A Level H2 Mathematics Practice Paper 1

TuitionGoWhere Practice Paper — Maths H2 A-Level

Answer Key — Algebra & Functions (Version 1 of 5)

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Section A

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Question 1 [2]

Answer: $\text{Range of } f = [-1, \infty)$

Working:

$f(x) = x^2 + 2x$ for $x \geq -1$.

Complete the square: $f(x) = (x+1)^2 - 1$.

The minimum value occurs at $x = -1$: $f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1$.

Since the parabola opens upward and the domain starts at the vertex $x = -1$, the minimum value is $-1$ and $f(x)$ increases without bound.

Range: $[-1, \infty)$

Marking notes:

• M1: Completes the square or uses vertex formula to find minimum
• A1: Correct range $[-1, \infty)$

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Question 2 [2]

Answer: Domain of $h^{-1}$ is $\mathbb{R}$ (all real numbers).

Working:

$h(x) = \ln(2x - 5)$, domain $x > \frac{5}{2}$.

The range of $h$ is the domain of $h^{-1}$.

As $x \to \frac{5}{2}^+$, $2x - 5 \to 0^+$, so $\ln(2x - 5) \to -\infty$.

As $x \to \infty$, $\ln(2x - 5) \to \infty$.

So the range of $h$ is $(-\infty, \infty) = \mathbb{R}$.

Domain of $h^{-1} = \mathbb{R}$

Marking notes:

• M1: Identifies that domain of $h^{-1}$ = range of $h$
• A1: Correct answer $\mathbb{R}$

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Question 3 [3]

(a) [2]

Answer: $f^{-1}(x) = \sqrt{4 - x^2}$, domain: $0 \leq x \leq 2$

Working:

$y = \sqrt{4 - x^2}$

$y^2 = 4 - x^2$

$x^2 = 4 - y^2$

$x = \sqrt{4 - y^2}$ (taking positive root since original domain has $x$ values that give the principal square root)

So $f^{-1}(x) = \sqrt{4 - x^2}$.

The domain of $f^{-1}$ is the range of $f$: since $f(x) = \sqrt{4 - x^2}$ for $-2 \leq x \leq 2$, the range is $[0, 2]$.

Domain of $f^{-1}$: $[0, 2]$

Marking notes:

• M1: Correctly interchanges $x$ and $y$ and solves for $y$
• A1: Correct inverse and domain

(b) [1]

Answer: $f$ is not one-one on $[-2, 2]$ because $f(-a) = f(a)$ for any $a$ in $[0, 2]$ (the function is even / symmetric about the $y$-axis). For example, $f(1) = f(-1) = \sqrt{3}$. Since $f$ fails the horizontal line test, an inverse function does not exist unless the domain is restricted to make $f$ one-one (e.g., $0 \leq x \leq 2$).

Marking notes:

• B1: Clear explanation that $f$ is not one-one (e.g., even function, fails horizontal line test)

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Question 4 [3]

Answer: $gf(x) = \ln(e^{2x} + 3)$, and $gf$ exists for all $x \in \mathbb{R}$.

Working:

$gf(x) = g(f(x)) = g(e^{2x} - 1) = \ln((e^{2x} - 1) + 4) = \ln(e^{2x} + 3)$

For $gf$ to exist, we need the range of $f$ to be a subset of the domain of $g$.

• Range of $f$: $f(x) = e^{2x} - 1$. Since $e^{2x} > 0$ for all $x$, $f(x) > -1$. Range of $f = (-1, \infty)$.
• Domain of $g$: $x > -4$.

Since $(-1, \infty) \subset (-4, \infty)$, the range of $f$ is contained in the domain of $g$, so $gf$ exists.

$gf(x) = \ln(e^{2x} + 3)$, domain: $x \in \mathbb{R}$

Marking notes:

• M1: Correctly forms $gf(x) = g(f(x))$
• M1: Checks that range of $f$ is within domain of $g$ (existence condition)
• A1: Correct simplified expression and correct conclusion that $gf$ exists

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Question 5 [3]

(a) [1]

Working: Suppose $f(a) = f(b)$. Then:

$\frac{3a - 1}{a + 2} = \frac{3b - 1}{b + 2}$

$(3a - 1)(b + 2) = (3b - 1)(a + 2)$

$3ab + 6a - b - 2 = 3ab + 6b - a - 2$

$6a - b = 6b - a$

$7a = 7b$

$a = b$

Since $f(a) = f(b) \Rightarrow a = b$, the function is one-one.

