If the range of operatorname{sech}^{-1} x + o | vedclass

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(C) Let $f(x) = \operatorname{sech}^{-1} x + \operatorname{cosech}^{-1} x$. The domain of $\operatorname{sech}^{-1} x$ is $x \in (0, 1]$. The domain o

Vedclass · Accepted Answer

$a=\log (1+\sqrt{2}), b=\infty$

Vedclass · Answer

(C) Let $f(x) = \operatorname{sech}^{-1} x + \operatorname{cosech}^{-1} x$. The domain of $\operatorname{sech}^{-1} x$ is $x \in (0, 1]$. The domain of $\operatorname{cosech}^{-1} x$ is $x \in \mathbb{R} \setminus \{0\}$. The domain of $f(x)$ is the intersection of these domains, which is $x \in (0, 1]$. We know that $\operatorname{sech}^{-1} x = \ln\left(\frac{1+\sqrt{1-x^2}}{x}ight)$ and $\operatorname{cosech}^{-1} x = \ln\left(\frac{1+\sqrt{1+x^2}}{x}ight)$. As $x 	o 0^+$, $f(x) 	o \infty$. At $x = 1$, $f(1) = \operatorname{sech}^{-1}(1) + \operatorname{cosech}^{-1}(1) = 0 + \ln(1+\sqrt{2}) = \ln(1+\sqrt{2})$. Since $f(x)$ is a strictly decreasing function on $(0, 1]$, the range is $[\ln(1+\sqrt{2}), \infty)$. Thus, $a = \ln(1+\sqrt{2})$ and $b = \infty$.

If the range of $\operatorname{sech}^{-1} x + \operatorname{cosech}^{-1} x$ is $[a, b]$, then

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