Question 13 - A-Level Maths

OCR MEI Further Pure Core 2020 November — Question 13 9 marks

Exam Board	OCR MEI
Module	Further Pure Core (Further Pure Core)
Year	2020
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Maclaurin series for inverse hyperbolics
Difficulty	Challenging +1.3 This is a structured Further Maths question on hyperbolic functions requiring multiple techniques (exponential definitions, differentiation, Maclaurin series), but each part follows logically from the previous with clear guidance. Part (a) is routine proof, (b) is standard differentiation, (c) requires understanding of even/odd functions, and (d) involves pattern recognition in series coefficients. While it's a multi-step question requiring FM knowledge, the scaffolding makes it more accessible than typical FM proof questions.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials 4.07d Differentiate/integrate: hyperbolic functions 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n

Using exponentials, prove that \(\sinh 2 x = 2 \cosh x \sinh x\).
Hence show that if \(\mathrm { f } ( x ) = \sinh ^ { 2 } x\), then \(\mathrm { f } ^ { \prime \prime } ( x ) = 2 \cosh 2 x\).
Explain why the coefficients of odd powers in the Maclaurin series for \(\sinh ^ { 2 } x\) are all zero.
Find the coefficient of \(x ^ { n }\) in this series when \(n\) is a positive even number.

Show mark scheme Show mark scheme source

Question 13:

Answer	Marks	Guidance
13	(a)	= sinh 2x
	M1

A1

Answer	Marks	Guidance
[2]	2.1
2.2a	substituting
13	(b)	f(x) = sinh2x

f′(x) = 2sinh x cosh x [= sinh 2x]

⇒f″(x) = 2cosh 2x*

Answer	Marks
	B1

B1

Answer	Marks
[2]	2.1
2.2a	SC B1 if other methods used

NB AG

Answer	Marks	Guidance
13	(c)	f‴(x) = 4sinh 2x, f(5)( x) = 16 sinh 2x, …

so all odd derivatives are multiples of sinh2x

so f‴(0) = f(5)( 0) = f(7)( 0) = … = 0

Answer	Marks
	M1

E1

Answer	Marks
[2]	2.1

2.4 PPMMTT

Y420/01 Mark SchemeNovember 2020

13 (d) f″(x) = 2cosh 2x, f(4)( x) = 8cosh 2x, .. M1 2.1

f( n )( 0) = 2n −1 [n even] A1 2.1 accept f(n)(x) = 2n−1cosh(2x)

coefft of xn = 2n− 1/ n! [n even] A1 2.2a

[3]

13

Answer	Marks	Guidance
13	(d)	f″(x) = 2cosh 2x, f(4)( x) = 8cosh 2x, ..

f( n )( 0) = 2n −1 [n even]

coefft of xn = 2n− 1/ n! [n even]

Answer	Marks
	M1

A1

Answer	Marks
[3]	2.1

2.1

Answer	Marks
2.2a	accept f(n)(x) = 2n−1cosh(2x)

PPMMTT

Y420/01 Mark SchemeNovember 2020

14 d2y +3dx = 5 dy

dt2 dt dt

M1 3.1a diff and subst for dy/dt or

dx/dt

1(5 dy − d2y )+ 2(5y− dy ) = 4y

⇒ M1 3.1a subst for y (or x ) 3 dt dt2 3 dt

⇒ A1 1.1 Must be simplified d2y −3 dy +2y = 0

dt2 dt

AE λ 2 − 3 λ + 2 = 0 AE λ 2 − 3 λ + 2 = 0

⇒λ = 1 or 2 ⇒λ = 1 or 2

GS x = A et + B e2 t B1ft 1.1 ft their values of λ GS y = C et + D e2 t

M1 2.1 subst for x, dx/dt

x = 4Cet+De2t

3 A1 2.2a

when t = 0, 4C + 3D = 0

M1 1.1 subst t = 0 in x , y to find eqns

when t = 0, x = A + B = 0

in A, B

C + D = 1

A1 1.1 both equations correct

M1 1.1 solving the equations to find

A and B C = -3 D = 4

⇒ A = −4, B = 4 A1 3.2a for both A and B

A1 3.2a for correct equations for both

so x = 4e2 t – 4et

[11] x and y

y = 4e2 t – 3et

15 (a)

