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Secondary 3 Additional Mathematics Calculus Quiz

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Questions

Secondary 3 Additional Mathematics Quiz - Calculus

Name: __________________________
Class: __________________________
Date: __________________________
Score: _________ / 60

Duration: 60 Minutes
Total Marks: 60

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly. No marks will be given for correct answers without working.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
The use of an approved scientific calculator is expected, where appropriate.

Section A: Differentiation Fundamentals (Questions 1–5)

[15 Marks]

1. Differentiate the following expressions with respect to $x$ : (a) $y = 4x^3 - 7x^2 + 5x - 2$ [2]

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

2. Given that $y = (2x - 1)(x + 3)$ , find $\frac{dy}{dx}$ by: (a) Expanding the brackets first. [2]

(b) Using the product rule. Verify that your answer matches part (a). [3]

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

4. The curve $y = 2x^2 + kx + 5$ has a stationary point at $x = -2$ . Find the value of $k$ . [3]

5. Given $y = \sin(3x) + \cos(2x)$ , find $\frac{dy}{dx}$ . [3]

Section B: Advanced Differentiation & Applications (Questions 6–10)

[15 Marks]

6. Differentiate $y = \frac{e^x}{x^2}$ with respect to $x$ , giving your answer in its simplest form. [4]

7. Given that $y = \ln(5x^2 + 1)$ , find the value of $\frac{dy}{dx}$ when $x = 1$ . [3]

8. A curve is defined by the parametric equations $x = t^2$ and $y = 2t - 1$ . (a) Find $\frac{dy}{dx}$ in terms of $t$ . [2]

(b) Hence, find the equation of the normal to the curve at the point where $t = 2$ . [4]

9. The volume $V$ cm $^3$ of a sphere is increasing at a constant rate of $10$ cm $^3$ s $^{-1}$ . Given that $V = \frac{4}{3}\pi r^3$ , find the rate of increase of the radius $r$ when $r = 5$ cm. [4]

10. Find the coordinates of the stationary points of the curve $y = x^3 - 6x^2 + 9x + 2$ and determine the nature of each point. [5]

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Section C: Integration Fundamentals (Questions 11–15)

[15 Marks]

11. Find the indefinite integrals: (a) $\int (6x^2 - 4x + 3) \, dx$ [2]

(b) $\int \left( \frac{1}{x} + 2e^{3x} \right) \, dx$ [3]

12. Given that $\frac{dy}{dx} = 3x^2 - 4x$ and that $y = 5$ when $x = 1$ , find $y$ in terms of $x$ . [4]

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13. Evaluate the definite integral $\int_{1}^{2} (3x^2 + 2x) \, dx$ . [3]

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14. Find the exact area of the region bounded by the curve $y = 4 - x^2$ , the x-axis, and the lines $x = 0$ and $x = 1$ . [4]

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15. The acceleration of a particle moving in a straight line is given by $a = 6t - 4$ m s $^{-2}$ . At $t = 0$ , the velocity is $3$ m s $^{-1}$ . Find an expression for the velocity $v$ in terms of $t$ . [3]

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Section D: Advanced Integration & Applications (Questions 16–20)

[15 Marks]

16. Find $\int (2x + 1)^5 \, dx$ . [3]

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17. Evaluate $\int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx$ . [3]

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18. The diagram shows the curve $y = x(x-2)(x-4)$ . (a) State the x-intercepts of the curve. [1]

(b) Calculate the total area of the finite regions bounded by the curve and the x-axis. [5]

19. A particle moves in a straight line such that its displacement $s$ metres from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 6t^2 + 9t$ . (a) Find the velocity of the particle when $t = 4$ . [2]

(b) Find the acceleration of the particle when $t = 4$ . [2]

20. The curve $y = \frac{1}{\sqrt{3x+1}}$ passes through the point $(1, 0.5)$ . (a) Find $\frac{dy}{dx}$ . [2]

(b) Hence, or otherwise, evaluate $\int_{0}^{1} \frac{1}{\sqrt{3x+1}} \, dx$ . [3]

End of Quiz

Answers

Secondary 3 Additional Mathematics Quiz - Calculus (Answer Key)

