| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 This is a standard Further Maths proof requiring manipulation of hyperbolic definitions and solving a quadratic in e^y, but it's well-rehearsed material. Part (b) tests understanding of arbitrary constants. Worth 6 marks total, requiring multiple algebraic steps but following a predictable method that Further Maths students practice extensively. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Expresses \(x\) in terms of \(y\) | M1 (3.1a) | |
| Recalls exponential form of cosh | B1 (1.1b) | |
| Forms quadratic equation in \(e^y\) | M1 (1.1a) | |
| Let \(y = \cosh^{-1}\left(\frac{x}{a}\right)\), then \(x = a\cosh y\); \(x = \frac{a}{2}(e^y + e^{-y})\); \(\frac{2x}{a} = e^y + e^{-y}\); \(\times e^y\): \(e^{2y} - \frac{2x}{a}e^y + 1 = 0\); \(e^y = \frac{\frac{2x}{a} \pm \sqrt{\frac{4x^2}{a^2}-4}}{2} = \frac{x \pm \sqrt{x^2 - a^2}}{a}\) | A1F (1.1b) | Solves quadratic in \(e^y\) to obtain two solutions |
| Product of roots \(= 1\), so one root is greater than 1 and other is less than 1; \(y \geq 0\) by definition of \(\cosh^{-1}\), so \(e^y \geq 1\); choose larger root; \(e^y = \frac{x + \sqrt{x^2 - a^2}}{a}\); therefore \(\cosh^{-1}\left(\frac{x}{a}\right) = \ln\left(\frac{x + \sqrt{x^2-a^2}}{a}\right)\) | M1 (2.4) | Explains inverse cosh is non-negative, OR shows \(\frac{x - \sqrt{x^2-a^2}}{a} < 1\), OR shows reciprocal relationship between roots |
| Completes rigorous argument with reference to larger root being valid with clear reason | R1 (2.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\ln\left(\frac{x+\sqrt{x^2-a^2}}{a}\right) = \ln(x + \sqrt{x^2-a^2}) - \ln(a)\) | E1 (2.4) | States this equivalence |
| \(\therefore c = -\ln(a)\) | E1 (2.3) |
## Question 12(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Expresses $x$ in terms of $y$ | M1 (3.1a) | |
| Recalls exponential form of cosh | B1 (1.1b) | |
| Forms quadratic equation in $e^y$ | M1 (1.1a) | |
| Let $y = \cosh^{-1}\left(\frac{x}{a}\right)$, then $x = a\cosh y$; $x = \frac{a}{2}(e^y + e^{-y})$; $\frac{2x}{a} = e^y + e^{-y}$; $\times e^y$: $e^{2y} - \frac{2x}{a}e^y + 1 = 0$; $e^y = \frac{\frac{2x}{a} \pm \sqrt{\frac{4x^2}{a^2}-4}}{2} = \frac{x \pm \sqrt{x^2 - a^2}}{a}$ | A1F (1.1b) | Solves quadratic in $e^y$ to obtain two solutions |
| Product of roots $= 1$, so one root is greater than 1 and other is less than 1; $y \geq 0$ by definition of $\cosh^{-1}$, so $e^y \geq 1$; choose larger root; $e^y = \frac{x + \sqrt{x^2 - a^2}}{a}$; therefore $\cosh^{-1}\left(\frac{x}{a}\right) = \ln\left(\frac{x + \sqrt{x^2-a^2}}{a}\right)$ | M1 (2.4) | Explains inverse cosh is non-negative, OR shows $\frac{x - \sqrt{x^2-a^2}}{a} < 1$, OR shows reciprocal relationship between roots |
| Completes rigorous argument with reference to larger root being valid with clear reason | R1 (2.1) | |
## Question 12(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\ln\left(\frac{x+\sqrt{x^2-a^2}}{a}\right) = \ln(x + \sqrt{x^2-a^2}) - \ln(a)$ | E1 (2.4) | States this equivalence |
| $\therefore c = -\ln(a)$ | E1 (2.3) | |
**Total: 8 marks**
---12
\begin{enumerate}[label=(\alph*)]
\item Use the definition of the cosh function to prove that
$$\cosh ^ { - 1 } \left( \frac { x } { a } \right) = \ln \left( \frac { x + \sqrt { x ^ { 2 } - a ^ { 2 } } } { a } \right) \quad \text { for } a > 0$$
[6 marks]\\
12
\item The formulae booklet gives the integral of $\frac { 1 } { \sqrt { x ^ { 2 } - a ^ { 2 } } }$ as
$$\cosh ^ { - 1 } \left( \frac { x } { a } \right) \text { or } \ln \left( x + \sqrt { x ^ { 2 } - a ^ { 2 } } \right) + c$$
Ronald says that this contradicts the result given in part (a).\\
Explain why Ronald is wrong.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q12 [8]}}