Secondary 3 Elementary Mathematics Calculus Quiz

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Secondary 3 Elementary Mathematics Quiz - Calculus

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Name: ____________________________
Class: ____________________________
Date: ____________________________
Score: ____ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer ALL questions in the spaces provided.
• Show clearly all working. Marks will be awarded for correct method even if the final answer is wrong.
• The use of calculators is allowed unless otherwise stated.
• Give non-exact answers correct to 3 significant figures unless otherwise stated.
• This quiz tests your understanding of the gradient of a curve, differentiation, and basic applications of rates of change.

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Section A: Gradient of a Curve and Differentiation (Questions 1–10)

Questions 1–5 are 2 marks each. Questions 6–10 are 3 marks each.

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1.
The equation of a curve is $y = x^2 + 3x - 4$.
(a) Copy and complete the table below by finding the value of $y$ when $x = 1$ and $x = 3$.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$-6$	$-6$	$-4$	____	$6$	____

(b) On the axes provided (not shown here), plot the points and sketch the curve for $-2 \leq x \leq 3$.

[2 marks]

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2.
A student draws a tangent to the curve $y = x^2$ at the point where $x = 2$. The tangent passes through the points $(1.5,\; 2.0)$ and $(3.0,\; 8.5)$.
Use these two points to estimate the gradient of the curve at $x = 2$.

[2 marks]

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3.
The gradient of a curve at a point is found to be $-3$. State, with a reason, whether the function is increasing or decreasing at that point.

[2 marks]

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4.
A curve has equation $y = 4x^2 - 7x + 2$.
Find the gradient of the curve at the point where $x = -1$.

[2 marks]

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5.
The displacement $s$ metres of a particle from a fixed point $O$ at time $t$ seconds is given by $s = 3t^2 - 5t + 1$.
Estimate the gradient of the $s$–$t$ graph when $t = 2$ by drawing a tangent through the points $(1.0,\; -1.0)$ and $(3.0,\; 13.0)$ on the curve.

[2 marks]

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6.
Differentiate the following with respect to $x$:

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

(b) $y = \dfrac{3}{x^2}$ (Hint: rewrite using index notation first)

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

[3 marks]

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7.
A curve is given by $y = 2x^3 - 9x^2 + 12x + 1$.

(a) Find $\dfrac{dy}{dx}$.

(b) Find the gradient of the curve at $x = 2$.

(c) Determine the coordinates of the point on the curve where the gradient is zero.

[3 marks]

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8.
The equation of a curve is $y = x^3 - 6x^2 + 9x$.

(a) Find $\dfrac{dy}{dx}$.

(b) Find the two values of $x$ for which $\dfrac{dy}{dx} = 0$.

(c) Hence state the coordinates of the two stationary points on the curve.

[3 marks]

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9.
A rectangular enclosure is to be built using 80 m of fencing along three sides, with the fourth side being a straight wall. Let the side perpendicular to the wall have length $x$ metres.

(a) Show that the area $A$ m² of the enclosure is given by $A = 80x - 2x^2$.

(b) Find the value of $x$ for which the area is a maximum.

(c) Hence find the maximum possible area.

[3 marks]

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10.
The velocity $v$ m/s of a car $t$ seconds after it starts moving is given by $v = 3t^2 - 12t + 9$, for $t \geq 0$.

(a) Find the acceleration of the car at time $t$.

(b) Find the time(s) when the velocity is zero.

(c) Find the acceleration when $t = 3$.

[3 marks]

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Section B: Applications of Differentiation (Questions 11–16)

All questions in this section are 3 marks each.

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11.
A curve has equation $y = x^3 - 3x + 2$.

(a) Find $\dfrac{dy}{dx}$.

(b) Find the gradient of the curve at the point $(2,\; 4)$.

(c) Find the equation of the tangent to the curve at the point $(2,\; 4)$.

[3 marks]

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12.
The cost $C$ dollars of producing $x$ items is modelled by $C = 0.01x^3 - 0.6x^2 + 15x + 200$.

(a) Find $\dfrac{dC}{dx}$.

(b) Find the rate at which the cost is changing when 10 items are being produced.

(c) Interpret your answer to part (b) in context.

[3 marks]

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13.
A ball is thrown vertically upwards. Its height $h$ metres above the ground after $t$ seconds is given by $h = 20t - 5t^2$, for $t \geq 0$.

