FAD1014 — Strategy Triage Drills

Problem: You stare at the problem and don't know the first step. This fixes that.

Method: Look at the problem → classify in 10 seconds → write the first step → done. Only then solve.


Quick Reference: Decision Trees

Integration Triage

∫ f(x) dx
│
├─ Is it a STANDARD FORM? (∫ uⁿ du, ∫ e^u du, ∫ 1/u du, ∫ sin u du, etc.)
│   → Direct integration (Part A)
│
├─ Do you see a function AND its derivative (or scalar multiple)?
│   → Substitution (u-sub). Let u = inner function (Part A)
│
├─ Is it a PRODUCT of two unrelated functions? (x·eˣ, x²·sin x, ln x, eˣ·sin x)
│   → Integration by Parts. ∫u dv = uv - ∫v du. (Part B)
│     LIATE: Log → Inverse trig → Algebraic → Trig → Exponential
│
├─ Does it contain √(a² - x²), √(a² + x²), or √(x² - a²)?
│   → Trigonometric Substitution (Part B)
│     √(a² - x²)  →  x = a sin θ
│     √(a² + x²)  →  x = a tan θ
│     √(x² - a²)  →  x = a sec θ
│
└─ Definite integral? Check for absolute values → split at roots

DE Triage

Given dy/dx = f(x,y) or M dx + N dy = 0
│
├─ Can you write g(y) dy = h(x) dx   (variables separate cleanly)?
│   → SEPARABLE (Part A & B)
│
├─ Does every term have the same total degree? (e.g., xy, x², y² all degree 2)
│   → HOMOGENEOUS. Substitute y = vx, dy = v dx + x dv (Part B)
│
├─ Is it in form dy/dx + P(x)y = Q(x)?
│   → LINEAR. Integrating factor μ = e^{∫ P dx} (NOT TESTED — skip)
│
└─ Is it in form dy/dx + P(x)y = Q(x)yⁿ?
    → BERNOULLI. v = y^{1-n} (NOT TESTED — skip)

Series Triage

Given a series or sequence
│
├─ Is it (a+b)ⁿ where n is a POSITIVE INTEGER?
│   → BINOMIAL I. Use nCr or Pascal's triangle (Part A)
│
├─ Need to find an expansion in powers of x?
│   → MACLAURIN SERIES. Use f(x) = Σ f⁽ⁿ⁾(0)/n! · xⁿ (Part B)
│     OR use standard expansions (eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ)
│     Substitution, multiplication, or term-by-term integration
│
├─ Given a sum like Σ [f(k) - f(k+1)] or Σ 1/(k(k+1))?
│   → METHOD OF DIFFERENCES. Write as telescoping series (Part A & B)
│
└─ Does the series converge?
    → Use nth term test, ratio test (Part A)

Geometry Triage

Given an equation in x and y
│
├─ No xy term, x² and y² have SAME coefficient?  → CIRCLE (NOT TESTED — skip)
│
├─ Only ONE squared term?  → PARABOLA (Part A & B)
│   Form: (x-h)² = 4a(y-k)  or  (y-k)² = 4a(x-h)
│   Key: vertex, focus, directrix
│
├─ x² and y² have SAME SIGN but DIFFERENT coefficients?  → ELLIPSE (Part A & B)
│   Form: (x-h)²/a² + (y-k)²/b² = 1
│   Key: centre, foci, vertices
│
├─ x² and y² have OPPOSITE SIGNS?  → HYPERBOLA (Part A & B)
│   Form: (x-h)²/a² - (y-k)²/b² = 1  (horizontal)
│   Form: (y-k)²/a² - (x-h)²/b² = 1  (vertical)
│   Key: centre, foci, vertices, asymptotes
│
└─ x and y both expressed in terms of t?
    → PARAMETRIC EQUATIONS (Part B)
    dy/dx = (dy/dt)/(dx/dt), convert to Cartesian

Round 1: Integration — Quick ID (5 min)

For each integral, state the method and first step (the substitution $u$, or $u$ and $dv$ for by-parts, or the trig substitution). Do NOT solve.

Q1

$$\int 3x^2 \cos(x^3) , dx$$

Q2

$$\int x^2 e^{3x} , dx$$

Q3

$$\int \frac{dx}{\sqrt{4 - x^2}}$$

Q4

$$\int \ln x , dx$$

Q5

$$\int \frac{2x}{1 + x^2} , dx$$

Q6

$$\int \frac{x^3}{\sqrt{16 - x^2}} , dx$$

Q7

$$\int x \sin(2x) , dx$$

Q8

$$\int \frac{dx}{\sqrt{x^2 + 9}}$$

Q9

$$\int e^x \cos x , dx$$

Q10

$$\int \frac{6x^2}{\sqrt{1 - 9x^2}} , dx$$


Round 2: DE — Quick ID (3 min)

State type: Separable, Homogeneous, Linear (if tested), Bernoulli (if tested), or None of the above.

