Question:

A real solution of the equation \[ \cosh x - 5 \sinh x - 5 = 0 \text{ is} \]

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Fission reactors produce energy by splitting heavy nuclei, releasing large amounts of energy.
Updated On: Mar 30, 2025
  • \(-\ln 2\)
  • \(\ln 2\)
  • \(-\ln 5\)
  • None of these
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The Correct Option is A

Solution and Explanation


Use identity: \[ \cosh x = \frac{e^x + e^{-x}}{2},\quad \sinh x = \frac{e^x - e^{-x}}{2} \] Then: \[ \cosh x - 5 \sinh x = \frac{e^x + e^{-x}}{2} - 5 \cdot \frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x} - 5e^x + 5e^{-x}}{2} = \frac{-4e^x + 6e^{-x}}{2} \] Set this equal to 5: \[ \frac{-4e^x + 6e^{-x}}{2} = 5 \Rightarrow -4e^x + 6e^{-x} = 10 \] Multiply by \(e^x\): \[ -4e^{2x} + 6 = 10e^x \Rightarrow 4e^{2x} + 10e^x - 6 = 0 \Rightarrow \text{Solve to find } x = -\ln 2 \]
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