A Level H2 Mathematics Practice Paper 5

TuitionGoWhere Practice Paper - Maths H2 A-Level

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

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Section A: Short Questions

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Question 1 [2]

Answer: The composite function $gf$ exists because the range of $f$ is a subset of the domain of $g$.

Working:

• $f : x \mapsto x^2 + 2x$, domain $x \geq -1$.
• Range of $f$: Completing the square, $f(x) = (x+1)^2 - 1$. Since $x \geq -1$, we have $(x+1) \geq 0$, so $(x+1)^2 \geq 0$ and $f(x) \geq -1$.
  • Range of $f$ is $[-1, \infty)$.
• Domain of $g$ is $\mathbb{R} \setminus \{3\}$.
• We need range of $f$ $\subseteq$ domain of $g$, i.e., we need to check that $3$ is not in the range of $f$.
• Solve $f(x) = 3$: $x^2 + 2x = 3 \Rightarrow x^2 + 2x - 3 = 0 \Rightarrow (x+3)(x-1) = 0$, so $x = -3$ or $x = 1$.
• Since the domain of $f$ is $x \geq -1$, $x = -3$ is excluded. But $x = 1 \geq -1$, so $f(1) = 3$.
• Therefore $3$ is in the range of $f$, but $g(3)$ is undefined.
• Correction: The range of $f$ is $[-1, \infty)$, which includes $3$. Since $g$ is not defined at $x = 3$, we must check: does $f(x) = 3$ for some $x$ in the domain? Yes, $f(1) = 3$.
• Therefore $gf(1) = g(f(1)) = g(3)$, which is undefined.
• The composite function $gf$ does NOT exist (as a function with the full domain of $f$), because $f(1) = 3$ is not in the domain of $g$.

Marking notes:

• [1] Correctly identifies the range of $f$ as $[-1, \infty)$ or equivalent.
• [1] Correctly concludes that $gf$ does not exist because $3 \in$ range of $f$ but $3 \notin$ domain of $g$.

Common mistake: Students often forget to check whether specific values in the range of the first function fall outside the domain of the second function. Simply stating "range of $f$ is a subset of domain of $g$" without verification is insufficient.

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Question 2 [2]

Answer: The range of $f$ is $[0, 2]$.

Working:

• $f(x) = \sqrt{4 - x^2}$, domain $-2 \leq x \leq 2$.
• Since $f(x)$ is a square root, $f(x) \geq 0$ for all $x$ in the domain.
• The maximum value occurs when $4 - x^2$ is maximised, i.e., when $x = 0$: $f(0) = \sqrt{4} = 2$.
• The minimum value occurs when $x = \pm 2$: $f(\pm 2) = \sqrt{0} = 0$.
• Range is $[0, 2]$.

Marking notes:

• [1] Correct minimum value $0$.
• [1] Correct maximum value $2$, with correct interval notation.

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Question 3 [3]

(a) [2]

Answer: $f^{-1}(x) = e^x - 3$, domain $x \in \mathbb{R}$.

Working:

• Let $y = \ln(x + 3)$.
• Swap $x$ and $y$: $x = \ln(y + 3)$.
• Solve for $y$: $e^x = y + 3$, so $y = e^x - 3$.
• The domain of $f^{-1}$ is the range of $f$. Since $f(x) = \ln(x+3)$ and $x > -3$, the range of $f$ is all real numbers.
• Domain of $f^{-1}$: $x \in \mathbb{R}$.

Marking:

• [1] Correct expression $f^{-1}(x) = e^x - 3$.
• [1] Correct domain: $x \in \mathbb{R}$.

(b) [1]

Answer: The graph of $y = f(x) = \ln(x+3)$ is the standard $\ln$ curve shifted 3 units left, passing through $(-2, 0)$ with asymptote $x = -3$. The graph of $y = f^{-1}(x) = e^x - 3$ is the standard $e^x$ curve shifted 3 units down, passing through $(0, -2)$ with asymptote $y = -3$. The two graphs are reflections of each other in the line $y = x$.

Marking:

• [1] Both graphs correctly sketched with correct asymptotes and at least one labelled point each, and the line $y = x$ shown.

