Question 7 - A-Level Maths

Edexcel FP2 2013 June — Question 7 13 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Resonance cases requiring modified PI
Difficulty	Challenging +1.2 This is a standard FP2 resonance case question requiring recognition that the RHS matches the complementary function (repeated root r=3), necessitating the modified PI form t²e^(3t). While requiring multiple steps (find λ, general solution, apply initial conditions), each step follows a well-established procedure taught explicitly in FP2. The algebraic manipulation is moderate but routine for Further Maths students.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

(a) Find the value of $\lambda$ for which $\lambda t ^ { 2 } \mathrm { e } ^ { 3 t }$ is a particular integral of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 9 y = 6 \mathrm { e } ^ { 3 t } , \quad t \geqslant 0$$ (b) Hence find the general solution of this differential equation. Given that when $t = 0 , y = 5$ and $\frac { \mathrm { d } y } { \mathrm {~d} t } = 4$ (c) find the particular solution of this differential equation, giving your solution in the form $y = \mathrm { f } ( t )$.

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Working	Mark	Guidance
$y = \lambda t^2 e^{3t}$, $\frac{dy}{dt} = 2\lambda te^{3t} + 3\lambda t^2 e^{3t}$	M1A1
$\frac{d^2y}{dt^2} = 2\lambda e^{3t} + 6\lambda te^{3t} + 6\lambda te^{3t} + 9\lambda t^2 e^{3t}$	A1
$2\lambda e^{3t} + 6\lambda te^{3t} + 6\lambda te^{3t} + 9\lambda t^2 e^{3t} - 12\lambda te^{3t} - 18\lambda t^2 e^{3t} + 9\lambda t^2 e^{3t} = 6e^{3t}$	M1dep
$\lambda = 3$	A1cso	Candidates giving $\lambda=3$ without working get 5/5 if no erroneous working seen

Part (b):

Answer	Marks	Guidance
Working	Mark	Guidance
$m^2 - 6m + 9 = 0$, $(m-3)^2 = 0$
C.F. $y = (A+Bt)e^{3t}$	M1A1
G.S. $y = (A+Bt)e^{3t} + 3t^2e^{3t}$	A1ft

Part (c):

Answer	Marks	Guidance
Working	Mark	Guidance
$t=0$, $y=5 \Rightarrow A=5$	B1
$\frac{dy}{dt} = Be^{3t} + 3(A+Bt)e^{3t} + 6te^{3t} + 9t^2e^{3t}$	M1
$\frac{dy}{dt} = 4$: $4 = B + 15$	M1dep
$B = -11$	A1
$y = (5-11t)e^{3t} + 3t^2e^{3t}$	A1ft

Question 7a:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Differentiating $y = \lambda t^2 e^{3t}$ with respect to $t$	M1	Product rule must be used
$\frac{dy}{dt} = 2\lambda t e^{3t} + 3\lambda t^2 e^{3t}$	A1	Correct first differentiation
$\frac{d^2y}{dt^2} = 2\lambda e^{3t} + 6\lambda t e^{3t} + 6\lambda t e^{3t} + 9\lambda t^2 e^{3t}$	A1	Correct second differential
Substituting differentials into equation to obtain numerical value for $\lambda$	M1dep	Dependent on first M mark
$\lambda = 3$	A1cso	No incorrect working seen

Question 7b:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Solving 3-term quadratic auxiliary equation for $m$	M1	Usual rules for solving quadratic
CF: $y = (A + Bt)e^{3t}$	A1
$y = (A + Bt)e^{3t} + 3t^2e^{3t}$	A1ft	Using their CF and their numerical $\lambda$; must have $y=...$ and rhs as function of $t$

Question 7c:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$A = 5$	B1	Deduced from initial conditions
Differentiating their GS to obtain $\frac{dy}{dt} = ...$	M1	Product rule must be used
Using $\frac{dy}{dt} = 4$ and their $A$ to find $B$	M1dep	Dependent on previous M mark
$B = -11$	A1cao cso
Particular solution using their $A$ and $B$	A1ft	Must have $y=...$ and rhs as function of $t$

Question 8a:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$A = \frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int \alpha\cos 2\theta\,d\theta$ with $\alpha = 3$ or $9$	M1	Limits not needed for this mark
$A = (4\times)\int_0^{\frac{\pi}{4}} \frac{9}{2}\cos 2\theta\,d\theta$	A1	Correct limits $\left(0,\frac{\pi}{4}\right)$ with multiple 4, or $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ with multiple 2
$= 18\left[\frac{\sin 2\theta}{2}\right]_0^{\frac{\pi}{4}}$	M1	$\cos 2\theta \to \pm\frac{1}{2}\sin 2\theta$; give M0 for $\pm 2\sin 2\theta$
$9\left[\sin\frac{\pi}{2} - 0\right] = 9$	A1cso	Using limits and factor 4 or 2 to obtain 9

