| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 13 |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Challenging +1.2 This is a structured multi-part question on hyperbolic functions requiring differentiation, solving equations, and finding stationary points. Part (i) is routine proof from definitions. Parts (ii)-(iv) involve standard calculus techniques with hyperbolic identities. Part (v) requires solving a cubic in sinh x and using inverse hyperbolic functions, but the question provides significant scaffolding throughout. While this is Further Maths content (inherently harder), the question is methodical and well-guided, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2\sinh x\cosh x = 2(\frac{e^x - e^{-x}}{2})(\frac{e^x+e^{-x}}{2})\) | M1 | Use of correct exponential form for sinh or cosh |
| \(= \frac{1}{2}(e^{2x} - e^{-2x} - 1 + 1) = \frac{1}{2}(e^{2x} - e^{-2x}) = \sinh 2x\) | A1 [2] | AG so evidence of cancellation or difference between squares required; Proof must be properly completed for A1; ie there must be complete reasoning from LHS = ... to ... = RHS |
| (ii) \(y = a\cosh x - \cosh 2x\); \(\Rightarrow \frac{dy}{dx} = a\sinh x - 2\sinh 2x\) | M1 | Diff'n and verify = 0 by substitution (which must be seen) oe |
| \(= a\sinh 0 - 2\sinh 0 = 0\) when \(x = 0\) | A1 [2] | |
| (iii) when \(x = 0, y = a - 1\) so \((0, a-1)\) | B1 [1] | |
| (iv) \(\frac{dy}{dx} = a\sinh x - 4\sinh x\cosh x = 0\) | M1 [4] | |
| when \(\sinh x = 0\) or \(\cosh x = \frac{a}{4}\) | A1 [4] | |
| \(\cosh x = \frac{a}{4}\) has two values if \(\frac{a}{4} > 1\) and no values if \(\frac{a}{4} < 1\) | M1 [4] | For considering the number of possible values of cosh x |
| \(\frac{a}{4} = 1\) gives the same stationary point as previously identified, so maximum value of \(a\) is \(4\) | A1 [4] | For full justification |
| (v) When \(a = 6, \cosh x = \frac{3}{2} \Rightarrow x = \cosh^{-1}\frac{3}{2}\) | M1 [4] | 3.1a |
| \(y = 6\cosh(\cosh^{-1}\frac{3}{2}) - \cosh(2\cosh^{-1}\frac{3}{2}) = \frac{11}{2}\) | M1, A1 [4] | 1.1; BC. Substituting in \(x\) value. Can be implied by correct answer. Accept awrt 5.500 |
| i.e. \((\cosh^{-1}\frac{3}{2}, \frac{11}{2})\) | A1 [4] | 1.1 |
**(i)** $2\sinh x\cosh x = 2(\frac{e^x - e^{-x}}{2})(\frac{e^x+e^{-x}}{2})$ | M1 | Use of correct exponential form for sinh or cosh
$= \frac{1}{2}(e^{2x} - e^{-2x} - 1 + 1) = \frac{1}{2}(e^{2x} - e^{-2x}) = \sinh 2x$ | A1 [2] | AG so evidence of cancellation or difference between squares required; Proof must be properly completed for A1; ie there must be complete reasoning from LHS = ... to ... = RHS
**(ii)** $y = a\cosh x - \cosh 2x$; $\Rightarrow \frac{dy}{dx} = a\sinh x - 2\sinh 2x$ | M1 | Diff'n and verify = 0 by substitution (which must be seen) oe
$= a\sinh 0 - 2\sinh 0 = 0$ when $x = 0$ | A1 [2]
**(iii)** when $x = 0, y = a - 1$ so $(0, a-1)$ | B1 [1]
**(iv)** $\frac{dy}{dx} = a\sinh x - 4\sinh x\cosh x = 0$ | M1 [4] |
when $\sinh x = 0$ or $\cosh x = \frac{a}{4}$ | A1 [4] |
$\cosh x = \frac{a}{4}$ has two values if $\frac{a}{4} > 1$ and no values if $\frac{a}{4} < 1$ | M1 [4] | For considering the number of possible values of cosh x
$\frac{a}{4} = 1$ gives the same stationary point as previously identified, so maximum value of $a$ is $4$ | A1 [4] | For full justification
**(v)** When $a = 6, \cosh x = \frac{3}{2} \Rightarrow x = \cosh^{-1}\frac{3}{2}$ | M1 [4] | 3.1a
$y = 6\cosh(\cosh^{-1}\frac{3}{2}) - \cosh(2\cosh^{-1}\frac{3}{2}) = \frac{11}{2}$ | M1, A1 [4] | 1.1; BC. Substituting in $x$ value. Can be implied by correct answer. Accept awrt 5.500
i.e. $(\cosh^{-1}\frac{3}{2}, \frac{11}{2})$ | A1 [4] | 1.1
---8 (i) Using the definitions of $\cosh x$ and $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that $\sinh 2 x = 2 \sinh x \cosh x$.
You are given the function $\mathrm { f } ( x ) = a \cosh x - \cosh 2 x$, where $a$ is a positive constant.\\
(ii) Verify that, for any value of $a$, the curve $y = \mathrm { f } ( x )$ has a stationary point on the $y$-axis.\\
(iii) Find the coordinates of the stationary point found in part (ii).\\
(iv) Determine the maximum value of $a$ for which the stationary point found in part (ii) is the only stationary point on the curve $y = \mathrm { f } ( x )$.
You are given that for any value of $a$ greater than the value found in part (iv) there are three stationary points, the one found in part (ii) and two others, one of which satisfies $x > 0$.\\
(v) Find the coordinates of this point when $a = 6$.
Give your answer in the form $\left( \cosh ^ { - 1 } p , q \right)$.
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q8 [13]}}