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Secondary 3 Elementary Mathematics Calculus Quiz

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Questions

Secondary 3 Elementary Mathematics Quiz - Calculus

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 50 minutes
Total Marks: 45

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, unless otherwise specified in the question.
The use of an approved scientific calculator is expected.

Section A: Gradient of Curves and Tangents (15 Marks)

1. The equation of a curve is $y = x^2 - 4x + 5$ .
Find the gradient of the curve at the point where $x = 3$ .
[2]

2. A curve has the equation $y = 3x^3 - 2x$ .
Find the coordinates of the stationary points on the curve.
[3]

3. The diagram shows the graph of $y = f(x)$ .
The tangent to the curve at point $P(2, 5)$ passes through the point $(0, 1)$ .
Find the value of $f'(2)$ .
[2]

4. Find the equation of the tangent to the curve $y = x^2 + 2x$ at the point where $x = 1$ .
Give your answer in the form $y = mx + c$ .
[3]

5. The gradient of a curve is given by $\frac{dy}{dx} = 6x - 4$ .
The curve passes through the point $(1, 3)$ .
Find the equation of the curve.
[3]

6. Determine whether the stationary point on the curve $y = x^2 - 6x + 10$ is a maximum or a minimum point.
[2]

Section B: Applications of Differentiation (15 Marks)

7. The volume $V$ cm $^3$ of a cube is increasing at a constant rate of 12 cm $^3$ /s.
Find the rate of increase of the side length $x$ cm when $x = 2$ .
[3]

8. 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:
$s = t^3 - 6t^2 + 9t$
Find the acceleration of the particle when $t = 2$ .
[3]

9. Using the same displacement equation as in Question 8 ( $s = t^3 - 6t^2 + 9t$ ):
Find the values of $t$ when the particle is instantaneously at rest.
[3]

10. The area $A$ cm $^2$ of a circle is increasing. The radius $r$ cm is increasing at a rate of 0.5 cm/s.
Find the rate of increase of the area when $r = 4$ .
[3]

11. A curve has equation $y = 2x^3 - 9x^2 + 12x$ .
Find the range of values of $x$ for which $y$ is an increasing function.
[3]

Section C: Integration and Area (15 Marks)

12. Find $\int (3x^2 - 4x + 5) \, dx$ .
[2]

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

14. Evaluate $\int_{1}^{3} (2x + 1) \, dx$ .
[3]

15. The diagram shows the curve $y = x^2$ and the line $y = 4$ .
The shaded region is bounded by the curve and the line.
Find the area of the shaded region.
[4]

16. Find the exact area of the region bounded by the curve $y = x^3$ , the x-axis, and the lines $x = 0$ and $x = 2$ .
[3]

Section D: Problem Solving (Mixed Concepts)

17. A rectangle has perimeter 20 cm. Let the length be $x$ cm and the width be $y$ cm.
(a) Express $y$ in terms of $x$ .
(b) Show that the area $A$ of the rectangle is given by $A = 10x - x^2$ .
(c) Find the maximum area of the rectangle.
[4]

18. The curve $y = x^2 - 4x$ intersects the x-axis at points $A$ and $B$ .
(a) Find the coordinates of $A$ and $B$ .
(b) Find the area of the region enclosed by the curve and the x-axis.
[4]

19. The gradient of a curve is given by $\frac{dy}{dx} = kx^2$ , where $k$ is a constant.
The curve passes through the points $(1, 2)$ and $(2, 10)$ .
Find the value of $k$ and the equation of the curve.
[4]

20. Water is poured into a cylindrical tank with base radius 2 m. The volume of water $V$ m $^3$ is increasing at a rate of 0.5 m $^3$ /min.
Find the rate at which the height $h$ of the water is rising.
[3]

Answers

Secondary 3 Elementary Mathematics Quiz - Calculus (Answer Key)

1. [2 marks]
$y = x^2 - 4x + 5$
$\frac{dy}{dx} = 2x - 4$
At $x = 3$ , Gradient $= 2(3) - 4 = 6 - 4 = 2$ .
Answer: 2

