The function f:(-infty,infty)rightarrow(-infty,1)$f:(-\infty,\infty)\rightarrow(-\infty,1)$, defined by f(x)=frac2^x-2^-x2^x+2^-x$f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}$ is: [cite: 3251, 3252]

### Related Formula A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain. ### Core Logic Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$: f(x) = frac2^2x - 12^2x + 1 = 1 - frac22^2x + 1$$f(x) = \frac{2^{2x} - 1}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$$ ### Step 1: Check One-One property Differentiating f(x)$f(x)$ with respect to x$x$ : f'(x) = frac2(2^2x + 1)^2 cdot 2 cdot 2^2x cdot ln 2 = frac4 cdot 2^2x cdot ln 2(2^2x + 1)^2$$f'(x) = \frac{2}{(2^{2x} + 1)^2} \cdot 2 \cdot 2^{2x} \cdot \ln 2 = \frac{4 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2}$$ Since 2^2x > 0$2^{2x} > 0$ and ln 2 > 0$\ln 2 > 0$, f'(x) > 0$f'(x) > 0$ always. Thus, f(x)$f(x)$ is strictly increasing, confirming it is a one-one function. ### Step 2: Check Onto property Analyze the limits at boundaries [cite: 3880, 3881]: lim_x to -infty f(x) = 1 - frac20 + 1 = -1$$\lim_{x \to -\infty} f(x) = 1 - \frac{2}{0 + 1} = -1$$ lim_x to infty f(x) = 1 - 0 = 1$$\lim_{x \to \infty} f(x) = 1 - 0 = 1$$ Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-infty, 1)$(-\infty, 1)$ and textRange neq textCo-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto. ### Pattern Recognition The expression given is a shifted form of the hyperbolic tangent function tanh(x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-infty, 1)$(-\infty, 1)$ non-surjective (not onto). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions