Secondary 3 Additional Mathematics Calculus Quiz

Secondary 3 Additional Mathematics Quiz - Calculus

Answer Key

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Section A: Differentiation (Questions 1–10)

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1. A) $12x^3 - 4x$ [2]

Working: $\frac{dy}{dx} = 3(4)x^3 - 2(2)x = 12x^3 - 4x$. The derivative of the constant $7$ is zero.

Common mistake: Choosing B — forgetting that the derivative of a constant is zero.

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2. B) $6(2x - 5)^2$ [2]

Working: Using the chain rule: $f'(x) = 3(2x - 5)^2 \times 2 = 6(2x - 5)^2$.

Common mistake: Choosing A — forgetting to multiply by the derivative of the inner function $(2x - 5)$, which is $2$.

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3. A) $1 - \frac{1}{x^2}$ [2]

Working: Rewrite $y = \frac{x^2}{x} + \frac{1}{x} = x + x^{-1}$. Then $\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$.

Alternative: Using the quotient rule: $\frac{dy}{dx} = \frac{(2x)(x) - (x^2+1)(1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2}$.

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4. C) $(1, 4)$ and $(3, 0)$ [2]

Working: $\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$. Setting $\frac{dy}{dx} = 0$: $x = 1$ or $x = 3$. When $x = 1$: $y = 1 - 6 + 9 = 4$. When $x = 3$: $y = 27 - 54 + 27 = 0$. Points are $(1, 4)$ and $(3, 0)$.

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5. A) $5 \text{ m/s}$ [2]

Working: $v = \frac{ds}{dt} = 3t^2 - 8t + 2$. When $t = 3$: $v = 3(9) - 8(3) + 2 = 27 - 24 + 2 = 5 \text{ m/s}$.

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6.

(a) $\frac{dy}{dx} = 15x^2 - 4$ [1]

Working: $\frac{d}{dx}(5x^3) - \frac{d}{dx}(4x) + \frac{d}{dx}(3) = 15x^2 - 4 + 0$.

(b) $\frac{dy}{dx} = 6x - 10$ [2]

Working: First expand: $y = 3x^2 - 12x + 2x - 8 = 3x^2 - 10x - 8$. Then $\frac{dy}{dx} = 6x - 10$.

Alternative using product rule: $\frac{dy}{dx} = (3)(x-4) + (3x+2)(1) = 3x - 12 + 3x + 2 = 6x - 10$.

(c) $\frac{dy}{dx} = 2 + \frac{2}{x^3}$ [2]

Working: Rewrite $y = \frac{2x^3}{x^2} - \frac{1}{x^2} = 2x - x^{-2}$. Then $\frac{dy}{dx} = 2 - (-2)x^{-3} = 2 + \frac{2}{x^3}$.

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7. $\frac{dy}{dx} = \frac{2}{\sqrt{4x+1}}$ [3]

Working: $y = (4x+1)^{1/2}$. Using the chain rule: $\frac{dy}{dx} = \frac{1}{2}(4x+1)^{-1/2} \times 4 = \frac{2}{\sqrt{4x+1}}$.

Marking: [1] for correct application of chain rule, [1] for correct simplification, [1] for final answer in surd form.

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8. Gradient $= 3$ [3]

Working: $\frac{dy}{dx} = 6x^2 - 10x + 3$. At $x = 2$: $\frac{dy}{dx} = 6(4) - 10(2) + 3 = 24 - 20 + 3 = 7$.

Correction: $\frac{dy}{dx} = 6(4) - 20 + 3 = 24 - 20 + 3 = 7$.

Gradient $= 7$ [3]

Marking: [1] for correct differentiation, [1] for correct substitution, [1] for correct evaluation.

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9. $f'(1) = -32$ [3]

Working: $f'(x) = 4(x^2 - 3)^3 \times 2x = 8x(x^2 - 3)^3$. At $x = 1$: $f'(1) = 8(1)(1 - 3)^3 = 8(-2)^3 = 8(-8) = -64$.

