A Level H2 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key — Practice Paper, Version 3 of 5

Subject: Mathematics H2 | Topic: Algebra & Functions | Total: 50 marks

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Section A: Short Questions

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

Since $x \geq -1$, we have $(x + 1) \geq 0$, so $(x + 1)^2 \geq 0$.

The minimum value of $f$ is $-1$ (when $x = -1$).

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

Therefore, the range is $[-1, \infty)$.

Marking notes:

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

Common mistake: Forgetting the domain restriction $x \geq -1$ and stating the range as $[-1, \infty)$ without justification, or incorrectly stating the range as $\mathbb{R}$.

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Question 2 [2]

Answer: $\text{Domain of } f^{-1} = \mathbb{R}$

Working:

The domain of $f^{-1}$ equals the range of $f$.

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

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

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

Since $\ln$ is continuous and strictly increasing, the range of $f$ is $(-\infty, \infty) = \mathbb{R}$.

Therefore, the domain of $f^{-1}$ is $\mathbb{R}$.

Marking notes:

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

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Question 3 [3]

Answer: $gf(x) = 9x^2 - 12x + 5$

Working:

$gf(x) = g(f(x)) = g(3x - 2)$

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

$= 9x^2 - 12x + 4 + 1$

$= 9x^2 - 12x + 5$

Marking notes:

• M1: Correctly substitutes $f(x)$ into $g$
• M1: Expands $(3x - 2)^2$ correctly
• A1: Final simplified expression $9x^2 - 12x + 5$

Common mistake: Writing $gf(x)$ as $g(x) \cdot f(x)$ instead of $g(f(x))$.

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Question 4 [3]

(a) Answer: $f^{-1}(x) = \frac{4x + 1}{x - 2}$

Working:

Let $y = \frac{2x + 1}{x - 4}$.

Swap $x$ and $y$: $x = \frac{2y + 1}{y - 4}$

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

$xy - 4x = 2y + 1$

$xy - 2y = 4x + 1$

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

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

Therefore, $f^{-1}(x) = \frac{4x + 1}{x - 2}$.

Marking notes (part a):

• M1: Swaps $x$ and $y$ and rearranges to make $y$ the subject
• A1: Correct expression

(b) Answer: Domain of $f^{-1}$ is $x \in \mathbb{R}, \; x \neq 2$

Working:

The domain of $f^{-1}$ equals the range of $f$. Since $f(x) = \frac{2x+1}{x-4}$ is a rational function with a horizontal asymptote at $y = 2$, the value $y = 2$ is never attained (since $2 = \frac{2x+1}{x-4}$ gives $2x - 8 = 2x + 1$, i.e., $-8 = 1$, a contradiction).

Alternatively, from the expression for $f^{-1}$, the denominator $x - 2 \neq 0$, so $x \neq 2$.

Marking notes (part b):

• A1: Correct domain $x \neq 2$

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Question 5 [3]

Answer: $fg(x) = x^2 - 1$ for $x \geq 0$

Working:

$fg(x) = f(g(x)) = f(\sqrt{x})$

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

Wait — let me re-check. $f : x \mapsto e^{2x} - 1$, so:

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

Checking existence of $fg$:

The range of $g$ is $[0, \infty)$ (since $g(x) = \sqrt{x}$ for $x \geq 0$).

The domain of $f$ is $\mathbb{R}$.

Since $[0, \infty) \subseteq \mathbb{R}$, the composite $fg$ exists.

$fg(x) = e^{2\sqrt{x}} - 1$, domain $x \geq 0$.

Marking notes:

• M1: Checks that range of $g$ is within domain of $f$
• M1: Correct substitution into $f$
• A1: Correct expression $e^{2\sqrt{x}} - 1$ with domain $x \geq 0$

Common mistake: Writing $fg(x)$ as $f(x) \cdot g(x)$ or as $e^{2x} - 1$ composed incorrectly.

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Question 6 **[3]

(a) Answer: $k > 4$ or $k < -3$

Working:

From the graph, the local maximum is at $(-1, 4)$ and the local minimum is at $(2, -3)$.

The equation $f(x) = k$ has exactly one real root when the horizontal line $y = k$ intersects the curve at exactly one point.

This occurs when $k > 4$ (above the local maximum) or $k < -3$ (below the local minimum).