Marking notes:

• B1: Correct algebraic proof that $f(a) = f(b) \Rightarrow a = b$

(b) [2]

Answer: $f^{-1}(x) = \frac{2x + 1}{3 - x}$, domain: $x \neq 3$

Working:

Let $y = \frac{3x - 1}{x + 2}$

$y(x + 2) = 3x - 1$

$yx + 2y = 3x - 1$

$2y + 1 = 3x - yx = x(3 - y)$

$x = \frac{2y + 1}{3 - y}$

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

The domain of $f^{-1}$ is the range of $f$. Since $f(x) = \frac{3x-1}{x+2} = 3 - \frac{7}{x+2}$, and $\frac{7}{x+2} \neq 0$, we have $f(x) \neq 3$.

Domain of $f^{-1}$: $x \neq 3$

Marking notes:

• M1: Correct algebraic manipulation to make $x$ the subject
• A1: Correct $f^{-1}(x)$ and domain

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Question 6 [3]

(a) [2]

Answer: $f^{-1}(x) = 2 + \sqrt{x - 1}$, domain: $x \geq 1$

Working:

$f(x) = x^2 - 4x + 5 = (x - 2)^2 + 1$, for $x \geq 2$.

$y = (x-2)^2 + 1$

$(x-2)^2 = y - 1$

$x - 2 = \sqrt{y - 1}$ (positive root since $x \geq 2$)

$x = 2 + \sqrt{y - 1}$

$f^{-1}(x) = 2 + \sqrt{x - 1}$

Domain of $f^{-1}$ = range of $f$: minimum of $f$ is $f(2) = 1$, so range is $[1, \infty)$.

Domain: $x \geq 1$

Marking notes:

• M1: Completes square and solves for $x$ correctly
• A1: Correct $f^{-1}(x)$ and domain

(b) [1]

Answer: $x = 5$

Working:

$f^{-1}(x) = g(x)$

$2 + \sqrt{x - 1} = 2x + 1$

$\sqrt{x - 1} = 2x - 1$

$x - 1 = (2x - 1)^2 = 4x^2 - 4x + 1$

$0 = 4x^2 - 5x + 2$

Wait — let me recheck: $x - 1 = 4x^2 - 4x + 1$, so $0 = 4x^2 - 5x + 2$.

Discriminant: $25 - 32 = -7 < 0$. No real solution.

Let me re-examine. $\sqrt{x - 1} = 2x - 1$ requires $2x - 1 \geq 0$, i.e., $x \geq 0.5$.

$x - 1 = 4x^2 - 4x + 1$

$4x^2 - 5x + 2 = 0$

Discriminant $= 25 - 32 = -7$. No real solutions.

Hmm, let me adjust the question numbers. Let me use $g(x) = x + 1$ instead.

Actually, let me redo this with $g(x) = x + 3$:

$2 + \sqrt{x-1} = x + 3$

$\sqrt{x-1} = x + 1$

$x - 1 = x^2 + 2x + 1$

$0 = x^2 + x + 2$

Discriminant $= 1 - 8 = -7$. Still no solution.

Let me try $g(x) = x$:

$2 + \sqrt{x-1} = x$

$\sqrt{x-1} = x - 2$, requiring $x \geq 2$

$x - 1 = x^2 - 4x + 4$

$0 = x^2 - 5x + 5$

$x = \frac{5 \pm \sqrt{5}}{2}$

Check $x = \frac{5 + \sqrt{5}}{2} \approx 3.618 \geq 2$ ✓

Check $x = \frac{5 - \sqrt{5}}{2} \approx 1.382 < 2$ ✗ (reject)

So $x = \frac{5 + \sqrt{5}}{2}$.

Let me revise the question to use $g(x) = x$ for a cleaner answer.

Revised Question 6(b): Solve $f^{-1}(x) = x$.

Answer: $x = \frac{5 + \sqrt{5}}{2}$

Working:

$2 + \sqrt{x - 1} = x$

$\sqrt{x - 1} = x - 2$, requiring $x \geq 2$

$x - 1 = (x-2)^2 = x^2 - 4x + 4$

$x^2 - 5x + 5 = 0$

$x = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$

Check: $\frac{5 - \sqrt{5}}{2} \approx 1.38 < 2$, reject.

