WFM02/01
June 2026 | Mark Scheme | Total: 75 marks
| Scheme | Marks | Notes |
|---|---|---|
| $z^3 = -8 = 8e^{i\pi}$ $\Rightarrow r = 2$, $3\theta = \pi + 2k\pi$ $\Rightarrow \theta = \pi/3, \pi, 5\pi/3$ | M1 A1 | M1 for modulus and argument method A1 for all three arguments correct |
| Roots: $z = -2,\; 1 \pm i\sqrt{3}$ | A1 | All three correct in $a+ib$ form |
| $|z_1-z_2| = |z_2-z_3| = |z_3-z_1| = 2\sqrt{3}$ $\therefore$ equilateral triangle | M1 A1 | M1 for showing sides equal A1 for correct conclusion |
| Area $= \frac{\sqrt{3}}{4}(2\sqrt{3})^2 = 3\sqrt{3}$ | A1 | Or equivalent exact value |
| Scheme | Marks | Notes |
|---|---|---|
| Recognises need to consider sign of denominator $x+2$ | M1 | |
| Case 1: $x > -2$ (denominator positive) $|2x-1| < x + 2$ | M1 | |
| Subcase $x \ge \frac12$: $2x-1 < x+2 \Rightarrow x < 3$ $\Rightarrow \frac12 \le x < 3$ Subcase $x < \frac12$: $-(2x-1) < x+2 \Rightarrow x > -\frac13$ $\Rightarrow -\frac13 < x < \frac12$ Union: $-\frac13 < x < 3$ | M1 A1 A1 | M1 for correct splitting at $x=\frac12$ A1 for first interval, A1 for second |
| Case 2: $x < -2$ (denominator negative) LHS: $|2x-1| \ge 0$, denominator $x+2 < 0$ $\Rightarrow$ LHS $\le 0 < 1$, automatically true. Solution: $x < -2$ | M1 A1 | M1 for recognising sign makes LHS non-positive A1 for $x < -2$ |
| Scheme | Marks | Notes |
|---|---|---|
| $\displaystyle \mathrm{f}(r) = \frac{1/4}{2r-1} - \frac{1/2}{2r+1} + \frac{1/4}{2r+3}$ | M1 A1 | M1 for correct partial fraction form A1 for all constants correct |
| $\displaystyle \sum \mathrm{f}(r) = \tfrac14\sum\frac{1}{2r-1} - \tfrac12\sum\frac{1}{2r+1} + \tfrac14\sum\frac{1}{2r+3}$ | M1 | Attempt to write out three series |
| After cancellation: $= \frac14 + (\frac14-\frac12)\frac13 + \frac14\frac{1}{2n+3} - \frac14\frac{1}{2n+1}$ $= \frac14 - \frac{1}{12} - \frac{1}{4(2n+1)} + \frac{1}{4(2n+3)}$ $= \frac16 - \frac{1}{4(2n+1)} + \frac{1}{4(2n+3)}$ | M1 A1 | M1 for correct cancellation pattern A1 for correct simplified expression with $1/6$ |
| $= \frac16 - \frac{(2n+3)-(2n+1)}{4(2n+1)(2n+3)}$ $= \frac16 - \frac{1}{2(2n+1)(2n+3)}$ $= \dfrac{2n(n+2)}{3(4n^2+8n+3)}$ | A1 | Correct final form as given |
| $\displaystyle \lim_{n\to\infty} = \frac16$ | B1 | Correct limit |
| Scheme | Marks | Notes |
|---|---|---|
| $\displaystyle \mu = \exp\!\left(\int\!\frac{2}{x}\,dx\right) = x^2$ | M1 A1 | M1 for integral of $2/x$ A1 for $x^2$ |
| $x^2\frac{dy}{dx} + 2xy = x^4 \Rightarrow \frac{d}{dx}(x^2 y) = x^4$ | M1 A1 | M1 for recognising LHS as derivative |
| $x^2 y = \int x^4\,dx = \frac{x^5}{5} + C$ | M1 | Correct integration, $+C$ required |
| $y = \dfrac{x^3}{5} + \dfrac{C}{x^2}$ | A1 | Correct general solution |
| $y=2$ at $x=1$: $2 = \frac15 + C \Rightarrow C = \frac95$ | M1 A1 | M1 for substituting conditions |
| $y = \dfrac{x^3}{5} + \dfrac{9}{5x^2}$ | — | |
| At $x=2$: $y = \frac85 + \frac{9}{20} = \frac{41}{20} = 2.05$ | M1 A1 | M1 for substitution A1 for $2.05$ or $41/20$ |
| Scheme | Marks | Notes |
|---|---|---|
| $f(0) = \ln(1) = 0$ | B1 | |
| $f'(x) = \dfrac{\cos x}{1+\sin x},\quad f'(0) = 1$ | M1 A1 | M1 for correct derivative A1 for correct value |