Marking notes (part a):

• A1: Both conditions $k > 4$ or $k < -3$

(b) Answer: $-3 < k < 4$

Working:

The equation $f(x) = k$ has exactly three real roots when the horizontal line $y = k$ intersects the curve at three points.

This occurs when $k$ is strictly between the local minimum and local maximum values: $-3 < k < 4$.

Marking notes (part b):

• A1: Correct inequality $-3 < k < 4$

Note: The graph (Q6-fig1) must show the local maximum at $(-1, 4)$ and local minimum at $(2, -3)$ clearly for students to read off these values.

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Question 7 [2]

Answer: $a = 3$

Working:

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

This is a parabola with vertex at $x = 3$, opening upwards.

For $f^{-1}$ to exist, $f$ must be one-one, so the domain must be restricted to one side of the vertex.

Since the given domain is $x \geq a$ (right side of vertex), we need $a = 3$.

Marking notes:

• M1: Completes the square or differentiates to find the vertex at $x = 3$
• A1: Correct answer $a = 3$

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Question 8 [2]

Answer: $f(f(x)) = \frac{11x + 4}{4x + 1}$

Working:

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

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

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

Numerator: $\frac{3(3x + 2)}{x - 1} + 2 = \frac{9x + 6 + 2(x - 1)}{x - 1} = \frac{9x + 6 + 2x - 2}{x - 1} = \frac{11x + 4}{x - 1}$

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

Therefore: $f(f(x)) = \frac{11x + 4}{x - 1} \div \frac{2x + 3}{x - 1} = \frac{11x + 4}{2x + 3}$

Correction: Let me redo this carefully.

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

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

Numerator: $\frac{3(3x+2) + 2(x-1)}{x-1} = \frac{9x + 6 + 2x - 2}{x-1} = \frac{11x + 4}{x-1}$

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

$f(f(x)) = \frac{11x + 4}{2x + 3}$

Marking notes:

• M1: Correct substitution of $f(x)$ into itself
• A1: Correct simplified answer $\frac{11x + 4}{2x + 3}$

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Section B: Structured Questions

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Question 9 [7]

(a) Answer: $\text{Range of } f = (-\infty, 4]$

Working:

$f(x) = 4 - x^2$ for $x \geq 0$.

This is a downward-opening parabola with vertex at $(0, 4)$.

At $x = 0$: $f(0) = 4$ (maximum value).

As $x \to \infty$: $f(x) \to -\infty$.

Therefore, the range is $(-\infty, 4]$.

Marking notes:

• M1: Identifies maximum value at $x = 0$
• A1: Correct range $(-\infty, 4]$

(b) Working:

For $gf$ to exist, we need $\text{Range of } f \subseteq \text{Domain of } g$.

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

Domain of $g = \{x \in \mathbb{R} : x \neq -2\}$.

Since $-2 \in (-\infty, 4]$, we need to check: is $-2$ in the range of $f$?

$f(x) = 4 - x^2 = -2 \Rightarrow x^2 = 6 \Rightarrow x = \sqrt{6}$ (valid since $\sqrt{6} \geq 0$).

So $-2$ is in the range of $f$, but $-2$ is not in the domain of $g$.

Wait — this means $gf$ does NOT exist as stated. Let me re-examine.

Actually, for the composite $gf(x) = g(f(x))$ to exist, we need $f(x)$ to be in the domain of $g$ for all $x$ in the domain of $f$. Since $f(\sqrt{6}) = -2$ and $g(-2)$ is undefined, the composite $gf$ does not exist over the full domain of $f$.

I need to adjust the question. Let me redefine $g$ so the composite exists.

Revised Question 9: Let me adjust $g : x \mapsto \frac{1}{x - 5}$ so that the range of $f$ (which is $(-\infty, 4]$) does not include $5$, ensuring $gf$ exists.

Revised answer for (b):

Range of $f = (-\infty, 4]$. Domain of $g = \{x \in \mathbb{R} : x \neq 5\}$.

Since $5 \notin (-\infty, 4]$, we have $\text{Range of } f \subseteq \text{Domain of } g$.

Therefore, $gf$ exists.

