A Level H1 Mathematics Calculus Quiz

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A-Level Maths H1 Quiz - Calculus

Name: ________________________________________
Class: ________________________________________
Date: ________________________________________
Score: ____ / 60

Duration: 90 minutes
Total Marks: 60

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Instructions:

• Answer ALL questions.
• Show all working clearly. Marks are awarded for correct method as well as final answers.
• Non-exact answers should be given correct to 3 significant figures unless otherwise stated.
• The use of a graphing calculator is permitted.
• This quiz covers Calculus topics only: differentiation, integration, and their applications.

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Section A: Differentiation (Questions 1–8)

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1. Differentiate the following with respect to $x$.
    (a) $y = 5x^3 - 4x^2 + 7x - 2$
    (b) $y = (3x + 1)(x^2 - 4)$
    (c) $y = \dfrac{2x^2 + 1}{x}$

[5 marks]

 

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2. Find $\dfrac{dy}{dx}$ for each of the following:
    (a) $y = \sqrt{4x + 1}$
    (b) $y = e^{3x} \cos(2x)$
    (c) $y = \ln(x^2 + 3x)$

[6 marks]

 

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3. A curve is defined parametrically by $x = 2t^2 + 1$, $y = t^3 - 4t$.
    (a) Find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$.
    (b) Hence find $\dfrac{dy}{dx}$ in terms of $t$.
    (c) Find the gradient of the curve at the point where $t = 2$.

[5 marks]

 

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4. Find the equation of the tangent to the curve $y = x^3 - 6x^2 + 9x + 1$ at the point where $x = 2$.

[4 marks]

 

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5. A function is defined by $f(x) = 2x^3 - 15x^2 + 36x + 10$.
    (a) Find $f'(x)$.
    (b) Find the coordinates of the stationary points and determine their nature.
    (c) Sketch the curve $y = f(x)$, clearly indicating the stationary points.

[8 marks]

 

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6. The cost $C$ (in dollars) of producing $x$ units of a product is given by
$C(x) = 0.01x^3 - 0.6x^2 + 15x + 200.$
    (a) Find the marginal cost function.
    (b) Find the production level $x$ at which the marginal cost is a minimum.
    (c) Interpret your answer in context.

[6 marks]

 

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7. Given that $y = x^2 e^{-x}$, show that $\dfrac{dy}{dx} = e^{-x}(2x - x^2)$. Hence find the exact coordinates of the stationary points of the curve.

[5 marks]

 

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8. A spherical balloon is being inflated. The radius $r$ (in cm) is increasing at a constant rate of 0.5 cm s$^{-1}$. Find the rate at which the volume $V$ is increasing when $r = 6$ cm.

[4 marks]

 

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Section B: Integration (Questions 9–14)

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9. Find the following integrals.
    (a) $\displaystyle\int (6x^2 - 4x + 3)\,dx$
    (b) $\displaystyle\int \left(\dfrac{3}{\sqrt{x}} + \dfrac{2}{x^2}\right)dx$
    (c) $\displaystyle\int (e^{2x} + 1)^2\,dx$

[6 marks]

 

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10. Evaluate the following definite integrals.
    (a) $\displaystyle\int_1^4 (2x + 3)\,dx$
    (b) $\displaystyle\int_0^{\pi/2} \sin x\,dx$
    (c) $\displaystyle\int_1^2 \dfrac{2x + 1}{x^2}\,dx$

[6 marks]

 

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11. Find the equation of the curve which passes through the point $(1, 5)$ and for which $\dfrac{dy}{dx} = 3x^2 - 6x + 2$.

[4 marks]

 

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12. The gradient of a curve is given by $\dfrac{dy}{dx} = 4x - \dfrac{1}{x^2}$, and the curve passes through the point $(1, 3)$. Find the equation of the curve.

[4 marks]

 

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13. Find the area of the region enclosed between the curve $y = x^2 - 4x + 3$ and the $x$-axis.

