FAC1004 — Full 2-3 Hour Mechanical Fluency Drill
6 sections, ~50 questions. Pen on paper. No peeking solutions. Each round has solutions below it.
SECTION A: COMPLEX NUMBERS — LOCI & INEQUALITIES
A1: Circles & Bisectors (6 min)
Q1
Find centre, radius, Cartesian equation: (a) $|z - (5 + 2i)| = 4$ (b) $|z + 3 - 4i| = 6$ (c) $|2z - 8 + 6i| = 10$
Q2
Perpendicular bisector equation: (a) $|z - (4 + 3i)| = |z - (2 + 5i)|$ (b) $|z + 2 - i| = |z - 4 + 3i|$ (c) $\left|\frac{z+1}{z-2}\right| = 1$
A2: Multiple Boundaries (10 min)
For each, find ALL Cartesian boundaries:
Q3
$$|z - 3i| \leq 5, \quad \text{Re}(z) \geq -1, \quad \text{Im}(z) \leq 4$$
Q4
$$|z - 2 + i| < |z + 4 - 3i|, \quad |z - 1| \leq 4, \quad \arg(z) \geq \frac{\pi}{6}$$
Q5
$$|z + 2 - i| \leq 3, \quad |z - i| < |z - 3i|, \quad \frac{\pi}{4} \leq \arg(z+1) \leq \frac{3\pi}{4}$$
Q6
$$|z - 2| \leq |z + 2|, \quad |z - 2i| \geq 3, \quad \text{Im}(z) \leq \text{Re}(z)$$
A3: Sketch & Test Points (8 min)
Q7
For Q4: sketch & shade. Is $z = 2i$ inside?
Q8
For Q5: sketch & shade. Is $z = 0$ inside?
Q9
For Q6: sketch & shade. Is $z = 1 + i$ inside?
Q10
A point $z$ satisfies: $$|z - 2 - 2i| \leq 2\sqrt{2}, \quad |z - 2i| \geq |z - 4i|, \quad \text{Re}(z) \geq 2$$ (a) Find all Cartesian boundaries. (b) Sketch & shade. (c) Is $z = 3 + i$ inside? Justify.
A4: Multiple Angles & Reverse Expansion (8 min)
Q11
Using $z + \frac{1}{z} = 2\cos x$ and $z - \frac{1}{z} = 2i\sin x$, express $\cos^3 x$ in terms of $\cos 3x$ and $\cos x$.
Q12
Express $\sin^5 x$ in terms of $\sin 5x$, $\sin 3x$ and $\sin x$.
Q13
Using Pascal's Triangle, expand $(\cos x + i\sin x)^5$ and hence prove: $$\cos 5x = 16\cos^5 x - 20\cos^3 x + 5\cos x$$
Q14
Using the compound angle formula (not $e^{ix}$), evaluate: $$\tan\left(\frac{\pi}{4} - i\right)$$
Solutions A
Q1
(a) Centre $(5,2)$, radius $4$: $(x-5)^2 + (y-2)^2 = 16$
(b) Centre $(-3,4)$, radius $6$: $(x+3)^2 + (y-4)^2 = 36$
(c) $2|z - 4 + 3i| = 10 \implies |z - (4-3i)| = 5$: Centre $(4,-3)$, radius $5$: $(x-4)^2 + (y+3)^2 = 25$
Q2
(a) Midpoint $(3,4)$. Seg slope $\frac{5-3}{2-4} = -1$. Perp slope $1$. $$y-4 = 1(x-3) \implies y = x+1$$
(b) $|z - (-2+i)| = |z - (4-3i)|$. Midpoint $(1,-1)$. Seg slope $\frac{-3-1}{4-(-2)} = -\frac{2}{3}$. Perp slope $\frac{3}{2}$. $$y+1 = \frac{3}{2}(x-1) \implies y = \frac{3}{2}x - \frac{5}{2}$$
(c) $|z+1| = |z-2| \implies$ bisector. Midpoint $(0.5,0)$. Vertical line $x = 0.5$.
Q3
- $|z-3i| = 5 \implies x^2 + (y-3)^2 = 25$
- $\text{Re}(z) = -1 \implies x = -1$
- $\text{Im}(z) = 4 \implies y = 4$
- Region: inside circle, right of $x=-1$, below $y=4$
Q4
- $|z-2+i| = |z+4-3i|$: bisector of $(2,-1)$ and $(-4,3)$. Midpoint $(-1,1)$. Seg slope $\frac{3-(-1)}{-4-2} = -\frac{2}{3}$. Perp slope $\frac{3}{2}$: $y-1 = \frac{3}{2}(x+1) \implies y = \frac{3}{2}x + \frac{5}{2}$
- $|z-1| = 4 \implies (x-1)^2 + y^2 = 16$ (circle centre $(1,0)$, r=4)
- $\arg(z) = \pi/6 \implies y = \frac{1}{\sqrt{3}}x$, $x>0$
- Region: side of bisector containing $(2,-1)$? inequality is $|z-2+i| < |z+4-3i|$ so closer to $(2,-1)$
- Inside/on circle, above $\arg = \pi/6$ ray
Q5
- $|z+2-i| = 3 \implies (x+2)^2 + (y-1)^2 = 9$ (circle centre $(-2,1)$, r=3)
- $|z-i| = |z-3i|$: Midpoint $(0,2)$. Horizontal $y=2$
- $\arg(z+1) = \pi/4$: line through $(-1,0)$ slope $1$: $y = x+1$
- $\arg(z+1) = 3\pi/4$: line through $(-1,0)$ slope $-1$: $y = -(x+1)$
- Region: inside circle, above $y=2$ (closer to $(0,1)$ than $(0,3)$ means below $y=2$... wait: $|z-i| < |z-3i|$ means closer to $(0,1)$ than $(0,3)$ → below $y=2$)
- Between the two rays from $(-1,0)$
Q6
- $|z-2| = |z+2|$: bisector of $(2,0)$ and $(-2,0)$. Midpoint $(0,0)$. Vertical line $x=0$. $|z-2| \leq |z+2|$ means closer to $(2,0)$ → right half-plane $x \geq 0$
- $|z-2i| = 3 \implies x^2 + (y-2)^2 = 9$, outside/on circle
- $\text{Im}(z) \leq \text{Re}(z) \implies y \leq x$
- Region: $x \geq 0$, outside/on circle $x^2+(y-2)^2=9$, below $y=x$
Q7
$z = 2i \implies (0,2)$. Check Q4 region.
