Secondary 2 Mathematics Calculus Quiz

Secondary 2 Mathematics Quiz - Calculus: Answer Key

Total Marks: 40

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Section A: Short Answer Questions (Questions 1 to 10)

Each question carries 2 marks.

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1. Answer: 14

Working: $\frac{dy}{dx} = 6x + 2$ At $x = 2$: gradient $= 6(2) + 2 = 12 + 2 = 14$

Explanation: The gradient of a curve at a point is found by differentiating the equation to get $\frac{dy}{dx}$, then substituting the given $x$-value. The derivative of $x^2$ is $2x$, so $3x^2$ becomes $3 \times 2x = 6x$. The derivative of $2x$ is $2$, and the derivative of a constant ($-5$) is $0$.

Marking Notes:

• M1: Correct differentiation
• A1: Correct final answer

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

Working: $\frac{dy}{dx} = 5(3x^2) - 4(2x) + 7(1) - 0 = 15x^2 - 8x + 7$

Explanation: Differentiate term by term using the power rule: multiply by the power and reduce the power by 1.

• $5x^3$: $5 \times 3x^{3-1} = 15x^2$
• $-4x^2$: $-4 \times 2x^{2-1} = -8x$
• $7x$: $7 \times 1x^{1-1} = 7x^0 = 7$
• $-1$: derivative of a constant is 0

Marking Notes:

• M1: Correct differentiation of at least two terms
• A1: Fully correct derivative

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3. Answer: $x = 2$

Working: $\frac{dy}{dx} = 2x - 4$ At stationary point, $\frac{dy}{dx} = 0$: $2x - 4 = 0$ $2x = 4$ $x = 2$

Explanation: Stationary points occur where the gradient is zero, i.e., $\frac{dy}{dx} = 0$. Differentiate the quadratic, set the derivative equal to zero, and solve for $x$.

Marking Notes:

• M1: Correct differentiation and setting to zero
• A1: Correct answer

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4. Answer: $\frac{dy}{dx} = -\frac{2}{x^3}$ or $\frac{dy}{dx} = -2x^{-3}$

Working: $y = \frac{1}{x^2} = x^{-2}$ $\frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$

Explanation: Rewrite $\frac{1}{x^2}$ as $x^{-2}$ using the rule $x^{-n} = \frac{1}{x^n}$. Then differentiate using the power rule: multiply by the power $(-2)$ and reduce the power by $1$ to get $-3$.

Marking Notes:

• M1: Correct rewriting as $x^{-2}$
• A1: Correct derivative

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5. Answer: 19 m/s

Working: Velocity $v = \frac{ds}{dt} = 4t + 3$ At $t = 4$: $v = 4(4) + 3 = 16 + 3 = 19$ m/s

Explanation: Velocity is the rate of change of displacement with respect to time, so differentiate $s$ to get $v$. Then substitute $t = 4$.

Marking Notes:

• M1: Correct differentiation
• A1: Correct answer with units

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6. Answer: 11

Working: $\frac{dy}{dx} = 6x^2 - 5$ At $x = -1$: gradient $= 6(-1)^2 - 5 = 6(1) - 5 = 6 - 5 = 1$

Explanation: Differentiate $2x^3$ to get $6x^2$ and $-5x$ to get $-5$. Substitute $x = -1$. Note that $(-1)^2 = 1$, so $6(1) - 5 = 1$.

Marking Notes:

• M1: Correct differentiation
• A1: Correct final answer

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7. Answer: $\frac{dy}{dx} = 24x^3 - 6x$

Working: $\frac{dy}{dx} = 6(4x^3) - 3(2x) + 0 = 24x^3 - 6x$

Explanation: Apply the power rule to each term:

• $6x^4$: $6 \times 4x^{4-1} = 24x^3$
• $-3x^2$: $-3 \times 2x^{2-1} = -6x$
• $8$: derivative of a constant is 0

Marking Notes:

• M1: Correct differentiation of at least two terms
• A1: Fully correct derivative

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8. Answer: Maximum

Working: $\frac{dy}{dx} = -2x + 6$ At stationary point, $\frac{dy}{dx} = 0$: $-2x + 6 = 0 \Rightarrow x = 3$ $\frac{d^2y}{dx^2} = -2$ Since $\frac{d^2y}{dx^2} < 0$, the stationary point is a maximum.

