A Level H1 Mathematics Algebra Functions Quiz

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A-Level Maths H1 Quiz - Algebra Functions

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 60 minutes
Total Marks: 45
Instructions:

1. Answer all 20 questions.
2. You are expected to use an approved graphing calculator (GC).
3. Where numerical answers are required, give non-exact answers correct to 3 significant figures, unless otherwise stated.
4. Show necessary working clearly. Unsupported answers from a calculator are generally allowed, but you must show the mathematical steps where reasoning is required.

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Section A: Basic Concepts and Manipulation (Questions 1–5)

Focus: Laws of indices, logarithms, and basic function definitions.

1. Solve the equation $3^{2x} - 10(3^x) + 9 = 0$. [2]

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2. Given that $\ln(x^2 y) = 3$ and $\ln(x y^2) = 4$, find the exact values of $x$ and $y$. [3]

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3. The function $f$ is defined by $f(x) = e^{2x} - 5$, for $x \in \mathbb{R}$. (a) Find the inverse function $f^{-1}(x)$ and state its domain. [2] (b) Sketch the graph of $y = f^{-1}(x)$, indicating any asymptotes and intercepts with the axes. [2]

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4. Simplify the expression $\frac{e^{3x} \cdot e^{-x}}{(e^x)^2}$, giving your answer in the form $e^{kx}$. [1]

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5. Solve the inequality $\ln(2x - 1) \leq 2$. Give your answer in exact form. [2]

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Section B: Graphs and Transformations (Questions 6–10)

Focus: Sketching, asymptotes, and transformations of exponential and logarithmic functions.

6. The diagram below shows the graph of $y = f(x)$, where $f(x) = \ln(x)$. (Note: Imagine a standard ln(x) graph passing through (1,0) with vertical asymptote x=0) On the same axes, sketch the graph of $y = 2 - \ln(x+1)$. Clearly label: (i) The equation of the vertical asymptote. (ii) The coordinates of the x-intercept. (iii) The coordinates of the point corresponding to $(e, 1)$ on the original graph. [3]

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7. Consider the function $g(x) = 3e^{-x} + 2$. (a) State the equation of the horizontal asymptote. [1] (b) Find the exact value of $x$ for which $g(x) = 5$. [2]

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8. The curve $C$ has equation $y = e^{2x} - 4e^x + 3$. (a) Find the coordinates of the points where $C$ crosses the x-axis. [2] (b) Find the coordinates of the stationary point on $C$ and determine its nature. [3]

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9. Explain why the equation $e^x = -2$ has no real solutions. [1]

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10. The function $h(x)$ is defined by $h(x) = |\ln(x)|$. (a) Sketch the graph of $y = h(x)$ for $0 < x \leq 5$. [2] (b) Hence, state the number of solutions to the equation $|\ln(x)| = 0.5$. [1]

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Section C: Applications and Modelling (Questions 11–15)

Focus: Exponential growth/decay, linearization, and context-based problems.

11. The population $P$ of a town $t$ years after 2020 is modelled by the equation $P = P_0 e^{kt}$, where $P_0$ and $k$ are constants. In 2020, the population was 50,000. In 2025, the population was 62,000. (a) Find the value of $k$ correct to 3 significant figures. [2] (b) Estimate the population in 2030. [1]

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12. The value $V$ of a car $t$ years after purchase is given by $V = A e^{-bt}$. (a) Show that $\ln V$ is a linear function of $t$. [1] (b) A plot of $\ln V$ against $t$ yields a straight line with gradient $-0.15$ and vertical intercept $9.2$. Find the values of $A$ and $b$. [2]

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13. A radioactive substance decays such that its mass $m$ grams at time $t$ hours is given by $m = 50(0.8)^t$. (a) Calculate the initial mass. [1] (b) Find the time taken for the mass to halve. [2]

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14. The temperature $T$ of a cup of coffee $t$ minutes after being poured is modelled by $T = 20 + 60e^{-0.05t}$. (a) What is the room temperature according to this model? [1] (b) Find the rate of change of the temperature when $t = 10$ minutes. [2]

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15. Solve the simultaneous equations: $y = e^x$ $y = 4e^{-x}$ Give your answers in exact form. [3]

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Section D: Advanced Algebraic Techniques (Questions 16–20)

Focus: Composite functions, domain/range implications (conceptual), and harder equations.