Question 8b:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$r\sin\theta = 3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$	M1	Or $r^2\sin^2\theta = 9\cos 2\theta\sin^2\theta$; 3 or 9 allowed
$\frac{d}{d\theta}(r\sin\theta) = \left\{-3\times\frac{1}{2}(\cos 2\theta)^{-\frac{1}{2}}\times 2\sin 2\theta\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta\right\}$	M1depA1	Product and chain rule must be used; correct differentiation
Setting $\frac{d}{d\theta}(r\sin\theta) = 0$	M1
Solving using appropriate trig formulae	M1dep	Must be correct formulae
$\sin\theta = \frac{1}{2}$ or $\cos\theta = \frac{\sqrt{3}}{2}$ or $\theta = \pm\frac{\pi}{6}$	A1	Ignore extra answers
$r\sin\frac{\pi}{6} = 3\left(\cos\frac{\pi}{3}\right)^{\frac{1}{2}}\times\frac{1}{2} = \frac{3\sqrt{2}}{4}$, so $PS = \frac{3\sqrt{2}}{2}$	B1	$\frac{1}{2}PS = \frac{3\sqrt{2}}{4}$; length of $PQ$ not needed
Shaded area $= 6\times\frac{3\sqrt{2}}{2} - 9 = 9\sqrt{2} - 9$	M1, A1	$PS\times 6 -$ answer to (a); evidence of $PS$ from their $\theta$

Mark Scheme Extraction

Question (Polar Curves / Differentiation):

Option 1 – Using $r^2 \cos^2\theta$ variants before differentiation

Working Step 1:

Answer	Marks	Guidance
Use of $3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\sin\theta$	M1	First M mark; use of $\cos 2\theta = \cos^2\theta - \sin^2\theta$ gives 4th M mark provided a value of $\sin\theta$ or alt is reached with no errors seen after the differentiation

Working Step 2:

Answer	Marks	Guidance
$3\left(\frac{1}{2}\right)(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$	DM1	Second (dependent) M mark for differentiating using the product rule
$-6(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$	A1	Correct derivative; M1 for setting derivative equal to 0

Working Step 3:

Answer	Marks	Guidance
$-6\sin^2\theta + 3(\cos^2\theta - \sin^2\theta) = 0$	—	Multiplication by $(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}$, division by $3\cos\theta$ and use of $\cos^2\theta = 1 - \sin^2\theta$ simplify the equation but do not provide specific M marks
$4\sin^2\theta = 1$	—

Working Step 4:

Answer	Marks	Guidance
$\sin\theta = \pm\frac{1}{2}$	M1	Value of $\sin\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned); gives final M mark
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark given here

Variant using $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$:

Answer	Marks	Guidance
Use of $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$	M1	First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark
$3\left(\frac{1}{2}\right)(2\cos^2\theta - 1)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$	DM1	Product rule differentiation
$-6(2\cos^2\theta - 1)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$	A1	Correct derivative; M1 for setting equal to 0
$-6\sin^2\theta + 3(2\cos^2\theta - 1) = 0$	—	Division by $3\cos\theta$; $\sin^2\theta = 1 - \cos^2\theta$ or vice versa simplifies but no specific M marks
$4\sin^2\theta = 1$ or $4\cos^2\theta = 3$	—
$\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$	M1	Final M mark
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark

Variant using $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$:

Answer	Marks	Guidance
Use of $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$	M1	First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark
$3\left(\frac{1}{2}\right)(1-2\sin^2\theta)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$	DM1	Product rule
$-6(1-2\sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$	A1	Correct derivative; M1 for setting equal to 0
$-6\sin^2\theta + 3(1-2\cos^2\theta) = 0$	—	Simplification steps; no specific M marks
$4\sin^2\theta = 1$ or $4\cos^2\theta = 3$	—
$\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$	M1	Final M mark
$\theta = \dfrac{\pi}{6}$	A1	Second A mark

Option 2 – Using $r^2\sin^2\theta$ with/without manipulation of $\cos 2\theta$ before differentiation

Variant using $9\cos 2\theta \sin^2\theta$:

