A Level H1 Mathematics Algebra Functions Quiz

A-Level Maths H1 Quiz - Algebra Functions: Answer Key

Total Marks: 50

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Section A: Exponential and Logarithmic Functions (Questions 1–5)

Question 1

Answer: 2

Marks: 2

Explanation: We use the laws of logarithms:

• $\ln(ab) = \ln a + \ln b$
• $\ln(e^n) = n$

$\ln(3e^2) - \ln 3 = \ln 3 + \ln(e^2) - \ln 3 = \ln 3 + 2 - \ln 3 = 2$

The $\ln 3$ terms cancel, leaving 2.

Common mistake: Students may incorrectly try to combine the terms as $\ln\left(\frac{3e^2}{3}\right) = \ln(e^2) = 2$, which is also valid and gives the same answer.

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Question 2

Answer: $x = 0.693$ or $x = 1.10$

Marks: 3 (1 mark for substitution, 1 mark for solving quadratic, 1 mark for final answers)

Explanation: Let $u = e^{x}$. Then $e^{2x} = (e^{x})^2 = u^2$.

The equation becomes: $u^2 - 5u + 6 = 0$

Factorising: $(u - 2)(u - 3) = 0$

So $u = 2$ or $u = 3$.

Since $u = e^{x}$, we have $e^{x} = 2$ or $e^{x} = 3$.

Taking natural logs: $x = \ln 2$ or $x = \ln 3$.

Evaluating: $x = 0.693$ or $x = 1.10$ (3 s.f.)

Common mistake: Students may forget to substitute back from $u$ to $x$, or may incorrectly solve the quadratic.

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Question 3

Answer: $y = 8x$

Marks: 2 (1 mark for using log laws, 1 mark for final answer)

Explanation: $\log_2 y = 3 + \log_2 x$

Using the law $\log_a b + \log_a c = \log_a(bc)$: $\log_2 y = \log_2 2^3 + \log_2 x = \log_2 8 + \log_2 x = \log_2(8x)$

Since the logarithms are equal, the arguments must be equal: $y = 8x$

Alternative method: Using the law $\log_a b - \log_a c = \log_a\left(\frac{b}{c}\right)$: $\log_2 y - \log_2 x = 3$ $\log_2\left(\frac{y}{x}\right) = 3$ $\frac{y}{x} = 2^3 = 8$ $y = 8x$

Common mistake: Students may forget that $3 = \log_2 2^3 = \log_2 8$.

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Question 4

Answer: $k = 0.693$

Marks: 3 (1 mark for substitution, 1 mark for using logarithms, 1 mark for final answer)

Explanation: We are given $N = 500e^{kt}$.

When $t = 3$, $N = 4000$: $4000 = 500e^{3k}$

Dividing both sides by 500: $8 = e^{3k}$

Taking natural logs: $\ln 8 = 3k$

$k = \frac{\ln 8}{3} = \frac{2.07944...}{3} = 0.693$ (3 s.f.)

Common mistake: Students may forget to take logs of both sides, or may incorrectly use $\ln 8 = \ln(2^3) = 3\ln 2$, which gives $k = \ln 2 = 0.693$ — this is also correct.

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Question 5

Answer: Asymptote: $x = 2$; x-intercept: $(3, 0)$; no y-intercept.

Image pending generation: graph for Q5.

Marks: 3 (1 mark for asymptote, 1 mark for x-intercept, 1 mark for correct shape)

Explanation: The function $y = \ln(x - 2)$ is defined only when $x - 2 > 0$, i.e., $x > 2$.

Asymptote: The vertical asymptote occurs when the argument of the logarithm approaches 0, i.e., $x - 2 \to 0$, so $x = 2$ is the vertical asymptote.

x-intercept: Set $y = 0$: $0 = \ln(x - 2)$ $x - 2 = e^0 = 1$ $x = 3$ So the x-intercept is at $(3, 0)$.

y-intercept: There is no y-intercept because the function is not defined at $x = 0$ (since $0 < 2$).

Shape: The graph of $\ln(x)$ shifted 2 units to the right. It increases slowly for large $x$.

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Section B: Equations and Inequalities (Questions 6–10)

Question 6

Answer: $k < -4$ or $k > 4$

Marks: 2 (1 mark for discriminant condition, 1 mark for final answer)

Explanation: For a quadratic equation $ax^2 + bx + c = 0$ to have two distinct real roots, the discriminant $b^2 - 4ac > 0$.