Correction: $f'(1) = 8(1)(1-3)^3 = 8(1)(-8) = -64$.

$f'(1) = -64$ [3]

Marking: [1] for correct chain rule application, [1] for correct substitution, [1] for correct evaluation.

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10. $a = \frac{13}{2}$, $b = -\frac{3}{2}$ [4]

Working: $y = ax^3 + bx$, so $\frac{dy}{dx} = 3ax^2 + b$. At $(1, 5)$: $y = a(1)^3 + b(1) = a + b = 5$ … (i). Gradient: $3a(1)^2 + b = 3a + b = 8$ … (ii). Subtracting (i) from (ii): $2a = 3$, so $a = \frac{3}{2}$. From (i): $\frac{3}{2} + b = 5$, so $b = \frac{7}{2}$.

Correction: From (ii) − (i): $(3a + b) - (a + b) = 8 - 5 = 3$, so $2a = 3$, $a = \frac{3}{2}$. Then $b = 5 - \frac{3}{2} = \frac{7}{2}$.

$a = \frac{3}{2}$, $b = \frac{7}{2}$ [4]

Marking: [1] for correct differentiation, [1] for forming equation from point on curve, [1] for forming equation from gradient, [1] for solving simultaneously.

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Section B: Applications of Differentiation (Questions 11–15)

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11.

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

(b) Stationary points: $(1, 6)$ and $(3, 2)$ [3]

Working: Set $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$, so $x^2 - 4x + 3 = 0$, giving $(x-1)(x-3) = 0$, so $x = 1$ or $x = 3$. When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$. When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$. Points are $(1, 6)$ and $(3, 2)$.

Marking: [1] for solving $\frac{dy}{dx}=0$, [1] for each correct point.

(c) At $x = 1$: maximum; at $x = 3$: minimum [3]

Working: $\frac{d^2y}{dx^2} = 6x - 12$. At $x = 1$: $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0$, so maximum. At $x = 3$: $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0$, so minimum.

Marking: [1] for correct second derivative, [1] for correct nature at $x=1$, [1] for correct nature at $x=3$.

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12.

(a) Shown. [2]

Working: Let the two sides perpendicular to the wall each have length $x$ and the side parallel to the wall have length $l$. Then $l + 2x = 120$, so $l = 120 - 2x$. Area $A = l \times x = (120 - 2x)x = 120x - 2x^2$.

Marking: [1] for correct expression for $l$, [1] for correct area expression.

(b) $x = 30$ [2]

Working: $\frac{dA}{dx} = 120 - 4x$. Set $\frac{dA}{dx} = 0$: $120 - 4x = 0$, so $x = 30$. Check: $\frac{d^2A}{dx^2} = -4 < 0$, confirming maximum.

Marking: [1] for differentiation and solving, [1] for confirming maximum.

(c) Maximum area $= 1800 \text{ m}^2$ [1]

Working: $A = 120(30) - 2(30)^2 = 3600 - 1800 = 1800$.

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13. $a = -3$, $b = 0$, $c = 3$ [5]

Working: $\frac{dy}{dx} = 3x^2 + 2ax + b$. At stationary point $(2, -5)$: $\frac{dy}{dx} = 0$ when $x = 2$, so $3(4) + 2a(2) + b = 0$, giving $12 + 4a + b = 0$ … (i). Also $y = -5$ when $x = 2$: $8 + 4a + 2b + c = -5$ … (ii). The curve passes through $(0, 3)$: $c = 3$ … (iii). From (iii) into (ii): $8 + 4a + 2b + 3 = -5$, so $4a + 2b = -16$, i.e. $2a + b = -8$ … (iv). From (i): $4a + b = -12$ … (i). Subtracting (iv) from (i): $2a = -4$, so $a = -2$. From (iv): $2(-2) + b = -8$, so $b = -4$.

Correction: From (i): $4a + b = -12$. From (iv): $2a + b = -8$. Subtracting: $2a = -4$, so $a = -2$. Then $b = -8 - 2(-2) = -8 + 4 = -4$. And $c = 3$.

