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Show that the function $f : R \rightarrow R$ defined by $f(x) = \frac{x}{x^{2}+1}, \forall x\in R$ is neither one-one nor onto.

Answers (1)

Given:

$f(x) = \frac{x}{x^{2}+1}, \forall x\in R$

$f : R \rightarrow R$

To prove: f(x) is neither one-one nor onto.

For one-one

$\\ $ Let $ x_{1}, x_{2} \in R$ and

$f\left(x_{1}\right)=f\left(x_{2}\right)$

$\frac{{x}_{1}}{1+{x}^2_{1}} =\frac{{x}_{2}}{1+x^2_{2}}$

$\\ {x}_{1}+{x}_{1} {x}^2_{2}={x}_{2}+{x}^2_{1}{x}_{2}$

$\\ {x}_{1}-{x}_{2}={x}^2_{1}{x}_{2}-{x}_{1} {x}^2_{2}$

$\\ {x}_{1}-{x}_{2}={x}_{1}{x}_{2}({x}_{1} -{x}_{2})$

$\\ {x}_{1}-{x}_{2}\neq 0$

$\\ {x}_{1}{x}_{2}=1$

f(X) is not one-one

For onto

$y \in R$

$y=\frac{x}{1+x^2}$

$y({1+x^2})={x}$

$y+yx^2=x$

$x^2y-x+y=0$

$x = \frac{1\pm\sqrt{1^2-4\times y \times y}}{2 \times y}$

for any value of y x has two values hence f(x) is not onto.

Posted by

Safeer PP

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