Secondary 4 Additional Mathematics Calculus Quiz

Secondary 4 Additional Mathematics Quiz - Calculus (Answer Key)

Total Marks: 60

Section A: Differentiation Techniques

1. $y = 3x^4 - 2x^{-3} + 5x^{1/2}$ $\frac{dy}{dx} = 12x^3 - 2(-3)x^{-4} + 5(\frac{1}{2})x^{-1/2}$ $\frac{dy}{dx} = 12x^3 + \frac{6}{x^4} + \frac{5}{2\sqrt{x}}$ [3 marks]: 1 mark for each term correct.

2. $y = (2x^2 + 1)(3x - 4) = 6x^3 - 8x^2 + 3x - 4$ (a) $\frac{dy}{dx} = 18x^2 - 16x + 3$ (b) Product Rule: $u = 2x^2+1, v = 3x-4$. $u' = 4x, v' = 3$. $\frac{dy}{dx} = (2x^2+1)(3) + (3x-4)(4x) = 6x^2 + 3 + 12x^2 - 16x = 18x^2 - 16x + 3$. [3 marks]: 1 mark for expansion/method, 1 mark for differentiation, 1 mark for correct final answer.

3. Quotient Rule: $u = e^{2x}, v = x^2+1$. $u' = 2e^{2x}, v' = 2x$. $\frac{dy}{dx} = \frac{(x^2+1)(2e^{2x}) - (e^{2x})(2x)}{(x^2+1)^2}$ $\frac{dy}{dx} = \frac{2e^{2x}(x^2 - x + 1)}{(x^2+1)^2}$ [3 marks]: 1 mark for correct quotient rule setup, 1 mark for simplification of numerator, 1 mark for final answer.

4. Chain Rule: Let $u = 3x^2+1$, then $y = \sin u$. $\frac{dy}{du} = \cos u$, $\frac{du}{dx} = 6x$. $\frac{dy}{dx} = 6x \cos(3x^2 + 1)$ [3 marks]: 1 mark for inner derivative, 1 mark for outer derivative, 1 mark for combination.

5. $y = x^3 - 6x^2 + 9x + 2$ $\frac{dy}{dx} = 3x^2 - 12x + 9$ (a) At $x=1$: $3(1)^2 - 12(1) + 9 = 3 - 12 + 9 = 0$. (Verified) (b) $\frac{d^2y}{dx^2} = 6x - 12$. At $x=1$: $6(1) - 12 = -6$. Since $\frac{d^2y}{dx^2} < 0$, it is a Maximum point. [3 marks]: 1 mark for substitution, 1 mark for 2nd derivative value, 1 mark for correct nature.

Section B: Applications of Differentiation

6. $y = x^2 - 4x + 5$. $\frac{dy}{dx} = 2x - 4$. At $x=3$, gradient $m = 2(3) - 4 = 2$. y-coordinate: $y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2$. Point $(3, 2)$. Equation: $y - 2 = 2(x - 3) \Rightarrow y = 2x - 6 + 2 \Rightarrow y = 2x - 4$. [3 marks]: 1 mark for gradient, 1 mark for point, 1 mark for equation.

7. $s = t^3 - 6t^2 + 9t$ (a) $v = \frac{ds}{dt} = 3t^2 - 12t + 9$. (b) $a = \frac{dv}{dt} = 6t - 12$. At $t=2$: $a = 6(2) - 12 = 0$ m/s$^2$. [3 marks]: 1 mark for v, 1 mark for a expression, 1 mark for evaluation.

8. $V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2$. Given $\frac{dV}{dt} = 10$. Chain Rule: $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$. $10 = 4\pi (5)^2 \times \frac{dr}{dt}$. $10 = 100\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}$ cm/s. [3 marks]: 1 mark for dV/dr, 1 mark for chain rule setup, 1 mark for final answer.

9. $f(x)$ increasing when $f'(x) > 0$. $f'(x) = 3x^2 - 6x - 9$. $3x^2 - 6x - 9 > 0 \Rightarrow x^2 - 2x - 3 > 0$. $(x - 3)(x + 1) > 0$. Critical values: $x = 3, x = -1$. Region: $x < -1$ or $x > 3$. [3 marks]: 1 mark for derivative, 1 mark for solving inequality/roots, 1 mark for correct range.

10. $\frac{dy}{dx} = e^x + 1$. For stationary point, $\frac{dy}{dx} = 0 \Rightarrow e^x + 1 = 0 \Rightarrow e^x = -1$. Since $e^x > 0$ for all real $x$, $e^x$ can never be $-1$. Therefore, there are no real solutions for $\frac{dy}{dx} = 0$, so no stationary points. [3 marks]: 1 mark for derivative, 1 mark for setting to 0/impossibility, 1 mark for explanation.

Section C: Integration Techniques

11. $\int (4x^3 - 6x^{-2} + x^{-1}) dx$ $= \frac{4x^4}{4} - \frac{6x^{-1}}{-1} + \ln|x| + C$ $= x^4 + \frac{6}{x} + \ln|x| + C$ [3 marks]: 1 mark for each term integrated correctly (including +C).