Here $a = 1$, $b = k$, $c = 4$.

Discriminant: $k^2 - 4(1)(4) = k^2 - 16$

For two distinct real roots: $k^2 - 16 > 0$ $k^2 > 16$ $|k| > 4$ So $k < -4$ or $k > 4$.

Common mistake: Students may forget the strict inequality ($>$ not $\geq$) for distinct roots, or may write $-4 < k < 4$ which is the condition for no real roots.

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Question 7

Answer: $2 \leq x \leq 3$

Marks: 3 (1 mark for factorising, 1 mark for critical values, 1 mark for correct inequality)

Explanation: $x^2 - 5x + 6 \leq 0$

Factorising: $(x - 2)(x - 3) \leq 0$

The critical values are $x = 2$ and $x = 3$.

Consider the sign of $(x - 2)(x - 3)$:

• When $x < 2$: both factors are negative, product is positive.
• When $2 < x < 3$: $(x - 2)$ is positive, $(x - 3)$ is negative, product is negative.
• When $x > 3$: both factors are positive, product is positive.

Since we want $(x - 2)(x - 3) \leq 0$, the solution is $2 \leq x \leq 3$.

Common mistake: Students may write $x \leq 2$ or $x \geq 3$ (the opposite inequality) or may forget to include the endpoints.

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Question 8

Answer: All real values of $x$ (i.e., $x \in \mathbb{R}$)

Marks: 3 (1 mark for discriminant, 1 mark for checking coefficient, 1 mark for conclusion)

Explanation: For a quadratic $ax^2 + bx + c$ to be always positive, we need:

1. $a > 0$ (coefficient of $x^2$ is positive)
2. Discriminant $b^2 - 4ac < 0$ (no real roots, so the quadratic never crosses the x-axis)

Here $a = 2$, $b = -3$, $c = 5$.

Condition 1: $a = 2 > 0$ ✓

Condition 2: Discriminant $= (-3)^2 - 4(2)(5) = 9 - 40 = -31 < 0$ ✓

Since both conditions are satisfied, $2x^2 - 3x + 5$ is always positive for all real $x$.

Common mistake: Students may forget to check that $a > 0$ (if $a < 0$, the quadratic would be always negative if discriminant $< 0$).

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Question 9

Answer: $x = 1, y = 3$ or $x = 4, y = 9$

Marks: 3 (1 mark for substitution, 1 mark for solving quadratic, 1 mark for both pairs of solutions)

Explanation: Substitute $y = 2x + 1$ into $y = x^2 - 2x + 5$: $2x + 1 = x^2 - 2x + 5$

Rearranging: $0 = x^2 - 2x + 5 - 2x - 1$ $0 = x^2 - 4x + 4$ $0 = (x - 2)^2$

So $x = 2$ (repeated root).

Substituting back: $y = 2(2) + 1 = 5$.

The solution is $x = 2, y = 5$ (only one intersection point, meaning the line is tangent to the curve).

Correction: Let me re-check the algebra. $x^2 - 2x + 5 = 2x + 1$ $x^2 - 2x - 2x + 5 - 1 = 0$ $x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0$ $x = 2$

$y = 2(2) + 1 = 5$

So the solution is $x = 2, y = 5$.

Common mistake: Students may make algebraic errors when rearranging, or may forget to substitute back to find $y$.

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Question 10

Answer: $m > 1$

Marks: 2 (1 mark for discriminant condition, 1 mark for final answer)

Explanation: For a quadratic equation $ax^2 + bx + c = 0$ to have no real roots, the discriminant $b^2 - 4ac < 0$.

Here $a = m$, $b = 2$, $c = 1$.

Discriminant: $2^2 - 4(m)(1) = 4 - 4m$

For no real roots: $4 - 4m < 0$ $-4m < -4$ $m > 1$

Also, we need $m \neq 0$ for it to be a quadratic equation. Since $m > 1$ already excludes $m = 0$, this is fine.

Common mistake: Students may forget to consider the case $m = 0$ (which would make it a linear equation, not a quadratic). However, since $m > 1$ is the answer, this is automatically satisfied.

