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Secondary 3 Additional Mathematics Calculus Quiz

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Secondary 3 Additional Mathematics Quiz - Calculus

Name: _________________________
Class: _________________________
Date: _________________________
Score: _______ / 60

Duration: 60 Minutes
Total Marks: 60

Instructions:

Answer all questions.
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 graphing calculator is expected.

Section A: Differentiation Techniques (Questions 1–5)

[20 Marks]

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

(b) $y = \sqrt{x}(3x - 2)$ [3]

2. Given that $y = (2x^2 + 1)(x - 3)$ , find $\frac{dy}{dx}$ using the product rule. Simplify your answer. [3]

3. Differentiate $y = \frac{3x + 1}{x^2 - 2}$ with respect to $x$ , giving your answer in the form $\frac{Ax^2 + Bx + C}{(x^2 - 2)^2}$ . [4]

4. Given $y = \sin(3x^2 + 1)$ , find $\frac{dy}{dx}$ . [3]

5. Find the equation of the tangent to the curve $y = e^{2x} - 4x$ at the point where $x = 0$ . [5]

Section B: Applications of Differentiation (Questions 6–10)

[20 Marks]

6. A curve has equation $y = x^3 - 6x^2 + 9x + 2$ . (a) Find the coordinates of the stationary points. [4]

(b) Determine the nature of each stationary point using the second derivative. [3]

7. The volume $V$ cm $^3$ of a sphere is increasing at a constant rate of 10 cm $^3$ s $^{-1}$ . Find the rate of increase of the radius $r$ when $r = 5$ cm. [4] (Note: $V = \frac{4}{3}\pi r^3$ )

8. A rectangular sheet of metal measuring 20 cm by 12 cm has squares of side $x$ cm cut from each corner. The sides are then folded up to form an open box. (a) Show that the volume of the box is given by $V = 4x^3 - 64x^2 + 240x$ . [2]

(b) Find the value of $x$ for which the volume is a maximum. [4]

9. Given that $y = x^2 \ln x$ , find the value of $x$ for which $\frac{dy}{dx} = 0$ . [3]

10. The displacement $s$ metres of a particle 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 $t = 2$ . [2]

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

Section C: Integration and Area (Questions 11–15)

[12 Marks]

11. Find the following indefinite integrals: (a) $\int (3x^2 - 4x + 5) \, dx$ [2]

(b) $\int (2\cos x - 3\sin x) \, dx$ [2]

12. Evaluate $\int_{1}^{2} \left( \frac{1}{x} + 2x \right) \, dx$ . [3]

13. Given that $\frac{dy}{dx} = 6x - 4$ and the curve passes through the point $(1, 5)$ , find the equation of the curve. [3]

14. Find $\int (3x + 1)^4 \, dx$ . [2]

15. The gradient of a curve is given by $\frac{dy}{dx} = e^{2x}$ . If the curve passes through $(0, 3)$ , find the equation of the curve. [3]

Section D: Definite Integrals and Area (Questions 16–20)

[8 Marks]

16. Calculate the area of the region bounded by the curve $y = x^2$ , the x-axis, and the lines $x = 1$ and $x = 3$ . [2]

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

18. Find the area of the region enclosed by the curve $y = 4 - x^2$ and the x-axis. [2]

19. Given $\int_{1}^{k} (2x + 1) \, dx = 15$ , find the positive value of $k$ . [2]

20. The curve $y = x^3 - 4x$ crosses the x-axis at $x = 0$ and $x = 2$ (for $x \ge 0$ ). Calculate the area of the finite region bounded by the curve and the x-axis between these points. [2] (Note: Consider the position of the curve relative to the x-axis)

*** End of Quiz ***

Answers

Secondary 3 Additional Mathematics Quiz - Calculus (Answer Key)

1. (a) $y = 4x^3 - 2x^{-2} + 5$ $\frac{dy}{dx} = 12x^2 - 2(-2)x^{-3} = 12x^2 + \frac{4}{x^3}$ [2]

