OCR Further Pure Core 1 2021 November — Question 2 8 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeTrigonometric method of differences
DifficultyStandard +0.3 This is a straightforward Further Maths question on Maclaurin series requiring standard differentiation of arctan (using chain rule), evaluation at x=0, and formula application. While it involves Further Pure content, the execution is mechanical with no problem-solving insight needed—just careful application of known techniques across 8 marks.
Spec1.07l Derivative of ln(x): and related functions4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n

You are given that \(\mathrm{f}(x) = \tan^{-1}(1 + x)\).
    1. Find the value of \(\mathrm{f}(0)\). [1]
    2. Determine the value of \(\mathrm{f}'(0)\). [2]
    3. Show that \(\mathrm{f}''(0) = -\frac{1}{2}\). [3]
  1. Hence find the Maclaurin series for \(\mathrm{f}(x)\) up to and including the term in \(x^2\). [2]

Question 2:
AnswerMarks Guidance
2(a) (i)
f(0)=
AnswerMarks Guidance
4B1 1.1
[1]
AnswerMarks
(ii)1 1
f '(x)= ⇒f '(0)=
AnswerMarks
1+( 1+x )2 2M1
A12.1
1.1Diffn – Must be seen
1
f '(x)= is M0
1+x2
[2]
AnswerMarks
(iii)1 1
f '(x)= =
1+( 1+x )2 2+2x+x2
1
⇒f ''(x)= ×(−1 )×( 2+2x )
( )2
2+2x+x2
−( 2+2x )
=
( )2
2+2x+x2
−2 1
⇒f ''(0)= =−
 
AnswerMarks
 4  2M1
A1
AnswerMarks
A12.1
2.1
AnswerMarks
2.1Diffn their f’(x)
2 ( 1+x )
oe, e.g. f′′( x )=−
( )2
1+( 1+x )2
f’’(0) must be seen. The substitution must be seen
2
(implied by − )
4
AG
[3]
AnswerMarks
(b)x2
f(x)=f(0)+f '(0)x+f ''(0)
2
π 1 1 x2
= + x− ×
4 2 2 2
π x x2
= + −
AnswerMarks
4 2 4M1
A11.1
2.2aUsing the formula and substituting their value for f;(0)
ft their values from (a)
[2]
Question 2:
2 | (a) | (i) | π
f(0)=
4 | B1 | 1.1 | Not for 450
[1]
(ii) | 1 1
f '(x)= ⇒f '(0)=
1+( 1+x )2 2 | M1
A1 | 2.1
1.1 | Diffn – Must be seen
1
f '(x)= is M0
1+x2
[2]
(iii) | 1 1
f '(x)= =
1+( 1+x )2 2+2x+x2
1
⇒f ''(x)= ×(−1 )×( 2+2x )
( )2
2+2x+x2
−( 2+2x )
=
( )2
2+2x+x2
−2 1
⇒f ''(0)= =−
 
 4  2 | M1
A1
A1 | 2.1
2.1
2.1 | Diffn their f’(x)
2 ( 1+x )
oe, e.g. f′′( x )=−
( )2
1+( 1+x )2
f’’(0) must be seen. The substitution must be seen
2
(implied by − )
4
AG
[3]
(b) | x2
f(x)=f(0)+f '(0)x+f ''(0)
2
π 1 1 x2
= + x− ×
4 2 2 2
π x x2
= + −
4 2 4 | M1
A1 | 1.1
2.2a | Using the formula and substituting their value for f;(0)
ft their values from (a)
[2]
You are given that $\mathrm{f}(x) = \tan^{-1}(1 + x)$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\mathrm{f}(0)$. [1]
\item Determine the value of $\mathrm{f}'(0)$. [2]
\item Show that $\mathrm{f}''(0) = -\frac{1}{2}$. [3]
\end{enumerate}
\item Hence find the Maclaurin series for $\mathrm{f}(x)$ up to and including the term in $x^2$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2021 Q2 [8]}}