A Level H1 Mathematics Algebra Functions Quiz

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A-Level Maths H1 Quiz - Algebra Functions

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 50

Duration: 60 minutes
Total Marks: 50
Instructions:

1. Answer all questions.
2. You are expected to use an approved graphing calculator (GC).
3. Unsupported GC answers are allowed unless stated otherwise.
4. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
5. Show clear mathematical working for all questions.

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Section A: Basic Concepts and Manipulation (15 Marks)

1. Solve the equation $e^{2x} - 5e^x + 6 = 0$, giving your answers in exact form. [3]

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2. Express $\ln(x^2 - 4) - \ln(x - 2)$ as a single logarithm in its simplest form, stating the range of values of $x$ for which the expression is defined. [3]

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3. The function $f$ is defined by $f(x) = 3e^{2x} + 1$ for $x \in \mathbb{R}$. Find the inverse function $f^{-1}(x)$ and state its domain. [3]

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4. Solve the inequality $\frac{2x - 1}{x + 3} \le 1$. [3]

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5. Given that $2^x = 3^{x-1}$, find the exact value of $x$. [3]

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Section B: Graphs and Transformations (15 Marks)

6. Sketch the graph of $y = |2x - 4|$. On your sketch, show the coordinates of any points where the graph meets the axes. [3]

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7. The diagram below shows the graph of $y = f(x)$. The graph has a maximum point at $A(2, 5)$ and crosses the x-axis at $B(-1, 0)$ and $C(4, 0)$. Sketch the graph of $y = f(x - 1) + 2$, showing the new coordinates of points $A$, $B$, and $C$. [3]

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8. Sketch the graph of $y = \ln(x + 2)$. State the equation of the asymptote and the coordinates of the x-intercept. [3]

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9. The curve $y = e^{-x}$ is transformed to the curve $y = 3e^{-x} - 1$. Describe fully the two geometric transformations that map the first curve to the second. [3]

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10. Sketch the graph of $y = 2^x$ and $y = 8 - x^2$ on the same axes. Hence, state the number of solutions to the equation $2^x + x^2 = 8$. [3]

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Section C: Applications and Problem Solving (20 Marks)

11. A radioactive substance decays such that its mass $M$ kg at time $t$ years is given by $M = M_0 e^{-kt}$, where $M_0$ and $k$ are positive constants. Initially, the mass is 10 kg. After 5 years, the mass is 8 kg. (a) Find the value of $k$ correct to 3 significant figures. [2] (b) Find the time taken for the mass to halve. [2]

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12. The population of a town is modelled by $P = 5000(1.03)^t$, where $t$ is the number of years after 2020. (a) Calculate the population in 2025. [2] (b) Find the year in which the population will first exceed 7000. [2]

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13. Find the set of values of $k$ for which the equation $x^2 + kx + (k + 3) = 0$ has no real roots. [3]

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14. Solve the simultaneous equations: $y = x + 2$ $y = x^2 - 4$ [3]

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15. The function $g$ is defined by $g(x) = \ln(3x - 1)$ for $x > \frac{1}{3}$. (a) Find $g^{-1}(x)$. [2] (b) Solve the equation $g(x) = 2$. [2]

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16. A company's profit $P$ (in thousands of dollars) is modelled by $P = 10 \ln(t + 1) - 2t$, where $t$ is the time in years since launch ($t \ge 0$). (a) Calculate the profit after 4 years. [2] (b) Find the time $t$ when the profit is maximized. [2]

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17. Given that $\log_2 x + \log_2 (x - 2) = 3$, find the value of $x$. [3]

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18. The curve $y = e^{2x} - 5e^x + 6$ crosses the x-axis at two points. Find the exact x-coordinates of these points. [3]

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19. Solve the inequality $e^{2x} - 4e^x + 3 < 0$. [3]

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20. The temperature $T$ of a cup of coffee at time $t$ minutes is given by $T = 20 + 80e^{-0.1t}$. (a) What is the initial temperature of the coffee? [1] (b) How long does it take for the coffee to cool to 50°C? Give your answer correct to 1 decimal place. [3]

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

1. [3 marks] Let $u = e^x$. Then $u^2 - 5u + 6 = 0$. $(u - 2)(u - 3) = 0$. $u = 2$ or $u = 3$. $e^x = 2 \Rightarrow x = \ln 2$. $e^x = 3 \Rightarrow x = \ln 3$. Answers: $x = \ln 2, \ln 3$.

