Secondary 4 Additional Mathematics Calculus Quiz

Secondary 4 Additional Mathematics Quiz - Calculus: Answer Key

Total Marks: 50

Section A: Differentiation Basics

Q1. [2 marks]
Differentiate term by term:
$\frac{d}{dx}(4x^3) = 12x^2$
$\frac{d}{dx}(-5x^2) = -10x$
$\frac{d}{dx}(7x) = 7$
$\frac{d}{dx}(-3) = 0$
Answer: $\frac{dy}{dx} = 12x^2 - 10x + 7$
Teaching note: Power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$. Constant term differentiates to 0.
Marking: 1 mark for each correct pair of terms or full expression.

Q2. [2 marks]
$f(x) = x^{-2}$
$f'(x) = -2x^{-3} = -\frac{2}{x^3}$
Teaching note: Rewrite reciprocal as negative power before differentiating.
Marking: 1 mark for rewrite, 1 mark for final answer.

Q3. [2 marks]
Product rule: $\frac{d}{dx}(uv) = u'v + uv'$
$u = 2x+1, u' = 2$; $v = x-3, v' = 1$
$\frac{dy}{dx} = 2(x-3) + (2x+1)(1) = 2x - 6 + 2x + 1 = 4x - 5$
Teaching note: Alternatively expand first: $(2x+1)(x-3)=2x^2-5x-3$, derivative $4x-5$.
Marking: 1 mark for correct rule application, 1 mark for simplified answer.

Q4. [3 marks]
$y = x^{1/2} + 2x^{-1}$
$\frac{dy}{dx} = \frac{1}{2}x^{-1/2} - 2x^{-2} = \frac{1}{2\sqrt{x}} - \frac{2}{x^2}$
Teaching note: $\sqrt{x}=x^{1/2}$; $\frac{1}{x}=x^{-1}$. Apply power rule.
Marking: 1 mark each term.

Q5. [3 marks]
Quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$
$u = 3x^2-1, u'=6x$; $v=x+2, v'=1$
$\frac{dy}{dx} = \frac{6x(x+2) - (3x^2-1)(1)}{(x+2)^2} = \frac{6x^2+12x-3x^2+1}{(x+2)^2} = \frac{3x^2+12x+1}{(x+2)^2}$
Teaching note: Common mistake: wrong sign in numerator.
Marking: 1 mark rule, 1 mark expansion, 1 mark simplified.

Section B: Stationary Points and Applications

Q6. [3 marks]
$\frac{dy}{dx} = 2x - 6 = 0 \Rightarrow x = 3$
$y = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$
$\frac{d^2y}{dx^2} = 2 > 0$ → minimum
Coordinates: $(3, -4)$, nature: minimum.
Marking: 1 mark stationary x, 1 mark y, 1 mark nature.

Q7. [4 marks]
$\frac{dy}{dx} = 3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1)=0$
$x = 3$: $y = 27 - 27 - 27 + 7 = -20$
$x = -1$: $y = -1 - 3 + 9 + 7 = 12$
$\frac{d^2y}{dx^2} = 6x - 6$
At $x=3$: $12 > 0$ minimum; at $x=-1$: $-12 < 0$ maximum.
Points: $(3, -20)$ min, $(-1, 12)$ max.
Marking: 1 mark solve, 1 mark coords, 2 marks nature.

Q8. [3 marks]
$\frac{dV}{dr} = 4\pi r^2$
$\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt} = 4\pi r^2 (0.2) = 0.8\pi r^2$
At $r=5$: $0.8\pi(25) = 20\pi \approx 62.8$ cm³/s.
Teaching note: Chain rule for related rates.
Marking: 1 mark derivative, 1 mark chain, 1 mark substitution.

Q9. [3 marks]
Perimeter $2(x + L) = 24 \Rightarrow L = 12 - x$
$A = x(12 - x) = 12x - x^2$
$\frac{dA}{dx} = 12 - 2x = 0 \Rightarrow x = 6$
$\frac{d^2A}{dx^2} = -2 < 0$ → max.
Marking: 1 mark area expr, 1 mark derivative, 1 mark x value.