12. $\int_{0}^{\frac{\pi}{2}} \cos(2x) dx = \left[ \frac{1}{2}\sin(2x) \right]_{0}^{\frac{\pi}{2}}$ $= \frac{1}{2}\sin(\pi) - \frac{1}{2}\sin(0)$ $= 0 - 0 = 0$. [3 marks]: 1 mark for integration, 1 mark for substitution, 1 mark for answer.

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

14. $u = x^2 + 1 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{1}{2} du$. Limits: $x=0 \to u=1$; $x=2 \to u=5$. $\int_{1}^{5} u^3 (\frac{1}{2}) du = \frac{1}{2} \left[ \frac{u^4}{4} \right]_{1}^{5} = \frac{1}{8} [5^4 - 1^4]$. $= \frac{1}{8} (625 - 1) = \frac{624}{8} = 78$. [3 marks]: 1 mark for substitution/limits, 1 mark for integration, 1 mark for evaluation.

15. Area $= \int_{0}^{\pi} \sin x dx = \left[ -\cos x \right]_{0}^{\pi}$. $= (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$. [3 marks]: 1 mark for integral, 1 mark for substitution, 1 mark for answer.

Section D: Advanced Integration & Applications

16. (a) Intersection: $x(4-x) = x \Rightarrow 4x - x^2 = x \Rightarrow 3x - x^2 = 0 \Rightarrow x(3-x)=0$. $x = 0$ and $x = 3$. (b) Area $= \int_{0}^{3} [(4x - x^2) - x] dx = \int_{0}^{3} (3x - x^2) dx$. $= \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3}$. $= (\frac{27}{2} - \frac{27}{3}) - 0 = 13.5 - 9 = 4.5$. [4 marks]: 1 mark for limits, 1 mark for correct integrand, 1 mark for integration, 1 mark for final answer.

17. (a) $v = \int (6t - 4) dt = 3t^2 - 4t + C_1$. At $t=0, v=3 \Rightarrow C_1 = 3$. $v = 3t^2 - 4t + 3$. (b) $s = \int (3t^2 - 4t + 3) dt = t^3 - 2t^2 + 3t + C_2$. At $t=0, s=2 \Rightarrow C_2 = 2$. $s = t^3 - 2t^2 + 3t + 2$. At $t=2$: $s = 2^3 - 2(2)^2 + 3(2) + 2 = 8 - 8 + 6 + 2 = 8$ m. [4 marks]: 1 mark for v expression, 1 mark for s expression, 1 mark for constants, 1 mark for final displacement.

18. Volume $V = \int_{0}^{4} \pi y^2 dx = \pi \int_{0}^{4} \left( \frac{1}{\sqrt{2x+1}} \right)^2 dx = \pi \int_{0}^{4} \frac{1}{2x+1} dx$. $= \pi \left[ \frac{1}{2} \ln|2x+1| \right]_{0}^{4}$. $= \frac{\pi}{2} (\ln(9) - \ln(1)) = \frac{\pi}{2} \ln 9$. (Note: $\ln 9 = 2 \ln 3$, so answer can be $\pi \ln 3$). [3 marks]: 1 mark for setup $\pi y^2$, 1 mark for integration, 1 mark for evaluation.

19. $\left[ x^2 + 3x \right]_{1}^{k} = 20$. $(k^2 + 3k) - (1^2 + 3(1)) = 20$. $k^2 + 3k - 4 = 20 \Rightarrow k^2 + 3k - 24 = 0$. Wait, let's re-calculate: $1+3=4$. So $k^2+3k - 4 = 20 \Rightarrow k^2+3k-24=0$. Using quadratic formula: $k = \frac{-3 \pm \sqrt{9 - 4(1)(-24)}}{2} = \frac{-3 \pm \sqrt{105}}{2}$. Since $k$ must be positive (and usually upper limit > lower limit in this context, though not strictly required if signed area, but "positive value" is asked): $k = \frac{-3 + \sqrt{105}}{2}$. Self-Correction Check: Did I copy the question numbers right? $\int (2x+3) = x^2+3x$. Limits 1 to k. Value 20. $(k^2+3k) - (1+3) = 20 \rightarrow k^2+3k-24=0$. Roots are not integers. Let's check if the question intended simpler numbers. If integral was 14: $k^2+3k-4=14 \rightarrow k^2+3k-18=0 \rightarrow (k+6)(k-3)=0 \rightarrow k=3$. Given the prompt asks for "exact" or standard practice, I will provide the exact surd form or assume a typo in my mental check. Let's stick to the math derived from the prompt text. $k = \frac{-3 + \sqrt{105}}{2}$. [2 marks]: 1 mark for forming equation, 1 mark for solving for k.

20. $\frac{dP}{dt} = 0.05P \Rightarrow \int \frac{1}{P} dP = \int 0.05 dt$. $\ln P = 0.05t + C$. $P = e^{0.05t+C} = A e^{0.05t}$. At $t=0, P=1000 \Rightarrow A=1000$. $P = 1000 e^{0.05t}$. [2 marks]: 1 mark for general exponential form, 1 mark for specific constant.