A Level H2 Mathematics Practice Paper 4

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper (Algebra & Functions Focus) — Version 4 of 5 Total Marks: 60

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Section A: Short Questions

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Question 1 [3]

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

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

Swap $x$ and $y$: $x = \dfrac{2y - 3}{y + 1}$

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

$xy + x = 2y - 3$

$xy - 2y = -3 - x$

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

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

Domain of $f^{-1}$: Since the denominator of $f^{-1}(x)$ cannot be zero, $x \neq 2$. Also, the domain of $f^{-1}$ equals the range of $f$. For $f(x) = \dfrac{2x-3}{x+1}$, the horizontal asymptote is $y = 2$, so the range of $f$ is $y \neq 2$.

Domain of $f^{-1}$: $x \in \mathbb{R}$, $x \neq 2$. [1]

(b) Range of $f^{-1}$ = Domain of $f$ = $x \in \mathbb{R}$, $x \neq -1$. [1]

[Total: 3 marks]

Common mistakes: Forgetting that the domain of $f^{-1}$ is the range of $f$. Students often state $x \neq -1$ as the domain of $f^{-1}$ (confusing it with the domain of $f$).

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Question 2 [3]

(a) For $gf$ to exist, the range of $f$ must be a subset of the domain of $g$.

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

Since $(x-2)^2 \geq 0$ for all $x$, we have $f(x) \geq 1$. At $x = 2$, $f(2) = 1$, and $f$ increases for $x \geq 2$. So the range of $f$ is $[1, \infty)$.

Domain of $g$ is $x > 3$. Since the range of $f$ is $[1, \infty)$, we need to check: is $[1, \infty) \subseteq (3, \infty)$? No — values of $f$ between 1 and 3 are not in the domain of $g$.

Wait — let me recheck. $f(x) = (x-2)^2 + 1$ for $x \geq 2$. At $x = 2$, $f(2) = 1$. At $x = 3$, $f(3) = 2$. At $x = 4$, $f(4) = 5$. So $f(x) = 3$ when $(x-2)^2 + 1 = 3$, i.e., $(x-2)^2 = 2$, $x = 2 + \sqrt{2} \approx 3.14$.

For $gf$ to exist, we need $f(x) > 3$ for all $x$ in the domain of $f$. But $f(2) = 1 < 3$, so $gf$ as stated does not exist for the full domain of $f$.

Hmm, this is a problem with the question design. Let me reconsider — the question says "show that $gf$ exists," so the numbers should work out. Let me adjust my answer to match the intended question.

Actually, re-reading: the domain of $g$ is $x > 3$, and range of $f$ for $x \geq 2$ is $[1, \infty)$. Since $[1, \infty) \not\subseteq (3, \infty)$, strictly $gf$ does not exist. But the question asks students to show it does exist. This suggests the domain of $g$ should be compatible. Let me proceed with the answer as the question intends, noting that for the composite to exist, we need the range of $f$ to lie within the domain of $g$.

Let me re-approach: Perhaps the intended interpretation is that $gf$ exists for the values where $f(x) > 3$, i.e., for $x > 2 + \sqrt{2}$. But the question says "show that $gf$ exists" implying it does.

I'll answer based on the question as written, showing the reasoning:

Range of $f$: For $x \geq 2$, $f(x) = (x-2)^2 + 1 \geq 1$. Range of $f$ is $[1, \infty)$.

Domain of $g$: $x > 3$.

For $gf$ to exist, we need $\text{Range}(f) \subseteq \text{Domain}(g)$, i.e., $[1, \infty) \subseteq (3, \infty)$. This is not satisfied.

However, if we interpret the question as asking whether $gf$ can be defined (perhaps with a restricted domain), then:

For $x$ such that $f(x) > 3$: $(x-2)^2 + 1 > 3 \Rightarrow (x-2)^2 > 2 \Rightarrow x > 2 + \sqrt{2}$ (since $x \geq 2$).

So $gf$ exists for $x \geq 2 + \sqrt{2}$. But this contradicts the question's premise.

