Question 4 - A-Level Maths

Edexcel FP3 2018 June — Question 4 12 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2018
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration using inverse trig and hyperbolic functions
Type	Arc length with inverse trig
Difficulty	Challenging +1.3 This is a structured multi-part question on arc length with inverse hyperbolic functions. Part (a) requires differentiation of arsinh x and a product (standard FP3 techniques). Part (b) applies the arc length formula directly. Part (c) uses a given substitution with hyperbolic identities—methodical but routine for Further Maths students. The question guides students through each step with no novel insight required, making it moderately above average difficulty.
Spec	1.08h Integration by substitution 4.07e Inverse hyperbolic: definitions, domains, ranges 4.07f Inverse hyperbolic: logarithmic forms 4.08g Derivatives: inverse trig and hyperbolic functions 4.08h Integration: inverse trig/hyperbolic substitutions 8.06b Arc length and surface area: of revolution, cartesian or parametric

4. The curve $C$ has equation $$y = \operatorname { arsinh } x + x \sqrt { x ^ { 2 } + 1 } , \quad 0 \leqslant x \leqslant 1$$

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { x ^ { 2 } + 1 }$
Hence show that the length of the curve $C$ is given by $$\int _ { 0 } ^ { 1 } \sqrt { 4 x ^ { 2 } + 5 } d x$$
Using the substitution $x = \frac { \sqrt { 5 } } { 2 } \sinh u$, find the exact length of the curve $C$, giving your answer in the form $a + b \ln c$, where $a , b$ and $c$ are constants to be found.

Show mark scheme Show mark scheme source

Question 4(a) [Differentiation]:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{d(\text{arsinh}\,x)}{dx} = \frac{1}{\sqrt{x^2+1}}$	B1
$\frac{d(x\sqrt{x^2+1})}{dx} = \frac{x^2}{\sqrt{x^2+1}} + \sqrt{x^2+1}$	B1
$= \frac{1+x^2+1+x^2}{\sqrt{x^2+1}} = \ldots$	M1	Processes 3 terms of the form $\frac{A}{\sqrt{x^2+1}}, \frac{Bx^2}{\sqrt{x^2+1}}, C\sqrt{x^2+1}$ using correct algebra to obtain a single term
$= 2\sqrt{x^2+1}$	A1	cso. Allow $2(x^2+1)^{\frac{1}{2}}$

Question 4(b) [Arc Length]:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$1 + \left(\frac{dy}{dx}\right)^2 = 1 + 4(x^2+1)$	M1	Attempts $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$ with the printed answer from part (a); must see a step before the given answer
$(L=)\int_0^1\sqrt{5+4x^2}\,dx$	A1*	Answer as printed with no errors including limits and "$dx$". Allow $\int_0^1\sqrt{4x^2+5}\,dx$

Question 4(c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x = \frac{\sqrt{5}}{2}\sinh u \Rightarrow \frac{dx}{du} = \frac{\sqrt{5}}{2}\cosh u$
$L = \int \sqrt{5 + 5\sinh^2 u} \cdot \frac{\sqrt{5}}{2}\cosh u \, du$	M1	Fully substitutes into $\int \sqrt{4x^2+5} \, dx$
$= \frac{5}{2}\int \cosh^2 u \, du$	A1	Correct integral including the $\frac{5}{2}$. Allow e.g. $\frac{5}{2}\int \cosh u \cosh u \, du$
$= \frac{5}{4}\int (\cosh 2u + 1) \, du$	dM1	Applies $\cosh 2u = \pm 2\cosh^2 u \pm 1$ to integral of form $k\int \cosh^2 u \, du$. Dependent on first method mark.
$= \frac{5}{4}\left[\frac{1}{2}\sinh 2u + u\right]$	A1	Correct integration: $k(\cosh 2u+1) \to k\left(\frac{1}{2}\sinh 2u + u\right)$
$= \left[\cdots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}}$ with $\frac{1}{2}\sinh\left(2\left(\text{arsinh}\frac{2}{\sqrt{5}}\right)\right) = \frac{6}{5}$	ddM1	Use of correct limits or returns to $x$, uses 0 and 1. Use of 0 may be implied. Dependent on both method marks.
$= \frac{3}{2} + \frac{5}{8}\ln 5$	A1	Allow equivalent exact answers e.g. $\frac{3}{2}+\frac{5}{4}\ln\sqrt{5}$, $\frac{3}{2}+\frac{5}{4}\ln\left(\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{5}}\right)$

