The number of roots of the equation sqrt{2}+e | vedclass

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(B) Identify the Domain and Use Logarithmic Forms:First, note that the function $\cosh^{-1} x$ is only defined for $x \geq 1$. Therefore, we must have

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$1$

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(B) Identify the Domain and Use Logarithmic Forms:First, note that the function $\cosh^{-1} x$ is only defined for $x \geq 1$. Therefore, we must have $x \geq 1$.We use the logarithmic definitions of inverse hyperbolic functions:$e^{\cosh^{-1} x} = x + \sqrt{x^2 - 1}$$e^{\sinh^{-1} x} = x + \sqrt{x^2 + 1}$Substitute and Simplify the Equation:Substitute these expressions back into the original equation:$\sqrt{2} + (x + \sqrt{x^2 - 1}) - (x + \sqrt{x^2 + 1}) = 0$The $x$ terms cancel out, leaving:$\sqrt{2} + \sqrt{x^2 - 1} - \sqrt{x^2 + 1} = 0$$\sqrt{2} + \sqrt{x^2 - 1} = \sqrt{x^2 + 1}$Solve for $x$:Square both sides of the equation to eliminate the radicals:$(\sqrt{2} + \sqrt{x^2 - 1})^2 = (\sqrt{x^2 + 1})^2$$2 + (x^2 - 1) + 2\sqrt{2}\sqrt{x^2 - 1} = x^2 + 1$$x^2 + 1 + 2\sqrt{2}\sqrt{x^2 - 1} = x^2 + 1$Subtract $x^2 + 1$ from both sides:$2\sqrt{2}\sqrt{x^2 - 1} = 0$$\sqrt{x^2 - 1} = 0 \implies x^2 = 1 \implies x = \pm 1$Since the domain requires $x \geq 1$, the only valid solution is $x = 1$. Thus, there is $1$ root.

The number of roots of the equation $\sqrt{2}+e^{\cosh^{-1} x}-e^{\sinh^{-1} x}=0$ is

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