FAD1014: Comprehensive Rapid-Fire Drill — Full Exam Scope

Objective: Diagnose weak spots across the entire confirmed exam scope through high-volume drilling.
Target: 2-3 min per problem. If stalled >3 min, skip and mark for review.
Total problems: 60
Estimated time: 120-180 min

Cheat Sheet (Memorize First)

Integration

Technique Form Method
Substitution $\int f(g(x))g'(x),dx$ $u = g(x)$, $du = g'(x),dx$
Parts $\int u,dv$ $uv - \int v,du$ (LIATE order)
Trig Sub $\sqrt{a^2 - x^2}$ $x = a\sin\theta$, $dx = a\cos\theta,d\theta$
Trig Sub $\sqrt{a^2 + x^2}$ $x = a\tan\theta$, $dx = a\sec^2\theta,d\theta$
Trig Sub $\sqrt{x^2 - a^2}$ $x = a\sec\theta$, $dx = a\sec\theta\tan\theta,d\theta$

Standard integrals to know: $\int x^n,dx = \frac{x^{n+1}}{n+1} + C;(n\neq -1)$, $\int \frac{1}{x},dx = \ln\vert x \vert + C$, $\int e^x,dx = e^x + C$, $\int \sin x,dx = -\cos x + C$, $\int \cos x,dx = \sin x + C$, $\int \sec^2 x,dx = \tan x + C$

Area Under Curve

$$A = \int_a^b f(x),dx \quad (f(x) \geq 0 \text{ on } [a,b])$$

Differential Equations

Type Form Method
Separable $\frac{dy}{dx} = f(x)g(y)$ $\int \frac{dy}{g(y)} = \int f(x),dx$
Homogeneous $\frac{dy}{dx} = f!\left(\frac{y}{x}\right)$ $y = vx$, $dy/dx = v + x,dv/dx$

Maclaurin Series

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$

Function Series Valid for
$e^x$ $\sum_{n=0}^\infty \frac{x^n}{n!}$ All $x$
$\sin x$ $\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ All $x$
$\cos x$ $\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$ All $x$
$\ln(1+x)$ $\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$ $-1 < x \leq 1$
$(1+x)^n$ $\sum_{r=0}^\infty \binom{n}{r} x^r$ $\vert x \vert < 1$

Method of Differences

$$\sum_{k=1}^n (f(k) - f(k+1)) = f(1) - f(n+1)$$

Binomial Expansion I

$$(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r,\quad n \in \mathbb{Z}^+$$

Conic Sections

Conic Standard Form (origin) Centre $(h,k)$ Key
Parabola (vert) $x^2 = 4py$ $(x-h)^2 = 4p(y-k)$ Focus $(h,k+p)$, directrix $y=k-p$
Parabola (horiz) $y^2 = 4px$ $(y-k)^2 = 4p(x-h)$ Focus $(h+p,k)$, directrix $x=h-p$
Ellipse (horiz) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a>b$ $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2-b^2$, foci $(\pm c,0)$, vertices $(\pm a,0)$
Ellipse (vert) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $b>a$ — $c^2 = b^2-a^2$, foci $(0,\pm c)$, vertices $(0,\pm b)$
Hyperbola (horiz) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2+b^2$, foci $(\pm c,0)$, asymptotes $y = \pm\frac{b}{a}x$
Hyperbola (vert) $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ $c^2 = a^2+b^2$, foci $(0,\pm c)$, asymptotes $y = \pm\frac{a}{b}x$

Parametric Equations

Curve Parametric Form Cartesian
Circle $x = h + r\cos t$, $y = k + r\sin t$ $(x-h)^2 + (y-k)^2 = r^2$
Parabola $x = h + at^2$, $y = k + 2at$ $(y-k)^2 = 4a(x-h)$
Ellipse $x = h + a\cos t$, $y = k + b\sin t$ $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
Hyperbola $x = h + a\sec t$, $y = k + b\tan t$ $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

Part A: Integration — Standard & Substitution

Target: 2 min per problem.

Set A1 — Standard Integrals (4 problems)

Evaluate each.