2. [3 marks] $\ln(x^2 - 4) - \ln(x - 2) = \ln\left(\frac{(x-2)(x+2)}{x-2}\right) = \ln(x + 2)$. For the original expression to be defined: $x^2 - 4 > 0 \Rightarrow x > 2$ or $x < -2$. $x - 2 > 0 \Rightarrow x > 2$. Intersection: $x > 2$. Answer: $\ln(x + 2)$, for $x > 2$.

3. [3 marks] Let $y = 3e^{2x} + 1$. $y - 1 = 3e^{2x}$. $\frac{y - 1}{3} = e^{2x}$. $\ln\left(\frac{y - 1}{3}\right) = 2x$. $x = \frac{1}{2} \ln\left(\frac{y - 1}{3}\right)$. $f^{-1}(x) = \frac{1}{2} \ln\left(\frac{x - 1}{3}\right)$. Domain of $f^{-1}$ is range of $f$. Since $e^{2x} > 0$, $3e^{2x} + 1 > 1$. Answer: $f^{-1}(x) = \frac{1}{2} \ln\left(\frac{x - 1}{3}\right)$, Domain: $x > 1$.

4. [3 marks] $\frac{2x - 1}{x + 3} - 1 \le 0$. $\frac{2x - 1 - (x + 3)}{x + 3} \le 0$. $\frac{x - 4}{x + 3} \le 0$. Critical values: $x = 4, x = -3$. Test intervals: $x < -3$: $(-)/(-) = (+)$ $-3 < x < 4$: $(-)/(+) = (-)$ $x > 4$: $(+)/(+) = (+)$ Inequality is $\le 0$, so we want the negative region and zero. Answer: $-3 < x \le 4$.

5. [3 marks] $2^x = 3^{x-1}$. Take $\ln$ of both sides: $x \ln 2 = (x - 1) \ln 3$. $x \ln 2 = x \ln 3 - \ln 3$. $\ln 3 = x \ln 3 - x \ln 2$. $\ln 3 = x(\ln 3 - \ln 2)$. $x = \frac{\ln 3}{\ln 3 - \ln 2} = \frac{\ln 3}{\ln(1.5)}$. Answer: $x = \frac{\ln 3}{\ln 3 - \ln 2}$.

6. [3 marks] $y = |2x - 4|$. x-intercept: $2x - 4 = 0 \Rightarrow x = 2$. Point $(2, 0)$. y-intercept: $x = 0 \Rightarrow y = |-4| = 4$. Point $(0, 4)$. V-shape graph with vertex at $(2, 0)$, passing through $(0, 4)$ and $(4, 4)$. Answer: Sketch showing V-shape, vertex $(2,0)$, y-int $(0,4)$.

7. [3 marks] Transformation: Translation by vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. $A(2, 5) \rightarrow (2+1, 5+2) = (3, 7)$. $B(-1, 0) \rightarrow (-1+1, 0+2) = (0, 2)$. $C(4, 0) \rightarrow (4+1, 0+2) = (5, 2)$. Answer: Sketch showing shifted graph with points $(3,7), (0,2), (5,2)$.

8. [3 marks] Graph of $\ln x$ shifted left by 2 units. Asymptote: $x = -2$. x-intercept: $\ln(x+2) = 0 \Rightarrow x+2=1 \Rightarrow x=-1$. Point $(-1, 0)$. Shape: Increasing curve, concave down, passing through $(-1,0)$ and $(e-2, 1) \approx (0.718, 1)$. Answer: Sketch, Asymptote $x = -2$, Intercept $(-1, 0)$.

9. [3 marks]

1. Stretch parallel to y-axis, scale factor 3.
2. Translation by vector $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$ (or 1 unit downwards). Answer: Stretch SF 3 in y-direction, then translate down by 1.

10. [3 marks] $y = 2^x$ is exponential growth passing through $(0,1)$. $y = 8 - x^2$ is inverted parabola with vertex $(0,8)$ and x-intercepts $\pm \sqrt{8} \approx \pm 2.82$. They intersect at one point with $x > 0$ (approx $x=2$) and one point with $x < 0$ (approx $x=-2.7$). Check $x=2$: $2^2=4, 8-4=4$. Intersection at $(2,4)$. Check $x=-2$: $2^{-2}=0.25, 8-4=4$. No. Check $x \approx -2.7$: $2^{-2.7} \approx 0.15$, $8-(-2.7)^2 \approx 0.7$. Close. Graphically, there are 2 solutions. Answer: 2 solutions.