16. Let $f(x) = e^x$ for $x \in \mathbb{R}$ and $g(x) = \ln(x-2)$ for $x > 2$. (a) Find the composite function $fg(x)$ in its simplest form. [2] (b) State the domain of $fg(x)$. [1]

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17. Solve the equation $2\ln(x) - \ln(x+3) = \ln(2)$. Check for extraneous roots. [3]

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18. The function $f(x) = \frac{e^x}{e^x + 1}$. (a) Show that $f(x)$ can be written as $1 - \frac{1}{e^x + 1}$. [1] (b) Hence, find the range of $f(x)$. [2]

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19. Given that $x = \log_2(y)$, express $y$ in terms of $x$ and hence solve $2^{2x} - 5(2^x) + 4 = 0$. [3]

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20. A bacteria culture grows according to the law $N = N_0 e^{rt}$. If the culture doubles every 3 hours, find the value of $r$ in the form $\frac{\ln a}{b}$. [2]

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A-Level Maths H1 Quiz - Algebra Functions (Answer Key)

1. Solve $3^{2x} - 10(3^x) + 9 = 0$. [2] Let $u = 3^x$. Then $u^2 - 10u + 9 = 0$. $(u-9)(u-1) = 0$. $u = 9$ or $u = 1$. If $3^x = 9$, then $x = 2$. If $3^x = 1$, then $x = 0$. Answer: $x = 0, x = 2$.

2. Given $\ln(x^2 y) = 3$ and $\ln(x y^2) = 4$. [3] $2\ln x + \ln y = 3$ --- (1) $\ln x + 2\ln y = 4$ --- (2) From (1), $\ln y = 3 - 2\ln x$. Substitute into (2): $\ln x + 2(3 - 2\ln x) = 4$ $\ln x + 6 - 4\ln x = 4$ $-3\ln x = -2 \implies \ln x = \frac{2}{3} \implies x = e^{2/3}$. $\ln y = 3 - 2(\frac{2}{3}) = 3 - \frac{4}{3} = \frac{5}{3} \implies y = e^{5/3}$. Answer: $x = e^{2/3}, y = e^{5/3}$.

3. $f(x) = e^{2x} - 5$. [4] (a) Let $y = e^{2x} - 5$. Swap $x$ and $y$: $x = e^{2y} - 5$. $x + 5 = e^{2y} \implies \ln(x+5) = 2y \implies y = \frac{1}{2}\ln(x+5)$. Domain of $f^{-1}$: Argument of log must be positive. $x+5 > 0 \implies x > -5$. Answer: $f^{-1}(x) = \frac{1}{2}\ln(x+5)$, Domain: $x > -5$.

(b) Graph of $y = \frac{1}{2}\ln(x+5)$. Vertical Asymptote: $x = -5$. x-intercept: $y=0 \implies \ln(x+5)=0 \implies x+5=1 \implies x=-4$. Point $(-4, 0)$. y-intercept: $x=0 \implies y = \frac{1}{2}\ln(5) \approx 0.8$. Point $(0, \frac{1}{2}\ln 5)$. Shape: Increasing logarithmic curve shifted left by 5 and scaled vertically by 0.5.

4. Simplify $\frac{e^{3x} \cdot e^{-x}}{(e^x)^2}$. [1] Numerator: $e^{3x-x} = e^{2x}$. Denominator: $e^{2x}$. Result: $\frac{e^{2x}}{e^{2x}} = 1 = e^0$. Answer: $1$ (or $e^{0x}$).

5. Solve $\ln(2x - 1) \leq 2$. [2] Domain condition: $2x - 1 > 0 \implies x > 0.5$. Exponentiate both sides: $2x - 1 \leq e^2$. $2x \leq e^2 + 1$. $x \leq \frac{e^2 + 1}{2}$. Combining with domain: Answer: $0.5 < x \leq \frac{e^2 + 1}{2}$.

