A Level H2 Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key & Marking Scheme

Subject: Mathematics H2 | Paper: Practice Paper — Algebra & Functions (Version 1 of 5) | Total Marks: 50

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Section A: Short Answer Questions [20 marks]

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Question 1 [4 marks]

(a) Find $f^{-1}(x)$ and state its domain. [3]

Method:

We have $f(x) = \dfrac{3x - 1}{x + 2}$, $x \neq -2$.

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

Solve for $x$ in terms of $y$:

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

Therefore:

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

The domain of $f^{-1}$ is the range of $f$. Since $f(x) = \dfrac{3x-1}{x+2}$ is a rational function with horizontal asymptote $y = 3$ (ratio of leading coefficients), and $f$ never equals 3 (since $\dfrac{3x-1}{x+2} = 3 \Rightarrow 3x - 1 = 3x + 6 \Rightarrow -1 = 6$, impossible), the domain of $f^{-1}$ is $x \neq 3$.

Mark breakdown:

• M1: Correct method to make $x$ the subject (cross-multiplication and collecting terms)
• A1: Correct expression for $f^{-1}(x) = \dfrac{2x+1}{3-x}$
• A1: Correct domain: $x \neq 3$ (or $x \in \mathbb{R}, x \neq 3$)

Common mistake: Students often forget that the domain of $f^{-1}$ equals the range of $f$, and may incorrectly state the domain as $x \neq -2$ (which is the domain of $f$, not $f^{-1}$).

(b) State the range of $f$. [1]

The range of $f$ is the domain of $f^{-1}$, which is all real values except 3.

Answer: $\text{Range of } f = \{y \in \mathbb{R} : y \neq 3\}$

[1 mark for correct range]

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Question 2 [4 marks]

(a) Show that the composite function $gf$ exists. [2]

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

$f(x) = x^2 - 4x + 5 = (x-2)^2 + 1$.

Since $(x-2)^2 \geq 0$ for all $x \in \mathbb{R}$, we have $f(x) \geq 1$.

So $\text{Range}(f) = [1, \infty)$.

The domain of $g$ is $x \geq 1$, i.e., $[1, \infty)$.

Since $\text{Range}(f) = [1, \infty) = \text{Domain}(g)$, we have $\text{Range}(f) \subseteq \text{Domain}(g)$.

Therefore $gf$ exists.

Mark breakdown:

• M1: Find the range of $f$ by completing the square or using vertex formula; range is $[1, \infty)$
• A1: Compare with domain of $g$ ($x \geq 1$) and conclude $gf$ exists

Common mistake: Students may state that $gf$ exists without explicitly showing that the range of $f$ is contained in the domain of $g$. Both conditions must be checked and stated.

(b) Find an expression for $gf(x)$ and state the range of $gf$. [2]

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

Since $x \in \mathbb{R}$, we have $|x - 2| \geq 0$.

The range of $gf$ is $[0, \infty)$.

Mark breakdown:

• M1: Correct substitution and simplification to get $|x - 2|$
• A1: Correct range $[0, \infty)$

Common mistake: Writing $\sqrt{(x-2)^2} = x - 2$ instead of $|x - 2|$. The square root of a squared expression gives the absolute value.

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Question 3 [3 marks]

(a) Find $f^{-1}(x)$. [2]

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

Let $y = e^{2x} + 3$.

$y - 3 = e^{2x}$

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

$x = \frac{1}{2}\ln(y - 3)$

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

Mark breakdown:

• M1: Correct method — isolate the exponential and take logarithms
• A1: Correct answer $f^{-1}(x) = \frac{1}{2}\ln(x - 3)$

(b) State the domain and range of $f^{-1}$. [1]

Domain of $f^{-1}$ = Range of $f$: Since $e^{2x} > 0$, we have $f(x) > 3$, so domain of $f^{-1}$ is $x > 3$.

Range of $f^{-1}$ = Domain of $f$: Since $f$ is defined for all real $x$, the range of $f^{-1}$ is $\mathbb{R}$.

Answer: Domain: $x > 3$; Range: $y \in \mathbb{R}$

[1 mark for both correct]

Common mistake: Students may write the domain as $x \geq 3$, but $e^{2x}$ is strictly positive (never zero), so $f(x) > 3$ strictly.

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Question 4 [2 marks]

$f^{-1}(0)$ means the value of $x$ such that $f(x) = 0$.

$\ln(2x - 5) = 0$

$2x - 5 = e^0 = 1$

$2x = 6$

$x = 3$

Answer: $f^{-1}(0) = 3$

Mark breakdown:

• M1: Set $f(x) = 0$ and solve (or find inverse first then substitute)
• A1: Correct answer $x = 3$

Teaching note: Finding $f^{-1}(0)$ does not require finding the full inverse function. It is often faster to solve $f(x) = 0$ directly.

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Question 5 [4 marks]

(a) Explain why $g$ has an inverse function. [1]

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

The domain is restricted to $x \leq 3$. On this interval, $g$ is strictly decreasing (since the parabola opens upward and we take the left branch, left of the vertex at $x = 3$).

Since $g$ is strictly monotonic (strictly decreasing) on its domain, it is one-one and therefore has an inverse.

[1 mark for stating that $g$ is one-one/strictly monotonic on the given domain]

(b) Find $g^{-1}(x)$ and state its domain. [3]

Let $y = (x - 3)^2 + 1$, with $x \leq 3$.

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

$x - 3 = \pm\sqrt{y - 1}$

Since $x \leq 3$, we have $x - 3 \leq 0$, so we take the negative square root:

$x = 3 - \sqrt{y - 1}$

Therefore $g^{-1}(x) = 3 - \sqrt{x - 1}$.

Domain of $g^{-1}$ = Range of $g$: Since $(x-3)^2 \geq 0$ for $x \leq 3$, the minimum value of $g$ is $1$ (at $x = 3$). As $x \to -\infty$, $g(x) \to \infty$. So the range of $g$ is $[1, \infty)$.

Domain of $g^{-1}$ is $x \geq 1$.

Mark breakdown:

• M1: Correct method to solve for $x$ in terms of $y$
• A1: Correct choice of negative root (justified by domain restriction $x \leq 3$)
• A1: Correct inverse $g^{-1}(x) = 3 - \sqrt{x - 1}$ with domain $x \geq 1$

Common mistake: Taking the positive square root instead of the negative one. The domain restriction $x \leq 3$ means $x - 3 \leq 0$, so the negative root must be chosen.

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Question 6 [3 marks]

$f(x) = \dfrac{ax + b}{cx + d}$, and $f(f(x)) = x$.

First compute $f(f(x))$:

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

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

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

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

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

For $f(f(x)) = x$, we need:

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

This means $(a^2 + bc)x + b(a+d) = x[(ac+cd)x + (bc+d^2)]$

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

Comparing coefficients:

• Coefficient of $x^2$: $0 = ac + cd = c(a + d)$. Since $c \neq 0$, we get $a + d = 0$.

Mark breakdown:

• M1: Correct computation of $f(f(x))$ as a single fraction
• M1: Setting $f(f(x)) = x$ and comparing coefficients
• A1: Conclusion that $a + d = 0$ (with justification that $c \neq 0$)

Teaching note: This shows that a rational function of the form $\dfrac{ax+b}{cx+d}$ is self-inverse (equal to its own inverse) if and only if $a + d = 0$. This connects to Question 9.

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Section B: Structured Questions [20 marks]

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Question 7 [6 marks]

(a) Determine whether $f$ is one-one. Justify your answer. [2]

For $x \leq 1$: $f(x) = x^2 - 2x + 3 = (x-1)^2 + 2$. This is a parabola opening upward with vertex at $x = 1$. On $(-\infty, 1]$, this is strictly decreasing, hence one-one on this interval.

For $x > 1$: $f(x) = 2x + 1$, which is strictly increasing, hence one-one on this interval.

However, we must check whether the function is one-one overall. At $x = 1$: $f(1) = 1 - 2 + 3 = 2$.

For $x > 1$: $f(x) = 2x + 1 > 3$. So the ranges of the two pieces don't overlap: the first piece gives values $\geq 2$ and the second piece gives values $> 3$.

Wait — let me recheck. For $x \leq 1$: as $x \to -\infty$, $f(x) \to \infty$, and at $x = 1$, $f(1) = 2$. So the range of the first piece is $[2, \infty)$.

For $x > 1$: $f(x) = 2x + 1$, and as $x \to 1^+$, $f(x) \to 3$. So the range of the second piece is $(3, \infty)$.

Since $(3, \infty) \subset [2, \infty)$, there is overlap. For example, $f(0) = 0 - 0 + 3 = 3$ and $f(1.1) = 2(1.1) + 1 = 3.2$... Let me check more carefully.

$f(x) = 3$ in the first piece: $x^2 - 2x + 3 = 3 \Rightarrow x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$. But $x = 2$ is not in $x \leq 1$, so only $x = 0$ gives $f(0) = 3$.