Answer	Marks	Guidance
Use of $9\cos 2\theta\sin^2\theta$	M1	First M mark even if slip on the 9 and use 3, but must be $\sin^2\theta$
$-9(2)\sin 2\theta\sin^2\theta + 9(2)\cos 2\theta\sin\theta\cos\theta = 0$	DM1	Product rule differentiation
(Correct derivative set to 0)	A1	Correct derivative; M1 for setting equal to 0
$-2\sin^2\theta + \cos 2\theta = 0$	—	Division by $9\sin 2\theta$ or $18\sin\theta$ and use of $\sin 2\theta = 2\sin\theta\cos\theta$ followed by division by $\cos\theta$; simplifies but no specific M marks
or $-\sin 2\theta\sin\theta + \cos 2\theta\cos\theta = 0$ leading to $-2\sin^2\theta + \cos 2\theta = 0$ or $\cos 3\theta = 1$ (compound angle formula)	—
$-2\sin^2\theta + 1 - 2\sin^2\theta = 0$	—	Use of $\cos 2\theta = 1 - 2\sin^2\theta$ gives next M mark
$4\sin^2\theta = 1$	M1	Provided value of $\sin\theta$ or alt is reached with no errors seen
$\sin\theta = \pm\frac{1}{2}$ or $3\theta = 2\pi$ (from $\cos 3\theta = 1$)	M1	Final M mark
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark

Variant using $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$:

Answer	Marks	Guidance
Use of $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$	M1	First M mark even if slip on 9; must be $\sin^2\theta$
Could be expanded to $9\cos^2\theta\sin^2\theta - 9\sin^4\theta$; derivative immediately: $-18\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 36\sin^3\theta\cos\theta$	—	Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark
$9(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin^2\theta + 9(\cos^2\theta - \sin^2\theta)2\sin\theta\cos\theta$	DM1	Product rule
$-36\sin^3\theta\cos\theta + 18(\cos^2\theta - \sin^2\theta)\sin\theta\cos\theta = 0$	A1	Correct derivative; M1 for setting equal to 0
$-36\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 18\sin^3\theta\cos\theta = 0$	—
$18\cos^3\theta\sin\theta - 54\sin^3\theta\cos\theta = 0$	—	Division by $18\cos\theta\sin\theta$; $\sin^2\theta = 1 - \cos^2\theta$ simplifies; no specific M marks
$\cos^2\theta - 3\sin^2\theta = 0$	—
$1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$	—
$\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$	M1	Final M mark
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark

Mark Scheme Content

Question (Differentiation involving $\cos 2\theta$ and $\sin^2\theta$):

Method 1: Using $9(2\cos^2\theta - 1)\sin^2\theta$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Use of $9(2\cos^2\theta - 1)\sin^2\theta$	M1	First M mark even if slip on the 9, but must have $\sin^2\theta$
Expanded form $18\cos^2\theta\sin^2\theta - 9\sin^2\theta$ before differentiation, giving derivative $-36\cos\theta\sin^3\theta + 36\cos^3\theta\sin\theta - 18\sin\theta\cos\theta$	M1	Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark, provided value of $\sin\theta$ or alt is reached with no errors after differentiation
$9(-4\cos\theta\sin\theta)\sin^2\theta + 9(2\cos^2\theta-1)2\sin\theta\cos\theta = 0$	DM1 A1	Second (dependent) M mark for differentiating using product rule; A1 for correct derivative
$-36\sin^3\theta\cos\theta + 36\cos^3\theta\sin\theta - 18\sin\theta\cos\theta = 0$	M1	M1 for setting derivative equal to 0
$-2\sin^2\theta + 2\cos^2\theta - 1 = 0$		Division by $18\cos\theta\sin\theta$ and use of $\sin^2\theta = 1 - \cos^2\theta$; also possible to use $\cos^2\theta - \sin^2\theta = \cos 2\theta$
$2\cos 2\theta = 1$ or $1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$	dM1
$\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ or $\cos 2\theta = \frac{1}{2}$	M1	Value of $\sin\theta$ or alt reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned)
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark

Method 2: Using $9(1 - 2\sin^2\theta)\sin^2\theta$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Use of $9(1-2\sin^2\theta)\sin^2\theta$	M1	First M mark even if slip on 9, but must be $\sin^2\theta$
Expanded: $9\sin^2\theta - 18\sin^4\theta$, derivative $18\sin\theta\cos\theta - 72\cos\theta\sin^3\theta$		Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark
$9(-4\cos\theta\sin\theta)\sin^2\theta + 9(1-2\sin^2\theta)2\sin\theta\cos\theta = 0$	DM1 A1	Dependent M for product rule; A1 for correct derivative; M1 for setting to 0
$-36\sin^3\theta\cos\theta - 36\sin^3\theta\cos\theta + 18\sin\theta\cos\theta = 0$
$1 - 4\sin^2\theta = 0$		Division by $18\cos\theta\sin\theta$
$\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ or $\cos 2\theta = \frac{1}{2}$	M1	Value of $\sin\theta$ or alt reached; arithmetic slips condoned
$\theta = \dfrac{\pi}{6}$	A1	Second accuracy mark