6. Sketch $y = 2 - \ln(x+1)$. [3] Original: $y=\ln x$. Transformations: Shift left 1 unit ($\ln(x+1)$), Reflect in x-axis ($-\ln(x+1)$), Shift up 2 units ($2-\ln(x+1)$). (i) Vertical Asymptote: $x = -1$ (since argument $x+1=0$). (ii) x-intercept: $0 = 2 - \ln(x+1) \implies \ln(x+1)=2 \implies x+1=e^2 \implies x=e^2-1$. Point $(e^2-1, 0)$. (iii) Point corresponding to $(e,1)$ on $\ln x$: On $\ln(x+1)$, input $x$ such that $x+1=e \implies x=e-1$. Value is $\ln(e)=1$. Reflect: $-1$. Shift up 2: $-1+2=1$. Point is $(e-1, 1)$. Answer: Sketch showing VA at $x=-1$, passing through $(e^2-1, 0)$ and $(e-1, 1)$, decreasing curve.

7. $g(x) = 3e^{-x} + 2$. [3] (a) As $x \to \infty$, $e^{-x} \to 0$. So $y \to 2$. Answer: $y = 2$. (b) $5 = 3e^{-x} + 2 \implies 3 = 3e^{-x} \implies 1 = e^{-x} \implies -x = \ln 1 = 0 \implies x = 0$. Answer: $x = 0$.

8. $y = e^{2x} - 4e^x + 3$. [5] (a) x-intercepts: $y=0$. Let $u=e^x$. $u^2 - 4u + 3 = 0 \implies (u-3)(u-1)=0$. $e^x = 3 \implies x = \ln 3$. $e^x = 1 \implies x = 0$. Points: $(0, 0)$ and $(\ln 3, 0)$. (b) Stationary point: $\frac{dy}{dx} = 2e^{2x} - 4e^x$. Set $\frac{dy}{dx} = 0 \implies 2e^x(e^x - 2) = 0$. Since $e^x \neq 0$, $e^x = 2 \implies x = \ln 2$. y-coordinate: $y = e^{2\ln 2} - 4e^{\ln 2} + 3 = (e^{\ln 2})^2 - 4(2) + 3 = 2^2 - 8 + 3 = 4 - 8 + 3 = -1$. Point: $(\ln 2, -1)$. Nature: $\frac{d^2y}{dx^2} = 4e^{2x} - 4e^x$. At $x=\ln 2$: $4(4) - 4(2) = 16 - 8 = 8 > 0$. Minimum. Answer: Min at $(\ln 2, -1)$.

9. Why $e^x = -2$ has no real solutions. [1] Answer: The exponential function $e^x$ is strictly positive for all real $x$ ($e^x > 0$). It can never equal a negative number.

10. $h(x) = |\ln(x)|$. [3] (a) Sketch: For $x \geq 1$, $y=\ln x$. For $0 < x < 1$, $y=-\ln x$ (reflection of the negative part of ln x above x-axis). V-shape touching x-axis at $(1,0)$. VA at $x=0$. (b) $|\ln x| = 0.5$. Two branches. $\ln x = 0.5 \implies x=e^{0.5}$ and $\ln x = -0.5 \implies x=e^{-0.5}$. Answer: 2 solutions.

11. Population Model $P = P_0 e^{kt}$. [3] (a) $t=0, P=50000 \implies P_0 = 50000$. $t=5, P=62000 \implies 62000 = 50000 e^{5k}$. $1.24 = e^{5k} \implies \ln(1.24) = 5k \implies k = \frac{\ln(1.24)}{5} \approx 0.0431$. Answer: $k \approx 0.0431$. (b) $t=10$ (2030). $P = 50000 e^{10(0.0431)} = 50000 e^{0.431} \approx 50000(1.5388) \approx 76940$. Alternatively, $P = 50000 (1.24)^2 = 50000(1.5376) = 76880$. (Using exact k gives ~76942). Answer: approx 76,900.

