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(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.Integrating both sides with respect to $x

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$\frac{e^2+1}{2 e}$

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(B) Given $f^{\prime}(x) = \sqrt{f^2(x)-1}$.We can write this as $\frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} = 1$.Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{\sqrt{f^2(x)-1}} dx = \int 1 dx$.Using the standard integral $\int \frac{1}{\sqrt{t^2-1}} dt = \log |t + \sqrt{t^2-1}| + C$,we have $\log |f(x) + \sqrt{f^2(x)-1}| = x + C$.Given $f(0) = 1$,substituting $x=0$ gives $\log |f(0) + \sqrt{f^2(0)-1}| = 0 + C$.$\log |1 + \sqrt{1-1}| = C \Rightarrow \log(1) = C \Rightarrow C = 0$.So,$\log |f(x) + \sqrt{f^2(x)-1}| = x$.At $x=1$,$\log |f(1) + \sqrt{f^2(1)-1}| = 1$.Taking the exponential of both sides,$f(1) + \sqrt{f^2(1)-1} = e^1 = e$.$\sqrt{f^2(1)-1} = e - f(1)$.Squaring both sides,$f^2(1) - 1 = e^2 + f^2(1) - 2ef(1)$.$-1 = e^2 - 2ef(1) \Rightarrow 2ef(1) = e^2 + 1$.Thus,$f(1) = \frac{e^2+1}{2e}$.

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

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