A Level H2 Mathematics Algebra Functions Quiz

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

Total Marks: 60
Topic: Algebra & Functions

Section A: Functions, Domain and Range

1. [2 marks]

• Domain: $x \ge 3$ (given).
• Range: Since $\sqrt{x-3} \ge 0$, range is $y \ge 0$ (or $[0, \infty)$).
  Teaching note: The square root function outputs non-negative values. Domain is the set of allowed $x$; range is the set of resulting $y$.
  Marks: 1 for domain, 1 for range.

2. [3 marks]
(a) Range of $g$: $g(x) = \frac{1}{x-2} \ne 0$ for all $x \ne 2$, so range is $y \ne 0$ (or $\mathbb{R} \setminus \{0\}$). [1]
(b) $g$ is not one-to-one on its natural domain $x \ne 2$ because, for example, $g(1) = -1$ and $g(3) = 1$ give distinct $x$ mapping to distinct $y$ but $g(2.5)=2$ and $g(1.5)=-2$; more directly, $g(x)=g(-x+4)$ shows many-to-one. An inverse requires a one-to-one function. [2]
Teaching note: A function has an inverse only if it is one-to-one (each $y$ from at most one $x$).

3. [2 marks]
$h(x)=x^2+1$ is decreasing for $x<0$ and increasing for $x>0$. Least $a = 0$ gives restricted domain $x \ge 0$ (or $[0,\infty)$) on which it is one-to-one. [2]
Marks: 1 for $a=0$, 1 for domain.

4. [3 marks]
Let $y = 2x - 5$. Then $x = \frac{y+5}{2}$, so $p^{-1}(x) = \frac{x+5}{2}$. [2]
Domain of $p^{-1}$ is range of $p$ = $\mathbb{R}$. [1]
Teaching note: Inverse swaps domain/range; solve $y=f(x)$ for $x$.

5. [3 marks]
$y = \ln(x+1) \Rightarrow e^y = x+1 \Rightarrow x = e^y - 1$, so $q^{-1}(x) = e^x - 1$. [1]
Domain of $q^{-1}$ = range of $q$ = $\mathbb{R}$ (since $\ln$ spans all reals). [1]
Range of $q^{-1}$ = domain of $q$ = $(-1, \infty)$. [1]

Section B: Composite Functions

6. [4 marks]
Domain of $g$ is $\mathbb{R}$, range of $g$ is $\mathbb{R}$. Domain of $f$ is $x \ge 0$. Since range of $g$ ($\mathbb{R}$) is not subset of domain of $f$, check: $fg(x)=f(g(x))=f(x+1)=(x+1)^2$ requires $x+1 \ge 0 \Rightarrow x \ge -1$. Thus $fg$ exists for domain $x \ge -1$. [2 for showing existence + domain]
$fg(x) = (x+1)^2$ for $x \ge -1$. [1]
Range: minimum at $x=-1$ gives 0, so range is $y \ge 0$. [1]

7. [4 marks]
$u(x)=1/x$ for $x>0$ (range $y>0$); $v(x)=x-3$ for $x>3$ (range $y>0$). For $uv$, compute $u(v(x)) = u(x-3) = 1/(x-3)$. Need $x-3 > 0$ (domain of $u$ is $x>0$) and $x>3$ (domain of $v$). Both give $x>3$. So $uv$ exists with domain $x>3$. [3]
$uv(x) = \frac{1}{x-3}$ for $x>3$. [1]

8. [3 marks]
$gf(x) = g(f(x)) = g(3x+2) = (3x+2)^2$. [2]
Range: $(3x+2)^2 \ge 0$ for all $x$, so range is $y \ge 0$. [1]

9. [4 marks]
$g$ domain $x>0$, range $\mathbb{R}$. $f$ domain $\mathbb{R}$, range $y>0$. For $fg$, $f(g(x)) = e^{\ln x} = x$ requires $x>0$ (domain of $g$). Since $g(x)=\ln x \in \mathbb{R}$ = domain of $f$, $fg$ exists. [2]
$fg(x) = x$ for $x>0$. [1]
Domain of $fg$ is $x>0$. [1]

10. [3 marks]
$ab(x) = a(b(x)) = a(x-4) = \sqrt{x-4}$ for $x \ge 4$. [2]
Range: $\sqrt{x-4} \ge 0$, so range $y \ge 0$. [1]

Section C: Graphs, Transformations and Equations

11. [2 marks]
Graph of $y=2/x+1$: vertical asymptote $x=0$, horizontal asymptote $y=1$. Sketch shows two branches in quadrants I and III relative to asymptotes. [2: 1 asymptote, 1 sketch]

12. [3 marks]
$y=f(x-2)+3$: translation of 2 units to the right, then 3 units upward. [3: 1 each direction, order noted]

13. [3 marks]
$\frac{x-1}{x+2}>0$. Critical points: $x=1, x=-2$. Sign chart: positive for $x<-2$ or $x>1$. [3: 1 critical, 1 chart, 1 answer]
Solution: $x < -2$ or $x > 1$.

14. [3 marks]
$|x-4|<3 \iff -3 < x-4 < 3 \iff 1 < x < 7$. [3]

15. [4 marks]
$x=2t \Rightarrow t = x/2$. Substitute: $y = (x/2)^2 + 1 = x^2/4 + 1$. [3]
Cartesian: $y = \frac{x^2}{4} + 1$ (or $x^2 = 4(y-1)$). [1]

Section D: Applications and Combined Skills

16. [3 marks]
$y = x^2 - 4x = (x-2)^2 - 4$, for $x \ge 2$. Solve: $(x-2)^2 = y+4 \Rightarrow x-2 = \sqrt{y+4}$ (since $x\ge2$). So $x = 2 + \sqrt{y+4}$, $f^{-1}(x) = 2 + \sqrt{x+4}$. [2]
Domain of $f^{-1}$ = range of $f$ = $[-4, \infty)$. [1]

17. [4 marks]
$g$ domain $x\ge0$, range $y\ge0$. $f$ domain $x>-1$, range $y>0$ (since $1/(x+1)>0$). For $gf$: $g(f(x)) = (1/(x+1))^2$ requires $f(x)\ge0$ (true) and $x>-1$. So $gf$ exists. [2]
$gf(x) = \frac{1}{(x+1)^2}$ for $x>-1$. [1]
Range: $>0$, so $y>0$. [1]

18. [3 marks]
$|2x+1|>5 \iff 2x+1 < -5$ or $2x+1 > 5 \iff 2x < -6$ or $2x > 4 \iff x < -3$ or $x > 2$. [3]

19. [3 marks]
Sketch $y=|x-3|$: V-shape with vertex at $(3,0)$, lines $y=3-x$ ($x<3$) and $y=x-3$ ($x\ge3$). [2]
Vertex: $(3,0)$. [1]

20. [4 marks]
Vertical asymptote $x=-c = -2 \Rightarrow c=2$. [1]
Horizontal asymptote $y = a/1 = 3 \Rightarrow a=3$. [1]
Passes through $(0,4)$: $f(0) = \frac{b}{2} = 4 \Rightarrow b=8$. [2]