Method 3: Using factor formulae after differentiating $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Use of $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$	M1	M1 awarded for using this form
$3\left(\frac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(-2\sin 2\theta)\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta = 0$	M1 A1	M1A1 for correct differentiation using product and chain rule
Setting derivative equal to zero	M1
Multiply by $(\cos 2\theta)^{\frac{1}{2}}$, divide by 3: $\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 0$	dM1	Using correct trig formulae to reduce to $\cos k\theta = \ldots$
$\cos 3\theta = 0$
$3\theta = \dfrac{\pi}{2}$, so $\theta = \dfrac{\pi}{6}$	A1	A mark requires $\cos\theta = \ldots$ or $\theta = \dfrac{\pi}{6}$

## Question 7:

### Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $y = \lambda t^2 e^{3t}$, $\frac{dy}{dt} = 2\lambda te^{3t} + 3\lambda t^2 e^{3t}$ | M1A1 | |
| $\frac{d^2y}{dt^2} = 2\lambda e^{3t} + 6\lambda te^{3t} + 6\lambda te^{3t} + 9\lambda t^2 e^{3t}$ | A1 | |
| $2\lambda e^{3t} + 6\lambda te^{3t} + 6\lambda te^{3t} + 9\lambda t^2 e^{3t} - 12\lambda te^{3t} - 18\lambda t^2 e^{3t} + 9\lambda t^2 e^{3t} = 6e^{3t}$ | M1dep | |
| $\lambda = 3$ | A1cso | Candidates giving $\lambda=3$ without working get 5/5 if no erroneous working seen |

### Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $m^2 - 6m + 9 = 0$, $(m-3)^2 = 0$ | | |
| C.F. $y = (A+Bt)e^{3t}$ | M1A1 | |
| G.S. $y = (A+Bt)e^{3t} + 3t^2e^{3t}$ | A1ft | |

### Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| $t=0$, $y=5 \Rightarrow A=5$ | B1 | |
| $\frac{dy}{dt} = Be^{3t} + 3(A+Bt)e^{3t} + 6te^{3t} + 9t^2e^{3t}$ | M1 | |
| $\frac{dy}{dt} = 4$: $4 = B + 15$ | M1dep | |
| $B = -11$ | A1 | |
| $y = (5-11t)e^{3t} + 3t^2e^{3t}$ | A1ft | |

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiating $y = \lambda t^2 e^{3t}$ with respect to $t$ | M1 | Product rule must be used |
| $\frac{dy}{dt} = 2\lambda t e^{3t} + 3\lambda t^2 e^{3t}$ | A1 | Correct first differentiation |
| $\frac{d^2y}{dt^2} = 2\lambda e^{3t} + 6\lambda t e^{3t} + 6\lambda t e^{3t} + 9\lambda t^2 e^{3t}$ | A1 | Correct second differential |
| Substituting differentials into equation to obtain numerical value for $\lambda$ | M1dep | Dependent on first M mark |
| $\lambda = 3$ | A1cso | No incorrect working seen |

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Solving 3-term quadratic auxiliary equation for $m$ | M1 | Usual rules for solving quadratic |
| CF: $y = (A + Bt)e^{3t}$ | A1 | |
| $y = (A + Bt)e^{3t} + 3t^2e^{3t}$ | A1ft | Using their CF and their numerical $\lambda$; must have $y=...$ and rhs as function of $t$ |

## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 5$ | B1 | Deduced from initial conditions |
| Differentiating their GS to obtain $\frac{dy}{dt} = ...$ | M1 | Product rule must be used |
| Using $\frac{dy}{dt} = 4$ and their $A$ to find $B$ | M1dep | Dependent on previous M mark |
| $B = -11$ | A1cao cso | |
| Particular solution using their $A$ and $B$ | A1ft | Must have $y=...$ and rhs as function of $t$ |

---

## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int \alpha\cos 2\theta\,d\theta$ with $\alpha = 3$ or $9$ | M1 | Limits not needed for this mark |
| $A = (4\times)\int_0^{\frac{\pi}{4}} \frac{9}{2}\cos 2\theta\,d\theta$ | A1 | Correct limits $\left(0,\frac{\pi}{4}\right)$ with multiple 4, or $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ with multiple 2 |
| $= 18\left[\frac{\sin 2\theta}{2}\right]_0^{\frac{\pi}{4}}$ | M1 | $\cos 2\theta \to \pm\frac{1}{2}\sin 2\theta$; give M0 for $\pm 2\sin 2\theta$ |
| $9\left[\sin\frac{\pi}{2} - 0\right] = 9$ | A1cso | Using limits and factor 4 or 2 to obtain 9 |

## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r\sin\theta = 3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | Or $r^2\sin^2\theta = 9\cos 2\theta\sin^2\theta$; 3 or 9 allowed |
| $\frac{d}{d\theta}(r\sin\theta) = \left\{-3\times\frac{1}{2}(\cos 2\theta)^{-\frac{1}{2}}\times 2\sin 2\theta\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta\right\}$ | M1depA1 | Product and chain rule must be used; correct differentiation |
| Setting $\frac{d}{d\theta}(r\sin\theta) = 0$ | M1 | |
| Solving using appropriate trig formulae | M1dep | Must be correct formulae |
| $\sin\theta = \frac{1}{2}$ or $\cos\theta = \frac{\sqrt{3}}{2}$ or $\theta = \pm\frac{\pi}{6}$ | A1 | Ignore extra answers |
| $r\sin\frac{\pi}{6} = 3\left(\cos\frac{\pi}{3}\right)^{\frac{1}{2}}\times\frac{1}{2} = \frac{3\sqrt{2}}{4}$, so $PS = \frac{3\sqrt{2}}{2}$ | B1 | $\frac{1}{2}PS = \frac{3\sqrt{2}}{4}$; length of $PQ$ not needed |
| Shaded area $= 6\times\frac{3\sqrt{2}}{2} - 9 = 9\sqrt{2} - 9$ | M1, A1 | $PS\times 6 -$ answer to (a); evidence of $PS$ from their $\theta$ |

# Mark Scheme Extraction

## Question (Polar Curves / Differentiation):

### Option 1 – Using $r^2 \cos^2\theta$ variants before differentiation

---

**Working Step 1:**

| Use of $3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = \cos^2\theta - \sin^2\theta$ gives 4th M mark provided a value of $\sin\theta$ or alt is reached with no errors seen after the differentiation |

**Working Step 2:**

| $3\left(\frac{1}{2}\right)(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | DM1 | Second (dependent) M mark for differentiating using the product rule |

| $-6(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | A1 | Correct derivative; M1 for setting derivative equal to 0 |

**Working Step 3:**

| $-6\sin^2\theta + 3(\cos^2\theta - \sin^2\theta) = 0$ | — | Multiplication by $(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}$, division by $3\cos\theta$ and use of $\cos^2\theta = 1 - \sin^2\theta$ simplify the equation but do not provide specific M marks |

| $4\sin^2\theta = 1$ | — | |

**Working Step 4:**

| $\sin\theta = \pm\frac{1}{2}$ | M1 | Value of $\sin\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned); gives final M mark |

| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark given here |

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**Variant using $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$:**

| Use of $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark |

| $3\left(\frac{1}{2}\right)(2\cos^2\theta - 1)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$ | DM1 | Product rule differentiation |

| $-6(2\cos^2\theta - 1)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$ | A1 | Correct derivative; M1 for setting equal to 0 |

| $-6\sin^2\theta + 3(2\cos^2\theta - 1) = 0$ | — | Division by $3\cos\theta$; $\sin^2\theta = 1 - \cos^2\theta$ or vice versa simplifies but no specific M marks |

| $4\sin^2\theta = 1$ or $4\cos^2\theta = 3$ | — | |

| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ | M1 | Final M mark |

| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |

---

**Variant using $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$:**

| Use of $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark |

| $3\left(\frac{1}{2}\right)(1-2\sin^2\theta)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | DM1 | Product rule |

| $-6(1-2\sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | A1 | Correct derivative; M1 for setting equal to 0 |

| $-6\sin^2\theta + 3(1-2\cos^2\theta) = 0$ | — | Simplification steps; no specific M marks |

| $4\sin^2\theta = 1$ or $4\cos^2\theta = 3$ | — | |

| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ | M1 | Final M mark |

| $\theta = \dfrac{\pi}{6}$ | A1 | Second A mark |

---

### Option 2 – Using $r^2\sin^2\theta$ with/without manipulation of $\cos 2\theta$ before differentiation

**Variant using $9\cos 2\theta \sin^2\theta$:**

| Use of $9\cos 2\theta\sin^2\theta$ | M1 | First M mark even if slip on the 9 and use 3, but must be $\sin^2\theta$ |

| $-9(2)\sin 2\theta\sin^2\theta + 9(2)\cos 2\theta\sin\theta\cos\theta = 0$ | DM1 | Product rule differentiation |

