| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Challenging +1.2 This is a standard Further Maths integration by parts question involving inverse hyperbolic functions. Part (a) requires product rule and knowledge of the derivative of arcosh (a formula students should know). Part (b) is a guided integration by parts with some algebraic manipulation of logarithms at the end. While inverse hyperbolic functions are a Further Maths topic (making it harder on an absolute scale than Core topics), the question itself follows a routine template with clear signposting ('hence') and no novel problem-solving required. |
| Spec | 1.08i Integration by parts4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{\frac{dy}{dx} =\right\} \operatorname{arcosh}(5x) + \frac{ax}{\sqrt{bx^2-1}}\) or \(\operatorname{arcosh}(5x) + \frac{cx}{\sqrt{x^2-d}}\) \(\Rightarrow \operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}\) | M1 | Differentiates to obtain expression of correct form \(a, b, c, d \neq 0\) |
| \(\operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}\) | A1 | Correct differentiation. Any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dx}(x\operatorname{arcosh}(5x)) = \operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}\) \(\Rightarrow \int \operatorname{arcosh}(5x)\, dx = x\operatorname{arcosh}(5x) - \int \frac{5x}{\sqrt{25x^2-1}}\, dx\) | M1 | Rearranges answer to (a) correctly and integrates, or uses correct formula to apply parts to \(1 \times \operatorname{arcosh}(5x)\) |
| \(\int \operatorname{arcosh}(5x)\, dx = x\operatorname{arcosh}(5x) - \int \frac{5x}{\sqrt{25x^2-1}}\, dx\) | A1 (limited ft) | Correct expression |
| \(= x\operatorname{arcosh}(5x) - \frac{1}{5}(25x^2-1)^{\frac{1}{2}}\ (+c)\) | M1 A1 (limited ft) | M1: \(\int \frac{Ax}{\sqrt{Bx^2-1}}\, dx \to C(Bx^2-1)^{\frac{1}{2}}\). A1: Fully correct expression with \(x\operatorname{arcosh}(5x)\) |
| \(\int_{\frac{1}{4}}^{\frac{3}{5}} \operatorname{arcosh}(5x)\, dx = \frac{3}{5}\operatorname{arcosh}(3) - \frac{1}{5}\sqrt{25\times\frac{9}{25}-1} - \left(\frac{1}{4}\operatorname{arcosh}\left(\frac{5}{4}\right) - \frac{1}{5}\sqrt{25\times\frac{1}{16}-1}\right)\) | M1 | Applies appropriate limits with subtraction the right way round, to obtain expression of form \(x\operatorname{arcosh}(5x) \pm f(x)\) where \(f(x)\) has come from integration |
| \(= \frac{3}{5}\operatorname{arcosh}(3) - \frac{2\sqrt{2}}{5} - \frac{1}{4}\operatorname{arcosh}\left(\frac{5}{4}\right) + \frac{3}{20}\) | A1 | Correct answer seen in any form. Must not follow clearly incorrect work |
| \(\operatorname{arcosh}(3) = \ln(3+\sqrt{3^2-1^2})\) or \(\operatorname{arcosh}\left(\frac{5}{4}\right) = \ln\left(\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^2-1^2}\right)\) \(\Rightarrow \frac{3}{5}\ln(3+\sqrt{8}) - \frac{2\sqrt{2}}{5} - \frac{1}{4}\ln 2 + \frac{3}{20}\) | M1 | Converts \(\operatorname{arcosh}(3)\) or \(\operatorname{arcosh}\left(\frac{5}{4}\right)\) to any correct log form. Independent mark but must have obtained \(x\operatorname{arcosh}(5x) \pm f(x)\) where \(f(x)\) has come from integration |
| \(= \frac{3}{20} - \frac{2\sqrt{2}}{5} + \ln\left(3+2\sqrt{2}\right)^{\frac{3}{5}} - \frac{1}{4}\ln 2\) | A1 | Correct answer. Terms in any order but otherwise written as shown. Allow values for \(p, q, r\) & \(k\). Must not follow clearly incorrect work |