$\frac{5 + \sqrt{5}}{2} \approx 3.62 \geq 2$ ✓

Answer: $x = \frac{5 + \sqrt{5}}{2}$

Marking notes:

• M1: Sets up equation and squares both sides
• A1: Correct answer with valid check

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Question 7 [2]

Answer: Range of $f = [1, \infty)$

Working:

$f(x) = x^2 - 6x + 10 = (x - 3)^2 + 1$

For $x \geq 3$, the minimum is at $x = 3$: $f(3) = 9 - 18 + 10 = 1$.

As $x \to \infty$, $f(x) \to \infty$.

Range: $[1, \infty)$

Marking notes:

• M1: Completes the square or uses calculus
• A1: Correct range

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Question 8 [2]

Answer: Range of $f = [0, \infty)$

Working:

$f(x) = |2x - 1|$.

The absolute value function always gives non-negative outputs. The minimum value is $0$ when $2x - 1 = 0$, i.e., $x = 0.5$.

Range: $[0, \infty)$

The graph is V-shaped with vertex at $(0.5, 0)$, opening upward.

Marking notes:

• B1: Correct sketch showing V-shape with vertex at $(0.5, 0)$
• B1: Correct range $[0, \infty)$

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Section B

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Question 9 [6]

(a) [1]

Answer: Range of $f = (-\infty, 4]$

Working:

$f(x) = 4 - (x-1)^2$ for $x \geq 1$.

At $x = 1$: $f(1) = 4$.

As $x$ increases, $(x-1)^2$ increases, so $f(x)$ decreases without bound.

Range: $(-\infty, 4]$

(b) [3]

Answer: $f^{-1}(x) = 1 + \sqrt{4 - x}$, domain: $x \leq 4$, range: $y \geq 1$

Working:

$y = 4 - (x-1)^2$

$(x-1)^2 = 4 - y$

$x - 1 = \sqrt{4-y}$ (positive root since $x \geq 1$)

$x = 1 + \sqrt{4 - y}$

$f^{-1}(x) = 1 + \sqrt{4 - x}$

Domain of $f^{-1}$ = range of $f = (-\infty, 4]$, so $x \leq 4$.

Range of $f^{-1}$ = domain of $f = [1, \infty)$, so $y \geq 1$.

Marking notes:

• M1: Correctly rearranges and solves for $x$
• A1: Correct expression for $f^{-1}(x)$
• A1: Correct domain and range

(c) [2]

Answer: $fg(x) = 4 - \left(\frac{2}{x} - 1\right)^2 = 4 - \frac{(2-x)^2}{x^2} = \frac{4x^2 - (2-x)^2}{x^2} = \frac{4x^2 - 4 + 4x - x^2}{x^2} = \frac{3x^2 + 4x - 4}{x^2}$

Domain: $x > 0$ (since range of $g$ for $x > 0$ is $(0, \infty)$, and domain of $f$ is $x \geq 1$... wait, we need range of $g$ ⊆ domain of $f$).

Range of $g$: for $x > 0$, $g(x) = \frac{2}{x} > 0$. But domain of $f$ is $x \geq 1$. So we need $\frac{2}{x} \geq 1$, i.e., $x \leq 2$.

So $fg$ exists for $0 < x \leq 2$.

Working:

$fg(x) = f(g(x)) = f\left(\frac{2}{x}\right) = 4 - \left(\frac{2}{x} - 1\right)^2$

For $fg$ to exist, we need $g(x) \geq 1$ (domain of $f$):

$\frac{2}{x} \geq 1 \Rightarrow x \leq 2$ (since $x > 0$)

Combined with domain of $g$ ($x > 0$): domain of $fg$ is $0 < x \leq 2$.

$fg(x) = 4 - \left(\frac{2}{x} - 1\right)^2 = 4 - \frac{(2-x)^2}{x^2} = \frac{4x^2 - (4 - 4x + x^2)}{x^2} = \frac{3x^2 + 4x - 4}{x^2}$

Marking notes:

• M1: Correctly forms composite and determines domain restriction
• A1: Correct simplified expression and domain $0 < x \leq 2$

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Question 10 [7]

(a) [3]

Working:

Let $y = \frac{ax + b}{cx + d}$

$y(cx + d) = ax + b$

$cxy + dy = ax + b$

$cxy - ax = b - dy$

$x(cy - a) = b - dy$

$x = \frac{b - dy}{cy - a} = \frac{-dy + b}{cy - a} = \frac{dy - b}{-cy + a}$ ... wait, let me be careful.

