OCR FP2 2015 June — Question 2 4 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeComposition of two standard series
DifficultyStandard +0.3 This is a straightforward application of composing two standard Maclaurin series: ln(1+u) and sin(x). Students substitute sin(x) into ln(1+u), expand sin(x) to x³, and collect terms. While it's Further Maths content, it's a routine textbook exercise requiring only careful algebraic manipulation of known series, making it slightly easier than average overall.
Spec4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n
2 It is given that \(\mathrm { f } ( x ) = \ln ( 1 + \sin x )\). Using standard series, find the Maclaurin series for \(\mathrm { f } ( x )\) up to and including the term in \(x ^ { 3 }\).
Question 2:
AnswerMarks Guidance
\(\ln(1+y) = y - \dfrac{y^2}{2} + \dfrac{y^3}{3} - \ldots\)B1 Soi; allow expansion in \(x\)
\(\sin x = x - \dfrac{x^3}{6} + \ldots\)B1 Soi
\(\ln(1+\sin x) = \left(x-\dfrac{x^3}{6}\right) - \dfrac{1}{2}\left(x-\dfrac{x^3}{6}\right)^2 + \dfrac{1}{3}\left(x-\dfrac{x^3}{6}\right)^3 - \ldots = x - \dfrac{1}{2}x^2 + x^3\left(\dfrac{1}{3}-\dfrac{1}{6}\right)\)M1 For combining series, even if wrong; must include at least the cubic bracket
\(= x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3\)A1 Ignore further terms; accept \(3!\) for \(6\)
Alternative (Maclaurin):
\(f(x) = \ln(1+\sin x)\), \(f(0)=0\)
AnswerMarks Guidance
\(f'(x) = \dfrac{\cos x}{1+\sin x}\), \(f'(0)=1\)B1 For \(f'(x)\)
\(f''(x) = \dfrac{-1}{1+\sin x}\), \(f''(0)=-1\)B1 For \(f''(x)\) and \(f''(0)\)
\(f'''(x) = \dfrac{\cos x}{(1+\sin x)^2}\), \(f'''(0)=1\)
AnswerMarks Guidance
Maclaurin: \(f(x) = f(0) + f'(0)x + \dfrac{f''(0)x^2}{2} + \dfrac{f'''(0)x^3}{6}\)M1 Correct formula up to 4th term and substituting their values
\(\Rightarrow f(x) = x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3\)A1 Accept \(3!\) for \(6\) 
# Question 2:

$\ln(1+y) = y - \dfrac{y^2}{2} + \dfrac{y^3}{3} - \ldots$ | B1 | Soi; allow expansion in $x$

$\sin x = x - \dfrac{x^3}{6} + \ldots$ | B1 | Soi

$\ln(1+\sin x) = \left(x-\dfrac{x^3}{6}\right) - \dfrac{1}{2}\left(x-\dfrac{x^3}{6}\right)^2 + \dfrac{1}{3}\left(x-\dfrac{x^3}{6}\right)^3 - \ldots = x - \dfrac{1}{2}x^2 + x^3\left(\dfrac{1}{3}-\dfrac{1}{6}\right)$ | M1 | For combining series, even if wrong; must include at least the cubic bracket

$= x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3$ | A1 | Ignore further terms; accept $3!$ for $6$

**Alternative (Maclaurin):**

$f(x) = \ln(1+\sin x)$, $f(0)=0$

$f'(x) = \dfrac{\cos x}{1+\sin x}$, $f'(0)=1$ | B1 | For $f'(x)$

$f''(x) = \dfrac{-1}{1+\sin x}$, $f''(0)=-1$ | B1 | For $f''(x)$ and $f''(0)$

$f'''(x) = \dfrac{\cos x}{(1+\sin x)^2}$, $f'''(0)=1$

Maclaurin: $f(x) = f(0) + f'(0)x + \dfrac{f''(0)x^2}{2} + \dfrac{f'''(0)x^3}{6}$ | M1 | Correct formula up to 4th term and substituting their values

$\Rightarrow f(x) = x - \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3$ | A1 | Accept $3!$ for $6$

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2 It is given that $\mathrm { f } ( x ) = \ln ( 1 + \sin x )$. Using standard series, find the Maclaurin series for $\mathrm { f } ( x )$ up to and including the term in $x ^ { 3 }$.

\hfill \mbox{\textit{OCR FP2 2015 Q2 [4]}}
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