  1. $\displaystyle \int (3x^5 - 2x^3 + 7),dx$
  2. $\displaystyle \int \frac{x^3 + 2}{x^2},dx$
  3. $\displaystyle \int (e^x + \cos x - \sec^2 x),dx$
  4. $\displaystyle \int \frac{1}{x},dx$

Score: ___/4

Set A2 — Integration by Substitution (4 problems)

  1. $\displaystyle \int (2x+1)^4,dx$
  2. $\displaystyle \int x e^{x^2},dx$
  3. $\displaystyle \int \frac{2x}{x^2+1},dx$
  4. $\displaystyle \int \sin^3 x \cos x,dx$

Score: ___/4

Set A3 — Definite Integrals (2 problems)

  1. $\displaystyle \int_0^1 x\sqrt{1+x^2},dx$
  2. $\displaystyle \int_1^e \frac{1}{x},dx$

Score: ___/2

Part B: Integration by Parts

Target: 3 min per problem. (6 problems)

  1. $\displaystyle \int x e^{2x},dx$
  2. $\displaystyle \int x \ln x,dx$
  3. $\displaystyle \int x^2 \ln x,dx$
  4. $\displaystyle \int e^x \sin x,dx$
  5. $\displaystyle \int \arcsin x,dx$
  6. $\displaystyle \int_0^1 x e^x,dx$

Score: ___/6

Part C: Trigonometric Substitution

Target: 3 min per problem. (4 problems)

  1. $\displaystyle \int \frac{dx}{\sqrt{4 - x^2}}$
  2. $\displaystyle \int \frac{dx}{\sqrt{x^2 + 1}}$
  3. $\displaystyle \int \frac{dx}{\sqrt{x^2 - 4}}$
  4. $\displaystyle \int \frac{dx}{(x^2+1)^{3/2}}$

Score: ___/4

Part D: Area Under Curve

Target: 3 min per problem. (3 problems)

  1. Find the area under $y = x^2$ from $x = 0$ to $x = 2$.
  2. Find the area under $y = \sin x$ from $x = 0$ to $x = \pi$.
  3. Find the area under $y = e^x$ from $x = \ln 2$ to $x = \ln 4$.

Score: ___/3

Part E: Separable DE

Target: 3 min per problem. (5 problems)

  1. $\displaystyle \frac{dy}{dx} = \frac{x}{y}$, $y(0) = 2$
  2. $\displaystyle \frac{dy}{dx} = 3x^2 e^{-y}$
  3. $\displaystyle \frac{dy}{dx} = \frac{y}{x}$, $y(1) = 1$
  4. $\displaystyle \frac{dy}{dx} = xy$, $y(0) = 3$
  5. $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{x}$, $y(1) = 0$

Score: ___/5

Part F: Homogeneous DE

Target: 3 min per problem. (6 problems)

  1. $\displaystyle \frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}$
  2. $\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{xy}$
  3. $\displaystyle \frac{dy}{dx} = \frac{y}{x} + \sin!\left(\frac{y}{x}\right)$
  4. $\displaystyle \frac{dy}{dx} = \frac{x + y}{x}$, $y(1) = 2$
  5. $\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$
  6. $\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{x^2}$, $y(1) = 0$

Score: ___/6

Part G: Maclaurin Series

Target: 2-3 min per problem. (10 problems)

Set G1 — Derivation from Definition (3 problems)

  1. $f(x) = \sec x$ up to $x^4$
  2. $f(x) = \ln(1 + \sin x)$ up to $x^4$
  3. $f(x) = e^x \cos x$ up to $x^4$

Score: ___/3

Set G2 — Substitution in Known Series (3 problems)

  1. $e^{-2x^2}$ up to $x^6$
  2. $\sin(x^3)$ up to $x^{15}$
  3. $\ln(1 + 3x)$ up to $x^4$

Score: ___/3

Set G3 — Term-by-Term Integration (4 problems)

Use series to find each indefinite integral (first 3 non-zero terms).

  1. $\displaystyle \int \frac{e^x - 1}{x},dx$
  2. $\displaystyle \int \cos(x^2),dx$
  3. $\displaystyle \int \frac{\sin x}{x},dx$
  4. $\displaystyle \int e^{-x^2},dx$

Score: ___/4

Part H: Method of Differences & Binomial I

Target: 2-3 min per problem. (5 problems)

  1. $\displaystyle \sum_{k=1}^{n} \frac{1}{k(k+1)}$ using method of differences. (Hint: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$)
  2. $\displaystyle \sum_{k=1}^{10} (2k-1)$ using method of differences.
  3. Expand $(2x + 3)^4$ using the binomial theorem.
  4. Find the coefficient of $x^3$ in $(1 + 2x)^5$.
  5. Find the term independent of $x$ in $\left(x + \frac{1}{x}\right)^6$.

