OCR Further Pure Core 2 2018 December — Question 3 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2018
SessionDecember
Marks6
TopicHyperbolic functions
TypeSolve using substitution u = cosh x or u = sinh x
DifficultyStandard +0.8 This is a Further Maths hyperbolic equation requiring substitution using cosh²x = 1 + sinh²x to form a quadratic, solving it, then converting sinh x values back to x using the definition sinh x = (e^x - e^{-x})/2, which leads to a quadratic in e^x. While systematic, it requires multiple non-trivial algebraic manipulations and careful handling of the logarithmic form with surds—more demanding than standard C3 work but follows a clear method once the approach is identified.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07e Inverse hyperbolic: definitions, domains, ranges

In this question you must show detailed reasoning. Solve the equation \(2\cosh^2 x + 5\sinh x - 5 = 0\) giving each answer in the form \(\ln(p + q\sqrt{r})\) where \(p\) and \(q\) are rational numbers, and \(r\) is an integer, whose values are to be determined. [6]

AnswerMarks Guidance
\(\cosh^2 x - \sinh^2 x = 1\)M1 Use of identity to leave an equation in either just coshx or just sinhx
\(2\sinh^2 x + 5\sinh x - 3 = 0\)M1 Reduction to 3 term quadratic in sinhx or coshx: \(4\cosh^4 x - 4\cosh^2 x + 50 = 0\)
\(\sinh x = \frac{1}{2}\) or \(-3\)A1
\(x = \ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right)\)A1 Use of In formula for \(\sinh^{-1}\) or \(\cosh^{-1}\): Do not allow eg \(\cosh^{-1}(5/4\) or 10) unless this is rejected (NB eg ln(3 - √10) is not real). If using \(\cosh^{-1}\) "rogue" solutions must be convincingly rejected. Most likely to see ln(3 + √10)
\(x = \ln\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)\)A1 Must be in the correct form but allow \(\ln\left(\frac{1 + \sqrt{5}}{2}\right)\).
\(p = \frac{1}{2}, q = \frac{1}{2}, r = 5\)A1
\(p = -3, q = 1, r = 10\)A1
\(x = \ln(-3 + \sqrt{10})\)A1
$\cosh^2 x - \sinh^2 x = 1$ | M1 | Use of identity to leave an equation in either just coshx or just sinhx

$2\sinh^2 x + 5\sinh x - 3 = 0$ | M1 | Reduction to 3 term quadratic in sinhx or coshx: $4\cosh^4 x - 4\cosh^2 x + 50 = 0$

$\sinh x = \frac{1}{2}$ or $-3$ | A1 |

$x = \ln\left(\frac{1}{2} + \sqrt{\frac{5}{4}}\right)$ | A1 | Use of In formula for $\sinh^{-1}$ or $\cosh^{-1}$: Do not allow eg $\cosh^{-1}(5/4$ or 10) unless this is rejected (NB eg ln(3 - √10) is not real). If using $\cosh^{-1}$ "rogue" solutions must be convincingly rejected. Most likely to see ln(3 + √10)

$x = \ln\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)$ | A1 | Must be in the correct form but allow $\ln\left(\frac{1 + \sqrt{5}}{2}\right)$.

$p = \frac{1}{2}, q = \frac{1}{2}, r = 5$ | A1 |
$p = -3, q = 1, r = 10$ | A1 |

$x = \ln(-3 + \sqrt{10})$ | A1 |

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\textbf{In this question you must show detailed reasoning.}

Solve the equation $2\cosh^2 x + 5\sinh x - 5 = 0$ giving each answer in the form $\ln(p + q\sqrt{r})$ where $p$ and $q$ are rational numbers, and $r$ is an integer, whose values are to be determined. [6]

\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q3 [6]}}