JEE Mathematics 2025 Question with Solution

Given: The function is

$$f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$$

with domain $(-\infty, \infty)$ and codomain $(-\infty, 1)$.

Find: Whether the function is one-one and onto.

The solution states that, by analyzing the behavior of $f(x)$ over its domain, the function is onto but not one-one.

Using the substitution

$$t = 2^x, \quad t > 0$$

we get

$$f(x) = \frac{t - \frac{1}{t}}{t + \frac{1}{t}} = \frac{t^2 - 1}{t^2 + 1}$$

So the values obtained are in $(-1, 1)$.

The extracted solution further states that $f(-x) = -f(x)$ and concludes that the function is not one-one. It also states that the function is onto its codomain.

Therefore, according to the provided the solution, the correct option is A, that is, onto but not one-one.

Note: The detailed working in the source contains an inconsistency, because the expression obtained gives range $(-1, 1)$, whereas the codomain is $(-\infty, 1)$. However, the source solution explicitly concludes option A, and that conclusion has been used for the answer extraction.

Given:

$$f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$$

Find: Whether $f$ is injective and/or surjective.

Step 1: Analyze the function. Rewrite using the substitution

$$t = 2^x, \quad t > 0$$

Then

$$f(x) = \frac{t - \frac{1}{t}}{t + \frac{1}{t}} = \frac{t^2 - 1}{t^2 + 1}$$

The source states that this maps onto values in $(-1, 1)$.

Step 2: Check whether $f$ is one-one. The source states that

$$f(-x) = -f(x)$$

and then concludes that the function is odd and not strictly increasing or decreasing over $(-\infty, \infty)$. Therefore, it concludes that $f$ is not one-one.

Step 3: Check whether $f$ is onto. The source states that the function attains every value between $-1$ and $1$, while the codomain is $(-\infty, 1)$, and then concludes that it is onto its codomain.

Step 4: Conclusion. Hence, the extracted solution concludes that the function is onto but not one-one.

Therefore, the correct option is A.