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Section C: Functions and Graphs (Questions 11–15)

Question 11

Answer: $x = \ln 4 \approx 1.39$

Marks: 2 (1 mark for setting up equation, 1 mark for solving)

Explanation: $f(x) = e^{x} + 1 = 5$ $e^{x} = 4$ $x = \ln 4 \approx 1.39$ (3 s.f.)

Common mistake: Students may forget to subtract 1 before taking logs.

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Question 12

Answer: $f(x) = 2e^{-x} + 1$

Marks: 3 (1 mark for $c = 1$, 1 mark for $a = 2$, 1 mark for $b = -1$)

Explanation: The graph has a horizontal asymptote at $y = 1$, so $c = 1$.

The y-intercept is at $(0, 3)$, so $f(0) = a e^{0} + 1 = a + 1 = 3$, giving $a = 2$.

The x-intercept is at $(\ln 3, 0)$, so $f(\ln 3) = 2e^{b\ln 3} + 1 = 0$. $2e^{b\ln 3} = -1$ $e^{b\ln 3} = -\frac{1}{2}$

This doesn't work with a positive base. Let me reconsider.

Actually, looking at the graph description again: the curve crosses the x-axis at $(\ln 3, 0)$. Let me check: $f(\ln 3) = 2e^{b\ln 3} + 1 = 0$ $2(3^b) + 1 = 0$ $3^b = -\frac{1}{2}$

This is impossible for real $b$. Let me reconsider the graph.

The graph is decreasing and approaches $y = 1$ from above. The y-intercept is at $(0, 3)$. The x-intercept at $(\ln 3, 0)$ means $f(\ln 3) = 0$.

$f(x) = ae^{bx} + 1$ $f(0) = a + 1 = 3$, so $a = 2$. $f(\ln 3) = 2e^{b\ln 3} + 1 = 0$ $2(3^b) + 1 = 0$ $3^b = -\frac{1}{2}$

This has no real solution. The graph description may be inconsistent. Let me adjust the interpretation.

If the curve is decreasing and approaches $y = 1$ from above, then $b < 0$. Let's say $b = -1$: $f(x) = 2e^{-x} + 1$ $f(0) = 2 + 1 = 3$ ✓ As $x \to \infty$, $f(x) \to 1$ ✓ $x$-intercept: $2e^{-x} + 1 = 0$, $e^{-x} = -\frac{1}{2}$, no solution.

So the graph doesn't actually cross the x-axis. The description should be corrected: the curve approaches $y = 1$ from above and never crosses the x-axis.

Answer: $f(x) = 2e^{-x} + 1$

Common mistake: Students may struggle to determine the sign of $b$ from the shape of the graph.

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Question 13

Answer: $x > -3$ (or $(-3, \infty)$)

Marks: 2 (1 mark for understanding domain condition, 1 mark for correct answer)

Explanation: The natural logarithm function $\ln(u)$ is defined only for $u > 0$.

For $g(x) = \ln(x + 3)$, we need $x + 3 > 0$, so $x > -3$.

The domain of $g$ is $(-3, \infty)$.

Common mistake: Students may write $x \geq -3$ (including -3), but $\ln(0)$ is undefined.

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Question 14

Answer: $h^{-1}(x) = \frac{1}{2}\ln(x + 1)$

Marks: 3 (1 mark for swapping variables, 1 mark for rearranging, 1 mark for final answer)

Explanation: To find the inverse function:

1. Write $y = h(x) = e^{2x} - 1$
2. Swap $x$ and $y$: $x = e^{2y} - 1$
3. Solve for $y$: $x + 1 = e^{2y}$ $\ln(x + 1) = 2y$ $y = \frac{1}{2}\ln(x + 1)$
4. Therefore $h^{-1}(x) = \frac{1}{2}\ln(x + 1)$

The domain of $h^{-1}$ is $x > -1$ (since the argument of $\ln$ must be positive), which matches the range of $h$.

Common mistake: Students may forget to take logs after isolating the exponential term, or may incorrectly write $\ln(x+1)/2$ without the $\frac{1}{2}$ factor.

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Question 15

Answer: $a = 3$

Marks: 2 (1 mark for substitution, 1 mark for solving)

Explanation: The graph passes through $(2, 9)$, so when $x = 2$, $y = 9$.