Given the question asks to "show that $gf$ exists," I'll proceed with the calculation assuming the composite is valid (perhaps the domain of $g$ was intended to be $x \geq 0$ or similar). For the purposes of this answer key:

(a) Range of $f$ is $[1, \infty)$. Domain of $g$ is $(3, \infty)$. Since range of $f$ is not a subset of domain of $g$, strictly $gf$ does not exist over the full domain of $f$. However, $gf$ exists for $x > 2 + \sqrt{2}$ where $f(x) > 3$. [1] (Award mark for correct reasoning about range/domain compatibility.)

(b) $gf(x) = g(f(x)) = g(x^2 - 4x + 5) = \dfrac{1}{(x^2 - 4x + 5) - 3} = \dfrac{1}{x^2 - 4x + 2}$ [1]

For the range: $x^2 - 4x + 2 = (x-2)^2 - 2$. For $x > 2 + \sqrt{2}$, $(x-2)^2 > 2$, so $(x-2)^2 - 2 > 0$. As $x \to (2+\sqrt{2})^+$, $(x-2)^2 - 2 \to 0^+$, so $gf(x) \to +\infty$. As $x \to \infty$, $(x-2)^2 - 2 \to \infty$, so $gf(x) \to 0^+$.

Range of $gf$: $(0, \infty)$. [1]

[Total: 3 marks]

Note to teacher: This question has a subtle domain issue. In an exam context, the domain of $g$ would typically be set to ensure the composite exists. Students who correctly identify the range/domain relationship should receive credit regardless.

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Question 3 [3]

(a) Let $y = \ln(2x - 5)$

$e^y = 2x - 5$

$x = \dfrac{e^y + 5}{2}$

$f^{-1}(x) = \dfrac{e^x + 5}{2}$ [1]

(b) Domain of $f^{-1}$ = Range of $f$: Since $2x - 5$ can take any positive value, $\ln(2x - 5)$ can take any real value. Domain of $f^{-1}$: $x \in \mathbb{R}$. [½]

Range of $f^{-1}$ = Domain of $f$: $x > \dfrac{5}{2}$. Range of $f^{-1}$: $y > \dfrac{5}{2}$. [½]

(c) Sketch: $y = f(x) = \ln(2x - 5)$ is a logarithmic curve with vertical asymptote $x = \dfrac{5}{2}$, passing through $(3, 0)$ since $f(3) = \ln(1) = 0$. The graph of $y = f^{-1}(x) = \dfrac{e^x + 5}{2}$ is the reflection of $y = f(x)$ in the line $y = x$, with horizontal asymptote $y = \dfrac{5}{2}$ (as $x \to -\infty$) and passing through $(0, 3)$. [1]

[Total: 3 marks]

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Question 4 [3]

$f(x) = e^{3x} + 2$

To find $f^{-1}(3)$: we need the value of $x$ such that $f(x) = 3$.

$e^{3x} + 2 = 3$

$e^{3x} = 1$

$3x = \ln 1 = 0$

$x = 0$

Therefore $f^{-1}(3) = 0$. [3]

Marking: M1 for setting $f(x) = 3$, M1 for solving $e^{3x} = 1$, A1 for $x = 0$.

Common mistake: Students may try to find the full expression for $f^{-1}(x)$ first, which is unnecessary and time-consuming. The direct method is much faster.

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Question 5 [4]

(a) $f(2) = 2^2 - 1 = 4 - 1 = 3$. [1]

$\displaystyle\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 3) = 3(2) - 3 = 3$. [1]

(b) For $f$ to be one-one, each value of $f(x)$ must correspond to exactly one value of $x$.

For $x \leq 2$: $f(x) = x^2 - 1$. This is a parabola. On $(-\infty, 2]$, $f(x) = x^2 - 1$ is decreasing on $(-\infty, 0]$ and increasing on $[0, 2]$. So $f(-1) = 0$ and $f(1) = 0$. Since $f(-1) = f(1) = 0$ with $-1 \neq 1$, $f$ is not one-one. [2]

Marking: M1 for identifying that the quadratic part is not monotonic / finding a counterexample, A1 for clear justification.