[5 marks]

 

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14. <image_placeholder> id: Q14-fig1 type: graph linked_question: Q14 description: Graph showing the curves y = x^2 and y = 2x, intersecting at (0,0) and (2,4). The region between the two curves is shaded. labels: y = x^2 (parabola), y = 2x (straight line), intersection points (0,0) and (2,4), shaded region between curves from x=0 to x=2 values: Curves: y = x^2, y = 2x; Intersections: (0,0) and (2,4); Shaded region bounded between x = 0 and x = 2 must_show: Both curves clearly labelled, intersection points marked, shaded region between curves visible, axes labelled </image_placeholder>

The diagram shows the curves $y = x^2$ and $y = 2x$.
    (a) Verify that the curves intersect at $(0, 0)$ and $(2, 4)$.
    (b) Find the area of the shaded region enclosed between the two curves.

[5 marks]

 

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Section C: Applications of Calculus (Questions 15–20)

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15. A particle moves in a straight line such that its displacement $s$ (in metres) from a fixed point $O$ at time $t$ seconds is given by
$s = t^3 - 6t^2 + 9t + 2, \quad t \geq 0.$
    (a) Find the velocity $v$ and acceleration $a$ of the particle at time $t$.
    (b) Find the times when the particle is instantaneously at rest.
    (c) Find the acceleration when the particle first comes to rest.

[6 marks]

 

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16. The revenue $R$ (in thousands of dollars) from selling $x$ hundred units of a product is modelled by
$R(x) = 50x - 2x^2, \quad 0 \leq x \leq 20.$
    (a) Find $\dfrac{dR}{dx}$.
    (b) Find the number of units that maximises revenue.
    (c) Calculate the maximum revenue.

[5 marks]

 

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17. A closed cylindrical can is to have a volume of $500\pi$ cm$^3$.
    (a) Show that the total surface area $A$ of the can is given by
$A = 2\pi r^2 + \frac{1000\pi}{r},$
where $r$ is the radius of the base.
    (b) Find the value of $r$ that minimises the surface area.
    (c) Hence find the minimum surface area.

[7 marks]

 

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18. The population $P$ of a town at time $t$ years is modelled by
$\frac{dP}{dt} = kP,$
where $k$ is a constant. At time $t = 0$, the population is 10,000, and at time $t = 5$, the population is 12,000.
    (a) Show that $P = 10\,000\,e^{kt}$.
    (b) Find the value of $k$, correct to 3 significant figures.
    (c) Find the population at time $t = 10$, correct to the nearest hundred.

[6 marks]

 

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19. <image_placeholder> id: Q19-fig1 type: graph linked_question: Q19 description: Graph of y = 4x - x^2, a downward parabola with vertex at (2,4), crossing the x-axis at (0,0) and (4,0). The region under the curve above the x-axis is shaded. labels: y = 4x - x^2, vertex (2,4), x-intercepts (0,0) and (4,0), shaded region under curve from x=0 to x=4 values: Curve: y = 4x - x^2; Vertex: (2,4); x-intercepts: x = 0 and x = 4; Shaded region: area under curve from x = 0 to x = 4 must_show: Parabola clearly drawn with vertex and intercepts labelled, shaded region under curve visible, axes labelled with scale </image_placeholder>

The diagram shows the curve $y = 4x - x^2$.
    (a) Use integration to find the area of the shaded region enclosed between the curve and the $x$-axis.
    (b) Verify your answer by using the formula for the area of a triangle as a check (note: the curve is not a triangle, so explain why the integral gives a different answer).

[5 marks]

 

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20. A company's profit $P$ (in thousands of dollars) from producing $x$ tonnes of a chemical is modelled by
$P(x) = -x^3 + 12x^2 + 144x - 500, \quad 0 \leq x \leq 20.$
    (a) Find $P'(x)$ and $P''(x)$.
    (b) Find the production level that gives maximum profit, and justify that it is a maximum.
    (c) The company can only produce up to 15 tonnes. Find the maximum profit achievable within this constraint.

