$I f f : R \to (0, 1] d e f i n e d b y f (x) = \frac{1}{x^{2} + 1}, t h e n f i s$

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Not one – one

Not onto

Not one – one but onto

One – one but not onto

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{x^{2} + 1} a n d f : R \to (0, 1] \\ L e t f (x_{1}) = f (x_{2}) \forall x_{1}, x_{2} \in R \\ \Rightarrow \frac{1}{x_{1}^{2} + 1} = \frac{1}{x_{2}^{2} + 1} \\ \Rightarrow x_{1}^{2} + 1 = x_{2}^{2} + 1 \\ \Rightarrow x_{1}^{2} = x_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} \\ ∴ f i s n o t o n e - o n e \\ L e t \frac{1}{x^{2} + 1} = y \\ \Rightarrow x^{2} + 1 = \frac{1}{y} \\ \Rightarrow x^{2} = \frac{1}{y} - 1 \\ \Rightarrow x = \sqrt{\frac{1}{y} - 1} \end{array}$

$\begin{array}{l} i s d e f i n e d i f \frac{1}{y} - 1 \geq 0 \\ \Rightarrow \frac{1}{y} \geq 1 \\ \Rightarrow \frac{1}{y} \geq 1 \\ \Rightarrow y \leq 1 a n d y > 0 \\ ∴ R a n g e = (0, 1] \\ ∴ R a n g e = C o d o m a i n \\ ∴ f i s o n t o \end{array}$