Question

The function \( f: (-\infty, \infty) \to (-\infty, 1) \), defined by \[ f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}, \] is:

• A. Onto but not one-one
• B. Both one-one and onto
• C. Neither one-one nor onto
• D. One-one but not onto

Hint:

To determine if a function is one-to-one or onto, analyze its behavior over its entire domain and check if each value of the range is mapped from exactly one value in the domain (one-to-one) or if every value in the range is achieved (onto).

Answer 1

The Correct Option is A

We analyze the function \( f(x) \) to determine its injectivity and surjectivity. By considering the behavior of the function for all values of \( x \), we determine that the function is onto but not one-one. 
Final Answer: Onto but not one-one.

Answer 2

Step 1: Analyze the function \( f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}} \).
Rewrite the function using substitution \( t = 2^x \), so \( t > 0 \):
\[ f(x) = \frac{t - \frac{1}{t}}{t + \frac{1}{t}} = \frac{t^2 - 1}{t^2 + 1} \] This means \( f(x) \) maps \( t^2 \in (0, \infty) \) onto values in \((-1, 1)\).

Step 2: Check if \( f \) is one-one (injective).
Since \( f(-x) = -f(x) \), the function is odd and symmetric about the origin, but it is not strictly increasing or decreasing over \((-\infty, \infty)\).
Therefore, \( f \) is not one-one.

Step 3: Check if \( f \) is onto (surjective).
The function attains every value between -1 and 1, while the codomain is \((-\infty, 1)\).
Thus, \( f \) attains all values less than 1, so it is onto its codomain.

Step 4: Conclusion.
Hence, the function \( f \) is <strong>onto but not one-one</strong>.