## Question 8:
### Part 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\frac{dy}{dx} =\right\} \operatorname{arcosh}(5x) + \frac{ax}{\sqrt{bx^2-1}}$ or $\operatorname{arcosh}(5x) + \frac{cx}{\sqrt{x^2-d}}$ $\Rightarrow \operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}$ | M1 | Differentiates to obtain expression of correct form $a, b, c, d \neq 0$ |
| $\operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}$ | A1 | Correct differentiation. Any equivalent form |
**Total: 2 marks**
### Part 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(x\operatorname{arcosh}(5x)) = \operatorname{arcosh}(5x) + \frac{5x}{\sqrt{25x^2-1}}$ $\Rightarrow \int \operatorname{arcosh}(5x)\, dx = x\operatorname{arcosh}(5x) - \int \frac{5x}{\sqrt{25x^2-1}}\, dx$ | M1 | Rearranges answer to (a) correctly and integrates, or uses correct formula to apply parts to $1 \times \operatorname{arcosh}(5x)$ |
| $\int \operatorname{arcosh}(5x)\, dx = x\operatorname{arcosh}(5x) - \int \frac{5x}{\sqrt{25x^2-1}}\, dx$ | A1 (limited ft) | Correct expression |
| $= x\operatorname{arcosh}(5x) - \frac{1}{5}(25x^2-1)^{\frac{1}{2}}\ (+c)$ | M1 A1 (limited ft) | M1: $\int \frac{Ax}{\sqrt{Bx^2-1}}\, dx \to C(Bx^2-1)^{\frac{1}{2}}$. A1: Fully correct expression with $x\operatorname{arcosh}(5x)$ |
| $\int_{\frac{1}{4}}^{\frac{3}{5}} \operatorname{arcosh}(5x)\, dx = \frac{3}{5}\operatorname{arcosh}(3) - \frac{1}{5}\sqrt{25\times\frac{9}{25}-1} - \left(\frac{1}{4}\operatorname{arcosh}\left(\frac{5}{4}\right) - \frac{1}{5}\sqrt{25\times\frac{1}{16}-1}\right)$ | M1 | Applies appropriate limits with subtraction the right way round, to obtain expression of form $x\operatorname{arcosh}(5x) \pm f(x)$ where $f(x)$ has come from integration |
| $= \frac{3}{5}\operatorname{arcosh}(3) - \frac{2\sqrt{2}}{5} - \frac{1}{4}\operatorname{arcosh}\left(\frac{5}{4}\right) + \frac{3}{20}$ | A1 | Correct answer seen in any form. Must not follow clearly incorrect work |
| $\operatorname{arcosh}(3) = \ln(3+\sqrt{3^2-1^2})$ or $\operatorname{arcosh}\left(\frac{5}{4}\right) = \ln\left(\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^2-1^2}\right)$ $\Rightarrow \frac{3}{5}\ln(3+\sqrt{8}) - \frac{2\sqrt{2}}{5} - \frac{1}{4}\ln 2 + \frac{3}{20}$ | M1 | Converts $\operatorname{arcosh}(3)$ or $\operatorname{arcosh}\left(\frac{5}{4}\right)$ to any correct log form. Independent mark but must have obtained $x\operatorname{arcosh}(5x) \pm f(x)$ where $f(x)$ has come from integration |
| $= \frac{3}{20} - \frac{2\sqrt{2}}{5} + \ln\left(3+2\sqrt{2}\right)^{\frac{3}{5}} - \frac{1}{4}\ln 2$ | A1 | Correct answer. Terms in any order but otherwise written as shown. Allow values for $p, q, r$ & $k$. Must not follow clearly incorrect work |
**Total: 8 marks**
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\begin{enumerate}
\item (a) Differentiate $x \operatorname { arcosh } 5 x$ with respect to $x$\\
(b) Hence, or otherwise, show that
\end{enumerate}
$$\int _ { \frac { 1 } { 4 } } ^ { \frac { 3 } { 5 } } \operatorname { arcosh } 5 x \mathrm {~d} x = \frac { 3 } { 20 } - \frac { 2 \sqrt { 2 } } { 5 } + \ln ( p + q \sqrt { 2 } ) ^ { k } - \frac { 1 } { 4 } \ln r$$
where $p , q , r$ and $k$ are rational numbers to be determined.
\hfill \mbox{\textit{Edexcel F3 2023 Q8 [10]}}