| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple proof techniques (exponential definitions, inverse hyperbolic forms, partial fractions) and connecting different representations. While each individual part uses standard FP2 techniques, the multi-step nature, requirement to prove logarithmic forms from first principles, and the synthesis across parts (connecting artanh to logarithms) elevates it above routine exercises. It's moderately challenging for Further Maths but not exceptionally difficult. |
| Spec | 1.02y Partial fractions: decompose rational functions4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cosh^2 x = \left[\tfrac{1}{2}(e^x+e^{-x})\right]^2 = \tfrac{1}{4}(e^{2x}+2+e^{-2x})\) | ||
| \(\sinh^2 x = \left[\tfrac{1}{2}(e^x-e^{-x})\right]^2 = \tfrac{1}{4}(e^{2x}-2+e^{-2x})\) | M1 | Both expressions (M0 if no "middle" term) and subtraction |
| \(\cosh^2 x - \sinh^2 x = \tfrac{1}{4}(2+2) = 1\) | A1 (ag) | www |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cosh x = \sqrt{1+\sinh^2 x} = \sqrt{1+\tan^2 y}\) | M1 | Use of \(\cosh^2 x = 1+\sinh^2 x\) and \(\sinh x = \tan y\) |
| \(= \sec y\) | A1 | |
| \(\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{\tan y}{\sec y} = \sin y\) | A1 (ag) | www |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{arsinh}\, x = \ln(x + \sqrt{1+x^2})\) | M1 | Attempt to use ln form of arsinh |
| \(\text{arsinh}(\tan y) = \ln(\tan y + \sqrt{1+\tan^2 y})\) | A1 | |
| \(x = \ln(\tan y + \sec y)\) | A1 (ag) | www |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \text{artanh}\, x \Rightarrow x = \tanh y\) | M1 | \(\tanh y =\) and attempt to differentiate |
| \(\dfrac{dx}{dy} = \text{sech}^2 y\) | Or \(\text{sech}^2 y\,\dfrac{dy}{dx} = 1\) | |
| \(\dfrac{dy}{dx} = \dfrac{1}{\text{sech}^2 y} = \dfrac{1}{1-\tanh^2 y} = \dfrac{1}{1-x^2}\) | A1 | Or B2 for \(\dfrac{1}{1-x^2}\) www |
| Integral \(= \left[\text{artanh}\, x\right]_{-\frac{1}{2}}^{\frac{1}{2}}\) | M1 | artanh or any tanh substitution |
| \(= 2\,\text{artanh}\,\dfrac{1}{2}\) | A1 (ag) | www |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{1}{1-x^2} = \dfrac{1}{(1-x)(1+x)} = \dfrac{A}{1-x} + \dfrac{B}{1+x}\) | Correct form of partial fractions | |
| \(1 = A(1+x) + B(1-x)\) | M1 | Attempt to evaluate constants |
| \(A = \tfrac{1}{2},\ B = \tfrac{1}{2}\) | A1 | |
| \(\int\dfrac{1}{1-x^2}\,dx = \int\dfrac{\frac{1}{2}}{1-x} + \dfrac{\frac{1}{2}}{1+x}\,dx\) | M1 | Log integrals |
| \(= -\tfrac{1}{2}\ln | 1-x | + \tfrac{1}{2}\ln |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{-\frac{1}{2}}^{\frac{1}{2}}\dfrac{1}{1-x^2}\,dx = \left[-\tfrac{1}{2}\ln | 1-x | +\tfrac{1}{2}\ln |
| \(2\,\text{artanh}\,\dfrac{1}{2} = \ln 3 \Rightarrow \text{artanh}\,\dfrac{1}{2} = \dfrac{1}{2}\ln 3\) | A1 (ag) | www |
| Total | 2 |
# Question 4:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cosh^2 x = \left[\tfrac{1}{2}(e^x+e^{-x})\right]^2 = \tfrac{1}{4}(e^{2x}+2+e^{-2x})$ | | |
| $\sinh^2 x = \left[\tfrac{1}{2}(e^x-e^{-x})\right]^2 = \tfrac{1}{4}(e^{2x}-2+e^{-2x})$ | M1 | Both expressions (M0 if no "middle" term) and subtraction |
| $\cosh^2 x - \sinh^2 x = \tfrac{1}{4}(2+2) = 1$ | A1 (ag) | www |
| **Total** | **2** | |