M1 3.1a calculating determinant or full attempt to solve

finding x, y or z in terms of λ

A1 1.1

= λ−3

det = 0 when λ = 3

B1ft 1.1 ft their λ

So unique point provided λ ≠ 3

[3]

14

Question 13:
13 | (a) | = sinh 2x
 | M1
A1
[2] | 2.1
2.2a | substituting
13 | (b) | f(x) = sinh2x
 
f′(x) = 2sinh x cosh x [= sinh 2x]
   
⇒f″(x) = 2cosh 2x*
  | B1
B1
[2] | 2.1
2.2a | SC B1 if other methods used
NB AG
13 | (c) | f‴(x) = 4sinh 2x, f(5)( x) = 16 sinh 2x, …
    
so all odd derivatives are multiples of sinh2x

so f‴(0) = f(5)( 0) = f(7)( 0) = … = 0
  | M1
E1
[2] | 2.1
2.4
PPMMTT
Y420/01 Mark SchemeNovember 2020
13 (d) f″(x) = 2cosh 2x, f(4)( x) = 8cosh 2x, .. M1 2.1
    
f(  n  )( 0) = 2n  −1 [n even] A1 2.1 accept f(n)(x) = 2n−1cosh(2x)
   
coefft of xn  = 2n−  1/ n! [n even] A1 2.2a
     
[3]
13
13 | (d) | f″(x) = 2cosh 2x, f(4)( x) = 8cosh 2x, ..
    
f(  n  )( 0) = 2n  −1 [n even]
   
coefft of xn  = 2n−  1/ n! [n even]
      | M1
A1
A1
[3] | 2.1
2.1
2.2a | accept f(n)(x) = 2n−1cosh(2x)
PPMMTT
Y420/01 Mark SchemeNovember 2020
14 d2y +3dx = 5 dy
dt2 dt dt
M1 3.1a diff and subst for dy/dt or
  
dx/dt

1(5 dy − d2y )+ 2(5y− dy ) = 4y
⇒ M1 3.1a subst for y (or x ) 3 dt dt2 3 dt
   
⇒ A1 1.1 Must be simplified d2y −3 dy +2y = 0
dt2 dt
AE λ 2 − 3 λ + 2 = 0 AE λ 2 − 3 λ + 2 = 0
       
⇒λ = 1 or 2 ⇒λ = 1 or 2
 
GS  x  =  A  et   +  B  e2   t B1ft 1.1 ft their values of λ GS  y  =  C  et   +  D  e2   t
M1 2.1 subst for x, dx/dt
  
x = 4Cet+De2t
3
A1 2.2a
when t = 0, 4C + 3D = 0
M1 1.1 subst t = 0 in x , y to find eqns
when t = 0, x = A + B = 0      
      in A, B
   C + D = 1
A1 1.1 both equations correct
M1 1.1 solving the equations to find
A and B C = -3 D = 4
⇒ A = −4, B = 4 A1 3.2a for both A and B
    A1 3.2a for correct equations for both
so x = 4e2  t  – 4et
   [11] x and y
y = 4e2  t  – 3et 
  
15 (a)
M1 3.1a calculating determinant or full attempt to solve
finding x, y or z in terms of λ
   
A1 1.1
= λ−3
det = 0 when λ = 3
 B1ft 1.1 ft their λ
So unique point provided λ ≠ 3
 [3]
14

Show LaTeX source

13
\begin{enumerate}[label=(\alph*)]
\item Using exponentials, prove that $\sinh 2 x = 2 \cosh x \sinh x$.
\item Hence show that if $\mathrm { f } ( x ) = \sinh ^ { 2 } x$, then $\mathrm { f } ^ { \prime \prime } ( x ) = 2 \cosh 2 x$.
\item Explain why the coefficients of odd powers in the Maclaurin series for $\sinh ^ { 2 } x$ are all zero.
\item Find the coefficient of $x ^ { n }$ in this series when $n$ is a positive even number.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q13 [9]}}

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Q1 6 Q2 6 Q3 4 Q4 8 Q5 8 Q6 4 Q7 6 Q8 9 Q9 8 Q10 7 Q11 8 Q12 8 Q13 9 Q14 11 Q15 17 Q16 25

OCR MEI Further Pure Core 2020 November — Question 13 9 marks

Question 13:

