A Level H1 Mathematics Algebra Functions Quiz

A-Level Maths H1 Quiz - Algebra Functions

Answer Key

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1.
(a) Domain of $f$: $-2 \leq x \leq 2$ or $[-2, 2]$
[1] — This is given in the definition of $f$. The domain is the set of all permissible input values of $x$.

(b) For $f(x) = \sqrt{4 - x^2}$:

• The expression under the square root, $4 - x^2$, ranges from $0$ (when $x = \pm 2$) to $4$ (when $x = 0$).
• So $\sqrt{4 - x^2}$ ranges from $\sqrt{0} = 0$ to $\sqrt{4} = 2$.
• Range: $0 \leq f(x) \leq 2$ or $[0, 2]$
  [2] — 1 mark for identifying the minimum value 0, 1 mark for identifying the maximum value 2 and stating the range.

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2.
(a) Let $y = g(x) = \frac{2x + 3}{x - 1}$.
Swap $x$ and $y$: $x = \frac{2y + 3}{y - 1}$
$x(y - 1) = 2y + 3$
$xy - x = 2y + 3$
$xy - 2y = x + 3$
$y(x - 2) = x + 3$
$y = \frac{x + 3}{x - 2}$

$g^{-1}(x) = \frac{x + 3}{x - 2}$
Domain of $g^{-1}$: $x \neq 2$ (i.e., $x \in \mathbb{R}, x \neq 2$)
[3] — 2 marks for correct algebraic manipulation to find $g^{-1}(x)$, 1 mark for correct domain.

(b) $g(3) = \frac{2(3) + 3}{3 - 1} = \frac{9}{2} = 4.5$
[1]

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3. $f(x) = x^2 - 4x + 7$
Complete the square: $f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$
Since $(x - 2)^2 \geq 0$, the minimum occurs when $(x - 2)^2 = 0$, i.e., $x = 2$.
Minimum value: $f(2) = 0 + 3 = 3$
[3] — 1 mark for completing the square, 1 mark for $x = 2$, 1 mark for minimum value 3.

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4.
(a) To show $f$ and $g$ are inverses, show $fg(x) = x$ and $gf(x) = x$:
$fg(x) = f(\ln x) = e^{\ln x} = x$ (for $x > 0$)
$gf(x) = g(e^x) = \ln(e^x) = x$ (for all $x \in \mathbb{R}$)
Since both compositions give the identity, $f$ and $g$ are inverse functions.
[2] — 1 mark for each composition shown correctly.

(b) $gf(x) = \ln(e^x) = x$

• Domain of $gf$: all real numbers, $x \in \mathbb{R}$ (since $e^x$ is always positive and $\ln$ is defined for all positive inputs)
• Range of $gf$: all real numbers, $y \in \mathbb{R}$
  [2] — 1 mark for domain, 1 mark for range.

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5.
(a) For all real $x$, $x^2 \geq 0$, so $x^2 + 1 \geq 1 > 0$. Therefore $\frac{1}{x^2 + 1} > 0$ for all $x \in \mathbb{R}$.
[1] — The denominator is always positive (minimum value 1), so the fraction is always positive.

(b) $h(x) = \frac{1}{x^2 + 1}$ is maximised when the denominator $x^2 + 1$ is minimised.
The minimum of $x^2 + 1$ is $1$ (when $x = 0$).
Maximum value of $h(x) = \frac{1}{1} = 1$.
[2] — 1 mark for identifying minimum denominator, 1 mark for maximum value 1.

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6.
(a) $f(1) = 3(1) - 2 = 1$
$gf(1) = g(1) = 1^2 + 1 = 2$
[1]

(b) $fg(x) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1$
[2] — 1 mark for correct substitution, 1 mark for simplification.

(c) $fg(x) = 25$: $3x^2 + 1 = 25$
$3x^2 = 24$
$x^2 = 8$
$x = \pm\sqrt{8} = \pm 2\sqrt{2}$
[2] — 1 mark for setting up equation, 1 mark for correct solutions.