(b) $y = 3x^{3/2} - 2x^{1/2}$ $\frac{dy}{dx} = 3(\frac{3}{2})x^{1/2} - 2(\frac{1}{2})x^{-1/2} = \frac{9}{2}\sqrt{x} - \frac{1}{\sqrt{x}}$ Or $\frac{9x - 2}{2\sqrt{x}}$ [3]

2. Let $u = 2x^2 + 1 \Rightarrow \frac{du}{dx} = 4x$ Let $v = x - 3 \Rightarrow \frac{dv}{dx} = 1$ $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ $= (2x^2 + 1)(1) + (x - 3)(4x)$ $= 2x^2 + 1 + 4x^2 - 12x$ $= 6x^2 - 12x + 1$ [3]

3. $u = 3x + 1 \Rightarrow u' = 3$ $v = x^2 - 2 \Rightarrow v' = 2x$ $\frac{dy}{dx} = \frac{v u' - u v'}{v^2} = \frac{(x^2 - 2)(3) - (3x + 1)(2x)}{(x^2 - 2)^2}$ $= \frac{3x^2 - 6 - (6x^2 + 2x)}{(x^2 - 2)^2}$ $= \frac{-3x^2 - 2x - 6}{(x^2 - 2)^2}$ [4] (A=-3, B=-2, C=-6)

4. Let $u = 3x^2 + 1$ , then $y = \sin u$ . $\frac{dy}{du} = \cos u$ , $\frac{du}{dx} = 6x$ . $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 6x \cos(3x^2 + 1)$ [3]

5. $y = e^{2x} - 4x$ $\frac{dy}{dx} = 2e^{2x} - 4$ At $x = 0$ : $y = e^0 - 0 = 1$ . Point is $(0, 1)$ . Gradient $m = 2e^0 - 4 = 2 - 4 = -2$ . Equation: $y - 1 = -2(x - 0) \Rightarrow y = -2x + 1$ [5]

6. (a) $\frac{dy}{dx} = 3x^2 - 12x + 9$ At stationary points, $\frac{dy}{dx} = 0$ : $3(x^2 - 4x + 3) = 0$ $3(x - 3)(x - 1) = 0$ $x = 1$ or $x = 3$ . When $x = 1, y = 1 - 6 + 9 + 2 = 6$ . Point $(1, 6)$ . When $x = 3, y = 27 - 54 + 27 + 2 = 2$ . Point $(3, 2)$ . [4]

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

7. $V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2$ Given $\frac{dV}{dt} = 10$ . Chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$ $10 = 4\pi (5)^2 \times \frac{dr}{dt}$ $10 = 100\pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}$ cm s $^{-1}$ ( $\approx 0.0318$ ) [4]

8. (a) Dimensions of box: Length $= 20 - 2x$ , Width $= 12 - 2x$ , Height $= x$ . $V = x(20 - 2x)(12 - 2x)$ $V = x(240 - 40x - 24x + 4x^2)$ $V = x(240 - 64x + 4x^2)$ $V = 240x - 64x^2 + 4x^3$ (Rearranged: $4x^3 - 64x^2 + 240x$ ) [2]

(b) $\frac{dV}{dx} = 12x^2 - 128x + 240$ For max/min, $\frac{dV}{dx} = 0$ : $12x^2 - 128x + 240 = 0$ Divide by 4: $3x^2 - 32x + 60 = 0$ $x = \frac{32 \pm \sqrt{32^2 - 4(3)(60)}}{6} = \frac{32 \pm \sqrt{1024 - 720}}{6} = \frac{32 \pm \sqrt{304}}{6}$ $x \approx \frac{32 \pm 17.436}{6}$ $x_1 \approx 8.24$ (Reject, as width $12-2x$ would be negative) $x_2 \approx 2.43$ Check second derivative or logic: $x \approx 2.43$ cm gives max volume. [4] (Exact form: $x = \frac{16 - 2\sqrt{19}}{3}$ )