| $f''(x) = -\dfrac{1}{1+\sin x},\quad f''(0) = -1$ | M1 A1 | M1 for correct second derivative |
| $f'''(x) = \dfrac{\cos x}{(1+\sin x)^2},\quad f'''(0) = 1$ | B1 | Correct third derivative value |
| Maclaurin: $f(x) = 0 + 1\cdot x + \frac{(-1)}{2!}x^2 + \frac{1}{3!}x^3 + \cdots$ $= x - \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$ | M1 A1 A1 | M1 for Maclaurin formula A1 for $x - x^2/2$ A1 for $+x^3/6$ |
| Scheme | Marks | Notes |
|---|---|---|
| Aux eq: $\lambda^2 - 4\lambda + 13 = 0$ $\lambda = \dfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i$ | M1 A1 | M1 for auxiliary equation |
| CF: $y_{\text{CF}} = e^{2x}(A\cos 3x + B\sin 3x)$ | M1 A1 | M1 for correct $e^{2x}$ form |
| PI: try $y = Pe^{2x}$ $\Rightarrow 9Pe^{2x} = 5e^{2x} \Rightarrow P = \frac59$ | M1 A1 | M1 for correct trial function |
| GS: $y = e^{2x}(A\cos 3x + B\sin 3x) + \frac59 e^{2x}$ | B1 | CF + PI |
| $y(0)=2$: $A + \frac59 = 2 \Rightarrow A = \frac{13}{9}$ | M1 A1 | |
| $y'(0)=3$: $2A + 3B + \frac{10}{9} = 3$ $\Rightarrow \frac{26}{9} + 3B + \frac{10}{9} = 3$ $\Rightarrow 4 + 3B = 3 \Rightarrow B = -\frac13$ | M1 A1 | |
| PS: $y = e^{2x}\!\left(\frac{13}{9}\cos 3x - \frac13\sin 3x + \frac59\right)$ | A1 | Correct particular solution |
| Scheme | Marks | Notes |
|---|---|---|
| $w = \dfrac{z}{z-1} \Rightarrow z = \dfrac{w}{w-1}$ | M1 A1 | M1 for rearranging |
| $|z| = 2 \Rightarrow \left|\dfrac{w}{w-1}\right| = 2 \Rightarrow |w| = 2|w-1|$ | M1 | |
| Let $w = u+iv$: $u^2+v^2 = 4[(u-1)^2+v^2]$ | M1 | |
| $\Rightarrow 3u^2+3v^2-8u+4=0$ $\Rightarrow 3(u-\frac43)^2+3v^2 = \frac43$ $\Rightarrow (u-\frac43)^2+v^2 = (\frac23)^2$ | M1 A1 | M1 for completing the square A1 for correct circle equation |
| Centre: $(\frac43, 0)$; Radius: $\frac23$ | B1 B1 | |
| Sketch: circle centre $(\frac43,0)$, radius $\frac23$, does not enclose origin (distance $>$ radius) | B1 B1 B1 | B1 for circle shape B1 for centre & radius B1 for position relative to origin |
| Scheme | Marks | Notes |
|---|---|---|
| Tangent $\parallel$ initial line when $\frac{dy}{d\theta}=0$ where $y = r\sin\theta = 3(1+\cos\theta)\sin\theta$ | M1 | |
| $\frac{dy}{d\theta} = 3(\cos\theta + \cos 2\theta) = 3(2\cos\theta-1)(\cos\theta+1)$ | M1 A1 | M1 for differentiation A1 for factorised form |
| $\frac{dy}{d\theta} = 0 \Rightarrow \cos\theta = \frac12$ or $\cos\theta = -1$ $\cos\theta = \frac12 \Rightarrow \theta = \pi/3,\; 5\pi/3,\; r = \frac92$ $\cos\theta = -1 \Rightarrow \theta = \pi,\; r = 0$ (pole) | M1 A1 | M1 for solving A1 for correct angles and $r$ values |
| Points (other than pole): $(\frac92, \pi/3),\; (\frac92, 5\pi/3)$ | A1 | Both required |
| $r^2 = 9(1+\cos\theta)^2 = 9(1+2\cos\theta+\cos^2\theta)$ $= 9\!\left(\frac32 + 2\cos\theta + \frac12\cos 2\theta\right)$ | M1 A1 | M1 for expansion A1 for correct expression |
| Area $= \frac12\int_0^{2\pi} r^2\,d\theta = \frac92\int_0^{2\pi}\!\left(\frac32+2\cos\theta+\frac12\cos 2\theta\right)d\theta$ $= \frac92\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi}$ $= \frac92(3\pi) = \dfrac{27\pi}{2}$ | M1 A1 | M1 for integration A1 for correct final answer shown |