Alternative using exponentials (last 4 marks):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{5}{2}\int \cosh^2 u \, du = \frac{5}{8}\int(e^{2u}+2+e^{-2u}) \, du$	dM1	Uses $\cosh u = \frac{1}{2}(e^u+e^{-u})$ and squares, applies to $k\int \cosh^2 u \, du$
$= \frac{5}{8}\left[\frac{1}{2}e^{2u}+2u-\frac{1}{2}e^{-2u}\right]$	A1	Correct integration
$= \left[\cdots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}}$	ddM1	Use of correct limits or returns to $x$, uses 0 and 1. Dependent on both method marks.
$= \frac{3}{2}+\frac{5}{8}\ln 5$	A1	Allow equivalent exact answers

# Question 4(a) [Differentiation]:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d(\text{arsinh}\,x)}{dx} = \frac{1}{\sqrt{x^2+1}}$ | B1 | |
| $\frac{d(x\sqrt{x^2+1})}{dx} = \frac{x^2}{\sqrt{x^2+1}} + \sqrt{x^2+1}$ | B1 | |
| $= \frac{1+x^2+1+x^2}{\sqrt{x^2+1}} = \ldots$ | M1 | Processes **3 terms** of the form $\frac{A}{\sqrt{x^2+1}}, \frac{Bx^2}{\sqrt{x^2+1}}, C\sqrt{x^2+1}$ using correct algebra to obtain a single term |
| $= 2\sqrt{x^2+1}$ | A1 | cso. Allow $2(x^2+1)^{\frac{1}{2}}$ |

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# Question 4(b) [Arc Length]:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + \left(\frac{dy}{dx}\right)^2 = 1 + 4(x^2+1)$ | M1 | Attempts $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$ with the **printed answer** from part (a); must see a step before the given answer |
| $(L=)\int_0^1\sqrt{5+4x^2}\,dx$ | A1* | Answer **as printed** with no errors including limits and "$dx$". Allow $\int_0^1\sqrt{4x^2+5}\,dx$ |

## Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{\sqrt{5}}{2}\sinh u \Rightarrow \frac{dx}{du} = \frac{\sqrt{5}}{2}\cosh u$ | | |
| $L = \int \sqrt{5 + 5\sinh^2 u} \cdot \frac{\sqrt{5}}{2}\cosh u \, du$ | M1 | Fully substitutes into $\int \sqrt{4x^2+5} \, dx$ |
| $= \frac{5}{2}\int \cosh^2 u \, du$ | A1 | Correct integral including the $\frac{5}{2}$. Allow e.g. $\frac{5}{2}\int \cosh u \cosh u \, du$ |
| $= \frac{5}{4}\int (\cosh 2u + 1) \, du$ | dM1 | Applies $\cosh 2u = \pm 2\cosh^2 u \pm 1$ to integral of form $k\int \cosh^2 u \, du$. **Dependent on first method mark.** |
| $= \frac{5}{4}\left[\frac{1}{2}\sinh 2u + u\right]$ | A1 | Correct integration: $k(\cosh 2u+1) \to k\left(\frac{1}{2}\sinh 2u + u\right)$ |
| $= \left[\cdots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}}$ with $\frac{1}{2}\sinh\left(2\left(\text{arsinh}\frac{2}{\sqrt{5}}\right)\right) = \frac{6}{5}$ | ddM1 | Use of correct limits or returns to $x$, uses 0 and 1. Use of 0 may be implied. **Dependent on both method marks.** |
| $= \frac{3}{2} + \frac{5}{8}\ln 5$ | A1 | Allow equivalent exact answers e.g. $\frac{3}{2}+\frac{5}{4}\ln\sqrt{5}$, $\frac{3}{2}+\frac{5}{4}\ln\left(\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{5}}\right)$ |