| (Correct derivative set to 0) | A1 | Correct derivative; M1 for setting equal to 0 |

| $-2\sin^2\theta + \cos 2\theta = 0$ | — | Division by $9\sin 2\theta$ or $18\sin\theta$ and use of $\sin 2\theta = 2\sin\theta\cos\theta$ followed by division by $\cos\theta$; simplifies but no specific M marks |

| or $-\sin 2\theta\sin\theta + \cos 2\theta\cos\theta = 0$ leading to $-2\sin^2\theta + \cos 2\theta = 0$ or $\cos 3\theta = 1$ (compound angle formula) | — | |

| $-2\sin^2\theta + 1 - 2\sin^2\theta = 0$ | — | Use of $\cos 2\theta = 1 - 2\sin^2\theta$ gives next M mark |

| $4\sin^2\theta = 1$ | M1 | Provided value of $\sin\theta$ or alt is reached with no errors seen |

| $\sin\theta = \pm\frac{1}{2}$ or $3\theta = 2\pi$ (from $\cos 3\theta = 1$) | M1 | Final M mark |

| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |

---

**Variant using $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$:**

| Use of $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$ | M1 | First M mark even if slip on 9; must be $\sin^2\theta$ |

| Could be expanded to $9\cos^2\theta\sin^2\theta - 9\sin^4\theta$; derivative immediately: $-18\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 36\sin^3\theta\cos\theta$ | — | Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark |

| $9(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin^2\theta + 9(\cos^2\theta - \sin^2\theta)2\sin\theta\cos\theta$ | DM1 | Product rule |

| $-36\sin^3\theta\cos\theta + 18(\cos^2\theta - \sin^2\theta)\sin\theta\cos\theta = 0$ | A1 | Correct derivative; M1 for setting equal to 0 |

| $-36\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 18\sin^3\theta\cos\theta = 0$ | — | |

| $18\cos^3\theta\sin\theta - 54\sin^3\theta\cos\theta = 0$ | — | Division by $18\cos\theta\sin\theta$; $\sin^2\theta = 1 - \cos^2\theta$ simplifies; no specific M marks |

| $\cos^2\theta - 3\sin^2\theta = 0$ | — | |

| $1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$ | — | |

| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ | M1 | Final M mark |

| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |

## Mark Scheme Content

---

### Question (Differentiation involving $\cos 2\theta$ and $\sin^2\theta$):

**Method 1: Using $9(2\cos^2\theta - 1)\sin^2\theta$**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $9(2\cos^2\theta - 1)\sin^2\theta$ | M1 | First M mark even if slip on the 9, but must have $\sin^2\theta$ |
| Expanded form $18\cos^2\theta\sin^2\theta - 9\sin^2\theta$ before differentiation, giving derivative $-36\cos\theta\sin^3\theta + 36\cos^3\theta\sin\theta - 18\sin\theta\cos\theta$ | M1 | Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark, provided value of $\sin\theta$ or alt is reached with no errors after differentiation |
| $9(-4\cos\theta\sin\theta)\sin^2\theta + 9(2\cos^2\theta-1)2\sin\theta\cos\theta = 0$ | DM1 A1 | Second (dependent) M mark for differentiating using product rule; A1 for correct derivative |
| $-36\sin^3\theta\cos\theta + 36\cos^3\theta\sin\theta - 18\sin\theta\cos\theta = 0$ | M1 | M1 for setting derivative equal to 0 |
| $-2\sin^2\theta + 2\cos^2\theta - 1 = 0$ | | Division by $18\cos\theta\sin\theta$ and use of $\sin^2\theta = 1 - \cos^2\theta$; also possible to use $\cos^2\theta - \sin^2\theta = \cos 2\theta$ |
| $2\cos 2\theta = 1$ or $1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$ | dM1 | |
| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ or $\cos 2\theta = \frac{1}{2}$ | M1 | Value of $\sin\theta$ or alt reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned) |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |

---

**Method 2: Using $9(1 - 2\sin^2\theta)\sin^2\theta$**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $9(1-2\sin^2\theta)\sin^2\theta$ | M1 | First M mark even if slip on 9, but must be $\sin^2\theta$ |
| Expanded: $9\sin^2\theta - 18\sin^4\theta$, derivative $18\sin\theta\cos\theta - 72\cos\theta\sin^3\theta$ | | Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark |
| $9(-4\cos\theta\sin\theta)\sin^2\theta + 9(1-2\sin^2\theta)2\sin\theta\cos\theta = 0$ | DM1 A1 | Dependent M for product rule; A1 for correct derivative; M1 for setting to 0 |
| $-36\sin^3\theta\cos\theta - 36\sin^3\theta\cos\theta + 18\sin\theta\cos\theta = 0$ | | |
| $1 - 4\sin^2\theta = 0$ | | Division by $18\cos\theta\sin\theta$ |
| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ or $\cos 2\theta = \frac{1}{2}$ | M1 | Value of $\sin\theta$ or alt reached; arithmetic slips condoned |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |

