Question 3 - A-Level Maths

Edexcel FP3 2010 June — Question 3 8 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2010
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Solve using double angle formulas
Difficulty	Standard +0.3 Part (a) is a straightforward proof from definitions requiring basic exponential manipulation. Part (b) uses the proven identity to convert to a quadratic in sinh x, then applies the inverse sinh formula—standard Further Maths technique with no novel insight required. The multi-step nature and Further Maths content places it slightly above average difficulty, but it's a textbook exercise.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

3. (a) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that $$\cosh 2 x = 1 + 2 \sinh ^ { 2 } x$$ (b) Solve the equation $$\cosh 2 x - 3 \sinh x = 15$$ giving your answers as exact logarithms.

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Question 3:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$rhs = 1 + 2\sinh^2 x = 1 + 2\left(\frac{e^x - e^{-x}}{2}\right)^2$	M1	Substituting exponential definition
$= \frac{2 + e^{2x} - 2 + e^{-2x}}{2}$	M1	Expanding
$= \frac{e^{2x} + e^{-2x}}{2} = \cosh 2x = lhs$ ✽	A1	(3)

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$1 + 2\sinh^2 x - 3\sinh x = 15 \Rightarrow 2\sinh^2 x - 3\sinh x - 14 = 0$	M1	Using part (a) result
$(\sinh x + 2)(2\sinh x - 7) = 0$	M1	Factorising
$\sinh x = -2, \frac{7}{2}$	A1	Both values
$x = \ln\left(-2 + \sqrt{(-2)^2+1}\right) = \ln\left(-2+\sqrt{5}\right)$	M1	Applying $\text{arcsinh}$ formula
$x = \ln\left(\frac{7}{2} + \sqrt{\left(\frac{7}{2}\right)^2+1}\right) = \ln\left(\frac{7+\sqrt{53}}{2}\right)$	A1	(5)

## Question 3:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $rhs = 1 + 2\sinh^2 x = 1 + 2\left(\frac{e^x - e^{-x}}{2}\right)^2$ | M1 | Substituting exponential definition |
| $= \frac{2 + e^{2x} - 2 + e^{-2x}}{2}$ | M1 | Expanding |
| $= \frac{e^{2x} + e^{-2x}}{2} = \cosh 2x = lhs$ ✽ | A1 | (3) |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 + 2\sinh^2 x - 3\sinh x = 15 \Rightarrow 2\sinh^2 x - 3\sinh x - 14 = 0$ | M1 | Using part (a) result |
| $(\sinh x + 2)(2\sinh x - 7) = 0$ | M1 | Factorising |
| $\sinh x = -2, \frac{7}{2}$ | A1 | Both values |
| $x = \ln\left(-2 + \sqrt{(-2)^2+1}\right) = \ln\left(-2+\sqrt{5}\right)$ | M1 | Applying $\text{arcsinh}$ formula |
| $x = \ln\left(\frac{7}{2} + \sqrt{\left(\frac{7}{2}\right)^2+1}\right) = \ln\left(\frac{7+\sqrt{53}}{2}\right)$ | A1 | (5) |

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3. (a) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that

$$\cosh 2 x = 1 + 2 \sinh ^ { 2 } x$$

(b) Solve the equation

$$\cosh 2 x - 3 \sinh x = 15$$

giving your answers as exact logarithms.\\

\hfill \mbox{\textit{Edexcel FP3 2010 Q3 [8]}}

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