Secondary 3 Elementary Mathematics Calculus Quiz

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Secondary 3 Elementary Mathematics Quiz - Calculus

Name: ____________________ Class: __________ Date: __________ Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50 Marks

Instructions:

• Answer all questions.
• Show all necessary working.
• Give your answers to 3 significant figures where appropriate.

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Section A: Basic Differentiation (1-10)

Focus: Power rule and basic gradients.

1. Differentiate $y = 5x^3$ with respect to $x$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

2. Find $\frac{dy}{dx}$ for $y = 4x^2 - 7x + 2$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

3. Differentiate $y = \frac{1}{3}x^3 + 2x$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

4. Find the derivative of $y = 8x^{-2}$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

5. Differentiate $y = 6\sqrt{x}$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

6. Find $\frac{dy}{dx}$ for $y = (2x + 3)^2$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

7. Differentiate $y = \frac{4}{x^2}$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

8. Find the derivative of $y = 10x^{1.5}$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

9. Differentiate $y = 3x^2 - \frac{2}{x}$.

   $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

10. Find $\frac{dy}{dx}$ for $y = \pi x^2$.

    $\text{Ans: } \underline{\hspace{3cm}}$ [2m]

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Section B: Gradients and Tangents (11-15)

Focus: Application of differentiation to coordinate geometry.

11. Find the gradient of the curve $y = x^2 - 4x$ at the point $(3, -3)$.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [3m]

12. A curve has the equation $y = 2x^3 - 5x$. Find the coordinates of the point where the gradient is 10.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [3m]

13. Find the equation of the tangent to the curve $y = x^2 + 2x$ at the point $(1, 3)$.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [4m]

14. The gradient of the curve $y = ax^2 + bx$ at $x=1$ is 5, and at $x=2$ is 9. Find the values of $a$ and $b$.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [4m]

15. Determine if the curve $y = -x^2 + 6x$ has a tangent parallel to the x-axis. If so, find the x-coordinate of that point.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [3m]

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Section C: Stationary Points and Optimization (16-20)

Focus: Finding maxima/minima and rates of change.

16. Find the stationary point of the curve $y = x^2 - 8x + 12$.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [3m]

17. For the curve $y = 2x^2 - 12x + 5$, determine whether the stationary point is a maximum or a minimum.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [3m]

18. A rectangle has a perimeter of 40 cm. Let the width be $x$. Express the area $A$ in terms of $x$ and find the value of $x$ that maximizes the area.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [4m]

19. Find the stationary points of the cubic function $y = x^3 - 3x$.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [4m]

20. A particle moves such that its displacement $s$ (in metres) at time $t$ (in seconds) is given by $s = t^3 - 6t^2 + 9t$. Find the acceleration of the particle at $t = 2$ seconds.

    $\text{Working: } \underline{\hspace{5cm}}$ $\text{Ans: } \underline{\hspace{3cm}}$ [4m]

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Secondary 3 Elementary Mathematics Quiz - Calculus (Answer Key)

1. $\frac{dy}{dx} = 15x^2$ [2m]

2. $\frac{dy}{dx} = 8x - 7$ [2m]

3. $\frac{dy}{dx} = x^2 + 2$ [2m]

4. $\frac{dy}{dx} = -16x^{-3}$ or $-\frac{16}{x^3}$ [2m]

5. $y = 6x^{1/2} \Rightarrow \frac{dy}{dx} = 3x^{-1/2}$ or $\frac{3}{\sqrt{x}}$ [2m]

6. $y = 4x^2 + 12x + 9 \Rightarrow \frac{dy}{dx} = 8x + 12$ [2m]

7. $y = 4x^{-2} \Rightarrow \frac{dy}{dx} = -8x^{-3}$ or $-\frac{8}{x^3}$ [2m]

8. $\frac{dy}{dx} = 15x^{0.5}$ or $15\sqrt{x}$ [2m]

9. $y = 3x^2 - 2x^{-1} \Rightarrow \frac{dy}{dx} = 6x + 2x^{-2}$ or $6x + \frac{2}{x^2}$ [2m]

10. $\frac{dy}{dx} = 2\pi x$ [2m]

11. $\frac{dy}{dx} = 2x - 4$. At $x=3$, gradient $= 2(3) - 4 = 2$. [3m]

12. $\frac{dy}{dx} = 6x^2 - 5$. Set $6x^2 - 5 = 10 \Rightarrow 6x^2 = 15 \Rightarrow x^2 = 2.5 \Rightarrow x = \pm \sqrt{2.5}$. Coordinates: $(\sqrt{2.5}, 2(2.5)^{1.5} - 5\sqrt{2.5})$ and $(-\sqrt{2.5}, \dots)$ [3m]

13. $\frac{dy}{dx} = 2x + 2$. At $x=1$, $m = 4$. Equation: $y - 3 = 4(x - 1) \Rightarrow y = 4x - 1$. [4m]

14. $\frac{dy}{dx} = 2ax + b$. $2a(1) + b = 5$ (i) $2a(2) + b = 9$ (ii) Subtract (i) from (ii): $2a = 4 \Rightarrow a = 2$. Substitute into (i): $4 + b = 5 \Rightarrow b = 1$. [4m]

15. Parallel to x-axis means $\frac{dy}{dx} = 0$. $\frac{dy}{dx} = -2x + 6$. $-2x + 6 = 0 \Rightarrow x = 3$. Yes, it exists. [3m]

16. $\frac{dy}{dx} = 2x - 8$. Set $2x - 8 = 0 \Rightarrow x = 4$. $y = 4^2 - 8(4) + 12 = 16 - 32 + 12 = -4$. Stationary point: $(4, -4)$. [3m]

17. $\frac{dy}{dx} = 4x - 12$. Stationary point at $x = 3$. $\frac{d^2y}{dx^2} = 4$. Since $4 > 0$, it is a minimum. [3m]

18. $2(x + w) = 40 \Rightarrow w = 20 - x$. $A = x(20 - x) = 20x - x^2$. $\frac{dA}{dx} = 20 - 2x$. Set $20 - 2x = 0 \Rightarrow x = 10$. [4m]

19. $\frac{dy}{dx} = 3x^2 - 3$. $3(x^2 - 1) = 0 \Rightarrow x = \pm 1$. If $x=1, y=1-3=-2$. If $x=-1, y=-1+3=2$. Points: $(1, -2)$ and $(-1, 2)$. [4m]

20. Velocity $v = \frac{ds}{dt} = 3t^2 - 12t + 9$. Acceleration $a = \frac{dv}{dt} = 6t - 12$. At $t=2$, $a = 6(2) - 12 = 0 \text{ m/s}^2$. [4m]