Question 3 - A-Level Maths

OCR MEI Further Pure with Technology Specimen — Question 3 20 marks

Exam Board	OCR MEI
Module	Further Pure with Technology (Further Pure with Technology)
Session	Specimen
Marks	20
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Iterative/numerical methods
Difficulty	Challenging +1.2 This is a multi-part question requiring tangent field sketching, domain analysis, Runge-Kutta numerical method implementation in a spreadsheet, and analytical solution of a separable differential equation. While it covers several techniques and has many parts (typical of Further Maths), each individual component is relatively standard: the numerical method is given explicitly, the analytical solution for a=0 is straightforward separation of variables, and verification is routine substitution. The question is more lengthy than conceptually demanding, placing it moderately above average difficulty.
Spec	1.09d Newton-Raphson method 4.10a General/particular solutions: of differential equations 4.10b Model with differential equations: kinematics and other contexts

3 This question explores the family of differential equations $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { 1 + a x + 2 y }$ for various values of the parameter $a$. Fig. 3 shows the tangent field in the case $a = 1$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{141c85ec-5749-4f24-9f6d-fe7a01567511-4_691_696_452_696} \captionsetup{labelformat=empty} \caption{Fig. 3}

\end{figure}

(A) Sketch the tangent field in the case $a = - 2$.
(B) Explain why the tangent field is not defined for the whole coordinate plane.
(C) Give an inequality which describes the region in which the tangent field is defined.
(D) Find a value of $a$ such that the region for which the tangent field is defined includes the entire $x$-axis.
(A) For the case $a = 1$, with $y = 1$ when $x = 0$, construct a spreadsheet for the Runge-Kutta method of order 2 with formulae as follows, where $\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }$. $$\begin{aligned} k _ { 1 } & = h \mathrm { f } \left( x _ { n } , y _ { n } \right) \\ k _ { 2 } & = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 1 } \right) \\ y _ { n + 1 } & = y _ { n } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right) \end{aligned}$$ State the formulae you have used in your spreadsheet.
(B) Use your spreadsheet to obtain the value of $y$ correct to 4 decimal places when $x = 1$ for
(A) For the case $a = 0$ find the analytical solution that passes through the point ( 0,1 ).
(B) Verify that the solution in part (iii) (A) is a solution to the differential equation.
(C) Use the solution in part (iii) (A) to find the value of $y$ correct to 4 decimal places when $x = 1$.
(A) Verify that $y = - \frac { a } { 2 } x + \frac { a ^ { 2 } } { 8 } - \frac { 1 } { 2 }$ is a solution for all cases when $a \leq 0$.
(B) Show that this is the only straight line solution in these cases. \section*{Copyright Information:} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
OCR is part of the

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
3	(i)	(A)

A1

Answer	Marks
[2]	1.1
1.1	Correct in at least one quadrant

N

Correct everywhere

E

Answer	Marks	Guidance
3	(i)	(B)

is not defined when 1(cid:14)ax(cid:14)2y(cid:31)0 because the

dx

Answer	Marks
square root is not real.	E1

C

Answer	Marks	Guidance
[1]	2.4
I	M
3	(i)	(C)

dy a 1

is defined when y(cid:116)(cid:16) x(cid:16) o.e.

Answer	Marks	Guidance
dx 2 2	B1
[1]	2.2a
3	(i)	(D)
a = 0 S	B1
[1]	1.1	may be obtained by using slider on their

tangent field – no evidence required – or by

using (i) (B)

Answer	Marks	Guidance
3	(ii)	(A)

B2=0, C2=1

D2= $A$2sqrt(1+B2+2C2),

E2= $A$2sqrt(1+(B2+$A$2)+2(C2+D2))

Answer	Marks
B3= B2+$A$2, C3 =C2+0.5*(D2+E2)	M1

M1

Answer	Marks
[3]	1.1

3.1a

2.5

Answer	Marks
I	Columns for x & y or equivalent

Columns for k & k or equivalent

N1 2

Formulae for x & y

n+1 n+1

E

M

or

define: h=0.1, f(x,y)=sqrt(1+x+2y)

A2=0, B2=1

C2=hf(A2, B2), D2=hf(A2 + h, B2 + C2)

A3=A2+h, B3=B2 + 0.5*(C2 + D2)

Answer	Marks	Guidance
3	(ii)	(B)
h = 0.05: y = 3.3500 when x = 1	C

A1

Answer	Marks
[2]	1.1

3.2a

Answer	Marks	Guidance
3	(iii)	(A)

y(cid:32) x2 (cid:14) 3x(cid:14)1

Answer	Marks	Guidance
2	B1
[1]	2.2a
3	(iii)	(B)

(cid:32) x(cid:14) 3

dx

(cid:32) x2 (cid:14)2 3x(cid:14)3

(cid:167)1 (cid:183)

(cid:32) 1(cid:14)2 x2 (cid:14) 3x(cid:14)1

(cid:168) (cid:184)

