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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Functions

The function f:(-∞, ∞) arrow(-∞, 1) , defined by f(x)= 2^x-2^ -x...

 The function f:(−∞,∞)→(−∞,1), defined by f(x)=2x−2−x2x+2−x is  : \text{ The function } f:(-\infty, \infty) \rightarrow(-\infty, 1) \text{, defined by } f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}} \text{\text{ is } : }

Options

  1. A Neither one-one nor onto \text{ Neither one-one nor onto }
  2. B Onto but not one-one \text{ Onto but not one-one }
  3. C Both one-one and onto \text{ Both one-one and onto }
  4. D One-one but not onto \text{ One-one but not onto }

Correct answer

D.  One-one but not onto \text{ One-one but not onto }

Step-by-step solution

f(x)=22x−122x+1=1−222x+1f′(x)=2(22x+1)2⋅2.22x⋅ln⁡2 i.e always +ve\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)f(x) is ↑\uparrow function ∴f(−∞)=−1f(∞)=1∴f(x)∈(−1,1)≠ co-domain \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
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