[7 marks]

 

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End of Quiz

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Mark Summary

Section	Questions	Marks
A: Differentiation	1–8	33
B: Integration	9–14	30
C: Applications	15–20	32
Total	20 questions	60

(Note: Some questions carry sub-part marks; the total across all 20 questions is 60 marks.)

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A-Level Maths H1 Quiz - Calculus

Answer Key

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Question 1 [5 marks]

(a) $y = 5x^3 - 4x^2 + 7x - 2$

$\frac{dy}{dx} = 15x^2 - 8x + 7$

[1 mark] — Apply the power rule: $\dfrac{d}{dx}(x^n) = nx^{n-1}$ to each term.

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(b) $y = (3x + 1)(x^2 - 4)$

First expand: $y = 3x^3 + x^2 - 12x - 4$

$\frac{dy}{dx} = 9x^2 + 2x - 12$

[2 marks] — 1 mark for correct expansion, 1 mark for correct differentiation.

Alternative: Use the product rule: $\dfrac{dy}{dx} = (3)(x^2 - 4) + (3x + 1)(2x) = 3x^2 - 12 + 6x^2 + 2x = 9x^2 + 2x - 12$.

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(c) $y = \dfrac{2x^2 + 1}{x} = 2x + x^{-1}$

$\frac{dy}{dx} = 2 - x^{-2} = 2 - \frac{1}{x^2}$

[2 marks] — 1 mark for simplifying the expression, 1 mark for correct differentiation.

Common mistake: Students who apply the quotient rule without simplifying should still get full marks if done correctly, but simplifying first is more efficient.

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Question 2 [6 marks]

(a) $y = \sqrt{4x + 1} = (4x + 1)^{1/2}$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{2}(4x + 1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x + 1}}$

[2 marks] — 1 mark for applying chain rule, 1 mark for correct simplification.

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(b) $y = e^{3x}\cos(2x)$

Using the product rule:

$\frac{dy}{dx} = 3e^{3x}\cos(2x) + e^{3x}(-2\sin(2x)) = e^{3x}(3\cos 2x - 2\sin 2x)$

[2 marks] — 1 mark for product rule set-up, 1 mark for correct answer.

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(c) $y = \ln(x^2 + 3x)$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{x^2 + 3x} \cdot (2x + 3) = \frac{2x + 3}{x^2 + 3x}$

[2 marks] — 1 mark for chain rule, 1 mark for correct answer.

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Question 3 [5 marks]

(a) $x = 2t^2 + 1 \implies \dfrac{dx}{dt} = 4t$

$y = t^3 - 4t \implies \dfrac{dy}{dt} = 3t^2 - 4$

[1 mark] — Both correct.

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(b) Using the chain rule for parametric equations:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 4}{4t}$

[2 marks] — 1 mark for formula, 1 mark for correct substitution.

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(c) At $t = 2$:

$\frac{dy}{dx} = \frac{3(4) - 4}{4(2)} = \frac{12 - 4}{8} = \frac{8}{8} = 1$

[2 marks] — 1 mark for substitution, 1 mark for correct answer.

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Question 4 [4 marks]

$y = x^3 - 6x^2 + 9x + 1$

Step 1: Find the $y$-coordinate at $x = 2$:

$y = 8 - 24 + 18 + 1 = 3$

So the point is $(2, 3)$.

Step 2: Find the gradient:

$\frac{dy}{dx} = 3x^2 - 12x + 9$

At $x = 2$: $\dfrac{dy}{dx} = 12 - 24 + 9 = -3$

Step 3: Equation of tangent using $y - y_1 = m(x - x_1)$:

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

$y = -3x + 6 + 3$

$\boxed{y = -3x + 9}$

[4 marks] — 1 mark for $y$-coordinate, 1 mark for derivative, 1 mark for gradient at $x = 2$, 1 mark for correct equation.