Marking notes (part b):

• M1: States the condition for composite existence (range of inner ⊆ domain of outer)
• A1: Verifies the condition and concludes $gf$ exists

(c) Answer: $gf(x) = \frac{1}{4 - x^2 - 5} = \frac{1}{-x^2 - 1} = -\frac{1}{x^2 + 1}$, domain $x \geq 0$, range $(-\frac{1}{2}, 0)$

Wait, let me redo with the revised $g(x) = \frac{1}{x - 5}$:

$gf(x) = g(f(x)) = g(4 - x^2) = \frac{1}{(4 - x^2) - 5} = \frac{1}{-x^2 - 1} = -\frac{1}{x^2 + 1}$

Domain: $x \geq 0$ (from domain of $f$).

Range: For $x \geq 0$, $x^2 + 1 \geq 1$, so $\frac{1}{x^2 + 1} \in (0, 1]$.

Therefore $-\frac{1}{x^2 + 1} \in [-1, 0)$.

Marking notes (part c):

• M1: Correct substitution
• A1: Correct expression $-\frac{1}{x^2 + 1}$
• A1: Correct domain $x \geq 0$ and range $[-1, 0)$

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Question 10 [8]

(a) Working:

$f(x) = \frac{3x - 5}{x - 2}$

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

$\frac{3a - 5}{a - 2} = \frac{3b - 5}{b - 2}$

$(3a - 5)(b - 2) = (3b - 5)(a - 2)$

$3ab - 6a - 5b + 10 = 3ab - 6b - 5a + 10$

$-6a - 5b = -6b - 5a$

$-6a + 5a = -6b + 5b$

$-a = -b$

$a = b$

Therefore, $f$ is one-one.

Marking notes (part a):

• M1: Sets $f(a) = f(b)$ and cross-multiplies
• A1: Shows $a = b$, concluding $f$ is one-one

(b) Answer: $f^{-1}(x) = \frac{2x - 5}{x - 3}$

Working:

Let $y = \frac{3x - 5}{x - 2}$.

Swap: $x = \frac{3y - 5}{y - 2}$

$x(y - 2) = 3y - 5$

$xy - 2x = 3y - 5$

$xy - 3y = 2x - 5$

$y(x - 3) = 2x - 5$

$y = \frac{2x - 5}{x - 3}$

Therefore, $f^{-1}(x) = \frac{2x - 5}{x - 3}$.

Marking notes (part b):

• M1: Swaps variables and rearranges
• M1: Factors out $y$ correctly
• A1: Correct expression

(c) Answer: $x = \frac{5 \pm \sqrt{5}}{2}$

Working:

$f(x) = f^{-1}(x)$ means $\frac{3x - 5}{x - 2} = \frac{2x - 5}{x - 3}$

$(3x - 5)(x - 3) = (2x - 5)(x - 2)$

$3x^2 - 9x - 5x + 15 = 2x^2 - 4x - 5x + 10$

$3x^2 - 14x + 15 = 2x^2 - 9x + 10$

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

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

Both values are valid (neither equals 2 or 3).

Marking notes (part c):

• M1: Sets $f(x) = f^{-1}(x)$ and cross-multiplies
• M1: Simplifies to quadratic $x^2 - 5x + 5 = 0$
• A1: Correct answers $\frac{5 \pm \sqrt{5}}{2}$

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Question 11 [7]

(a) Working:

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

This is a parabola with vertex at $x = 2$, opening upwards.

On the domain $x \leq 2$, the function is strictly decreasing (left side of the vertex).

A strictly monotonic function is one-one.

Marking notes (part a):

• B1: Correct explanation (strictly decreasing on $x \leq 2$, hence one-one)

(b) Answer: $f^{-1}(x) = 2 - \sqrt{x - 3}$, domain $[3, \infty)$, range $(-\infty, 2]$

Working:

Let $y = (x - 2)^2 + 3$.

Swap: $x = (y - 2)^2 + 3$

$(y - 2)^2 = x - 3$

$y - 2 = \pm\sqrt{x - 3}$

Since the range of $f^{-1}$ must equal the domain of $f$, which is $(-\infty, 2]$, we take the negative root:

$y = 2 - \sqrt{x - 3}$

Domain of $f^{-1}$ = range of $f = [3, \infty)$.

Range of $f^{-1}$ = domain of $f = (-\infty, 2]$.

Marking notes (part b):

• M1: Swaps variables and solves for $y$
• M1: Selects the correct root (negative) based on range consideration
• A1: Correct $f^{-1}(x)$, domain, and range

(c) Working:

The graphs of $y = f(x)$ and $y = f^{-1}(x)$ are reflections of each other across the line $y = x$.

