The solution set of frac{x^{2}}{x^{2}+1}

Question

(C) For any real number $x$,$x^{2} \ge 0$.Since $x^{2} \ge 0$,it follows that $x^{2} + 1 \ge 1 > 0$.Thus,the denominator $x^{2} + 1$ is always positiv

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(C) For any real number $x$,$x^{2} \ge 0$.Since $x^{2} \ge 0$,it follows that $x^{2} + 1 \ge 1 > 0$.Thus,the denominator $x^{2} + 1$ is always positive for all $x \in R$.The fraction $\frac{x^{2}}{x^{2}+1}$ is a ratio of a non-negative number and a positive number,which implies $\frac{x^{2}}{x^{2}+1} \ge 0$ for all $x \in R$.Therefore,there is no real value of $x$ for which $\frac{x^{2}}{x^{2}+1} Hence,the solution set is the empty set,denoted by $\phi$.

The solution set of $\frac{x^{2}}{x^{2}+1} < 0$ is

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