The function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is ........

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\frac{x}{\sqrt{1+x^2}} \ & f\left(x_1\right)=f(x)_2 \Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}} \end{aligned}$ $\begin{array}{rlrl} \Rightarrow & & x_1^2\left(1+x_2^2\right) & =x_2^2\left(1+x_1^2\right) \ \Rightarrow & & x_1^2 & =x_2^2 \ x_1 & =x_2 \end{array}$ $\therefore F$ is injective in nature Also, Let $f(x)=y=\frac{x}{\sqrt{1+x^2}}$ $\begin{aligned} & \Rightarrow \quad y^2\left(1+x^2\right)=x^2 \ & \Rightarrow \quad \frac{y^2}{1-y^2}=x^2 \Rightarrow x=\frac{y}{\sqrt{1-y^2}} \end{aligned}$ As $y^2 \leq 1 \Rightarrow-1 \leq|y| \leq 1$ So, $f$ is non surjective.

The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is ........

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