[Total: 4 marks]

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Question 6 [4]

$g(x) = -3f(x-2) = -3 \cdot \dfrac{1}{x-2} = \dfrac{-3}{x-2}$

Asymptotes:

• Vertical asymptote: $x - 2 = 0 \Rightarrow x = 2$. [1]
• Horizontal asymptote: As $x \to \pm\infty$, $g(x) \to 0$. So $y = 0$. [1]

Intercepts:

• $x$-intercept: Set $g(x) = 0$: $\dfrac{-3}{x-2} = 0$. No solution (numerator is never zero). No $x$-intercept. [1]
• $y$-intercept: $g(0) = \dfrac{-3}{0-2} = \dfrac{3}{2}$. $y$-intercept at $\left(0, \dfrac{3}{2}\right)$. [1]

[Total: 4 marks]

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Section B: Structured Questions

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Question 7 [6]

(a) $f(x) = \dfrac{ax + b}{cx + d}$

If $f$ is self-inverse, then $f(f(x)) = x$ for all $x$ in the domain.

$f(f(x)) = f\left(\dfrac{ax+b}{cx+d}\right) = \dfrac{a\cdot\dfrac{ax+b}{cx+d} + b}{c\cdot\dfrac{ax+b}{cx+d} + d}$

$= \dfrac{a(ax+b) + b(cx+d)}{c(ax+b) + d(cx+d)}$

$= \dfrac{a^2x + ab + bcx + bd}{acx + cb + dcx + d^2}$

$= \dfrac{(a^2 + bc)x + b(a+d)}{(ac + cd)x + (bc + d^2)}$

For this to equal $x = \dfrac{x}{1}$, we need:

$\dfrac{(a^2 + bc)x + b(a+d)}{(ac + cd)x + (bc + d^2)} = \dfrac{x}{1}$

Cross-multiplying: $(a^2 + bc)x + b(a+d) = x[(ac + cd)x + (bc + d^2)]$

$(a^2 + bc)x + b(a+d) = (ac + cd)x^2 + (bc + d^2)x$

Comparing coefficients:

• $x^2$: $0 = ac + cd = c(a + d)$. Since $c \neq 0$, we get $a + d = 0$. [2] (for correct derivation)
• $x^1$: $a^2 + bc = bc + d^2 \Rightarrow a^2 = d^2$. This is consistent with $a + d = 0$ (i.e., $d = -a$).
• Constant: $b(a + d) = 0$. Consistent with $a + d = 0$. [1] (for concluding $a + d = 0$)

(b) $f(x) = \dfrac{3x - 5}{2x + k}$. Here $a = 3$, $d = k$. For self-inverse: $a + d = 0 \Rightarrow 3 + k = 0 \Rightarrow k = -3$. [1]

(c) With $k = -3$: $f(x) = \dfrac{3x - 5}{2x - 3}$

Domain: $2x - 3 \neq 0 \Rightarrow x \neq \dfrac{3}{2}$. Domain: $x \in \mathbb{R}$, $x \neq \dfrac{3}{2}$. [½]

Since $f$ is self-inverse, the range of $f$ equals the domain of $f^{-1}$, which equals the domain of $f$. So the range is $y \neq \dfrac{3}{2}$. [½]

Alternatively, the horizontal asymptote is $y = \dfrac{3}{2}$, so range is $y \neq \dfrac{3}{2}$. [1] (for both domain and range)

[Total: 6 marks]

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Question 8 [7]

(a) $f(x) = 4 - (x-1)^2$, domain $x \geq 1$.

At $x = 1$: $f(1) = 4$. As $x$ increases, $(x-1)^2$ increases, so $f(x)$ decreases. As $x \to \infty$, $f(x) \to -\infty$.

Range of $f$: $(-\infty, 4]$. [1]

(b) For $fg$ to exist, we need $\text{Range}(g) \subseteq \text{Domain}(f)$.

$g(x) = \sqrt{x} + 2$, domain $x \geq 0$. Range of $g$: $[2, \infty)$.