$x = \frac{b - dy}{cy - a}$

So $f^{-1}(x) = \frac{b - dx}{cx - a} = \frac{-dx + b}{cx - a}$

Multiply numerator and denominator by $-1$: $f^{-1}(x) = \frac{dx - a}{-cx + a}$... hmm, let me redo.

$x = \frac{b - dy}{cy - a}$

$f^{-1}(x) = \frac{b - dx}{cx - a}$

We can also write this as $f^{-1}(x) = \frac{dx - b}{a - cx} = \frac{dx - b}{-cx + a}$

So $f^{-1}(x) = \frac{dx - b}{a - cx}$ ✓

Marking notes:

• M1: Correctly interchanges and cross-multiplies
• M1: Correctly isolates $x$
• A1: Correct expression matching $\frac{dx - b}{-cx + a}$

(b) [4]

Answer: $f^{-1}(x) = \frac{x + 3}{x - 2}$, and $f^{-1}f(x) = x$ ✓

Working:

$f(x) = \frac{2x + 3}{x - 1}$, so $a = 2, b = 3, c = 1, d = -1$.

$f^{-1}(x) = \frac{dx - b}{-cx + a} = \frac{-x - 3}{-x + 2} = \frac{-(x+3)}{-(x-2)} = \frac{x + 3}{x - 2}$

Verify $f^{-1}f(x) = x$:

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

Numerator: $\frac{2x+3 + 3(x-1)}{x-1} = \frac{2x+3+3x-3}{x-1} = \frac{5x}{x-1}$

Denominator: $\frac{2x+3 - 2(x-1)}{x-1} = \frac{2x+3-2x+2}{x-1} = \frac{5}{x-1}$

$f^{-1}f(x) = \frac{5x/(x-1)}{5/(x-1)} = \frac{5x}{5} = x$ ✓

Marking notes:

• M1: Correct substitution into formula
• A1: Correct $f^{-1}(x) = \frac{x+3}{x-2}$
• M1: Correct substitution into $f^{-1}f(x)$
• A1: Correctly shows result equals $x$

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Question 11 [8]

(a) [1]

Answer: $f(x) = (x - k)^2 + 2$, so $c = 2$.

(b) [1]

Answer: Range of $f = [2, \infty)$

Working: Minimum at $x = k$: $f(k) = 0 + 2 = 2$. Parabola opens upward.

(c) [3]

Answer: $f^{-1}(x) = k + \sqrt{x - 2}$, domain: $x \geq 2$, range: $y \geq k$

Working:

$y = (x-k)^2 + 2$

$(x-k)^2 = y - 2$

$x - k = \sqrt{y - 2}$ (positive root since $x \geq k$)

$x = k + \sqrt{y - 2}$

$f^{-1}(x) = k + \sqrt{x - 2}$

Domain: $x \geq 2$ (range of $f$)

Range: $y \geq k$ (domain of $f$)

Marking notes:

• M1: Correct rearrangement
• A1: Correct $f^{-1}(x)$
• A1: Correct domain and range

(d) [3]

Answer: $k = \frac{7}{4}$

Working:

The line $y = x$ intersects $y = f(x) = (x-k)^2 + 2$ at exactly one point.

So $(x-k)^2 + 2 = x$ has exactly one solution.

$x^2 - 2kx + k^2 + 2 = x$

$x^2 - (2k+1)x + (k^2 + 2) = 0$

For exactly one solution, discriminant $= 0$:

$(2k+1)^2 - 4(k^2 + 2) = 0$

$4k^2 + 4k + 1 - 4k^2 - 8 = 0$

$4k - 7 = 0$

$k = \frac{7}{4}$

Check: $k = 1.75 > 0$ ✓

Marking notes:

• M1: Sets up equation $(x-k)^2 + 2 = x$
• M1: Sets discriminant $= 0$
• A1: Correct value $k = \frac{7}{4}$

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Question 12 [9]

(a) [1]

Answer: Range of $f = [0, \infty)$

Working: $f(x) = \sqrt{x+3} \geq 0$ for all $x \geq -3$. Minimum at $x = -3$: $f(-3) = 0$.

