$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \
& =1-\frac{2}{2^{2 \mathrm{x}}+1} \
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2.2^{2 \mathrm{x}} \cdot \ln 2 \text { i.e always }+\mathrm{ve}
\end{aligned}sof(x)is\uparrowfunction\begin{aligned}
& \therefore \mathrm{f}(-\infty)=-1 \
& \mathrm{f}(\infty)=1 \
& \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text { co-domain }
\end{aligned}$ so function is one-one but not onto
The function f:(-∞, ∞) →(-∞, 1) , defined by f(x)= 2^x-2^-x 2^x+2^-x is :
JEE Main 2025 — Mathematics Algebra
2025mcqeasy
The function f:(−∞,∞)→(−∞,1), defined by f(x)=2x+2−x2x−2−x is :
Official previous-year question
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
Neither one-one nor onto
- B
Onto but not one-one
- C
Both one-one and onto
- D
One-one but not onto
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