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 =

 

 

विकल्प

• R

• [0, 1]

• [0, 1]

• [0, 1]

Answer 1

\[\text{As f is surjective, range of f = co - domain of}f\] 
\[ \Rightarrow A = \text{range of f} \] 
\[ \because f\left( x \right) = \frac{x^2}{x^2 + 1}, \] 
\[ y = \frac{x^2}{x^2 + 1}\] 
\[ \Rightarrow y\left( x^2 + 1 \right) = x^2 \] 
\[ \Rightarrow \left( y - 1 \right) x^2 + y = 0\] 
\[ \Rightarrow x^2 = \frac{- y}{\left( y - 1 \right)}\] 
\[ \Rightarrow x = \sqrt{\frac{y}{\left( 1 - y \right)}}\] 
\[ \Rightarrow \frac{y}{\left( 1 - y \right)} \geq 0\] 
\[ \Rightarrow y \in [0, 1)\] 
\[ \Rightarrow \text{Range of f} = [0, 1)\] 
\[ \Rightarrow A = [0, 1)\] 

So, the answer is (d) .