A2 June 2023 Paper 1 Q12
12
(a) Starting from the identities for \(\sinh 2x\) and \(\cosh 2x\), prove the identity\[\tanh 2x = \frac{2\tanh x}{1 + \tanh^2 x}\] [2 marks]
(b)
(i) The function \(\mathrm{f}\) is defined by\[\mathrm{f}(x) = \tanh x \qquad (x \gt 0)\]
State the range of \(\mathrm{f}\) [1 mark]
(ii) Use part (a) and part (b)(i) to prove that \(\tanh 2x \gt \tanh x\) if \(x \gt 0\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses hyperbolic identities correctly to express \(\tanh 2x\) as a quotient | M1 | 1.1a |
| Completes a reasoned argument to obtain the required result | R1 | 2.1 |
| (2) |
Typical solution
\[\tanh 2x = \frac{\sinh 2x}{\cosh 2x} = \frac{2\sinh x\cosh x}{\cosh^2 x + \sinh^2 x}\]\[= \frac{\dfrac{2\sinh x\cosh x}{\cosh^2 x}}{\dfrac{\cosh^2 x + \sinh^2 x}{\cosh^2 x}} = \frac{2\tanh x}{1 + \tanh^2 x}\]| Scheme | Marks | AO |
|---|---|---|
| (i) Correctly states range ACF | B1 | 1.2 |
| (1) | ||
| (ii) Uses range of f to show that \(1 + \tanh^2 x \lt 2\) | M1 | 3.1a |
| (ii) Deduces that \(\dfrac{2}{1 + \tanh^2 x} \gt 1\) | M1 | 2.2a |
| (ii) Completes a reasoned argument to obtain the required result Working backwards from the result to the range of \(\tanh x\) scores 0/3 | R1 | 2.1 |
| (3) | ||
| (6 marks) |
Typical solution
(i)
\[0 \lt \mathrm{f}(x) \lt 1\](ii)
For \(x \gt 0\),
\[0 \lt \tanh x \lt 1 \Rightarrow 1 \lt 1 + \tanh^2 x \lt 2\]\[\therefore \frac{2}{1 + \tanh^2 x} \gt 1\]and
\[\frac{2\tanh x}{1 + \tanh^2 x} \gt \tanh x\]\[\therefore \tanh 2x \gt \tanh x\]