$a = -2$, $b = -4$, $c = 3$ [5]

Marking: [1] for $\frac{dy}{dx}=0$ at $x=2$, [1] for point $(2,-5)$ on curve, [1] for $c=3$, [1] for solving for $a$, [1] for solving for $b$.

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14.

(a) $v = 6t^2 - 30t + 24$ [1]

Working: $v = \frac{ds}{dt} = 6t^2 - 30t + 24$.

(b) $t = 1$ and $t = 4$ [3]

Working: Set $v = 0$: $6t^2 - 30t + 24 = 0$, so $t^2 - 5t + 4 = 0$, giving $(t-1)(t-4) = 0$. So $t = 1$ or $t = 4$.

Marking: [1] for setting $v=0$, [1] for correct factorisation, [1] for both values.

(c) Acceleration $= -6 \text{ m/s}^2$ [2]

Working: $a = \frac{dv}{dt} = 12t - 30$. At $t = 2$: $a = 24 - 30 = -6 \text{ m/s}^2$.

Marking: [1] for correct differentiation, [1] for correct substitution.

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15. $Q = \left(\frac{1}{2}, \frac{13}{4}\right)$ [5]

Working: $\frac{dy}{dx} = 2x - 4$. At $x = 3$: gradient of tangent $= 2(3) - 4 = 2$. Gradient of normal $= -\frac{1}{2}$. Equation of normal at $(3, 4)$: $y - 4 = -\frac{1}{2}(x - 3)$, so $y = -\frac{1}{2}x + \frac{3}{2} + 4 = -\frac{1}{2}x + \frac{11}{2}$. To find intersection with curve: $x^2 - 4x + 7 = -\frac{1}{2}x + \frac{11}{2}$. Multiply by 2: $2x^2 - 8x + 14 = -x + 11$, so $2x^2 - 7x + 3 = 0$. Factorise: $(2x - 1)(x - 3) = 0$. So $x = 3$ (point $P$) or $x = \frac{1}{2}$. When $x = \frac{1}{2}$: $y = \frac{1}{4} - 2 + 7 = \frac{21}{4} = \frac{13}{4}$... Let me recalculate: $y = \left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 7 = \frac{1}{4} - 2 + 7 = \frac{1}{4} + 5 = \frac{21}{4}$.

$Q = \left(\frac{1}{2}, \frac{21}{4}\right)$ [5]

Marking: [1] for gradient of tangent, [1] for gradient of normal, [1] for equation of normal, [1] for solving simultaneous equations, [1] for correct coordinates of $Q$.

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Section C: Integration (Questions 16–20)

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16.

(a) $\int(6x^2 - 4x + 1)\,dx = 2x^3 - 2x^2 + x + c$ [2]

Marking: [1] for correct integration of each term, [1] for including constant of integration.

(b) $\int(2x-3)^2\,dx = \frac{(2x-3)^3}{6} + c$ (or expanded form) [2]

Working (substitution): Let $u = 2x - 3$, $du = 2\,dx$. $\int u^2 \cdot \frac{du}{2} = \frac{1}{2} \cdot \frac{u^3}{3} = \frac{(2x-3)^3}{6} + c$.

Working (expansion): $(2x-3)^2 = 4x^2 - 12x + 9$. $\int(4x^2 - 12x + 9)\,dx = \frac{4x^3}{3} - 6x^2 + 9x + c$.

Marking: Either method accepted. [1] for correct method, [1] for correct answer with constant.

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

Working: $\frac{x^3-1}{x^2} = x - x^{-2}$. $\int(x - x^{-2})\,dx = \frac{x^2}{2} - \frac{x^{-1}}{-1} + c = \frac{x^2}{2} + \frac{1}{x} + c$.

Marking: [1] for correct simplification, [1] for correct integration with constant.

