Paper 1 · November 2010 · hyperbolic functions

Let $y = \sinh^{-1} 2x$ . Writing $\sinh y$ in exponential form and clearing the fraction gives which equation in $e^y$ ?

e^{2y} + 4xe^y - 1 = 0

e^{2y} - 4xe^y - 1 = 0

e^{2y} - 4xe^y + 1 = 0

e^{2y} - 2xe^y - 1 = 0

Explanation

From

y=\sinh^{-1}2x

2x=\sinh y=\dfrac{e^y-e^{-y}}{2}

, so

e^y-e^{-y}=4x

. Multiplying every term by

e^y

gives

e^{2y}-4xe^y-1=0

, a quadratic in

e^y

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q2

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