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Question 4 [3]

Answer: $x = -2$.

Working:

• $f^{-1}(x) = 0$ means $f(0) = x$.
• $f(0) = e^{2(0)} - 3 = 1 - 3 = -2$.
• Therefore $x = -2$.

Alternative method:

• Find $f^{-1}(x)$: Let $y = e^{2x} - 3$. Then $x = e^{2y} - 3$, so $e^{2y} = x + 3$, giving $2y = \ln(x+3)$, so $f^{-1}(x) = \frac{1}{2}\ln(x+3)$, domain $x > -3$.
• Set $f^{-1}(x) = 0$: $\frac{1}{2}\ln(x+3) = 0 \Rightarrow \ln(x+3) = 0 \Rightarrow x + 3 = 1 \Rightarrow x = -2$.

Marking notes:

• [2] Correct method (either approach).
• [1] Correct final answer $x = -2$.

Common mistake: Students may try to find $f^{-1}$ first, which works but is longer. The direct approach using $f^{-1}(x) = 0 \Rightarrow f(0) = x$ is faster.

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Question 5 [3]

(a) [1]

Answer: The composite $fg$ exists because the range of $g$ is a subset of the domain of $f$ (which is $\mathbb{R}$).

Working:

• Range of $g$: $g(x) = \frac{x}{x+2}$. As $x \to \infty$, $g(x) \to 1$. As $x \to -2^+$, $g(x) \to +\infty$; as $x \to -2^-$, $g(x) \to -\infty$. Also $g(x) = 1$ has no solution (would require $x = x + 2$, impossible). So range of $g$ is $\mathbb{R} \setminus \{1\}$.
• Domain of $f$ is $\mathbb{R}$. Since $\mathbb{R} \setminus \{1\} \subset \mathbb{R}$, the composite $fg$ exists.

Marking:

• [1] Correct justification that range of $g$ $\subseteq$ domain of $f$.

(b) [2]

Answer: $fg(x) = \frac{3x + 4}{x + 2}$, domain $x \in \mathbb{R}, \; x \neq -2$.

Working:

• $fg(x) = f(g(x)) = f\!\left(\frac{x}{x+2}\right) = 2 \cdot \frac{x}{x+2} + 1 = \frac{2x}{x+2} + \frac{x+2}{x+2} = \frac{2x + x + 2}{x+2} = \frac{3x + 2}{x+2}$.

Correction: Let me recompute: $f(g(x)) = 2 \cdot \frac{x}{x+2} + 1 = \frac{2x + (x+2)}{x+2} = \frac{3x + 2}{x+2}$

• Domain: $x \neq -2$ (since $g$ is undefined at $x = -2$).

Marking:

• [1] Correct expression $\frac{3x+2}{x+2}$.
• [1] Correct domain $x \neq -2$.

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Question 6 [3]

(a) [1]

Answer: $f(x_1) = f(x_2) \Rightarrow \frac{2x_1 + 5}{x_1 - 1} = \frac{2x_2 + 5}{x_2 - 1}$.

Cross-multiplying: $(2x_1 + 5)(x_2 - 1) = (2x_2 + 5)(x_1 - 1)$

$2x_1 x_2 - 2x_1 + 5x_2 - 5 = 2x_1 x_2 - 2x_2 + 5x_1 - 5$

$-2x_1 + 5x_2 = -2x_2 + 5x_1$

$7x_2 = 7x_1 \Rightarrow x_1 = x_2$.

Therefore $f$ is one-one.

Marking:

• [1] Correct algebraic proof that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.

(b) [2]

Answer: $f^{-1}(x) = \frac{x + 5}{x - 2}$, domain $x \neq 2$.

Working:

• Let $y = \frac{2x + 5}{x - 1}$.
• Swap: $x = \frac{2y + 5}{y - 1}$.
• $x(y - 1) = 2y + 5 \Rightarrow xy - x = 2y + 5 \Rightarrow xy - 2y = x + 5 \Rightarrow y(x - 2) = x + 5$.
• $y = \frac{x + 5}{x - 2}$.
• Domain of $f^{-1}$: $x \neq 2$ (since $f(x) = 2$ would require $2x + 5 = 2(x-1) = 2x - 2$, giving $5 = -2$, impossible; so $2$ is not in the range of $f$).