---

**Method 3: Using factor formulae after differentiating $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | M1 awarded for using this form |
| $3\left(\frac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(-2\sin 2\theta)\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta = 0$ | M1 A1 | M1A1 for correct differentiation using product and chain rule |
| Setting derivative equal to zero | M1 | |
| Multiply by $(\cos 2\theta)^{\frac{1}{2}}$, divide by 3: $\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 0$ | dM1 | Using correct trig formulae to reduce to $\cos k\theta = \ldots$ |
| $\cos 3\theta = 0$ | | |
| $3\theta = \dfrac{\pi}{2}$, so $\theta = \dfrac{\pi}{6}$ | A1 | A mark requires $\cos\theta = \ldots$ or $\theta = \dfrac{\pi}{6}$ |

Show LaTeX source

\begin{enumerate}
  \item (a) Find the value of $\lambda$ for which $\lambda t ^ { 2 } \mathrm { e } ^ { 3 t }$ is a particular integral of the differential equation
\end{enumerate}

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 9 y = 6 \mathrm { e } ^ { 3 t } , \quad t \geqslant 0$$

(b) Hence find the general solution of this differential equation.

Given that when $t = 0 , y = 5$ and $\frac { \mathrm { d } y } { \mathrm {~d} t } = 4$\\
(c) find the particular solution of this differential equation, giving your solution in the form $y = \mathrm { f } ( t )$.\\

\hfill \mbox{\textit{Edexcel FP2 2013 Q7 [13]}}

This paper (8 questions)

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Q1 4 Q2 7 Q3 8 Q4 9 Q5 10 Q6 11 Q7 13 Q8 13

Edexcel FP2 2013 June — Question 7 13 marks

Question 7:

Part (a):

Part (b):

Part (c):

Question 7a:

Question 7b:

Question 7c:

Question 8a:

Question 8b:

Mark Scheme Extraction

Question (Polar Curves / Differentiation):

Option 1 – Using \(r^2 \cos^2\theta\) variants before differentiation

Working Step 1:

Working Step 2:

Working Step 3:

Working Step 4:

Variant using \(3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta\):

Variant using \(3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta\):

Option 2 – Using \(r^2\sin^2\theta\) with/without manipulation of \(\cos 2\theta\) before differentiation

Variant using \(9\cos 2\theta \sin^2\theta\):

Variant using \(9(\cos^2\theta - \sin^2\theta)\sin^2\theta\):

Mark Scheme Content

Question (Differentiation involving \(\cos 2\theta\) and \(\sin^2\theta\)):

Method 1: Using \(9(2\cos^2\theta - 1)\sin^2\theta\)

Method 2: Using \(9(1 - 2\sin^2\theta)\sin^2\theta\)

Method 3: Using factor formulae after differentiating \(3(\cos 2\theta)^{\frac{1}{2}}\sin\theta\)

This paper (8 questions)

Answer	Marks	Guidance
\(3\left(\frac{1}{2}\right)(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\)	DM1	Second (dependent) M mark for differentiating using the product rule
\(-6(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\)	A1	Correct derivative; M1 for setting derivative equal to 0

Answer	Marks	Guidance
\(-6\sin^2\theta + 3(\cos^2\theta - \sin^2\theta) = 0\)	—	Multiplication by \((\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\), division by \(3\cos\theta\) and use of \(\cos^2\theta = 1 - \sin^2\theta\) simplify the equation but do not provide specific M marks
\(4\sin^2\theta = 1\)	—

Answer	Marks	Guidance
\(\sin\theta = \pm\frac{1}{2}\)	M1	Value of \(\sin\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned); gives final M mark
\(\theta = \dfrac{\pi}{6}\)	A1	Second accuracy mark given here

Answer	Marks	Guidance
Use of \(3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta\)	M1	First M mark; use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark
\(3\left(\frac{1}{2}\right)(2\cos^2\theta - 1)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0\)	DM1	Product rule differentiation
\(-6(2\cos^2\theta - 1)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0\)	A1	Correct derivative; M1 for setting equal to 0
\(-6\sin^2\theta + 3(2\cos^2\theta - 1) = 0\)	—	Division by \(3\cos\theta\); \(\sin^2\theta = 1 - \cos^2\theta\) or vice versa simplifies but no specific M marks
\(4\sin^2\theta = 1\) or \(4\cos^2\theta = 3\)	—
\(\sin\theta = \pm\frac{1}{2}\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\)	M1	Final M mark
\(\theta = \dfrac{\pi}{6}\)	A1	Second accuracy mark

