Question 3 - A-Level Maths

OCR FP2 2013 January — Question 3 6 marks

Exam Board	OCR
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	January
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Solve mixed sinh/cosh linear combinations
Difficulty	Standard +0.3 This is a straightforward Further Maths question requiring substitution of standard exponential definitions (cosh x = (e^x + e^{-x})/2, sinh x = (e^x - e^{-x})/2), algebraic manipulation to form a quadratic in e^x, and solving to get a logarithmic answer. The method is explicitly guided ('by first expressing...'), making it a routine application of technique rather than requiring problem-solving insight. While Further Maths content, it's mechanically simpler than average A-level questions.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials

3 By first expressing $\cosh x$ and $\sinh x$ in terms of exponentials, solve the equation $$3 \cosh x - 4 \sinh x = 7$$ giving your answer in an exact logarithmic form.

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Question 3:

Answer	Marks	Guidance
Answer	Marks	Guidance
$3\frac{e^x+e^{-x}}{2} - 4\frac{e^x-e^{-x}}{2} = 7$	M1	Use of formulae
$\Rightarrow 3(e^x+e^{-x}) - 4(e^x-e^{-x}) = 14$	A1	Correct equation
$\Rightarrow -e^x + 7e^{-x} = 14 \Rightarrow e^{2x} + 14e^x - 7 = 0$	A1	Correct quadratic equation in $e^x$
$\Rightarrow e^x = \frac{-14 \pm \sqrt{196+28}}{2}$	M1	Solve quadratic
$[e^x > 0]$ so $e^x = \frac{-14+\sqrt{196+28}}{2} = -7+\sqrt{56}$	A1	Correct value for $e^x$ (ignore negative value)
$\Rightarrow x = \ln(2\sqrt{14}-7)$	A1	One value only with statement of rejection of invalid value for $e^x$
[6]
Alternative: Make sinh or cosh the subject, square, use $c^2-s^2=1$	M1, A1
Gives $7s^2+56s+40=0$ or $7c^2+42c-65=0$	A1

## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\frac{e^x+e^{-x}}{2} - 4\frac{e^x-e^{-x}}{2} = 7$ | M1 | Use of formulae |
| $\Rightarrow 3(e^x+e^{-x}) - 4(e^x-e^{-x}) = 14$ | A1 | Correct equation |
| $\Rightarrow -e^x + 7e^{-x} = 14 \Rightarrow e^{2x} + 14e^x - 7 = 0$ | A1 | Correct quadratic equation in $e^x$ |
| $\Rightarrow e^x = \frac{-14 \pm \sqrt{196+28}}{2}$ | M1 | Solve quadratic |
| $[e^x > 0]$ so $e^x = \frac{-14+\sqrt{196+28}}{2} = -7+\sqrt{56}$ | A1 | Correct value for $e^x$ (ignore negative value) |
| $\Rightarrow x = \ln(2\sqrt{14}-7)$ | A1 | One value only with statement of rejection of invalid value for $e^x$ |
| **[6]** | | |
| **Alternative:** Make sinh or cosh the subject, square, use $c^2-s^2=1$ | M1, A1 | |
| Gives $7s^2+56s+40=0$ or $7c^2+42c-65=0$ | A1 | |

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3 By first expressing $\cosh x$ and $\sinh x$ in terms of exponentials, solve the equation

$$3 \cosh x - 4 \sinh x = 7$$

giving your answer in an exact logarithmic form.

\hfill \mbox{\textit{OCR FP2 2013 Q3 [6]}}

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