Question

If \[ \sinh x=-\frac12,\qquad \cosh y=2, \] and \[ x+y=\log p, \] then \(p=\)

• A. \(\dfrac{4+2\sqrt3}{\sqrt5+1}\)
• B. \(\dfrac{\sqrt5-1}{4\sqrt3+2}\)
• C. \((\sqrt5-1)(2+\sqrt3)\)
• D. \(\dfrac{4-2\sqrt3}{\sqrt5+1}\)

Hint:

For hyperbolic function equations, substitute \(e^x=t\) or \(e^y=t\) to obtain a quadratic equation. Then use \(e^{x+y}=e^x e^y\) whenever logarithms are involved.

Answer 1

The Correct Option is A

Concept: Use the definitions \[ \sinh x=\frac{e^x-e^{-x}}{2}, \qquad \cosh y=\frac{e^y+e^{-y}}{2}, \] to find \(e^x\) and \(e^y\). Then use \[ x+y=\log p \] which implies \[ p=e^{x+y}=e^x e^y. \]

Step 1: Find \(e^x\). Given \[ \sinh x=-\frac12. \] So, \[ \frac{e^x-e^{-x}}{2} = -\frac12. \] \[ e^x-e^{-x}=-1. \] Let \[ t=e^x. \] Then \[ t-\frac1t=-1. \] \[ t^2+t-1=0. \] Since \(t=e^x\gt 0\), \[ t=\frac{\sqrt5-1}{2}. \] Hence, \[ e^x=\frac{\sqrt5-1}{2}. \]

Step 2: Find \(e^y\). Given \[ \cosh y=2. \] Thus, \[ \frac{e^y+e^{-y}}{2}=2. \] \[ e^y+e^{-y}=4. \] Let \[ u=e^y. \] Then \[ u+\frac1u=4. \] \[ u^2-4u+1=0. \] \[ u=2\pm\sqrt3. \] Taking the principal value \(y\gt 0\), \[ e^y=2+\sqrt3. \]

Step 3: Find \(p\). Since \[ x+y=\log p, \] \[ p=e^{x+y}=e^x e^y. \] Therefore, \[ p= \frac{\sqrt5-1}{2}(2+\sqrt3). \] Rationalizing, \[ p= \frac{(\sqrt5-1)(2+\sqrt3)(\sqrt5+1)} {2(\sqrt5+1)}. \] Using \[ (\sqrt5-1)(\sqrt5+1)=4, \] \[ p= \frac{4(2+\sqrt3)} {2(\sqrt5+1)} = \frac{4+2\sqrt3}{\sqrt5+1}. \] Therefore, \[ \boxed{p=\frac{4+2\sqrt3}{\sqrt5+1}} \] \[ \boxed{\text{Answer = (A)}} \]