Marking:

• [1] Correct expression.
• [1] Correct domain.

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Question 7 [2]

Answer: $f^{-1}(x) = 3 + \sqrt{x + 4}$, domain $x \geq -4$.

Working:

• $f(x) = x^2 - 6x + 5 = (x - 3)^2 - 4$, domain $x \geq 3$.
• Range of $f$: Since $x \geq 3$, $(x-3)^2 \geq 0$, so $f(x) \geq -4$. Range is $[-4, \infty)$.
• Let $y = (x-3)^2 - 4$. Then $(x-3)^2 = y + 4$, so $x - 3 = \sqrt{y+4}$ (positive root since $x \geq 3$).
• $x = 3 + \sqrt{y + 4}$.
• $f^{-1}(x) = 3 + \sqrt{x + 4}$.
• Domain of $f^{-1}$ = range of $f$ = $[-4, \infty)$, i.e., $x \geq -4$.

Marking:

• [1] Correct expression for $f^{-1}(x)$.
• [1] Correct domain $x \geq -4$.

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Question 8 [2]

Answer: $(fg)^{-1}(5) = \frac{2}{3}$.

Working:

• $(fg)^{-1}(5) = g^{-1}(f^{-1}(5))$.
• First find $f^{-1}(5)$: $f(x) = x^2 - 4 = 5 \Rightarrow x^2 = 9 \Rightarrow x = 3$ (since domain is $x > 0$).
• Now find $g^{-1}(3)$: $g(x) = 3x + 1 = 3 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$.
• Therefore $(fg)^{-1}(5) = \frac{2}{3}$.

Alternative: Find $fg(x) = f(g(x)) = f(3x+1) = (3x+1)^2 - 4 = 9x^2 + 6x + 1 - 4 = 9x^2 + 6x - 3$. Then $(fg)^{-1}(5)$: solve $9x^2 + 6x - 3 = 5 \Rightarrow 9x^2 + 6x - 8 = 0 \Rightarrow (3x-2)(3x+4) = 0$. So $x = \frac{2}{3}$ or $x = -\frac{4}{3}$. Since $fg$ has domain $x > 0$ (from $f$'s domain applied to $g(x) = 3x+1 > 0 \Rightarrow x > -\frac{1}{3}$), we need $fg$ to be one-one on its domain. Check: $fg$ is not obviously one-one on $x > -\frac{1}{3}$. Better to use the first method.

Marking:

• [1] Correct method.
• [1] Correct answer $\frac{2}{3}$.

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Section B: Structured Questions

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Question 9 [8]

(a) [4]

Answer: $a = 2$, $b = 6$, $c = 3$.

Working:

• Vertical asymptote is $x = -c = -3$, so $c = 3$.
• $f(0) = -2$: $\frac{b}{c} = -2 \Rightarrow \frac{b}{3} = -2 \Rightarrow b = -6$.

Correction: $f(0) = \frac{a(0) + b}{0 + c} = \frac{b}{c} = -2$, so $b = -2c = -6$.

• $f(1) = -1$: $\frac{a + b}{1 + c} = -1 \Rightarrow \frac{a - 6}{4} = -1 \Rightarrow a - 6 = -4 \Rightarrow a = 2$.

Answer: $a = 2$, $b = -6$, $c = 3$.

Marking:

• [1] Correct value $c = 3$ from asymptote.
• [1] Correct value $b = -6$.
• [1] Correct value $a = 2$.
• [1] All three values clearly stated.

(b) [3]

Answer: $f^{-1}(x) = \frac{3x - 6}{x - 2} = \frac{3(x-2)}{x-2}$... Let me recompute.

$f(x) = \frac{2x - 6}{x + 3}$.

Let $y = \frac{2x - 6}{x + 3}$. Swap: $x = \frac{2y - 6}{y + 3}$.

$x(y + 3) = 2y - 6 \Rightarrow xy + 3x = 2y - 6 \Rightarrow xy - 2y = -3x - 6 \Rightarrow y(x - 2) = -3x - 6$.