(a) Find $\dfrac{dh}{dt}$ and interpret what it represents.

(b) Find the time when the ball reaches its maximum height.

(c) Find the maximum height reached by the ball.

[3 marks]

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14.
The equation of a curve is $y = \dfrac{4}{x} + x$, for $x > 0$.

(a) Express $y$ in index form and find $\dfrac{dy}{dx}$.

(b) Find the gradient of the curve at $x = 2$.

(c) Find the equation of the normal to the curve at the point where $x = 2$.

[3 marks]

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15.
A curve has equation $y = x^3 - 6x^2 + 12x - 5$.

(a) Find $\dfrac{dy}{dx}$.

(b) Show that $\dfrac{dy}{dx} \geq 0$ for all values of $x$, and hence explain why the curve has no stationary points.

(c) Determine whether the curve is always increasing or always decreasing.

[3 marks]

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16.
The volume $V$ cm³ of water in a container at time $t$ minutes is given by $V = 2t^3 - 15t^2 + 24t + 50$, for $t \geq 0$.

(a) Find $\dfrac{dV}{dt}$.

(b) Find the rate at which the volume is changing when $t = 1$.

(c) Find the two times at which the rate of change of volume is zero.

[3 marks]

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Section C: Structured Problem Solving (Questions 17–20)

All questions in this section are 4 marks each.

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17.
A closed cylindrical can is to have a volume of $500\pi$ cm³. Let the radius of the base be $r$ cm and the height be $h$ cm.

(a) Show that $h = \dfrac{500}{r^2}$.

(b) Show that the total surface area $A$ cm² of the can is given by $A = 2\pi r^2 + \dfrac{1000\pi}{r}$.

(c) Find $\dfrac{dA}{dr}$.

(d) Find the value of $r$ for which the surface area is a minimum, and verify that this gives a minimum.

[4 marks]

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18.
A curve has equation $y = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. The curve passes through the point $(0,\; -2)$ and has a stationary point at $(1,\; 0)$.

(a) Use the point $(0,\; -2)$ to find the value of $c$.

(b) Find $\dfrac{dy}{dx}$ in terms of $a$ and $b$.

(c) Use the stationary point at $(1,\; 0)$ to form two simultaneous equations in $a$ and $b$.

(d) Solve your equations to find the values of $a$ and $b$.

[4 marks]

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19.
The displacement $s$ metres of a particle moving in a straight line from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 9t^2 + 24t$, for $t \geq 0$.

(a) Find an expression for the velocity $v$ of the particle at time $t$.

(b) Find the times when the particle is instantaneously at rest.

(c) Find an expression for the acceleration of the particle at time $t$.

(d) Find the displacement of the particle from $O$ at the first time it comes to rest.

[4 marks]

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20.
A rectangular sheet of cardboard measuring 60 cm by 40 cm has four identical squares of side $x$ cm cut from each corner. The flaps are then folded up to form an open box.

(a) Show that the volume $V$ cm³ of the box is given by $V = 4x^3 - 200x^2 + 2400x$.

(b) Find $\dfrac{dV}{dx}$.

(c) Find the value of $x$ that gives the maximum volume. (You may assume $0 < x < 20$.)

(d) Hence find the maximum volume of the box.

[4 marks]

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End of Quiz

This quiz was generated as syllabus-aligned practice content. While informed by assessment patterns, individual questions are not reproduced from past examination papers.

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Secondary 3 Elementary Mathematics Quiz - Calculus

Answer Key and Marking Scheme

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Question 1 — [2 marks]

(a)
When $x = 1$: $y = (1)^2 + 3(1) - 4 = 1 + 3 - 4 = \boxed{0}$
When $x = 3$: $y = (3)^2 + 3(3) - 4 = 9 + 9 - 4 = \boxed{14}$

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$-6$	$-6$	$-4$	$\mathbf{0}$	$6$	$\mathbf{14}$

(b) Plot points and draw a smooth U-shaped parabola through them.

Marking: 1 mark for both $y$-values correct; 1 mark for correct plot and smooth curve.

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Question 2 — [2 marks]

Gradient $= \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{8.5 - 2.0}{3.0 - 1.5} = \dfrac{6.5}{1.5} = \dfrac{13}{3} \approx \boxed{4.33}$

Marking: 1 mark for correct method (rise/run); 1 mark for correct answer (4.33 or 13/3).

