FAC1004 Tonight Full Drill — SOLUTIONS ONLY

Do NOT look at this until you have attempted all questions.

SECTION A: COMPLEX NUMBERS — SOLUTIONS

Solutions A1–A3

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 $(2,-1)$ (closer to $(2,-1)$ than $(-4,3)$)
  • 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, below $y=2$ (closer to $(0,1)$ than $(0,3)$), 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: closer to $(2,-1)$ than $(-4,3)$. Distance to $(2,-1)$: $\sqrt{13}$. Distance to $(-4,3)$: $\sqrt{17}$. $\sqrt{13} < \sqrt{17}$ ✓
  • Circle: $(0-1)^2+2^2 = 5 \leq 16$ ✓
  • Ray: $\arg(2i) = \pi/2 \geq \pi/6$ ✓
  • Inside.

Q8

$z = 0 \implies (0,0)$. Check Q5.

  • Circle: $5 \leq 9$ ✓
  • $|0-i| = 1 < |0-3i| = 3$ ✓
  • $\arg(1) = 0$. 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: $1^2+(1-2)^2 = 2 < 9$. 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, $y \geq 3$ (further from $(0,2)$ than $(0,4)$), $x \geq 2$

(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 $\sin^2 x = 1 - \cos^2 x$:

$$= \cos^5 x - 10\cos^3 x(1-\cos^2 x) + 5\cos x(1-\cos^2 x)^2$$ $= 16\cos^5 x - 20\cos^3 x + 5\cos x = \cos 5x$ ✓

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 $\tan(i) = -\tanh(1) = -\frac{e^2 - 1}{e^2 + 1}$:

$$= \frac{1 + \tanh(1)}{1 - \tanh(1)} = \frac{e^2 + 1 + e^2 - 1}{e^2 + 1 - e^2 + 1} = \frac{2e^2}{2} = e^2$$

SECTION B: BERNOULLI & NON-HOMOGENEOUS DEs — SOLUTIONS

Solutions B1–B5

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: $= xe^{-2x} + \frac{1}{2}e^{-2x} + C$

$$e^{-2x}v = xe^{-2x} + \frac{1}{2}e^{-2x} + C$$

$$v = x + \frac{1}{2} + Ce^{2x}$$

$$\boxed{y^{-2} = x + \frac{1}{2} + Ce^{2x}}$$

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: $= (x^2+1)\tan^{-1}x - x + C$

$$xy^2 = (x^2+1)\tan^{-1}x - x + C$$

$$\boxed{y^2 = \frac{(x^2+1)\tan^{-1}x - x + C}{x}}$$

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: $4(2y) - 3y + 2 = 0 \implies y = -\frac{2}{5}$, $x = -\frac{4}{5}$

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)

$$\frac{dy}{dx} = -\frac{2x + y - 1}{4x + 2y + 3} = -\frac{z - 1}{2z + 3}$$

$\frac{dz}{dx} = 2 + \frac{dy}{dx} \cdot \frac{dy}{dz}$... Actually: $z = 2x + y$, $\frac{dz}{dx} = 2 + \frac{dy}{dx}$

$$\frac{dz}{dx} = 2 - \frac{z-1}{2z+3} = \frac{4z+6-z+1}{2z+3} = \frac{3z+7}{2z+3}$$

Separable: $\frac{2z+3}{3z+7},dz = dx$

$$\frac{2z}{3} + \frac{5}{9}\ln|3z-7| = x + C$$

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: $4x - 2 = 0$ → $x = \frac{1}{2}$, $y = \frac{5}{2}$

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

$M = 4x^3 - 8xy^2 - y\sin xy$, $N = 16y^3 - 8x^2y - x\sin xy$

$M_y = N_x = -16xy - \sin xy - xy\cos xy$ → Exact ✓

$$F = \int M,dx = x^4 - 4x^2y^2 + \cos(xy) + g(y)$$

$$F_y = -8x^2y - x\sin(xy) + g'(y) = N = 16y^3 - 8x^2y - x\sin xy$$

$$g'(y) = 16y^3 \implies g(y) = 4y^4$$

$$(x^2 - 2y^2)^2 + \cos(xy) = C$$

$$\boxed{A(x,y) = x^2 - 2y^2}$$

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: $\frac{dQ}{dt} = -\frac{20}{1000}Q = -\frac{Q}{50}$, $Q(0)=50$

$Q(t) = Ce^{-t/50}$, $Q(0)=50 \implies C=50$

$$\boxed{Q(t) = 50e^{-t/50}}$$

(b) $V(t) = 300 + 3t$, $\frac{dQ}{dt} = 9.6 - \frac{5Q}{300+3t}$, $Q(0)=60$

(c) Constant volume: $\frac{dQ}{dt} = 20 - \frac{Q}{50}$, $Q(0)=0$

Q34

$V_0 = 200$, $Q_0 = 20$, $R_{in}=5$, $C_{in}=2$, $R_{out}=3$

$$V(t) = 200 + 2t$$

$$\frac{dQ}{dt} = 10 - \frac{3Q}{2(100+t)}$$

Linear DE. $\mu = (100+t)^{3/2}$

$$(100+t)^{3/2}Q = 4(100+t)^{5/2} + C$$

$$Q = 4(100+t) + C(100+t)^{-3/2}$$

$$Q(0)=20 \implies C = -380,000$$

$$\boxed{Q(t) = 4(100+t) - 380,000(100+t)^{-3/2}}$$

SECTION C: INVERSE HYPERBOLIC — SOLUTIONS

Solutions C1–C3

Q35

(a) $\frac{dy}{dx} = \frac{5}{\sqrt{(5x-7)^2 - 1}} = \frac{5}{\sqrt{25x^2 - 70x + 48}}$