Answer	Marks	Guidance
Use of \(3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta\)	M1	First M mark; use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark
\(3\left(\frac{1}{2}\right)(1-2\sin^2\theta)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\)	DM1	Product rule
\(-6(1-2\sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(1-2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\)	A1	Correct derivative; M1 for setting equal to 0
\(-6\sin^2\theta + 3(1-2\cos^2\theta) = 0\)	—	Simplification steps; no specific M marks
\(4\sin^2\theta = 1\) or \(4\cos^2\theta = 3\)	—
\(\sin\theta = \pm\frac{1}{2}\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\)	M1	Final M mark
\(\theta = \dfrac{\pi}{6}\)	A1	Second A mark

Answer	Marks	Guidance
Use of \(9\cos 2\theta\sin^2\theta\)	M1	First M mark even if slip on the 9 and use 3, but must be \(\sin^2\theta\)
\(-9(2)\sin 2\theta\sin^2\theta + 9(2)\cos 2\theta\sin\theta\cos\theta = 0\)	DM1	Product rule differentiation
(Correct derivative set to 0)	A1	Correct derivative; M1 for setting equal to 0
\(-2\sin^2\theta + \cos 2\theta = 0\)	—	Division by \(9\sin 2\theta\) or \(18\sin\theta\) and use of \(\sin 2\theta = 2\sin\theta\cos\theta\) followed by division by \(\cos\theta\); simplifies but no specific M marks
or \(-\sin 2\theta\sin\theta + \cos 2\theta\cos\theta = 0\) leading to \(-2\sin^2\theta + \cos 2\theta = 0\) or \(\cos 3\theta = 1\) (compound angle formula)	—
\(-2\sin^2\theta + 1 - 2\sin^2\theta = 0\)	—	Use of \(\cos 2\theta = 1 - 2\sin^2\theta\) gives next M mark
\(4\sin^2\theta = 1\)	M1	Provided value of \(\sin\theta\) or alt is reached with no errors seen
\(\sin\theta = \pm\frac{1}{2}\) or \(3\theta = 2\pi\) (from \(\cos 3\theta = 1\))	M1	Final M mark
\(\theta = \dfrac{\pi}{6}\)	A1	Second accuracy mark

Answer	Marks	Guidance
Use of \(9(\cos^2\theta - \sin^2\theta)\sin^2\theta\)	M1	First M mark even if slip on 9; must be \(\sin^2\theta\)
Could be expanded to \(9\cos^2\theta\sin^2\theta - 9\sin^4\theta\); derivative immediately: \(-18\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 36\sin^3\theta\cos\theta\)	—	Use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark
\(9(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin^2\theta + 9(\cos^2\theta - \sin^2\theta)2\sin\theta\cos\theta\)	DM1	Product rule
\(-36\sin^3\theta\cos\theta + 18(\cos^2\theta - \sin^2\theta)\sin\theta\cos\theta = 0\)	A1	Correct derivative; M1 for setting equal to 0
\(-36\cos\theta\sin^3\theta + 18\cos^3\theta\sin\theta - 18\sin^3\theta\cos\theta = 0\)	—
\(18\cos^3\theta\sin\theta - 54\sin^3\theta\cos\theta = 0\)	—	Division by \(18\cos\theta\sin\theta\); \(\sin^2\theta = 1 - \cos^2\theta\) simplifies; no specific M marks
\(\cos^2\theta - 3\sin^2\theta = 0\)	—
\(1 - 4\sin^2\theta = 0\) or \(4\cos^2\theta - 3 = 0\)	—
\(\sin\theta = \pm\frac{1}{2}\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\)	M1	Final M mark
\(\theta = \dfrac{\pi}{6}\)	A1	Second accuracy mark

Answer	Marks	Guidance
Working	Mark	Guidance
\(t=0\), \(y=5 \Rightarrow A=5\)	B1
\(\frac{dy}{dt} = Be^{3t} + 3(A+Bt)e^{3t} + 6te^{3t} + 9t^2e^{3t}\)	M1
\(\frac{dy}{dt} = 4\): \(4 = B + 15\)	M1dep
\(B = -11\)	A1
\(y = (5-11t)e^{3t} + 3t^2e^{3t}\)	A1ft