Secondary 3 Additional Mathematics Calculus Quiz

Secondary 3 Additional Mathematics Quiz - Calculus

Answer Key

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Question 1 [5 marks]

(a) $y = 5x^3$
$\dfrac{dy}{dx} = 5 \times 3x^2 = 15x^2$

(b) $y = 4x^2 - 3x + 7$
$\dfrac{dy}{dx} = 8x - 3$

(c) $y = \sqrt{x} = x^{1/2}$
$\dfrac{dy}{dx} = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$

(d) $y = \dfrac{1}{x^2} = x^{-2}$
$\dfrac{dy}{dx} = -2x^{-3} = -\dfrac{2}{x^3}$

Marking: 1 mark each for (a), (b), (c), (d); 1 mark for overall correct notation and presentation.

Common mistakes: Forgetting to reduce the exponent by 1; writing $\sqrt{x}$ as $x^2$ instead of $x^{1/2}$; sign errors in (d).

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Question 2 [3 marks]

Expand: $y = (2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$

Differentiate: $\dfrac{dy}{dx} = 4x - 5$

Marking: 1 mark for correct expansion, 1 mark for correct differentiation, 1 mark for final simplified answer.

Common mistake: Attempting to use product rule without being asked; expanding incorrectly (e.g., $2x \times (-3) = -5x$).

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Question 3 [3 marks]

$f(x) = 3x^4 - 2x^3 + x - 5$

$f'(x) = 12x^3 - 6x^2 + 1$

$f'(1) = 12(1)^3 - 6(1)^2 + 1 = 12 - 6 + 1 = 7$

Marking: 2 marks for correct $f'(x)$, 1 mark for correct evaluation of $f'(1) = 7$.

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Question 4 [3 marks]

$y = 2x^3 - 6x^2 + 4x - 1$

$\dfrac{dy}{dx} = 6x^2 - 12x + 4$

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

The gradient is 4.

Marking: 1 mark for correct derivative, 1 mark for correct substitution, 1 mark for final answer.

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Question 5 [4 marks]

$y = x^3 - 3x$

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

Set gradient = 0: $3x^2 - 3 = 0$

$3x^2 = 3$

$x^2 = 1$

$x = 1$ or $x = -1$

When $x = 1$: $y = 1 - 3 = -2$, so point is $(1, -2)$

When $x = -1$: $y = -1 + 3 = 2$, so point is $(-1, 2)$

Marking: 1 mark for correct derivative, 1 mark for solving $3x^2 - 3 = 0$, 1 mark for each correct point.

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Question 6 [4 marks]

$y = x^2 - 4x + 3$

At $x = 1$: $y = 1 - 4 + 3 = 0$, so the point is $(1, 0)$

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

At $x = 1$: gradient $= 2(1) - 4 = -2$

Equation of tangent: $y - 0 = -2(x - 1)$

$y = -2x + 2$

Marking: 1 mark for correct $y$-coordinate, 1 mark for correct gradient, 1 mark for using point-slope form, 1 mark for correct final equation.

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Question 7 [4 marks]

$y = \dfrac{2}{x} = 2x^{-1}$

At $x = 1$: $y = 2$, so the point is $(1, 2)$

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

At $x = 1$: gradient of tangent $= -2$

Gradient of normal $= \dfrac{1}{2}$ (negative reciprocal)

Equation of normal: $y - 2 = \dfrac{1}{2}(x - 1)$

$y = \dfrac{1}{2}x - \dfrac{1}{2} + 2$

$y = \dfrac{1}{2}x + \dfrac{3}{2}$

Marking: 1 mark for correct point, 1 mark for correct gradient of tangent, 1 mark for correct gradient of normal, 1 mark for correct equation.

Common mistake: Forgetting to take the negative reciprocal for the normal gradient.

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Question 8 [8 marks]

(a) $\dfrac{dy}{dx} = 3x^2 - 12x + 9$ [1 mark]

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

When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$, point is $(1, 6)$
When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$, point is $(3, 2)$ [1 mark]

Second derivative: $\dfrac{d^2y}{dx^2} = 6x - 12$

At $x = 1$: $\dfrac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → maximum at $(1, 6)$
At $x = 3$: $\dfrac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → minimum at $(3, 2)$ [2 marks]

(c) Sketch: cubic with positive leading coefficient, maximum at $(1, 6)$, minimum at $(3, 2)$, $y$-intercept at $(0, 2)$. [2 marks] (1 mark for correct shape, 1 mark for labelled stationary points)

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Question 9 [6 marks]

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

(b) Set $v = 0$:
$3t^2 - 12t + 9 = 0$
$3(t^2 - 4t + 3) = 0$
$3(t - 1)(t - 3) = 0$
$t = 1$ s or $t = 3$ s [2 marks]

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

At $t = 3$: $a = 18 - 12 = 6$ m/s² [2 marks]

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Question 10 [4 marks]

$V = \dfrac{4}{3}\pi r^3$

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

Using the chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$12\pi = 4\pi r^2 \times \dfrac{dr}{dt}$

When $r = 2$: $12\pi = 4\pi(4) \times \dfrac{dr}{dt}$

$12\pi = 16\pi \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{12\pi}{16\pi} = \dfrac{3}{4} = 0.75$ cm/s

Marking: 1 mark for differentiating $V$, 1 mark for applying chain rule, 1 mark for correct substitution, 1 mark for final answer.