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Question 5 [8 marks]

(a) $f(x) = 2x^3 - 15x^2 + 36x + 10$

$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$

[1 mark]

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(b) Stationary points occur where $f'(x) = 0$:

$6(x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3$

At $x = 2$: $f(2) = 16 - 60 + 72 + 10 = 38$, so point is $(2, 38)$.

At $x = 3$: $f(3) = 54 - 135 + 108 + 10 = 37$, so point is $(3, 37)$.

Nature: Use $f''(x) = 12x - 30$.

At $x = 2$: $f''(2) = 24 - 30 = -6 < 0$ → maximum at $(2, 38)$.

At $x = 3$: $f''(3) = 36 - 30 = 6 > 0$ → minimum at $(3, 37)$.

[5 marks] — 1 mark for $f'(x)$, 1 mark for solving $f'(x) = 0$, 1 mark for $y$-coordinates, 1 mark for second derivative, 1 mark for correct nature.

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(c) <image_placeholder> id: Q5-fig1 type: graph linked_question: Q5 description: Cubic curve y = 2x^3 - 15x^2 + 36x + 10 with local maximum at (2,38) and local minimum at (3,37). y-intercept at (0,10). Curve rises from left, peaks at (2,38), dips to (3,37), then rises to the right. labels: Local maximum (2,38), local minimum (3,37), y-intercept (0,10) values: Maximum at (2,38), minimum at (3,37), y-intercept at (0,10) must_show: Both turning points labelled, y-intercept shown, correct cubic shape (rising-falling-rising) </image_placeholder>

[2 marks] — 1 mark for correct shape, 1 mark for labelled stationary points.

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Question 6 [6 marks]

(a) Marginal cost = $C'(x)$:

$C'(x) = 0.03x^2 - 1.2x + 15$

[1 mark]

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(b) To minimise marginal cost, differentiate $C'(x)$:

$C''(x) = 0.06x - 1.2$

Set $C''(x) = 0$: $0.06x = 1.2 \implies x = 20$

Check: $C'''(x) = 0.06 > 0$, confirming a minimum.

The marginal cost is minimum at $x = 20$ units.

[3 marks] — 1 mark for differentiating again, 1 mark for solving, 1 mark for confirming minimum.

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(c) At a production level of 20 units, the rate at which cost is increasing is at its lowest. This means the cost function is increasing most slowly at this point — producing the 20th unit adds the least additional cost compared to nearby production levels.

[2 marks] — 1 mark for contextual interpretation, 1 mark for clarity.

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Question 7 [5 marks]

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

Using the product rule:

$\frac{dy}{dx} = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) = 2xe^{-x} - x^2e^{-x} = e^{-x}(2x - x^2) \quad \checkmark$

[2 marks] — 1 mark for product rule, 1 mark for factorising.

For stationary points, set $\dfrac{dy}{dx} = 0$:

$e^{-x}(2x - x^2) = 0$

Since $e^{-x} > 0$ for all $x$:

$2x - x^2 = 0 \implies x(2 - x) = 0 \implies x = 0 \text{ or } x = 2$

At $x = 0$: $y = 0 \cdot e^0 = 0$, so $(0, 0)$.

At $x = 2$: $y = 4e^{-2} = \dfrac{4}{e^2}$, so $\left(2, \dfrac{4}{e^2}\right)$.

[3 marks] — 1 mark for setting derivative to zero, 1 mark for noting $e^{-x} \neq 0$, 1 mark for both coordinates.

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Question 8 [4 marks]

Volume of a sphere: $V = \dfrac{4}{3}\pi r^3$

Differentiate with respect to $t$:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Given $\dfrac{dr}{dt} = 0.5$ and $r = 6$:

$\frac{dV}{dt} = 4\pi(36)(0.5) = 72\pi \approx 226 \text{ cm}^3\text{s}^{-1}$

[4 marks] — 1 mark for formula, 1 mark for differentiating, 1 mark for substitution, 1 mark for correct answer with units.

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Question 9 [6 marks]

(a) $\displaystyle\int (6x^2 - 4x + 3)\,dx = \frac{6x^3}{3} - \frac{4x^2}{2} + 3x + c = 2x^3 - 2x^2 + 3x + c$

[2 marks] — 1 mark for integration, 1 mark for constant of integration.