(cid:169)2 (cid:185)

Answer	Marks
(cid:32) 1(cid:14)2y	M1

A1

Answer	Marks
[2]	1.1
I2.1	N

E

M

Clear reasoning must be seen throughout

Answer	Marks	Guidance
3	(iii)	(C)

B1

Answer	Marks	Guidance
[1]	1.1
3	(iv)	(A)

1(cid:14)ax(cid:14)2(cid:168)(cid:16) x(cid:14) (cid:16) (cid:184) (cid:32)

(cid:169) 2 8 2(cid:185) 4

a

(cid:32)(cid:16)

2 dy

(cid:32)

Answer	Marks
dx	M1

A1

Answer	Marks
[2]	2.1
2.4	a

As a<0N positive root is (cid:16) .

2 Must give convincing reason for negative

sign

Answer	Marks	Guidance
3	(iv)	(B)

m(cid:32) 1(cid:14)ax(cid:14)2mx(cid:14)2c

m independent of x requires ax(cid:14)2mx(cid:32)0

a

therefore m(cid:32)(cid:16) .

2 a

(cid:16) (cid:32) 1(cid:14)2c E

2 a2 1

(cid:159)c(cid:32) (cid:16)

Answer	Marks
8 2	M1

A1

C

M1

A1

Answer	Marks
[4]	3.1a

I

2.2a

1.1

Answer	Marks
2.2a	E

M

a (cid:167) a (cid:183)

Accept (cid:16) (cid:32) 1(cid:14)ax(cid:14)2 (cid:168) (cid:16) x(cid:14)c (cid:184)

2 (cid:169) 2 (cid:185)

Answer	Marks	Guidance
Question	AO1	AO2
1iA	2	0
1iB	1	0
1iC	0	1
1iiA	2	0
1iiB	2	0
1iiiA	1	0
1iiiB	1	1
1iv	2	2
1v	1	1
2iA	1	2
2iB	2	0
2ii	1	0
2iiiA	0	1
2iiiB	1	1

2

Answer	Marks	Guidance
2iiiC	0	1
2iiiD	1	0
0	2
2ivA	2	0
2ivB	1	0
2ivC	0	0
1	0	1
3iA	2	0
3iB	0	1
0	0	1
3iC	0	1
3iD	1	C
0	0	0
3iiA	1	1
3iiB	E
1	0	1
3iii	2	2
3iv	1	4
totals	P
29	19	9