Domain of $f$: $[1, \infty)$. Since $[2, \infty) \subseteq [1, \infty)$, the composite $fg$ exists. [1]

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

$= 4 - (x + 2\sqrt{x} + 1) = 3 - x - 2\sqrt{x}$ [1] (for correct simplification)

Domain of $fg$: Domain of $g$ is $x \geq 0$, and all outputs of $g$ are in domain of $f$. So domain of $fg$ is $x \geq 0$. [½]

Range of $fg$: At $x = 0$: $fg(0) = 3 - 0 - 0 = 3$. As $x$ increases, $3 - x - 2\sqrt{x}$ decreases (both $x$ and $\sqrt{x}$ increase). As $x \to \infty$, $fg(x) \to -\infty$.

Range of $fg$: $(-\infty, 3]$. [½]

(d) Sketch: The graph of $y = 3 - x - 2\sqrt{x}$ starts at $(0, 3)$ and decreases, curving downward. It crosses the $x$-axis when $3 - x - 2\sqrt{x} = 0$. Let $u = \sqrt{x}$: $3 - u^2 - 2u = 0 \Rightarrow u^2 + 2u - 3 = 0 \Rightarrow (u+3)(u-1) = 0 \Rightarrow u = 1$ (since $u \geq 0$). So $x = 1$. The graph crosses the $x$-axis at $(1, 0)$. [2] (M1 for correct shape/direction, A1 for correct intercepts and turning point)

[Total: 7 marks]

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Question 9 [6]

(a) $\dfrac{x^2 + 4x + 7}{x + 2}$

Polynomial long division: $x^2 + 4x + 7$ divided by $x + 2$.

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

So $f(x) = x + 2 + \dfrac{3}{x + 2}$

Therefore $A = 1$, $B = 2$, $C = 3$. [2] (M1 for attempting division, A1 for correct answer)

(b) Oblique asymptote: $y = x + 2$. [1]

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

$f'(x) = 1 - 3(x+2)^{-2} = 1 - \dfrac{3}{(x+2)^2}$

Set $f'(x) = 0$: $1 = \dfrac{3}{(x+2)^2} \Rightarrow (x+2)^2 = 3 \Rightarrow x = -2 \pm \sqrt{3}$

At $x = -2 + \sqrt{3}$: $f(-2+\sqrt{3}) = (-2+\sqrt{3}) + 2 + \dfrac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3}$

At $x = -2 - \sqrt{3}$: $f(-2-\sqrt{3}) = (-2-\sqrt{3}) + 2 + \dfrac{3}{-\sqrt{3}} = -\sqrt{3} - \sqrt{3} = -2\sqrt{3}$

$f''(x) = 6(x+2)^{-3} = \dfrac{6}{(x+2)^3}$

At $x = -2 + \sqrt{3}$: $x + 2 = \sqrt{3} > 0$, so $f'' > 0$ → minimum at $(-2+\sqrt{3}, 2\sqrt{3})$.

At $x = -2 - \sqrt{3}$: $x + 2 = -\sqrt{3} < 0$, so $f'' < 0$ → maximum at $(-2-\sqrt{3}, -2\sqrt{3})$.

[3] (M1 for differentiating, M1 for solving $f'(x) = 0$, A1 for correct coordinates and nature)

[Total: 6 marks]

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Question 10 [6]

(a) $f(x) = \dfrac{px + q}{x + r}$

Vertical asymptote at $x = -3$: $x + r = 0$ when $x = -3$, so $r = 3$. [1]

Horizontal asymptote at $y = 4$: $\dfrac{p}{1} = 4$, so $p = 4$. [1]

Passes through $(0, 2)$: $f(0) = \dfrac{q}{r} = \dfrac{q}{3} = 2$, so $q = 6$. [1]

$p = 4$, $q = 6$, $r = 3$.

(b) $f(x) = \dfrac{4x + 6}{x + 3}$

Set $f(x) = x$: $\dfrac{4x + 6}{x + 3} = x$

$4x + 6 = x(x + 3) = x^2 + 3x$

$x^2 + 3x - 4x - 6 = 0$

$x^2 - x - 6 = 0$

$(x - 3)(x + 2) = 0$

$x = 3$ or $x = -2$

When $x = 3$: $y = 3$. When $x = -2$: $y = -2$.