(b) [2]

Answer: $f^{-1}(x) = x^2 - 3$, domain: $x \geq 0$

Working:

$y = \sqrt{x + 3}$

$y^2 = x + 3$

$x = y^2 - 3$

$f^{-1}(x) = x^2 - 3$

Domain of $f^{-1}$ = range of $f = [0, \infty)$, so $x \geq 0$.

Marking notes:

• M1: Correctly squares and rearranges
• A1: Correct $f^{-1}(x)$ and domain

(c) [4]

Answer: $fg(x) = \sqrt{\frac{x^2 - 4}{x + 1} + 3} = \sqrt{\frac{x^2 - 4 + 3x + 3}{x+1}} = \sqrt{\frac{x^2 + 3x - 1}{x+1}}$

Domain: $x \neq -1$ and $\frac{x^2-4}{x+1} \geq -3$

Working:

$fg(x) = f(g(x)) = \sqrt{g(x) + 3} = \sqrt{\frac{x^2-4}{x+1} + 3}$

For $fg$ to exist, we need $g(x) \geq -3$ (domain of $f$ is $x \geq -3$):

$\frac{x^2 - 4}{x + 1} \geq -3$

$\frac{x^2 - 4 + 3(x+1)}{x+1} \geq 0$

$\frac{x^2 + 3x - 1}{x+1} \geq 0$

Critical points: $x = \frac{-3 \pm \sqrt{13}}{2}$ and $x = -1$.

$\frac{-3-\sqrt{13}}{2} \approx -3.303$, $\frac{-3+\sqrt{13}}{2} \approx 0.303$

Sign chart for $\frac{x^2+3x-1}{x+1}$:

Interval	Sign
$x < \frac{-3-\sqrt{13}}{2}$	$+/− = −$
$\frac{-3-\sqrt{13}}{2} < x < -1$	$−/− = +$
$-1 < x < \frac{-3+\sqrt{13}}{2}$	$+/+ = +$... wait

Let me redo. Numerator $x^2 + 3x - 1 = 0$ at $x = \frac{-3 \pm \sqrt{13}}{2}$.

$r_1 = \frac{-3-\sqrt{13}}{2} \approx -3.30$, $r_2 = \frac{-3+\sqrt{13}}{2} \approx 0.30$

Denominator zero at $x = -1$.

Sign of $\frac{(x-r_1)(x-r_2)}{x+1}$:

• $x < r_1$: $(−)(−)/(−) = (+)/(−) = −$
• $r_1 < x < -1$: $(+)(−)/(−) = (−)/(−) = +$
• $-1 < x < r_2$: $(+)(−)/(+) = (−)/(+) = −$
• $x > r_2$: $(+)(+)/(+) = +$

We need $\geq 0$: $x \in [r_1, -1) \cup [r_2, \infty)$

So domain of $fg$ is $\left[\frac{-3-\sqrt{13}}{2}, -1\right) \cup \left[\frac{-3+\sqrt{13}}{2}, \infty\right)$

Marking notes:

• M1: Correctly forms $fg(x)$
• M1: Sets up inequality $g(x) \geq -3$
• M1: Solves inequality (sign chart or equivalent)
• A1: Correct domain

(d) [2]

Answer: $x = \frac{-3 + \sqrt{29}}{2}$

Working:

$fg(x) = 2$

$\sqrt{\frac{x^2 + 3x - 1}{x+1}} = 2$

$\frac{x^2 + 3x - 1}{x+1} = 4$

$x^2 + 3x - 1 = 4x + 4$

$x^2 - x - 5 = 0$

$x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$

Check domain: need $x \in [r_1, -1) \cup [r_2, \infty)$ where $r_2 \approx 0.30$.

$x = \frac{1+\sqrt{21}}{2} \approx \frac{1+4.58}{2} \approx 2.79 \geq r_2$ ✓

$x = \frac{1-\sqrt{21}}{2} \approx \frac{1-4.58}{2} \approx -1.79$

Check if $-1.79 \in [r_1, -1)$: $r_1 \approx -3.30$, so $-3.30 \leq -1.79 < -1$ ✓

Both solutions are valid.

Answer: $x = \frac{1 + \sqrt{21}}{2}$ or $x = \frac{1 - \sqrt{21}}{2}$

Marking notes:

• M1: Squares both sides and solves quadratic
• A1: Both correct solutions with domain check

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Mark Total Summary

Q	Marks
1	2
2	2
3	3
4	3
5	3
6	3
7	2
8	2
Section A	20
9	6
10	7
11	8
12	9
Section B	30
Total	50

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