The function f: (-infty, infty) rightarrow (- | vedclass

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(A) Given $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$.Multiplying numerator and denominator by $2^x$,we get $f(x) = \frac{2^{2x} - 1}{2^{2x} + 1}$.We c

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One-one but not onto

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(A) Given $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$.Multiplying numerator and denominator by $2^x$,we get $f(x) = \frac{2^{2x} - 1}{2^{2x} + 1}$.We can rewrite this as $f(x) = \frac{(2^{2x} + 1) - 2}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$.To check for one-one,we find the derivative: $f'(x) = \frac{d}{dx} (1 - 2(2^{2x} + 1)^{-1}) = 0 - 2(-1)(2^{2x} + 1)^{-2} \cdot (2^{2x} \cdot \ln 2 \cdot 2) = \frac{4 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2}$.Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function is strictly increasing,hence it is one-one.To check for onto,we find the range: As $x \rightarrow -\infty$,$2^{2x} \rightarrow 0$,so $f(x) \rightarrow 1 - \frac{2}{0+1} = -1$. As $x \rightarrow \infty$,$2^{2x} \rightarrow \infty$,so $f(x) \rightarrow 1 - 0 = 1$.The range of $f(x)$ is $(-1, 1)$.Since the range $(-1, 1)$ is not equal to the codomain $(-\infty, \infty)$,the function is not onto.Therefore,the function is one-one but not onto.

List-$I$	List-$II$
$(A)$ $f(-5) + f(-4)$	$(i)$ $14$
$(B)$ $f(\|f(-8)\|)$	$(ii)$ $4$
$(C)$ $f(f(-7) + f(3))$	$(iii)$ $-11$
$(D)$ $f(f(f(f(0)))) + 1$	$(iv)$ $-1$
	$(v)$ $1$
	$(vi)$ $0$

The function $f: (-\infty, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is :

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