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(b) $\displaystyle\int \left(3x^{-1/2} + 2x^{-2}\right)dx = \frac{3x^{1/2}}{1/2} + \frac{2x^{-1}}{-1} + c = 6\sqrt{x} - \frac{2}{x} + c$

[2 marks] — 1 mark for rewriting powers, 1 mark for correct integration.

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(c) $\displaystyle\int (e^{2x} + 1)^2\,dx = \int (e^{4x} + 2e^{2x} + 1)\,dx = \frac{e^{4x}}{4} + e^{2x} + x + c$

[2 marks] — 1 mark for expanding, 1 mark for correct integration.

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Question 10 [6 marks]

(a) $\displaystyle\int_1^4 (2x + 3)\,dx = \left[x^2 + 3x\right]_1^4 = (16 + 12) - (1 + 3) = 28 - 4 = 24$

[2 marks] — 1 mark for antiderivative, 1 mark for evaluation.

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(b) $\displaystyle\int_0^{\pi/2} \sin x\,dx = \left[-\cos x\right]_0^{\pi/2} = -\cos(\pi/2) + \cos(0) = 0 + 1 = 1$

[2 marks] — 1 mark for antiderivative, 1 mark for evaluation.

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(c) $\displaystyle\int_1^2 \frac{2x + 1}{x^2}\,dx = \int_1^2 \left(\frac{2}{x} + \frac{1}{x^2}\right)dx = \int_1^2 (2x^{-1} + x^{-2})\,dx$

$= \left[2\ln|x| - \frac{1}{x}\right]_1^2 = \left(2\ln 2 - \frac{1}{2}\right) - \left(2\ln 1 - 1\right)$

$= 2\ln 2 - \frac{1}{2} + 1 = 2\ln 2 + \frac{1}{2}$

[2 marks] — 1 mark for splitting the fraction, 1 mark for correct evaluation.

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Question 11 [4 marks]

$\dfrac{dy}{dx} = 3x^2 - 6x + 2$

Integrate:

$y = x^3 - 3x^2 + 2x + c$

The curve passes through $(1, 5)$:

$5 = 1 - 3 + 2 + c \implies c = 5$

$\boxed{y = x^3 - 3x^2 + 2x + 5}$

[4 marks] — 1 mark for integration, 1 mark for constant, 1 mark for substitution, 1 mark for final answer.

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Question 12 [4 marks]

$\dfrac{dy}{dx} = 4x - x^{-2}$

Integrate:

$y = 2x^2 + x^{-1} + c = 2x^2 + \frac{1}{x} + c$

The curve passes through $(1, 3)$:

$3 = 2 + 1 + c \implies c = 0$

$\boxed{y = 2x^2 + \frac{1}{x}}$

[4 marks] — 1 mark for integration, 1 mark for substitution, 1 mark for $c = 0$, 1 mark for final answer.

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Question 13 [5 marks]

$y = x^2 - 4x + 3 = (x - 1)(x - 3)$

The curve crosses the $x$-axis at $x = 1$ and $x = 3$.

Between $x = 1$ and $x = 3$, the parabola opens upward and lies below the $x$-axis (since the vertex is at $x = 2$, $y = -1$).

Area = $-\displaystyle\int_1^3 (x^2 - 4x + 3)\,dx$ (negative because the region is below the axis)

$= -\left[\frac{x^3}{3} - 2x^2 + 3x\right]_1^3$

At $x = 3$: $\dfrac{27}{3} - 18 + 9 = 9 - 18 + 9 = 0$

At $x = 1$: $\dfrac{1}{3} - 2 + 3 = \dfrac{1}{3} + 1 = \dfrac{4}{3}$

Area $= -(0 - \dfrac{4}{3}) = \dfrac{4}{3}$ square units.

