Question

If

\[ \cosh^{-1}\left(\frac{5}{3}\right) + \sinh^{-1}\left(\frac{3}{4}\right) = k, \]

then \( e^k \) is:

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{2}\)
• C. \(6\)
• D. \(5\)

Hint:

When combining exponential terms, especially with hyperbolic functions, ensure proper manipulation of exponential equations to accurately compute values.

Answer 1

The Correct Option is C

Step 1: Simplify the expression for \(k\).
Using the definition of hyperbolic inverse functions: \[ \cosh^{-1}\left(\frac{5}{3}\right) \text{ and } \sinh^{-1}\left(\frac{3}{4}\right) \] \[ \cosh y = \frac{5}{3}, \sinh x = \frac{3}{4}. \] Recall that \(\cosh y = \frac{e^y + e^{-y}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\). Step 2: Determine the values of \(e^y\) and \(e^x\).
Solving these equations, we find: \[ e^y = \frac{5 + \sqrt{25 - 9}}{3} = \frac{5 + 4}{3} = 3, \quad e^{-y} = \frac{1}{3} \] \[ e^x = \frac{3 + \sqrt{9 + 16}}{4} = \frac{3 + 5}{4} = 2, \quad e^{-x} = \frac{1}{2} \] Step 3: Evaluate \(e^k\) using \(k = x + y\).
\[ e^k = e^{x+y} = e^x \times e^y = 2 \times 3 = 6. \]