12. Car Value $V = A e^{-bt}$. [3] (a) $\ln V = \ln(A e^{-bt}) = \ln A + \ln(e^{-bt}) = \ln A - bt$. This is linear form $Y = C + mX$ with $Y=\ln V, X=t$. (b) Gradient $m = -b = -0.15 \implies b = 0.15$. Intercept $C = \ln A = 9.2 \implies A = e^{9.2} \approx 9897$. Answer: $A = e^{9.2}$ (or 9900), $b = 0.15$.

13. Radioactive Decay $m = 50(0.8)^t$. [3] (a) Initial mass at $t=0$: $m = 50(0.8)^0 = 50$ g. Answer: 50 g. (b) Half mass = 25 g. $25 = 50(0.8)^t \implies 0.5 = 0.8^t$. $\ln(0.5) = t \ln(0.8) \implies t = \frac{\ln 0.5}{\ln 0.8} \approx 3.106$. Answer: 3.11 hours.

14. Coffee Temperature $T = 20 + 60e^{-0.05t}$. [3] (a) Room temp is the asymptote as $t \to \infty$. $T \to 20$. Answer: $20^\circ$C. (b) Rate of change $\frac{dT}{dt} = 60(-0.05)e^{-0.05t} = -3e^{-0.05t}$. At $t=10$: $\frac{dT}{dt} = -3e^{-0.5} \approx -3(0.6065) \approx -1.82$. Answer: $-1.82^\circ$C/min.

15. Simultaneous: $y = e^x$ and $y = 4e^{-x}$. [3] $e^x = 4e^{-x} \implies e^x \cdot e^x = 4 \implies e^{2x} = 4$. $2x = \ln 4 \implies x = \frac{1}{2}\ln 4 = \ln(4^{1/2}) = \ln 2$. $y = e^{\ln 2} = 2$. Answer: $x = \ln 2, y = 2$.

16. $f(x) = e^x, g(x) = \ln(x-2)$. [3] (a) $fg(x) = f(g(x)) = e^{\ln(x-2)}$. Since $e^{\ln u} = u$, $fg(x) = x - 2$. Answer: $x - 2$. (b) Domain of $g$ is $x > 2$. Range of $g$ is $\mathbb{R}$, which is domain of $f$. So domain of $fg$ is domain of $g$. Answer: $x > 2$.

17. $2\ln(x) - \ln(x+3) = \ln(2)$. [3] $\ln(x^2) - \ln(x+3) = \ln(2)$. $\ln\left(\frac{x^2}{x+3}\right) = \ln(2)$. $\frac{x^2}{x+3} = 2 \implies x^2 = 2(x+3) \implies x^2 - 2x - 6 = 0$. $x = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$. Check validity: Argument of log must be positive. $x > 0$ and $x+3 > 0 \implies x > 0$. $1 - \sqrt{7} \approx -1.65$ (Reject). $1 + \sqrt{7} \approx 3.65$ (Accept). Answer: $x = 1 + \sqrt{7}$.

18. $f(x) = \frac{e^x}{e^x + 1}$. [3] (a) $\frac{e^x}{e^x+1} = \frac{e^x+1-1}{e^x+1} = \frac{e^x+1}{e^x+1} - \frac{1}{e^x+1} = 1 - \frac{1}{e^x+1}$. Shown. (b) Range: $e^x > 0 \implies e^x + 1 > 1$. $0 < \frac{1}{e^x+1} < 1$. $-1 < -\frac{1}{e^x+1} < 0$. $0 < 1 - \frac{1}{e^x+1} < 1$. Answer: $0 < f(x) < 1$.

19. $x = \log_2(y) \implies y = 2^x$. [3] Equation: $2^{2x} - 5(2^x) + 4 = 0$. Let $u = 2^x$. $u^2 - 5u + 4 = 0$. $(u-4)(u-1) = 0$. $u=4 \implies 2^x=4 \implies x=2$. $u=1 \implies 2^x=1 \implies x=0$. Answer: $x=0, x=2$.

20. Doubling time 3 hours. $N = N_0 e^{rt}$. [2] $2N_0 = N_0 e^{r(3)} \implies 2 = e^{3r}$. $\ln 2 = 3r \implies r = \frac{\ln 2}{3}$. Answer: $r = \frac{\ln 2}{3}$.