Common mistake: Forgetting to use the chain rule and simply dividing $12\pi$ by $4\pi r^2$ without setting up the relationship properly.

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Question 11 [6 marks]

(a) $\displaystyle\int 6x^2 \, dx = \dfrac{6x^3}{3} + c = 2x^3 + c$ [1 mark]

(b) $\displaystyle\int (3x^2 - 4x + 1) \, dx = \dfrac{3x^3}{3} - \dfrac{4x^2}{2} + x + c = x^3 - 2x^2 + x + c$ [1.5 marks]

(c) $\displaystyle\int \dfrac{1}{x^3} \, dx = \displaystyle\int x^{-3} \, dx = \dfrac{x^{-2}}{-2} + c = -\dfrac{1}{2x^2} + c$ [1.5 marks]

(d) $\displaystyle\int \sqrt{x} \, dx = \displaystyle\int x^{1/2} \, dx = \dfrac{x^{3/2}}{3/2} + c = \dfrac{2}{3}x^{3/2} + c$ [2 marks]

Common mistakes: Forgetting the constant of integration $c$; adding 1 to the exponent but forgetting to divide by the new exponent.

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Question 12 [4 marks]

$\dfrac{dy}{dx} = 4x^3 - 6x$

$y = \displaystyle\int (4x^3 - 6x) \, dx = \dfrac{4x^4}{4} - \dfrac{6x^2}{2} + c = x^4 - 3x^2 + c$ [2 marks]

When $x = 1$, $y = 10$:
$10 = 1 - 3 + c$
$c = 12$ [1 mark]

$\boxed{y = x^4 - 3x^2 + 12}$ [1 mark]

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Question 13 [4 marks]

$\dfrac{dy}{dx} = 3x^2 - 2x + 1$

$y = \displaystyle\int (3x^2 - 2x + 1) \, dx = x^3 - x^2 + x + c$ [2 marks]

When $x = 2$, $y = 5$:
$5 = 8 - 4 + 2 + c$
$5 = 6 + c$
$c = -1$ [1 mark]

$\boxed{y = x^3 - x^2 + x - 1}$ [1 mark]

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Question 14 [6 marks]

(a) $\displaystyle\int_1^4 2x \, dx = \left[x^2\right]_1^4 = 16 - 1 = 15$ [1.5 marks]

(b) $\displaystyle\int_0^2 (x^2 + 1) \, dx = \left[\dfrac{x^3}{3} + x\right]_0^2 = \left(\dfrac{8}{3} + 2\right) - 0 = \dfrac{8}{3} + \dfrac{6}{3} = \dfrac{14}{3}$ [2 marks]

(c) $\displaystyle\int_{-1}^3 (3x^2 - 2x) \, dx = \left[x^3 - x^2\right]_{-1}^3$
$= (27 - 9) - ((-1) - 1)$
$= 18 - (-2)$
$= 20$ [2.5 marks]

Common mistake: Sign errors when substituting the lower limit, especially with negative values as in (c).

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Question 15 [6 marks]

$\dfrac{dy}{dx} = 6x^2 - 4x + k$

Stationary point at $x = \dfrac{1}{3}$, so $\dfrac{dy}{dx} = 0$ when $x = \dfrac{1}{3}$:

$6\left(\dfrac{1}{3}\right)^2 - 4\left(\dfrac{1}{3}\right) + k = 0$

$6 \times \dfrac{1}{9} - \dfrac{4}{3} + k = 0$

$\dfrac{6}{9} - \dfrac{4}{3} + k = 0$

$\dfrac{2}{3} - \dfrac{4}{3} + k = 0$

$-\dfrac{2}{3} + k = 0$

$k = \dfrac{2}{3}$ [3 marks]

Now integrate: $y = \displaystyle\int \left(6x^2 - 4x + \dfrac{2}{3}\right) dx = 2x^3 - 2x^2 + \dfrac{2}{3}x + c$

Curve passes through $(1, 3)$:
$3 = 2 - 2 + \dfrac{2}{3} + c$
$c = 3 - \dfrac{2}{3} = \dfrac{7}{3}$ [2 marks]

$\boxed{y = 2x^3 - 2x^2 + \dfrac{2}{3}x + \dfrac{7}{3}}$ [1 mark]

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Question 16 [3 marks]

Area $= \displaystyle\int_0^3 x^2 \, dx = \left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{27}{3} - 0 = 9$ square units

Marking: 1 mark for correct integral setup, 1 mark for correct antiderivative, 1 mark for correct evaluation.

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Question 17 [4 marks]

First, find where the curve crosses the $x$-axis:
$x^2 - 2x = 0$
$x(x - 2) = 0$
$x = 0$ or $x = 2$

Between $x = 0$ and $x = 2$, the curve is below the $x$-axis (since the parabola opens upward and the vertex is at $x = 1$, $y = -1$).