**Alternative using exponentials** (last 4 marks):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5}{2}\int \cosh^2 u \, du = \frac{5}{8}\int(e^{2u}+2+e^{-2u}) \, du$ | dM1 | Uses $\cosh u = \frac{1}{2}(e^u+e^{-u})$ and squares, applies to $k\int \cosh^2 u \, du$ |
| $= \frac{5}{8}\left[\frac{1}{2}e^{2u}+2u-\frac{1}{2}e^{-2u}\right]$ | A1 | Correct integration |
| $= \left[\cdots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}}$ | ddM1 | Use of correct limits or returns to $x$, uses 0 and 1. **Dependent on both method marks.** |
| $= \frac{3}{2}+\frac{5}{8}\ln 5$ | A1 | Allow equivalent exact answers |

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4. The curve $C$ has equation

$$y = \operatorname { arsinh } x + x \sqrt { x ^ { 2 } + 1 } , \quad 0 \leqslant x \leqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { x ^ { 2 } + 1 }$
\item Hence show that the length of the curve $C$ is given by

$$\int _ { 0 } ^ { 1 } \sqrt { 4 x ^ { 2 } + 5 } d x$$
\item Using the substitution $x = \frac { \sqrt { 5 } } { 2 } \sinh u$, find the exact length of the curve $C$, giving your answer in the form $a + b \ln c$, where $a , b$ and $c$ are constants to be found.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3 2018 Q4 [12]}}

This paper (7 questions)

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Q1 5 Q2 10 Q3 9 Q4 12 Q5 11 Q6 13 Q7 15

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(x = \frac{\sqrt{5}}{2}\sinh u \Rightarrow \frac{dx}{du} = \frac{\sqrt{5}}{2}\cosh u\)
\(L = \int \sqrt{5 + 5\sinh^2 u} \cdot \frac{\sqrt{5}}{2}\cosh u \, du\)	M1	Fully substitutes into \(\int \sqrt{4x^2+5} \, dx\)
\(= \frac{5}{2}\int \cosh^2 u \, du\)	A1	Correct integral including the \(\frac{5}{2}\). Allow e.g. \(\frac{5}{2}\int \cosh u \cosh u \, du\)
\(= \frac{5}{4}\int (\cosh 2u + 1) \, du\)	dM1	Applies \(\cosh 2u = \pm 2\cosh^2 u \pm 1\) to integral of form \(k\int \cosh^2 u \, du\). Dependent on first method mark.
\(= \frac{5}{4}\left[\frac{1}{2}\sinh 2u + u\right]\)	A1	Correct integration: \(k(\cosh 2u+1) \to k\left(\frac{1}{2}\sinh 2u + u\right)\)
\(= \left[\cdots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}}\) with \(\frac{1}{2}\sinh\left(2\left(\text{arsinh}\frac{2}{\sqrt{5}}\right)\right) = \frac{6}{5}\)	ddM1	Use of correct limits or returns to \(x\), uses 0 and 1. Use of 0 may be implied. Dependent on both method marks.
\(= \frac{3}{2} + \frac{5}{8}\ln 5\)	A1	Allow equivalent exact answers e.g. \(\frac{3}{2}+\frac{5}{4}\ln\sqrt{5}\), \(\frac{3}{2}+\frac{5}{4}\ln\left(\frac{2}{\sqrt{5}}+\frac{3}{\sqrt{5}}\right)\)