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The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is ........
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Verified Answer
The correct answer is:
Injective but not surjective
\(\begin{aligned}
& f(x)=\frac{x}{\sqrt{1+x^2}} \\
& f\left(x_1\right)=f(x)_2 \Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}}
\end{aligned}\)
\(\begin{array}{rlrl}
\Rightarrow & & x_1^2\left(1+x_2^2\right) & =x_2^2\left(1+x_1^2\right) \\
\Rightarrow & & x_1^2 & =x_2^2 \\
x_1 & =x_2
\end{array}\)
\(\therefore F\) 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.
& 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.
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