Question

If \( f(x) \) is a function such that \( f'(x) = \sqrt{f^2(x) - 1} \) and \( f(0) = 1 \), then \( f(1) = \):

• A. \( \frac{e^{-2} + 1}{2e} \)
• B. \( \frac{e^{2} + 1}{2e} \)
• C. \( \frac{e^{2} - 1}{2e} \)
• D. \( \frac{e^{-2} - 1}{2e} \)

Hint:

The integral \( \int \frac{du}{\sqrt{u^2 - 1}} = \cosh^{-1}(u) + C \). The hyperbolic cosine function is \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

Answer 1

The Correct Option is B

Step 1: Separate the variables.
\( \frac{df}{\sqrt{f^2 - 1}} = dx \)
Step 2: Integrate both sides.
\( \int \frac{df}{\sqrt{f^2 - 1}} = \int dx \implies \cosh^{-1}(f) = x + C \)
Step 3: Use the initial condition \( f(0) = 1 \).
\( \cosh^{-1}(1) = 0 + C \implies 0 = C \)
So, \( \cosh^{-1}(f(x)) = x \implies f(x) = \cosh(x) \)
Step 4: Find \( f(1) \). \( f(1) = \cosh(1) = \frac{e^1 + e^{-1}}{2} = \frac{e + \frac{1}{e}}{2} = \frac{e^2 + 1}{2e} \)