Match the functions given in List-I with thei | vedclass

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(D) $\sinh x = \frac{e^x - e^{-x}}{2}$. Since $\sinh(-x) = \frac{e^{-x} - e^x}{2} = -\sinh x$, it is an odd function with range $\mathbb{R}$. Thus, $A

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$A-IV, B-III, C-V, D-II$

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(D) $\sinh x = \frac{e^x - e^{-x}}{2}$. Since $\sinh(-x) = \frac{e^{-x} - e^x}{2} = -\sinh x$, it is an odd function with range $\mathbb{R}$. Thus, $A-IV$.$(B)$ $	ext{sech } x = \frac{2}{e^x + e^{-x}}$. Since $	ext{sech}(-x) = \frac{2}{e^{-x} + e^x} = 	ext{sech } x$, it is an even function. Thus, $B-III$.$(C)$ $	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. Since $	anh(-x) = -	anh x$, it is an odd function with range $(-1, 1)$. Thus, $C-V$.$(D)$ $	ext{cosech}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}ight)$. The domain is $x 
eq 0$ and it is neither even nor odd. Thus, $D-II$.Therefore, the correct matching is $A-IV, B-III, C-V, D-II$.

List-$I$	List-$II$
$(A)$ $\sinh x$	$(I)$ Domain is $(-1, 1)$, even function
$(B)$ $\text{sech } x$	$(II)$ Domain is $[1, \infty)$, neither even nor odd function
$(C)$ $\tanh x$	$(III)$ Even function
$(D)$ $\text{cosech}^{-1} x$	$(IV)$ Range is $\mathbb{R}$, odd function
	$(V)$ Range is $(-1, 1)$, odd function

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