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

$= -\left[\left(\dfrac{8}{3} - 4\right) - 0\right]$

$= -\left[\dfrac{8}{3} - \dfrac{12}{3}\right]$

$= -\left[-\dfrac{4}{3}\right]$

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

Marking: 1 mark for finding limits, 1 mark for recognising the curve is below the axis (or taking absolute value), 1 mark for correct integration, 1 mark for correct final answer.

Common mistake: Forgetting to take the negative (or absolute value) when the area is below the $x$-axis, giving $-\dfrac{4}{3}$.

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Question 18 [6 marks]

(a) Set $x^2 + 2 = 6$:
$x^2 = 4$
$x = -2$ or $x = 2$

Points of intersection: $(-2, 6)$ and $(2, 6)$ [2 marks]

(b) Area $= \displaystyle\int_{-2}^{2} \left[6 - (x^2 + 2)\right] dx = \displaystyle\int_{-2}^{2} (4 - x^2) \, dx$

$= \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2}$

$= \left(8 - \dfrac{8}{3}\right) - \left(-8 + \dfrac{8}{3}\right)$

$= \left(\dfrac{24}{3} - \dfrac{8}{3}\right) - \left(-\dfrac{24}{3} + \dfrac{8}{3}\right)$

$= \dfrac{16}{3} - \left(-\dfrac{16}{3}\right)$

$= \dfrac{16}{3} + \dfrac{16}{3} = \dfrac{32}{3}$ square units [4 marks]

Marking: 1 mark for correct integrand (line minus curve), 1 mark for correct antiderivative, 1 mark for correct substitution of limits, 1 mark for correct final answer.

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Question 19 [7 marks]

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

The curve crosses the $x$-axis at $x = 0$, $x = 1$, and $x = 3$.

Points: $(0, 0)$, $(1, 0)$, $(3, 0)$ [2 marks]

(b) Check the sign of $y$ in each interval:

• For $0 < x < 1$: test $x = 0.5$, $y = 0.5(−0.5)(−2.5) = 0.625 > 0$ (above axis)
• For $1 < x < 3$: test $x = 2$, $y = 2(1)(−1) = −2 < 0$ (below axis)

Area $= \displaystyle\int_0^1 (x^3 - 4x^2 + 3x) \, dx - \displaystyle\int_1^3 (x^3 - 4x^2 + 3x) \, dx$

First integral: $\left[\dfrac{x^4}{4} - \dfrac{4x^3}{3} + \dfrac{3x^2}{2}\right]_0^1 = \dfrac{1}{4} - \dfrac{4}{3} + \dfrac{3}{2} = \dfrac{3}{12} - \dfrac{16}{12} + \dfrac{18}{12} = \dfrac{5}{12}$

Second integral: $\left[\dfrac{x^4}{4} - \dfrac{4x^3}{3} + \dfrac{3x^2}{2}\right]_1^3$

At $x = 3$: $\dfrac{81}{4} - \dfrac{108}{3} + \dfrac{27}{2} = \dfrac{81}{4} - 36 + \dfrac{27}{2} = \dfrac{81}{4} - \dfrac{144}{4} + \dfrac{54}{4} = -\dfrac{9}{4}$

At $x = 1$: $\dfrac{5}{12}$ (from above)

Second integral $= -\dfrac{9}{4} - \dfrac{5}{12} = -\dfrac{27}{12} - \dfrac{5}{12} = -\dfrac{32}{12} = -\dfrac{8}{3}$

Area $= \dfrac{5}{12} - \left(-\dfrac{8}{3}\right) = \dfrac{5}{12} + \dfrac{32}{12} = \dfrac{37}{12}$ square units [5 marks]

Marking: 1 mark for each root, 1 mark for identifying which region is above/below, 1 mark for correct integral setup, 1 mark for correct evaluation, 1 mark for correct total area.

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Question 20 [6 marks]

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

Set equal to $\dfrac{16}{3}$:

$2k^2 - \dfrac{k^3}{3} = \dfrac{16}{3}$

Multiply through by 3:

$6k^2 - k^3 = 16$

$-k^3 + 6k^2 - 16 = 0$

$k^3 - 6k^2 + 16 = 0$ ✓ (as required) [3 marks]

To solve $k^3 - 6k^2 + 16 = 0$:

Test $k = 2$: $8 - 24 + 16 = 0$ ✓

So $k = 2$ is a root.

Factor: $(k - 2)(k^2 - 4k - 8) = 0$

$k^2 - 4k - 8 = 0$ gives $k = \dfrac{4 \pm \sqrt{16 + 32}}{2} = \dfrac{4 \pm \sqrt{48}}{2} = \dfrac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$

$2 + 2\sqrt{3} \approx 5.46$ (outside the range $0 < k < 4$, reject)
$2 - 2\sqrt{3} \approx -1.46$ (outside the range, reject)

$\boxed{k = 2}$ [3 marks]

Marking: 2 marks for setting up and evaluating the integral correctly, 1 mark for deriving the equation, 1 mark for testing $k = 2$, 1 mark for factorising and checking other roots, 1 mark for final answer.

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END OF ANSWER KEY