$f(x) = 3$ in the second piece: $2x + 1 = 3 \Rightarrow x = 1$, but $x = 1$ is not in $x > 1$. So $f(x) = 3$ only at $x = 0$.

Let me check $f(x) = 4$: First piece: $x^2 - 2x + 3 = 4 \Rightarrow x^2 - 2x - 1 = 0 \Rightarrow x = 1 \pm \sqrt{2}$. Only $x = 1 - \sqrt{2} \approx -0.414$ is in $x \leq 1$. Second piece: $2x + 1 = 4 \Rightarrow x = 1.5$, which is in $x > 1$.

So $f(1-\sqrt{2}) = 4$ and $f(1.5) = 4$. Since two different $x$-values give the same output, $f$ is not one-one.

Answer: $f$ is not one-one. For example, $f(1-\sqrt{2}) = f(1.5) = 4$ but $1-\sqrt{2} \neq 1.5$.

Mark breakdown:

• M1: Find a counterexample or show that two different inputs give the same output
• A1: Correct conclusion that $f$ is not one-one, with valid justification

(b) State the range of $f$. [2]

For $x \leq 1$: $f(x) = (x-1)^2 + 2$. The minimum is at $x = 1$ where $f(1) = 2$. As $x \to -\infty$, $f(x) \to \infty$. Range: $[2, \infty)$.

For $x > 1$: $f(x) = 2x + 1 > 3$. Range: $(3, \infty)$.

The union is $[2, \infty)$.

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

Mark breakdown:

• M1: Find the range of each piece
• A1: Correct overall range $[2, \infty)$

(c) Find the value of $f^{-1}(3)$, if it exists. Justify your answer. [2]

Since $f$ is not one-one, $f^{-1}$ does not exist as a function.

However, we can still ask: for which $x$ is $f(x) = 3$?

From the first piece: $x^2 - 2x + 3 = 3 \Rightarrow x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$. Only $x = 0$ satisfies $x \leq 1$.

From the second piece: $2x + 1 = 3 \Rightarrow x = 1$, but $x = 1$ is not in $x > 1$.

So $f(0) = 3$, meaning if we were to define an inverse on a restricted domain containing $x = 0$, we would have $f^{-1}(3) = 0$.

But since $f$ is not one-one on its full domain, $f^{-1}$ does not exist as a function.

Answer: $f^{-1}$ does not exist because $f$ is not one-one. However, the unique solution to $f(x) = 3$ is $x = 0$.

Mark breakdown:

• M1: State that $f^{-1}$ does not exist (or explain the issue)
• A1: Note that $f(x) = 3$ has the unique solution $x = 0$

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Question 8 [7 marks]

(a) Find an expression for $fg(x)$. State whether $fg$ exists and justify your answer. [3]

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

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

$g(x) = \frac{1}{x-2}$ for $x \neq 2$. The range of $g$ is all real values except 0 (since $\frac{1}{x-2} \neq 0$ for any $x$).

$\text{Domain of } f = \mathbb{R}$.

Since $\mathbb{R} \setminus \{0\} \subseteq \mathbb{R}$, the range of $g$ is a subset of the domain of $f$.

Therefore $fg$ exists.

Answer: $fg(x) = 3 - \dfrac{1}{(x-2)^2}$, $x \neq 2$. Yes, $fg$ exists because $\text{Range}(g) = \mathbb{R} \setminus \{0\} \subseteq \mathbb{R} = \text{Domain}(f)$.

Mark breakdown:

• M1: Correct substitution into $f$
• A1: Correct simplified expression for $fg(x)$
• A1: Correct justification that $fg$ exists

(b) Find the range of $fg$. [2]

$fg(x) = 3 - \frac{1}{(x-2)^2}$

Since $(x-2)^2 > 0$ for all $x \neq 2$, we have $\frac{1}{(x-2)^2} > 0$.

Therefore $fg(x) = 3 - \text{(positive number)} < 3$.

As $(x-2)^2 \to \infty$, $\frac{1}{(x-2)^2} \to 0^+$, so $fg(x) \to 3^-$.

As $(x-2)^2 \to 0^+$, $\frac{1}{(x-2)^2} \to \infty$, so $fg(x) \to -\infty$.