Score: ___/5

Part I: Parabola

Target: 2-3 min per problem. (4 problems)

  1. Find vertex, focus, directrix, and latus rectum length for $y^2 = 12x$.
  2. Find vertex, focus, directrix for $(x-3)^2 = 8(y+1)$.
  3. Write the equation of a parabola with vertex at $(0,0)$ and focus at $(0,4)$.
  4. Convert $x^2 + 4x - 8y + 12 = 0$ to standard form and find vertex, focus, directrix.

Score: ___/4

Part J: Ellipse

Target: 2-3 min per problem. (4 problems)

  1. For $\frac{x^2}{25} + \frac{y^2}{9} = 1$, find centre, vertices, foci, major/minor axes.
  2. For $\frac{(x-2)^2}{16} + \frac{(y+3)^2}{25} = 1$, find centre, vertices, foci, orientation.
  3. Convert $x^2 + 4y^2 + 6x - 16y + 21 = 0$ to standard form. Find centre, $a$, $b$, $c$.
  4. Write the equation of an ellipse with centre $(0,0)$, vertices $(\pm 5, 0)$, foci $(\pm 3, 0)$.

Score: ___/4

Part K: Hyperbola

Target: 2-3 min per problem. (4 problems)

  1. For $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find centre, vertices, foci, asymptotes.
  2. For $\frac{y^2}{25} - \frac{x^2}{4} = 1$, find centre, vertices, foci, asymptotes, orientation.
  3. Write the equation of a hyperbola with centre $(0,0)$, vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$.

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A Standard/Substitution/Definite Integration 1-10 ___/10
B Integration by Parts 11-16 ___/6
C Trigonometric Substitution 17-20 ___/4
D Area Under Curve 21-23 ___/3
E Separable DE 24-28 ___/5
F Homogeneous DE 29-34 ___/6
G Maclaurin Series 35-44 ___/10
H Method of Differences & Binomial I 45-49 ___/5
I Parabola 50-53 ___/4
J Ellipse 54-57 ___/4
K Hyperbola 58-60 ___/3
TOTAL 1-60 ___/60

Per-Topic Weak-Spot Diagnosis

Topic Score Verdict
Standard Integration ___/4 if <3, drill basic power/trig integrals
Substitution ___/4 if <3, practise $u$-sub patterns
Definite Integrals ___/2 if <1, review FTC and limit changing
Integration by Parts ___/6 if <4, drill LIATE ordering
Trig Substitution ___/4 if <2, memorise the 3 forms
Area Under Curve ___/3 if <2, review $A = \int_a^b f(x),dx$
Separable DE ___/5 if <3, drill separation + integration
Homogeneous DE ___/6 if <4, drill $y=vx$ substitution
Maclaurin Series ___/10 if <7, memorise standard expansions cold
Method of Differences ___/2 if <1, review telescoping sums
Binomial I ___/3 if <2, drill $\binom{n}{r}$ expansion
Parabola ___/4 if <2, memorise $4p$ sign conventions
Ellipse ___/4 if <2, drill $c^2 = a^2-b^2$ and completing square
Hyperbola ___/3 if <2, drill $c^2 = a^2+b^2$ and asymptotes

Proficiency Benchmarks

  • 42/60 (70%) — Proficient. You can handle standard exam problems.
  • 51/60 (85%) — Solid. Fast and accurate.
  • 56/60 (93%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <90 min: Excellent mechanical fluency.
  • 90-135 min: Good. Review missed patterns.
  • >135 min: Drill the specific sections you scored lowest on again tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Topic Reason
e.g. 4 Substitution wrong $u$

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Set A1 — Standard Integrals

  1. $\int (3x^5 - 2x^3 + 7),dx = \frac{3x^6}{6} - \frac{2x^4}{4} + 7x + C = \frac{x^6}{2} - \frac{x^4}{2} + 7x + C$

  2. $\int \frac{x^3 + 2}{x^2},dx = \int (x + 2x^{-2}),dx = \frac{x^2}{2} - \frac{2}{x} + C$

  3. $\int (e^x + \cos x - \sec^2 x),dx = e^x + \sin x - \tan x + C$

  4. $\int \frac{1}{x},dx = \ln\vert x \vert + C$

Set A2 — Substitution

  1. $\int (2x+1)^4,dx$. $u=2x+1$, $du=2dx$. $\frac12\int u^4,du = \frac12\cdot\frac{u^5}{5} + C = \frac{(2x+1)^5}{10} + C$