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7.
(a) Let $y = \frac{x + 4}{x - 2}$.
$x = \frac{y + 4}{y - 2}$
$x(y - 2) = y + 4$
$xy - 2x = y + 4$
$xy - y = 2x + 4$
$y(x - 1) = 2x + 4$
$y = \frac{2x + 4}{x - 1}$

$f^{-1}(x) = \frac{2x + 4}{x - 1}$, for $x \neq 1$
[3] — 2 marks for correct algebra, 1 mark for correct expression.

(b) Set $f(x) = f^{-1}(x)$:
$\frac{x + 4}{x - 2} = \frac{2x + 4}{x - 1}$
$(x + 4)(x - 1) = (2x + 4)(x - 2)$
$x^2 - x + 4x - 4 = 2x^2 - 4x + 4x - 8$
$x^2 + 3x - 4 = 2x^2 - 8$
$0 = x^2 - 3x - 4$
$(x - 4)(x + 1) = 0$
$x = 4$ or $x = -1$

Both values are valid (neither equals 2 or 1).
[2] — 1 mark for setting up equation, 1 mark for correct solutions.

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8. $fg(x) = f(x^2 - 3) = 2(x^2 - 3) + 5 = 2x^2 - 6 + 5 = 2x^2 - 1$
$gf(x) = g(2x + 5) = (2x + 5)^2 - 3 = 4x^2 + 20x + 25 - 3 = 4x^2 + 20x + 22$

Set $fg(x) = gf(x)$:
$2x^2 - 1 = 4x^2 + 20x + 22$
$0 = 2x^2 + 20x + 23$

Using the quadratic formula:
$x = \frac{-20 \pm \sqrt{400 - 4(2)(23)}}{2(2)} = \frac{-20 \pm \sqrt{400 - 184}}{4} = \frac{-20 \pm \sqrt{216}}{4}$
$= \frac{-20 \pm 6\sqrt{6}}{4} = \frac{-10 \pm 3\sqrt{6}}{2}$

$x = \frac{-10 + 3\sqrt{6}}{2} \approx -2.68$ or $x = \frac{-10 - 3\sqrt{6}}{2} \approx -7.32$
[4] — 1 mark for each of $fg(x)$ and $gf(x)$, 1 mark for setting up equation, 1 mark for correct solutions.

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9.
(a) Let $y = \sqrt{x + 3}$.
$x = \sqrt{y + 3}$
$x^2 = y + 3$
$y = x^2 - 3$

$f^{-1}(x) = x^2 - 3$
Domain of $f^{-1}$: $x \geq 0$ (since the range of $f$ is $[0, \infty)$)
Range of $f^{-1}$: $y \geq -3$
[3] — 1 mark for correct $f^{-1}(x)$, 1 mark for domain, 1 mark for range.

(b) The graph of $y = f(x) = \sqrt{x + 3}$ is the standard square root curve shifted 3 units left, starting at $(-3, 0)$ and passing through $(1, 2)$.
The graph of $y = f^{-1}(x) = x^2 - 3$ is a parabola with vertex at $(0, -3)$, but only the portion for $x \geq 0$ is plotted (since the domain of $f^{-1}$ is $x \geq 0$).
The two graphs are reflections of each other across the line $y = x$.
[2] — 1 mark for correct shape of each graph, 1 mark for showing reflection symmetry about $y = x$.

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10.
(a) Let $y = \frac{3x - 1}{2}$.
$2y = 3x - 1$
$3x = 2y + 1$
$x = \frac{2y + 1}{3}$

$f^{-1}(x) = \frac{2x + 1}{3}$
[1]

(b) $f(f^{-1}(x)) = f\!\left(\frac{2x + 1}{3}\right) = \frac{3\left(\frac{2x+1}{3}\right) - 1}{2} = \frac{2x + 1 - 1}{2} = \frac{2x}{2} = x$ ✓
[1]

(c) $f^{-1}f(5) = 5$ (since $f^{-1}f(x) = x$ for all $x$ in the domain of $f$)
[2] — 1 mark for correct answer, 1 mark for reasoning (or direct computation: $f(5) = 7$, $f^{-1}(7) = 5$).

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11.
(a) From the graph, at $x = 3$, the curve is approximately at $y \approx 1$.
Answer: $f(3) \approx 1$ (accept values in range 0.5 to 1.5)
[1] — Reading from the graph.

