Pearson Edexcel International Advanced Level

$-8 = 8(\cos\pi + i\sin\pi) = 8e^{i\pi}$

By De Moivre: $z = 8^{1/3} \cdot \exp[i(\pi+2k\pi)/3],\; k = 0,1,2$

Modulus: $r = 2$

$k = 0$: $\theta = \pi/3 \to z = 2(\tfrac12 + i\tfrac{\sqrt{3}}{2}) =$ $1 + i\sqrt{3}$

$k = 1$: $\theta = \pi \to z = -2 =$ $-2$

$k = 2$: $\theta = 5\pi/3 \to z = 2(\tfrac12 - i\tfrac{\sqrt{3}}{2}) =$ $1 - i\sqrt{3}$

$z_1 = 1 + i\sqrt{3},\; z_2 = -2,\; z_3 = 1 - i\sqrt{3}$

$|z_1-z_2| = |(1+i\sqrt{3})-(-2)| = |3+i\sqrt{3}| = \sqrt{12} = 2\sqrt{3}$

$|z_2-z_3| = |-2-(1-i\sqrt{3})| = |-3+i\sqrt{3}| = 2\sqrt{3}$

$|z_3-z_1| = |(1-i\sqrt{3})-(1+i\sqrt{3})| = |-2i\sqrt{3}| = 2\sqrt{3}$

All three sides equal $\to$ equilateral triangle ✓

Area $= \frac{\sqrt{3}}{4} \times (2\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 12 =$ $3\sqrt{3}$

Denominator positive, multiply without flipping:

$|2x-1| < x + 2$

Subcase 1a: $x \ge \frac12$: $2x-1 < x+2 \to x < 3 \to \frac12 \le x < 3$

Subcase 1b: $x < \frac12$: $-(2x-1) < x+2 \to -2x+1 < x+2 \to -1 < 3x \to x > -\frac13$

Combined: $-\frac13 < x < \frac12$

Union of subcases: $-\frac13 < x < 3$

Denominator is negative, so LHS $= |2x-1|/(x+2) \le 0$ (since numerator $\ge 0$, denominator $< 0$).

Since the RHS is $1$ and the LHS is non-positive, the inequality LHS $< 1$ automatically holds.

Solution: $x < -2$

Let $\displaystyle \frac{2}{(2r-1)(2r+1)(2r+3)} = \frac{A}{2r-1} + \frac{B}{2r+1} + \frac{C}{2r+3}$

Multiply through: $2 = A(2r+1)(2r+3) + B(2r-1)(2r+3) + C(2r-1)(2r+1)$

$r = \frac12$: $2 = A(2)(4) = 8A \to A = \frac14$

$r = -\frac12$: $2 = B(-2)(2) = -4B \to B = -\frac12$

$r = -\frac32$: $2 = C(-4)(-2) = 8C \to C = \frac14$

$\therefore$ $\displaystyle \mathrm{f}(r) = \frac{1/4}{2r-1} - \frac{1/2}{2r+1} + \frac{1/4}{2r+3}$

$\displaystyle \sum_{r=1}^n \mathrm{f}(r) = \tfrac14\sum\frac{1}{2r-1} - \tfrac12\sum\frac{1}{2r+1} + \tfrac14\sum\frac{1}{2r+3}$

Write out terms ($r = 1$ to $n$) and align by denominator:

$\frac11$: $\frac14$ (from 1st series only)

$\frac13$: $\frac14 - \frac12 = -\frac14$

$\frac15$ to $\frac{1}{2n-1}$: $\frac14 - \frac12 + \frac14 = 0$ (all cancel)

$\frac{1}{2n+1}$: $-\frac12 + \frac14 = -\frac14$ (2nd series term $n$, 3rd series term $n-1$)

$\frac{1}{2n+3}$: $\frac14$ (3rd series term $n$ only)

$\displaystyle = \frac14 + \left(-\tfrac14\right)\!\left(\tfrac13\right) + \left(-\tfrac14\right)\!\left(\tfrac{1}{2n+1}\right) + \tfrac14\!\left(\tfrac{1}{2n+3}\right)$

