Q39 / 75 · JEE Main 2025 (24 Jan Shift 2) · Mathematics

$\text { The function } f:(-\infty, \infty) \rightarrow(-\infty, 1) \text {, defined by } f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}} \text { is : }$

  • A
    $\text { Neither one-one nor onto }$
  • B
    $\text { Onto but not one-one }$
  • C
    $\text { Both one-one and onto }$
  • Correct answer
    D
    $\text { One-one but not onto }$

Solution

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \\ & =1-\frac{2}{2^{2 \mathrm{x}}+1} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2.2^{2 \mathrm{x}} \cdot \ln 2 \text { i.e always }+\mathrm{ve} \end{aligned}$ so $f(x)$ is $\uparrow$ function $\begin{aligned} & \therefore \mathrm{f}(-\infty)=-1 \\ & \mathrm{f}(\infty)=1 \\ & \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text { co-domain } \end{aligned}$ so function is one-one but not onto

Hence, the correct answer is option: 'D'.

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