Q11

$$\frac{dy}{dx} = \frac{x}{y}$$

Q12

$$\frac{dy}{dx} = \frac{y^2 + xy}{x^2}$$

Q13

$$\frac{dy}{dx} + 2xy = x$$

Q14

$$x \frac{dy}{dx} - y = x^3$$

Q15

$$\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$$

Q16

$$(x + 1) \frac{dy}{dx} = x(y + 3)$$


Round 3: Series — Quick ID (3 min)

State method: Binomial I, Maclaurin (standard), Maclaurin (substitution), Method of Differences.

Q17

Expand $(2 + x)^5$ in ascending powers of $x$.

Q18

Find the Maclaurin series for $f(x) = e^{3x}$ up to $x^3$.

Q19

Find $\sum_{r=1}^{n} \frac{1}{r(r+1)}$.

Q20

Expand $\left(x^3 - \frac{1}{x^2}\right)^9$ and find coefficient of $x^{15}$.

Q21

Find $\int_0^{0.5} e^{-x^2} , dx$ correct to 4 d.p. using series.

Q22

Expand $\sqrt{1 - x}$ up to $x^3$ and approximate $\sqrt{0.98}$.


Round 4: Geometry — Quick ID (3 min)

State type: Parabola, Ellipse, Hyperbola, Circle (skip), Parametric.

Q23

$$(x+2)^2 = 8(y-3)$$

Q24

$$\frac{x^2}{16} + \frac{y^2}{9} = 1$$

Q25

$$\frac{y^2}{4} - \frac{x^2}{5} = 1$$

Q26

$$x = 2t + 1,; y = t^2 - 3$$

Q27

$$16x^2 + 4y^2 - 64x - 40y + 100 = 0$$

Q28

$$2x^2 - 3y^2 = 6$$


Round 5: Mixed Solve — Full Mechanics (20 min)

Now actually SOLVE. Triage first, then execute.

Q29

$$\int \frac{dx}{\sqrt{4 - x^2}}$$

Q30

$$\int x^2 \sin x , dx$$

Q31

Solve: $$\frac{dy}{dx} = \frac{y}{x}, \quad y(1) = 3$$

Q32

$$\int \frac{dx}{\sqrt{x^2 + 9}}$$

Q33

Find ellipse centre, foci, vertices: $16x^2 + 4y^2 - 64x - 40y + 100 = 0$

Q34

Solve: $$(x+1) \frac{dy}{dx} = x(y+3)$$

Q35

$$\int x e^{3x} , dx$$

Q36

Approximate $\int_0^{0.5} e^{-x^2} , dx$ correct to 4 d.p.


Solutions

R1: Integration ID

Q Method First Step
Q1 Substitution $u = x^3$, $du = 3x^2,dx$
Q2 By Parts $u = x^2$, $dv = e^{3x},dx$
Q3 Standard form $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a) + C$, $a=2$
Q4 By Parts $u = \ln x$, $dv = dx$
Q5 Substitution $u = 1 + x^2$, $du = 2x,dx$
Q6 Trig Sub $x = 4\sin\theta$, $dx = 4\cos\theta,d\theta$
Q7 By Parts $u = x$, $dv = \sin 2x,dx$
Q8 Standard form $\int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}(x/a) + C$, $a=3$
Q9 By Parts (×2) $u = e^x$, $dv = \cos x,dx$ (loop method)
Q10 Trig Sub or Standard $3x = \sin\theta$, or recognize $\int \frac{6x^2}{\sqrt{1-9x^2}},dx$

R2: DE ID

Q Type Triage Cue
Q11 Separable $\frac{dy}{dx} = \frac{x}{y} \implies y,dy = x,dx$
Q12 Homogeneous All terms degree 2: $y^2, xy, x^2$
Q13 Linear Form $y' + P(x)y = Q(x)$
Q14 Linear (or separable after rewrite) $y' - \frac{1}{x}y = x^2$ — linear
Q15 Homogeneous $y^2, x^2, xy$ all degree 2
Q16 Separable $(x+1)\frac{dy}{dx} = x(y+3) \implies \frac{dy}{y+3} = \frac{x}{x+1},dx$

R3: Series ID

Q Method Key Insight
Q17 Binomial I Positive integer $n=5$, use Pascal/nCr
Q18 Maclaurin sub $e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots$
Q19 Method of Differences $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$
Q20 Binomial I $(x^3)^? (-\frac{1}{x^2})^{9-?}$ → solve exponents
Q21 Maclaurin integration $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \cdots$, integrate term by term
Q22 Binomial II (negative power) $\sqrt{1-x} = (1-x)^{1/2}$, expand

