Standard +0.8 This is a standard Further Maths proof requiring students to manipulate hyperbolic functions using exponential definitions and solve for y. While it involves multiple algebraic steps (substitution, rearranging, taking logarithms), it follows a well-established method taught in FP2 with no novel insight required. It's moderately harder than average due to being Further Maths content and requiring careful algebraic manipulation, but it's a textbook-style proof that students specifically prepare for.
# Question 1:
$\tanh^{-1} x = y \Rightarrow x = \tanh y = \dfrac{e^{2y}-1}{e^{2y}+1}$ | M1 | Correct use of exponential form; a muddle of $x$ and $y$ unless recovered is M0
$(e^{2y}+1)x = e^{2y}-1$, then $e^{2y}(1-x) = (1+x)$, giving $e^{2y} = \dfrac{1+x}{1-x}$ | A1 | Correct expression for $e^{2y}$
$2y = \ln\left(\dfrac{1+x}{1-x}\right) \Rightarrow \left(y = \tanh^{-1}x\right) = \dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)$ | A1 | ag
---