A Level H1 Mathematics Calculus Quiz

Answer Key - A-Level Maths H1 Quiz (Calculus)

1. $\frac{dy}{dx} = 12x^2 - x^{-1/2}$ or $12x^2 - \frac{1}{\sqrt{x}}$

   • (2 marks: 1 for $12x^2$, 1 for $-x^{-1/2}$)

2. $\frac{dy}{dx} = 5e^{5x} + \frac{1}{x}$

   • (2 marks: 1 for $5e^{5x}$, 1 for $1/x$)

3. $y' = 2x e^{2x} + x^2(2e^{2x}) = (2x + 2x^2)e^{2x}$

   • (3 marks: 1 for product rule setup, 1 for differentiation, 1 for simplification)

4. $y' = \frac{x^2(1/x) - (\ln x)(2x)}{x^4} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}$

   • (3 marks: 1 for quotient rule, 1 for substitution, 1 for simplification)

5. $y' = 2e^{2x} + 3$. At $x=0, y' = 2(1) + 3 = 5$.

   • (2 marks: 1 for derivative, 1 for final value)

6. $y' = -4(x+2)^{-2}$. At $x=2, m = -4(4)^{-2} = -1/4$. Point $(2, 1)$. $y - 1 = -1/4(x - 2) \Rightarrow y = -1/4x + 1.5$ or $y = -0.25x + 1.5$.

   • (3 marks: 1 for $m$, 1 for point, 1 for equation)

7. $y' = 2x e^{-x} + x^2(-e^{-x}) = xe^{-x}(2-x)$. Set $y' = 0 \Rightarrow x = 0$ or $x = 2$. Points: $(0, 0)$ and $(2, 4e^{-2})$.

   • (4 marks: 1 for derivative, 1 for solving $y'=0$, 2 for coordinates)

8. $y'' = e^{-x}(2-2x) - xe^{-x}(1) = e^{-x}(2 - 3x)$. At $x=0, y'' = 2 > 0$ (Minimum). At $x=2, y'' = e^{-2}(2-6) = -4e^{-2} < 0$ (Maximum).

   • (3 marks: 1 for $y''$, 1 for testing $x=0$, 1 for testing $x=2$)

9. $y = \ln(2x) - \ln(5-x) \Rightarrow y' = \frac{2}{2x} - \frac{-1}{5-x} = \frac{1}{x} + \frac{1}{5-x} = \frac{5}{x(5-x)}$.

   • (3 marks: 1 for log laws, 1 for differentiation, 1 for simplification)

10. $y' = \frac{1}{x+1}$. At $(0,0), m_{tangent} = 1$. $m_{normal} = -1$. Equation: $y - 0 = -1(x - 0) \Rightarrow y = -x$.

    • (3 marks: 1 for $y'$, 1 for $m_{normal}$, 1 for equation)

11. $2x^3 - 2x^2 + 5x + C$

    • (2 marks)

12. $\frac{1}{3}e^{3x-1} + C$

    • (2 marks)

13. $[\frac{1}{3}x^3 + \ln x]_1^2 = (\frac{8}{3} + \ln 2) - (\frac{1}{3} + 0) = \frac{7}{3} + \ln 2 \approx 3.03$

    • (3 marks: 1 for antiderivative, 1 for substitution, 1 for final value)

14. $\int_0^1 e^{2x} \, dx = [\frac{1}{2}e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1) \approx 3.19$ units²

    • (3 marks: 1 for antiderivative, 1 for limits, 1 for value)

15. $[\frac{1}{2 \cdot 5}(2x+1)^5]_0^1 = \frac{1}{10}(3^5 - 1^5) = \frac{242}{10} = 24.2$

    • (3 marks: 1 for antiderivative, 1 for limits, 1 for value)

16. $\int_1^4 x^{-1/2} \, dx = [2x^{1/2}]_1^4 = 2(2) - 2(1) = 2$ units²

    • (3 marks: 1 for antiderivative, 1 for limits, 1 for value)

17. $\int_0^2 (3x^2 - 6x) \, dx = [x^3 - 3x^2]_0^2 = (8 - 12) - 0 = -4$. Area = $|-4| = 4$ units².

    • (4 marks: 1 for antiderivative, 1 for limits, 1 for calculation, 1 for absolute value)

18. $\int_0^{\ln 2} k e^{-x} \, dx = [-k e^{-x}]_0^{\ln 2} = -k(e^{-\ln 2} - e^0) = -k(1/2 - 1) = \frac{1}{2}k$. $\frac{1}{2}k = 1 \Rightarrow k = 2$.

    • (4 marks: 1 for antiderivative, 1 for substitution, 1 for solving $k$, 1 for final answer)

19. Let $u = \ln x, du = \frac{1}{x}dx$. $\int_0^1 u \, du = [\frac{1}{2}u^2]_0^1 = \frac{1}{2}$.

    • (4 marks: 1 for substitution, 1 for new limits, 1 for integration, 1 for final value)

20. $C(x) = \int (20 + 0.5x) \, dx = 20x + 0.25x^2 + C$. Since $C(0) = 500$, then $C = 500$. $C(x) = 0.25x^2 + 20x + 500$.

    • (4 marks: 1 for integration, 1 for constant $C$, 1 for using fixed cost, 1 for final function)