- Bisector: $y = \frac{3}{2}x + \frac{5}{2}$. At $x=0$: $2 < 2.5$? Wait, which side? Closer to $(2,-1)$ than $(-4,3)$. Distance to $(2,-1)$: $\sqrt{(0-2)^2+(2+1)^2} = \sqrt{4+9}=\sqrt{13}$. Distance to $(-4,3)$: $\sqrt{(0+4)^2+(2-3)^2} = \sqrt{16+1}=\sqrt{17}$. $\sqrt{13} < \sqrt{17}$ ✓
- Circle: $(0-1)^2+2^2 = 1+4=5 \leq 16$ ✓
- Ray: $\arg(z) = \arg(2i) = \pi/2 \geq \pi/6$ ✓
- Inside.
Q8
$z = 0 \implies (0,0)$. Check Q5.
- Circle: $(0+2)^2+(0-1)^2 = 4+1=5 \leq 9$ ✓
- $|0-i| = 1$, $|0-3i| = 3$. $1 < 3$ ✓
- $\arg(0+1) = \arg(1) = 0$. Is $0$ between $\pi/4$ and $3\pi/4$? No. ✗
- Not inside.
Q9
$z = 1+i \implies (1,1)$. Check Q6.
- $x \geq 0$ ✓
- Circle: $1^2+(1-2)^2 = 1+1=2 < 9$. Condition is $\geq$ so... $2 \geq 9$? No ✗
- Not inside.
Q10
(a) Boundaries:
- $|z-2-2i| = 2\sqrt{2} \implies (x-2)^2 + (y-2)^2 = 8$ (circle centre $(2,2)$, r=$2\sqrt{2}$)
- $|z-2i| = |z-4i|$: Midpoint $(0,3)$. Horizontal $y=3$
- $\text{Re}(z) = 2 \implies x=2$
- Region: inside circle, above $y=3$ (closer to $(0,2)$ than $(0,4)$ means below $y=3$... wait $|z-2i| \geq |z-4i|$ means further from $(0,2)$ than $(0,4)$ → $y \geq 3$), right of $x=2$
(b) Region: inside/on circle $(x-2)^2+(y-2)^2 \leq 8$, $y \geq 3$, $x \geq 2$
(c) $z=3+i \implies (3,1)$: $y=1 < 3$ ✗. Not inside.
Q11
$z + \frac{1}{z} = 2\cos x$, so $\left(z + \frac{1}{z}\right)^3 = 8\cos^3 x$
$$(z + z^{-1})^3 = z^3 + 3z + 3z^{-1} + z^{-3} = (z^3 + z^{-3}) + 3(z + z^{-1})$$
Using identities: $z^3 + z^{-3} = 2\cos 3x$ and $z + z^{-1} = 2\cos x$
$$8\cos^3 x = 2\cos 3x + 3(2\cos x) = 2\cos 3x + 6\cos x$$
$$\cos^3 x = \frac{1}{4}\cos 3x + \frac{3}{4}\cos x$$
Q12
$z - \frac{1}{z} = 2i\sin x$, so $\left(z - z^{-1}\right)^5 = (2i)^5\sin^5 x = 32i\sin^5 x$
$$(z - z^{-1})^5 = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5}$$ $$= (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1})$$
Using identities: $z^n - z^{-n} = 2i\sin nx$
$$32i\sin^5 x = 2i\sin 5x - 5(2i\sin 3x) + 10(2i\sin x)$$
$$\sin^5 x = \frac{1}{16}\sin 5x - \frac{5}{16}\sin 3x + \frac{5}{8}\sin x$$
Q13
$$(\cos x + i\sin x)^5 = \cos^5 x + 5i\cos^4 x\sin x + 10i^2\cos^3 x\sin^2 x + 10i^3\cos^2 x\sin^3 x + 5i^4\cos x\sin^4 x + i^5\sin^5 x$$
Real part: $\cos^5 x - 10\cos^3 x\sin^2 x + 5\cos x\sin^4 x$
Substitute $\sin^2 x = 1 - \cos^2 x$:
$$= \cos^5 x - 10\cos^3 x(1-\cos^2 x) + 5\cos x(1-\cos^2 x)^2$$ $$= \cos^5 x - 10\cos^3 x + 10\cos^5 x + 5\cos x(1 - 2\cos^2 x + \cos^4 x)$$ $$= 11\cos^5 x - 10\cos^3 x + 5\cos x - 10\cos^3 x + 5\cos^5 x$$ $$= 16\cos^5 x - 20\cos^3 x + 5\cos x$$
By De Moivre: Real part $= \cos 5x$ ✓
Q14
Compound angle (not $e^{ix}$): $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$$
With $A = \frac{\pi}{4}$ and $B = i$: $$\tan\left(\frac{\pi}{4} - i\right) = \frac{1 - \tan(i)}{1 + \tan(i)}$$
Using $\tan(i) = \frac{e^{-1} - e^{1}}{e^{-1} + e^{1}} = -\tanh(1) = -\frac{e^2 - 1}{e^2 + 1}$:
$$= \frac{1 + \tanh(1)}{1 - \tanh(1)} = \frac{e^2 + 1 + e^2 - 1}{e^2 + 1 - e^2 + 1} = \frac{2e^2}{2} = e^2$$
SECTION B: BERNOULLI & NON-HOMOGENEOUS DEs
B1: DE Identification (3 min)
State type and method:
Q15
$$\frac{dy}{dx} + \frac{y}{x} = x^2 y^3$$
Q16
$$(x - 2y),dx + (4x - 3y + 2),dy = 0$$
Q17
$$\frac{dy}{dx} = \frac{y^2 + 2xy}{x^2}$$
Q18
$$3x\frac{dy}{dx} + y = \frac{x^2}{y^2}$$
Q19
$$(2x + y - 1),dx + (4x + 2y + 3),dy = 0$$
B2: Bernoulli — Full Solve (15 min)
KEY INSIGHT: Coefficient of $\frac{dy}{dx}$ must be 1 before starting. Never simplify $\frac{C}{x^{-2}}$ as a new constant.