Answer: Range of $fg = (-\infty, 3)$

Mark breakdown:

• M1: Analyze the behaviour of $\frac{1}{(x-2)^2}$ and its effect on $fg(x)$
• A1: Correct range $(-\infty, 3)$

(c) Sketch the graph of $y = fg(x)$, clearly labelling any asymptotes and intercepts. [2]

Key features:

• Vertical asymptote: $x = 2$ (denominator of $g(x)$ is zero)
• Horizontal asymptote: $y = 3$ (as $x \to \pm\infty$, $fg(x) \to 3$ from below)
• $y$-intercept: $fg(0) = 3 - \frac{4}{1} = 3 - \frac{1}{4} = \frac{11}{4} = 2.75$. Point: $(0, 2.75)$
• $x$-intercepts: $3 - \frac{1}{(x-2)^2} = 0 \Rightarrow (x-2)^2 = \frac{1}{3} \Rightarrow x = 2 \pm \frac{1}{\sqrt{3}}$. Points: $(2 - \frac{\sqrt{3}}{3}, 0)$ and $(2 + \frac{\sqrt{3}}{3}, 0)$

The graph lies entirely below $y = 3$, approaching the asymptotes.

Mark breakdown:

• M1: Correct asymptotes and intercepts identified
• A1: Reasonable sketch showing correct shape, branches, and labelled features

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Question 9 [7 marks]

(a) Find $f^{-1}(x)$. [2]

$f(x) = \dfrac{2x + 3}{x - 1}$, $x \neq 1$.

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

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

$xy - y = 2x + 3$

$xy - 2x = y + 3$

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

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

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

Mark breakdown:

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

(b) Show that $f^{-1}(x) = f(x)$. What property does this tell you about $f$? [2]

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

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

These are not obviously the same. Let me check by computing $f(f(x))$:

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

$= \frac{\frac{4x+6+3(x-1)}{x-1}}{\frac{2x+3-(x-1)}{x-1}} = \frac{4x+6+3x-3}{2x+3-x+1} = \frac{7x+3}{x+4}$

This is not equal to $x$. So $f^{-1} \neq f$ in this case. Let me re-examine.

Actually, let me recheck the inverse calculation:

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

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

$xy - y = 2x + 3$

$xy - 2x = y + 3$

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

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

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

Now $f(x) = \dfrac{2x+3}{x-1}$ and $f^{-1}(x) = \dfrac{x+3}{x-2}$. These are different functions.

Let me verify: $f(f^{-1}(x))$ should equal $x$.

$f(f^{-1}(x)) = f\!\left(\frac{x+3}{x-2}\right) = \frac{2\cdot\frac{x+3}{x-2}+3}{\frac{x+3}{x-2}-1} = \frac{\frac{2x+6+3(x-2)}{x-2}}{\frac{x+3-(x-2)}{x-2}} = \frac{2x+6+3x-6}{x+3-x+2} = \frac{5x}{5} = x$. ✓

So the inverse is correct, but $f^{-1} \neq f$. The question as stated has an issue. Let me adjust the question to make it work.

Actually, for a function to equal its own inverse, we need $a + d = 0$ (from Question 6). Here $a = 2$, $d = -1$, so $a + d = 1 \neq 0$. So $f$ is not self-inverse.

Let me revise the question to use a function that IS self-inverse. Let me use $f(x) = \dfrac{x+3}{x+1}$ instead, where $a = 1, d = 1$, so $a + d = 2 \neq 0$... that doesn't work either.

For self-inverse: $a + d = 0$. So we need $d = -a$. Let's use $f(x) = \dfrac{2x+3}{x-2}$. Then $a = 2, d = -2$, so $a + d = 0$.

Let me redo the question with $f(x) = \dfrac{2x+3}{x-2}$:

Revised Question 9: $f(x) = \dfrac{2x+3}{x-2}$, $x \neq 2$.

(a) Find $f^{-1}(x)$. [2]

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

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

$xy - 2y = 2x + 3$

$xy - 2x = 2y + 3$

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

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

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

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

Mark breakdown:

• M1: Correct algebraic manipulation
• A1: Correct inverse

(b) Show that $f^{-1}(x) = f(x)$. What property does this tell you about $f$? [2]

From part (a), $f^{-1}(x) = \dfrac{2x+3}{x-2} = f(x)$.

This means $f$ is a self-inverse function (or an involution). Applying $f$ twice returns the original input: $f(f(x)) = x$.

Mark breakdown:

• M1: Show algebraically that $f^{-1}(x) = f(x)$
• A1: Correct identification of the property (self-inverse/involution)

(c) On the same diagram, sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$. State the equation of the line of symmetry between them. [3]

Since $f^{-1} = f$, the two graphs are identical — they are the same curve.

The graph of $y = f(x)$ is symmetric about the line $y = x$ (since a self-inverse function's graph is always symmetric about $y = x$).

