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JEE MAIN 2019
09-04-2019 - S1
Question
If the function $f: R-\{1,-1\} \rightarrow A$ defined by $f(x)=\frac{x^2}{1-x^2}$, is surjective, then $A$ is equal to :
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A
B
C
D
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Question Tags
JEE Main
Mathematics
Medium
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JEE Advanced 2020
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Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined by $\mathrm{f}(\mathrm{x})=\frac{4^{\mathrm{x}}}{4^{\mathrm{x}}+2}$ Then the value of $\mathrm{f}\left(\frac{1}{40}\right)+\mathrm{f}\left(\frac{2}{40}\right)+\mathrm{f}\left(\frac{3}{40}\right)+\ldots \ldots \ldots+\mathrm{f}\left(\frac{39}{40}\right)-\mathrm{f}\left(\frac{1}{2}\right)$ is
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Let 𝒇:ℝ→ℝ and 𝒈:ℝ→ℝ be functions satisfying 𝑓(𝑥+𝑦)=𝑓(𝑥)+𝑓(𝑦)+𝑓(𝑥)𝑓(𝑦) and 𝑓(𝑥)=𝑥𝑔(𝑥) for all 𝑥,𝑦∈ℝ. If $\lim _{x \rightarrow 0} g(x)=1$, then...
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