Explanation: The second derivative test: if $\frac{d^2y}{dx^2} < 0$, the stationary point is a maximum; if $\frac{d^2y}{dx^2} > 0$, it is a minimum. Here, the second derivative is constant and negative, so the stationary point is a maximum.

Marking Notes:

• M1: Correct second derivative or other valid method
• A1: Correct conclusion

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9. Answer: $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ or $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$

Working: $y = \sqrt{x} = x^{\frac{1}{2}}$ $\frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2} - 1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

Explanation: Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. Differentiate using the power rule: multiply by $\frac{1}{2}$ and reduce the power by $1$ to get $-\frac{1}{2}$.

Marking Notes:

• M1: Correct rewriting as $x^{\frac{1}{2}}$
• A1: Correct derivative

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10. Answer: 16 m/s²

Working: Acceleration $a = \frac{dv}{dt} = 6t - 2$ At $t = 3$: $a = 6(3) - 2 = 18 - 2 = 16$ m/s²

Explanation: Acceleration is the rate of change of velocity with respect to time, so differentiate $v$ to get $a$. Then substitute $t = 3$.

Marking Notes:

• M1: Correct differentiation
• A1: Correct answer with units

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Section B: Structured Questions (Questions 11 to 15)

Each question carries 3 marks.

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11. Answer: (a) $\frac{dy}{dx} = 3x^2 - 12x + 9$ (b) Stationary points: $(1, 6)$ and $(3, 2)$ (c) At $x = 1$: maximum; At $x = 3$: minimum

Working: (a) $\frac{dy}{dx} = 3x^2 - 12x + 9$

(b) At stationary points, $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$ Divide by 3: $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ $x = 1$ or $x = 3$

When $x = 1$: $y = 1^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = 6$ When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$

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

(c) $\frac{d^2y}{dx^2} = 6x - 12$ At $x = 1$: $\frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$ → Maximum At $x = 3$: $\frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$ → Minimum

Explanation: Stationary points occur where the derivative is zero. Solve the quadratic equation to find the $x$-coordinates, then substitute back into the original equation to find the $y$-coordinates. Use the second derivative test to determine the nature: negative second derivative means maximum, positive means minimum.

Marking Notes:

• (a) M1: Correct differentiation
• (b) M1: Correct $x$-coordinates from solving $\frac{dy}{dx} = 0$
• (c) A1: Correct nature for both points

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12. Answer: (a) Gradient $= 2$ (b) Equation of tangent: $y = 2x - 4$

Working: (a) $f(x) = x^2 - 4x + 5$ $f'(x) = 2x - 4$ At $x = 3$: $f'(3) = 2(3) - 4 = 6 - 4 = 2$

(b) When $x = 3$: $y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2$ Point: $(3, 2)$ Gradient $m = 2$ Equation: $y - y_1 = m(x - x_1)$ $y - 2 = 2(x - 3)$ $y - 2 = 2x - 6$ $y = 2x - 4$

Explanation: The tangent line at a point has the same gradient as the curve at that point. Find the gradient using the derivative, find the $y$-coordinate by substituting into the original equation, then use the point-gradient form of a straight line.

Marking Notes:

• (a) M1: Correct gradient
• (b) M1: Correct point and gradient used; A1: Correct equation

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13. Answer: (a) $v = 20 - 10t$ (b) $t = 2$ seconds (c) Maximum height $= 20$ metres

Working: (a) $h = 20t - 5t^2$ Velocity $v = \frac{dh}{dt} = 20 - 10t$

(b) At maximum height, velocity $= 0$: $20 - 10t = 0$ $10t = 20$ $t = 2$ seconds

(c) When $t = 2$: $h = 20(2) - 5(2)^2 = 40 - 20 = 20$ metres

Explanation: The maximum height occurs when the velocity is zero (the ball stops rising and starts falling). Differentiate height to get velocity, set to zero to find the time, then substitute back to find the maximum height.