Note: The actual gradient of $y = x^2$ at $x = 2$ is $2x = 4$. The estimate of 4.33 is reasonable given the tangent drawn through approximate points.

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Question 3 — [2 marks]

The function is decreasing at that point.
Reason: A negative gradient means that as $x$ increases, $y$ decreases.

Marking: 1 mark for "decreasing"; 1 mark for correct reason linking negative gradient to decreasing function.

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Question 4 — [2 marks]

$y = 4x^2 - 7x + 2$
$\dfrac{dy}{dx} = 8x - 7$

At $x = -1$: $\dfrac{dy}{dx} = 8(-1) - 7 = -8 - 7 = \boxed{-15}$

Marking: 1 mark for correct differentiation; 1 mark for correct substitution and answer.

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Question 5 — [2 marks]

Gradient $= \dfrac{13.0 - (-1.0)}{3.0 - 1.0} = \dfrac{14.0}{2.0} = \boxed{7}$

Marking: 1 mark for correct method; 1 mark for correct answer.

Note: The actual value of $\dfrac{ds}{dt} = 6t - 5$ at $t = 2$ gives $12 - 5 = 7$, confirming the estimate is exact in this case.

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Question 6 — [3 marks]

(a) $y = 5x^3 - 2x^2 + 7x - 4$
$\dfrac{dy}{dx} = \boxed{15x^2 - 4x + 7}$

(b) $y = \dfrac{3}{x^2} = 3x^{-2}$
$\dfrac{dy}{dx} = 3 \times (-2)x^{-3} = \boxed{-6x^{-3}}$ or $\boxed{\dfrac{-6}{x^3}}$

(c) First expand: $y = (2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$
$\dfrac{dy}{dx} = \boxed{4x + 5}$

Marking: 1 mark for each correct derivative.

Common mistake: Forgetting to expand or rewrite before differentiating in parts (b) and (c).

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Question 7 — [3 marks]

(a) $\dfrac{dy}{dx} = \boxed{6x^2 - 18x + 12}$

(b) At $x = 2$: $\dfrac{dy}{dx} = 6(4) - 18(2) + 12 = 24 - 36 + 12 = \boxed{0}$

(c) Set $\dfrac{dy}{dx} = 0$:
$6x^2 - 18x + 12 = 0$
$x^2 - 3x + 2 = 0$
$(x - 1)(x - 2) = 0$
$x = 1$ or $x = 2$

When $x = 1$: $y = 2(1) - 9(1) + 12(1) + 1 = 2 - 9 + 12 + 1 = 6$
When $x = 2$: $y = 2(8) - 9(4) + 12(2) + 1 = 16 - 36 + 24 + 1 = 5$

Coordinates: $\boxed{(1,\; 6)}$ and $\boxed{(2,\; 5)}$

Marking: 1 mark for (a); 1 mark for (b); 1 mark for both coordinates in (c).

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Question 8 — [3 marks]

(a) $\dfrac{dy}{dx} = \boxed{3x^2 - 12x + 9}$

(b) Set $\dfrac{dy}{dx} = 0$:
$3x^2 - 12x + 9 = 0$
$x^2 - 4x + 3 = 0$
$(x - 1)(x - 3) = 0$
$x = \boxed{1}$ or $x = \boxed{3}$

(c) When $x = 1$: $y = 1 - 6 + 9 = 4$ → $(1,\; 4)$
When $x = 3$: $y = 27 - 54 + 27 = 0$ → $(3,\; 0)$

Stationary points: $\boxed{(1,\; 4)}$ and $\boxed{(3,\; 0)}$

Marking: 1 mark each for (a), (b), and (c).

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Question 9 — [3 marks]

(a) Let the side parallel to the wall have length $y$ metres.
Then $y + 2x = 80$, so $y = 80 - 2x$.
Area $A = xy = x(80 - 2x) = \boxed{80x - 2x^2}$

(b) $\dfrac{dA}{dx} = 80 - 4x$
Set $\dfrac{dA}{dx} = 0$: $80 - 4x = 0$ → $x = \boxed{20}$

(c) Maximum area $= 80(20) - 2(20)^2 = 1600 - 800 = \boxed{800 \text{ m}^2}$

Marking: 1 mark for (a); 1 mark for (b); 1 mark for (c).

Common mistake: Forgetting to verify it is a maximum (second derivative $= -4 < 0$, confirming maximum).