## Part (ii)(A)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cosh x = \sqrt{1+\sinh^2 x} = \sqrt{1+\tan^2 y}$ | M1 | Use of $\cosh^2 x = 1+\sinh^2 x$ and $\sinh x = \tan y$ |
| $= \sec y$ | A1 | |
| $\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{\tan y}{\sec y} = \sin y$ | A1 (ag) | www |
| **Total** | **3** | |
## Part (ii)(B)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{arsinh}\, x = \ln(x + \sqrt{1+x^2})$ | M1 | Attempt to use ln form of arsinh |
| $\text{arsinh}(\tan y) = \ln(\tan y + \sqrt{1+\tan^2 y})$ | A1 | |
| $x = \ln(\tan y + \sec y)$ | A1 (ag) | www |
| **Total** | **3** | |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \text{artanh}\, x \Rightarrow x = \tanh y$ | M1 | $\tanh y =$ and attempt to differentiate |
| $\dfrac{dx}{dy} = \text{sech}^2 y$ | | Or $\text{sech}^2 y\,\dfrac{dy}{dx} = 1$ |
| $\dfrac{dy}{dx} = \dfrac{1}{\text{sech}^2 y} = \dfrac{1}{1-\tanh^2 y} = \dfrac{1}{1-x^2}$ | A1 | Or B2 for $\dfrac{1}{1-x^2}$ www |
| Integral $= \left[\text{artanh}\, x\right]_{-\frac{1}{2}}^{\frac{1}{2}}$ | M1 | artanh or any tanh substitution |
| $= 2\,\text{artanh}\,\dfrac{1}{2}$ | A1 (ag) | www |
| **Total** | **4** | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{1}{1-x^2} = \dfrac{1}{(1-x)(1+x)} = \dfrac{A}{1-x} + \dfrac{B}{1+x}$ | | Correct form of partial fractions |
| $1 = A(1+x) + B(1-x)$ | M1 | Attempt to evaluate constants |
| $A = \tfrac{1}{2},\ B = \tfrac{1}{2}$ | A1 | |
| $\int\dfrac{1}{1-x^2}\,dx = \int\dfrac{\frac{1}{2}}{1-x} + \dfrac{\frac{1}{2}}{1+x}\,dx$ | M1 | Log integrals |
| $= -\tfrac{1}{2}\ln|1-x| + \tfrac{1}{2}\ln|1+x| + c$ or $\tfrac{1}{2}\ln\left|\dfrac{1+x}{1-x}\right| + c$ o.e. | A1 | www. Condone omitted modulus signs and constant. After 0 scored, SC1 for correct answer |
| **Total** | **4** | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{-\frac{1}{2}}^{\frac{1}{2}}\dfrac{1}{1-x^2}\,dx = \left[-\tfrac{1}{2}\ln|1-x|+\tfrac{1}{2}\ln|1+x|\right]_{-\frac{1}{2}}^{\frac{1}{2}} = \ln 3$ | M1 | Substitution of $\tfrac{1}{2}$ and $-\tfrac{1}{2}$ seen anywhere (or correct use of $0,\ \tfrac{1}{2}$) |
| $2\,\text{artanh}\,\dfrac{1}{2} = \ln 3 \Rightarrow \text{artanh}\,\dfrac{1}{2} = \dfrac{1}{2}\ln 3$ | A1 (ag) | www |
| **Total** | **2** | |
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Prove, from definitions involving exponentials, that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
\item Given that $\sinh x = \tan y$, where $- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi$, show that\\
(A) $\tanh x = \sin y$,\\
(B) $x = \ln ( \tan y + \sec y )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \operatorname { artanh } x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.
Hence show that $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x = 2 \operatorname { artanh } \frac { 1 } { 2 }$.
\item Express $\frac { 1 } { 1 - x ^ { 2 } }$ in partial fractions and hence find an expression for $\int \frac { 1 } { 1 - x ^ { 2 } } \mathrm {~d} x$ in terms of logarithms.
\item Use the results in parts (i) and (ii) to show that $\operatorname { artanh } \frac { 1 } { 2 } = \frac { 1 } { 2 } \ln 3$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q4 [18]}}