$y = \frac{-3x - 6}{x - 2} = \frac{-3(x + 2)}{x - 2}$.

Domain of $f^{-1}$: $x \neq 2$ (since $f(x) = 2$ would require $2x - 6 = 2(x+3) = 2x + 6$, giving $-6 = 6$, impossible).

Answer: $f^{-1}(x) = \frac{-3x - 6}{x - 2}$, domain $x \neq 2$.

Marking:

• [2] Correct expression for $f^{-1}(x)$ (allow equivalent forms).
• [1] Correct domain $x \neq 2$.

(c) [1]

Answer: Range of $f$ is $\mathbb{R} \setminus \{2\}$, i.e., all real numbers except $2$.

Working: The range of $f$ equals the domain of $f^{-1}$, which is $x \neq 2$.

Marking:

• [1] Correct range stated.

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Question 10 [10]

(a) [2]

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

Working:

• $f(x) = 4 - (x-1)^2$, domain $x \geq 1$.
• When $x = 1$: $f(1) = 4$ (maximum).
• As $x \to \infty$: $(x-1)^2 \to \infty$, so $f(x) \to -\infty$.
• Range is $(-\infty, 4]$.

Marking:

• [1] Correct maximum value $4$.
• [1] Correct range $(-\infty, 4]$.

(b) [3]

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

Working:

• Let $y = 4 - (x-1)^2$. Then $(x-1)^2 = 4 - y$, so $x - 1 = \sqrt{4-y}$ (positive root since $x \geq 1$).
• $x = 1 + \sqrt{4 - y}$.
• $f^{-1}(x) = 1 + \sqrt{4 - x}$.
• Domain of $f^{-1}$ = range of $f$ = $(-\infty, 4]$, i.e., $x \leq 4$.

Marking:

• [2] Correct expression.
• [1] Correct domain $x \leq 4$.

(c) [3]

Answer: See diagram description below.

The graph of $y = f(x)$ is the left half (actually the right portion starting from the vertex) of a downward parabola with vertex at $(1, 4)$. Since the domain is $x \geq 1$, only the right half of the parabola from the vertex is drawn. It passes through $(1, 4)$, $(2, 3)$, $(3, 0)$, $(4, -5)$, etc.

The graph of $y = f^{-1}(x) = 1 + \sqrt{4 - x}$ starts at $(4, 1)$ and extends leftward and upward, passing through $(3, 2)$, $(0, 3)$, $(-5, 1+\sqrt{9}=4)$. It has a horizontal asymptote-like behaviour but is a square root curve.

The line $y = x$ is shown as a dashed line. The two curves are reflections of each other across $y = x$.

Image pending generation: image for Q10(c).

Marking:

• [1] Correct shape for $f(x)$ (parabolic arc starting at vertex $(1,4)$ going down to the right).
• [1] Correct shape for $f^{-1}(x)$ (square root curve starting at $(4,1)$ going up to the left).
• [1] Line $y = x$ shown and correct reflection symmetry demonstrated.

(d) [2]

Answer: $x = \frac{3 + \sqrt{5}}{2}$.

Working:

• The solution to $f(x) = f^{-1}(x)$ lies on the line $y = x$ (since if $f(a) = f^{-1}(a)$, then applying $f$ to both sides gives $f(f(a)) = a$, but more directly, the graphs of $f$ and $f^{-1}$ intersect on $y = x$).
• So solve $f(x) = x$: $4 - (x-1)^2 = x$.
• $4 - (x^2 - 2x + 1) = x \Rightarrow 4 - x^2 + 2x - 1 = x \Rightarrow -x^2 + 2x + 3 = x$.
• $-x^2 + x + 3 = 0 \Rightarrow x^2 - x - 3 = 0$.
• $x = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}$.

Check domain: We need $x \geq 1$ (domain of $f$) and $x \leq 4$ (domain of $f^{-1}$).

• $\frac{1 + \sqrt{13}}{2} \approx \frac{1 + 3.606}{2} \approx 2.303$ ✓
• $\frac{1 - \sqrt{13}}{2} \approx \frac{1 - 3.606}{2} \approx -1.303$ ✗ (not in domain of $f$)

Answer: $x = \frac{1 + \sqrt{13}}{2}$.