Q20
Solve: $$x\frac{dy}{dx} + y = x^2 y^2$$ (First: divide by $x$ to make coefficient 1. Then identify $n,P,Q$.)
Q21
Solve: $$\frac{dy}{dx} + y = xy^3$$
Q22
Solve: $$\frac{dy}{dx} - \frac{y}{x} = \frac{x}{y}$$
Q23
Solve: $$2x\frac{dy}{dx} + y = \frac{2x}{y}\tan^{-1} x$$ (First: divide by $2x$ to make coefficient 1.)
Q24
Solve IVP: $$\frac{dy}{dx} + \frac{2}{x}y = x^3 y^2, \quad y(1) = 1$$
Q25
Solve: $$\frac{dy}{dx} - \frac{2y}{x} = x^3 y^2$$
B3: Non-Homogeneous DE — Dependent & Independent (10 min)
Q26
Given $(x - 2y),dx + (4x - 3y + 2),dy = 0$: (a) Are the coefficients linearly dependent or independent? Use the ratio test. (b) If dependent, find the substitution $z = a_2 x + b_2 y$ and reduce to a separable DE.
Q27
Given $(2x + y - 1),dx + (4x + 2y + 3),dy = 0$: (a) Are the coefficients linearly dependent or independent? (b) If dependent, find the substitution and solve.
Q28
Given $(3x - y + 1),dx + (x + y - 3),dy = 0$: (a) Are the coefficients linearly dependent or independent? (b) If independent, solve the simultaneous equations to eliminate constants, then solve as a homogeneous DE.
B4: Exact DE — Test & Solve (8 min)
Q29
$$(2xy - 3x^2),dx + (x^2 - 2y),dy = 0$$
Q30
$$(3x^2 + y),dx + (x - 2y),dy = 0$$
Q31
$$(4x^3 - 8xy^2 - y\sin xy),dx + (16y^3 - 8x^2y - x\sin xy),dy = 0$$ Given solution form $A(x,y)^2 + \cos(xy) = C$, find $A(x,y)$.
Q32
$$(x^2 + y^2),dx + (2xy),dy = 0$$
B5: Mixing Tank (6 min)
Q33
For each, write DE for $Q(t)$ and solve (a):
| $V_0$ (L) | $Q_0$ (g) | $R_{in}$ (L/min) | $C_{in}$ (g/L) | $R_{out}$ (L/min) | |
|---|---|---|---|---|---|
| (a) | 1000 | 50 | 20 | 0 | 20 |
| (b) | 300 | 60 | 8 | 1.2 | 5 |
| (c) | 500 | 0 | 10 | 2 | 10 |
Q34
A tank contains 200 L of brine with 20 kg salt. Brine with concentration 2 kg/L enters at 5 L/min. Mixture exits at 3 L/min. Find $Q(t)$.
Solutions B
Q15–Q19 ID
| Q | Type | Method |
|---|---|---|
| 15 | Bernoulli ($n=3$) | Make coeff 1, then $v = y^{-2}$ |
| 16 | Non-Homogeneous | Test ratio: $\frac{1}{4} \neq \frac{-2}{-3}$ → independent |
| 17 | Homogeneous | $y = vx$ |
| 18 | Bernoulli ($n=-2$) | Make coeff 1, then $v = y^3$ |
| 19 | Non-Homogeneous | Test ratio: $\frac{2}{4} = \frac{1}{2}$ → dependent |
Q20
Divide by $x$: $\frac{dy}{dx} + \frac{1}{x}y = x y^2$
$n=2$, $P=\frac{1}{x}$, $Q=x$
$v = y^{-1}$, $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$
Multiply by $-y^{-2}$: $-y^{-2}\frac{dy}{dx} - \frac{1}{x}y^{-1} = -x$
$$\frac{dv}{dx} - \frac{1}{x}v = -x$$
$$\mu = e^{\int -\frac{1}{x},dx} = e^{-\ln x} = \frac{1}{x}$$
$$\frac{d}{dx}\left(\frac{v}{x}\right) = -1$$
$$\frac{v}{x} = -x + C \implies v = -x^2 + Cx$$
$$y^{-1} = Cx - x^2 \implies \boxed{y = \frac{1}{Cx - x^2}}$$
Q21
$\frac{dy}{dx} + y = xy^3$, $n=3$, $P=1$, $Q=x$
$v = y^{-2}$, $\frac{dv}{dx} = -2y^{-3}\frac{dy}{dx}$
Multiply by $-2y^{-3}$: $-2y^{-3}\frac{dy}{dx} - 2y^{-2} = -2x$
$$\frac{dv}{dx} - 2v = -2x$$
$$\mu = e^{\int -2,dx} = e^{-2x}$$
$$\frac{d}{dx}(e^{-2x}v) = -2xe^{-2x}$$
Integrate RHS by parts: $u=-2x$, $dv=e^{-2x}dx$ → $du=-2dx$, $v=-\frac{1}{2}e^{-2x}$
$$\int -2xe^{-2x},dx = (-2x)(-\frac{1}{2}e^{-2x}) - \int (-\frac{1}{2}e^{-2x})(-2)dx = xe^{-2x} - \int e^{-2x},dx = xe^{-2x} + \frac{1}{2}e^{-2x} + C$$
$$e^{-2x}v = xe^{-2x} + \frac{1}{2}e^{-2x} + C$$
$$v = x + \frac{1}{2} + Ce^{2x}$$
$$\boxed{y^{-2} = x + \frac{1}{2} + Ce^{2x}}$$
Q22
$\frac{dy}{dx} - \frac{y}{x} = \frac{x}{y}$, $n=-1$, $P=-\frac{1}{x}$, $Q=x$
$v = y^2$, $\frac{dv}{dx} = 2y\frac{dy}{dx}$
Multiply by $2y$: $2y\frac{dy}{dx} - \frac{2}{x}y^2 = 2x$
$$\frac{dv}{dx} - \frac{2}{x}v = 2x$$