Points of intersection: $(3, 3)$ and $(-2, -2)$. [3] (M1 for setting $f(x) = x$, M1 for solving the quadratic, A1 for both points)

[Total: 6 marks]

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Section C: Application and Extension

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Question 11 [7]

(a) $P(0) = \dfrac{1200}{5 + 7e^0} = \dfrac{1200}{5 + 7} = \dfrac{1200}{12} = 100$.

Initial population is 100. [1]

(b) $P(5) = \dfrac{1200}{5 + 7e^{-1.5}} = \dfrac{1200}{5 + 7(0.2231)} = \dfrac{1200}{5 + 1.5617} = \dfrac{1200}{6.5617} \approx 182.9$

$P(5) \approx 183$ (to nearest whole number). [1]

(c) As $t \to \infty$, $e^{-0.3t} \to 0$, so $P(t) \to \dfrac{1200}{5 + 0} = 240$.

The limiting population is 240. The population grows and approaches 240 asymptotically. [2] (B1 for limit value, B1 for description)

(d) 90% of limiting value: $0.9 \times 240 = 216$.

$\dfrac{1200}{5 + 7e^{-0.3t}} = 216$

$5 + 7e^{-0.3t} = \dfrac{1200}{216} = \dfrac{50}{9}$

$7e^{-0.3t} = \dfrac{50}{9} - 5 = \dfrac{50 - 45}{9} = \dfrac{5}{9}$

$e^{-0.3t} = \dfrac{5}{63}$

$-0.3t = \ln\left(\dfrac{5}{63}\right) = \ln 5 - \ln 63$

$t = \dfrac{\ln 63 - \ln 5}{0.3} = \dfrac{\ln(12.6)}{0.3} = \dfrac{2.5337}{0.3} \approx 8.45$

$t \approx 8.45$ hours (to 3 s.f.). [3] (M1 for setting up equation, M1 for algebraic manipulation, A1 for correct answer)

[Total: 7 marks]

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Question 12 [8]

(a) $x^2 - 6x + 13 = x^2 - 6x + 9 + 4 = (x-3)^2 + 4$

So $f(x) = \ln[(x-3)^2 + 4]$. [1]

(b) Since $\ln$ is an increasing function, $f(x)$ is minimised when $(x-3)^2 + 4$ is minimised.

$(x-3)^2 \geq 0$ for all $x$, with minimum 0 at $x = 3$.

Minimum value of $(x-3)^2 + 4$ is $4$.

Minimum value of $f(x) = \ln 4 = 2\ln 2$, occurring at $x = 3$. [2] (B1 for $x = 3$, B1 for $\ln 4$ or $2\ln 2$)

(c) Since $(x-3)^2 + 4 \geq 4$ for all $x$, and $\ln$ is increasing:

$f(x) \geq \ln 4$. As $x \to \pm\infty$, $(x-3)^2 + 4 \to \infty$, so $f(x) \to \infty$.

Range of $f$: $[\ln 4, \infty)$ or $[2\ln 2, \infty)$. [1]

(d) $g(x) = e^{f(x)} = e^{\ln(x^2 - 6x + 13)} = x^2 - 6x + 13$. [1]

$g(x) = (x-3)^2 + 4$ is a parabola with minimum at $x = 3$. Since a parabola is not one-one over $\mathbb{R}$ (e.g., $g(2) = g(4) = 5$), $g$ is not one-one. [1]

(e) The student is partially correct. Since $f$ is not one-one over its full domain $\mathbb{R}$, an inverse function $f^{-1}$ does not exist without restricting the domain. However, if we restrict the domain of $f$ to either $[3, \infty)$ or $(-\infty, 3]$, then $f$ becomes one-one on that restricted domain, and $f^{-1}$ can be defined. [2] (B1 for agreeing that $f^{-1}$ doesn't exist over full domain, B1 for explaining domain restriction)

[Total: 8 marks]

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Mark Summary

Section	Marks
Section A: Questions 1–6	20
Section B: Questions 7–10	25
Section C: Questions 11–12	15
Total	60