$\displaystyle = \frac14 - \frac{1}{12} - \frac{1}{4(2n+1)} + \frac{1}{4(2n+3)}$

$\displaystyle = \frac16 - \left[\frac{1}{4(2n+1)} - \frac{1}{4(2n+3)}\right]$

$\displaystyle = \frac16 - \frac{(2n+3)-(2n+1)}{4(2n+1)(2n+3)}$

$\displaystyle = \frac16 - \frac{2}{4(2n+1)(2n+3)}$

$\displaystyle = \frac16 - \frac{1}{2(2n+1)(2n+3)}$

Converting to the required form:

$\displaystyle = \frac{2(2n+1)(2n+3) - 6}{12(2n+1)(2n+3)} = \frac{8n^2+16n+6-6}{12(4n^2+8n+3)} = \frac{8n(n+2)}{12(4n^2+8n+3)}$

$\displaystyle =$ $\displaystyle \frac{2n(n+2)}{3(4n^2+8n+3)}$ ✓

As $n \to \infty$, the fraction $\frac{1}{2(2n+1)(2n+3)} \to 0$.

$\displaystyle S_{\infty} = \frac16$

$\displaystyle \mu = \exp\!\left(\int\!\frac{2}{x}\,dx\right) = \exp(2\ln x) =$ $x^2$

Multiplying DE by $x^2$: $x^2\frac{dy}{dx} + 2xy = x^4$

The LHS is $\frac{d}{dx}(x^2 y)$, so: $\frac{d}{dx}(x^2 y) = x^4$

Integrate: $x^2 y = \int x^4\,dx = \frac{x^5}{5} + C$

$\therefore\; y = \frac{x^3}{5} + \frac{C}{x^2}$

Using $y(1)=2$: $2 = \frac15 + C \to C = \frac95$

$\displaystyle y = \frac{x^3}{5} + \frac{9}{5x^2}$

$y(2) = \frac{8}{5} + \frac{9}{20} = \frac{32}{20} + \frac{9}{20} = \frac{41}{20}$

$= 2.05$ (3 s.f.)

$f(0) = \ln(1+0) =$ $0$

$f'(x) = \frac{\cos x}{1+\sin x} \to f'(0) =$ $1$

$f''(x) = \frac{d}{dx}\!\left[\frac{\cos x}{1+\sin x}\right] = \frac{-\sin x(1+\sin x)-\cos^2 x}{(1+\sin x)^2}$

$= -\frac{\sin x+\sin^2 x+\cos^2 x}{(1+\sin x)^2} = -\frac{1+\sin x}{(1+\sin x)^2} = -\frac{1}{1+\sin x}$

$f''(0) =$ $-1$

$f'''(x) = \frac{\cos x}{(1+\sin x)^2} \to f'''(0) =$ $1$

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

$= 0 + 1\cdot x + \frac{(-1)}{2}x^2 + \frac{1}{6}x^3 + \cdots$

$\displaystyle = x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

Auxiliary equation: $\lambda^2 - 4\lambda + 13 = 0$

$\lambda = \dfrac{4 \pm \sqrt{16-52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i$

Since $\lambda = \alpha \pm i\beta$ with $\alpha = 2$, $\beta = 3$:

$y_{\text{CF}} = e^{2x}(A\cos 3x + B\sin 3x)$

Try $y = Pe^{2x}$ (RHS is $5e^{2x}$, not in CF)

$y' = 2Pe^{2x},\; y'' = 4Pe^{2x}$

Substitute: $4Pe^{2x} - 8Pe^{2x} + 13Pe^{2x} = 5e^{2x}$

$9Pe^{2x} = 5e^{2x} \to P = \frac59$

$y_{\text{PI}} = \frac59 e^{2x}$

$y = y_{\text{CF}} + y_{\text{PI}} = e^{2x}(A\cos 3x + B\sin 3x) + \frac59 e^{2x}$

$y(0)=2$: $A + \frac59 = 2 \to A = \frac{13}{9}$

$y' = 2e^{2x}(A\cos 3x + B\sin 3x) + e^{2x}(-3A\sin 3x + 3B\cos 3x) + \frac{10}{9}e^{2x}$