9. $y = x^2 \ln x$ Product rule: $u=x^2, v=\ln x$ . $\frac{dy}{dx} = 2x \ln x + x^2 (\frac{1}{x}) = 2x \ln x + x$ Set $\frac{dy}{dx} = 0$ : $x(2 \ln x + 1) = 0$ Since $x>0$ for $\ln x$ , $2 \ln x + 1 = 0$ $\ln x = -0.5$ $x = e^{-0.5} = \frac{1}{\sqrt{e}}$ [3]

10. $s = t^3 - 6t^2 + 9t$ $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ $a = \frac{dv}{dt} = 6t - 12$

(a) At $t=2$ : $v = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ m s $^{-1}$ . [2] (b) At $t=2$ : $a = 6(2) - 12 = 0$ m s $^{-2}$ . [2]

11. (a) $\int (3x^2 - 4x + 5) \, dx = x^3 - 2x^2 + 5x + C$ [2] (b) $\int (2\cos x - 3\sin x) \, dx = 2\sin x + 3\cos x + C$ [2]

12. $\int_{1}^{2} (x^{-1} + 2x) \, dx = [\ln|x| + x^2]_{1}^{2}$ $= (\ln 2 + 2^2) - (\ln 1 + 1^2)$ $= \ln 2 + 4 - 0 - 1$ $= 3 + \ln 2$ [3]

13. $y = \int (6x - 4) \, dx = 3x^2 - 4x + C$ Substitute $(1, 5)$ : $5 = 3(1)^2 - 4(1) + C$ $5 = 3 - 4 + C \Rightarrow 5 = -1 + C \Rightarrow C = 6$ Equation: $y = 3x^2 - 4x + 6$ [3]

14. Let $u = 3x + 1$ , then $du = 3 dx \Rightarrow dx = \frac{1}{3} du$ . $\int u^4 \frac{1}{3} du = \frac{1}{3} \frac{u^5}{5} + C = \frac{(3x+1)^5}{15} + C$ [2]

15. $y = \int e^{2x} \, dx = \frac{1}{2}e^{2x} + C$ Substitute $(0, 3)$ : $3 = \frac{1}{2}e^0 + C \Rightarrow 3 = 0.5 + C \Rightarrow C = 2.5$ Equation: $y = \frac{1}{2}e^{2x} + 2.5$ [3]

16. Area $= \int_{1}^{3} x^2 \, dx = [\frac{x^3}{3}]_{1}^{3}$ $= \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ or $8.67$ [2]

17. $\int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx = [-\frac{1}{2}\cos(2x)]_{0}^{\frac{\pi}{2}}$ $= -\frac{1}{2}(\cos \pi - \cos 0)$ $= -\frac{1}{2}(-1 - 1) = -\frac{1}{2}(-2) = 1$ [2]

18. Intercepts: $4 - x^2 = 0 \Rightarrow x = \pm 2$ . Area $= \int_{-2}^{2} (4 - x^2) \, dx$ Due to symmetry: $2 \int_{0}^{2} (4 - x^2) \, dx$ $= 2 [4x - \frac{x^3}{3}]_{0}^{2}$ $= 2 [(8 - \frac{8}{3}) - 0] = 2 [\frac{16}{3}] = \frac{32}{3}$ or $10.67$ [2]

19. $\int_{1}^{k} (2x + 1) \, dx = [x^2 + x]_{1}^{k}$ $= (k^2 + k) - (1^2 + 1) = k^2 + k - 2$ $k^2 + k - 2 = 15$ $k^2 + k - 17 = 0$ $k = \frac{-1 \pm \sqrt{1 - 4(1)(-17)}}{2} = \frac{-1 \pm \sqrt{69}}{2}$ Since $k$ is positive (and upper limit > lower limit 1 usually implied, but strictly $k>0$ ): $k = \frac{-1 + \sqrt{69}}{2} \approx 3.65$ [2]

20. Curve $y = x(x^2 - 4)$ . Between $x=0$ and $x=2$ , test $x=1 \Rightarrow y = -3$ . Curve is below axis. Area $= |\int_{0}^{2} (x^3 - 4x) \, dx|$ $= |[\frac{x^4}{4} - 2x^2]_{0}^{2}|$ $= |(\frac{16}{4} - 2(4)) - 0| = |4 - 8| = |-4| = 4$ [2]