$$\mu = e^{\int -\frac{2}{x},dx} = e^{-2\ln x} = \frac{1}{x^2}$$
$$\frac{d}{dx}\left(\frac{v}{x^2}\right) = \frac{2}{x}$$
$$\frac{v}{x^2} = 2\ln|x| + C \implies v = 2x^2\ln|x| + Cx^2$$
$$\boxed{y^2 = x^2(2\ln|x| + C)}$$
Q23
Divide by $2x$: $\frac{dy}{dx} + \frac{1}{2x}y = \frac{\tan^{-1} x}{y}$
$n=-1$, $P=\frac{1}{2x}$, $Q=\tan^{-1} x$
$v = y^2$, $\frac{dv}{dx} = 2y\frac{dy}{dx}$
Multiply by $2y$: $2y\frac{dy}{dx} + \frac{1}{x}y^2 = 2\tan^{-1} x$
$$\frac{dv}{dx} + \frac{1}{x}v = 2\tan^{-1} x$$
$$\mu = e^{\int \frac{1}{x},dx} = x$$
$$\frac{d}{dx}(xv) = 2x\tan^{-1} x$$
RHS integrate by parts: $u=\tan^{-1}x$, $dv=2x,dx$ $du = \frac{1}{1+x^2}dx$, $v = x^2$
$$\int 2x\tan^{-1}x,dx = x^2\tan^{-1}x - \int\frac{x^2}{1+x^2},dx$$ $$= x^2\tan^{-1}x - \int(1 - \frac{1}{1+x^2})dx$$ $$= x^2\tan^{-1}x - x + \tan^{-1}x + C$$ $$= (x^2+1)\tan^{-1}x - x + C$$
$$xv = (x^2+1)\tan^{-1}x - x + C$$
$$xy^2 = (x^2+1)\tan^{-1}x - x + C$$
$$\boxed{y^2 = \frac{(x^2+1)\tan^{-1}x - x + C}{x}}$$
Q24
$\frac{dy}{dx} + \frac{2}{x}y = x^3 y^2$, $n=2$, $P=\frac{2}{x}$, $Q=x^3$
$v = y^{-1}$, $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$
Multiply by $-y^{-2}$: $-y^{-2}\frac{dy}{dx} - \frac{2}{x}y^{-1} = -x^3$
$$\frac{dv}{dx} - \frac{2}{x}v = -x^3$$
$$\mu = e^{\int -\frac{2}{x},dx} = e^{-2\ln x} = \frac{1}{x^2}$$
$$\frac{d}{dx}\left(\frac{v}{x^2}\right) = -x$$
$$\frac{v}{x^2} = -\frac{x^2}{2} + C \implies v = -\frac{x^4}{2} + Cx^2$$
$$\frac{1}{y} = -\frac{x^4}{2} + Cx^2$$
IVP $y(1)=1$: $1 = -\frac{1}{2} + C \implies C = \frac{3}{2}$
$$\boxed{\frac{1}{y} = -\frac{x^4}{2} + \frac{3}{2}x^2}$$
Q25
$\frac{dy}{dx} - \frac{2y}{x} = x^3 y^2$, $n=2$, $P=-\frac{2}{x}$, $Q=x^3$
$v = y^{-1}$, $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$
Multiply by $-y^{-2}$: $-y^{-2}\frac{dy}{dx} + \frac{2}{x}y^{-1} = -x^3$
$$\frac{dv}{dx} + \frac{2}{x}v = -x^3$$
$$\mu = e^{\int \frac{2}{x},dx} = e^{2\ln x} = x^2$$
$$\frac{d}{dx}(x^2 v) = -x^5$$
$$x^2 v = -\frac{x^6}{6} + C \implies v = -\frac{x^4}{6} + Cx^{-2}$$
$$\boxed{\frac{1}{y} = -\frac{x^4}{6} + \frac{C}{x^2}}$$
Q26
$$(x - 2y),dx + (4x - 3y + 2),dy = 0$$
(a) Ratio test: $\frac{a_1}{a_2} = \frac{1}{4}$, $\frac{b_1}{b_2} = \frac{-2}{-3} = \frac{2}{3}$
$\frac{1}{4} \neq \frac{2}{3}$ → Linearly Independent
(b) Set up simultaneous equations for $x$ and $y$: $x - 2y = 0$ and $4x - 3y + 2 = 0$
From first: $x = 2y$. Substitute into second: $4(2y) - 3y + 2 = 0 \implies 8y - 3y + 2 = 0 \implies y = -\frac{2}{5}$, $x = -\frac{4}{5}$
Substitute $X = x + \frac{4}{5}$, $Y = y + \frac{2}{5}$ → reduces to homogeneous DE in $X,Y$.
Q27
$$(2x + y - 1),dx + (4x + 2y + 3),dy = 0$$
(a) Ratio test: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$
$\frac{1}{2} = \frac{1}{2}$ → Linearly Dependent ✓
(b) Substitution: $z = 4x + 2y$ (or $z = 2x + y$ — same ratio)
Then $dz = 2,dx + dy$, or express $y = \frac{z - 4x}{2}$
Substitute into original: $(2x + \frac{z-4x}{2} - 1),dx + z,dy = 0$
$$(\frac{z - 2}{2}),dx + z,dy = 0$$
Separate: $\frac{z - 2}{2z},dx = -dy$
Or better: from $z = 2x + y$, $\frac{dy}{dx} = \frac{dz}{dx} - 2$
Original as $\frac{dy}{dx} = -\frac{2x + y - 1}{4x + 2y + 3} = -\frac{z - 1}{2z + 3}$
$$\frac{dz}{dx} - 2 = -\frac{z - 1}{2z + 3}$$
$$\frac{dz}{dx} = 2 - \frac{z - 1}{2z + 3} = \frac{2(2z+3) - (z-1)}{2z+3} = \frac{4z + 6 - z + 1}{2z+3} = \frac{3z + 7}{2z + 3}$$
Separable: $\frac{2z+3}{3z+7},dz = dx$
$$\int \frac{2z+3}{3z+7},dz = \int dx$$
$$x + C = \int \left(\frac{2}{3} - \frac{5/3}{3z+7}\right)dz = \frac{2z}{3} - \frac{5}{9}\ln|3z+7|$$
Substitute back $z = 2x + y$.