2. [3 marks]
$y = 3x^3 - 2x$
$\frac{dy}{dx} = 9x^2 - 2$
At stationary points, $\frac{dy}{dx} = 0$ .
$9x^2 - 2 = 0 \Rightarrow x^2 = \frac{2}{9} \Rightarrow x = \pm \frac{\sqrt{2}}{3}$ .
When $x = \frac{\sqrt{2}}{3}$ , $y = 3(\frac{2\sqrt{2}}{27}) - 2(\frac{\sqrt{2}}{3}) = \frac{2\sqrt{2}}{9} - \frac{6\sqrt{2}}{9} = -\frac{4\sqrt{2}}{9}$ .
When $x = -\frac{\sqrt{2}}{3}$ , $y = 3(-\frac{2\sqrt{2}}{27}) - 2(-\frac{\sqrt{2}}{3}) = -\frac{2\sqrt{2}}{9} + \frac{6\sqrt{2}}{9} = \frac{4\sqrt{2}}{9}$ .
Answer: $(\frac{\sqrt{2}}{3}, -\frac{4\sqrt{2}}{9})$ and $(-\frac{\sqrt{2}}{3}, \frac{4\sqrt{2}}{9})$

3. [2 marks]
Gradient of tangent $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 1}{2 - 0} = \frac{4}{2} = 2$ .
$f'(2)$ is the gradient of the tangent at $x=2$ .
Answer: 2

4. [3 marks]
$y = x^2 + 2x$
$\frac{dy}{dx} = 2x + 2$
At $x = 1$ , $y = 1^2 + 2(1) = 3$ . Point is $(1, 3)$ .
Gradient $m = 2(1) + 2 = 4$ .
Equation: $y - 3 = 4(x - 1) \Rightarrow y = 4x - 4 + 3 \Rightarrow y = 4x - 1$ .
Answer: $y = 4x - 1$

5. [3 marks]
$\frac{dy}{dx} = 6x - 4$
$y = \int (6x - 4) \, dx = 3x^2 - 4x + c$
Substitute $(1, 3)$ : $3 = 3(1)^2 - 4(1) + c \Rightarrow 3 = 3 - 4 + c \Rightarrow 3 = -1 + c \Rightarrow c = 4$ .
Answer: $y = 3x^2 - 4x + 4$

6. [2 marks]
$y = x^2 - 6x + 10$
$\frac{dy}{dx} = 2x - 6$ . Stationary point at $2x - 6 = 0 \Rightarrow x = 3$ .
$\frac{d^2y}{dx^2} = 2$ .
Since $\frac{d^2y}{dx^2} > 0$ , it is a minimum point.
Answer: Minimum point

7. [3 marks]
$V = x^3$
$\frac{dV}{dx} = 3x^2$
Given $\frac{dV}{dt} = 12$ .
Chain rule: $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$
$12 = 3(2)^2 \times \frac{dx}{dt} \Rightarrow 12 = 12 \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = 1$ .
Answer: 1 cm/s

8. [3 marks]
$s = t^3 - 6t^2 + 9t$
Velocity $v = \frac{ds}{dt} = 3t^2 - 12t + 9$
Acceleration $a = \frac{dv}{dt} = 6t - 12$
At $t = 2$ , $a = 6(2) - 12 = 12 - 12 = 0$ .
Answer: 0 m/s $^2$

9. [3 marks]
Particle at rest when $v = 0$ .
$3t^2 - 12t + 9 = 0$
Divide by 3: $t^2 - 4t + 3 = 0$
$(t - 3)(t - 1) = 0$
$t = 1$ or $t = 3$ .
Answer: $t = 1, 3$

10. [3 marks]
$A = \pi r^2$
$\frac{dA}{dr} = 2\pi r$
Given $\frac{dr}{dt} = 0.5$ .
$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 2\pi r \times 0.5 = \pi r$ .
When $r = 4$ , $\frac{dA}{dt} = 4\pi$ .
Answer: $4\pi$ cm $^2$ /s (or approx 12.6 cm $^2$ /s)

11. [3 marks]
$y = 2x^3 - 9x^2 + 12x$
$\frac{dy}{dx} = 6x^2 - 18x + 12$
Increasing when $\frac{dy}{dx} > 0$ .
$6x^2 - 18x + 12 > 0$
Divide by 6: $x^2 - 3x + 2 > 0$
$(x - 2)(x - 1) > 0$
Critical values $x = 1, 2$ . Since quadratic opens upward, positive outside roots.
Answer: $x < 1$ or $x > 2$