(b) The line $y = 2$ intersects the curve at approximately 3 points (once between $x = 0$ and $x = 2$, once between $x = 2$ and $x = 4$, and possibly once more depending on curve shape).
Given the curve passes through $(0,1)$, $(2,5)$, and $(4,-3)$, the curve rises from $y = 1$ to $y = 5$ then falls to $y = -3$. The horizontal line $y = 2$ crosses the curve twice (once on the way up, once on the way down).
Number of solutions: 2
[2] — 1 mark for correct number, 1 mark for reasoning.

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12. $y = x^2 - 2x - 3$
Complete the square: $y = (x - 1)^2 - 1 - 3 = (x - 1)^2 - 4$

This represents a translation of $y = x^2$ by 1 unit in the positive $x$-direction and 4 units in the negative $y$-direction.
Translation vector: $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$
[3] — 1 mark for correct completed square form, 1 mark for identifying horizontal shift, 1 mark for identifying vertical shift.

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13.
(a) From the graph, the vertex is at $(1, -2)$. The parabola passes through $(0, 0)$.
$f(x) = a(x - 1)^2 - 2$
Substitute $(0, 0)$: $0 = a(0 - 1)^2 - 2 = a - 2$, so $a = 2$.
$f(x) = 2(x - 1)^2 - 2$
[2] — 1 mark for identifying vertex form, 1 mark for correct value of $a$.

(b)(i) $y = f(x + 2) = 2(x + 2 - 1)^2 - 2 = 2(x + 1)^2 - 2$
This is a translation of $f(x)$ by 2 units to the left.
Vertex: $(-1, -2)$; $y$-intercept: $f(2) = 2(3)^2 - 2 = 16$, so $(0, 16)$.
[2] — 1 mark for correct vertex, 1 mark for correct sketch/intercepts.

(b)(ii) $y = -f(x) = -[2(x - 1)^2 - 2] = -2(x - 1)^2 + 2$
This is a reflection of $f(x)$ in the $x$-axis.
Vertex: $(1, 2)$; $y$-intercept: $-f(0) = -0 = 0$, so $(0, 0)$.
[2] — 1 mark for correct vertex, 1 mark for correct sketch/intercepts.

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14. $f(x) = |2x - 4| = 2|x - 2|$
This is a V-shaped graph with vertex at $x = 2$, where $f(2) = 0$.

• Vertex: $(2, 0)$
• $y$-intercept: $f(0) = |0 - 4| = 4$, so $(0, 4)$
• At $x = 5$: $f(5) = |10 - 4| = 6$, so $(5, 6)$
• At $x = -1$: $f(-1) = |-2 - 4| = 6$, so $(-1, 6)$

The graph is a V-shape with the vertex at $(2, 0)$, rising linearly on both sides.
[3] — 1 mark for correct shape (V-shape), 1 mark for vertex, 1 mark for intercepts.

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15.
(a) The graph of $y = \frac{1}{x}$ is a rectangular hyperbola in the first and third quadrants, with asymptotes $x = 0$ (y-axis) and $y = 0$ (x-axis).
[1]

(b) $y = f(x - 2) + 1 = \frac{1}{x - 2} + 1$
This is a translation of $y = \frac{1}{x}$ by 2 units right and 1 unit up.
[1]

(c) The vertical asymptote moves from $x = 0$ to $x = 2$.
The horizontal asymptote moves from $y = 0$ to $y = 1$.
Equations: $x = 2$ and $y = 1$
[3] — 1 mark for vertical asymptote, 1 mark for horizontal asymptote, 1 mark for both stated as equations.

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16.
(a) $P(x) = -2x^2 + 80x - 300$
This is a downward-opening parabola. The maximum occurs at $x = -\frac{b}{2a} = -\frac{80}{2(-2)} = \frac{80}{4} = 20$.
Number of units: 20
[2] — 1 mark for formula, 1 mark for correct answer.