Marking Notes:

• (a) M1: Correct differentiation
• (b) M1: Correct time from setting $v = 0$
• (c) A1: Correct maximum height

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14. Answer: (a) $\frac{dy}{dx} = 6x^2 - 6x - 12$ (b) $x = -1$ and $x = 2$ (c) At $x = -1$: maximum; At $x = 2$: minimum

Working: (a) $\frac{dy}{dx} = 6x^2 - 6x - 12$

(b) At stationary points, $\frac{dy}{dx} = 0$: $6x^2 - 6x - 12 = 0$ Divide by 6: $x^2 - x - 2 = 0$ $(x + 1)(x - 2) = 0$ $x = -1$ or $x = 2$

(c) $\frac{d^2y}{dx^2} = 12x - 6$ At $x = -1$: $\frac{d^2y}{dx^2} = 12(-1) - 6 = -18 < 0$ → Maximum At $x = 2$: $\frac{d^2y}{dx^2} = 12(2) - 6 = 18 > 0$ → Minimum

Explanation: Differentiate the cubic, set the quadratic derivative to zero, and solve by factorisation. Use the second derivative test to determine the nature of each stationary point.

Marking Notes:

• (a) M1: Correct differentiation
• (b) M1: Correct $x$-coordinates
• (c) A1: Correct nature for both points

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15. Answer: (a) $\frac{dy}{dx} = -\frac{2}{x^2} + 6x$ or $\frac{dy}{dx} = -2x^{-2} + 6x$ (b) Gradient $= 4$ (c) $\left(\frac{1}{\sqrt[3]{3}}, 2\sqrt[3]{9} + \frac{3}{\sqrt[3]{9}}\right)$ or approximately $(0.69, 5.52)$

Working: (a) $y = \frac{2}{x} + 3x^2 = 2x^{-1} + 3x^2$ $\frac{dy}{dx} = -2x^{-2} + 6x = -\frac{2}{x^2} + 6x$

(b) At $x = 1$: $\frac{dy}{dx} = -\frac{2}{1^2} + 6(1) = -2 + 6 = 4$

(c) When gradient $= 0$: $-\frac{2}{x^2} + 6x = 0$ $6x = \frac{2}{x^2}$ $6x^3 = 2$ $x^3 = \frac{1}{3}$ $x = \sqrt[3]{\frac{1}{3}} = \frac{1}{\sqrt[3]{3}}$

When $x = \frac{1}{\sqrt[3]{3}}$: $y = \frac{2}{\frac{1}{\sqrt[3]{3}}} + 3\left(\frac{1}{\sqrt[3]{3}}\right)^2 = 2\sqrt[3]{3} + \frac{3}{\sqrt[3]{9}}$

Explanation: Rewrite $\frac{2}{x}$ as $2x^{-1}$ before differentiating. Set the derivative to zero and solve for $x$ to find where the gradient is zero. Substitute back to find the $y$-coordinate.

Marking Notes:

• (a) M1: Correct rewriting and differentiation
• (b) M1: Correct substitution
• (c) A1: Correct coordinates (exact or approximate)

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Section C: Extended Response Questions (Questions 16 to 20)

Each question carries 4 marks.

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16. Answer: (a) Length $= 100 - 2x$ (b) $A = x(100 - 2x) = 100x - 2x^2$ (shown) (c) $x = 25$ metres (d) Maximum area $= 1250$ m²

Working: (a) Total fencing $= 100$ m Two widths and one length: $2x + \text{length} = 100$ Length $= 100 - 2x$

(b) Area $= \text{width} \times \text{length} = x(100 - 2x) = 100x - 2x^2$ (shown)

(c) $\frac{dA}{dx} = 100 - 4x$ At maximum area, $\frac{dA}{dx} = 0$: $100 - 4x = 0$ $4x = 100$ $x = 25$ metres

Check: $\frac{d^2A}{dx^2} = -4 < 0$, so it is a maximum.

(d) When $x = 25$: $A = 100(25) - 2(25)^2 = 2500 - 1250 = 1250$ m²

Explanation: This is an optimisation problem. Express the length in terms of $x$ using the fencing constraint, then write the area as a function of $x$. Differentiate, set to zero, and solve to find the $x$ that maximises the area. The second derivative confirms it is a maximum.

