Function – One one – Onto – JEE Main 24 January 2025 Shift 2

Solution:

First, we simplify the given function by multiplying the numerator and denominator by $2^x$:
$$ f(x) = \frac{2^x(2^x – 2^{-x})}{2^x(2^x + 2^{-x})} = \frac{2^{2x} – 1}{2^{2x} + 1} $$

To check for injectivity (one-one), we find the derivative $f'(x)$. Using the quotient rule:
$$ f'(x) = \frac{(2^{2x}+1)(2 \cdot 2^{2x} \ln 2) – (2^{2x}-1)(2 \cdot 2^{2x} \ln 2)}{(2^{2x}+1)^2} $$

$$ f'(x) = \frac{2 \cdot 2^{2x} \ln 2 \left[ (2^{2x}+1) – (2^{2x}-1) \right]}{(2^{2x}+1)^2} $$

$$ f'(x) = \frac{4 \cdot 2^{2x} \ln 2}{(2^{2x}+1)^2} $$

Since $2^{2x} > 0$ and $\ln 2 > 0$ for all $x \in \mathbb{R}$, we have $f'(x) > 0$. Therefore, $f(x)$ is strictly increasing and **one-one**.

To check for surjectivity (onto), we determine the range of the function. Let $y = f(x)$:
$$ y = \frac{2^{2x} – 1}{2^{2x} + 1} $$

Solving for $2^{2x}$ in terms of $y$:
$$ y(2^{2x} + 1) = 2^{2x} – 1 $$
$$ y \cdot 2^{2x} + y = 2^{2x} – 1 $$
$$ 1 + y = 2^{2x}(1 – y) $$
$$ 2^{2x} = \frac{1+y}{1-y} $$

We know that for any real $x$, the exponential term $2^{2x}$ must be strictly positive ($>0$). Thus:
$$ \frac{1+y}{1-y} > 0 $$

Using the sign scheme for this inequality, we find that $y$ must lie between $-1$ and $1$:
$$ y \in (-1, 1) $$

The calculated Range is $(-1, 1)$. However, the given Co-domain is $(-\infty, 1)$. Since the Range is a proper subset of the Co-domain (Range $\neq$ Co-domain), the function is **not onto**.

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