(b) $P(20) = -2(400) + 80(20) - 300 = -800 + 1600 - 300 = 500$
Maximum daily profit: $500
[1]

(c) Set $P(x) = 0$: $-2x^2 + 80x - 300 = 0$
$x^2 - 40x + 150 = 0$
$x = \frac{40 \pm \sqrt{1600 - 600}}{2} = \frac{40 \pm \sqrt{1000}}{2} = \frac{40 \pm 10\sqrt{10}}{2} = 20 \pm 5\sqrt{10}$

$x = 20 - 5\sqrt{10} \approx 4.19$ and $x = 20 + 5\sqrt{10} \approx 35.8$
Break-even points: $x \approx 4.19$ and $x \approx 35.8$
[2] — 1 mark for setting up equation, 1 mark for correct solutions.

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17.
(a) Initial temperature: $T(0) = 20 + 80e^0 = 20 + 80 = 100$°C
[1]

(b) As $t \to \infty$, $e^{-0.1t} \to 0$, so $T(t) \to 20$°C.
The object approaches 20°C (room/ambient temperature).
[1]

(c) $T(t) = 40$:
$20 + 80e^{-0.1t} = 40$
$80e^{-0.1t} = 20$
$e^{-0.1t} = 0.25$
$-0.1t = \ln(0.25) = -\ln 4$
$t = \frac{\ln 4}{0.1} = 10\ln 4 = 10 \times 1.3863 \approx 13.9$ minutes
[3] — 1 mark for setting up equation, 1 mark for correct logarithmic step, 1 mark for correct answer to 3 s.f.

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18.
(a) Let $y = \ln(3x + 6)$.
$e^y = 3x + 6$
$3x = e^y - 6$
$x = \frac{e^y - 6}{3}$

$f^{-1}(x) = \frac{e^x - 6}{3}$
[2] — 1 mark for correct exponential step, 1 mark for correct expression.

(b) Domain of $f^{-1}$: all real numbers ($x \in \mathbb{R}$), since $e^x$ is defined for all $x$.
Range of $f^{-1}$: $y > -2$, since $e^x > 0$, so $\frac{e^x - 6}{3} > \frac{-6}{3} = -2$.
[2] — 1 mark for domain, 1 mark for range.

(c) $f(x) = 2$: $\ln(3x + 6) = 2$
$3x + 6 = e^2$
$x = \frac{e^2 - 6}{3} = \frac{7.389 - 6}{3} = \frac{1.389}{3} \approx 0.463$
[1]

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19.
(a) $h(t) = -5t^2 + 20t + 1.5$
Maximum occurs at $t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2$ seconds.
$h(2) = -5(4) + 20(2) + 1.5 = -20 + 40 + 1.5 = 21.5$ m
Maximum height: 21.5 m
[2] — 1 mark for correct time, 1 mark for correct height.

(b) Ball hits ground when $h(t) = 0$:
$-5t^2 + 20t + 1.5 = 0$
$5t^2 - 20t - 1.5 = 0$
$t = \frac{20 \pm \sqrt{400 + 30}}{10} = \frac{20 \pm \sqrt{430}}{10}$

$\sqrt{430} \approx 20.736$
$t = \frac{20 + 20.736}{10} \approx 4.07$ s (rejecting the negative root)
[2] — 1 mark for setting up equation, 1 mark for correct positive solution.

(c) The ball starts at $h(0) = 1.5$ m, rises to 21.5 m, then falls back to 0 m.
Range: $0 \leq h(t) \leq 21.5$
[1]

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20.
(a) $f(0) = 2$: $\frac{b}{c} = 2$, so $b = 2c$ ... (i)
$f(1) = 3$: $\frac{a + b}{1 + c} = 3$, so $a + b = 3(1 + c) = 3 + 3c$ ... (ii)
$f(-1) = 1$: $\frac{-a + b}{-1 + c} = 1$, so $-a + b = c - 1$ ... (iii)

From (i): $b = 2c$. Substitute into (ii): $a + 2c = 3 + 3c$, so $a = 3 + c$ ... (iv)
Substitute into (iii): $-(3 + c) + 2c = c - 1$
$-3 - c + 2c = c - 1$
$-3 + c = c - 1$
$-3 = -1$ — contradiction. Let me recheck.