[5 marks] — 1 mark for finding $x$-intercepts, 1 mark for recognising curve is below axis, 1 mark for setting up integral with negative sign, 1 mark for antiderivative, 1 mark for correct answer.

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Question 14 [5 marks]

(a) Set $x^2 = 2x$:

$x^2 - 2x = 0 \implies x(x - 2) = 0 \implies x = 0$ or $x = 2$

At $x = 0$: $y = 0$. At $x = 2$: $y = 4$.

Intersection points: $(0, 0)$ and $(2, 4)$. $\checkmark$

[2 marks] — 1 mark for equating, 1 mark for both points.

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(b) Area = $\displaystyle\int_0^2 \left(\text{top} - \text{bottom}\right)dx = \int_0^2 (2x - x^2)\,dx$

$= \left[x^2 - \frac{x^3}{3}\right]_0^2 = \left(4 - \frac{8}{3}\right) - 0 = \frac{12 - 8}{3} = \frac{4}{3}$

Area $= \dfrac{4}{3}$ square units.

[3 marks] — 1 mark for correct integrand, 1 mark for antiderivative, 1 mark for correct answer.

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Question 15 [6 marks]

(a) Velocity: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9$

Acceleration: $a = \dfrac{dv}{dt} = 6t - 12$

[2 marks] — 1 mark each.

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(b) Particle at rest when $v = 0$:

$3t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t - 1)(t - 3) = 0$

$t = 1$ or $t = 3$ seconds.

[2 marks] — 1 mark for setting $v = 0$, 1 mark for solving.

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(c) The particle first comes to rest at $t = 1$.

$a = 6(1) - 12 = -6$ m s$^{-2}$.

[2 marks] — 1 mark for identifying $t = 1$, 1 mark for correct acceleration.

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Question 16 [5 marks]

(a) $\dfrac{dR}{dx} = 50 - 4x$

[1 mark]

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(b) Maximum revenue when $\dfrac{dR}{dx} = 0$:

$50 - 4x = 0 \implies x = 12.5$

Since $\dfrac{d^2R}{dx^2} = -4 < 0$, this is a maximum.

$x = 12.5$ hundred units = 1250 units.

[2 marks] — 1 mark for solving, 1 mark for confirming maximum.

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(c) $R(12.5) = 50(12.5) - 2(12.5)^2 = 625 - 312.5 = 312.5$

Maximum revenue = $312,500 (since $R$ is in thousands of dollars).

[2 marks] — 1 mark for substitution, 1 mark for correct answer with units.

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Question 17 [7 marks]

(a) Volume: $V = \pi r^2 h = 500\pi \implies h = \dfrac{500}{r^2}$

Surface area: $A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{500}{r^2} = 2\pi r^2 + \frac{1000\pi}{r} \quad \checkmark$

[2 marks] — 1 mark for finding $h$, 1 mark for surface area formula.

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(b) $\dfrac{dA}{dr} = 4\pi r - \dfrac{1000\pi}{r^2}$

Set $\dfrac{dA}{dr} = 0$:

$4\pi r = \dfrac{1000\pi}{r^2} \implies 4r^3 = 1000 \implies r^3 = 250 \implies r = \sqrt[3]{250} = 5\sqrt[3]{2} \approx 6.30$ cm

Check: $\dfrac{d^2A}{dr^2} = 4\pi + \dfrac{2000\pi}{r^3} > 0$ for $r > 0$, confirming a minimum.

[3 marks] — 1 mark for differentiating, 1 mark for solving, 1 mark for confirming minimum.

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(c) $A = 2\pi(250)^{2/3} + \dfrac{1000\pi}{(250)^{1/3}}$

With $r = \sqrt[3]{250}$: $r^2 = (250)^{2/3}$ and $\dfrac{1}{r} = (250)^{-1/3}$

$A = 2\pi(250)^{2/3} + 1000\pi(250)^{-1/3}$

Numerically: $r \approx 6.2996$

$A \approx 2\pi(39.685) + 1000\pi(0.15874) \approx 249.33 + 498.66 \approx 748$ cm$^2$

[2 marks] — 1 mark for substitution, 1 mark for correct answer.