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17. $y = x^3 - 3x^2 + 2x + 5$ [4]

Working: $y = \int(3x^2 - 6x + 2)\,dx = x^3 - 3x^2 + 2x + c$. When $x = 1$, $y = 5$: $1 - 3 + 2 + c = 5$, so $c = 5$. Therefore $y = x^3 - 3x^2 + 2x + 5$.

Marking: [1] for correct integration, [1] for including constant, [1] for correct substitution, [1] for correct value of $c$ and final expression.

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18. $y = x^4 - 3x^2 + 2$ [4]

Working: $y = \int(4x^3 - 6x)\,dx = x^4 - 3x^2 + c$. When $x = 2$, $y = 10$: $16 - 12 + c = 10$, so $c = 6$. Therefore $y = x^4 - 3x^2 + 6$.

Correction: $16 - 12 + c = 10$, so $4 + c = 10$, $c = 6$.

$y = x^4 - 3x^2 + 6$ [4]

Marking: [1] for correct integration, [1] for including constant, [1] for correct substitution, [1] for correct final equation.

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19.

(a) $x = 2$ and $x = -3$ [2]

Working: At stationary points, $\frac{dy}{dx} = 0$: $(x-2)(x+3) = 0$, so $x = 2$ or $x = -3$.

Marking: [1] for each correct value.

(b) $y = \frac{x^3}{3} + \frac{x^2}{2} - 6x + \frac{83}{6}$ [4]

Working: $y = \int(x^2 + x - 6)\,dx = \frac{x^3}{3} + \frac{x^2}{2} - 6x + c$. When $x = 1$, $y = 6$: $\frac{1}{3} + \frac{1}{2} - 6 + c = 6$. So $\frac{5}{6} - 6 + c = 6$, giving $c = 6 + 6 - \frac{5}{6} = 12 - \frac{5}{6} = \frac{67}{6}$.

Correction: $\frac{1}{3} + \frac{1}{2} = \frac{5}{6}$. So $\frac{5}{6} - 6 + c = 6$, meaning $c = 12 - \frac{5}{6} = \frac{72-5}{6} = \frac{67}{6}$.

$y = \frac{x^3}{3} + \frac{x^2}{2} - 6x + \frac{67}{6}$ [4]

Marking: [1] for expanding the integrand, [1] for correct integration, [1] for correct substitution, [1] for correct final equation.

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20.

(a) $s = t^3 - 6t^2 + 9t + 5$ [2]

Working: $s = \int(3t^2 - 12t + 9)\,dt = t^3 - 6t^2 + 9t + c$. When $t = 0$, $s = 5$: $c = 5$. So $s = t^3 - 6t^2 + 9t + 5$.

Marking: [1] for correct integration with constant, [1] for correct value of $c$.

(b) Displacement $= 5 \text{ m}$ [2]

Working: $s = 27 - 54 + 27 + 5 = 5 \text{ m}$.

Marking: [1] for correct substitution, [1] for correct answer.

(c) Total distance $= 12 \text{ m}$ [4]

Working: $v = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$. The velocity changes sign at $t = 1$ and $t = 3$. For $0 < t < 1$: $v > 0$ (particle moves forward). For $1 < t < 3$: $v < 0$ (particle moves backward). For $3 < t < 4$: $v > 0$ (particle moves forward). Displacement from $t = 0$ to $t = 1$: $s(1) - s(0) = (1 - 6 + 9 + 5) - 5 = 9 - 5 = 4 \text{ m}$. Displacement from $t = 1$ to $t = 3$: $s(3) - s(1) = (27 - 54 + 27 + 5) - 9 = 5 - 9 = -4 \text{ m}$ (distance $= 4 \text{ m}$). Displacement from $t = 3$ to $t = 4$: $s(4) - s(3) = (64 - 96 + 36 + 5) - 5 = 9 - 5 = 4 \text{ m}$. Total distance $= 4 + 4 + 4 = 12 \text{ m}$.

Marking: [1] for finding when $v = 0$, [1] for determining sign changes/direction, [1] for calculating each stage of distance, [1] for correct total.

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END OF ANSWER KEY