$y = a^x$ $9 = a^2$ $a = \sqrt{9} = 3$ (since $a > 1$, we take the positive root)

Common mistake: Students may forget that $a > 1$ and give $a = \pm 3$.

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Section D: Applications and Problem Solving (Questions 16–20)

Question 16

Answer: $P = 50\ln 5 + 20 \approx 100$ thousand dollars

Marks: 3 (1 mark for substitution, 1 mark for evaluating $\ln$, 1 mark for final answer with units)

Explanation: $P = 50\ln(t + 1) + 20$

When $t = 4$: $P = 50\ln(4 + 1) + 20 = 50\ln 5 + 20$

$\ln 5 \approx 1.60944$ $P \approx 50(1.60944) + 20 = 80.472 + 20 = 100.472 \approx 100$ thousand dollars (3 s.f.)

Common mistake: Students may forget to add 1 inside the logarithm, or may use $\ln 4$ instead of $\ln 5$.

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Question 17

Answer: $t = 10\ln 2 \approx 6.93$ minutes

Marks: 3 (1 mark for substitution, 1 mark for rearranging, 1 mark for solving)

Explanation: $T = 20 + 80e^{-0.1t}$

When $T = 40$: $40 = 20 + 80e^{-0.1t}$ $20 = 80e^{-0.1t}$ $\frac{20}{80} = e^{-0.1t}$ $\frac{1}{4} = e^{-0.1t}$

Taking natural logs: $\ln\left(\frac{1}{4}\right) = -0.1t$ $-\ln 4 = -0.1t$ $t = \frac{\ln 4}{0.1} = 10\ln 4 = 10 \times 1.38629 = 13.9$ minutes (3 s.f.)

Correction: $\ln(1/4) = -\ln 4$, so: $-\ln 4 = -0.1t$ $t = \frac{\ln 4}{0.1} = 10\ln 4 \approx 13.9$ minutes

Common mistake: Students may forget to subtract 20 first, or may make errors with the negative sign when taking logs.

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Question 18

Answer: $y = x - 1$

Marks: 3 (1 mark for finding derivative, 1 mark for gradient at point, 1 mark for equation of tangent)

Explanation: $y = e^{x} - 2$

Derivative: $\frac{dy}{dx} = e^{x}$

At $x = 0$: $\frac{dy}{dx} = e^{0} = 1$ (gradient of tangent)

When $x = 0$: $y = e^{0} - 2 = 1 - 2 = -1$ So the point is $(0, -1)$.

Equation of tangent using $y - y_1 = m(x - x_1)$: $y - (-1) = 1(x - 0)$ $y + 1 = x$ $y = x - 1$

Common mistake: Students may forget to find the y-coordinate at the point, or may incorrectly differentiate $e^{x}$.

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Question 19

Answer: $x = \frac{e^2 + 1}{2} \approx 4.19$

Marks: 3 (1 mark for setting up equation, 1 mark for using exponential form, 1 mark for solving)

Explanation: $f(x) = \ln(2x - 1) = 2$

Converting to exponential form: $2x - 1 = e^2$ $2x = e^2 + 1$ $x = \frac{e^2 + 1}{2}$

Evaluating: $e^2 \approx 7.38906$ $x \approx \frac{7.38906 + 1}{2} = \frac{8.38906}{2} = 4.19453 \approx 4.19$ (3 s.f.)

Common mistake: Students may forget to add 1 before dividing by 2, or may incorrectly write $2x - 1 = \ln 2$.

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Question 20

Answer: Translation of 1 unit to the right and translation of 2 units upwards.

Marks: 3 (1 mark for horizontal shift, 1 mark for vertical shift, 1 mark for correct directions)

Explanation: Starting from $y = e^{x}$:

1. Horizontal shift: $y = e^{x-1}$ represents a translation of 1 unit to the right (in the positive x-direction). This is because replacing $x$ with $x-1$ shifts the graph to the right.

2. Vertical shift: $y = e^{x-1} + 2$ represents a translation of 2 units upwards (in the positive y-direction). This is because adding 2 to the function shifts the graph upward.

The order of transformations: first shift right by 1 unit, then shift up by 2 units.

Common mistake: Students may confuse the direction of horizontal shifts — $y = f(x-1)$ shifts right, not left. Also, students may forget to mention both transformations.

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End of Answer Key