WJEC Further Unit 4 2022 June — Question 8 6 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeProve inverse hyperbolic logarithmic form
DifficultyStandard +0.8 This is a standard Further Maths derivation requiring substitution of the exponential definition of sinh, solving a quadratic in e^y, and careful algebraic manipulation to reach the logarithmic form. While it involves multiple steps and hyperbolic function knowledge, it's a well-known result that follows a predictable method once the exponential definitions are applied, making it moderately challenging but not requiring novel insight.
Spec4.07e Inverse hyperbolic: definitions, domains, ranges
By writing \(x = \sinh y\), show that \(\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)\). [6]
AnswerMarks Guidance
\(x = \sinh y\)B1
\(x = \frac{e^y - e^{-y}}{2}\)B1
\(2xe^y = (e^y)^2 - 1\)B1
\(\therefore (e^y)^2 - 2xe^y - 1 = 0\)B1
Using quadratic formula,M1
\(e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \left(x \pm \sqrt{x^2 + 1}\right)\)A1 Allow omission of \(\pm\)
\(y = \ln\left(x + \sqrt{x^2 + 1}\right)\)A1
As \(x - \sqrt{x^2 + 1} < 0\),B1 Justification may be seen earlier
\(\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)\) 
$x = \sinh y$ | B1 | 
$x = \frac{e^y - e^{-y}}{2}$ | B1 | 

$2xe^y = (e^y)^2 - 1$ | B1 | 
$\therefore (e^y)^2 - 2xe^y - 1 = 0$ | B1 | 

Using quadratic formula, | M1 | 
$e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \left(x \pm \sqrt{x^2 + 1}\right)$ | A1 | Allow omission of $\pm$

$y = \ln\left(x + \sqrt{x^2 + 1}\right)$ | A1 | 

As $x - \sqrt{x^2 + 1} < 0$, | B1 | Justification may be seen earlier
$\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)$ | | 

---
By writing $x = \sinh y$, show that $\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)$. [6]

\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q8 [6]}}
This paper (14 questions)
View full paper
Q1 8 Q2 4 Q3 9 Q4 5 Q5 5 Q6 6 Q7 8 Q8 6 Q9 12 Q10 9 Q11 15 Q12 12 Q13 11 Q14 10