O Level Additional Mathematics Calculus Quiz

O-Level Additional Mathematics Quiz - Calculus (Answer Key)

1. $y = 3x^4 - 2x^{-2} + 5x^{1/2}$ $\frac{dy}{dx} = 12x^3 - 2(-2)x^{-3} + 5(\frac{1}{2})x^{-1/2}$ $\frac{dy}{dx} = 12x^3 + \frac{4}{x^3} + \frac{5}{2\sqrt{x}}$ [3 marks] (1 mark for each term correct)

2. Let $u = 2x^2 - 1$, then $y = u^5$. $\frac{du}{dx} = 4x, \quad \frac{dy}{du} = 5u^4$ $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 5(2x^2 - 1)^4 (4x)$ $\frac{dy}{dx} = 20x(2x^2 - 1)^4$ [3 marks] (1 mark for chain rule setup, 1 mark for derivatives, 1 mark for final answer)

3. Product Rule: $u = x^2, v = e^{3x}$. $u' = 2x, \quad v' = 3e^{3x}$ $\frac{dy}{dx} = u'v + uv' = 2x(e^{3x}) + x^2(3e^{3x})$ $\frac{dy}{dx} = e^{3x}(2x + 3x^2) \quad \text{or} \quad xe^{3x}(2 + 3x)$ [3 marks] (1 mark for rule, 1 mark for components, 1 mark for simplification)

4. Quotient Rule: $u = \ln x, v = x^2$. $u' = \frac{1}{x}, \quad v' = 2x$ $\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(\frac{1}{x})(x^2) - (\ln x)(2x)}{(x^2)^2}$ $\frac{dy}{dx} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2\ln x}{x^3}$ [3 marks] (1 mark for rule, 1 mark for substitution, 1 mark for simplification)

5. $y = \tan(2x + \frac{\pi}{4})$. $\frac{dy}{dx} = \sec^2(2x + \frac{\pi}{4}) \cdot \frac{d}{dx}(2x + \frac{\pi}{4}) = 2\sec^2(2x + \frac{\pi}{4})$ At $x = 0$: $\frac{dy}{dx} = 2\sec^2(\frac{\pi}{4}) = 2(\sqrt{2})^2 = 2(2) = 4$ [3 marks] (1 mark for derivative, 1 mark for substitution, 1 mark for final value)

6. (a) At $x = 2$, $y = 2^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$. Coordinates: $(2, -2)$. (b) $\frac{dy}{dx} = 3x^2 - 6x$. $\frac{d^2y}{dx^2} = 6x - 6$. At $x = 2$, $\frac{d^2y}{dx^2} = 6(2) - 6 = 6 > 0$. Since second derivative is positive, it is a minimum point. [4 marks] (1 mark for y-coord, 1 mark for 1st deriv, 1 mark for 2nd deriv, 1 mark for conclusion)

7. $y = 2x^2 - 5x + 1$. At $x = 1$, $y = 2(1)^2 - 5(1) + 1 = -2$. Point: $(1, -2)$. $\frac{dy}{dx} = 4x - 5$. Gradient $m$ at $x = 1$: $m = 4(1) - 5 = -1$. Equation: $y - y_1 = m(x - x_1) \Rightarrow y - (-2) = -1(x - 1)$. $y + 2 = -x + 1 \Rightarrow y = -x - 1$. [4 marks] (1 mark for point, 1 mark for gradient, 1 mark for formula, 1 mark for final eq)

8. $s = t^3 - 6t^2 + 9t + 4$. Velocity $v = \frac{ds}{dt} = 3t^2 - 12t + 9$. Acceleration $a = \frac{dv}{dt} = 6t - 12$. At $t = 2$, $a = 6(2) - 12 = 0$ m s$^{-2}$. [3 marks] (1 mark for v, 1 mark for a, 1 mark for substitution)

9. $V = \frac{4}{3}\pi r^3$. $\frac{dV}{dr} = 4\pi r^2$. Given $\frac{dV}{dt} = 10$. Chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$. $10 = 4\pi r^2 \times \frac{dr}{dt}$. When $r = 5$: $10 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}$. $\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}$ cm s$^{-1}$ (or approx 0.0318). [4 marks] (1 mark for dV/dr, 1 mark for chain rule setup, 1 mark for substitution, 1 mark for answer)

10. (a) $y = x^3 + ax^2 + bx$. $\frac{dy}{dx} = 3x^2 + 2ax + b$. At stationary point $(1, 4)$:

1. Curve passes through $(1,4)$: $4 = 1 + a + b \Rightarrow a + b = 3$.
2. Gradient is 0 at $x=1$: $0 = 3(1)^2 + 2a(1) + b \Rightarrow 2a + b = -3$. Subtract eq 1 from eq 2: $(2a+b) - (a+b) = -3 - 3 \Rightarrow a = -6$. Substitute $a = -6$ into eq 1: $-6 + b = 3 \Rightarrow b = 9$. $a = -6, b = 9$. (b) Equation: $y = x^3 - 6x^2 + 9x$. $\frac{dy}{dx} = 3x^2 - 12x + 9 = 0$. Divide by 3: $x^2 - 4x + 3 = 0$. $(x - 3)(x - 1) = 0$. $x = 1$ or $x = 3$. When $x = 3$, $y = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0$. Other stationary point: $(3, 0)$. [6 marks] (2 marks for finding a,b, 2 marks for solving quadratic, 2 marks for coords)

11. $\int (4x^3 - 6x + x^{-1/2}) \, dx$ $= 4(\frac{x^4}{4}) - 6(\frac{x^2}{2}) + \frac{x^{1/2}}{1/2} + C$ $= x^4 - 3x^2 + 2\sqrt{x} + C$ [3 marks] (1 mark per term integrated correctly, including C)