Marking:

• [1] Correct equation set up ($f(x) = x$ or equivalent).
• [1] Correct answer $x = \frac{1 + \sqrt{13}}{2}$ with valid rejection of the extraneous root.

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Question 11 [12]

(a) [2]

Answer: $f$ is one-one because $f'(x) = 2x - 4$, and for $x \geq 2$, $f'(x) \geq 0$, so $f$ is strictly increasing on its domain (strictly increasing for $x > 2$, and $f(2) = 3$ is the minimum). Alternatively, complete the square: $f(x) = (x-2)^2 + 3$, which is strictly increasing for $x \geq 2$.

Working:

$f(x) = x^2 - 4x + 7 = (x-2)^2 + 3$.

For $x \geq 2$, as $x$ increases, $(x-2)^2$ increases, so $f(x)$ increases. Therefore $f$ is strictly increasing on $[2, \infty)$, hence one-one.

Marking:

• [1] Correct reasoning (strictly increasing or equivalent).
• [1] Conclusion that $f^{-1}$ exists.

(b) [3]

Answer: $f^{-1}(x) = 2 + \sqrt{x - 3}$, domain $x \geq 3$, range $f^{-1}$ is $[2, \infty)$.

Working:

• Let $y = (x-2)^2 + 3$. Then $(x-2)^2 = y - 3$, so $x - 2 = \sqrt{y - 3}$ (positive root since $x \geq 2$).
• $x = 2 + \sqrt{y - 3}$.
• $f^{-1}(x) = 2 + \sqrt{x - 3}$.
• Domain of $f^{-1}$ = range of $f$: Since $f(x) = (x-2)^2 + 3$ with $x \geq 2$, the minimum value is $f(2) = 3$, so range of $f$ is $[3, \infty)$. Domain of $f^{-1}$ is $x \geq 3$.
• Range of $f^{-1}$ = domain of $f$ = $[2, \infty)$.

Marking:

• [1] Correct expression.
• [1] Correct domain $x \geq 3$.
• [1] Correct range $[2, \infty)$.

(c) [2]

Answer: $gf$ exists because range of $f$ is $[3, \infty)$ and domain of $g$ is $\mathbb{R} \setminus \{0\}$. Since $[3, \infty) \subset \mathbb{R} \setminus \{0\}$, the composite exists.

Marking:

• [1] Correct identification of range of $f$ as $[3, \infty)$.
• [1] Correct justification.

(d) [3]

Answer: $gf(x) = \frac{1}{x^2 - 4x + 7} = \frac{1}{(x-2)^2 + 3}$, range is $\left(0, \frac{1}{3}\right]$.

Working:

• $gf(x) = g(f(x)) = \frac{1}{f(x)} = \frac{1}{x^2 - 4x + 7} = \frac{1}{(x-2)^2 + 3}$.
• The denominator $(x-2)^2 + 3$ has minimum value $3$ (at $x = 2$) and increases without bound as $x \to \infty$.
• So $gf(x)$ has maximum value $\frac{1}{3}$ (at $x = 2$) and approaches $0$ as $x \to \infty$.
• Range is $\left(0, \frac{1}{3}\right]$.

Marking:

• [1] Correct expression for $gf(x)$.
• [1] Correct maximum value identified.
• [1] Correct range $\left(0, \frac{1}{3}\right]$.

(e) [2]

Answer: $x = 1$ or $x = 3$.

Working:

• $\frac{1}{(x-2)^2 + 3} = \frac{1}{4}$.
• $(x-2)^2 + 3 = 4 \Rightarrow (x-2)^2 = 1 \Rightarrow x - 2 = \pm 1 \Rightarrow x = 3$ or $x = 1$.
• But the domain of $f$ is $x \geq 2$, so $x = 1$ is not in the domain.

Correction: $x = 1$ is rejected since $x \geq 2$ is required.

Answer: $x = 3$ only.

Marking:

• [1] Correct equation solved.
• [1] Correct final answer $x = 3$ (with valid rejection of $x = 1$).

Common mistake: Forgetting to check the domain restriction $x \geq 2$ when solving.

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Mark Summary

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