Question

If x $ϵ$ R, and $\alpha =\frac{x^2}{(1+x^4)}$, then:

• A. $0\le \alpha \le 2$
• B. $0\le \alpha \le 1$
• C. $0\le \alpha \le \frac{1}{4}$
• D. $0\le \alpha \le \frac{1}{2}$

Answer 1

The correct option is D $0\le \alpha \le \frac{1}{2}$

For real x, we have $\frac{x^2}{(1+x^4)}\ge 0$
Also,
$1+x^4-2x^2=(1-x^2{)}^2\ge 0\Rightarrow 1+x^4\ge 2x^2\therefore \left(\frac{x^2}{1+x^4}\right)\le \frac{1}{2}So0\le \frac{x^2}{1+x^4}\le \frac{1}{2}$