Key Concepts and Formulas

• One-one (Injective) Function: A function $f: A \rightarrow B$ is one-one if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Alternatively, if $f'(x) > 0$ or $f'(x) < 0$ for all $x$ in the domain, the function is one-one.
• Onto (Surjective) Function: A function $f: A \rightarrow B$ is onto if its range is equal to its codomain. The range is the set of all possible output values of the function.
• Limits at Infinity: To find the range of a continuous function over an infinite domain, we often evaluate the limits of the function as the variable approaches $\pm \infty$.

Step-by-Step Solution

Step 1: Simplify the function and analyze its domain and codomain.

The given function is $f:(-\infty, \infty) \rightarrow(-\infty, 1)$, defined by $f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}$. The domain is $(-\infty, \infty)$ and the codomain is $(-\infty, 1)$. To simplify, multiply the numerator and denominator by $2^x$: $f(x) = \frac{2^x-2^{-x}}{2^x+2^{-x}} \times \frac{2^x}{2^x} = \frac{2^{2x} - 1}{2^{2x} + 1}$ This simplified form is easier to work with. We can also rewrite it as: $f(x) = \frac{(2^{2x} + 1) - 2}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$

Step 2: Check for Injectivity (One-one).

We will use the derivative test. First, we find the derivative of $f(x)$ using the simplified form $f(x) = 1 - \frac{2}{2^{2x} + 1}$. $f'(x) = \frac{d}{dx} \left( 1 - 2(2^{2x} + 1)^{-1} \right)$ $f'(x) = 0 - 2 \cdot (-1)(2^{2x} + 1)^{-2} \cdot \frac{d}{dx}(2^{2x} + 1)$ Using the chain rule for $2^{2x}$: $\frac{d}{dx}(a^{u(x)}) = a^{u(x)} \ln a \cdot u'(x)$. $\frac{d}{dx}(2^{2x} + 1) = 2^{2x} \ln 2 \cdot \frac{d}{dx}(2x) + 0 = 2^{2x} \ln 2 \cdot 2 = 2 \cdot 2^{2x} \ln 2$ Substituting this back into the expression for $f'(x)$: $f'(x) = 2 (2^{2x} + 1)^{-2} \cdot (2 \cdot 2^{2x} \ln 2)$ $f'(x) = \frac{4 \cdot 2^{2x} \ln 2}{(2^{2x} + 1)^2}$ Now, let's analyze the sign of $f'(x)$. For any real number $x$, $2^{2x} > 0$. Therefore, $4 > 0$, $2^{2x} > 0$, and $\ln 2 > 0$. The denominator $(2^{2x} + 1)^2$ is also always positive since $2^{2x} + 1 > 1$. Thus, $f'(x) > 0$ for all $x \in (-\infty, \infty)$. Since the derivative is always positive, the function $f(x)$ is strictly increasing over its entire domain. A strictly increasing function is always one-one. Therefore, $f(x)$ is one-one.

Step 3: Check for Surjectivity (Onto).

To determine if $f(x)$ is onto, we need to find its range and compare it with the codomain $(-\infty, 1)$. Since $f(x)$ is strictly increasing, its range will be determined by the limits as $x \to -\infty$ and $x \to \infty$. We use the simplified form $f(x) = \frac{2^{2x} - 1}{2^{2x} + 1}$.

• Limit as $x \to -\infty$: As $x \to -\infty$, $2x \to -\infty$. Let $y = 2x$. Then as $x \to -\infty$, $y \to -\infty$. We need to evaluate $\lim_{y \to -\infty} \frac{2^y - 1}{2^y + 1}$. As $y \to -\infty$, $2^y \to 0$. So, $\lim_{x \to -\infty} f(x) = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1$.

• Limit as $x \to \infty$: As $x \to \infty$, $2x \to \infty$. Let $y = 2x$. Then as $x \to \infty$, $y \to \infty$. We need to evaluate $\lim_{y \to \infty} \frac{2^y - 1}{2^y + 1}$. To evaluate this, we can divide the numerator and denominator by $2^y$: $\lim_{y \to \infty} \frac{1 - 2^{-y}}{1 + 2^{-y}}$ As $y \to \infty$, $2^{-y} \to 0$. So, $\lim_{x \to \infty} f(x) = \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1$.

Since $f(x)$ is strictly increasing, its range is the interval $(-1, 1)$. The given codomain is $(-\infty, 1)$. The range of the function is $(-1, 1)$, which is a subset of the codomain $(-\infty, 1)$, but not equal to it. Specifically, the values in $(-\infty, -1]$ are not included in the range. Therefore, the function $f(x)$ is not onto.

Common Mistakes & Tips

• Simplification Errors: Be careful when multiplying by $2^x$ to ensure all terms are handled correctly, especially $2^{-x} \times 2^x = 2^0 = 1$.
• Derivative of Exponential Functions: Remember the formula for the derivative of $a^{u(x)}$ is $a^{u(x)} \ln a \cdot u'(x)$. For $2^{2x}$, this is $2^{2x} \ln 2 \cdot 2$.
• Range vs. Codomain: The range must be exactly equal to the codomain for the function to be onto. If the range is a proper subset of the codomain, the function is not onto.

Summary

We analyzed the function $f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}$ by first simplifying it to $f(x) = \frac{2^{2x}-1}{2^{2x}+1}$. We then used the derivative test to prove that $f'(x) > 0$ for all $x$, establishing that the function is strictly increasing and therefore one-one. To check for surjectivity, we computed the limits of $f(x)$ as $x \to -\infty$ and $x \to \infty$, which gave us the range $(-1, 1)$. Since this range is not equal to the given codomain $(-\infty, 1)$, the function is not onto. Thus, the function is one-one but not onto.

The final answer is \boxed{A}.