11. [4 marks] (a) $M_0 = 10$. At $t=5, M=8$. $8 = 10 e^{-5k}$. $0.8 = e^{-5k}$. $\ln 0.8 = -5k$. $k = -\frac{\ln 0.8}{5} \approx 0.0446$. Answer: $k = 0.0446$.

(b) Halve mass: $M = 5$. $5 = 10 e^{-0.0446t}$. $0.5 = e^{-0.0446t}$. $\ln 0.5 = -0.0446t$. $t = \frac{\ln 0.5}{-0.0446} \approx 15.5$ years. Answer: 15.5 years.

12. [4 marks] (a) $t = 2025 - 2020 = 5$. $P = 5000(1.03)^5 \approx 5796$. Answer: 5796.

(b) $7000 < 5000(1.03)^t$. $1.4 < 1.03^t$. $\ln 1.4 < t \ln 1.03$. $t > \frac{\ln 1.4}{\ln 1.03} \approx 11.53$. Year: $2020 + 12 = 2032$. Answer: 2032.

13. [3 marks] No real roots $\Rightarrow b^2 - 4ac < 0$. $k^2 - 4(1)(k + 3) < 0$. $k^2 - 4k - 12 < 0$. $(k - 6)(k + 2) < 0$. Critical values: $6, -2$. Parabola opens upward, so negative between roots. Answer: $-2 < k < 6$.

14. [3 marks] $x + 2 = x^2 - 4$. $x^2 - x - 6 = 0$. $(x - 3)(x + 2) = 0$. $x = 3$ or $x = -2$. If $x = 3, y = 5$. If $x = -2, y = 0$. Answer: $(3, 5)$ and $(-2, 0)$.

15. [4 marks] (a) $y = \ln(3x - 1)$. $e^y = 3x - 1$. $3x = e^y + 1$. $x = \frac{e^y + 1}{3}$. $g^{-1}(x) = \frac{e^x + 1}{3}$. Answer: $g^{-1}(x) = \frac{e^x + 1}{3}$.

(b) $\ln(3x - 1) = 2$. $3x - 1 = e^2$. $3x = e^2 + 1$. $x = \frac{e^2 + 1}{3}$. Answer: $x = \frac{e^2 + 1}{3}$.

16. [4 marks] (a) $t = 4$. $P = 10 \ln(5) - 2(4) = 10(1.609) - 8 = 16.09 - 8 = 8.09$. Answer: $8.09$ thousand dollars.

(b) Maximize $P$. Differentiate w.r.t $t$. $\frac{dP}{dt} = \frac{10}{t + 1} - 2$. Set $\frac{dP}{dt} = 0$. $\frac{10}{t + 1} = 2$. $10 = 2(t + 1)$. $5 = t + 1 \Rightarrow t = 4$. Check second derivative: $\frac{d^2P}{dt^2} = -\frac{10}{(t+1)^2}$. At $t=4$, this is negative, so maximum. Answer: $t = 4$ years.

17. [3 marks] $\log_2 (x(x - 2)) = 3$. $x(x - 2) = 2^3 = 8$. $x^2 - 2x - 8 = 0$. $(x - 4)(x + 2) = 0$. $x = 4$ or $x = -2$. Since arguments of logs must be positive: $x > 0$ and $x - 2 > 0 \Rightarrow x > 2$. Reject $x = -2$. Answer: $x = 4$.

18. [3 marks] Crosses x-axis when $y = 0$. $e^{2x} - 5e^x + 6 = 0$. Same as Q1. $(e^x - 2)(e^x - 3) = 0$. $e^x = 2 \Rightarrow x = \ln 2$. $e^x = 3 \Rightarrow x = \ln 3$. Answer: $x = \ln 2, \ln 3$.

19. [3 marks] Let $u = e^x$. $u^2 - 4u + 3 < 0$. $(u - 3)(u - 1) < 0$. $1 < u < 3$. $1 < e^x < 3$. $\ln 1 < x < \ln 3$. $0 < x < \ln 3$. Answer: $0 < x < \ln 3$.

20. [4 marks] (a) $t = 0$. $T = 20 + 80e^0 = 20 + 80 = 100$. Answer: 100°C.

(b) $50 = 20 + 80e^{-0.1t}$. $30 = 80e^{-0.1t}$. $\frac{30}{80} = e^{-0.1t}$. $0.375 = e^{-0.1t}$. $\ln 0.375 = -0.1t$. $t = \frac{\ln 0.375}{-0.1} \approx 9.808$. Answer: 9.8 minutes.