  2. $\int x e^{x^2},dx$. $u=x^2$, $du=2x,dx$. $\frac12\int e^u,du = \frac12 e^{x^2} + C$

  3. $\int \frac{2x}{x^2+1},dx$. $u=x^2+1$, $du=2x,dx$. $\int \frac{du}{u} = \ln\vert x^2+1 \vert + C$

  4. $\int \sin^3 x \cos x,dx$. $u=\sin x$, $du=\cos x,dx$. $\int u^3,du = \frac{\sin^4 x}{4} + C$

Set A3 — Definite Integrals

  1. $\int_0^1 x\sqrt{1+x^2},dx$. $u=1+x^2$, $du=2x,dx$. $x=0\to u=1$, $x=1\to u=2$. $$\frac12\int_1^2 u^{1/2},du = \frac12\cdot\frac23\big[u^{3/2}\big]_1^2 = \frac13(2\sqrt2 - 1)$$

  2. $\int_1^e \frac{1}{x},dx = \big[\ln x\big]_1^e = \ln e - \ln 1 = 1 - 0 = 1$

Part B — Integration by Parts

  1. $\int x e^{2x},dx$. $u=x$, $dv=e^{2x}dx$. $du=dx$, $v=\frac12 e^{2x}$. $$= \frac{x}{2}e^{2x} - \int \frac12 e^{2x},dx = \frac{x}{2}e^{2x} - \frac14 e^{2x} + C$$

  2. $\int x \ln x,dx$. $u=\ln x$, $dv=x,dx$. $du=\frac1x dx$, $v=\frac{x^2}{2}$. $$= \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac1x,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

  3. $\int x^2 \ln x,dx$. $u=\ln x$, $dv=x^2,dx$. $du=\frac1x dx$, $v=\frac{x^3}{3}$. $$= \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac1x,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C$$

  4. $\int e^x \sin x,dx$. $I = \int e^x\sin x,dx$. $u=\sin x$, $dv=e^x,dx$. $du=\cos x,dx$, $v=e^x$. $I = e^x\sin x - \int e^x\cos x,dx$. For $\int e^x\cos x,dx$: $u=\cos x$, $dv=e^x,dx$. $du=-\sin x,dx$, $v=e^x$. $= e^x\cos x + \int e^x\sin x,dx = e^x\cos x + I$. So $I = e^x\sin x - (e^x\cos x + I) \implies 2I = e^x(\sin x - \cos x) \implies I = \frac12 e^x(\sin x - \cos x) + C$

  5. $\int \arcsin x,dx$. $u=\arcsin x$, $dv=dx$. $du=\frac{1}{\sqrt{1-x^2}}dx$, $v=x$. $= x\arcsin x - \int \frac{x}{\sqrt{1-x^2}},dx$. $t=1-x^2$, $dt=-2x,dx$. $$= x\arcsin x + \frac12\int t^{-1/2},dt = x\arcsin x + \sqrt{1-x^2} + C$$

  6. $\int_0^1 x e^x,dx$. $u=x$, $dv=e^x,dx$. $du=dx$, $v=e^x$. $$= \big[x e^x\big]_0^1 - \int_0^1 e^x,dx = e - (e-1) = 1$$

Part C — Trig Substitution

  1. $\int \frac{dx}{\sqrt{4-x^2}}$. $x=2\sin\theta$, $dx=2\cos\theta,d\theta$. $$\int \frac{2\cos\theta,d\theta}{\sqrt{4-4\sin^2\theta}} = \int \frac{2\cos\theta}{2\cos\theta},d\theta = \int d\theta = \theta + C = \arcsin!\left(\frac{x}{2}\right) + C$$

  2. $\int \frac{dx}{\sqrt{x^2+1}}$. $x=\tan\theta$, $dx=\sec^2\theta,d\theta$. $$\int \frac{\sec^2\theta,d\theta}{\sqrt{\tan^2\theta+1}} = \int \frac{\sec^2\theta}{\sec\theta},d\theta = \int \sec\theta,d\theta = \ln\vert \sec\theta + \tan\theta \vert + C = \ln\vert \sqrt{x^2+1} + x \vert + C$$