From (iii): $-a + b = c - 1$. Substitute $a = 3 + c$ and $b = 2c$:
$-(3 + c) + 2c = c - 1$
$-3 - c + 2c = c - 1$
$-3 + c = c - 1$
$-3 = -1$ — this is inconsistent. Let me re-derive.

From (ii): $a + b = 3 + 3c$. From (i): $b = 2c$, so $a = 3 + 3c - 2c = 3 + c$.
From (iii): $-a + b = c - 1$, so $-(3+c) + 2c = c - 1$, giving $-3 + c = c - 1$, so $-3 = -1$.

Let me re-examine. Perhaps $f(-1) = 1$ means $\frac{-a + b}{-1 + c} = 1$, so $-a + b = -1 + c$.

Then: $-(3+c) + 2c = -1 + c$
$-3 - c + 2c = -1 + c$
$-3 + c = -1 + c$
$-3 = -1$. Still inconsistent.

Let me try a different approach. Set $c = 2$ (so $b = 4$ from (i)). Then from (ii): $a + 4 = 3(3) = 9$, so $a = 5$. Check (iii): $\frac{-5 + 4}{-1 + 2} = \frac{-1}{1} = -1 \neq 1$.

Try $c = -2$ (so $b = -4$). From (ii): $a - 4 = 3(-1) = -3$, so $a = 1$. Check (iii): $\frac{-1 + (-4)}{-1 + (-2)} = \frac{-5}{-3} = \frac{5}{3} \neq 1$.

Let me re-read the problem. $f(x) = \frac{ax + b}{x + c}$.

$f(0) = \frac{b}{c} = 2 \Rightarrow b = 2c$
$f(1) = \frac{a + b}{1 + c} = 3 \Rightarrow a + b = 3 + 3c$
$f(-1) = \frac{-a + b}{-1 + c} = 1 \Rightarrow -a + b = c - 1$

Adding (ii) and (iii): $(a + b) + (-a + b) = (3 + 3c) + (c - 1)$
$2b = 2 + 4c$
$b = 1 + 2c$

But from (i): $b = 2c$. So $2c = 1 + 2c$, giving $0 = 1$. Contradiction.

The system as stated is inconsistent. Let me adjust the question values to make it consistent. I'll use $f(0) = 2$, $f(1) = 3$, $f(2) = \frac{5}{2}$ instead.

Revised Q20: The function $f$ is defined by $f(x) = \frac{ax + b}{x + c}$, where $a$, $b$, and $c$ are constants. It is given that $f(0) = 2$, $f(1) = 3$, and $f(2) = \frac{5}{2}$.

$f(0) = 2$: $\frac{b}{c} = 2$, so $b = 2c$ ... (i)
$f(1) = 3$: $\frac{a + b}{1 + c} = 3$, so $a + b = 3 + 3c$ ... (ii)
$f(2) = \frac{5}{2}$: $\frac{2a + b}{2 + c} = \frac{5}{2}$, so $2(2a + b) = 5(2 + c)$, giving $4a + 2b = 10 + 5c$ ... (iii)

From (i) and (ii): $a = 3 + 3c - 2c = 3 + c$.
Substitute into (iii): $4(3 + c) + 2(2c) = 10 + 5c$
$12 + 4c + 4c = 10 + 5c$
$12 + 8c = 10 + 5c$
$3c = -2$
$c = -\frac{2}{3}$

$b = 2c = -\frac{4}{3}$
$a = 3 + c = 3 - \frac{2}{3} = \frac{7}{3}$

[4] — 1 mark for each equation set up, 1 mark for correct solution.

(b) $f(x) = \frac{\frac{7}{3}x - \frac{4}{3}}{x - \frac{2}{3}} = \frac{7x - 4}{3x - 2}$

Let $y = \frac{7x - 4}{3x - 2}$.
$y(3x - 2) = 7x - 4$
$3xy - 2y = 7x - 4$
$3xy - 7x = 2y - 4$
$x(3y - 7) = 2y - 4$
$x = \frac{2y - 4}{3y - 7}$

$f^{-1}(x) = \frac{2x - 4}{3x - 7}$, for $x \neq \frac{7}{3}$
[2] — 1 mark for correct algebraic manipulation, 1 mark for correct final answer.

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