CAIE Further Paper 2 2021 June — Question 2 7 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeFind series for logarithmic function
DifficultyChallenging +1.2 This requires finding derivatives of ln(cosh x) up to fourth order, which involves repeated application of chain rule and quotient rule with hyperbolic functions. While the technique is standard for Further Maths students, the algebraic manipulation of hyperbolic derivatives (tanh x, sech²x) and simplification to identify coefficients requires careful work and is more demanding than routine Maclaurin series questions for simpler functions like e^x or sin x.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08a Maclaurin series: find series for function
2 Find the Maclaurin's series for \(\ln \cosh x\) up to and including the term in \(x ^ { 4 }\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = \tanh x\)B1 Finds first derivative
\(\frac{d^2y}{dx^2} = \text{sech}^2 x\)B1 Finds second derivative
\(\frac{d^3y}{dx^3} = -2\tanh x\,\text{sech}^2 x = -2\frac{\sinh x}{\cosh^3 x}\)M1 A1 Finds third derivative
\(\frac{d^4y}{dx^4} = -2\frac{\cosh^2(x)-3\sinh^2(x)}{\cosh^4(x)} = -2\text{sech}^4 x + 4\tanh^2 x\,\text{sech}^2 x\)B1 Finds fourth derivative. Alternative: \(6\tanh^2 x\,\text{sech}^2 x - 2\text{sech}^2 x\)
\(y(0) = y^{(1)}(0) = y^{(3)}(0) = 0\), \(y^{(2)}(0) = 1\), \(y^{(4)}(0) = -2\)M1 Evaluates derivatives at \(x=0\)
\(y = \frac{1}{2}x^2 - \frac{1}{12}x^4\)A1 CWO
Total: 7 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \tanh x$ | B1 | Finds first derivative |
| $\frac{d^2y}{dx^2} = \text{sech}^2 x$ | B1 | Finds second derivative |
| $\frac{d^3y}{dx^3} = -2\tanh x\,\text{sech}^2 x = -2\frac{\sinh x}{\cosh^3 x}$ | M1 A1 | Finds third derivative |
| $\frac{d^4y}{dx^4} = -2\frac{\cosh^2(x)-3\sinh^2(x)}{\cosh^4(x)} = -2\text{sech}^4 x + 4\tanh^2 x\,\text{sech}^2 x$ | B1 | Finds fourth derivative. Alternative: $6\tanh^2 x\,\text{sech}^2 x - 2\text{sech}^2 x$ |
| $y(0) = y^{(1)}(0) = y^{(3)}(0) = 0$, $y^{(2)}(0) = 1$, $y^{(4)}(0) = -2$ | M1 | Evaluates derivatives at $x=0$ |
| $y = \frac{1}{2}x^2 - \frac{1}{12}x^4$ | A1 | CWO |
| **Total: 7** | | |

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2 Find the Maclaurin's series for $\ln \cosh x$ up to and including the term in $x ^ { 4 }$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q2 [7]}}
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Q1 7 Q2 7 Q3 10 Q4 7 Q5 10 Q6 10 Q7 11 Q8 13