  3. $\int \frac{dx}{\sqrt{x^2-4}}$. $x=2\sec\theta$, $dx=2\sec\theta\tan\theta,d\theta$. $$\int \frac{2\sec\theta\tan\theta,d\theta}{\sqrt{4\sec^2\theta-4}} = \int \frac{2\sec\theta\tan\theta}{2\tan\theta},d\theta = \int \sec\theta,d\theta = \ln\vert \sec\theta+\tan\theta \vert + C = \ln\left\vert \frac{x}{2} + \frac{\sqrt{x^2-4}}{2} \right\vert + C$$

  4. $\int \frac{dx}{(x^2+1)^{3/2}}$. $x=\tan\theta$, $dx=\sec^2\theta,d\theta$. $$\int \frac{\sec^2\theta,d\theta}{(\tan^2\theta+1)^{3/2}} = \int \frac{\sec^2\theta}{\sec^3\theta},d\theta = \int \cos\theta,d\theta = \sin\theta + C = \frac{x}{\sqrt{x^2+1}} + C$$

Part D — Area Under Curve

  1. $A = \int_0^2 x^2,dx = \big[\frac{x^3}{3}\big]_0^2 = \frac{8}{3}$

  2. $A = \int_0^\pi \sin x,dx = \big[-\cos x\big]_0^\pi = -(-1) - (-1) = 1 + 1 = 2$

  3. $A = \int_{\ln 2}^{\ln 4} e^x,dx = \big[e^x\big]_{\ln 2}^{\ln 4} = 4 - 2 = 2$

Part E — Separable DE

  1. $\frac{dy}{dx} = \frac{x}{y}$, $y(0)=2$. $y,dy = x,dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C \implies y^2 = x^2 + 2C$. $y(0)=2 \implies 4 = 2C \implies C=2$. So $y^2 = x^2 + 4$, $y = \sqrt{x^2+4}$

  2. $\frac{dy}{dx} = 3x^2 e^{-y}$. $$e^y,dy = 3x^2,dx \implies e^y = x^3 + C \implies y = \ln(x^3 + C)$$

  3. $\frac{dy}{dx} = \frac{y}{x}$, $y(1)=1$. $\frac{dy}{y} = \frac{dx}{x} \implies \ln\vert y \vert = \ln\vert x \vert + C \implies y = C_1 x$. $y(1)=1 \implies C_1 = 1$. So $y = x$

  4. $\frac{dy}{dx} = xy$, $y(0)=3$. $\frac{dy}{y} = x,dx \implies \ln\vert y \vert = \frac{x^2}{2} + C \implies y = C_1 e^{x^2/2}$. $y(0)=3 \implies C_1 = 3$. So $y = 3e^{x^2/2}$

  5. $\frac{dy}{dx} = \frac{1+y^2}{x}$, $y(1)=0$. $\frac{dy}{1+y^2} = \frac{dx}{x} \implies \arctan y = \ln\vert x \vert + C$. $y(1)=0 \implies 0 = 0 + C \implies C=0$. So $y = \tan(\ln\vert x \vert)$

Part F — Homogeneous DE

  1. $\frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}$. $y=vx$, $dy/dx = v + x,dv/dx$. $v + x\frac{dv}{dx} = v + \frac1v \implies x\frac{dv}{dx} = \frac1v \implies v,dv = \frac{dx}{x}$. $$\frac{v^2}{2} = \ln\vert x \vert + C \implies \frac{y^2}{2x^2} = \ln\vert x \vert + C \implies y^2 = 2x^2\ln\vert x \vert + C_1 x^2$$

  2. $\frac{dy}{dx} = \frac{x^2+y^2}{xy} = \frac{x}{y} + \frac{y}{x}$. $y=vx$. $v + x\frac{dv}{dx} = \frac1v + v \implies x\frac{dv}{dx} = \frac1v \implies v,dv = \frac{dx}{x}$. $$\frac{v^2}{2} = \ln\vert x \vert + C \implies y^2 = 2x^2\ln\vert x \vert + C_1 x^2$$

  3. $\frac{dy}{dx} = \frac{y}{x} + \sin!\left(\frac{y}{x}\right)$. $y=vx$. $v + x\frac{dv}{dx} = v + \sin v \implies x\frac{dv}{dx} = \sin v \implies \csc v,dv = \frac{dx}{x}$. $$\ln\vert \csc v - \cot v \vert = \ln\vert x \vert + C \implies \csc!\left(\frac{y}{x}\right) - \cot!\left(\frac{y}{x}\right) = C_1 x$$