1. (a) $\frac{dy}{dx} = 12x^2 - 14x + 5$ [B1 for $12x^2$ , B1 for $-14x+5$ ] (b) Rewrite $y = 3x^{-2} + 4x^{1/2}$ . $\frac{dy}{dx} = -6x^{-3} + 2x^{-1/2}$ $= -\frac{6}{x^3} + \frac{2}{\sqrt{x}}$ [M1 for power rule application, A1 for simplification]

2. (a) $y = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$ . $\frac{dy}{dx} = 4x + 5$ [M1 for expansion, A1 for differentiation] (b) Let $u = 2x-1, v = x+3$ . $u' = 2, v' = 1$ . $\frac{dy}{dx} = u v' + v u' = (2x-1)(1) + (x+3)(2) = 2x - 1 + 2x + 6 = 4x + 5$ . [M1 for product rule setup, A1 for correct result]

3. When $x=1, y = 1 - 2 + 4 = 3$ . Point $(1,3)$ . $\frac{dy}{dx} = 3x^2 - 4x$ . Gradient $m = 3(1)^2 - 4(1) = -1$ . Equation: $y - 3 = -1(x - 1) \Rightarrow y = -x + 4$ or $x + y = 4$ . [M1 for y-coord, M1 for gradient, A1 for equation]

4. $\frac{dy}{dx} = 4x + k$ . At stationary point, $\frac{dy}{dx} = 0$ . $4(-2) + k = 0 \Rightarrow -8 + k = 0 \Rightarrow k = 8$ . [M1 for derivative, M1 for setting to 0, A1 for k]

5. $\frac{dy}{dx} = 3\cos(3x) - 2\sin(2x)$ . [B1 for chain rule on sin, B1 for chain rule on cos, A1 for signs]

6. Using Quotient Rule: $u = e^x, v = x^2$ . $u' = e^x, v' = 2x$ . $\frac{dy}{dx} = \frac{x^2(e^x) - e^x(2x)}{(x^2)^2} = \frac{e^x(x^2 - 2x)}{x^4} = \frac{e^x(x-2)}{x^3}$ . [M1 for quotient rule, M1 for simplification, A1 for final form]

7. $\frac{dy}{dx} = \frac{1}{5x^2+1} \times (10x) = \frac{10x}{5x^2+1}$ . When $x=1$ , $\frac{dy}{dx} = \frac{10(1)}{5(1)+1} = \frac{10}{6} = \frac{5}{3}$ . [M1 for chain rule, M1 for substitution, A1 for value]

8. (a) $\frac{dx}{dt} = 2t, \frac{dy}{dt} = 2$ . $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$ . [M1 for derivatives, A1 for ratio] (b) When $t=2$ , $x=4, y=3$ . Gradient of tangent $m = 1/2$ . Gradient of normal $m_{\perp} = -2$ . Equation: $y - 3 = -2(x - 4) \Rightarrow y = -2x + 11$ . [M1 for coords, M1 for normal gradient, A1 for equation]

9. $\frac{dV}{dt} = 10$ . $V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2$ . $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \Rightarrow 10 = 4\pi r^2 \frac{dr}{dt}$ . When $r=5$ , $10 = 4\pi(25) \frac{dr}{dt} = 100\pi \frac{dr}{dt}$ . $\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}$ cm s $^{-1}$ . [M1 for chain rule setup, M1 for substitution, A1 for answer]

10. $\frac{dy}{dx} = 3x^2 - 12x + 9$ . Set $\frac{dy}{dx} = 0 \Rightarrow 3(x^2 - 4x + 3) = 0 \Rightarrow 3(x-3)(x-1) = 0$ . $x = 1, x = 3$ . Points: $(1, 6)$ and $(3, 2)$ . $\frac{d^2y}{dx^2} = 6x - 12$ . At $x=1$ , $\frac{d^2y}{dx^2} = -6 < 0$ (Maximum). At $x=3$ , $\frac{d^2y}{dx^2} = 6 > 0$ (Minimum). [M1 for solving quadratic, A1 for coords, M1 for 2nd derivative, A1 for nature]

11. (a) $2x^3 - 2x^2 + 3x + C$ . [B1 for powers, B1 for constant] (b) $\ln|x| + \frac{2}{3}e^{3x} + C$ . [B1 for ln, B1 for exponential, B1 for constant]