12. $\int_0^1 (3x^2 + 2e^x) \, dx = [x^3 + 2e^x]_0^1$ Upper limit ($x=1$): $1^3 + 2e^1 = 1 + 2e$. Lower limit ($x=0$): $0^3 + 2e^0 = 0 + 2(1) = 2$. Value: $(1 + 2e) - 2 = 2e - 1$. [3 marks] (1 mark for integration, 1 mark for substitution, 1 mark for final answer)

13. $\int \sin(3x - \frac{\pi}{2}) \, dx$ $= -\frac{1}{3}\cos(3x - \frac{\pi}{2}) + C$ [2 marks] (1 mark for cos, 1 mark for factor -1/3 and C)

14. $y = \int (6x - 4) \, dx = 3x^2 - 4x + C$. Given $y = 5$ when $x = 1$: $5 = 3(1)^2 - 4(1) + C \Rightarrow 5 = 3 - 4 + C \Rightarrow 5 = -1 + C \Rightarrow C = 6$. $y = 3x^2 - 4x + 6$. [3 marks] (1 mark for integration, 1 mark for finding C, 1 mark for final eq)

15. $\int_1^2 (2x+1)^{-2} \, dx$. Let $u = 2x+1$, or use reverse chain rule. Integral is $[\frac{(2x+1)^{-1}}{-1} \cdot \frac{1}{2}]_1^2 = [-\frac{1}{2(2x+1)}]_1^2$. Upper ($x=2$): $-\frac{1}{2(5)} = -\frac{1}{10}$. Lower ($x=1$): $-\frac{1}{2(3)} = -\frac{1}{6}$. Value: $-\frac{1}{10} - (-\frac{1}{6}) = \frac{1}{6} - \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15}$. [4 marks] (1 mark for integration form, 1 mark for factor 1/2, 1 mark for limits, 1 mark for answer)

16. Area $= |\int_0^4 (x^2 - 4x) \, dx|$. Note: Curve crosses x-axis at $x(x-4)=0 \Rightarrow x=0, 4$. Between 0 and 4, $y$ is negative. $\int_0^4 (x^2 - 4x) \, dx = [\frac{x^3}{3} - 2x^2]_0^4$. At $x=4$: $\frac{64}{3} - 2(16) = \frac{64}{3} - \frac{96}{3} = -\frac{32}{3}$. At $x=0$: $0$. Area $= |-\frac{32}{3}| = \frac{32}{3}$ or $10.67$ units$^2$. [4 marks] (1 mark for integral setup, 1 mark for integration, 1 mark for evaluation, 1 mark for positive area)

17. $v = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$. Velocity changes sign at $t=1$ and $t=3$. Distance $= \int_0^1 v \, dt + |\int_1^3 v \, dt| + \int_3^4 v \, dt$. $\int v \, dt = t^3 - 6t^2 + 9t$. $s(0) = 0$. $s(1) = 1 - 6 + 9 = 4$. Dist $0 \to 1 = 4$. $s(3) = 27 - 54 + 27 = 0$. Dist $1 \to 3 = |0 - 4| = 4$. $s(4) = 64 - 96 + 36 = 4$. Dist $3 \to 4 = |4 - 0| = 4$. Total Distance $= 4 + 4 + 4 = 12$ m. [5 marks] (1 mark for finding roots, 1 mark for splitting intervals, 1 mark for integration, 1 mark for absolute values, 1 mark for sum)

18. (a) $\sqrt{x} = x \Rightarrow x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x-1)=0$. $x=0, y=0$ and $x=1, y=1$. Points: $(0,0)$ and $(1,1)$. (b) Area $= \int_0^1 (\sqrt{x} - x) \, dx$ (Curve is above line in this interval). $= [\frac{2}{3}x^{3/2} - \frac{x^2}{2}]_0^1$. $= (\frac{2}{3}(1) - \frac{1}{2}) - 0 = \frac{4-3}{6} = \frac{1}{6}$. [5 marks] (2 marks for intersection, 1 mark for setup, 1 mark for integration, 1 mark for answer)

19. $\frac{dy}{dx} = 3x^2 + 3$. For stationary points, $\frac{dy}{dx} = 0 \Rightarrow 3x^2 + 3 = 0 \Rightarrow 3x^2 = -3 \Rightarrow x^2 = -1$. Since $x^2 \ge 0$ for all real $x$, there are no real solutions. Thus, the curve has no stationary points. [2 marks] (1 mark for derivative/equation, 1 mark for reasoning)

20. (a) $y = \int (2x - x^{-2}) \, dx = x^2 - \frac{x^{-1}}{-1} + C = x^2 + \frac{1}{x} + C$. Passes through $(1,3)$: $3 = 1^2 + \frac{1}{1} + C \Rightarrow 3 = 2 + C \Rightarrow C = 1$. Equation: $y = x^2 + \frac{1}{x} + 1$. (b) Stationary point when $\frac{dy}{dx} = 0 \Rightarrow 2x - \frac{1}{x^2} = 0 \Rightarrow 2x = \frac{1}{x^2} \Rightarrow 2x^3 = 1 \Rightarrow x^3 = 0.5 \Rightarrow x = \sqrt[3]{0.5}$. $\frac{d^2y}{dx^2} = 2 + 2x^{-3} = 2 + \frac{2}{x^3}$. At $x = \sqrt[3]{0.5}$, $x^3 = 0.5$. $\frac{d^2y}{dx^2} = 2 + \frac{2}{0.5} = 2 + 4 = 6 > 0$. Minimum point. [6 marks] (2 marks for integration/C, 1 mark for eq, 1 mark for x-coord, 1 mark for 2nd deriv test, 1 mark for nature)