Standard +0.8 This is a standard Further Maths inverse hyperbolic function proof requiring substitution of the exponential definition, manipulation into a quadratic in e^y, and solving using the quadratic formula. While methodical, it requires careful algebraic manipulation and selecting the correct root based on the constraint y < 0, making it moderately challenging but still a textbook-style derivation.
2. Given that
$$\cosh y = x \quad \text { and } \quad y < 0$$
use the definition of coshy in terms of exponential functions to prove that
$$y = \ln \left( x - \sqrt { x ^ { 2 } - 1 } \right)$$
Uses quadratic formula or completing the square to solve for \(e^y\)
\(= x \pm \sqrt{x^2-1}\)
A1
Correct solution(s) for \(e^y\). Accept if only negative one is given. Accept \(\frac{2x\pm\sqrt{4x^2-4}}{2}\)
So \(y = \ln\left[x-\sqrt{x^2-1}\right]\)*
A1*
Completely correct work leading to given answer. Must be no errors seen.
since \(y<0 \Rightarrow e^y<1\) so need \(x-\sqrt{x^2-1}\) (as \(x>1\) so must subtract)
B1
Suitable justification for taking negative root. Can only be awarded if all previous marks awarded.
(6)
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cosh y = x \Rightarrow x = \frac{e^y+e^{-y}}{2}$ | B1 | Correct statement for $x$ in terms of exponentials |
| $\Rightarrow 2xe^y = e^{2y}+1$ | M1 | Multiplies through by $e^y$ to achieve a quadratic in $e^y$ |
| $\Rightarrow e^{2y}-2xe^y+1=0 \Rightarrow e^y = \frac{2x\pm\sqrt{(2x)^2-4\times1\times1}}{2}$ | M1 | Uses quadratic formula or completing the square to solve for $e^y$ |
| $= x \pm \sqrt{x^2-1}$ | A1 | Correct solution(s) for $e^y$. Accept if only negative one is given. Accept $\frac{2x\pm\sqrt{4x^2-4}}{2}$ |
| So $y = \ln\left[x-\sqrt{x^2-1}\right]$* | A1* | Completely correct work leading to given answer. Must be no errors seen. |
| since $y<0 \Rightarrow e^y<1$ so need $x-\sqrt{x^2-1}$ (as $x>1$ so must subtract) | B1 | Suitable justification for taking negative root. **Can only be awarded if all previous marks awarded.** |
| | **(6)** | |
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2. Given that
$$\cosh y = x \quad \text { and } \quad y < 0$$
use the definition of coshy in terms of exponential functions to prove that
$$y = \ln \left( x - \sqrt { x ^ { 2 } - 1 } \right)$$
\hfill \mbox{\textit{Edexcel F3 2021 Q2 [6]}}