OCR FP2 2006 June — Question 9 13 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2006
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeHyperbolic function reduction
DifficultyChallenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques: proving the inverse hyperbolic sine formula (standard but non-trivial), deriving a reduction formula using integration by parts with hyperbolic functions, and applying it recursively. While the individual steps follow established methods, the combination of hyperbolic function manipulation, reduction formula derivation, and multi-step calculation places this well above average difficulty, though it remains a structured exam question rather than requiring deep insight.
Spec4.07e Inverse hyperbolic: definitions, domains, ranges4.08f Integrate using partial fractions

9
  1. Given that \(y = \sinh ^ { - 1 } x\), prove that \(y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)\).
  2. It is given that, for non-negative integers \(n\), $$I _ { n } = \int _ { 0 } ^ { \alpha } \sinh ^ { n } \theta \mathrm {~d} \theta$$ where \(\alpha = \sinh ^ { - 1 } 1\). Show that $$n I _ { n } = \sqrt { 2 } - ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2 .$$
  3. Evaluate \(I _ { 4 }\), giving your answer in terms of \(\sqrt { 2 }\) and logarithms.

AnswerMarks Guidance
(i) Rewrite as quad. in \(e^y\)M1 Any form
Solve to \(e^y = (x \pm \sqrt{(x^2 + 1)})\)A1 Allow \(y = \ln(\) \()\)
Justify one solution onlyB1 \(x - \sqrt{(x^2 + 1)} < 0\) for all real \(x\)
SCUse \(C^2 - S^2 = 1\) for \(C = \pm\sqrt{(1+x^2)}\) M1
Use/state cosh \(y + \sinh y = e^y\)A1
Justify one solution onlyB1
(ii) Attempt parts on sinh \(x, \sinh^{n-1} x\)M1
Get correct answerA1 \((\cosh x.\sinh^{n-1}x - \int\cosh^2 x.(n-1)\sinh^{n-2}x\,dx)\)
Justify \(\sqrt{2}\) by \(\sqrt{(1+\sinh^2 x)}\) for cosh \(x\) when limits insertedB1 Must be clear
Replace \(\cosh^2 = 1 + \sinh^2\); tidy at this stageM1
Produce \(I_{n,2}\)A1
Gain A.G. clearlyA1
(iii) Attempt \(4I_4 = \sqrt{2} - 3I_2 ; 2I_2 = \sqrt{2} - I_0\)M1 Clear attempt at iteration (one at least seen)
Work out \(I_0 = \sinh^{-1}1 = \ln(1 + \sqrt{2}) = a\)B1 Allow \(I_2\)
Sub. back completely for \(I_4\)M1
Get \(\frac{I_6}{6}(3\ln(1+\sqrt{2}) - \sqrt{2})\)A1 AEEF
**(i)** Rewrite as quad. in $e^y$ | M1 | Any form
Solve to $e^y = (x \pm \sqrt{(x^2 + 1)})$ | A1 | Allow $y = \ln($ $)$
Justify one solution only | B1 | $x - \sqrt{(x^2 + 1)} < 0$ for all real $x$
| SC | Use $C^2 - S^2 = 1$ for $C = \pm\sqrt{(1+x^2)}$ | M1
| | Use/state cosh $y + \sinh y = e^y$ | A1
| | Justify one solution only | B1

**(ii)** Attempt parts on sinh $x, \sinh^{n-1} x$ | M1 |
Get correct answer | A1 | $(\cosh x.\sinh^{n-1}x - \int\cosh^2 x.(n-1)\sinh^{n-2}x\,dx)$
Justify $\sqrt{2}$ by $\sqrt{(1+\sinh^2 x)}$ for cosh $x$ when limits inserted | B1 | Must be clear
Replace $\cosh^2 = 1 + \sinh^2$; tidy at this stage | M1 |
Produce $I_{n,2}$ | A1 |
Gain A.G. clearly | A1 |

**(iii)** Attempt $4I_4 = \sqrt{2} - 3I_2 ; 2I_2 = \sqrt{2} - I_0$ | M1 | Clear attempt at iteration (one at least seen)
Work out $I_0 = \sinh^{-1}1 = \ln(1 + \sqrt{2}) = a$ | B1 | Allow $I_2$
Sub. back completely for $I_4$ | M1 |
Get $\frac{I_6}{6}(3\ln(1+\sqrt{2}) - \sqrt{2})$ | A1 | AEEF
9 (i) Given that $y = \sinh ^ { - 1 } x$, prove that $y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)$.\\
(ii) It is given that, for non-negative integers $n$,

$$I _ { n } = \int _ { 0 } ^ { \alpha } \sinh ^ { n } \theta \mathrm {~d} \theta$$

where $\alpha = \sinh ^ { - 1 } 1$. Show that

$$n I _ { n } = \sqrt { 2 } - ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2 .$$

(iii) Evaluate $I _ { 4 }$, giving your answer in terms of $\sqrt { 2 }$ and logarithms.

\hfill \mbox{\textit{OCR FP2 2006 Q9 [13]}}