Q28
$$(3x - y + 1),dx + (x + y - 3),dy = 0$$
(a) Ratio test: $\frac{a_1}{a_2} = \frac{3}{1} = 3$, $\frac{b_1}{b_2} = \frac{-1}{1} = -1$
$3 \neq -1$ → Linearly Independent
(b) Solve: $3x - y + 1 = 0$ and $x + y - 3 = 0$
Add: $4x - 2 = 0 \implies x = \frac{1}{2}$, $y = \frac{5}{2}$
Substitute $X = x - \frac{1}{2}$, $Y = y - \frac{5}{2}$ → homogeneous DE.
Q29
$M = 2xy - 3x^2$, $N = x^2 - 2y$
$M_y = 2x$, $N_x = 2x$ → Exact ✓
$$F = \int M,dx = x^2y - x^3 + g(y)$$
$$F_y = x^2 + g'(y) = N = x^2 - 2y \implies g'(y) = -2y \implies g(y) = -y^2$$
$$\boxed{x^2y - x^3 - y^2 = C}$$
Q30
$M = 3x^2 + y$, $N = x - 2y$
$M_y = 1$, $N_x = 1$ → Exact ✓
$$F = \int M,dx = x^3 + xy + g(y)$$
$$F_y = x + g'(y) = N = x - 2y \implies g'(y) = -2y \implies g(y) = -y^2$$
$$\boxed{x^3 + xy - y^2 = C}$$
Q31
$M = 4x^3 - 8xy^2 - y\sin xy$, $N = 16y^3 - 8x^2y - x\sin xy$
$$M_y = -16xy - \sin xy - xy\cos xy$$
$N_x = -16xy - \sin xy - xy\cos xy$ → Exact ✓
$$F = \int M,dx = x^4 - 4x^2y^2 + \cos(xy) + g(y)$$
$$F_y = -8x^2y - x\sin(xy) + g'(y) = N = 16y^3 - 8x^2y - x\sin xy$$
$$g'(y) = 16y^3 \implies g(y) = 4y^4$$
$$x^4 - 4x^2y^2 + 4y^4 + \cos(xy) = C \implies (x^2 - 2y^2)^2 + \cos(xy) = C$$
$$\boxed{A(x,y) = x^2 - 2y^2}$$
Q32
$M = x^2 + y^2$, $N = 2xy$
$M_y = 2y$, $N_x = 2y$ → Exact ✓
$$F = \int M,dx = \frac{x^3}{3} + xy^2 + g(y)$$
$$F_y = 2xy + g'(y) = 2xy \implies g'(y) = 0$$
$$\boxed{\frac{x^3}{3} + xy^2 = C}$$
Q33
General: $\frac{dQ}{dt} = R_{in}C_{in} - R_{out}\frac{Q}{V(t)}$, $V(t) = V_0 + (R_{in}-R_{out})t$
(a) Constant volume: $\frac{dQ}{dt} = -\frac{20}{1000}Q = -\frac{Q}{50}$, $Q(0)=50$
Separable: $\frac{dQ}{Q} = -\frac{dt}{50} \implies \ln|Q| = -\frac{t}{50} + C$
$Q(t) = Ce^{-t/50}$, $Q(0)=50 \implies C=50$
$$\boxed{Q(t) = 50e^{-t/50}}$$
(b) $V(t) = 300 + 3t$, $\frac{dQ}{dt} = 8(1.2) - 5\cdot\frac{Q}{300+3t} = 9.6 - \frac{5Q}{300+3t}$, $Q(0)=60$
(c) Constant volume: $\frac{dQ}{dt} = 10(2) - 10\cdot\frac{Q}{500} = 20 - \frac{Q}{50}$, $Q(0)=0$
Q34
$V_0 = 200$, $Q_0 = 20$, $R_{in}=5$, $C_{in}=2$, $R_{out}=3$
$$V(t) = 200 + (5-3)t = 200 + 2t$$
$$\frac{dQ}{dt} = 5\cdot2 - 3\cdot\frac{Q}{200+2t} = 10 - \frac{3Q}{2(100+t)}$$
$$\frac{dQ}{dt} + \frac{3}{2(100+t)}Q = 10$$
This is a linear DE (integrating factor method). $\mu = e^{\int \frac{3}{2(100+t)},dt} = e^{\frac{3}{2}\ln(100+t)} = (100+t)^{3/2}$
$$\frac{d}{dt}\left[(100+t)^{3/2}Q\right] = 10(100+t)^{3/2}$$
$$(100+t)^{3/2}Q = 10\cdot\frac{2}{5}(100+t)^{5/2} + C = 4(100+t)^{5/2} + C$$
$$Q = 4(100+t) + C(100+t)^{-3/2}$$
$$Q(0)=20 \implies 20 = 400 + C(100)^{-3/2} \implies C = -380 \cdot 1000 = -380,000$$
$$\boxed{Q(t) = 4(100+t) - 380,000(100+t)^{-3/2}}$$
SECTION C: INVERSE HYPERBOLIC — DERIVATIVES & INTEGRALS
C1: Derivatives (6 min)
Q35
Find $\frac{dy}{dx}$: (a) $y = \cosh^{-1}(5x - 7)$ (b) $y = \sinh^{-1}(e^{2x})$ (c) $y = \tanh^{-1}(\ln x)$ (d) $y = \operatorname{sech}^{-1}(x^3)$
Q36
Differentiate: (a) $y = x^2 \cosh^{-1}(6x^2 - 7x^{-2})$ (b) $y = \cos(\sinh^{-1} x^6)$ (c) $y = \frac{\sinh^{-1}(2x)}{\tanh^2(4x)}$
C2: Standard Integral Forms (6 min)
Q37
Evaluate: (a) $\int \frac{dx}{\sqrt{x^2 + 16}}$ (b) $\int \frac{dx}{\sqrt{x^2 - 9}}$ (c) $\int \frac{dx}{\sqrt{1 + 4x^2}}$ (d) $\int \frac{dx}{\sqrt{9x^2 - 25}}$ (e) $\int \frac{dx}{16 - x^2}$
Q38
Evaluate (with substitution): (a) $\int \frac{dx}{\sqrt{1 + 9x^2}}$ (let $u=3x$) (b) $\int \frac{dx}{\sqrt{4x^2 - 9}}$ (c) $\int \frac{dx}{\sqrt{5 + x^2}}$
C3: Definite Integrals (6 min)
Q39
$$\int_4^6 \frac{dx}{\sqrt{x^2 - 9}}$$
Q40
$$\int_0^{1/3} \frac{dx}{\sqrt{1 + 9x^2}}$$
Q41
$$\int_0^{1/2} \frac{\sin^{-1}(2x)}{\sqrt{1 - 4x^2}},dx$$
Q42
$$\int_{1/2}^1 \frac{\cosh^{-1}(2x)}{\sqrt{4x^2 - 1}},dx$$
Solutions C
Q35
Derivative formulas: $$\frac{d}{dx}\sinh^{-1}u = \frac{1}{\sqrt{1+u^2}}\frac{du}{dx}$$ $$\frac{d}{dx}\cosh^{-1}u = \frac{1}{\sqrt{u^2-1}}\frac{du}{dx}$$ $$\frac{d}{dx}\tanh^{-1}u = \frac{1}{1-u^2}\frac{du}{dx}$$ $$\frac{d}{dx}\operatorname{sech}^{-1}u = -\frac{1}{u\sqrt{1-u^2}}\frac{du}{dx}$$