12. [2 marks]
$\int (3x^2 - 4x + 5) \, dx = \frac{3x^3}{3} - \frac{4x^2}{2} + 5x + c$
Answer: $x^3 - 2x^2 + 5x + c$

13. [3 marks]
$y = \int (4x^3 - 2) \, dx = x^4 - 2x + c$
When $x = 1, y = 5$ :
$5 = 1^4 - 2(1) + c \Rightarrow 5 = 1 - 2 + c \Rightarrow 5 = -1 + c \Rightarrow c = 6$ .
Answer: $y = x^4 - 2x + 6$

14. [3 marks]
$\int_{1}^{3} (2x + 1) \, dx = [x^2 + x]_{1}^{3}$
Upper limit ( $x=3$ ): $3^2 + 3 = 12$
Lower limit ( $x=1$ ): $1^2 + 1 = 2$
Area $= 12 - 2 = 10$ .
Answer: 10

15. [4 marks]
Intersection: $x^2 = 4 \Rightarrow x = \pm 2$ .
Area $= \int_{-2}^{2} (4 - x^2) \, dx$
Due to symmetry, $= 2 \int_{0}^{2} (4 - x^2) \, dx$
$= 2 [4x - \frac{x^3}{3}]_{0}^{2}$
$= 2 [(8 - \frac{8}{3}) - 0] = 2 [\frac{24-8}{3}] = 2 [\frac{16}{3}] = \frac{32}{3}$ .
Answer: $\frac{32}{3}$ or $10.67$

16. [3 marks]
Area $= \int_{0}^{2} x^3 \, dx = [\frac{x^4}{4}]_{0}^{2}$
$= \frac{2^4}{4} - 0 = \frac{16}{4} = 4$ .
Answer: 4

17. [4 marks]
(a) Perimeter $2x + 2y = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$ .
(b) Area $A = xy = x(10 - x) = 10x - x^2$ .
(c) $\frac{dA}{dx} = 10 - 2x$ .
Max when $\frac{dA}{dx} = 0 \Rightarrow 10 - 2x = 0 \Rightarrow x = 5$ .
Max Area $A = 10(5) - 5^2 = 50 - 25 = 25$ .
Answer: 25 cm $^2$

18. [4 marks]
(a) $x^2 - 4x = 0 \Rightarrow x(x - 4) = 0$ . Points $A(0,0)$ and $B(4,0)$ .
(b) Area $= |\int_{0}^{4} (x^2 - 4x) \, dx|$
$= [\frac{x^3}{3} - 2x^2]_{0}^{4}$
$= (\frac{64}{3} - 2(16)) - 0 = \frac{64}{3} - 32 = \frac{64 - 96}{3} = -\frac{32}{3}$ .
Area is positive magnitude.
Answer: $\frac{32}{3}$ or $10.67$

19. [4 marks]
$y = \int kx^2 \, dx = \frac{kx^3}{3} + c$
Point $(1, 2)$ : $2 = \frac{k}{3} + c$ (Eq 1)
Point $(2, 10)$ : $10 = \frac{8k}{3} + c$ (Eq 2)
(Eq 2) - (Eq 1): $8 = \frac{7k}{3} \Rightarrow 24 = 7k \Rightarrow k = \frac{24}{7}$ .
Sub $k$ into Eq 1: $2 = \frac{24/7}{3} + c \Rightarrow 2 = \frac{8}{7} + c \Rightarrow c = 2 - \frac{8}{7} = \frac{6}{7}$ .
Answer: $k = \frac{24}{7}$ , Equation: $y = \frac{8}{7}x^3 + \frac{6}{7}$

20. [3 marks]
$V = \pi r^2 h$ . Since $r=2$ is constant, $V = \pi (2^2) h = 4\pi h$ .
$\frac{dV}{dh} = 4\pi$ .
Given $\frac{dV}{dt} = 0.5$ .
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$
$0.5 = 4\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{0.5}{4\pi} = \frac{1}{8\pi}$ .
Answer: $\frac{1}{8\pi}$ m/min (or approx 0.0398 m/min)