| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on hyperbolic functions with clear scaffolding through parts (i)-(iv). Parts (i) and (ii) are standard proofs from definitions that FP2 students practice routinely. Part (iii) follows a given substitution with straightforward application of hyperbolic identities. Part (iv) applies the result with definite integration. While this requires knowledge beyond standard A-level and multiple techniques, the heavy scaffolding and standard nature of each component makes it moderately above average difficulty rather than genuinely challenging for the target FP2 cohort. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07e Inverse hyperbolic: definitions, domains, ranges4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{1}{2}\cosh u \Rightarrow \frac{dx}{du} = \frac{1}{2}\sinh u\) | M1 | Reaching integrand equivalent to \(k\sinh^2 u\) |
| \(\int\sqrt{4x^2 - 1}\,dx = \int\sqrt{\cosh^2 u - 1} \times \frac{1}{2}\sinh u\,du\) | A1 | |
| \(= \int\frac{1}{2}\sinh^2 u\,du\) | M1 | Simplifying to integrable form. Dependent on M1 above; \(\frac{1}{8}e^{2u} - \frac{1}{4} + \frac{1}{8}e^{-2u}\) Or \(\frac{1}{2}\sinh 2u\) o.e. and \(-\frac{1}{4}u\) seen; Or \(\frac{1}{8}e^{2u} - \frac{1}{4}u - \frac{1}{16}e^{-2u} + c\); Condone omission of \(+ c\) throughout |
| \(= \int\frac{1}{4}\cosh 2u - \frac{1}{4}\,du\) | M1 | |
| \(= \frac{1}{8}\sinh 2u - \frac{1}{4}u + c\) | A1 A1 | For \(\frac{1}{8}\sinh 2u\) o.e. and \(-\frac{1}{4}u\) seen |
| \(= \frac{1}{4}\sinh u \cosh u - \frac{1}{4}u + c\) | M1 | Clear use of \(\sinh 2u = 2\sinh u \cosh u\); Dependent on M1 M1 above |
| \(= \frac{1}{4}\sqrt{4x^2 - 1} \times 2x - \frac{1}{4}\text{arcosh}\,2x + c\) | M1 | |
| \(= \frac{1}{2}x\sqrt{4x^2 - 1} - \frac{1}{4}\text{arcosh}\,2x + c\) | A1 [7] | \(a = \frac{1}{2}\); \(a\), \(b\) need not be written separately |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\sqrt{3}}\sqrt{4x^2 - 1}\,dx = \left[\frac{1}{2}x\sqrt{4x^2 - 1} - \frac{1}{4}\text{arcosh}\,2x\right]\) | M1 | Using their (iii) and using limits correctly |
| \(= \frac{\sqrt{3}}{2} - \frac{1}{4}\text{arcosh}\,2 + \frac{1}{4}\text{arcosh}\,1\) | A1ft | May be implied; Must have obtained values for \(a\) and \(b\) in (iii); F.t. values of \(a\) and \(b\) in (iii) |
| \(= \frac{\sqrt{3}}{2} - \frac{1}{4}\ln(2 + \sqrt{3}) + \frac{1}{4}\ln 1\) | M1 | Using (ii) accurately; Dependent on M1 above; c.a.o. A0 if ln 1 retained; Mark final answer; \(a\sqrt{3} - b\text{arcosh}\,2\). No decimals. Must have obtained values for \(a\) and \(b\) |
| \(= \frac{\sqrt{3}}{2} - \frac{1}{4}\ln(2 + \sqrt{3})\) | A1 [4] | Correct answer www scores 4/4 |
### (iii)
$x = \frac{1}{2}\cosh u \Rightarrow \frac{dx}{du} = \frac{1}{2}\sinh u$ | M1 | Reaching integrand equivalent to $k\sinh^2 u$
$\int\sqrt{4x^2 - 1}\,dx = \int\sqrt{\cosh^2 u - 1} \times \frac{1}{2}\sinh u\,du$ | A1 |
$= \int\frac{1}{2}\sinh^2 u\,du$ | M1 | Simplifying to integrable form. Dependent on M1 above; $\frac{1}{8}e^{2u} - \frac{1}{4} + \frac{1}{8}e^{-2u}$ Or $\frac{1}{2}\sinh 2u$ o.e. and $-\frac{1}{4}u$ seen; Or $\frac{1}{8}e^{2u} - \frac{1}{4}u - \frac{1}{16}e^{-2u} + c$; Condone omission of $+ c$ throughout
$= \int\frac{1}{4}\cosh 2u - \frac{1}{4}\,du$ | M1 |
$= \frac{1}{8}\sinh 2u - \frac{1}{4}u + c$ | A1 A1 | For $\frac{1}{8}\sinh 2u$ o.e. and $-\frac{1}{4}u$ seen
$= \frac{1}{4}\sinh u \cosh u - \frac{1}{4}u + c$ | M1 | Clear use of $\sinh 2u = 2\sinh u \cosh u$; Dependent on M1 M1 above
$= \frac{1}{4}\sqrt{4x^2 - 1} \times 2x - \frac{1}{4}\text{arcosh}\,2x + c$ | M1 |
$= \frac{1}{2}x\sqrt{4x^2 - 1} - \frac{1}{4}\text{arcosh}\,2x + c$ | A1 [7] | $a = \frac{1}{2}$; $a$, $b$ need not be written separately
### (iv)
$\int_0^{\sqrt{3}}\sqrt{4x^2 - 1}\,dx = \left[\frac{1}{2}x\sqrt{4x^2 - 1} - \frac{1}{4}\text{arcosh}\,2x\right]$ | M1 | Using their (iii) and using limits correctly
$= \frac{\sqrt{3}}{2} - \frac{1}{4}\text{arcosh}\,2 + \frac{1}{4}\text{arcosh}\,1$ | A1ft | May be implied; Must have obtained values for $a$ and $b$ in (iii); F.t. values of $a$ and $b$ in (iii)
$= \frac{\sqrt{3}}{2} - \frac{1}{4}\ln(2 + \sqrt{3}) + \frac{1}{4}\ln 1$ | M1 | Using (ii) accurately; Dependent on M1 above; c.a.o. A0 if ln 1 retained; Mark final answer; $a\sqrt{3} - b\text{arcosh}\,2$. No decimals. Must have obtained values for $a$ and $b$
$= \frac{\sqrt{3}}{2} - \frac{1}{4}\ln(2 + \sqrt{3})$ | A1 [4] | Correct answer www scores 4/4
---4 (i) Prove, from definitions involving exponential functions, that
$$\cosh 2 u = 2 \sinh ^ { 2 } u + 1$$
(ii) Prove that, if $y \geqslant 0$ and $\cosh y = u$, then $y = \ln \left( u + \sqrt { } \left( u ^ { 2 } - 1 \right) \right)$.\\
(iii) Using the substitution $2 x = \cosh u$, show that
$$\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x = a x \sqrt { 4 x ^ { 2 } - 1 } - b \operatorname { arcosh } 2 x + c$$
where $a$ and $b$ are constants to be determined and $c$ is an arbitrary constant.\\
(iv) Find $\int _ { \frac { 1 } { 2 } } ^ { 1 } \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x$, expressing your answer in an exact form involving logarithms.
\hfill \mbox{\textit{OCR MEI FP2 2012 Q4 [18]}}