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Question 10 — [3 marks]

(a) Acceleration $a = \dfrac{dv}{dt} = \boxed{6t - 12}$ m/s²

(b) Set $v = 0$: $3t^2 - 12t + 9 = 0$
$t^2 - 4t + 3 = 0$
$(t - 1)(t - 3) = 0$
$t = \boxed{1 \text{ s}}$ or $t = \boxed{3 \text{ s}}$

(c) At $t = 3$: $a = 6(3) - 12 = 18 - 12 = \boxed{6 \text{ m/s}^2}$

Marking: 1 mark each for (a), (b), and (c).

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Question 11 — [3 marks]

(a) $\dfrac{dy}{dx} = \boxed{3x^2 - 3}$

(b) At $x = 2$: $\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = \boxed{9}$

(c) Tangent at $(2,\; 4)$ with gradient $9$:
$y - 4 = 9(x - 2)$
$y = 9x - 18 + 4$
$\boxed{y = 9x - 14}$

Marking: 1 mark each for (a), (b), and (c).

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Question 12 — [3 marks]

(a) $\dfrac{dC}{dx} = \boxed{0.03x^2 - 1.2x + 15}$

(b) At $x = 10$: $\dfrac{dC}{dx} = 0.03(100) - 1.2(10) + 15 = 3 - 12 + 15 = \boxed{6}$

(c) When 10 items are being produced, the cost is increasing at a rate of $6 per additional item.

Marking: 1 mark for (a); 1 mark for (b); 1 mark for correct interpretation in (c).

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Question 13 — [3 marks]

(a) $\dfrac{dh}{dt} = \boxed{20 - 10t}$ m/s
This represents the velocity of the ball (rate of change of height with respect to time).

(b) At maximum height, $\dfrac{dh}{dt} = 0$:
$20 - 10t = 0$ → $t = \boxed{2 \text{ s}}$

(c) Maximum height $= 20(2) - 5(4) = 40 - 20 = \boxed{20 \text{ m}}$

Marking: 1 mark for derivative and interpretation; 1 mark for time; 1 mark for maximum height.

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Question 14 — [3 marks]

(a) $y = 4x^{-1} + x$
$\dfrac{dy}{dx} = -4x^{-2} + 1 = \boxed{1 - \dfrac{4}{x^2}}$

(b) At $x = 2$: $\dfrac{dy}{dx} = 1 - \dfrac{4}{4} = 1 - 1 = \boxed{0}$

(c) When the gradient of the tangent is $0$, the normal is a vertical line.
At $x = 2$: $y = \dfrac{4}{2} + 2 = 4$, so the point is $(2,\; 4)$.
Equation of the normal: $\boxed{x = 2}$

Marking: 1 mark for (a); 1 mark for (b); 1 mark for (c).

Common mistake: Students may try to use the negative reciprocal of 0, which is undefined. The normal to a horizontal tangent is vertical.

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Question 15 — [3 marks]

(a) $\dfrac{dy}{dx} = \boxed{3x^2 - 12x + 12}$

(b) $\dfrac{dy}{dx} = 3x^2 - 12x + 12 = 3(x^2 - 4x + 4) = 3(x - 2)^2$
Since $(x - 2)^2 \geq 0$ for all real $x$, we have $\dfrac{dy}{dx} = 3(x - 2)^2 \geq 0$ for all $x$.
The gradient is zero only at $x = 2$ but does not change sign, so there are no stationary points that are turning points (it is a stationary point of inflection).

(c) Since $\dfrac{dy}{dx} \geq 0$ for all $x$, the curve is always increasing (non-decreasing).

Marking: 1 mark for (a); 1 mark for completing the square and explaining; 1 mark for correct conclusion.

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Question 16 — [3 marks]

(a) $\dfrac{dV}{dt} = \boxed{6t^2 - 30t + 24}$ cm³/min

(b) At $t = 1$: $\dfrac{dV}{dt} = 6 - 30 + 24 = \boxed{0 \text{ cm}^3/\text{min}}$

(c) Set $\dfrac{dV}{dt} = 0$:
$6t^2 - 30t + 24 = 0$
$t^2 - 5t + 4 = 0$
$(t - 1)(t - 4) = 0$
$t = \boxed{1 \text{ min}}$ or $t = \boxed{4 \text{ min}}$

Marking: 1 mark each for (a), (b), and (c).