(a) $\frac{dy}{dx} = \frac{1}{\sqrt{(5x-7)^2 - 1}}\cdot 5 = \frac{5}{\sqrt{25x^2 - 70x + 48}}$
(b) $\frac{dy}{dx} = \frac{1}{\sqrt{1 + e^{4x}}}\cdot 2e^{2x} = \frac{2e^{2x}}{\sqrt{1+e^{4x}}}$
(c) $\frac{dy}{dx} = \frac{1}{1 - (\ln x)^2}\cdot\frac{1}{x} = \frac{1}{x(1 - \ln^2 x)}$
(d) $\frac{dy}{dx} = -\frac{1}{x^3\sqrt{1 - x^6}}\cdot 3x^2 = -\frac{3}{x\sqrt{1-x^6}}$
Q36
(a) $y' = 2x\cosh^{-1}(6x^2 - 7x^{-2}) + x^2\cdot\frac{1}{\sqrt{(6x^2-7x^{-2})^2-1}}\cdot(12x + 14x^{-3})$
(b) $y' = -\sin(\sinh^{-1}x^6)\cdot\frac{1}{\sqrt{1+x^{12}}}\cdot 6x^5$
(c) Quotient rule: $u = \sinh^{-1}(2x)$, $v = \tanh^2(4x)$ $u' = \frac{2}{\sqrt{1+4x^2}}$, $v' = 2\tanh(4x)\cdot\operatorname{sech}^2(4x)\cdot 4 = 8\tanh(4x)\operatorname{sech}^2(4x)$ $$y' = \frac{u'v - uv'}{v^2}$$
Q37
(a) $\int \frac{dx}{\sqrt{x^2 + 16}} = \sinh^{-1}\left(\frac{x}{4}\right) + C = \ln\left|x + \sqrt{x^2+16}\right| + C$
(b) $\int \frac{dx}{\sqrt{x^2 - 9}} = \cosh^{-1}\left(\frac{x}{3}\right) + C = \ln\left|x + \sqrt{x^2-9}\right| + C$
(c) $\int \frac{dx}{\sqrt{1 + 4x^2}} = \frac{1}{2}\sinh^{-1}(2x) + C = \frac{1}{2}\ln\left|2x + \sqrt{1+4x^2}\right| + C$
(d) $\int \frac{dx}{\sqrt{9x^2 - 25}} = \frac{1}{3}\cosh^{-1}\left(\frac{3x}{5}\right) + C = \frac{1}{3}\ln\left|3x + \sqrt{9x^2-25}\right| + C$
(e) $\int \frac{dx}{16 - x^2} = \frac{1}{8}\ln\left|\frac{4+x}{4-x}\right| + C$ (using $\int dx/(a^2-x^2) = \frac{1}{2a}\ln|(a+x)/(a-x)| + C$)
Q38
(a) Let $u = 3x$, $du = 3dx$: $$\int \frac{dx}{\sqrt{1+9x^2}} = \frac{1}{3}\int \frac{du}{\sqrt{1+u^2}} = \frac{1}{3}\sinh^{-1}(3x) + C$$
(b) Let $u = 2x$, $du = 2dx$: $$\int \frac{dx}{\sqrt{4x^2-9}} = \frac{1}{2}\int \frac{du}{\sqrt{u^2-9}} = \frac{1}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C$$
(c) $\int \frac{dx}{\sqrt{5+x^2}} = \sinh^{-1}\left(\frac{x}{\sqrt{5}}\right) + C$
Q39
$$\int_4^6 \frac{dx}{\sqrt{x^2 - 9}} = \left[\cosh^{-1}\left(\frac{x}{3}\right)\right]_4^6 = \cosh^{-1}(2) - \cosh^{-1}\left(\frac{4}{3}\right)$$
In log form: $\ln(2+\sqrt{3}) - \ln\left(\frac{4}{3}+\sqrt{\frac{16}{9}-1}\right) = \ln(2+\sqrt{3}) - \ln\left(\frac{4}{3}+\frac{\sqrt{7}}{3}\right)$
$$= \ln\frac{2+\sqrt{3}}{(4+\sqrt{7})/3} = \ln\frac{3(2+\sqrt{3})}{4+\sqrt{7}}$$
Q40
$$\int_0^{1/3} \frac{dx}{\sqrt{1+9x^2}} = \left[\frac{1}{3}\sinh^{-1}(3x)\right]_0^{1/3} = \frac{1}{3}\sinh^{-1}(1) - 0 = \frac{1}{3}\ln(1+\sqrt{2})$$
Q41
Let $u = \sin^{-1}(2x)$, $du = \frac{2}{\sqrt{1-4x^2}},dx$
$$\int \frac{\sin^{-1}(2x)}{\sqrt{1-4x^2}},dx = \frac{1}{2}\int u,du = \frac{1}{4}u^2 + C = \frac{1}{4}[\sin^{-1}(2x)]^2 + C$$
$$\int_0^{1/2} = \frac{1}{4}[\sin^{-1}(1)]^2 - \frac{1}{4}[\sin^{-1}(0)]^2 = \frac{1}{4}\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{16}$$
Q42
Let $u = \cosh^{-1}(2x)$, $du = \frac{2}{\sqrt{4x^2-1}},dx$
$$\int \frac{\cosh^{-1}(2x)}{\sqrt{4x^2-1}},dx = \frac{1}{2}\int u,du = \frac{1}{4}u^2 + C = \frac{1}{4}[\cosh^{-1}(2x)]^2 + C$$
$$\int_{1/2}^1 = \frac{1}{4}[\cosh^{-1}(2)]^2 - \frac{1}{4}[\cosh^{-1}(1)]^2 = \frac{1}{4}[\ln(2+\sqrt{3})]^2 - 0 = \frac{1}{4}[\ln(2+\sqrt{3})]^2$$
SECTION D: TANGENT EXPANSION (MACLAURIN)
D1: Standard Series You MUST Know (2 min)
Write the first 3-4 non-zero terms:
Q43
(a) $e^x =$ (b) $\sin x =$ (c) $\cos x =$ (d) $\ln(1+x) =$ (e) $\frac{1}{1-x} =$ (f) $\arctan x =$ (g) $\sinh x =$ (h) $\cosh x =$
D2: Substitution & Multiplication (6 min)
Use standard series (from above) to find expansions up to $x^4$:
Q44
$$e^{-x^2}$$
Q45
$$\sin(2x^2)$$
Q46
$$x^3\cos(x^2)$$
Q47
$e^x\sin x$ (up to $x^3$)
Q48
Term-by-term integration: $\int e^{x^2},dx$ (up to $x^5$)
D3: Tangent Expansion — The Leak Topic (8 min)
Q49
Find Maclaurin series for $\tan x$ up to $x^5$ by:
Method A — Direct differentiation: Compute $f(0), f'(0), f''(0), f'''(0), f^{(4)}(0), f^{(5)}(0)$
Q50
Find Maclaurin series for $\tan x$ up to $x^5$ by:
Method B — Division of series: Use $\tan x = \frac{\sin x}{\cos x}$ with known series for $\sin x$ and $\cos x$.
Q51
Use the series $\tan x \approx x + \frac{x^3}{3} + \frac{2x^5}{15}$ to estimate $\tan(0.2)$ correct to 4 d.p.
Q52
Find Maclaurin series for $\sec x$ up to $x^4$ by using $\sec x = \frac{1}{\cos x}$.
Q53
Use term-by-term differentiation of the $\tan x$ series to verify that $\frac{d}{dx}\tan x = \sec^2 x$.
Q54
Find Maclaurin series for $f(x) = \sinh x$ up to $x^5$. Then use it to approximate $\sinh(0.3)$.
Solutions D
Q43
(a) $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$ (b) $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ (c) $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$ (d) $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$, $|x| < 1$ (e) $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$, $|x| < 1$ (f) $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ (g) $\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$ (h) $\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$
Q44
$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$, let $u = -x^2$: $$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$
Q45
$\sin u = u - \frac{u^3}{3!} + \cdots$, let $u = 2x^2$: $$\sin(2x^2) = 2x^2 - \frac{(2x^2)^3}{3!} + \cdots = 2x^2 - \frac{8x^6}{6} + \cdots$$ Up to $x^4$: just $\boxed{2x^2}$
Q46
$$x^3\cos(x^2) = x^3\left(1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots\right) = x^3 - \frac{x^7}{2} + \cdots$$ Up to $x^4$: just $\boxed{x^3}$
Q47
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ $$\sin x = x - \frac{x^3}{3!} + \cdots$$
$$(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots)(x - \frac{x^3}{6} + \cdots)$$ $$= x - \frac{x^3}{6} + x^2 - \frac{x^4}{6} + \frac{x^3}{2} - \frac{x^5}{12} + \frac{x^4}{6} + \cdots$$ $$= x + x^2 + \left(-\frac{1}{6}+\frac{1}{2}\right)x^3 + \left(-\frac{1}{6}+\frac{1}{6}\right)x^4 + \cdots$$ $$= \boxed{x + x^2 + \frac{x^3}{3} + \cdots}$$
Q48
$$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots$$
$$\int e^{x^2},dx = \int \left(1 + x^2 + \frac{x^4}{2} + \cdots\right)dx = x + \frac{x^3}{3} + \frac{x^5}{10} + \cdots + C$$
Q49
$f(x) = \tan x$, $f(0) = 0$ $f'(x) = \sec^2 x$, $f'(0) = 1$ $f''(x) = 2\sec^2 x \tan x$, $f''(0) = 0$ $f'''(x) = 4\sec^2 x \tan^2 x + 2\sec^4 x$, $f'''(0) = 2$ $f^{(4)}(x) = 8\sec^2 x \tan^3 x + 8\sec^4 x \tan x$, $f^{(4)}(0) = 0$ $f^{(5)}(x)$ → evaluate at $0$: $f^{(5)}(0) = 16$
$$\tan x = 0 + 1\cdot x + \frac{0}{2!}x^2 + \frac{2}{3!}x^3 + \frac{0}{4!}x^4 + \frac{16}{5!}x^5 + \cdots$$ $$= \boxed{x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots}$$
Q50
$$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$
Long division: $$1-\frac{x^2}{2}+\frac{x^4}{24} \enclose{longdiv}{x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots}$$ First term: $x$
Multiply: $x(1-\frac{x^2}{2}+\frac{x^4}{24}) = x - \frac{x^3}{2} + \frac{x^5}{24}$
Subtract: $(x - \frac{x^3}{6} + \frac{x^5}{120}) - (x - \frac{x^3}{2} + \frac{x^5}{24}) = \frac{x^3}{3} - \frac{x^5}{30} + \cdots$
Next term: $\frac{x^3}{3} \div 1 = \frac{x^3}{3}$
Multiply: $\frac{x^3}{3}(1-\frac{x^2}{2}+\cdots) = \frac{x^3}{3} - \frac{x^5}{6} + \cdots$
Subtract: $(\frac{x^3}{3} - \frac{x^5}{30}) - (\frac{x^3}{3} - \frac{x^5}{6}) = \frac{2x^5}{15}$
$$\boxed{\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots}$$
Q51
$$\tan(0.2) \approx 0.2 + \frac{0.2^3}{3} + \frac{2(0.2)^5}{15}$$
$0.2^3 = 0.008$ → $0.008/3 = 0.0026667$ $0.2^5 = 0.00032$ → $2 \times 0.00032/15 = 0.00004267$
$$\tan(0.2) \approx 0.2 + 0.002667 + 0.000043 = 0.20271$$
$$\boxed{\approx 0.2027}$$
Q52
$\sec x = 1/\cos x$. Use $1/(1-u) = 1+u+u^2+\cdots$ with $u = 1-\cos x = \frac{x^2}{2} - \frac{x^4}{24} + \cdots$