R4: Geometry ID

Q Type Why
Q23 Parabola Only $x$ squared, form $(x-h)^2 = 4a(y-k)$, opens up
Q24 Ellipse Both squared, same sign, different coefficients
Q25 Hyperbola (vertical) Opposite signs, $y^2$ positive = vertical axis
Q26 Parametric $x$ and $y$ in terms of $t$
Q27 Ellipse Need to complete squares: $16(x^2-4x) + 4(y^2-10y) = -100$
Q28 Hyperbola (horizontal) $x^2$ positive, $y^2$ negative

R5: Mixed Solve

Q29

$$\int \frac{dx}{\sqrt{4 - x^2}} = \sin^{-1}\left(\frac{x}{2}\right) + C$$ (Standard form: $\int dx/\sqrt{a^2 - x^2} = \sin^{-1}(x/a) + C$, $a=2$)

Q30

$$\int x^2 \sin x , dx$$

By parts: $u = x^2$, $dv = \sin x,dx$ $du = 2x,dx$, $v = -\cos x$

$$\int x^2 \sin x , dx = -x^2\cos x + \int 2x\cos x,dx$$

Second by parts: $u = 2x$, $dv = \cos x,dx$ $du = 2,dx$, $v = \sin x$

$$= -x^2\cos x + 2x\sin x - \int 2\sin x,dx$$ $$= -x^2\cos x + 2x\sin x + 2\cos x + C$$

Q31

$$\frac{dy}{dx} = \frac{y}{x},\quad y(1) = 3$$

Separable: $\frac{dy}{y} = \frac{dx}{x}$

$$\ln|y| = \ln|x| + C \implies y = Cx$$

$$y(1) = 3 \implies C = 3$$

$$\boxed{y = 3x}$$

Q32

$$\int \frac{dx}{\sqrt{x^2 + 9}} = \sinh^{-1}\left(\frac{x}{3}\right) + C = \ln\left|x + \sqrt{x^2+9}\right| + C$$

Q33

$$16x^2 + 4y^2 - 64x - 40y + 100 = 0$$

Complete squares: $$16(x^2 - 4x) + 4(y^2 - 10y) = -100$$ $$16[(x-2)^2 - 4] + 4[(y-5)^2 - 25] = -100$$ $$16(x-2)^2 + 4(y-5)^2 = 64 + 100 - 100 = 64$$ $$\frac{(x-2)^2}{4} + \frac{(y-5)^2}{16} = 1$$

Centre: $(2, 5)$ $a^2 = 16 \implies a = 4$ (vertical major axis), $b^2 = 4 \implies b = 2$ $$c^2 = a^2 - b^2 = 16 - 4 = 12 \implies c = 2\sqrt{3}$$

Vertices: $(2, 5\pm4) = (2, 9)$ and $(2, 1)$ Foci: $(2, 5\pm2\sqrt{3})$

Q34

$$(x+1)\frac{dy}{dx} = x(y+3)$$

Separable: $\frac{dy}{y+3} = \frac{x}{x+1},dx$

RHS: $\frac{x}{x+1} = 1 - \frac{1}{x+1}$

$$\int \frac{dy}{y+3} = \int \left(1 - \frac{1}{x+1}\right)dx$$ $$\ln|y+3| = x - \ln|x+1| + C$$ $$\ln|y+3| + \ln|x+1| = x + C$$ $$\ln|(y+3)(x+1)| = x + C$$ $$(y+3)(x+1) = Ce^x$$ $$y+3 = \frac{Ce^x}{x+1}$$

Q35

$$\int x e^{3x} , dx$$

By parts: $u = x$, $dv = e^{3x},dx$ $du = dx$, $v = \frac{1}{3}e^{3x}$

$$= \frac{x}{3}e^{3x} - \frac{1}{3}\int e^{3x},dx = \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C$$

Q36

$$\int_0^{0.5} e^{-x^2} , dx$$

Standard series: $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$

Let $u = -x^2$: $$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots$$ $$= 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \frac{x^8}{24} - \cdots$$

Integrate term by term: $$\int_0^{0.5} e^{-x^2},dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots\right]_0^{0.5}$$

$x = 0.5$:

  • $x = 0.5$
  • $-\frac{x^3}{3} = -\frac{0.125}{3} = -0.0416667$
  • $+\frac{x^5}{10} = \frac{0.03125}{10} = 0.003125$
  • $-\frac{x^7}{42} = -\frac{0.0078125}{42} = -0.000186$
  • $+\frac{x^9}{216} = \frac{0.001953125}{216} \approx 0.000009$

Sum: $0.5 - 0.041667 + 0.003125 - 0.000186 + 0.000009 = 0.461281$

$$\boxed{\approx 0.4613 \text{ (4 d.p.)}}$$