Q10. [4 marks]
$x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3)=0$ → $A(1,0), B(3,0)$
Stationary: $\frac{dy}{dx}=2x-4=0 \Rightarrow x=2, y = 4-8+3=-1$ → $(2,-1)$
Marking: 1 mark A, 1 mark B, 2 marks stationary.

Section C: Integration

Q11. [2 marks]
$\int (3x^2 - 2x + 1) dx = x^3 - x^2 + x + c$
Marking: 1 mark terms, 1 mark +c.

Q12. [2 marks]
$\int x^{-3} dx = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c$
Marking: 1 mark integration, 1 mark +c.

Q13. [3 marks]
$\int_0^2 (x^2+2x)dx = \left[\frac{x^3}{3} + x^2\right]_0^2 = \left(\frac{8}{3}+4\right) - 0 = \frac{20}{3}$
Marking: 1 mark antiderivative, 1 mark sub, 1 mark value.

Q14. [3 marks]
$y = \int (6x-4)dx = 3x^2 - 4x + c$
$5 = 3(1)^2 - 4(1) + c = -1 + c \Rightarrow c = 6$
$y = 3x^2 - 4x + 6$
Marking: 1 mark integral, 1 mark c, 1 mark equation.

Q15. [4 marks]
$y = \int (2x-3)dx = x^2 - 3x + c$
$1 = 4 - 6 + c \Rightarrow c = 3$
$y = x^2 - 3x + 3$
At $x=0$: $y = 3$
Marking: 1 mark integral, 1 mark c, 1 mark eq, 1 mark y(0).

Section D: Mixed Calculus

Q16. [3 marks]
At $x=2$: $y = 5$, point $(2,5)$
$\frac{dy}{dx} = 2x = 4$ (tangent gradient)
Normal gradient $= -\frac{1}{4}$
Eq: $y - 5 = -\frac{1}{4}(x - 2) \Rightarrow y = -\frac{1}{4}x + \frac{11}{2}$
Marking: 1 mark point, 1 mark normal grad, 1 mark equation.

Q17. [3 marks]
$v = \frac{ds}{dt} = 6t^2 - 18t + 12$
At $t=3$: $v = 54 - 54 + 12 = 12$ m/s
Marking: 1 mark derivative, 1 mark sub, 1 mark answer.

Q18. [4 marks]
Area = $\int_0^2 (x^2 - 1) dx = \left[\frac{x^3}{3} - x\right]_0^2 = \frac{8}{3} - 2 = \frac{2}{3}$ sq units
Note: Curve below axis from 0 to1, above from1 to2; net signed area = $\frac{2}{3}$. For enclosed magnitude, split: $\int_0^1(1-x^2)dx + \int_1^2(x^2-1)dx = \frac{2}{3}+\frac{2}{3}=\frac{4}{3}$. Question asks area enclosed with x-axis → $\frac{4}{3}$.*
Correct: $\frac{4}{3}$ units².
Marking: 1 mark setup, 2 marks split or correct, 1 mark final.

Q19. [3 marks]
$\frac{d}{dx}(e^{2x}) = 2e^{2x}$; $\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{dy}{dx} = 2e^{2x} + \frac{1}{x}$
Marking: 1 mark each term.

Q20. [4 marks]
$\frac{dy}{dx} = \int (12x-6)dx = 6x^2 - 6x + c$
$2 = 6 - 6 + c \Rightarrow c = 2$
$y = \int (6x^2 - 6x + 2)dx = 2x^3 - 3x^2 + 2x + d$
$3 = 0 - 0 + 0 + d \Rightarrow d = 3$
$y = 2x^3 - 3x^2 + 2x + 3$
Marking: 1 mark first int, 1 mark c, 1 mark second int, 1 mark d.