Question 1 - A-Level Maths

AQA FP2 2010 January — Question 1 9 marks

Exam Board	AQA
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	January
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Solve using substitution u = cosh x or u = sinh x
Difficulty	Standard +0.3 This is a straightforward Further Maths question testing standard hyperbolic function techniques: proving the fundamental identity from definitions (routine algebra), using the identity to simplify an expression, sketching a standard curve, and solving a quadratic equation in cosh x. All steps are textbook exercises requiring no novel insight, though it's multi-part and involves Further Maths content which places it slightly above average A-level difficulty.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials 4.07b Hyperbolic graphs: sketch and properties 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

Use the definitions $\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)$ and $\sinh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right)$ to show that $$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$$
1. Express $$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x$$ in terms of $\cosh x$.
2. Sketch the curve $y = \cosh x$.
3. Hence solve the equation $$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x = 9.5$$ giving your answers in logarithmic form.

Show mark scheme Show mark scheme source

Part (a)

Answer	Marks	Guidance
LHS = $\frac{1}{4}(e^x + e^{-x})^2 - \frac{1}{4}(e^x - e^{-x})^2$	M1	Correct expansion of either square
Shown equal to 1	A1, A1	3 marks total

Part (b)(i)

Answer	Marks	Guidance
$8\cosh^2 x - 3$	B1	1 mark

Part (b)(ii)

Answer	Marks	Guidance
Sketch of $y = \cosh x$	B1	1 mark

Part (b)(iii)

Answer	Marks	Guidance
$\cosh x = (\pm)1.25$	B1F	OE; ft errors in (b)(i); allow ± missing
$x = \ln(1.25 + \sqrt{1.25^2 - 1})$	M1
$= \ln 2$	A1F
$\ln \frac{1}{2}$ by symmetry	A1F	Accept $-\ln 2$ written straight down. Alternatively, if solved by using $e^{2x} - 2.5e^x + 1 = 0$, allow M1 for $x = \ln\left(\frac{2.5 \pm \sqrt{2.5^2 - 4}}{2}\right)$

### Part (a)
LHS = $\frac{1}{4}(e^x + e^{-x})^2 - \frac{1}{4}(e^x - e^{-x})^2$ | M1 | Correct expansion of either square
Shown equal to 1 | A1, A1 | 3 marks total | AG

### Part (b)(i)
$8\cosh^2 x - 3$ | B1 | 1 mark

### Part (b)(ii)
Sketch of $y = \cosh x$ | B1 | 1 mark | Must cross y-axis above x-axis

### Part (b)(iii)
$\cosh x = (\pm)1.25$ | B1F | OE; ft errors in (b)(i); allow ± missing
$x = \ln(1.25 + \sqrt{1.25^2 - 1})$ | M1 | 
$= \ln 2$ | A1F | 
$\ln \frac{1}{2}$ by symmetry | A1F | Accept $-\ln 2$ written straight down. Alternatively, if solved by using $e^{2x} - 2.5e^x + 1 = 0$, allow M1 for $x = \ln\left(\frac{2.5 \pm \sqrt{2.5^2 - 4}}{2}\right)$ | 4 marks total

Show LaTeX source

1 (a) Use the definitions $\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)$ and $\sinh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right)$ to show that

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

(b) (i) Express

$$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x$$

in terms of $\cosh x$.\\
(ii) Sketch the curve $y = \cosh x$.\\
(iii) Hence solve the equation

$$5 \cosh ^ { 2 } x + 3 \sinh ^ { 2 } x = 9.5$$

giving your answers in logarithmic form.

\hfill \mbox{\textit{AQA FP2 2010 Q1 [9]}}

This paper (8 questions)

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Q1 9 Q2 8 Q3 14 Q4 10 Q5 8 Q6 6 Q7 8 Q8 12

Answer	Marks	Guidance
LHS = \(\frac{1}{4}(e^x + e^{-x})^2 - \frac{1}{4}(e^x - e^{-x})^2\)	M1	Correct expansion of either square
Shown equal to 1	A1, A1	3 marks total

Answer	Marks	Guidance
\(\cosh x = (\pm)1.25\)	B1F	OE; ft errors in (b)(i); allow ± missing
\(x = \ln(1.25 + \sqrt{1.25^2 - 1})\)	M1
\(= \ln 2\)	A1F
\(\ln \frac{1}{2}\) by symmetry	A1F	Accept \(-\ln 2\) written straight down. Alternatively, if solved by using \(e^{2x} - 2.5e^x + 1 = 0\), allow M1 for \(x = \ln\left(\frac{2.5 \pm \sqrt{2.5^2 - 4}}{2}\right)\)