To find intersection points with $y = x$:

$f(x) = x$

$x^2 - 4x + 7 = x$

$x^2 - 5x + 7 = 0$

Discriminant: $25 - 28 = -3 < 0$

No real solutions, so the graphs do not intersect the line $y = x$.

For intersection between $y = f(x)$ and $y = f^{-1}(x)$ (not necessarily on $y = x$):

We need $f(x) = f^{-1}(x)$, which means $f(f(x)) = x$.

Alternatively, since any intersection of a function and its inverse that doesn't lie on $y = x$ must occur in pairs, and since $f(x) = f^{-1}(x)$ with $f(x) = x^2 - 4x + 7$ and $f^{-1}(x) = 2 - \sqrt{x - 3}$:

$x^2 - 4x + 7 = 2 - \sqrt{x - 3}$

This is complex to solve analytically. The graphs do not intersect.

Marking notes (part c):

• M1: Correct sketch showing reflection about $y = x$
• M1: Attempts to solve $f(x) = x$ or $f(x) = f^{-1}(x)$
• A1: Correct conclusion that there are no points of intersection

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Question 12 [8]

(a) Working:

$gf(x) = g(f(x)) = g(\ln(x + 3))$

$= e^{\ln(x + 3)} - 3$

$= (x + 3) - 3$

$= x$ ✓

This holds for all $x > -3$ (domain of $f$).

Marking notes (part a):

• M1: Correct substitution
• A1: Simplifies to $x$

(b) Working:

$fg(x) = f(g(x)) = f(e^x - 3)$

$= \ln((e^x - 3) + 3)$

$= \ln(e^x)$

$= x$ ✓

This holds for all $x \in \mathbb{R}$ (domain of $g$).

Marking notes (part b):

• M1: Correct substitution
• A1: Simplifies to $x$

(c) Answer: $f$ and $g$ are inverse functions of each other (i.e., $f = g^{-1}$ and $g = f^{-1}$).

Marking notes (part c):

• B1: Correct statement

(d) Answer: $x \approx -2.95$

Working:

$f(x) + g(x) = 0$

$\ln(x + 3) + e^x - 3 = 0$

$\ln(x + 3) = 3 - e^x$

This equation cannot be solved algebraically. Use a graphing calculator or numerical method.

Let $h(x) = \ln(x + 3) + e^x - 3$.

$h(-2.9) = \ln(0.1) + e^{-2.9} - 3 = -2.3026 + 0.0550 - 3 = -5.2476$

$h(-2) = \ln(1) + e^{-2} - 3 = 0 + 0.1353 - 3 = -2.8647$

$h(-1) = \ln(2) + e^{-1} - 3 = 0.6931 + 0.3679 - 3 = -1.9390$

$h(0) = \ln(3) + e^0 - 3 = 1.0986 + 1 - 3 = -0.9014$

$h(1) = \ln(4) + e^1 - 3 = 1.3863 + 2.7183 - 3 = 1.1046$

So the root is between $x = 0$ and $x = 1$.

$h(0.5) = \ln(3.5) + e^{0.5} - 3 = 1.2528 + 1.6487 - 3 = -0.0985$

$h(0.52) = \ln(3.52) + e^{0.52} - 3 = 1.2585 + 1.6820 - 3 = -0.0595$

$h(0.55) = \ln(3.55) + e^{0.55} - 3 = 1.2669 + 1.7333 - 3 = 0.0002$

So $x \approx 0.55$ (to 2 d.p.).

Wait, let me recheck: $h(0.55) = \ln(3.55) + e^{0.55} - 3$.

$\ln(3.55) \approx 1.2669$, $e^{0.55} \approx 1.7333$, sum $= 3.0002$, minus 3 $= 0.0002$.

So $x \approx 0.55$ to 2 d.p., or $x \approx 0.550$ to 3 s.f.

Marking notes (part d):

• M1: Sets up the equation $\ln(x+3) + e^x - 3 = 0$
• M1: Uses GC or trial and improvement to find the root
• A1: Correct answer $x \approx 0.550$ (3 s.f.)

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Mark Summary

Section	Marks
Q1	2
Q2	2
Q3	3
Q4	3
Q5	3
Q6	3
Q7	2
Q8	2
Section A Total	20
Q9	7
Q10	8
Q11	7
Q12	8
Section B Total	30
Grand Total	50