OCR Further Pure Core 2 2021 November — Question 10 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeUse series to approximate numerical value
DifficultyChallenging +1.2 This is a structured multi-part question requiring Maclaurin series expansion and inequality manipulation. Part (a) is straightforward (truncate e^x series), part (b) requires recognizing the substitution x = t-1, and part (c) needs setting t = π/e. While it requires careful algebraic manipulation and some insight into choosing the right substitution, the question provides significant scaffolding through its parts, making it moderately above average difficulty but not requiring exceptional problem-solving skills.
Spec4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n
10 In this question you must show detailed reasoning.
  1. By using an appropriate Maclaurin series prove that if \(x > 0\) then \(\mathrm { e } ^ { x } > 1 + x\).
  2. Hence, by using a suitable substitution, deduce that \(\mathrm { e } ^ { t } > \mathrm { e } t\) for \(t > 1\).
  3. Using the inequality in part (b), and by making a suitable choice for \(t\), determine which is greater, \(\mathrm { e } ^ { \pi }\) or \(\pi ^ { \mathrm { e } }\).
Question 10:
AnswerMarks Guidance
10(a) DR
x2 x3 x2 x3 
ex =1+x+ + +...=(1+x)+ + +...
2! 3!  2! 3! 
x2 x3
x>0⇒ + +...>0
2! 3!
AnswerMarks
⇒ex >1+xM1
A1
AnswerMarks
[2]1.1
2.2aQuoting and using the Maclaurin
series
AG. Result with sufficient
justification
AnswerMarks
(b)DR
et
t =x+1⇒et−1 >t⇒ >t⇒et >et
AnswerMarks Guidance
eB1
[1]3.1a AG
(c)DR
π
t = >1 since 2 < e < 3 and π > 3
e
π π
ee >e× (=π)
e
AnswerMarks
⇒eπ >πe (ie eπ is greater)B1
M1
AnswerMarks
A13.1a
3.1a
AnswerMarks
1.1Some justification that t >1 is
required
Substituting their choice into the
inequality
Answer without use of inequality
in part (b) scores M0A0
Alternative method
AnswerMarks Guidance
t =lnπt =lnπ B1
required
AnswerMarks
elnπ>elnπM1
elnπ >elnπ
π>ln(πe)
AnswerMarks
eπ >πeA1
A1
[3]
PMT
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Question 10:
10 | (a) | DR
x2 x3 x2 x3 
ex =1+x+ + +...=(1+x)+ + +...
2! 3!  2! 3! 
x2 x3
x>0⇒ + +...>0
2! 3!
⇒ex >1+x | M1
A1
[2] | 1.1
2.2a | Quoting and using the Maclaurin
series
AG. Result with sufficient
justification
(b) | DR
et
t =x+1⇒et−1 >t⇒ >t⇒et >et
e | B1
[1] | 3.1a | AG
(c) | DR
π
t = >1 since 2 < e < 3 and π > 3
e
π π
ee >e× (=π)
e
⇒eπ >πe (ie eπ is greater) | B1
M1
A1 | 3.1a
3.1a
1.1 | Some justification that t >1 is
required
Substituting their choice into the
inequality
Answer without use of inequality
in part (b) scores M0A0
Alternative method
t =lnπ | t =lnπ | B1 | B1 | Some justification that t >1 is
required
elnπ>elnπ | M1
elnπ >elnπ
π>ln(πe)
eπ >πe | A1
A1
[3]
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
10 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item By using an appropriate Maclaurin series prove that if $x > 0$ then $\mathrm { e } ^ { x } > 1 + x$.
\item Hence, by using a suitable substitution, deduce that $\mathrm { e } ^ { t } > \mathrm { e } t$ for $t > 1$.
\item Using the inequality in part (b), and by making a suitable choice for $t$, determine which is greater, $\mathrm { e } ^ { \pi }$ or $\pi ^ { \mathrm { e } }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q10 [6]}}
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