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Question
If the function\[f : R \to \text{A given by} f\left( x \right) = \frac{x^2}{x^2 + 1}\] is a surjection, then A =
Options
R
[0, 1]
[0, 1]
[0, 1]
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Solution
\[\text{As f is surjective, range of f = co - domain of}f\]
\[ \Rightarrow A = \text{range of f} \]
\[ \because f\left( x \right) = \frac{x^2}{x^2 + 1}, \]
\[ y = \frac{x^2}{x^2 + 1}\]
\[ \Rightarrow y\left( x^2 + 1 \right) = x^2 \]
\[ \Rightarrow \left( y - 1 \right) x^2 + y = 0\]
\[ \Rightarrow x^2 = \frac{- y}{\left( y - 1 \right)}\]
\[ \Rightarrow x = \sqrt{\frac{y}{\left( 1 - y \right)}}\]
\[ \Rightarrow \frac{y}{\left( 1 - y \right)} \geq 0\]
\[ \Rightarrow y \in [0, 1)\]
\[ \Rightarrow \text{Range of f} = [0, 1)\]
\[ \Rightarrow A = [0, 1)\]
So, the answer is (d) .
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