Question 3:
3 | (i) | (A) | a = –2 | M1
A1
[2] | 1.1
1.1 | Correct in at least one quadrant
N
Correct everywhere
E
3 | (i) | (B) | dy
is not defined when 1(cid:14)ax(cid:14)2y(cid:31)0 because the
dx
square root is not real. | E1
C
[1] | 2.4
I | M
3 | (i) | (C) | E
dy a 1
is defined when y(cid:116)(cid:16) x(cid:16) o.e.
dx 2 2 | B1
[1] | 2.2a
3 | (i) | (D) | P
a = 0 S | B1
[1] | 1.1 | may be obtained by using slider on their
tangent field – no evidence required – or by
using (i) (B)
3 | (ii) | (A) | A2=0.1
B2=0, C2=1
D2= $A$2*sqrt(1+B2+2*C2),
E2= $A$2*sqrt(1+(B2+$A$2)+2*(C2+D2))
B3= B2+$A$2, C3 =C2+0.5*(D2+E2) | M1
M1
M1
[3] | 1.1
3.1a
2.5
I | Columns for x & y or equivalent
Columns for k & k or equivalent
N1 2
Formulae for x & y
n+1 n+1
E
M
or
define: h=0.1, f(x,y)=sqrt(1+x+2y)
A2=0, B2=1
C2=h*f(A2, B2), D2=h*f(A2 + h, B2 + C2)
A3=A2+h, B3=B2 + 0.5*(C2 + D2)
3 | (ii) | (B) | h = 0.1: y = 3.3488 when x = 1 E
h = 0.05: y = 3.3500 when x = 1 | C
A1
A1
[2] | 1.1
3.2a
3 | (iii) | (A) | 1
y(cid:32) x2 (cid:14) 3x(cid:14)1
2 | B1
[1] | 2.2a
3 | (iii) | (B) | dy
(cid:32) x(cid:14) 3
dx
(cid:32) x2 (cid:14)2 3x(cid:14)3
(cid:167)1 (cid:183)
(cid:32) 1(cid:14)2 x2 (cid:14) 3x(cid:14)1
(cid:168) (cid:184)
(cid:169)2 (cid:185)
(cid:32) 1(cid:14)2y | M1
A1
[2] | 1.1
I2.1 | N
E
M
Clear reasoning must be seen throughout
3 | (iii) | (C) | y = 3.2321 when x = 1 | C
B1
[1] | 1.1
3 | (iv) | (A) | (cid:167) a a2 1(cid:183) a2
1(cid:14)ax(cid:14)2(cid:168)(cid:16) x(cid:14) (cid:16) (cid:184) (cid:32)
(cid:169) 2 8 2(cid:185) 4
a
(cid:32)(cid:16)
2
dy
(cid:32)
dx | M1
A1
[2] | 2.1
2.4 | a
As a<0N positive root is (cid:16) .
2
Must give convincing reason for negative
sign
3 | (iv) | (B) | Substituting in y(cid:32)mx(cid:14)c gives
m(cid:32) 1(cid:14)ax(cid:14)2mx(cid:14)2c
m independent of x requires ax(cid:14)2mx(cid:32)0
a
therefore m(cid:32)(cid:16) .
2
a
(cid:16) (cid:32) 1(cid:14)2c E
2
a2 1
(cid:159)c(cid:32) (cid:16)
8 2 | M1
A1
C
M1
A1
[4] | 3.1a
I
2.2a
1.1
2.2a | E
M
a (cid:167) a (cid:183)
Accept (cid:16) (cid:32) 1(cid:14)ax(cid:14)2 (cid:168) (cid:16) x(cid:14)c (cid:184)
2 (cid:169) 2 (cid:185)
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
1iA | 2 | 0 | 0 | 0 | 2
1iB | 1 | 0 | 0 | 0 | 1
1iC | 0 | 1 | 0 | 0 | 1
1iiA | 2 | 0 | 0 | 0 | 2
1iiB | 2 | 0 | 0 | 0 | 2
1iiiA | 1 | 0 | 0 | 0 | 1
1iiiB | 1 | 1 | 0 | 0 | 2
1iv | 2 | 2 | 2 | 0 | 6
1v | 1 | 1 | 0 | 0 | 2
2iA | 1 | 2 | 0 | 2 | 5
2iB | 2 | 0 | 0 | 1 | 3
2ii | 1 | 0 | 0 | 0 | 1
2iiiA | 0 | 1 | 0 | 0 | 1
2iiiB | 1 | 1 | 0 | 0 | N
2
2iiiC | 0 | 1 | 0 | 0 | 1
2iiiD | 1 | 0 | 1 | E
0 | 2
2ivA | 2 | 0 | 1 | 0 | 3
2ivB | 1 | 0 | 1 | 0 | 2
2ivC | 0 | 0 | M
1 | 0 | 1
3iA | 2 | 0 | 0 | 0 | 2
3iB | 0 | 1 | I
0 | 0 | 1
3iC | 0 | 1 | 0 | 0 | 1
3iD | 1 | C
0 | 0 | 0 | 1
3iiA | 1 | 1 | 1 | 0 | 3
3iiB | E
1 | 0 | 1 | 0 | 2
3iii | 2 | 2 | 0 | 0 | 4
3iv | 1 | 4 | 1 | 0 | 6
totals | P
29 | 19 | 9 | 3 | 60

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3 This question explores the family of differential equations $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { 1 + a x + 2 y }$ for various values of the parameter $a$. Fig. 3 shows the tangent field in the case $a = 1$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{141c85ec-5749-4f24-9f6d-fe7a01567511-4_691_696_452_696}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item (A) Sketch the tangent field in the case $a = - 2$.\\
(B) Explain why the tangent field is not defined for the whole coordinate plane.\\
(C) Give an inequality which describes the region in which the tangent field is defined.\\
(D) Find a value of $a$ such that the region for which the tangent field is defined includes the entire $x$-axis.
\item (A) For the case $a = 1$, with $y = 1$ when $x = 0$, construct a spreadsheet for the Runge-Kutta method of order 2 with formulae as follows, where $\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }$.

$$\begin{aligned}
k _ { 1 } & = h \mathrm { f } \left( x _ { n } , y _ { n } \right) \\
k _ { 2 } & = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 1 } \right) \\
y _ { n + 1 } & = y _ { n } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)
\end{aligned}$$

State the formulae you have used in your spreadsheet.\\
(B) Use your spreadsheet to obtain the value of $y$ correct to 4 decimal places when $x = 1$ for

\begin{itemize}
  \item $h = 0.1$\\
and
  \item $h = 0.05$.
\end{itemize}
\item (A) For the case $a = 0$ find the analytical solution that passes through the point ( 0,1 ).\\
(B) Verify that the solution in part (iii) (A) is a solution to the differential equation.\\
(C) Use the solution in part (iii) (A) to find the value of $y$ correct to 4 decimal places when $x = 1$.
\item (A) Verify that $y = - \frac { a } { 2 } x + \frac { a ^ { 2 } } { 8 } - \frac { 1 } { 2 }$ is a solution for all cases when $a \leq 0$.\\
(B) Show that this is the only straight line solution in these cases.

\section*{Copyright Information:}
}{www.ocr.org.uk}) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the 
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure with Technology  Q3 [20]}}

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