$y'(0)=3$: $2A + 3B + \frac{10}{9} = 3$

$2(\frac{13}{9}) + 3B + \frac{10}{9} = 3 \to \frac{36}{9} + 3B = 3 \to 4 + 3B = 3 \to B = -\frac13$

$\displaystyle y = e^{2x}\!\left(\frac{13}{9}\cos 3x - \frac13\sin 3x + \frac59\right)$

$w = \frac{z}{z-1} \to w(z-1) = z \to wz - w = z \to z(w-1) = w$

$\therefore\; z = \frac{w}{w-1}$

Since $|z| = 2$: $\displaystyle \left|\frac{w}{w-1}\right| = 2 \to |w| = 2|w-1|$

Let $w = u + iv$:

$|w|^2 = 4|w-1|^2$

$u^2+v^2 = 4[(u-1)^2+v^2]$

$u^2+v^2 = 4(u^2-2u+1+v^2)$

$u^2+v^2 = 4u^2-8u+4+4v^2$

$0 = 3u^2+3v^2-8u+4$

$3(u^2-\frac{8u}{3}) + 3v^2 + 4 = 0$

$3[(u-\frac43)^2 - \frac{16}{9}] + 3v^2 + 4 = 0$

$3(u-\frac43)^2 - \frac{16}{3} + 3v^2 + 4 = 0$

$3(u-\frac43)^2 + 3v^2 = \frac{4}{3}$

$(u-\frac43)^2 + v^2 = \frac{4}{9}$

This is the equation of a circle $\to$ image is a circle. ✓

From $(u-\frac43)^2 + v^2 = (\frac23)^2$:

Centre: $(\frac43, 0)$ in the $w$-plane, Radius: $\frac23$

The circle has centre at $(\frac43, 0)$ and radius $\frac23$.

Distance from centre to origin $= \frac43 > \frac23$, so the circle does not enclose the origin.

The origin $w = 0$ corresponds to $z = 0$ (which lies inside $|z| = 2$), confirming the origin is outside the image circle.

The circle passes through $(2, 0)$ and $(\frac23, 0)$ on the real axis.

Tangent $\parallel$ initial line when $\frac{dy}{d\theta}=0$ (where $y = r\sin\theta$)

$y = 3(1+\cos\theta)\sin\theta = 3\sin\theta + 3\sin\theta\cos\theta$

$\frac{dy}{d\theta} = 3\cos\theta + 3(\cos^2\theta - \sin^2\theta) = 3\cos\theta + 3\cos 2\theta$

$= 3(\cos\theta + \cos 2\theta) = 3(\cos\theta + 2\cos^2\theta - 1)$

$= 3(2\cos^2\theta + \cos\theta - 1) =$ $3(2\cos\theta-1)(\cos\theta+1)$

$\frac{dy}{d\theta} = 0 \Rightarrow \cos\theta = \frac12$ or $\cos\theta = -1$

$\cos\theta = \frac12$: $\theta = \pi/3,\; 5\pi/3 \to r = 3(1+\frac12) = \frac92$

$\cos\theta = -1$: $\theta = \pi \to r = 0$ (the pole)

Polar coordinates (other than pole): $(\frac92, \pi/3),\; (\frac92, 5\pi/3)$

Area $\displaystyle = \frac12\int_0^{2\pi} r^2\,d\theta = \frac12\int_0^{2\pi} 9(1+\cos\theta)^2\,d\theta$

$\displaystyle = \frac92\int_0^{2\pi}(1 + 2\cos\theta + \cos^2\theta)\,d\theta$

$\displaystyle = \frac92\int_0^{2\pi}\!\left(1 + 2\cos\theta + \tfrac12(1+\cos 2\theta)\right)d\theta$

$\displaystyle = \frac92\int_0^{2\pi}\!\left(\tfrac32 + 2\cos\theta + \tfrac12\cos 2\theta\right)d\theta$

$\displaystyle = \frac92\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi}$

$\displaystyle = \frac92\left[(3\pi + 0 + 0) - (0 + 0 + 0)\right]$

$\displaystyle = \frac92(3\pi) =$ $\displaystyle \frac{27\pi}{2}$