FAD1014 — 25-Question Minimum Viable Drill — SOLUTIONS
Attempt all 25 questions first before looking here.
Block 1: Maclaurin Standard Forms
Q1
$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
Q2
$$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
Q3
$$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$
Q4
$$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \quad (-1 < x \leq 1)$$
Q5
$$1 + x + x^2 + x^3 + \cdots \quad (|x| < 1)$$
Q6
$$x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
Q7
$$1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$$
Block 2: Maclaurin Substitution
Q8
$e^{-3x^2}$: substitute $u = -3x^2$ into $e^u = 1 + u + \frac{u^2}{2!} + \cdots$
$$1 - 3x^2 + \frac{9x^4}{2} + \cdots$$
Q9
$\ln(1+5x)$: substitute $u = 5x$ into $\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots$
$$5x - \frac{25x^2}{2} + \frac{125x^3}{3} - \frac{625x^4}{4} + \cdots$$
Q10
$x^2 e^{-x}$: $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$
Multiply by $x^2$: $$x^2 - x^3 + \frac{x^4}{2} + \cdots$$
Block 3: Ellipse — Completing Square
Q11
$$x^2 - 4x + 9y^2 + 36y = -4$$
$$(x^2 - 4x + 4) + 9(y^2 + 4y + 4) = -4 + 4 + 36$$
$$(x-2)^2 + 9(y+2)^2 = 36$$
$$\frac{(x-2)^2}{36} + \frac{(y+2)^2}{4} = 1$$
Centre $(2,-2)$. $a = 6$ (horizontal), $b = 2$. $c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$.
Foci: $(2 \pm 4\sqrt{2}, -2)$.
Q12
$$4x^2 - 8x + y^2 + 4y = 8$$
$$4(x^2 - 2x + 1) + (y^2 + 4y + 4) = 8 + 4 + 4$$
$$4(x-1)^2 + (y+2)^2 = 16$$
$$\frac{(x-1)^2}{4} + \frac{(y+2)^2}{16} = 1$$
Centre $(1,-2)$. $a = 2$, $b = 4$ (vertical). $c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.
Vertices: $(1, 2)$ and $(1, -6)$. Foci: $(1, -2 \pm 2\sqrt{3})$.
Block 4: Ellipse — Write Equation
Q13
$c = 3$, $a = 5$. $b^2 = a^2 - c^2 = 25 - 9 = 16$.
$$\frac{x^2}{25} + \frac{y^2}{16} = 1$$
Q14
Centre $(1,-2)$. Focus $(1,1)$ → $c = 3$ (vertical). Vertex $(1,3)$ → $b = 5$. $a^2 = b^2 - c^2 = 25 - 9 = 16$.
$$\frac{(x-1)^2}{16} + \frac{(y+2)^2}{25} = 1$$
Block 5: DE Identification
| Q | Type | Method |
|---|---|---|
| 15 | Separable | $y^2,dy = (x^2+1),dx$ |
| 16 | Homogeneous | $y = vx$ (all degree 2) |
| 17 | Non-Homogeneous Dependent | $\frac11 = \frac11$, $z = x + y$ |
| 18 | Non-Homogeneous Independent | $\frac21 \neq \frac{1}{-1}$, solve $h,k$ |
Block 6: Separable DE
Q19
$y,dy = x,dx$ → $\frac{y^2}{2} = \frac{x^2}{2} + C$ → $y^2 = x^2 + C$
$y(0) = 3$ → $9 = C$
$$\boxed{y^2 = x^2 + 9}$$
Q20
$\frac{dy}{y^2} = e^x,dx$ → $-y^{-1} = e^x + C$
$y(0) = \frac12$ → $-2 = 1 + C$ → $C = -3$
$-y^{-1} = e^x - 3$ → $y^{-1} = 3 - e^x$
$$\boxed{y = \frac{1}{3 - e^x}}$$
(Horizontal asymptote at $x = \ln 3$ — function blows up there.)
Block 7: Growth/Decay
Q21
$P(t) = P_0 e^{kt}$, $P_0 = 1000$
$P(2) = 1000 e^{2k} = 3000$ → $e^{2k} = 3$ → $k = \frac{\ln 3}{2}$
(a) $P(6) = 1000 e^{6k} = 1000(e^{2k})^3 = 1000(3^3) = \boxed{27,!000}$
(b) $1000 e^{kt} = 9000$ → $e^{kt} = 9$ → $kt = \ln 9$
$$t = \frac{\ln 9}{k} = \frac{2\ln 3}{\ln 3 / 2} = \boxed{4\text{ hours}}$$
Q22
$m(t) = m_0 e^{-kt}$, $m_0 = 50$
$m(10) = 50 e^{-10k} = 40$ → $e^{-10k} = 0.8$ → $k = -\frac{\ln 0.8}{10}$
Half-life when $e^{-kt} = \frac12$: $t_{1/2} = \frac{\ln 2}{k} = \frac{10\ln 2}{-\ln 0.8}$
$$= \frac{10\ln 2}{\ln(5/4)} \approx \frac{6.9315}{0.2231} \approx \boxed{31.1\text{ years}}$$
Block 8: Part A Quickies
Q23
$$\int_0^{\pi/4} \sec^2 x,dx = \left[\tan x\right]_0^{\pi/4} = 1 - 0 = \boxed{1}$$
Q24
$$A = \int_1^3 (x^2 + 1),dx = \left[\frac{x^3}{3} + x\right]_1^3 = (9 + 3) - \left(\frac13 + 1\right) = 12 - \frac43 = \boxed{\frac{32}{3}}$$
Q25
$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$$
$$\sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1} = \boxed{\frac{n}{n+1}}$$
Quick Reference Card
| Concept | Key Formula / Step |
|---|---|
| Maclaurin | $f(x) = \sum \frac{f^{(n)}(0)}{n!} x^n$ |
| Series sub | Replace $x$ with argument, collect terms |
| Ellipse CS | Group $x$'s and $y$'s, complete square, divide |
| Ellipse $c^2$ | $c^2 = a^2 - b^2$ (subtract!) |
| DE: Sep | $g(y),dy = f(x),dx$ |
| DE: Homo | $y = vx$ |
| DE: Non-Homo | Ratio test → dependent ($z$) or independent ($h,k$) |
| Growth | $P = P_0 e^{kt}$ |
| Decay | $m = m_0 e^{-kt}$ |
| Area | $\int_a^b f(x),dx$ |
| Method of diff | Decompose → telescope → first minus last |