A1

​​ ​

f′(x) = 2sinh x cosh x [= sinh 2x]

​​ ​​ ​​ ​​

⇒f″(x) = 2cosh 2x*

B1

NB AG

​​ ​​ ​ ​​ ​​

so all odd derivatives are multiples of sinh2x

​

so f‴(0) = f(5)(​ 0) = f(7)(​ 0) = … = 0

E1

2.4

PPMMTT

Y420/01 Mark SchemeNovember 2020

13 (d) f″(x) = 2cosh 2x, f(4)(​ x) = 8cosh 2x, .. M1 2.1

​​ ​​ ​ ​​ ​​

f( ​ n ​ )(​ 0) = 2n ​ −1 ​[n even] A1 2.1 accept f(n)(x) = 2n−1cosh(2x)

​ ​ ​ ​

coefft of xn ​ = 2n− ​ 1/​ n! [n even] A1 2.2a

​​ ​ ​ ​ ​ ​

[3]

13

​​ ​​ ​ ​​ ​​

f( ​ n ​ )(​ 0) = 2n ​ −1 ​[n even]

​ ​ ​ ​

coefft of xn ​ = 2n− ​ 1/​ n! [n even]

A1

A1

2.1

PPMMTT

Y420/01 Mark SchemeNovember 2020

14 d2y +3dx = 5 dy

dt2 dt dt

M1 3.1a diff and subst for dy/dt or

​​ ​ ​

dx/dt

​​

1(5 dy − d2y )+ 2(5y− dy ) = 4y

⇒ M1 3.1a subst for y (or x ) 3 dt dt2 3 dt

​ ​ ​ ​

⇒ A1 1.1 Must be simplified d2y −3 dy +2y = 0

dt2 dt

AE λ 2 ​− 3 λ + 2 = 0 AE λ 2 ​− 3 λ + 2 = 0

​ ​ ​ ​ ​ ​ ​ ​

⇒λ = 1 or 2 ⇒λ = 1 or 2

​​ ​​

GS ​ x ​ = ​ A ​ et ​ ​ + ​ B ​ e2 ​ ​ t B1ft 1.1 ft their values of λ GS ​ y ​ = ​ C ​ et ​ ​ + ​ D ​ e2 ​ ​ t

M1 2.1 subst for x, dx/dt

​​ ​​ ​

x = 4Cet+De2t

3

A1 2.2a

when t = 0, 4C + 3D = 0

M1 1.1 subst t = 0 in x , y to find eqns

when t = 0, x = A + B = 0 ​ ​ ​ ​ ​ ​

​​ ​​ ​ ​ ​ ​ in A, B

​ ​ ​ C + D = 1

A1 1.1 both equations correct

M1 1.1 solving the equations to find

A and B C = -3 D = 4

⇒ A = −4, B = 4 A1 3.2a for both A and B

​ ​ ​ ​ A1 3.2a for correct equations for both

so x = 4e2 ​ t ​ – 4et

​​ ​ ​ [11] x and y

y = 4e2 ​ t ​ – 3et ​

​​ ​ ​

15 (a)

M1 3.1a calculating determinant or full attempt to solve

finding x, y or z in terms of λ

​​ ​​ ​​ ​

A1 1.1

= λ−3

det = 0 when λ = 3

​​ B1ft 1.1 ft their λ

So unique point provided λ ≠ 3

​​ [3]

14

so f‴(0) = f(5)( 0) = f(7)( 0) = … = 0

13 (d) f″(x) = 2cosh 2x, f(4)( x) = 8cosh 2x, .. M1 2.1

f( n )( 0) = 2n −1 [n even] A1 2.1 accept f(n)(x) = 2n−1cosh(2x)

coefft of xn = 2n− 1/ n! [n even] A1 2.2a

f( n )( 0) = 2n −1 [n even]

coefft of xn = 2n− 1/ n! [n even]

AE λ 2 − 3 λ + 2 = 0 AE λ 2 − 3 λ + 2 = 0

GS x = A et + B e2 t B1ft 1.1 ft their values of λ GS y = C et + D e2 t

when t = 0, x = A + B = 0

in A, B

C + D = 1

A1 3.2a for correct equations for both

so x = 4e2 t – 4et

[11] x and y

y = 4e2 t – 3et

B1ft 1.1 ft their λ

[3]