  4. $\frac{dy}{dx} = \frac{x+y}{x} = 1 + \frac{y}{x}$, $y(1)=2$. $y=vx$. $v + x\frac{dv}{dx} = 1 + v \implies x\frac{dv}{dx} = 1 \implies dv = \frac{dx}{x}$. $v = \ln\vert x \vert + C \implies y = x\ln\vert x \vert + Cx$. $y(1)=2 \implies 2 = 0 + C \implies C=2$. So $y = x\ln\vert x \vert + 2x$

  5. $\frac{dy}{dx} = \frac{x^2+y^2}{2xy} = \frac12!\left(\frac{x}{y} + \frac{y}{x}\right)$. $y=vx$. $v + x\frac{dv}{dx} = \frac12!\left(\frac1v + v\right) \implies x\frac{dv}{dx} = \frac{1}{2v} - \frac{v}{2} = \frac{1-v^2}{2v}$. $\frac{2v}{1-v^2},dv = \frac{dx}{x} \implies -\ln\vert 1-v^2 \vert = \ln\vert x \vert + C \implies \frac{1}{1-v^2} = C_1 x$. $$\frac{x^2}{x^2-y^2} = C_1 x \implies x = C_1(x^2 - y^2)$$

  6. $\frac{dy}{dx} = \frac{x^2+y^2}{x^2} = 1 + \frac{y^2}{x^2}$, $y(1)=0$. $y=vx$. $v + x\frac{dv}{dx} = 1 + v^2 \implies x\frac{dv}{dx} = 1 + v^2 - v$. $\frac{dv}{v^2 - v + 1} = \frac{dx}{x}$. $v^2 - v + 1 = (v-\frac12)^2 + \frac34$. $\frac{2}{\sqrt3}\arctan!\left(\frac{2v-1}{\sqrt3}\right) = \ln\vert x \vert + C$. $y(1)=0 \implies v(1)=0 \implies \frac{2}{\sqrt3}\arctan!\left(-\frac{1}{\sqrt3}\right) = C \implies C = \frac{2}{\sqrt3}\cdot(-\frac{\pi}{6}) = -\frac{\pi}{3\sqrt3}$. $$\frac{2}{\sqrt3}\arctan!\left(\frac{2y-x}{\sqrt3,x}\right) = \ln\vert x \vert - \frac{\pi}{3\sqrt3}$$

Set G1 — Maclaurin from Definition

  1. $f(x)=\sec x$. $f(0)=1$, $f'(0)=0$, $f''(0)=1$, $f'''(0)=0$, $f^{(4)}(0)=5$. $$\sec x = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$

  2. $f(x)=\ln(1+\sin x)$. $f(0)=0$, $f'(0)=1$, $f''(0)=-1$, $f'''(0)=1$, $f^{(4)}(0)=-2$. $$= x - \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^4}{12} + \cdots$$

  3. $f(x)=e^x\cos x$. $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$, $\cos x = 1 - \frac{x^2}{2} + \cdots$. Multiply: $= 1 + x + \left(\frac12-\frac12\right)x^2 + \left(\frac16-\frac12\right)x^3 + \cdots = 1 + x - \frac{x^3}{3} + \cdots$

Set G2 — Substitution

  1. $e^{-2x^2} = 1 - 2x^2 + \frac{4x^4}{2} - \frac{8x^6}{6} + \cdots = 1 - 2x^2 + 2x^4 - \frac43 x^6 + \cdots$

  2. $\sin(x^3) = x^3 - \frac{x^9}{6} + \frac{x^{15}}{120} - \cdots$

  3. $\ln(1+3x) = 3x - \frac{9x^2}{2} + 9x^3 - \frac{81x^4}{4} + \cdots$

Set G3 — Integration

  1. $\frac{e^x-1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$ $$\int = C + x + \frac{x^2}{4} + \frac{x^3}{18} + \frac{x^4}{96} + \cdots$$

  2. $\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \frac{x^{12}}{720} + \cdots$ $$\int = C + x - \frac{x^5}{10} + \frac{x^9}{216} - \frac{x^{13}}{9360} + \cdots$$

  3. $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$ $$\int = C + x - \frac{x^3}{18} + \frac{x^5}{600} - \frac{x^7}{35280} + \cdots$$