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Question 17 — [4 marks]

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

(b) Surface area $A = 2\pi r^2$ (top and bottom) $+ 2\pi r h$ (curved surface)
$A = 2\pi r^2 + 2\pi r \times \dfrac{500}{r^2} = 2\pi r^2 + \dfrac{1000\pi}{r}$
$\boxed{A = 2\pi r^2 + \dfrac{1000\pi}{r}}$

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

(d) Set $\dfrac{dA}{dr} = 0$:
$4\pi r = \dfrac{1000\pi}{r^2}$
$4r^3 = 1000$
$r^3 = 250$
$r = \sqrt[3]{250} = \boxed{5\sqrt[3]{2} \approx 6.30 \text{ cm}}$

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

Marking: 1 mark each for (a), (b), (c), and (d) including verification.

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Question 18 — [4 marks]

(a) At $(0,\; -2)$: $y = 0 + 0 + 0 + c = -2$ → $\boxed{c = -2}$

(b) $\dfrac{dy}{dx} = \boxed{3x^2 + 2ax + b}$

(c) Using the point $(1,\; 0)$ on the curve:
$0 = 1 + a + b + c = 1 + a + b - 2$
$a + b = 1$ ... (i)

Using the stationary point (gradient $= 0$ at $x = 1$):
$0 = 3(1) + 2a(1) + b = 3 + 2a + b$
$2a + b = -3$ ... (ii)

(d) Subtract (i) from (ii):
$(2a + b) - (a + b) = -3 - 1$
$a = -4$

From (i): $-4 + b = 1$ → $b = 5$

$\boxed{a = -4}$ and $\boxed{b = 5}$

Marking: 1 mark each for (a), (b), (c) (both equations), and (d).

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Question 19 — [4 marks]

(a) Velocity $v = \dfrac{ds}{dt} = \boxed{3t^2 - 18t + 24}$ m/s

(b) Set $v = 0$:
$3t^2 - 18t + 24 = 0$
$t^2 - 6t + 8 = 0$
$(t - 2)(t - 4) = 0$
$t = \boxed{2 \text{ s}}$ or $t = \boxed{4 \text{ s}}$

(c) Acceleration $a = \dfrac{dv}{dt} = \boxed{6t - 18}$ m/s²

(d) At $t = 2$: $s = (2)^3 - 9(2)^2 + 24(2) = 8 - 36 + 48 = \boxed{20 \text{ m}}$

Marking: 1 mark each for (a), (b), (c), and (d).

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Question 20 — [4 marks]

(a) After cutting squares of side $x$ from each corner:
Length of box base $= 60 - 2x$
Width of box base $= 40 - 2x$
Height of box $= x$

$V = x(60 - 2x)(40 - 2x)$
$= x(2400 - 120x - 80x + 4x^2)$
$= x(2400 - 200x + 4x^2)$
$= 4x^3 - 200x^2 + 2400x$
$\boxed{V = 4x^3 - 200x^2 + 2400x}$

(b) $\dfrac{dV}{dx} = \boxed{12x^2 - 400x + 2400}$

(c) Set $\dfrac{dV}{dx} = 0$:
$12x^2 - 400x + 2400 = 0$
$3x^2 - 100x + 600 = 0$
Using the quadratic formula:
$x = \dfrac{100 \pm \sqrt{10000 - 7200}}{6} = \dfrac{100 \pm \sqrt{2800}}{6} = \dfrac{100 \pm 52.92}{6}$

$x = \dfrac{100 + 52.92}{6} \approx 25.5$ (reject, since $x < 20$)
$x = \dfrac{100 - 52.92}{6} \approx \boxed{7.85 \text{ cm}}$

(d) Maximum volume $= 4(7.85)^3 - 200(7.85)^2 + 2400(7.85)$
$\approx 4(483.7) - 200(61.62) + 18840$
$\approx 1934.8 - 12324 + 18840$
$\approx \boxed{8451 \text{ cm}^3}$ (to 3 s.f.)

Marking: 1 mark each for (a), (b), (c), and (d).

Common mistake: Not rejecting the extraneous root $x \approx 25.5$ which exceeds the constraint $x < 20$.

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Total: 40 marks

This answer key was generated as syllabus-aligned practice content. While informed by assessment patterns, individual questions are not reproduced from past examination papers.