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Question 18 [6 marks]

(a) $\dfrac{dP}{dt} = kP \implies \dfrac{1}{P}\,dP = k\,dt$

Integrating: $\ln P = kt + c \implies P = Ae^{kt}$ where $A = e^c$.

At $t = 0$, $P = 10\,000$: $10\,000 = Ae^0 = A$

$\boxed{P = 10\,000\,e^{kt}} \quad \checkmark$

[2 marks] — 1 mark for separation and integration, 1 mark for finding $A$.

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(b) At $t = 5$, $P = 12\,000$:

$12\,000 = 10\,000\,e^{5k} \implies e^{5k} = 1.2 \implies 5k = \ln 1.2$

$k = \dfrac{\ln 1.2}{5} = \dfrac{0.18232...}{5} \approx 0.0365$ (3 s.f.)

[2 marks] — 1 mark for substitution, 1 mark for correct $k$.

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(c) At $t = 10$: $P = 10\,000\,e^{10(0.036464)} = 10\,000\,e^{0.36464} \approx 10\,000 \times 1.440 = 14\,400$

To the nearest hundred: 14,400.

[2 marks] — 1 mark for substitution, 1 mark for correct answer.

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Question 19 [5 marks]

(a) $y = 4x - x^2 = x(4 - x)$, so $x$-intercepts at $x = 0$ and $x = 4$.

Area = $\displaystyle\int_0^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4$

$= 2(16) - \dfrac{64}{3} = 32 - \dfrac{64}{3} = \dfrac{96 - 64}{3} = \dfrac{32}{3}$

Area $= \dfrac{32}{3} \approx 10.67$ square units.

[3 marks] — 1 mark for limits, 1 mark for antiderivative, 1 mark for correct answer.

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(b) The region under the parabola is not a triangle. If one were to approximate using a triangle with base 4 and height 4, the area would be $\dfrac{1}{2}(4)(4) = 8$, which is less than $\dfrac{32}{3} \approx 10.67$. The integral accounts for the curved boundary, giving the exact area, whereas a triangle approximation underestimates it because the parabola bulges above the straight line connecting $(0,0)$ to $(4,0)$ through the vertex.

[2 marks] — 1 mark for triangle area calculation, 1 mark for explanation of difference.

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Question 20 [7 marks]

(a) $P(x) = -x^3 + 12x^2 + 144x - 500$

$P'(x) = -3x^2 + 24x + 144 = -3(x^2 - 8x - 48) = -3(x - 12)(x + 4)$

$P''(x) = -6x + 24$

[2 marks] — 1 mark each for $P'(x)$ and $P''(x)$.

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(b) Set $P'(x) = 0$: $-3(x - 12)(x + 4) = 0 \implies x = 12$ or $x = -4$

Since $x \geq 0$, we consider $x = 12$.

$P''(12) = -72 + 24 = -48 < 0$, confirming a maximum.

Maximum profit at $x = 12$ tonnes.

$P(12) = -1728 + 1728 + 1728 - 500 = 1228$

Maximum profit = $1,228,000.

[3 marks] — 1 mark for solving $P'(x) = 0$, 1 mark for second derivative test, 1 mark for profit value.

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(c) On the interval $0 \leq x \leq 15$, the critical point $x = 12$ lies within the domain.

Check endpoints:

• $P(0) = -500$
• $P(12) = 1228$
• $P(15) = -3375 + 2700 + 2160 - 500 = 985$

Maximum on $[0, 15]$ occurs at $x = 12$ with profit $1,228,000.

[2 marks] — 1 mark for checking endpoints, 1 mark for correct conclusion.

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Mark Summary

Q	Marks		Q	Marks
1	5		11	4
2	6		12	4
3	5		13	5
4	4		14	5
5	8		15	6
6	6		16	5
7	5		17	7
8	4		18	6
9	6		19	5
10	6		20	7
			Total	60