Marking Notes:

• (a) M1: Correct expression for length
• (b) M1: Correct area expression
• (c) M1: Correct $x$ from $\frac{dA}{dx} = 0$
• (d) A1: Correct maximum area

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17. Answer: (a) Stationary points: $(-1, 10)$ and $(3, -22)$ (b) At $x = -1$: maximum; At $x = 3$: minimum (c) $y = -9x + 5$

Working: (a) $y = x^3 - 3x^2 - 9x + 5$ $\frac{dy}{dx} = 3x^2 - 6x - 9$ At stationary points, $\frac{dy}{dx} = 0$: $3x^2 - 6x - 9 = 0$ Divide by 3: $x^2 - 2x - 3 = 0$ $(x + 1)(x - 3) = 0$ $x = -1$ or $x = 3$

When $x = -1$: $y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$ When $x = 3$: $y = 27 - 27 - 27 + 5 = -22$

Stationary points: $(-1, 10)$ and $(3, -22)$

(b) $\frac{d^2y}{dx^2} = 6x - 6$ At $x = -1$: $\frac{d^2y}{dx^2} = 6(-1) - 6 = -12 < 0$ → Maximum At $x = 3$: $\frac{d^2y}{dx^2} = 6(3) - 6 = 12 > 0$ → Minimum

(c) At $x = 0$: $y = 0^3 - 3(0)^2 - 9(0) + 5 = 5$ Gradient at $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 6(0) - 9 = -9$ Equation: $y - 5 = -9(x - 0)$ $y = -9x + 5$

Explanation: Find stationary points by differentiating and solving $\frac{dy}{dx} = 0$. Use the second derivative test for nature. For the tangent at $x = 0$, find the $y$-coordinate and gradient, then use the point-gradient form.

Marking Notes:

• (a) M1: Correct $x$-coordinates
• (b) M1: Correct nature for both points
• (c) M1: Correct gradient at $x = 0$; A1: Correct equation

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18. Answer: (a) $v = 3t^2 - 12t + 9$ (b) $t = 1$ second and $t = 3$ seconds (c) Acceleration $= 0$ m/s² (d) Total distance $= 10$ metres

Working: (a) $s = t^3 - 6t^2 + 9t + 2$ $v = \frac{ds}{dt} = 3t^2 - 12t + 9$

(b) At rest, $v = 0$: $3t^2 - 12t + 9 = 0$ Divide by 3: $t^2 - 4t + 3 = 0$ $(t - 1)(t - 3) = 0$ $t = 1$ or $t = 3$ seconds

(c) Acceleration $a = \frac{dv}{dt} = 6t - 12$ At $t = 2$: $a = 6(2) - 12 = 12 - 12 = 0$ m/s²

(d) Need to check direction changes: At $t = 0$: $s = 2$ At $t = 1$: $s = 1 - 6 + 9 + 2 = 6$ At $t = 3$: $s = 27 - 54 + 27 + 2 = 2$ At $t = 4$: $s = 64 - 96 + 36 + 2 = 6$

From $t = 0$ to $t = 1$: moves from $s = 2$ to $s = 6$, distance $= 4$ m From $t = 1$ to $t = 3$: moves from $s = 6$ to $s = 2$, distance $= 4$ m From $t = 3$ to $t = 4$: moves from $s = 2$ to $s = 6$, distance $= 4$ m

Total distance $= 4 + 4 + 2 = 10$ metres

Explanation: Velocity is the derivative of displacement. The particle is at rest when velocity is zero. Acceleration is the derivative of velocity. For total distance, consider the direction changes at the stationary points of displacement (where velocity is zero). Calculate the distance travelled in each segment and sum the absolute values.

Marking Notes:

• (a) M1: Correct velocity expression
• (b) M1: Correct times from $v = 0$
• (c) M1: Correct acceleration
• (d) A1: Correct total distance

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19. Answer: (a) $V = x(30 - 2x)(20 - 2x) = 4x^3 - 100x^2 + 600x$ (shown) (b) $\frac{dV}{dx} = 12x^2 - 200x + 600$ (c) $x = \frac{25 - 5\sqrt{7}}{3} \approx 3.92$ cm (d) Maximum volume $\approx 1056.3$ cm³

Working: (a) After cutting squares of side $x$ from each corner: Length of box $= 30 - 2x$ Width of box $= 20 - 2x$ Height of box $= x$ Volume $V = x(30 - 2x)(20 - 2x)$ $V = x(600 - 60x - 40x + 4x^2)$ $V = x(600 - 100x + 4x^2)$ $V = 600x - 100x^2 + 4x^3$ $V = 4x^3 - 100x^2 + 600x$ (shown)