(b) $\frac{dy}{dx} = \frac{2e^{2x}}{\sqrt{1 + e^{4x}}}$

(c) $\frac{dy}{dx} = \frac{1}{x(1 - \ln^2 x)}$

(d) $\frac{dy}{dx} = -\frac{3}{x\sqrt{1-x^6}}$

Q36

(a) $y' = 2x\cosh^{-1}(6x^2 - 7x^{-2}) + x^2\cdot\frac{12x + 14x^{-3}}{\sqrt{(6x^2-7x^{-2})^2-1}}$

(b) $y' = -\sin(\sinh^{-1}x^6)\cdot\frac{6x^5}{\sqrt{1+x^{12}}}$

(c) Quotient rule with $u = \sinh^{-1}(2x)$, $v = \tanh^2(4x)$

Q37

(a) $\sinh^{-1}\left(\frac{x}{4}\right) + C = \ln\left|x + \sqrt{x^2+16}\right| + C$

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

(c) $\frac{1}{2}\sinh^{-1}(2x) + C = \frac{1}{2}\ln\left|2x + \sqrt{1+4x^2}\right| + C$

(d) $\frac{1}{3}\cosh^{-1}\left(\frac{3x}{5}\right) + C = \frac{1}{3}\ln\left|3x + \sqrt{9x^2-25}\right| + C$

(e) $\frac{1}{8}\ln\left|\frac{4+x}{4-x}\right| + C$

Q38

(a) $\frac{1}{3}\sinh^{-1}(3x) + C$

(b) $\frac{1}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C$

(c) $\sinh^{-1}\left(\frac{x}{\sqrt{5}}\right) + C$

Q39

$$\left[\cosh^{-1}\left(\frac{x}{3}\right)\right]_4^6 = \cosh^{-1}(2) - \cosh^{-1}\left(\frac{4}{3}\right) = \ln\frac{3(2+\sqrt{3})}{4+\sqrt{7}}$$

Q40

$$\left[\frac{1}{3}\sinh^{-1}(3x)\right]_0^{1/3} = \frac{1}{3}\sinh^{-1}(1) = \frac{1}{3}\ln(1+\sqrt{2})$$

Q41

$$\frac{1}{4}[\sin^{-1}(2x)]^2\Big|_0^{1/2} = \frac{1}{4}\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{16}$$

Q42

$$\frac{1}{4}[\cosh^{-1}(2x)]^2\Big|_{1/2}^1 = \frac{1}{4}[\ln(2+\sqrt{3})]^2$$

SECTION D: TANGENT EXPANSION (MACLAURIN) — SOLUTIONS

Solutions D1–D3

Q43

(a) $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$ (b) $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ (c) $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ (d) $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$ (e) $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$ (f) $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ (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

$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$

Q45

$\sin(2x^2) = 2x^2 - \frac{8x^6}{6} + \cdots$; up to $x^4$: $\boxed{2x^2}$

Q46

$x^3\cos(x^2) = x^3 - \frac{x^7}{2} + \cdots$; up to $x^4$: $\boxed{x^3}$

Q47

$e^x\sin x = x + x^2 + \frac{x^3}{3} + \cdots$ (up to $x^3$)

Q48

$$\int e^{x^2},dx = x + \frac{x^3}{3} + \frac{x^5}{10} + \cdots + C$$

Q49

$f(0) = 0$, $f'(0) = 1$, $f''(0) = 0$, $f'''(0) = 2$, $f^{(4)}(0) = 0$, $f^{(5)}(0) = 16$

$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$

Q50

Long division of $\sin x / \cos x$: $$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$

Q51

$$\tan(0.2) \approx 0.2 + 0.002667 + 0.0000427 = \boxed{0.2027}$$

Q52

$$\sec x = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$

Q53

$\frac{d}{dx}\tan x = 1 + x^2 + \frac{2x^4}{3} + \cdots$; $\sec^2 x = 1 + x^2 + \frac{2x^4}{3} + \cdots$. Match ✓

Q54

$$\sinh x = x + \frac{x^3}{6} + \frac{x^5}{120} + \cdots$$

$$\sinh(0.3) \approx 0.3045$$

SECTION E: PART A SPEED ROUND — SOLUTIONS

Solutions E

Q55

(a) Yes, $\pi = \pi + 0i$ is a complex number with $a=\pi$, $b=0$. (b) $\arg(z) = \pi$ (negative real axis). (c) $r = 2$, $\arg(z) = \frac{2\pi}{3}$. (d) $n = 2$ and $n = 6$ (any $n = 4k + 2$).

Q56

(a) TRUE (b) TRUE (c) FALSE — $\cosh x$ and $\operatorname{sech} x$ need $x \geq 0$ restriction. (d) TRUE (e) FALSE — hyperbolic functions model catenaries, not triangle angles.

Q57

(a) $\frac{3}{5}$ (b) $\frac{12}{13}$ (c) $\frac{4}{3}$ (d) $\frac{12}{13}$

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 $z = a_2 x + b_2 y$. (c) Solve simultaneous equations for $x$ and $y$ to eliminate constants $c_1, c_2$, then solve as homogeneous DE.