  4. $e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$ $$\int = C + x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots$$

Part H — Method of Differences & Binomial I

  1. $\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \left(\frac1k - \frac1{k+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$

  2. $\sum_{k=1}^{10} (2k-1) = \sum_{k=1}^{10} (k^2 - (k-1)^2) = 10^2 - 0^2 = 100$

  3. $(2x+3)^4 = \sum_{r=0}^4 \binom{4}{r}(2x)^{4-r}3^r = 16x^4 + 96x^3 + 216x^2 + 216x + 81$

  4. $(1+2x)^5$. Term $r+1$: $\binom{5}{r}1^{5-r}(2x)^r = \binom{5}{r}2^r x^r$. $x^3$ term: $r=3$, $\binom{5}{3}2^3 = 10 \cdot 8 = 80$

  5. $(x + \frac1x)^6$. General term: $\binom{6}{r} x^{6-r} x^{-r} = \binom{6}{r} x^{6-2r}$. Independent: $6-2r=0 \implies r=3$. Term: $\binom{6}{3} = 20$

Part I — Parabola

  1. $y^2 = 12x$. $4p = 12 \implies p = 3$. Vertex $(0,0)$, focus $(3,0)$, directrix $x=-3$, latus rectum $12$. Opens right.

  2. $(x-3)^2 = 8(y+1)$. $4p = 8 \implies p = 2$, $h=3$, $k=-1$. Vertex $(3,-1)$, focus $(3,1)$, directrix $y=-3$.

  3. Vertex $(0,0)$, focus $(0,4)$. $p=4$, opens up. $x^2 = 4py = 16y$.

  4. $x^2 + 4x - 8y + 12 = 0$. $(x^2+4x+4) - 4 - 8y + 12 = 0 \implies (x+2)^2 = 8y-8 = 8(y-1)$. $4p=8 \implies p=2$, $h=-2$, $k=1$. Vertex $(-2,1)$, focus $(-2,3)$, directrix $y=-1$.

Part J — Ellipse

  1. $\frac{x^2}{25} + \frac{y^2}{9} = 1$. $a=5$, $b=3$. Horizontal ($a>b$). $c^2 = 25-9 = 16$, $c=4$. Centre $(0,0)$. Vertices $(\pm5,0)$. Foci $(\pm4,0)$. Major $10$, minor $6$.

  2. $\frac{(x-2)^2}{16} + \frac{(y+3)^2}{25} = 1$. $a=4$, $b=5$. Vertical ($b>a$). $c^2 = 25-16 = 9$, $c=3$. Centre $(2,-3)$. Vertices $(2,2)$ and $(2,-8)$. Foci $(2,0)$ and $(2,-6)$.

  3. $x^2 + 4y^2 + 6x - 16y + 21 = 0$. $(x^2+6x) + 4(y^2-4y) = -21$. $(x^2+6x+9) + 4(y^2-4y+4) = -21 + 9 + 16$. $(x+3)^2 + 4(y-2)^2 = 4$. $\frac{(x+3)^2}{4} + \frac{(y-2)^2}{1} = 1$. Centre $(-3,2)$, $a=2$, $b=1$, $c^2 = 4-1 = 3$, $c = \sqrt3$. Horizontal.

  4. Centre $(0,0)$, vertices $(\pm5,0)$ so $a=5$, foci $(\pm3,0)$ so $c=3$. $b^2 = a^2 - c^2 = 25-9 = 16$, $b=4$. $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$

Part K — Hyperbola

  1. $\frac{x^2}{16} - \frac{y^2}{9} = 1$. $a=4$, $b=3$. Horizontal. $c^2 = 16+9 = 25$, $c=5$. Centre $(0,0)$. Vertices $(\pm4,0)$. Foci $(\pm5,0)$. Asymptotes $y = \pm\frac34 x$.

  2. $\frac{y^2}{25} - \frac{x^2}{4} = 1$. $a=5$, $b=2$. Vertical. $c^2 = 25+4 = 29$, $c = \sqrt{29}$. Centre $(0,0)$. Vertices $(0,\pm5)$. Foci $(0,\pm\sqrt{29})$. Asymptotes $y = \pm\frac52 x$.

  3. Centre $(0,0)$, vertices $(\pm3,0)$ so $a=3$, foci $(\pm5,0)$ so $c=5$. $b^2 = c^2 - a^2 = 25-9 = 16$, $b=4$. $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$

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