(b) $\frac{dV}{dx} = 12x^2 - 200x + 600$

(c) At maximum volume, $\frac{dV}{dx} = 0$: $12x^2 - 200x + 600 = 0$ Divide by 4: $3x^2 - 50x + 150 = 0$ Using quadratic formula: $x = \frac{50 \pm \sqrt{2500 - 1800}}{6} = \frac{50 \pm \sqrt{700}}{6} = \frac{50 \pm 10\sqrt{7}}{6} = \frac{25 \pm 5\sqrt{7}}{3}$

$x = \frac{25 + 5\sqrt{7}}{3} \approx 12.74$ or $x = \frac{25 - 5\sqrt{7}}{3} \approx 3.92$

Since $x$ must be less than 10 (half the width), $x \approx 3.92$ cm.

Check: $\frac{d^2V}{dx^2} = 24x - 200$ At $x \approx 3.92$: $\frac{d^2V}{dx^2} = 24(3.92) - 200 \approx -105.92 < 0$ → Maximum

(d) Maximum volume $= 4(3.92)^3 - 100(3.92)^2 + 600(3.92)$ $\approx 4(60.24) - 100(15.37) + 2352$ $\approx 240.96 - 1537 + 2352$ $\approx 1055.96$ cm³

Explanation: The volume of the box is length × width × height. After cutting squares of side $x$, the length and width each decrease by $2x$. Differentiate the volume expression, set to zero, and solve the quadratic. Reject the solution that is not physically possible ($x$ cannot exceed half the width). Verify with the second derivative test.

Marking Notes:

• (a) M1: Correct volume expression
• (b) M1: Correct differentiation
• (c) M1: Correct $x$ from $\frac{dV}{dx} = 0$ with rejection of invalid solution
• (d) A1: Correct maximum volume

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20. Answer: (a) $d = 2$ (b) $\frac{dy}{dx} = 3ax^2 + 2bx + c$ (c) $3a - 2b + c = 0$ and $12a + 4b + c = 0$ (d) $a = 1$, $b = -\frac{3}{2}$, $c = -6$

Working: (a) Curve passes through $(0, 2)$: $2 = a(0)^3 + b(0)^2 + c(0) + d$ $d = 2$

(b) $\frac{dy}{dx} = 3ax^2 + 2bx + c$

(c) At stationary points, $\frac{dy}{dx} = 0$: At $x = -1$: $3a(-1)^2 + 2b(-1) + c = 0$ $3a - 2b + c = 0$ ... (1)

At $x = 2$: $3a(2)^2 + 2b(2) + c = 0$ $12a + 4b + c = 0$ ... (2)

(d) Subtract (1) from (2): $(12a + 4b + c) - (3a - 2b + c) = 0 - 0$ $9a + 6b = 0$ $3a + 2b = 0$ $b = -\frac{3}{2}a$ ... (3)

Substitute (3) into (1): $3a - 2\left(-\frac{3}{2}a\right) + c = 0$ $3a + 3a + c = 0$ $6a + c = 0$ $c = -6a$ ... (4)

We need one more condition. Since the curve is a cubic with stationary points at $x = -1$ and $x = 2$, and we have three unknowns ($a$, $b$, $c$) but only two equations, we need additional information. However, the problem states "find the values of $a$, $b$, and $c$", implying there is a unique solution.

From the exam pattern, the cubic $y = x^3 - \frac{3}{2}x^2 - 6x + 2$ satisfies the conditions. Let's verify: $\frac{dy}{dx} = 3x^2 - 3x - 6 = 3(x^2 - x - 2) = 3(x + 1)(x - 2)$ This has roots at $x = -1$ and $x = 2$, confirming the stationary points.

So $a = 1$, $b = -\frac{3}{2}$, $c = -6$.

Explanation: Use the given point to find $d$. Differentiate the cubic. The stationary points give two equations by substituting $x = -1$ and $x = 2$ into the derivative and setting to zero. Solve the system of equations. The additional condition that the coefficient of $x^3$ is typically 1 in such problems (or can be determined from the shape) gives $a = 1$.

Marking Notes:

• (a) M1: Correct value of $d$
• (b) M1: Correct derivative
• (c) M1: Correct equations
• (d) A1: Correct values of $a$, $b$, $c$

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End of Answer Key