Key features of $y = \dfrac{2x+3}{x-2}$:

• Vertical asymptote: $x = 2$
• Horizontal asymptote: $y = 2$ (ratio of leading coefficients)
• $x$-intercept: $2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}$. Point: $(-1.5, 0)$
• $y$-intercept: $f(0) = \dfrac{3}{-2} = -1.5$. Point: $(0, -1.5)$

The graph is a rectangular hyperbola with two branches, symmetric about $y = x$.

Answer: The line of symmetry is $y = x$. Since $f = f^{-1}$, the two graphs coincide.

Mark breakdown:

• M1: Correct identification that the graphs are the same
• M1: Correct asymptotes and intercepts
• A1: Correct line of symmetry $y = x$

Expected visual for Q9(c): The graph should show a rectangular hyperbola with centre at the intersection of the asymptotes $(2, 2)$, branches in the top-right and bottom-left regions relative to the centre, passing through $(-1.5, 0)$ and $(0, -1.5)$, with the line $y = x$ shown as a dashed line. The graph should be clearly symmetric about $y = x$.

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Section C: Application & Extended Reasoning [10 marks]

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Question 10 [10 marks]

(a) State the initial temperature of the object. [1]

At $t = 0$: $T(0) = 22 + 78e^0 = 22 + 78 = 100$.

Answer: The initial temperature is $100°C$.

[1 mark]

(b) Find the value of $t$ when the temperature reaches 60°C, giving your answer correct to 2 decimal places. [3]

$22 + 78e^{-0.05t} = 60$

$78e^{-0.05t} = 38$

$e^{-0.05t} = \dfrac{38}{78} = \dfrac{19}{39}$

$-0.05t = \ln\!\left(\dfrac{19}{39}\right)$

$t = \dfrac{\ln(19/39)}{-0.05} = \dfrac{\ln(39/19)}{0.05}$

$t = \dfrac{\ln(39) - \ln(19)}{0.05} = \dfrac{3.6636 - 2.9444}{0.05} = \dfrac{0.7191}{0.05} \approx 14.38$

Answer: $t \approx 14.38$ minutes

Mark breakdown:

• M1: Correct equation setup and isolation of the exponential term
• M1: Correct use of logarithms to solve for $t$
• A1: Correct answer $t \approx 14.38$ minutes (to 2 d.p.)

(c) Explain, in the context of the model, what happens to the temperature as $t \to \infty$. [2]

As $t \to \infty$, $e^{-0.05t} \to 0$.

Therefore $T(t) \to 22 + 0 = 22$.

In context: The temperature of the object approaches $22°C$ as time increases. This is the ambient (surrounding) temperature — the object cools down and approaches room temperature but never goes below it.

Mark breakdown:

• M1: Correct mathematical limit ($T \to 22$)
• A1: Correct contextual interpretation (approaches ambient temperature of 22°C)

(d) The function $T$ is one-one for $t \geq 0$. Find $T^{-1}(t)$ and explain what $T^{-1}$ represents in this context. [4]

Let $T = 22 + 78e^{-0.05t}$.

$T - 22 = 78e^{-0.05t}$

$\dfrac{T - 22}{78} = e^{-0.05t}$

$-0.05t = \ln\!\left(\dfrac{T - 22}{78}\right)$

$t = \dfrac{1}{-0.05}\ln\!\left(\dfrac{T - 22}{78}\right) = -20\ln\!\left(\dfrac{T - 22}{78}\right)$

Therefore:

$T^{-1}(t) = -20\ln\!\left(\dfrac{t - 22}{78}\right)$

Domain of $T^{-1}$: Since $T(t) = 22 + 78e^{-0.05t}$ and $e^{-0.05t} \in (0, 1]$ for $t \geq 0$, we have $T(t) \in (22, 100]$. So the domain of $T^{-1}$ is $22 < t \leq 100$.

Interpretation: $T^{-1}(t)$ gives the time (in minutes) at which the object reaches a temperature of $t°C$. While $T(t)$ tells us the temperature at a given time, $T^{-1}(t)$ tells us the time needed to reach a given temperature.

Mark breakdown:

• M1: Correct method to make $t$ the subject
• A1: Correct expression for $T^{-1}(t)$
• M1: Correct contextual interpretation (inverse gives time for a given temperature)
• A1: Correct domain stated or implied

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Mark Summary

Section	Marks
Q1	4
Q2	4
Q3	3
Q4	2
Q5	4
Q6	3
Section A Total	20
Q7	6
Q8	7
Q9	7
Section B Total	20
Q10	10
Section C Total	10
Grand Total	50

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This practice paper was AI-generated by TuitionGoWhere. It is syllabus-aligned but not derived from any specific past-year examination paper.