$$\sec x = \frac{1}{1 - (\frac{x^2}{2} - \frac{x^4}{24} + \cdots)} = 1 + (\frac{x^2}{2} - \frac{x^4}{24}) + (\frac{x^2}{2})^2 + \cdots$$ $$= 1 + \frac{x^2}{2} - \frac{x^4}{24} + \frac{x^4}{4} + \cdots = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$
Or differentiate $\tan x$: $\sec^2 x = \frac{d}{dx}(x + \frac{x^3}{3} + \frac{2x^5}{15}) = 1 + x^2 + \frac{2x^4}{3} + \cdots$
$$\boxed{\sec x = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots}$$
Q53
$$\frac{d}{dx}\tan x = \frac{d}{dx}\left(x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots\right) = 1 + x^2 + \frac{2x^4}{3} + \cdots$$
$\sec^2 x = 1 + x^2 + \frac{2x^4}{3} + \cdots$ (from Q52). They match ✓
Q54
$f(x) = \sinh x$, $f(0) = 0$ $f'(x) = \cosh x$, $f'(0) = 1$ $f''(x) = \sinh x$, $f''(0) = 0$ $f'''(x) = \cosh x$, $f'''(0) = 1$ $f^{(4)}(x) = \sinh x$, $f^{(4)}(0) = 0$ $f^{(5)}(x) = \cosh x$, $f^{(5)}(0) = 1$
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = x + \frac{x^3}{6} + \frac{x^5}{120} + \cdots$$
$$\sinh(0.3) \approx 0.3 + \frac{0.027}{6} + \frac{0.00243}{120} = 0.3 + 0.0045 + 0.00002025 = 0.30452$$
$$\boxed{\approx 0.3045}$$
SECTION E: PART A SPEED ROUND (Quick Recall)
E1: Complex Basics (3 min)
Q55
(a) Is $\pi$ a complex number? Justify. (b) If $\text{Im}(z) = 0$ and $\text{Re}(z) < 0$, what is $\arg(z)$? (c) Find modulus and argument of $z = -1 + i\sqrt{3}$. (d) If $i^n = -1$, give two possible positive integer values of $n$.
E2: Hyperbolic T/F (3 min)
Q56
True or False: (a) $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ (b) $\cosh^2 x - 1 = \sinh^2 x$ (c) All hyperbolic functions are invertible without domain restriction. (d) $\text{sech},x = \frac{2}{e^x + e^{-x}}$ (e) Hyperbolic functions are used to determine angles in a triangle.
E3: Inverse Trig Evaluations (3 min)
Q57
Find exact value: (a) $\sin(\tan^{-1}(3/4))$ (b) $\cos(\sin^{-1}(5/13))$ (c) $\tan(\sec^{-1}(5/3))$ (d) $\sin(2\tan^{-1}(2/3))$
E4: Non-Homogeneous DE — Part A Style (3 min)
Q58
Given $(a_1 x + b_1 y + c_1),dx + (a_2 x + b_2 y + c_2),dy = 0$: (a) What is the ratio test for linear dependence? (b) If $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, what substitution reduces the DE to separable form? (c) If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, what two steps eliminate the constants?
Solutions E
Q55
(a) Yes, $\pi = \pi + 0i$ is a complex number with $a=\pi$, $b=0$. (b) $\arg(z) = \pi$ (point lies on negative real axis). (c) $r = \sqrt{1+3} = 2$, $\arg(z) = \frac{2\pi}{3}$ (Q2: $\tan^{-1}(-\sqrt{3}) = -\pi/3$, so $\pi - \pi/3 = 2\pi/3$) (d) $i^n = -1$. $i^2 = -1$, $i^6 = i^4\cdot i^2 = 1\cdot(-1) = -1$. So $n=2$ and $n=6$.
Q56
(a) TRUE — $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ (b) TRUE — $\cosh^2 x - \sinh^2 x = 1 \implies \cosh^2 x - 1 = \sinh^2 x$ (c) FALSE — $\cosh x$ and $\operatorname{sech} x$ require domain restriction ($x \geq 0$) to invert. (d) TRUE — $\operatorname{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$ (e) FALSE — Hyperbolic functions model catenaries, not triangle angles (that's trig functions).
Q57
(a) $\theta = \tan^{-1}(3/4)$, $\sin\theta = \frac{3}{5}$ (b) $\theta = \sin^{-1}(5/13)$, $\cos\theta = \frac{12}{13}$ (c) $\theta = \sec^{-1}(5/3)$, $\sec\theta = 5/3$, $\cos\theta = 3/5$, $\tan\theta = \frac{4}{3}$ (d) $\theta = \tan^{-1}(2/3)$, $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{4/3}{1+4/9} = \frac{4/3}{13/9} = \frac{12}{13}$
Q58
(a) $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ → dependent; $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ → independent.
(b) Substitution $z = a_2 x + b_2 y$ reduces to separable DE.
(c) Solve simultaneous equations for $x$ and $y$ to eliminate constants $c_1, c_2$, then solve as homogeneous DE.