OCR MEI Further Pure Core 2022 June — Question 9 12 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks12
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Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeSecond derivative relations with hyperbolics
DifficultyChallenging +1.2 This is a structured multi-part question on hyperbolic functions requiring differentiation, algebraic manipulation, and Taylor approximation. Part (a) requires understanding domain restrictions (sinh x > -1), parts (b)(i)-(ii) involve routine chain rule and quotient rule differentiation with hyperbolic identities, part (c) applies standard Maclaurin series formula, and part (d) is straightforward percentage error calculation. While it requires multiple techniques and careful algebra, each step follows standard procedures without requiring novel insightβ€”slightly above average difficulty for Further Maths due to the multi-step nature and hyperbolic function manipulation.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08a Maclaurin series: find series for function
9 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \ln ( 1 + \sinh x )\).
  1. Given that \(k\) lies in the domain of this function, explain why \(k\) must be greater than \(\ln ( \sqrt { 2 } - 1 )\).
    1. Find \(\mathrm { f } ^ { \prime } ( x )\).
    2. Show that \(\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = \frac { \mathrm { a } \sinh \mathrm { x } + \mathrm { b } } { ( 1 + \sinh \mathrm { x } ) ^ { 2 } }\), where \(a\) and \(b\) are integers to be determined.
  3. Hence find a quadratic approximation to \(\mathrm { f } ( x )\) for small values of \(x\).
  4. Find the percentage error in this approximation when \(x = 0.1\).
Question 9:
AnswerMarks Guidance
9(a) sinhπ‘˜ > βˆ’1
β‡’ π‘˜ > sinhβˆ’1(βˆ’1) = ln(βˆ’1+√2)M1
A1
AnswerMarks
[2]2.1
2.2aMay be in exponential form
AG
AnswerMarks Guidance
9(b) (i)
𝑓′(π‘₯) =
AnswerMarks
1+sinhπ‘₯M1
A1
AnswerMarks Guidance
[2]1.1
1.1chain rule
9(b) (ii)
𝑓″(π‘₯) =
(1+sinhπ‘₯)2
sinhπ‘₯βˆ’1
𝑓″(π‘₯) =
AnswerMarks
(1+sinhπ‘₯)2M1
M1
A1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
1.1quotient or product rule
π‘π‘œπ‘ β„Ž 2π‘₯βˆ’π‘ π‘–π‘›β„Ž 2π‘₯ = 1 used
AnswerMarks Guidance
9(c) 𝑓(0) = 0, 𝑓′(0) = 1, 𝑓″(0) = βˆ’1
1
𝑓(π‘₯) = π‘₯βˆ’ π‘₯2
AnswerMarks
2B1ft
M1
A1cao
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1soi
Maclaurin expansion attempted, must see their values substituted
in
AnswerMarks Guidance
9(d) ln(1+sinh(0.1))βˆ’0.095
Γ—100
ln(1+sinh(0.1))
AnswerMarks
= 0.48%M1
A1
AnswerMarks
[2]1.1
1.1Allow –0.48% 
Question 9:
9 | (a) | sinhπ‘˜ > βˆ’1
β‡’ π‘˜ > sinhβˆ’1(βˆ’1) = ln(βˆ’1+√2) | M1
A1
[2] | 2.1
2.2a | May be in exponential form
AG
9 | (b) | (i) | coshπ‘₯
𝑓′(π‘₯) =
1+sinhπ‘₯ | M1
A1
[2] | 1.1
1.1 | chain rule
9 | (b) | (ii) | (1+sinhπ‘₯)sinhπ‘₯βˆ’cosh2π‘₯
𝑓″(π‘₯) =
(1+sinhπ‘₯)2
sinhπ‘₯βˆ’1
𝑓″(π‘₯) =
(1+sinhπ‘₯)2 | M1
M1
A1
[3] | 1.1
3.1a
1.1 | quotient or product rule
π‘π‘œπ‘ β„Ž 2π‘₯βˆ’π‘ π‘–π‘›β„Ž 2π‘₯ = 1 used
9 | (c) | 𝑓(0) = 0, 𝑓′(0) = 1, 𝑓″(0) = βˆ’1
1
𝑓(π‘₯) = π‘₯βˆ’ π‘₯2
2 | B1ft
M1
A1cao
[3] | 1.1
1.1
1.1 | soi
Maclaurin expansion attempted, must see their values substituted
in
9 | (d) | ln(1+sinh(0.1))βˆ’0.095
Γ—100
ln(1+sinh(0.1))
= 0.48% | M1
A1
[2] | 1.1
1.1 | Allow –0.48%
9 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = \ln ( 1 + \sinh x )$.
\begin{enumerate}[label=(\alph*)]
\item Given that $k$ lies in the domain of this function, explain why $k$ must be greater than $\ln ( \sqrt { 2 } - 1 )$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ^ { \prime } ( x )$.
\item Show that $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = \frac { \mathrm { a } \sinh \mathrm { x } + \mathrm { b } } { ( 1 + \sinh \mathrm { x } ) ^ { 2 } }$, where $a$ and $b$ are integers to be determined.
\end{enumerate}\item Hence find a quadratic approximation to $\mathrm { f } ( x )$ for small values of $x$.
\item Find the percentage error in this approximation when $x = 0.1$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q9 [12]}}
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