12. $y = \int (3x^2 - 4x) dx = x^3 - 2x^2 + C$ . Sub $x=1, y=5$ : $5 = 1 - 2 + C \Rightarrow C = 6$ . $y = x^3 - 2x^2 + 6$ . [M1 for integration, M1 for finding C, A1 for equation]

13. $\left[ x^3 + x^2 \right]_1^2 = (2^3 + 2^2) - (1^3 + 1^2) = (8+4) - (1+1) = 12 - 2 = 10$ . [M1 for integration, M1 for substitution, A1 for answer]

14. Area $= \int_0^1 (4-x^2) dx = \left[ 4x - \frac{x^3}{3} \right]_0^1$ . $= (4(1) - \frac{1}{3}) - 0 = 3\frac{2}{3}$ or $\frac{11}{3}$ . [M1 for setup, M1 for integration, A1 for value]

15. $v = \int (6t - 4) dt = 3t^2 - 4t + C$ . At $t=0, v=3 \Rightarrow 3 = 0 - 0 + C \Rightarrow C = 3$ . $v = 3t^2 - 4t + 3$ . [M1 for integration, M1 for C, A1 for expression]

16. Let $u = 2x+1$ , then $du = 2dx$ or $dx = du/2$ . $\int u^5 \frac{du}{2} = \frac{1}{2} \frac{u^6}{6} + C = \frac{1}{12}(2x+1)^6 + C$ . [M1 for reverse chain rule, A1 for coefficient, A1 for constant]

17. $\left[ -\frac{1}{2}\cos(2x) \right]_0^{\pi/2}$ . Upper: $-\frac{1}{2}\cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2}$ . Lower: $-\frac{1}{2}\cos(0) = -\frac{1}{2}(1) = -\frac{1}{2}$ . Result: $\frac{1}{2} - (-\frac{1}{2}) = 1$ . [M1 for integration, M1 for limits, A1 for answer]

18. (a) $x = 0, 2, 4$ . [B1] (b) Area $= \int_0^2 (x^3 - 6x^2 + 8x) dx + |\int_2^4 (x^3 - 6x^2 + 8x) dx|$ . $\int (x^3 - 6x^2 + 8x) dx = \frac{x^4}{4} - 2x^3 + 4x^2$ . At $x=2$ : $\frac{16}{4} - 16 + 16 = 4$ . At $x=0$ : $0$ . Area 1 = 4. At $x=4$ : $\frac{256}{4} - 2(64) + 4(16) = 64 - 128 + 64 = 0$ . Area 2 = $|0 - 4| = 4$ . Total Area = $4 + 4 = 8$ . [M1 for splitting areas, M1 for integration, M1 for evaluation, A1 for total]

19. (a) $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ . $v(4) = 3(16) - 48 + 9 = 48 - 48 + 9 = 9$ m s $^{-1}$ . [M1 for diff, A1 for value] (b) $a = \frac{dv}{dt} = 6t - 12$ . $a(4) = 24 - 12 = 12$ m s $^{-2}$ . [M1 for diff, A1 for value] (c) Check for turning points in $0 \le t \le 4$ . $v = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow t=1, 3$ . $s(0) = 0$ . $s(1) = 1 - 6 + 9 = 4$ . Dist = 4. $s(3) = 27 - 54 + 27 = 0$ . Dist from $t=1$ to $3$ is $|0-4|=4$ . $s(4) = 64 - 96 + 36 = 4$ . Dist from $t=3$ to $4$ is $|4-0|=4$ . Total distance = $4 + 4 + 4 = 12$ m. [M1 for finding turning points, M1 for calculating positions, A1 for summing distances]

20. (a) $y = (3x+1)^{-1/2}$ . $\frac{dy}{dx} = -\frac{1}{2}(3x+1)^{-3/2} \times 3 = -\frac{3}{2}(3x+1)^{-3/2}$ . [M1 for chain rule, A1 for answer] (b) $\int (3x+1)^{-1/2} dx = \frac{(3x+1)^{1/2}}{1/2 \times 3} = \frac{2}{3}\sqrt{3x+1}$ . $\left[ \frac{2}{3}\sqrt{3x+1} \right]_0^1 = \frac{2}{3}(\sqrt{4} - \sqrt{1}) = \frac{2}{3}(2-1) = \frac{2}{3}$ . [M1 for integration, M1 for limits, A1 for answer]