{x+1}{x} = 1 + \frac{1}{x}$. Horizontal asymptote $y=1$, vertical asymptote $x=0$. Curve is a hyperbola in the first quadrant. (4 marks)